Sets, Functions, Measures: Volume 2 Fundamentals of Functions and Measure Theory 9783110550962, 9783110550092

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Valeriy K. Zakharov, Timofey V. Rodionov, Alexander V. Mikhalev Sets, Functions, Measures

De Gruyter Studies in Mathematics

| Edited by Carsten Carstensen, Berlin, Germany Gavril Farkas, Berlin, Germany Nicola Fusco, Napoli, Italy Fritz Gesztesy, Waco, Texas, USA Niels Jacob, Swansea, United Kingdom Zenghu Li, Beijing, China Karl-Hermann Neeb, Erlangen, Germany

Volume 68/2

Valeriy K. Zakharov, Timofey V. Rodionov, Alexander V. Mikhalev

Sets, Functions, Measures | Volume II: Fundamentals of Functions and Measure Theory

Mathematics Subject Classification 2010 26-02, 28-02, 26A21, 26A30, 26A42, 28A05, 28A25, 28C05, 28C15, 46E25, 46J10, 54A05, 54C30 Author Prof. Dr. Valeriy K. Zakharov Lomonosov Moscow State University Faculty of Mathematics and Mechanics Department of Mathematical Analysis Leninskie Gory b.1, GSP-1 119991 Moscow Russia Coauthors Prof. Dr. Timofey V. Rodionov Lomonosov Moscow State University Faculty of Mathematics and Mechanics Department of Mathematical Analysis Leniniskie Gory b.1, GSP-1 119991 Moscow Russia

Prof. Dr. Alexander V. Mikhalev Lomonosov Moscow State University Faculty of Mathematics & Mechanics Department of Theoretical Informatics Leninskie Gory b.1, GSP-1 119991 Moscow Russia

ISBN 978-3-11-055009-2 e-ISBN (PDF) 978-3-11-055096-2 e-ISBN (EPUB) 978-3-11-055022-1 Set-ISBN 978-3-11-055097-9 ISSN 0179-0986 Library of Congress Cataloging-in-Publication Data A CIP catalog record for this book has been applied for at the Library of Congress. Bibliographic information published by the Deutsche Nationalbibliothek The Deutsche Nationalbibliothek lists this publication in the Deutsche Nationalbibliografie; detailed bibliographic data are available on the Internet at http://dnb.dnb.de. © 2018 Walter de Gruyter GmbH, Berlin/Boston Typesetting: Compuscript Ltd., Shannon, Ireland Printing and binding: CPI books GmbH, Leck ♾ Printed on acid-free paper Printed in Germany www.degruyter.com

| The authors dedicate their book to the centenary of Felix Hausdorff’s outstanding book “Set Theory”

Contents Historical foreword on the centenary after Felix Hausdorff’s classic Set Theory | xi Preface | xv 2 Fundamentals of the theory of functions | 1 Introduction | 1 2.1 Descriptive and prescriptive spaces | 2 2.1.1 Ensembles and their envelopes | 2 2.1.2 The four transfinite collections of extensions of an ensemble | 20 2.1.3 Classification of Borel sets for arbitrary and perfect ensembles | 31 2.1.4 Descriptive spaces with negligence | 42 2.1.5 Prescriptive spaces | 47 2.2 Families of real-valued functions on a set | 49 2.2.1 Real-valued functions and pointwise operations over them | 49 2.2.2 The pointwise order between functions | 52 2.2.3 The pointwise and uniform convergences of nets and sequences of functions | 55 2.2.4 Some useful functional families | 59 2.2.5 Zero-sets and cozero-sets of functions | 71 2.2.6 The equivalence of functions with respect to ideal ensembles | 74 2.2.7 Seminorms and norms of the uniform convergence on families and factor-families of functions | 77 2.2.8 Pointwise continuous linear functionals on lattice-ordered linear spaces of functions | 92 2.2.9 Truncatable lattice-ordered linear spaces of functions | 98 2.3 Families of measurable and distributable functions on a descriptive space | 100 2.3.1 Measurable and distributable functions | 100 2.3.2 Pointwise operations over measurable and distributable functions | 104 2.3.3 The pointwise order between measurable and distributable functions | 108 2.3.4 The pointwise and uniform convergences of sequences of measurable and distributable functions | 109 2.3.5 Separability of sets by measurable and distributable functions | 116

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2.3.6 Description of normal and completely normal families and envelopes. Naturalness of the family of measurable functions | 119 2.3.7 Correlations between Baire’s and Borel’s functional collections | 124 2.3.8 Families of semimeasurable functions on a space with an ensemble | 134 2.4 Families of uniform functions on a prescriptive space | 140 2.4.1 Uniform functions and their properties | 140 2.4.2 Pointwise operations over uniform functions | 142 2.4.3 The uniform convergence of sequences of uniform functions | 144 2.4.4 Separability of sets by uniform functions | 146 2.4.5 Symmetrizable functions on a space with an ensemble | 150 2.4.6 Descriptions of boundedly normal families and envelopes. Naturalness of the family of uniform functions | 153 2.4.7 Fine correlations between Baire’s and Borel’s functional collections | 158 2.5 Families of functions on a descriptive space with negligence | 162 2.5.1 Almost measurable, almost distributable, and almost uniform functions | 162 2.5.2 Quasimeasurable, quasidistributable, and quasiuniform functions | 165 3 Fundamentals of the measure theory | 179 Introduction | 179 3.1 Spaces with semimeasures and measures | 180 3.1.1 Spaces with evaluations, semimeasures and measures | 180 3.1.2 Families of evaluations, semimeasures, and measures on a descriptive space | 185 3.1.3 The total variation of a natural evaluation | 187 3.1.4 Some extensions of additive evaluations defined on semirings and rings | 193 3.1.5 Extension of a positive measure to a wide complete saturated measure | 208 3.1.6 Properties of the extended Borel – Lebesgue measure on Rn | 222 3.2 Decompositions of semimeasures and measures | 226 3.2.1 The Hahn and Jordan decompositions of measures on a 𝛿-ring | 226 3.2.2 The Riesz decomposition of overfinite semimeasures and measures on a ring | 231 3.2.3 Norms on linear spaces of bounded semimeasures and measures | 239

Contents | ix

3.3

3.4

3.5

3.6

3.2.4 Absolute continuity and singularity and the Lebesgue decomposition | 240 The Lebesgue integral | 247 3.3.1 Measurable functions on a space with a positive wide measure | 247 3.3.2 The Lebesgue integral over a space with a positive measure | 255 3.3.3 Sequential properties of the Lebesgue integral | 263 3.3.4 Properties of density and completeness for the family and the factor-family of integrable functions | 266 3.3.5 Comparison of some Lebesgue integrals over spaces with positive wide measures | 273 3.3.6 The Lebesgue integral over a space with an arbitrary wide measure. The problem of characterization of Lebesgue integrals as linear functionals | 276 3.3.7 Wide measures defined by densities | 283 3.3.8 The Lebesgue – Radon – Nikodym theorem | 289 3.3.9 Dual to the factor-space of integrable functions | 296 Representation of a functional by the Lebesgue integral | 300 3.4.1 Regularity and continuity of evaluations. The key theorem for integral representations | 300 3.4.2 Representation of pointwise 𝜎-continuous functionals by Lebesgue integrals. The solution of the problem of characterization of Lebesgue integrals as linear functionals | 303 3.4.3 Representation of pointwise continuous functionals by Lebesgue integrals | 311 Topological spaces with measures. The Radon integral | 316 3.5.1 Topological spaces with evaluations, semimeasures, and measures | 316 3.5.2 Measurable and integrable functions on topological spaces with measures | 323 3.5.3 Wide Radon measures on Hausdorff spaces. The problem of characterization of Radon integrals as linear functionals | 328 3.5.4 Narrow Radon measures on Hausdorff spaces | 334 3.5.5 Radon bimeasures on Hausdorff spaces | 339 3.5.6 The Radon integral over a Hausdorff space with a Radon bimeasure | 352 Representation of a functional by the Radon integral | 355 3.6.1 𝜎-Exact linear functionals on spaces of symmetrizable functions | 355 3.6.2 Extensions of 𝜎-exact functionals on spaces of symmetrizable functions by the Young – Daniell method | 362

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3.6.3 Characterizations of Radon integrals with respect to positive Radon measures on a Hausdorff space as linear functionals | 381 3.6.4 The solution of the problem of characterization of Radon integrals as linear functionals | 388 3.7 The Riemann integral | 394 3.7.1 The Riemann integral over a topological space with a positive bounded Radon measure | 394 3.7.2 Description of the family of Riemann integrable functions on a Tychonoff space and its consequences | 396 3.7.3 The Riemann integral for Rn | 405 D Historical notes on the Riesz – Radon – Fréchet problem of characterization of Radon integrals as linear functionals | 411 D.1 The original Riesz representation theorem | 411 D.1.1 Functions of bounded variation | 411 D.1.2 The Riemann – Stiltjes integral | 412 D.1.3 The Hadamard – Fréchet problem | 412 D.1.4 The Riesz theorem | 413 D.1.5 Extension of positive functionals by Young’s method | 413 D.1.6 The property of norm preserving | 414 D.1.7 Construction of the integral corresponding to a given functional | 416 D.2 Transition from functions of bounded variation to bounded measures on compact spaces | 417 D.3 Stage of unbounded positive measures on locally compact spaces | 418 D.4 Stage of bounded measures on non-compact spaces. Tight functionals | 419 D.5 Transition to unbounded measures on arbitrary Hausdorff spaces | 420 Index of terms | 421 Index of notations | 443 Bibliography | 451

Historical foreword on the centenary after Felix Hausdorff’s classic Set Theory From Euclid’s “Elements” (III c. BC), the most extensive and influential mathematical treatise of the antiquity, mathematics was presented as the totality of several separate domains such as arithmetic, algebra, and geometry. Until the nineteenth century, these domains (and analysis, appearing in XVII c.) were developed rather independently. There was no general foundation connecting them in any integrity. This peculiarity distinguished mathematics disadvantageously from the most part of natural sciences, since each of them were united up to this time in some special integrities on the base of some uniting concepts going back to philosophy of the antiquity. For physics, the notion of atom, for chemistry the notion of chemical element, and for biology the notion of biological cell became such uniting concepts. This situation in mathematics changed cardinally when Georg Cantor (1845–1918) developed the theory of abstract sets consisting of abstract elements, i.e. connected with each other by only one membership relation. Unfortunately, such a general idea allowed the use such boundless and indicationless notions as the set of all sets. This brought to the discovery of paradoxes in Cantor’s set theory and induced the distrust to it among mathematicians. But Set Theory as a dream has been defended by David Hilbert (1862–1943). In his famous expression, “Aus dem Paradies, das Cantor uns geschaffen, soll uns niemand vertreiben können”, or in English, “From the paradise, that Cantor created for us, no-one can expel us” (D. Hilbert, Grundlagen der Geometry, Teubner, Leipzig-Berlin, 1930, p. 274). Hilbert used the word paradise because he excellently understood that some well-postulated set theory can be that general foundation, which will give the opportunity to unite arithmetic, algebra, geometry, analysis, and other domains of mathematics in a unique integrity. Felix Hausdorff (1868–1942) was one of those mathematicians who were occupied with the creating and forming of this mathematical set-theoretical paradise. In his famous book, Grundzüge der Mengenlehre [Vien, Leipzig, 1914; 2nd ed. Mengenlehre, Walter de Gruyter, Berlin, 1927], F. Hausdorff described the architecture of contemporary mathematics in the form of a tree with the set theory as a trunk and all separate domains of mathematics as its branches. Hausdorff himself laid there the foundations of two such main domains of mathematics, function theory and measure and integration theory. This outstanding book became a model for all subsequent authors who certainly built their books dealing with any branch of mathematics on the basis of the set theory. Starting at Hausdorff’s initial architecture of mathematics, a group of mathematicians acting under the pseudonym Nicolas Bourbaki described the final architecture of contemporary mathematics. On the basis of set theory and formal logic, they introduced a general concept of a mathematical structure and a more substantial concept

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xii | Historical foreword on the centenary after Felix Hausdorff’s classic Set Theory

of a mathematical system consisting of a principle carrier set and a totality of some mathematical structures on this set connected by certain logical axioms [N. Bourbaki, Eléments de Mathématique. Livre I. Théorie des ensembles. Chapitres 1–4, Hermann, Paris, 1956–1960]. This allowed presenting every branch of the “mathematical tree” as a mathematical theory studying some separated totality of mathematical systems with their own special structures and axioms. Thus, the books of Hausdorff and Bourbaki have played the leading role in the consolidation of mathematics as well as mathematicians on the basis of just a few general ideas: element, set, structure, system (with an indispensable and indiscernible involvement of logical tools). Unfortunately, Bourbaki’s book on sets and structures turned out to be so formalized and difficult that it could not eclipse Hausdorff’s book and become an acknowledged introduction to contemporary mathematics. The aforementioned famous book by Hausdorff expounding on the set theory, the theory of (real-valued) functions, and the measure and integration theory as foundations of mathematics plays this role up to our days, of course, along with remarkable later books of other authors such as K. Kuratowski and A. Mostovski [Set theory. North-Holland Publishing Company, Amsterdam, 1967], K. Kuratowski [Topology. Volume 1, Academic Press, New York-London, 1966], and so on. Surely, for the past centenary after the first edition of Hausdorff’s book in 1914, these domains (as the trunk and two main branches) of mathematics developed swiftly. Therefore, in our time, they differ considerably from those in the beginning of the twentieth century. The discovery of paradoxes in Cantor’s set theory forced mathematicians to bring the strictness up to a higher level. Therefore, mathematics advanced from the naive set theory expounded by Hausdorff to axiomatic set theories with strict logical language and adjusted axioms restricting the bounds of mathematical creation. The Zermelo – Fraenkel set theory and the Neumann – Bernays – Gödel set theory became the most well-known axiomatic set theories. At present, there are remarkable texts on this subject, although almost all of them are rather intended for particular specialists than for a wide circle of mathematicians. Further, using von Neumann’s approach to the construction of ordinal and cardinal numbers, mathematicians were able to construct at first the set of natural numbers and then the following sets of integer, rational, and real numbers within the aforementioned axiomatic set theories. This achievement allowed overcoming the gaps in Hausdorff’s book, where (a) ordinal and cardinal numbers were introduced not as some special sets but on the naive level by means of extended notions such as “thing”, “symbol”, and others, and (b) number theory was lacking since it was considered as a prolegomenon to the naive set theory. The enormous virtue of Hausdorff’s book is the general theory of measurable (realvalued) functions on descriptive spaces (a descriptive space is a set with a fixed set of its subsets). The extreme importance and naturalness of the family of all measurable functions follows from the famous Borel – Lebesgue – Hausdorff theorem asserting

Historical foreword on the centenary after Felix Hausdorff’s classic Set Theory

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that (a) this family is closed under all natural mathematical operations (in particular, addition, multiplication, division, and so on) and uniform convergence and (b) every family of (real-valued) functions on a set with the mentioned properties is some family of all measurable functions on some descriptive space. This family is extraordinary abundant in its concrete forms. However, it turned out that the concept of a measurable function is not sufficient for the solution of some problems which arose later in function theory. Thus, in 2006, the family of uniform functions on a prescriptive space was discovered (a prescriptive space is a set with a fixed set of its finite covers). The importance and naturalness of the family of all uniform functions follows from the characterization theorem (proved in 2008) asserting that (a) this family is closed under all natural mathematical operations (in particular, addition, multiplication, bounded division, and so on) and uniform convergence and (b) every family of (real-valued bounded) functions on a set with the mentioned properties is some family of all uniform functions on some prescriptive (in particular, descriptive) space. This family also became sufficiently abundant in its concrete forms (for example the family of all Riemann integrable functions on the real interval was described as some family of all uniform functions on it). Further, the most important concrete family of measurable functions considered in Hausdorff’s book is the family of Borel functions on a descriptive space. The outstanding result about this family presented in his book is the Lebesgue-Hausdorff classification describing the family of Borel functions on a metric space by means of the transfinite application of the Baire operation of addition of the pointwise limits of sequences of functions from the preceding families. This result essentially uses the remarkable transfinite construction of Borel sets given by William Henry Young and Hausdorff himself. However, these classifications and constructions are not valid for an arbitrary descriptive space. Thus, in 2002, new more general and more complicated constructions and classifications were created. It is remarkable that the finest classification (2014) uses, in the capacity of the initial functional family, some narrow family of uniform functions. In the first edition of his book published in 1914, Hausdorff also expounded on an important branch of mathematics, the theory of the Lebesgue integral. Naturally, for the past centenary, the measure and integration theory had an enormous development. This is reflected in the large number of excellent books on this domain with different degrees of generality and profundity. From the times of Lebesgue and Young, two parallel points of view were developed in integration theory: the first considers the integral as a special structure over a descriptive space with some measure; the second considers the integral as a superstructure over a functional linear space with some linear functional on it. For many years, the efforts of many outstanding mathematicians were devoted to the proof of the parallelism of these points of view. For the most popular topological space with a Radon measure, this supposed parallelism is known as the Riesz – Radon – Fréchet problem of characterization of Radon integrals as linear functionals. The solution of this problem as well as of the

xiv | Historical foreword on the centenary after Felix Hausdorff’s classic Set Theory

problem of the general parallelism in the most general and complete form took up almost one hundred years. Finally, although the Lebesgue integral substantially darkened the Riemann integral, the latter continued to develop and in result has been generalized onto an arbitrary Tychonoff topological space with some bounded positive Radon measure. It is remarkable that the description of Riemann integrable functions requires involving some family of uniform functions. By the same token, the Riemann integrable functions was characterized in 2006 and this characterization is completely different from the famous Lebesgue characterization (as almost everywhere continuous functions) even for the real interval. All previously described (along with many other) changes and achievements happening in the centenary after the first edition of Hausdorff’s book are reflected (in detail and in up-to-date mathematical language) in the present comprehensive two-volume book Sets, Functions, Measures published by Walter de Gruyter in 2018. The present work expounds set theory, the theory of (real-valued) functions, and the measure and integration theory as the fundamental domains of contemporary mathematics successively built on each other. It may be said that the authors of this book have attempted to solve one hundred years later the problem solved successfully by Hausdorff at the beginning of the twentieth century. In particular, continuing Hausdorff’s line, the material of the book is presented in such a way that there is no need for references to other sources. V. K. Zakharov and T. V. Rodionov October 2017

Preface The book’s title Sets, Functions, Measures shows that it is devoted to the exposition of the most general fundamentals of mathematics. It may be said that the book goes back to the famous Set Theory by Felix Hausdorff [1914], where he expounded the set theory, the function theory, and the function theory as the general fundamentals of mathematics. The authors of this book have attempted to solve a hundred years later the problem solved successfully by F. Hausdorff in the beginning of 20th century. The manner of exposition also goes back to the manner of F. Hausdorff. As in “Set Theory”, we set as an object to expound the most general results of the set theory, the function theory, and the measure theory in such a way that there was no need for references to other sources. Since the number theory was not included by F. Hausdorff in his book, we, following the indicated line, considered it necessary to eliminate this minor defect and to expound the theory of natural, integer, rational, and real numbers, deducing it from the set theory itself. It follows from the above that this book is addressed to a wide range of mathematicians, and it can be useful both to mature mathematicians and to students and young mathematicians, who would like to be acquainted with the fundamentals of the listed theories. According to its title, the book is divided into three chapters, each leaning on the other subsequently and is supplied with special appendices. The content of the book and the motivations of the authors are explicitly given in the introduction to each chapter. Here, we shall touch only some of the peculiarities of the presented material. The first chapter is devoted to the theory of classes, sets, and numbers. This theory is expounded in the framework of Neumann – Bernays – Gödel axiomatics with generality, completeness, and thoroughness. It is called the Neumann – Bernays – Gödel set theory (NBG). The summit of this chapter is the theory of real and extended real numbers, including the theory of series in the extended real line [−∞, ∞]. The other version of the theory of sets (the Zermelo – Fraenkel set theory with the choice axiom (ZF)) is presented in Appendix A (see A.2). The second chapter is devoted to the theory of functions. It is based on the first chapter, especially on its last section. It contains together with the more or less standard material a lot of new and non-trivial materials such as the theory of uniform functions on prescriptive spaces. The third chapter is devoted to the general theory of measure and integral. It is based on two preceding chapters, especially on the theory of series in the extended real line. This allows us to consider measures taking their values in [−∞, ∞] at the very beginning. Note that in the last chapter, the authors widely used ideas and methods expounded in the remarkable books of Fremlin [1974], Jacobs [1978], and Konig [1997]. The particular value of this chapter consists of the solution of the problem of characterization of Lebesgue integrals and the problem of characterization https://doi.org/10.1515/9783110550962-205

xvi | Preface

of Radon integrals in the most general cases, and the key tool in the solution of the second problem is some concrete family of uniform functions. Appendix A is devoted to the characterization of all natural models of the NBG and ZF set theories. These models are extremely important in virtue of their simplicity. For the reader’s convenience, the first section of this appendix describes the structure of an arbitrary first-order language (theory) and contains necessary notions from mathematical logic. In the second and third sections, the proper axioms and axiom schemes of the ZF set theory are formulated, the notions of ordinals, cardinals, and inaccessible cardinals are introduced, and properties of cumulative Mirimanov – Neumann sets are described. Thus, these sections are good supplements to the first chapter, giving a more profound presentation about contemporary mathematical logic and axiomatic set theories. Appendix B is devoted to the local theory of sets giving the solution of Maclane’s problem of constructing a new and more flexible axiomatic set theory that could serve as an adequate logical foundation for all the naive category theory. Appendix C is devoted to the proof of the compactness theorem for some generalized second-order language. The compactness theorem is valid for the first-order language, but it is not valid for the usual second-order language. Appendix D contains historical notes on the famous Riesz – Radon – Fréchet problem of characterization of Radon integrals as linear functionals. Each chapter C is divided into sections with two-valued numeration C.P. Each section is divided into subsections with three-valued numeration C.P.S. Important statements in each subsection such as lemmas, propositions, and theorems are numbered in a subsection by natural numbers in the manner Lemma N, Proposition N, Theorem N. When referring in some subsection to statements from another subsection C.P.S, we use number C.P.S in round brackets in the manner N (C.P.S). The symbol is used throughout the text to indicate the end of a proof. If some notion has several names, then the other names are written in parentheses after the name chosen by the authors as the main name. To shorten writings, we use the method of parallel writing in the following form. A short writing “A 𝜋 [𝜘, 𝜌, . . .] B p [q, r, . . .] C.” is equivalent to the following full writing: “(1) A 𝜋 B p C; (2) A 𝜘 B q C; (3) A 𝜌 B r C; . . . ”, where the capital letters denote some texts and the small letters denote some words, word combinations, or formulae. By the recommendation of the publisher, the book is divided into two volumes. For the convenience of readers, each volume is equipped by the same index of terms, index of notations, and bibliography. The authors express their profound gratitude to the Rector of the Lomonosov Moscow State University, Professor Victor Antonovich Sadovnichy for his continuous support during the twenty-year work on this book under the aegis of the University.

2 Fundamentals of the theory of functions Introduction The outstanding mathematicians of the twentieth century, E. Borel, H. Lebesgue, R. Baire, W. H. Young, F. Hausdorff, N. Lusin, et al., have laid down the foundations of the classical theory of functions. Its initial notions are the notion of a descriptive space ⟮T, S⟯ with the ensemble S on the set T and the notion of a measurable function f : T → R on ⟮T, S⟯. Its initial operational tools are an element S ∈ S and the preimage f −1 [S] of a function f on S. The described classical language can be briefly called the preimage language. The theory of measurable functions on descriptive spaces is expounded in 2.3. The culmination of this theory is the Borel – Lebesgue – Hausdorff theorem about the naturalness of the family of measurable functions first published in Hausdorff’s book “Set Theory” [Hausdorff, 1927, 37, I – VIII] and presented below in modernized form (see Theorem 1 (2.3.6)). However, in the mentioned theory, a range of problems appeared, the solution of which necessitated usage of the cardinally wider postclassical theory of functions. Its initial notions are the notion of a prescriptive space ⟮T, C⟯ with the covering C and the notion of a uniform function f : T → R on ⟮T, C⟯. Its initial operational tools are a cover 𝜋 ≡ ⟮S i | i ∈ I⟯ of the set T and the oscillation 𝜔(f , 𝜋) of a function f on 𝜋. This postclassical language can be called the cover language. It was introduced in papers [Zakharov, 1984; 1989a; 1991; 2006a]. The prefix “post” means that using the cover language hugely enhances the expressive possibilities in comparison with the preimage language. The postclassical theory of uniform functions on prescriptive spaces is expounded in 2.4. The analogous culmination of this theory is the Zakharov theorem about the naturalness of the family of uniform functions (see Theorem 1 (2.4.6)). It is clear that all the structures on the set R of real numbers considered in the last section of Chapter 1 automatically generate the corresponding structures on families of real-valued functions. In particular, the family M(T, S) of measurable functions on a classical descriptive space ⟮T, S⟯ is closed under multiplication by real numbers, finite addition, finite multiplication, inversion, finite exact bounds, and uniform convergence of sequences (see Theorem 1 (2.3.6)). Analogously, the family of uniform functions U(T, C) on a good prescriptive space ⟮T, C⟯ is closed under multiplication by real numbers, finite addition, finite multiplication, bounded inversion, finite exact bounds, and uniform convergence of sequences (see Theorem 1 (2.4.6)). As pointed above, the presence of these natural structures completely characterizes the mentioned families M(T, S) and U(T, C). The most remarkable and useful of all classical descriptive spaces ⟮T, S⟯ are quite classical descriptive spaces ⟮T, A⟯ with 𝜎-algebras A. Namely, the closeness https://doi.org/10.1515/9783110550962-001

2 | 2.1 Descriptive and prescriptive spaces

of A under the complement implies the closeness of M(T, A) even under pointwise convergence of sequences. The Baire – Borel correlation between the structural enrichment of the ensemble S by the transfinite usage of complements and the structural enrichment of the corresponding family M(T, S) by the transfinite usage of pointwise limits was investigated for the classical descriptive space ⟮T, S⟯ by H. Lebesgue and F. Hausdorff and first published in Hausdorff’s book “Set Theory” [Hausdorff, 1927, 39, I – IV]. In our book, this famous correlation is generalized to the case of an arbitrary descriptive space ⟮T, S⟯ (see Corollary 1 to Theorem 5 (2.3.7)). Moreover, the usage of the initial family of uniform functions instead of the initial family of measurable functions lets us to obtain some generalized correlation in a finer form (see Corollary 1 to Theorem 1 (2.4.7)). Finally, note that the concrete non-degenerative family of uniform functions (the family of symmetrizable functions S(T, G) ≠ {r1 | r ∈ R}) took the place of classical family of continuous functions C(T, G) and played the key role in solving the Riesz – Radon – Fréchet problem of characterization of Radon integrals on arbitrary Hausdorff (topological) spaces ⟮T, G⟯ as linear functionals expounded in 3.6 of Chapter 3. In addition, the usage of uniform functions let us obtain some new characterizations of Riemann integrable functions (different from the famous Lebesgue one) not only for the classical Riemann integral on the interval T = [a, b], but also for its N. Bourbaki’s generalization to an arbitrary Tychonoff topological measurable space ⟮T, G, M, 𝜇⟯ with the topology G, the 𝜎-algebra M ⊃ G, and the bounded Radon measure 𝜇 : M → R+ (see Theorem 3 (3.7.2)).

2.1 Descriptive and prescriptive spaces 2.1.1 Ensembles and their envelopes Let T be a fixed set, P(T) be the complete (full) ensemble on T, i. e. the set of all subsets of T (see 1.1.5), S be an ensemble on T, i. e. a non-empty subset of P(T) (see 1.2.11). Then, the pair ⟮T, S⟯ will be called a descriptive space or a space with an ensemble. A set P ⊂ T will be called S-compact or compact with respect to the ensemble S if, for every collection, (S i ∈ S | i ∈ I), such that P ⊂ ⋃⟮S i | i ∈ I⟯, there is a finite set J ⊂ I, such that P ⊂ ⋃⟮S i | i ∈ J⟯. If the whole set T is compact, then the descriptive space ⟮T, S⟯ will be called compact. An ensemble S is called closed (or stable) with respect to or (under) the complement if T\S ∈ S for every S ∈ S. An ensemble S is called finitely [countably, completely] additive if ⋃⟮S i | i ∈ I⟯ ∈ S for every finite [countable, arbitrary] collection (S i ∈ S | i ∈ I) (see 1.1.9, 1.2.6, 1.3.1, and 1.3.2). Similarly, an ensemble S is called finitely [countably, completely] multiplicative if ⋂⟮S i | i ∈ I⟯ ∈ S for every finite [countable, arbitrary] collection (S i ∈ S | i ∈ I).

2.1.1 Ensembles and their envelopes | 3

A finitely [countably, completely] additive ensemble S will be called also 𝜑additive [𝜎-additive, 𝜏-additive]. A finitely [countably, completely] multiplicative ensemble S will be called also 𝜂-multiplicative [𝛿-multiplicative, 𝜀-multiplicative]. Finitely additive and finitely multiplicative ensembles will be sometimes called additive and multiplicative, respectively. An ensemble S is called binary additive [binary multiplicative] if Q ∪ R ∈ S [Q ∩ R ∈ S] for every pair (Q, R) in S. Proposition 1. Let S be an ensemble on a set T. Then, the following conclusions are equivalent: 1) S is binary additive [binary multiplicative]; 2) S is (finitely) additive [multiplicative]. Proof. (1) ⊢ (2). Consider the subset X of 𝜔, consisting of all numbers n such that ⋃⟮S i | i ∈ I⟯ ∈ S for all collections (S i ∈ S | i ∈ I) such that card I = n + 2 (see 1.3.2). Let I = {j, k} and j = / k. Then, ⋃⟮S i | i ∈ I⟯ = S j ∪ S k ∈ S. This means that 0 ∈ X. Let n ∈ X and card I = n + 3. Take some j ∈ I and consider the set K ≡ I\{j}. Since card K = n + 2 we have S ≡ ⋃⟮S k | k ∈ K⟯ ∈ S. By Corollary 2 to Proposition 1 (1.1.10), ⋃⟮S i | i ∈ I⟯ = S j ∪ S ∈ S. Consequently, n + 1 ∈ X. By the principle of natural induction (Theorem 1 (1.2.6)), X = 𝜔. (2) ⊢ (1). Let Q, R ∈ S. Consider collection (S i ∈ S | i ∈ I) such that S0 ≡ Q and S1 ≡ R. Then, Q ∪ R = S0 ∪ S1 = ⋃⟮S i | i ∈ I⟯ ∈ S. An ensemble S will be called finitely [countably, completely] disjointly additive if ⋃⟮S i | i ∈ I⟯ ∈ S for every finite [countable, arbitrary] disjoint collection (S i ∈ S | i ∈ I). An ensemble S will be called binary disjointly additive if Q ∪ R ∈ S for every disjoint pair (Q, R) in S. The same terminology is used for multiplicativity. Remark. For such a particular form of additivity and multiplicativity, the complete analogue of Proposition 1 is valid. The set ⌀ will be called the left edge (or the zero element) of the full ensemble P(T). The set T will be called the right edge (or the unity element or the unit) of the full ensemble P(T). If an ensemble S contains ⌀ and T we will say that S is an edge ensemble or an ensemble with the edges or that S has the edges. A multiplicative edge ensemble on T will be called a foundation on T; an additive edge ensemble on T will be called a co-foundation on T. If a foundation is 𝜑-, 𝜎-, or 𝜏-additive, then it is called 𝜑-, 𝜎-, or 𝜏-foundation, respectively. In the same way, we define 𝜂-, 𝛿-, 𝜀-co-foundations. A 𝜏-foundation G on T is usually called an open topology on T or simply a topology on T, every element of G is called an open set, and the pair ⟮T, G⟯ is called a topological space. Fix some set T and some ensemble S on T. Consider the co-ensemble co-S on T consisting of all elements P ∈ P(T) such that P = T\S for some S ∈ S. In the case of a

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topological space ⟮T, G⟯, the co-ensemble F ≡ co-G is called the ensemble of closed sets. The mapping S 󳨃→ co-S form P(P(T)) into P(P(T)) will be called the operation of complementation on the set P(P(T)). Consider also the ensembles S𝜑 , S𝜎 , and S𝜏 on T consisting of all elements P ∈ P(T) such that P = ⋃⟮S i | i ∈ I⟯ for some finite, countable, and arbitrary collection (S i ∈ S | i ∈ I), respectively. These ensembles are also called 𝜑-, 𝜎-, and 𝜏-hulls of the ensemble S. It is clear that S ⊂ S𝜑 ⊂ S𝜎 ⊂ S𝜏 . In a similar way, we consider the ensembles S𝜂 , S𝛿 , and S𝜀 on T consisting of all elements P ∈ P(T) such that P = ⋂⟮S i | i ∈ I⟯ for some finite, countable, and arbitrary collection (S i ∈ S | i ∈ I), respectively. These ones are also called 𝜂-, 𝛿-, and 𝜀-hulls of the ensemble S. As above, S ⊂ S𝜂 ⊂ S𝛿 ⊂ S𝜀 . Lemma 1. Let S be an ensemble on a set T. Then, 1) the ensembles S𝜑 , S𝜎 , and S𝜏 are finitely, countably, and completely additive, respectively; the ensembles S𝜂 , S𝛿 , and S𝜀 are finitely, countably, and completely multiplicative, respectively; 2) (S𝜑 )𝜎 = S𝜎 , (S𝜂 )𝛿 = S𝛿 ; 3) if S is multiplicative, then S𝜑 , S𝜎 , and S𝜏 are also multiplicative; if S is additive, then S𝜂 , S𝛿 , and S𝜀 are also additive. Proof. 1. Take any countable collection (P i ∈ S𝜎 | i ∈ I) and the set P ≡ ⋃⟮P i | i ∈ I⟯. By definition, P i ≡ ⋃⟮S ij | j ∈ J i ⟯ for some countable collections (S ij ∈ S | j ∈ J i ). Consider the set K ≡ ⋃d ⟮J i | i ∈ I⟯ ≡ ⋃⟮J i ∗ {i} | i ∈ I⟯, it is countable by Theorem 1 (1.3.9). Define the collection (S k ∈ S | k ∈ K) setting S k ≡ S ij for every k = ⟨j, i⟩ such that j ∈ J i . Using assertion 2 of Proposition 1 (1.1.10), we check that P = ⋃⟮S k | k ∈ K⟯. Thus, P ∈ S𝜎 . This means that S𝜎 is countably additive. All the other assertions are proven in a similar way. 2. Since S ⊂ S𝜑 ⊂ S𝜎 , we get S𝜎 ⊂ (S𝜑 )𝜎 ⊂ (S𝜎 )𝜎 . By (1), the ensemble S𝜎 is 𝜎-additive. Therefore, we get (S𝜎 )𝜎 = S𝜎 . The second equality is checked in the same way. 3. Suppose that S is multiplicative. Take two elements P ≡ ⋃⟮P i | i ∈ I⟯ and Q ≡ ⋃⟮S j | j ∈ J⟯ of S∙ , where ∙ means 𝜑, 𝜎, or 𝜏. Then, by Corollary 2 to Theorem 1 (1.1.13), P ∩ Q = ⋃⟮S i ∩ S j | (i, j) ∈ I × J⟯. Since S i ∩ S j ∈ S and I × J are finite, countable or arbitrary, respectively, we conclude that P ∩ Q ∈ S∙ . Therefore, S∙ is multiplicative by virtue of Proposition 1. The last assertion is checked in a similar way. Let T be a set, S be an ensemble on T, and E ⊂ T. Consider the ensemble SE ≡ {P ∈ P(T) | ∃ S ∈ S (P = S ∩ E)} on E. It is called the trace of the ensemble S on the set E. If ⟮T, S⟯ is a descriptive space, then ⟮E, SE ⟯ is a descriptive space as well. It is easily seen that for multiplicative S the trace SE is also multiplicative and T ∈ S implies E ∈ SE . If

2.1.1 Ensembles and their envelopes |

5

S is multiplicative and E ∈ S, then SE = {S ∈ S | S ⊂ E}. It is clear that (S𝜑 )E = (SE )𝜑 , (S𝜎 )E = (SE )𝜎 . It is also clear that if D ⊂ E ⊂ T, then (SE )D = SD . An additive ensemble S will be called perfect if S contains the edges ⌀ and T and S ⊂ (co-S)𝜎 . A topological space ⟮T, G⟯ with perfect G is also called perfect. A multiplicative ensemble S will be called co-perfect if S contains the edges ⌀ and T and S ⊂ (co-S)𝛿 . An ensemble S is latticed (is a lattice) if it is additive and multiplicative; S is a 𝜎-lattice if it is 𝜎-additive and multiplicative. If S is an additive ensemble with edges closed under the complement, then S is called an algebra (or a field). If an algebra is 𝜎-additive, then it is called a 𝜎-algebra. It is clear that algebra[ 𝜎-algebra] is multiplicative [𝛿-multiplicative]. For the most part of this section, the co-ensemble co-S of an arbitrary ensemble S will be denoted simply by R. Proposition 2. Let S be a perfect ensemble on a set T. Then, 1) S𝜂 is a perfect foundation and R𝜑 = co-S𝜂 is a co-perfect co-foundation; 2) R𝜎 is a perfect 𝜎-foundation and S𝛿 = co-R𝜎 is a co-perfect 𝛿-co-foundation; 3) R𝜎 = (R𝜑 )𝜎 and S𝛿 = (S𝜂 )𝛿 . Proof. 1. It is clear that S𝜂 has the edges. By Lemma 1, S𝜂 is latticed. Since R is multiplicative, by Lemma 1 R𝜎 is also multiplicative. Therefore, using the inclusion S ⊂ R𝜎 and Lemma 1, we get S𝜂 ⊂ (R𝜎 )𝜂 = R𝜎 = (R𝜑 )𝜎 . Going to the corresponding co-ensembles, we get R𝜑 ⊂ (S𝜂 )𝛿 . Besides, R𝜑 has the edges and is latticed. 2. By Lemma 1, R𝜎 is a 𝜎-foundation and S𝛿 is a 𝛿-co-foundation. Using the condition, we get R𝜎 ⊂ (S𝛿 )𝜎 and S𝛿 ⊂ (R𝜎 )𝛿 . 3. The equalities follow from Lemma 1. The derivative ensembles K and L Consider for an ensemble S its initial derivative ensembles: 1) S ∨ co-S ≡ S ∨ R ≡ {S ∪ R | S ∈ {⌀} ∪ S ∧ R ∈ {⌀} ∪ R}; 2) S ∧ co-S ≡ S ∧ R ≡ {S ∩ R | S ∈ {T} ∪ S ∧ R ∈ {T} ∪ R}. It follows directly from the definition that {⌀}∪S∪R ⊂ S ∨ R and {T}∪S∪R ⊂ S ∧ R. Since S is non-empty, there is S ∈ S. Take R ≡ T\S. Then, T = S ∪ R ∈ S ∨ R and ⌀ = S ∩ R ∈ S ∧ R. Thus, the initial derivative ensembles have the edges. It is clear that the initial derivative ensembles are the co-ensembles to each other. Now, connect with an ensemble S the derivative ensembles L(T, S) and K(T, S): 1) the additive derivative ensemble L(T, S) consists of all elements P ∈ P(T) such that P = ⋃⟮S m ∪ R m | m ∈ M⟯ for some finite simple collections (S m ∈ {⌀} ∪ S | m ∈ M) and (R m ∈ {⌀} ∪ R | m ∈ M);

6 | 2.1 Descriptive and prescriptive spaces

2) the multiplicative derivative ensemble K(T, S) consists of all elements P ∈ P(T) such that P = ⋂⟮S m ∩ R m | m ∈ M⟯ for some finite simple collections (S m ∈ {T} ∪ S | m ∈ M) and (R m ∈ {T} ∪ R | m ∈ M). These ensembles were introduced in [Zakharov, 1989a; 1990]. The elements of the ensemble K are called symmetrizable sets. It follows from this definition that K(T, S) = co-L(T, S) and vice versa. By definition, L(T, S) = (S ∨ R)𝜑 and K(T, S) = (S ∧ R)𝜂 . Therefore, it follows from Lemma 1 that L(T, S) is additive and K(T, S) is multiplicative. Besides, S ∨ R ⊂ L(T, S) and S ∧ R ⊂ K(T, S). Consequently, the derivative ensembles have edges and contain S and R. Thus, K is a foundation and L is a co-foundation. It easily follows from the definitions that for an algebra S we have K(T, S) = L(T, S) = S. Lemma 2. Let S be an ensemble on a set T. Then, 1) L ∈ L(T, S) if and only if L = ⋃⟮R m | m ∈ M⟯ ∪ ⋃⟮S n | n ∈ N⟯ for some finite collections (R m ∈ {⌀} ∪ R | m ∈ M) and (S n ∈ {⌀} ∪ S | n ∈ N); 2) K ∈ K(T, S) if and only if K = ⋂⟮R m | m ∈ M⟯ ∩ ⋂⟮S n | n ∈ N⟯ for some finite collections (R m ∈ {T} ∪ R | m ∈ M) and (S n ∈ {T} ∪ S | n ∈ N). Proof. If L ∈ L, then L = ⋃⟮S m ∪ R m | m ∈ M⟯ = ⋃⟮S m | m ∈ M⟯ ∪ ⋃⟮R m | m ∈ M⟯. Conversely, if P = ⋃⟮S m | m ∈ M⟯ ∪ ⋃⟮R n | n ∈ N⟯, then by Corollary 2 to Theorem 2 (1.1.13), P = ⋃⟮S m ∪ R n | (m, n) ∈ M × N⟯. Denote the finite set M × N by K. Define finite collections (R󸀠k | k ∈ K) and (S󸀠k | k ∈ K) setting R󸀠(m,n) ≡ R m for every n ∈ N and S󸀠(m,n) ≡ S n for every m ∈ M. It is obvious that P = ⋃⟮S󸀠k ∪ R󸀠k | k ∈ K⟯. Assertion 2 is checked in the same way. Corollary 1. Let S be a latticed ensemble on a set T. Then, 1) L ∈ L(T, S) iff L = S ∪ R for some S ∈ S and R ∈ R; 2) K ∈ K(T, S) iff K = S ∩ R for some S ∈ S and R ∈ R. In particular, if ⟮T, G⟯ is a topological space with an open topology G and the ensemble of closed sets F ≡ co-G, then L(T, G) = {G ∪ F | G ∈ G ∧ F ∈ F} and K(T, G) = {G ∩ F | G ∈ G ∧ F ∈ F}. Lemma 3. Let S be a latticed ensemble on a set T. Then, the ensembles L(T, S) and K(T, S) are latticed as well. Proof. If S is latticed, then R is also latticed. Thus, the ensemble S ∨ R is multiplicative and the ensemble S ∧ R is additive. By virtue of Lemma 1, the ensembles L(T, S) = (S ∨ R)𝜑 and K(T, S) = (S ∧ R)𝜂 are latticed.

2.1.1 Ensembles and their envelopes |

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Lemma 4. Let S be an ensemble on a set T. Then, K(T, S) ⊂ L(T, S)𝛿 and L(T, S) ⊂ K(T, S)𝜎 , i. e. the ensemble L(T, S) is perfect and the ensemble K(T, S) is co-perfect. Proof. Let K ∈ K, i. e. K = ⋂⟮S m ∩ R m | m ∈ M⟯. Since {T} ∪ S ∪ R ⊂ L ⊂ L𝛿 , we infer that S m and R m belong to L𝛿 for all m. By Lemma 1, L𝛿 is multiplicative. Therefore, S m ∩ R m ∈ L𝛿 for every m, and so, K ∈ L𝛿 . The second inclusion follows immediately from the first one. Corollary 1. Let S be a perfect ensemble on a set T the derivative ensembles L ≡ L(T, S) and K ≡ K(T, S). Then, K𝜎 is a perfect 𝜎-foundation and L𝛿 is a co-perfect 𝛿-co-foundation. Proof. By Lemma 4, K𝜎 ⊂ L𝛿𝜎 and L𝛿 ⊂ K𝜎𝛿 . In addition, L𝛿 = co-K𝜎 . By Lemma 1, K𝜎 is a 𝜎-foundation and L𝛿 is a 𝛿-co-foundation. Corollary 2. Let S be a perfect ensemble on a set T with the co-ensemble R and the derivative ensembles L ≡ L(T, S) and K ≡ K(T, S). Then, K𝜎 = R𝜎 and L𝛿 = S𝛿 . Proof. Clearly, R ⊂ K implies R𝜎 ⊂ K𝜎 . By Lemma 1, R𝜎 is multiplicative and 𝜎-additive. Therefore, S ⊂ R𝜎 implies S ∧ R ⊂ R𝜎 and we have K = (S ∧ R)𝜂 ⊂ R𝜎 ; the last inclusion implies K𝜎 ⊂ R𝜎 . Using the derivative ensembles for an ensemble S, consider on T the ensembles S𝜆 ≡ K(T, S)𝜎 and S𝛾 ≡ L(T, S)𝛿 . By definition, the ensemble S𝜆 consists of all elements P ∈ P(T) such that P = ⋃ {⋂⟮S im ∩ R im | m ∈ M i ⟯ | i ∈ I } { } for some countable collection ((S im ∩ R im | m ∈ M i ) | i ∈ I) of finite collections (S im ∩ R im ∈ S ∧ R | m ∈ M i ). And the ensemble S𝛾 consists of all elements P ∈ P(T) such that P = ⋂ {⋃⟮S im ∪ R im | m ∈ M i ⟯ | i ∈ I } } { for some countable collection ((S im ∪ R im | m ∈ M i ) | i ∈ I) of finite collections (S im ∪ R im ∈ S ∨ R | m ∈ M i ). The ensembles S𝜆 and S𝛾 are called 𝜆-hull and 𝛾-hull of an ensemble S, respectively. Lemma 5. Let S be an ensemble on a set T. Then, the ensemble S𝜆 is countably additive and the ensemble S𝛾 is countably multiplicative. Moreover, S𝜆 is a perfect 𝜎-foundation and S𝛾 is a co-perfect 𝛿-co-foundation.

8 | 2.1 Descriptive and prescriptive spaces

Proof. All assertion follow from Lemma 1, Corollary 1 to Lemma 4, and Proposition 2 since S𝜆 ≡ K𝜎 , S𝛾 ≡ L𝛿 , K is multiplicative, and L is additive. Let us remark that S𝜆 = (co-S)𝜆 = co-S𝛾 and S𝛾 = (co-S)𝛾 = co-S𝜆 . Some simple envelopes of an ensemble For an ensemble S on a set T, there exists in P(T) the smallest latticed ensemble ̂ containing the ensemble S. In fact, consider the set L of all latticed ensembles S on T containing the ensemble S. Since P(T) ∈ L, the set L is not empty. Consider ̂ = S󸀠 . S󸀠 ≡ ⋂⟮L | L ∈ L⟯. It is easy to check that S󸀠 ∈ L. It is clear that S The lattice ̂ S is called the lattice generated by the ensemble S in P(T) or the lattice envelope of S. In the similar way, we can define the algebra A(T, S) and the 𝜎-algebra B(T, S) generated by the ensemble S as the smallest algebra and the smallest 𝜎-algebra containing S. The 𝜎-algebra B(T, S) is also called the Borel envelope of the ensemble S or the family of Borel sets with respect to the ensemble S. The following lemma gives the full description of the lattice envelope. Lemma 6. Let S be an ensemble on a set T and ̂ S be the lattice generated by S. Then, ̂ S = S𝜑𝜂 = S𝜂𝜑 . Proof. Using Theorem 1 (1.1.13), it is easy to check that S𝜑𝜂 = S𝜂𝜑 . By Lemma 1, S𝜑 is additive and S𝜑𝜂 is additive and multiplicative. If X is a latticed ensemble on T and S ⊂ X, then evidently, S𝜑𝜂 ⊂ X. Thus, S𝜑𝜂 = ̂ S. ̂ = co-S. ̂ Corollary 1. Let S be an ensemble on a set T with the co-ensemble R. Then, R ̂ Proof. We have co-̂ S = co-S𝜂𝜑 = (co-S𝜂 )𝜂 = R𝜑𝜂 = R. It follows directly from the definitions that A(T, S) = A(T, co-S) and B(T, S) = B(T, co-S). Now, we shall give the full description of the algebra A(T, S). Proposition 3. Let S be an ensemble on a set T. Then, 1) K𝜑 = L𝜂 ; 2) K𝜑 is an algebra; 3) A(T, S) = K𝜑 ; 4) S𝜆 ≡ K𝜎 = K𝜑𝜎 = L𝜂𝜎 = A𝜎 and S𝛾 ≡ L𝛿 = L𝜂𝛿 = K𝜑𝛿 = A𝛿 ; 5) K𝜑 is a (co-)perfect (co-)foundation. Proof. 1. From S ⊂ K, R ⊂ K, ⌀ ∈ K, and T ∈ K we see that L ⊂ K𝜑 . Since K is multiplicative, K𝜑 is also multiplicative by virtue of Lemma 1. Therefore, L𝜂 ⊂ K𝜑 . Taking the corresponding co-ensembles, we get K𝜑 = co-L𝜂 ⊂ co-K𝜑 = L𝜂 .

2.1.1 Ensembles and their envelopes | 9

2. Thus, the additive and multiplicative ensemble K𝜑 coincides with its coensemble L𝜂 and, by this reason, is an algebra. 3. On the one hand, since A is an algebra containing S, we have that S ⊂ A, R ≡ co-S ⊂ A. According to the definition of K, it follows from S ∩ R ∈ A for any S ∈ S and R ∈ R that K ⊂ A. Therefore, K𝜑 ⊂ A. On the other hand, since A is the smallest such an algebra, we see that A ⊂ K𝜑 . 4. By (2) from Lemma 1, K𝜑𝜎 = K𝜎 and L𝜂𝛿 = L𝛿 . 5. This follows from (2) and (4). The algebra A(T, S) = K𝜑 (T, S) was widely used by A. D. Alexandrov in his papers [Alexandrov, 1940; 1941; 1943]; therefore, it will be also called the algebra of Alexandrov sets. The description of the Borel envelope B(T, S) of an ensemble S is not so simple and we shall consider it in 2.1.3.

The reduction, separation, and improvement theorems To get the improvement theorem it is necessary to prove at first that the pair of ensembles K𝜎 and L𝛿 have very significant properties of reduction and separation. An ensemble S will be called separable if every disjoint sets R, R󸀠 ∈ R are contained in disjoint sets S, S󸀠 ∈ S. A topological space ⟮T, G⟯ with a separable ensemble G is called normal if {t} ∈ F for every t ∈ T. An ensemble S will be called reducible if every sets S, S󸀠 ∈ S covering the set T contain sets R, R󸀠 ∈ R also covering T. It is easy to see that an ensemble S is reducible if and only if co-S is separable. Lemma 7. Let S be a separable ensemble on a set T with the co-ensemble R. Then, for every set P ∈ R and Q ∈ S such that P ⊂ Q, there are sets X ∈ S and Y ∈ R such that P ⊂ X ⊂ Y ⊂ Q. Proof. By condition for disjoint sets P and R ≡ T\Q from R, there are disjoint sets X and Z form S such that P ⊂ X and R ⊂ Z. Take Y ≡ T\Z, then we get the necessary inclusions. Following Theorems 1 and 2, go back to the theorems of reduction and separation by W. Sierpiński and K. Kuratowski (see [Kuratowski, 1966, 30.VII, 26.II]). Theorem 1 (the reduction theorem in the perfect case). Let S be a perfect ensemble on a set T. Then, for every countable collection (Z i ∈ R𝜎 | i ∈ I), there is a disjoint collection (X i ∈ R𝜎 | i ∈ I) such that X i ⊂ Z i and ⋃⟮Z i | i ∈ I⟯ = ⋃⟮X i | i ∈ I⟯. Proof. Let S󸀠 ≡ S𝜂 and R󸀠 ≡ co-S󸀠 = R𝜑 , then by Lemma 1 R𝜎 = R󸀠𝜎 . By the above Z i = ⋃⟮R ip | p ∈ 𝜔⟯, for some collection, (R ip ∈ R󸀠 | p ∈ 𝜔).

10 | 2.1 Descriptive and prescriptive spaces

Since by Lemma 3 (1.3.9) the set I × 𝜔 is countable, there is a bijective mapping 𝜔. Denote R ip such that f (i, p) = n simply by R n . f : I×𝜔 Define by induction (see Theorem 2 (1.2.7)) a sequence (R∗n ∈ P(T) | n ∈ 𝜔), setting R∗0 ≡ R0 and R∗n+1 ≡ R n+1 \ ⋃⟮R∗m | m ∈ n + 1⟯ for every n ∈ 𝜔. If t ∈ ⋃⟮R m | m ∈ n⟯ ≡ L n , then t ∈ R m for some m ∈ n. Thus, t ∈ R∗m or t ∈ ⋃⟮R∗l | l ∈ m⟯. This means that t ∈ ⋃⟮R∗m | m ∈ n⟯. Consequently, L n = ⋃⟮R∗m | m ∈ n⟯. This implies R∗n = R n \L n and ⋃⟮R n | n ∈ 𝜔⟯ = ⋃⟮R∗n | n ∈ 𝜔⟯. Denote R∗n such that f (i, p) = n by R∗ip . Consider the sets X i ≡ ⋃⟮R∗ip | p ∈ 𝜔⟯. Since R󸀠 is additive, we infer that L n ∈ R󸀠 . Then, T\L n ∈ S󸀠 ⊂ R󸀠𝜎 , where the inclusion is true by virtue of Proposition 2. Since by Lemma 1 R󸀠 is multiplicative, we conclude that R∗n = R n ∩ (T\L n ) ∈ R󸀠𝜎 . Finally, by Lemma 1, R󸀠𝜎 is 𝜎-additive. Therefore, X i ∈ R󸀠𝜎 = R𝜎 . Using Corollary 1 to Proposition 1 (1.1.13), by the above, we have the equalities ⋃⟮Z i | i ∈ I⟯ = ⋃ {⋃⟮R ip | p ∈ 𝜔⟯ | i ∈ I } = ⋃⟮R ip | (i, p) ∈ I × 𝜔⟯ = { } = ⋃⟮R∗ip | (i, p) ∈ I × 𝜔⟯ = ⋃ {⋃⟮R∗ip | p ∈ 𝜔⟯ | i ∈ I } = ⋃⟮X i | i ∈ I⟯. { } From R∗ip ⊂ R ip , we infer that X i ⊂ Z i . Let i ≠ j. If t ∈ X i , then t ∈ R∗ip for some p. From i ≠ j, we see that (i, p) ≠ (j, q) for every q. Therefore, f (i, p) ≠ f (j, q). If f (j, q) < f (i, p), then t ∈ ̸ R∗jq . If f (j, q) > f (i, p), then again t ∈ ̸ R∗jq . In both cases, t ∈ ̸ X j . Consequently, X i ∩ X j = ⌀. Corollary 1 (the reduction theorem in the general case). Let S be an ensemble on a set T with the derivative ensembles L ≡ L(T, S) and K ≡ K(T, S). Then, for every countable collection (Z i ∈ K𝜎 | i ∈ I), there is a disjoint collection (X i ∈ K𝜎 | i ∈ I) such that X i ⊂ Z i and ⋃⟮Z i | i ∈ I⟯ = ⋃⟮X i | i ∈ I⟯. Proof. By Lemma 4, L is perfect. Therefore, we can apply Theorem 1 to the ensemble L with co-L = K. Corollary 2. Under the conditions of Theorem 1, if ⋃⟮Z i | i ∈ I⟯ = T, then X i ∈ R𝜎 ∩ S𝛿 . Under the conditions of Corollary 1, if ⋃⟮Z i | i ∈ I⟯ = T, then X i ∈ K𝜎 ∩ L𝛿 . Proof. Since R𝜎 is 𝜎-additive, we infer that A i ≡ ⋃⟮X j | j ∈ I\{i}⟯ ∈ R𝜎 . Consequently, X i = T\A i ∈ co-R𝜎 = S𝛿 . The second case is completely analogous. Theorem 2 (the separation theorem in the perfect case). Let S be a perfect ensemble on a set T. Then, for every countable collection (X i ∈ S𝛿 | i ∈ I) such that ⋂⟮X i | i ∈ I⟯ = ⌀, there is a collection (Z i ∈ R𝜎 ∩ S𝛿 | i ∈ I) such that X i ⊂ Z i and ⋂⟮Z i | i ∈ I⟯ = ⌀. Proof. Consider the sets C i ≡ T\X i ∈ R𝜎 . Then, ⋃⟮C i | i ∈ I⟯ = T. By Corollary 2 to Theorem 1, there is a collection (A i ∈ R𝜎 ∩ S𝛿 | i ∈ I) such that A i ⊂ C i and

2.1.1 Ensembles and their envelopes |

11

⋃⟮A i | i ∈ I⟯ = T. Consider the sets Z i ≡ T\A i . Then, the collection (Z i | i ∈ I) has the necessary properties. Corollary 1 (the separation theorem in the general case). Let S be an ensemble on a set T with the derivative ensembles L ≡ L(T, S) and K ≡ K(T, S). Then, for every countable collection, (X i ∈ L𝛿 | i ∈ I) such that ⋂⟮X i | i ∈ I⟯ = ⌀, there is a collection (Z i ∈ K𝜎 ∩ L𝛿 | i ∈ I) such that X i ⊂ Z i and ⋂⟮Z i | i ∈ I⟯ = ⌀. Proof. By Lemma 4, the ensemble L is perfect. Therefore, we can apply Theorem 2 to the ensemble L with co-ensemble co-L = K. Corollary 2. If in Theorem 2 and its Corollary 1, the set I consists of only two different elements i and j, then the collection (Z i | i ∈ I) so that in addition Z j = T\Z i , i. e. ⋃⟮Z i | i ∈ I⟯ = T. Combining some results proven above, we obtain the following important theorem. Theorem 3 (the improvement theorem in the perfect case). Let S be a perfect ensemble on a set T with the co-ensemble R. Then, R𝜎 is a separable perfect 𝜎-foundation and S𝛿 is a reducible co-perfect 𝛿-co-foundation. Proof. By Proposition 2, R𝜎 is a perfect 𝜎-foundation. It follows from Theorem 2 that R𝜎 is separable. Corollary 1 (the improvement theorem in the general case). Let S be an ensemble on a set T with the derivative ensembles L ≡ L(T, S) and K ≡ K(T, S). Then, K𝜎 is a separable perfect 𝜎-foundation and L𝛿 is a reducible co-perfect 𝛿-co-foundation. Proof. By Corollary 1 to Lemma 4, the ensemble K𝜎 is a perfect 𝜎-foundation. It follows from Corollary 1 to Theorem 2 that K𝜎 is separable. A separable perfect 𝜎-foundation was introduced by A. D. Alexandrov [Alexandrov, 1941] as “a completely normal base”. In the paper [Zakharov, 1991], it was called an a-foundation and the pair ⟮T, S⟯ with a-foundation S was called an a-space or an Alexandrov space. Note that the ensembles R𝜎 and K𝜎 are widely used in 2.3, especially in 2.3.7.

Rings of sets Rings of sets are of great importance in the descriptive theory of sets along with algebras of sets. An ensemble S is called closed under the difference if R\S ∈ S for all S, R ∈ S. A lattice closed under the difference is called a ring. It is clear that a ring contains

12 | 2.1 Descriptive and prescriptive spaces

the zero element ⌀. It is easy to check that an ensemble is a ring if and only if it is additive and closed under the difference. It is also easy to check that an ensemble is an algebra if and only if it is a ring with the unit T. Clearly, a ring is an algebra if and only if it is closed under the complement. A 𝜎-additive [𝛿-multiplicative] ring is called a 𝜎-ring [𝛿-ring]. Obviously, every 𝜎-ring is a 𝛿-ring. Henceforth and up to the end of this subsection, R denotes a ring not the co-ensemble of some ensemble. Lemma 8. Let R be a ring on T. Then, for every finite [countable] collection (R i ∈ R | i ∈ I), there is a finite [countable] collection (S n ∈ R | n ∈ N ⊂ 𝜔) of disjoint subsets such that every S n is contained in a certain R i and ⋃⟮R i | i ∈ I⟯ = ⋃⟮S n | n ∈ N⟯. I Proof. Take the cardinal number N ≡ card I ⊂ 𝜔 and some bijection u : N (see 1.3.2). Put Q k ≡ R u(k) for every k ∈ N. Consider the sets S n ≡ Q n \ ⋃⟮Q k | k ∈ n⟯ ∈ R for all n ∈ N. It is clear that S n ⊂ Q n , and therefore, ⋃⟮S n | n ∈ N⟯ ⊂ ⋃⟮Q k | k ∈ N⟯. If t ∈ S n , then t ∈ ̸ Q k for all k ∈ n, hence t ∈ ̸ S k for all k ∈ n. Thus, the sets S n are disjoint. Let k ∈ N and t ∈ Q k , then there is n0 ≡ sm (n ∈ N | t ∈ Q n ) ⩽ k. Therefore, t ∈ S n0 . This implies ⋃⟮Q k | k ∈ N⟯ ⊂ ⋃⟮S n | n ∈ N⟯. Thus, ⋃⟮S n | n ∈ N⟯ = ⋃⟮Q k | k ∈ N⟯ = ⋃⟮R i | i ∈ I⟯. Corollary 1. Let R be a countably disjointly additive ring on T. Then, R is countably additive. Lemma 9. Let R be a ring. Then, the following statements are equivalent: 1) R is a 𝛿-ring; 2) if (R i ∈ R | i ∈ I) is a countable collection, R ∈ R, and ⋃⟮R i | i ∈ I⟯ ⊂ R, then ⋃⟮R i | i ∈ I⟯ ∈ R. Proof. (1) ⊢ (2). For every i ∈ I, we have R\R i ∈ R. Therefore, P ≡ ⋂⟮R\R i | i ∈ I⟯ ∈ R. This implies ⋃⟮R i | i ∈ I⟯ = R\P ∈ R. (2) ⊢ (1). Let (R i ∈ R | i ∈ I) be a countable collection. Fix some k ∈ I. For every i ∈ I, we have S i ≡ R k \R i ∈ R and S i ⊂ R k . Therefore, S ≡ ⋃⟮S i | i ∈ I⟯ ∈ R. Then, ⋂⟮R i | i ∈ I⟯ = R k \S ∈ R. Lemma 10. For every ensemble S on T, there are, in P(T), the smallest ring R(T, S), the smallest 𝛿-ring R𝛿 (T, S), and the smallest 𝜎-ring R𝜎 (T, S) such that S ⊂ R(T, S) ⊂ R𝛿 (T, S) ⊂ R𝜎 (T, S). The proof is analogous to the proof for lattices given above. Corollary 1. R𝛿 (T, R(T, S)) = R𝛿 (T, S), R𝜎 (T, R(T, S)) = R𝜎 (T, S), R𝜎 (T, R𝛿 (T, S)) = R𝜎 (T, S).

2.1.1 Ensembles and their envelopes | 13

These smallest rings R(T, S), R𝛿 (T, S), and R𝜎 (T, S) are called the ring, the 𝛿-ring, and the 𝜎-ring generated by the ensemble S, respectively. The descriptions of these rings are complicated and we give them only for some particular cases. Proposition 4. Let S be a 𝛿-ring. Then, R𝜎 (T, S) = S𝜎 . Proof. By Lemma 1, S𝜎 is 𝜎-additive. It is clear that S ⊂ S𝜎 ⊂ R𝜎 (T, S). Let X, Y ∈ S𝜎 . Then, X ≡ ⋃⟮R󸀠i | i ∈ I⟯ and Y ≡ ⋃⟮R󸀠󸀠j | j ∈ J⟯ for some countable collections (R󸀠i ∈ S | i ∈ I) and (R󸀠󸀠j ∈ S | j ∈ J). Therefore, Z i ≡ ⋂⟮R󸀠i \R󸀠󸀠j | j ∈ J⟯ ∈ S for each i ∈ I implies X\Y = ⋃⟮Z i | i ∈ I⟯ ∈ S𝜎 . Thus, S𝜎 is a 𝜎-ring. This guarantees that R𝜎 (T, S) ⊂ S𝜎 . Corollary 1. Let S be a 𝛿-ring. Then, R𝜎 (T, S) consists of all subsets X ⊂ T such that X = ⋃⟮S i | i ∈ I⟯ for some countable pairwise disjoint collection (S i ∈ S | i ∈ I). Proof. Let X ∈ R𝜎 (T, S). By Proposition 4, there is a sequence (R i ∈ S | i ∈ 𝜔) such that X = ⋃⟮R i | i ∈ 𝜔⟯. Consider the pairwise disjoint sets S i ≡ R i \ ⋃⟮R j | j ∈ i⟯ ∈ R. Then, X = ⋃⟮S i | i ∈ 𝜔⟯. Corollary 2. Let S be an ensemble. Then, R𝜎 (T, S) consists of all subsets X ⊂ T such that X = ⋃⟮S i | i ∈ I⟯ for some countable pairwise disjoint collection (S i ∈ R𝛿 (T, S) | i ∈ I). Proof. By Corollary 1 to Lemma 10, R𝜎 (T, R𝛿 (T, S)) = R𝜎 (T, S). Corollary 3. Let S be a 𝛿-ring. If S ∈ S, X ∈ R𝜎 (T, S), and X ⊂ S, then X ∈ S. Proof. By Proposition 4, X = ⋃⟮R i | i ∈ I⟯ (R i ∈ S | i ∈ I). Hence, by Lemma 9, X ∈ S.

for

some

countable

collection

Corollary 4. Let S be a 𝛿-ring. Then, X ∩ S ∈ S for every X ∈ R𝜎 (T, S) and S ∈ S. Let S be an ensemble on T and H𝛿 (T, S) be the ensemble of all subsets H of T such that there exists a finite collection (S i ∈ S | i ∈ I), such that H ⊂ ⋃⟮S i | i ∈ I⟯. Lemma 11. For every ensemble S on T, the ensemble H𝛿 (T, S) is a 𝛿-ring containing S. Proof. Let H 󸀠 , H 󸀠󸀠 ∈ H𝛿 and H 󸀠 ⊂ ⋃⟮S󸀠i | i ∈ I⟯. Since H 󸀠 \H 󸀠󸀠 ⊂ ⋃⟮S󸀠i | i ∈ I⟯, we get H 󸀠 \H 󸀠󸀠 ∈ H𝛿 . Therefore, H𝛿 is closed under the difference. By the same reason, H𝛿 is multiplicative. Now, let (H m | m ∈ M) is a finite collection and H m ⊂ ⋃⟮S i m | i m ∈ I m ⟯ for some finite sets I m . Consider the finite set I ≡ ⋃d ⟮I m | m ∈ M⟯ ≡ ⋃⟮I m ∗ {m} | m ∈ M⟯ (see 1.1.10) and the collection (X i | i ∈ I) such that X i ≡ S i m for i = (i m , m). Then, by

14 | 2.1 Descriptive and prescriptive spaces

Proposition 1 (1.1.10), we get ⋃⟮H m | m ∈ M⟯ ⊂ ⋃⟮X i ∈ S | i ∈ I⟯ ∈ S. Hence, H𝛿 is finitely additive. Corollary 1. Let S be an ensemble. Then, for every E from R𝛿 (T, S) (and, in particular, from R(T, S)), there is a finite collection (S i ∈ S | i ∈ I) such that E ⊂ ⋃⟮S i | i ∈ I⟯. Corollary 2. Let S be a finitely additive ensemble on T. Then, every element from R𝛿 (T, S) (and, in particular, from R(T, S)) is contained in some element from S. Corollary 3. Let S be an ensemble. Then, H𝛿 (T, S) = {H ⊂ T | ∃ E ∈ R𝛿 (T, S) (H ⊂ E)}. Let S be an ensemble on T. Consider the ensemble H𝜎 (T, S) of all subsets H of T such that there is a countable collection (S i ∈ S | i ∈ I), such that H ⊂ ⋃⟮S i | i ∈ I⟯. Lemma 12. For every ensemble S, the ensemble H𝜎 (T, S) is a 𝜎-ring containing S. The proof is similar to the proof of Lemma 11, where we need to consider countable collections and sets instead of finite ones. By Theorem 1 (1.3.9) the set I is countable. Corollary 1. Let S be an ensemble. Then, for every F from R𝜎 (T, S), there is a countable collection (S i ∈ S | i ∈ I) such that F ⊂ ⋃⟮S i | i ∈ I⟯. Corollary 2. Let S be an ensemble. Then, H𝜎 (T, S) = {H ⊂ T | ∃ F ∈ R𝜎 (T, S) (H ⊂ F)}. Proposition 5. Let S be an ensemble. Then, for every X ∈ R𝛿 (T, S) [X ∈ R𝜎 (T, S)], there is a countable collection s ≡ (S i ∈ S | i ∈ I) such that X ∈ R𝛿 (T, Ss ) [X ∈ R𝜎 (T, Ss )], where Ss ≡ rng s ≡ ⋃⟮S i | i ∈ I⟯. Proof. Let us consider the set Δ of all 𝛿-rings R ≡ R𝛿 (T, Ss ) [𝜎-rings R ≡ R𝜎 (T, Ss )] generated by ensembles Ss ≡ ⋃⟮S i | i ∈ I⟯. Consider the ensemble D ≡ ⋃⟮R | R ∈ Δ⟯. Let (D m ∈ D | m ∈ M) be a countable collection. Then, every D m , is contained in some Rm ∈ Δ such that Rm = R𝛿 (T, ⋃⟮S i m | i m ∈ I m ⟯) for some countable sets I m . Consider the set I ≡ ⋃d ⟮I m | m ∈ M⟯ ≡ ⋃⟮I m × {m} | m ∈ M⟯ (see 1.1.10) and the collection (S i | i ∈ I) such that S i ≡ S i m for i = (i m , m). From Theorem 1 (1.3.9), it follows that I is countable. Therefore, R ≡ R𝛿 (T, ⋃⟮S i | i ∈ I⟯) ∈ Δ and ⋃⟮Rm | m ∈ M⟯ ⊂ R. Hence, {D m | m ∈ M} ⊂ R. Since R is a 𝛿-ring, we get ⋂⟮D m | m ∈ M⟯ ∈ R ⊂ D. Thus, D is countably multiplicative. If M is finite, then ⋃⟮D m | m ∈ M⟯ ∈ R ⊂ D, i. e. D is finitely additive. If we consider 𝜎-rings, then we conclude that ⋃⟮D m | m ∈ M⟯ ∈ R ⊂ D for every countable M, i. e. D is countably additive.

2.1.1 Ensembles and their envelopes | 15

Finally, let E, F ∈ D. Consider the collection (D m ∈ {E, F} | m ∈ 2) such that D0 ≡ E, D1 ≡ F. As was proven above, {D m | m ∈ 2} ⊂ R for some R ∈ Δ. Therefore, E\F = D0 \D1 ∈ R ⊂ D. Thus, D is a 𝛿-ring [𝜎-ring] containing S. Hence, R𝛿 (T, S) ⊂ D [R𝜎 (T, S) ⊂ D]. Semirings of sets Rings of sets are closely related to semirings. An ensemble S with the zero element ⌀ is called a semiring if it is finitely multiplicative and for every pair (Q, R) of elements of S, there exists a finite collection (S i ∈ S | i ∈ I) of pairwise disjoint elements such that R\Q = ⋃⟮S i | i ∈ I⟯. Without loss of generality, it can be assumed that Q ⊂ R. Lemma 13. Let S be a finitely multiplicative ensemble on T. Then, the following statements are equivalent: 1) S is a semiring; 2) for every pair (P, Q) of elements of S such that P ⊂ Q, there exists a finite sequence (R k | k ∈ n + 2) ↑ such that P = R0 ⊂ R1 ⊂ . . . ⊂ R n+1 = Q and R k+1 \R k ∈ S for every k ∈ n + 1. Proof. (1) ⊢ (2). For (P, Q), there is a finite disjoint sequence (S i | i ∈ n + 1) such that Q\P = ⋃⟮S i | i ∈ n + 1⟯. Take R k ≡ P ∪ ⋃⟮S i | i ∈ k⟯ for every k ∈ n + 2. Then, P = R0 ⊂ R1 ⊂ . . . ⊂ R n+1 = Q and R k+1 \R k = S k ∈ S. (2) ⊢ (1). Take some S ∈ S. Then, for (S, S), there is a sequence (R k | k ∈ n + 2) ↑ such that S = R0 ⊂ R1 ⊂ . . . ⊂ R n+1 = S and R k+1 \R k ∈ S for every k ∈ n + 1. Therefore, ⌀ = R k+1 \R k ∈ S. Let (Q, R) be a pair of elements of S such that Q ⊂ R. By (2), there is a sequence (R k | k ∈ n + 2) ↑ such that Q = R0 , R n+1 = R, and S k ≡ R k+1 \R k ∈ S for every k ∈ n + 1. Then, the collection (S k | k ∈ n + 1) consists of pairwise disjoint elements and R\Q = ⋃⟮S k | k ∈ n + 1⟯. Let us cite the most popular examples of semirings. Example 1. The ensemble of all intervals of the real line R of the form [a, b[ is a semiring on R (see 1.4.3). In fact, for any finite collection ([a m , b m [| m ∈ M) we have ⋂⟮[a m , b m [| m ∈ M⟯ = [a, b[, where a = gr (a m | m ∈ M) and b = sm (b m | m ∈ M) (see 1.1.15). If [a, b[⊂ [a󸀠 , b󸀠 [, then we take S1 ≡ [a󸀠 , a[ and S2 ≡ [b, b󸀠 [. Thus, [a󸀠 , b󸀠 [\[a, b[= S1 ∪ S2 . Example 2. Consider the set Rn of all finite sequences s ≡ (a i | i ∈ n) with the order s ⩽ t ≡ (b i | i ∈ n) such that a i ⩽ b i for every i ∈ n (see 1.2.6). According to 1.1.15, |a, b| denotes an interval of a general kind in R determined by the end-points a ⩽ b. Similarly, any set P|s, t| ≡ ∏⟮|a i , b i | | i ∈ n⟯ for any points

16 | 2.1 Descriptive and prescriptive spaces

s ≡ (a i | i ∈ n) and t ≡ (b i | i ∈ n) such that a i ⩽ b i will be called a parallelepiped of general kind in Rn determined by the main corner points s ⩽ t. (In case n = 2, it is called also a rectangle.) The ensemble of all parallelepipeds P|s, t| will be denoted by Ppar . It is multiplicative. The parallelepipeds P[s, t] ≡ ∏⟮[a i , b i ] | i ∈ n⟯, P]s, t[≡ ∏⟮]a i , b i [| i ∈ n⟯, and P[s, t[≡ ∏⟮[a i , b i [| i ∈ n⟯ are called closed, open, and half-open, respectively. Consider in the ensemble Ppar the subensemble Spar of all half-open parallelepipeds P[s, t[. We now check that Spar is a semiring on Rn . In fact, if P[s, t[⊂ P[s󸀠 , t󸀠 [, then [a i , b i [⊂ [a󸀠i , b󸀠i [ for every i. Consider the intervals S i1 ≡ [a󸀠i , a i [, S i2 ≡ [a i , b i [, and S i3 ≡ [b i , b󸀠i [. Consider the set K of all collections 𝜘 ≡ (k i ∈ {1, 2, 3} | i ∈ n) and the parallelepipeds P𝜘 ≡ ∏⟮S ik i | i ∈ n⟯ ∈ Spar . For 𝜘2 ≡ (k i ∈ {2} | i ∈ n) we have P𝜘2 = [s, t[. Let 𝜌 ≡ (l i ∈ {1, 2, 3} | i ∈ n) and 𝜘 ≠ 𝜌. Then, k j ≠ l j for some j implies S jk j ∩ S jl j = ⌀. Suppose that there is r ≡ (x i | i ∈ n) ∈ P𝜘 ∩ P𝜌 ; then, x j ∈ S jk j ∩ S jl j = ⌀. It follows from this contradiction that P𝜘 ∩ P𝜌 = ⌀ for 𝜘 ≠ 𝜌. Besides, 󸀠 󸀠 P[s , t [\P[s, t[= ⋃⟮P𝜘 | 𝜘 ∈ K\{𝜘2 }⟯. Let (P[s m , t m [∈ Spar | m ∈ M) be a finite collection, where s m ≡ (a mi | i ∈ n), t m ≡ (b mi | i ∈ n). According to Example 1 for every i, there is an interval [a i , b i [= ⋂⟮[a mi , b mi [| m ∈ M⟯. Put s ≡ (a i | i ∈ n) and t ≡ (b i | i ∈ n). Then, ⋂⟮P[s m , t m [∈ Spar | m ∈ M⟯ = P[s, t[. Thus, Spar is finitely multiplicative. Finally, ⌀ = [s, s[∈ Spar . Lemma 14. Let L be a lattice on T. Then, the ensemble S ≡ {L\M | L, M ∈ L} is a semiring. Proof. We remark first that every element S = L\M of S can be written in the form S = L\N, where N ≡ L ∩ M ∈ L and N ⊂ L. Let (S i ∈ S | i ∈ I) be a finite collection and S i = L i \M i . Therefore, we get ⋂⟮S i | i ∈ I⟯ = (⋂⟮L i | i ∈ I⟯)\(⋂⟮M i | i ∈ I⟯) ∈ S, i. e. S is finitely multiplicative. Suppose now that R = K\L, S = M\N, L ⊂ K, N ⊂ M, and R ⊂ S. Take X0 ≡ R, X2 ≡ S, and X1 ≡ (K ∩ M)\N ∈ S. Then, X0 ⊂ X1 ⊂ X2 , X1 \X0 = (M ∩ L)\N ∈ S, and X2 \X1 = M\(K ∪ N) ∈ S. By Lemma 13 S is a semiring. Corollary 1. Let ⟮T, G⟯ is a topological space. Then, the ensembles S(G) ≡ {G\H | G, H ∈ G} and S(F) ≡ {E\F | E, F ∈ F}, where F ≡ co-G}, are semirings. If ⟮T, G⟯ is a Hausdorff space, then for the ensemble C of compact subsets of T, the ensemble S(C) ≡ {C\D | C, D ∈ C} is a semiring as well. For a semiring S, there exists a good description of R(T, S). Theorem 4. Let S be a semiring on T. Then, the ring R(T, S) generated by S consists of all sets R ≡ ⋃⟮S i | i ∈ I⟯ for all finite collection (S i | i ∈ I) of pairwise disjoint elements.

2.1.1 Ensembles and their envelopes | 17

Proof. Denote the ensemble of such elements R = ⋃⟮S i | i ∈ I⟯ by R. It is clear that R ⊂ R(T, S). Let (R m ∈ R | m ∈ M) be a finite collection and R m = ⋃⟮S mi m | i m ∈ I m ⟯. Consider the set U = ∏⟮I m | m ∈ M⟯. By Lemma 5 (1.3.3), U is finite. By Theorem 1 (1.1.13), we get ⋂⟮R m | m ∈ M⟯ = ⋃⟮X u | u ∈ U⟯, where X u ≡ ⋂⟮S mu(m) | m ∈ M⟯ ∈ S. Suppose u, v ∈ U and u ≠ v; then, u(n) ≠ v(n) for some n ∈ M implies S nu(n) ∩ S nv(n) = ⌀. Hence, X u ∩ X v = ⌀. Thus, ⋂⟮R m | m ∈ M⟯ ∈ R. Take a finite collection (R m ∈ R | m ∈ M), where R m = ⋃⟮S mi m | i m ∈ I m ⟯ are pairwise disjoint. Consider the finite set I ≡ ⋃d ⟮I m | m ∈ M⟯ ≡ ⋃⟮I m ∗ {m} | m ∈ M⟯ (see 1.1.10) and the collection (X i | i ∈ I) such that X i ≡ S i m for i = (i m , m). Then, by Proposition 1 (1.1.10), we get ⋃⟮R m | m ∈ M⟯ = ⋃⟮X i ∈ S | i ∈ I⟯ and all X i are pairwise disjoint. Hence, ⋃⟮R m | m ∈ M⟯ ∈ R. Let X, Y ∈ S. Then, X\Y = X\(X ∩ Y) = ⋃⟮S i | i ∈ I⟯ for some finite collection (S i ∈ S | i ∈ I) of pairwise disjoint elements. Hence, X\Y ∈ R. Let R󸀠 = ⋃⟮S i ∈ S | i ∈ I⟯, R󸀠󸀠 = ⋃⟮S j ∈ S | j ∈ J⟯ belong to R and R󸀠󸀠 ⊂ R󸀠 . Then, 󸀠 R \R󸀠󸀠 = ⋃⟮⋂⟮S󸀠i \S󸀠󸀠j | j ∈ J⟯ | i ∈ I⟯. It follows from the properties of R proven above that S󸀠i \S󸀠󸀠j ∈ R, X i ≡ ⋂⟮S󸀠i \S󸀠󸀠j | j ∈ J⟯ ∈ R, and R󸀠 \R󸀠󸀠 = ⋃⟮X i | i ∈ I⟯ ∈ R because all X i are pairwise disjoint. Finally, let R󸀠 , R󸀠󸀠 ∈ R. Consider the pairwise disjoint sets R0 ≡ R󸀠 \R󸀠󸀠 , R1 ≡ R󸀠 ∩ 󸀠󸀠 R , and R2 ≡ R󸀠󸀠 \R󸀠 from R. Then, R󸀠 ∪ R󸀠󸀠 = ⋃⟮R i | i ∈ 3⟯ ∈ R. By Proposition 1, R is finitely additive. Thus, R is a ring. Therefore, R(T, S) ⊂ R. Corollary 1. Let S be a semiring on T. Then, R(T, S) = S𝜑 . Proof. It is clear that S𝜑 ⊂ R(T, S). By Theorem 4, R(T, S) ⊂ S𝜑 . Corollary 2. Let L be a lattice on T. Then, the ring R(T, L) consists of all sets ⋃⟮L i \M i | i ∈ I⟯ for all finite collections (L i \M i | i ∈ I) of pairwise disjoint differences with L i , M i ∈ L. Proof. This follows from Lemma 14 and Theorem 4. Saturation of an ensemble Finally, we shall consider a construction of saturation of an ensemble, which is very important for measure theory (see 3.1.4). Let S be an ensemble on T and E ⊂ S. Consider the ensemble UE (T, S) of all subsets U ⊂ T such that U ∩ E ∈ S for every E ∈ E. It will be called the E-saturation of S. If S = UE (T, S), then S will be called E-saturated. The ensemble US (T, S) will be simply called the saturation of S and will be also denoted by U(T, S). An S-saturated ensemble S will be simply called saturated. It is clear that if F ⊂ E ⊂ S, then U(T, S) ⊂ UE (T, S) ⊂ UF (T, S).

18 | 2.1 Descriptive and prescriptive spaces

Lemma 15. Let S be an ensemble on T, E ⊂ S, and UE (T, S) be its E-saturation. Then, the following assertions hold: 1) T ∈ UE (T, S); if ⌀ ∈ E, then ⌀ ∈ UE (T, S); 2) if UE (T, S) ⊂ UE (T, E), then UE (T, S) is multiplicative; 3) if U ∈ UE (T, S) and U ⊂ E ∈ E, then U ∈ S; 4) if T ∈ E, then UE (T, S) ⊂ S; 5) if S is multiplicative, then S ⊂ UE (T, S). 6) if S is additive, then UE (T, S) is additive as well. Proof. 1. For every set E ∈ E, we have T ∩ E = E ∈ E ⊂ S and ⌀ ∩ E = ⌀. 2. By the condition, we have U ∩ E ∈ E for every U ∈ UE (T, S) and E ∈ E. Then, for every U, V ∈ UE (T, S) and E ∈ E, we obtain (U ∩ V) ∩ E = V∩ (U ∩ E) ∈ S. 3. In this case, U = U ∩ E ∈ S. 4. For every U ∈ UE (T, S), we have U = U ∩ T ∈ S. 5. If U ∈ S, then U ∩ S ∈ S for every S ∈ S. Therefore, U ∈ UE (T, S). 6. For every U, V ∈ UE (T, S) and E ∈ E, we get (U ∪ V) ∩ E = (U ∩ E) ∪ (V ∩ E) ∈ S. Hence, U ∪ V ∈ UE (T, S). Corollary 1. Let S be an ensemble on T and U(T, S) be its saturation. Then, the following assertions hold: 1) T ∈ U(T, S); if ⌀ ∈ S, then ⌀ ∈ U(T, S); 2) U(T, S) is multiplicative; 3) if U ∈ U(T, S) and U ⊂ S ∈ S, then U ∈ S; 4) if T ∈ S, then U(T, S) ⊂ S; 5) if S is multiplicative, then S ⊂ U(T, S). 6) if S is additive, then U(T, S) is additive as well. Corollary 2. If S is multiplicative and T ∈ E, then UE (T, S) = S, i. e. S is E-saturated. Corollary 3. If S is multiplicative and T ∈ S, then U(T, S) = S, i. e. S is saturated. In particular, every foundation is saturated and every algebra is saturated. Proposition 6. Let R be a ring, E ⊂ R, and UE (T, S) ⊂ UE (T, E). Then, UE (T, R) is an algebra containing R. Proof. The relations T ∈ U ≡ UE (T, R) and R ⊂ U follow from assertions 1 and 5 of Lemma 15. By assertions 2 and 6, U is multiplicative and additive. Let U, V ∈ U. Then, (U\V) ∩ R = (U ∩ R)\(V ∩ R) ∈ R and we obtain U\V ∈ U. Thus, U is a ring with the unit T, i. e., an algebra. Corollary 1. If R is a ring, then U(T, R) is an algebra containing R.

2.1.1 Ensembles and their envelopes |

19

Remark. U(T, R) is not necessarily the smallest algebra containing R. Proposition 7. Let R be a 𝛿-ring, E ⊂ R, and UE (T, S) ⊂ UE (T, E). Then, UE (T, R) is a 𝜎-algebra containing R. Proof. By Proposition 6, U ≡ UE (T, R) is an algebra. Suppose (U i ∈ U | i ∈ I) is a countable collection and U ≡ ⋃⟮U i | i ∈ I⟯. For every set E ∈ E, we have U i ∩ E ∈ R for all i ∈ I and U ∩ E ⊂ E ∈ R. By Lemma 9, the set U ∩ E = ⋃⟮U i ∩ E | i ∈ I⟯ ∈ R. Therefore, U ∈ U. Corollary 1. If R is a 𝛿-ring, then U(T, R) is a 𝜎-algebra containing R. Lemma 16. Let S be an ensemble and R be a ring such that S ⊂ R ⊂ H𝛿 (T, S). Then, U(T, S) ⊂ U(T, R) and U ∈ U(T, R) iff U ∩ S ∈ R for every S ∈ S. Proof. If U ∈ U(T, R), then U ∩ S ∈ R for every S ∈ R, in particular for every S ∈ S. Conversely, suppose that U ∩ S ∈ R for every S ∈ S. Take R ∈ R ⊂ H𝛿 (T, S), then R ⊂ ⋃⟮S i | i ∈ I⟯ for some finite collection (S i ∈ S | i ∈ I). Therefore, U ∩ R = ⋃⟮U ∩ S i | i ∈ I⟯ ∩ R ∈ R implies U ∈ U(T, R). If now U ∈ U(T, S), then U ∩ S ∈ S ⊂ R for every S ∈ S, where U ∈ U(T, R). Corollary 1. Let S be an ensemble on T. Then, U(T, S) ⊂ U(T, R(T, S)) ⊂ U(T, R𝛿 (T, S)). Moreover, U ∈ U(T, R(T, S)) [U ∈ U(T, R𝛿 (T, S))] iff U ∩S ∈ R(T, S) [U ∩ S ∈ R𝛿 (T, S)] for every S ∈ S. Proof. The statement follows from Lemma 16 and Corollary 1 to Lemma 11. Lemma 17. Let S be an ensemble and R be a 𝜎-ring such that S ⊂ R ⊂ H𝜎 (T, S). Then, U(T, S) ⊂ U(T, R) and U ∈ U(T, R) iff U ∩ S ∈ R for every S ∈ S. The proof is similar to the proof of Lemma 16. Corollary 1. Let R be a 𝛿-ring. Then, U(T, R) = U(T, R𝜎 (T, R)). Proof. By Corollary 1 to Lemma 12, U(T, R) ⊂ U(T, R𝜎 (T, R)). On the other hand, if U ∈ U(T, R𝜎 (T, R)) and R ∈ R, then U ∩ R ∈ R𝜎 (T, R) and U ∩ R ⊂ R. By Corollary 3 to Proposition 5, U ∩ R ∈ R. Consequently, U ∈ U(T, R). Lemma 18. Let S be a multiplicative ensemble. Then, U(T, U(T, S)) = U(T, S). Proof. By Corollary 1 to Lemma 15, S ⊂ U(T, S) and U(T, S) ⊂ U(T, U(T, S)). Let U ∈ U(T, U(T, S)) and S ∈ S. Then, U ∩ S ∈ U(T, S) implies (U ∩ S)∩ S ∈ S. Therefore, U ∈ U(T, S).

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Corollary 1. Let S be a multiplicative ensemble. Then, the ensemble U(T, S) is saturated.

2.1.2 The four transfinite collections of extensions of an ensemble In this subsection, we shall consider some important transfinite schemes of extension of a fixed ensemble S on a fixed set T by means of the operations 𝜎, 𝛿, 𝜆, and 𝛾 defined in 2.1.1. Recall that according to Theorem 1 (1.2.10), for every ordinal number 𝛼 ⩾ 𝜔, there is a unique limit ordinal number 𝛾 ⩾ 𝜔 and a unique natural number n such that 𝛼 = 𝛾 +o n. According to the definition from 1.3.6, if the natural number n is even [odd], then the ordinal number 𝛼 is called even [odd]. Since n = 0 is even, all limit ordinal numbers are even as well. Σ, Δ, Λ, and Γ collections Fix a set T and an ensemble S on T. For a ordinal number 𝜔1 and the set B = P(P(T)), define a mapping V : ⋃⟮Map(𝛼, B) | 𝛼 ∈ 𝜔1 ⟯ → B, setting V(0) = S, V(v) = (⋃⟮v(𝛾) | 𝛾 ∈ 𝛼⟯)𝜎 for every odd ordinal number 𝛼 ∈ 𝜔1 and every mapping v : 𝛼 → B, and V(v) = (⋃⟮v(𝛾) | 𝛾 ∈ 𝛼⟯)𝛿 for every even non-zero ordinal number 𝛼 ∈ 𝜔1 and every mapping v : 𝛼 → B. According to the scheme of construction of mappings by transfinite induction (see Theorem 1 (1.2.8)), there is a unique mapping u : 𝜔1 + 1 → B such that u(0) = V(0) and u(𝛼) = V(u|𝛼) for every 𝛼 ∈ 𝜔1 + 1. Denote u(𝛼) by Σ𝛼 (T, S). Then, we get the transfinite collection ⟮Σ𝛼 (T, S) ⊂ P(T) | 𝛼 ∈ 𝜔1 + 1⟯ of the ensembles on the set T such that 1) Σ0 (T, S) = S, Σ1 (T, S) = S𝜎 , Σ2 (T, S) = (S𝜎 )𝛿 ≡ S𝜎𝛿 , Σ3 (T, S) = ((S𝜎 )𝛿 )𝜎 ≡ S𝜎𝛿𝜎 , ...; 2) Σ𝛼 (T, S) = (⋃⟮Σ𝛽 (T, S) | 𝛽 ∈ 𝛼⟯)𝜎 for every odd ordinal number 𝛼 < 𝜔1 ; 3) Σ𝛼 (T, S) = (⋃⟮Σ𝛽 (T, S) | 𝛽 ∈ 𝛼⟯)𝛿 for every even ordinal number 0 < 𝛼 ⩽ 𝜔1 . In a similar way, we construct the transfinite collection ⟮Δ 𝛼 (T, S) ⊂ P(T) | 𝛼 ∈ 𝜔1 + 1⟯ of the ensembles on the set T such that 1) Δ 0 (T, S) = S, Δ 1 (T, S) = S𝛿 , Δ 2 (T, S) = (S𝛿 )𝜎 ≡ S𝛿𝜎 , . . . ; 2) Δ 𝛼 (T, S) = (⋃⟮Δ 𝛽 (T, S) | 𝛽 ∈ 𝛼⟯)𝛿 for every odd ordinal number 𝛼 < 𝜔1 ; 3) Δ 𝛼 (T, S) = (⋃⟮Δ 𝛽 (T, S) | 𝛽 ∈ 𝛼⟯)𝜎 for every even ordinal number 0 < 𝛼 ⩽ 𝜔1 .

2.1.2 The four transfinite collections of extensions of an ensemble |

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The collections ⟮Σ𝛼 (T, S) | 𝛼 ∈ 𝜔1 + 1⟯ and ⟮Δ 𝛼 (T, S) | 𝛼 ∈ 𝜔1 + 1⟯ will be called the Young – Hausdorff collections of extensions of the ensemble S of the types Σ and Δ, respectively. The ensembles Σ𝛼 (T, S) and Δ 𝛼 (T, S) are called the Young – Hausdorff ensembles of the class 𝛼 for the ensemble S. Now, for 𝛼 and B, define another mapping V : ⋃⟮Map(𝛼, B) | 𝛼 ∈ 𝜔1 ⟯ → B setting V(0) = S, V(v) = (⋃⟮v(𝛾) | 𝛾 ∈ 𝛼⟯)𝜆 for every non-zero ordinal number 𝛼 ∈ 𝜔1 and every mapping v : 𝛼 → B. According to Theorem 1 (1.2.8), there is a unique mapping u : 𝜔1 + 1 → B such that u(0) = V(0) and u(𝛼) = V(u|𝛼) for every 𝛼 ∈ 𝜔1 + 1. Denote u(𝛼) by Λ 𝛼 (T, S). Then, we get the transfinite collection ⟮Λ 𝛼 (T, S) ⊂ P(T) | 𝛼 ∈ 𝜔1 + 1⟯ of the ensembles on the set T such that 1) Λ 0 (T, S) = S, Λ 1 (T, S) = S𝜆 , Λ 2 (T, S) = (S𝜆 )𝜆 ≡ S𝜆𝜆 , . . . ; 2) Λ 𝛼 (T, S) = (⋃⟮Λ 𝛽 (T, S) | 𝛽 ∈ 𝛼⟯)𝜆 for every non-zero ordinal number 𝛼 ⩽ 𝜔1 . In a similar manner, we construct the transfinite collection ⟮Γ𝛼 (T, S) ⊂ P(T) | 𝛼 ∈ 𝜔1 + 1⟯ of the ensembles on the set T such that 1) Γ0 (T, S) = S, Γ1 (T, S) = S𝛾 , Γ2 (T, S) = (S𝛾 )𝛾 ≡ S𝛾𝛾 , . . . ; 2) Γ𝛼 (T, S) = (⋃⟮Γ𝛽 (T, S) | 𝛽 ∈ 𝛼⟯)𝛾 for every non-zero ordinal number 𝛼 ⩽ 𝜔1 . The collections ⟮Λ 𝛼 (T, S) | 𝛼 ∈ 𝜔1 + 1⟯ and ⟮Γ𝛼 (T, S) | 𝛼 ∈ 𝜔1 + 1⟯ will be called the collections of extensions of the ensemble S of the types Λ and Γ, respectively. They were introduced by V. K. Zakharov and A. V. Koldunov [1982a; 1982b]. The ensembles Λ 𝛼 (T, S) and Γ𝛼 (T, S) will be called the Zakharov – Koldunov ensembles of the class 𝛼 for the ensemble S. According to Lemma 1 (2.1.1), the ensembles Σ𝛼 (T, S) for odd ordinal numbers 𝛼 < 𝜔1 and the ensembles Δ 𝛼 (T, S) for even ordinal numbers 0 < 𝛼 ⩽ 𝜔1 are countably additive. Respectively, the ensembles Σ𝛼 (T, S) for even ordinal numbers 0 < 𝛼 ⩽ 𝜔1 and the ensembles Δ 𝛼 (T, S) for odd ordinal numbers 𝛼 < 𝜔1 are countably multiplicative. Lemma 1. Let S be an ensemble on a set T. Then, Σ- and Δ-ensembles have the following properties: 1) Σ𝛼 (T, S) ⊂ Σ𝛽 (T, S) and Δ 𝛼 (T, S) ⊂ Δ 𝛽 (T, S) for all ordinal numbers 𝛼 ⩽ 𝛽 ⩽ 𝜔1 ; 2) Σ𝛼+1 (T, S) = (Σ𝛼 (T, S))𝜎 and Δ 𝛼+1 (T, S) = (Δ 𝛼 (T, S))𝛿 for every even ordinal number 𝛼 ∈ 𝜔1 ; 3) Σ𝛼+1 (T, S) = (Σ𝛼 (T, S))𝛿 and Δ 𝛼+1 (T, S) = (Δ 𝛼 (T, S))𝜎 for every odd ordinal number 𝛼 ∈ 𝜔1 .

22 | 2.1 Descriptive and prescriptive spaces

Proof. The first property follows directly from the definitions. If 𝛼 is even, then 𝛼 + 1 is odd. Therefore, Σ𝛼+1 = (⋃⟮Σ𝛾 | 𝛾 ∈ 𝛼 + 1⟯)𝜎 . Using (1) we infer from 𝛼 ∈ 𝛼+1 that ⋃⟮Σ𝛾 | 𝛾 ∈ 𝛼+1⟯ = Σ𝛼 . This gives the necessary equality from (2). All the other equalities checked in a similar way. Lemma 2. Let S be an ensemble on T with the co-ensemble R ≡ co-S. Then, Λ- and Γ-ensembles have the following properties: 1) Λ 𝛼 (T, S) = Λ 𝛼 (T, R) and Γ𝛼 (T, S) = Γ𝛼 (T, R) for every ordinal number 1 ⩽ 𝛼 < 𝜔1 ; 2) Λ 𝛼 (T, S) ⊂ Λ 𝛽 (T, S) and Γ𝛼 (T, S) ⊂ Γ𝛽 (T, S) for all ordinal numbers 𝛼 ⩽ 𝛽 ⩽ 𝜔1 ; 3) Λ 𝛼+1 (T, S) = (Λ 𝛼 (T, S))𝜆 and Γ𝛼+1 (T, S) = (Γ𝛼 (T, S))𝛾 for every ordinal number 𝛼 < 𝜔1 . Proof. 1. According to 2.1.1, Λ 1 (T, S) ≡ S𝜆 = R𝜆 ≡ Λ 1 (T, R) and Γ1 (T, S) ≡ S𝛾 = R𝛾 ≡ Γ1 (T, R). Applying the transfinite induction, we prove the necessary equalities for all 2 ⩽ 𝛼 < 𝜔1 . The proof of (2) and (3) is similar to proof of Lemma 1. Corollary 1. Let S be an edge ensemble on T. Then, co-Λ 𝛼 (T, S) ⊂ Λ 𝛼+1 (T, S) and co-Γ𝛼 (T, S) ⊂ Γ𝛼+1 (T, S) for every 𝛼 ∈ 𝜔1 . Proof. Since S is an edge ensemble, we have co-Λ 𝛼 ⊂ (Λ 𝛼 )𝜆 = Λ 𝛼+1 and co-Γ𝛼 ⊂ (Γ𝛼 )𝛾 = Γ𝛼+1 . Proposition 1. Let S be an ensemble on a set T. Then, 1) Σ𝛼 (T, S) is latticed for every 2 ⩽ 𝛼 < 𝜔1 ; if S is multiplicative, then 1 ⩽ 𝛼 < 𝜔1 ; 2) Δ 𝛼 (T, S) is latticed for every 2 ⩽ 𝛼 < 𝜔1 ; if S is additive, then 1 ⩽ 𝛼 < 𝜔1 . Proof. 1. Take a subset B1 of the set 𝜔1 consisting of all ordinal numbers 𝛼 ∈ [2, 𝜔1 [ such that Σ𝛼 is latticed. It follows from Lemma 1 (2.1.1) that Σ2 = S𝜎𝛿 is multiplicative and additive, i. e. latticed. Thus, 2 ∈ B1 . Consider the set B ≡ B1 ∪ 2. Let 𝛽 ∈ [3, 𝜔1 [ and 𝛽 ⊂ B. Consider the ensemble E ≡ ⋃⟮Σ𝛾 | 𝛾 ∈ 𝛽⟯. If 𝛽 is odd, then Σ𝛽 = E𝜎 . Take any finite collection (E m | m ∈ M). Therefore, E m ∈ Σ𝛾 for some 𝛾 ∈ 𝛽. Therefore, the set X m ≡ {𝛾 ∈ 𝛽 | E m ∈ Σ𝛾 } is non-empty. By the choice axiom from 1.1.12, there exists a mapping p : P(𝛽)\{⌀} → 𝛽 such that p(P) ∈ P. Consider the element 𝛾m ≡ p(X m ) ∈ X m ⊂ 𝛽. Then, E m ∈ Σ𝛾m . Consider the number 𝛾 ≡ gr(𝛾 ⊻ 2 | m ∈ M) ∈ [2, 𝛽[. By assumption, Σ𝛾 is latticed. It follows from Lemma 1 E m ∈ Σ𝛾 ⊂ Σ𝛾 for every m ∈ M. Hence, ⋂⟮E m | m ∈ M⟯ and ⋃⟮E m | m ∈ M⟯ belong to Σ𝛾 and, therefore, to E. This means that E is latticed. Consequently, by Lemma 1 (2.1.1), Σ𝛽 = E𝜎 is latticed. If 𝛽 is even, then Σ𝛽 = E𝛿 . Again by the same reason, Σ𝛽 is latticed. Thus, 𝛽 ∈ B. Using the principle of transfinite induction from 1.2.8, we conclude that B = 𝜔1 .

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If S is multiplicative, then by Lemma 1 (2.1.1), Σ1 = S𝜎 is latticed. 2. For Δ-ensembles, all the arguments are analogous. Corollary 1. Let S be a perfect ensemble on a set T. Then, Σ𝛼 (T, S) and Δ 𝛼 (T, S) are latticed foundations for every 1 ⩽ 𝛼 < 𝜔1 . Corollary 2. Let S be an ensemble on a set T with the derivative ensembles L ≡ L(T, S) and K ≡ K(T, S). Then, Σ𝛼 (T, K) and Δ 𝛼 (T, L) are latticed foundations for every 1 ⩽ 𝛼 < 𝜔1 . Proof. According to their definitions in 2.1.1, K is multiplicative and L is additive. Proposition 2. Let S be an ensemble on a set T. Then, 1) Σ𝜔1 (T, S) and Δ 𝜔1 (T, S) are countably additive and countably multiplicative; 2) Σ𝜔1 (T, S) = ⋃⟮Σ𝛼 (T, S) | 𝛼 ∈ 𝜔1 ⟯ and Δ 𝜔1 (T, S) = ⋃⟮Δ 𝛼 (T, S) | 𝛼 ∈ 𝜔1 ⟯. Proof. Since 𝜔1 is limit ordinal, it is even. Therefore, Σ𝜔1 = E𝛿 , where E = ⋃⟮Σ𝛼 | 𝛼 ∈ 𝜔1 ⟯. Take any countable collection ⟮E m ∈ E | m ∈ M⟯. By definition, E m ∈ Σ𝛼 for some 𝛼 ∈ 𝜔1 . Therefore, the set X m ≡ {𝛼 ∈ 𝜔1 | E m ∈ Σ𝛼 } is non-empty. By the choice axiom from 1.1.12, there exists a mapping p : P(𝜔1 )\{⌀} → 𝜔1 such that p(P) ∈ P. Consider the element 𝛼m ≡ p(X m ) ∈ X m . Then, E m ∈ Σ𝛼m . By Lemma 1 (1.3.10), 𝛼 ≡ ⋃⟮𝛼m | m ∈ M⟯ ∈ 𝜔1 . Therefore, by Corollary 1 to Lemma 3 (1.3.10), 𝛼+1 ∈ 𝜔1 and 𝛼+2 ∈ 𝜔1 . Besides, 𝛼m ∈ 𝛼+1 and 𝛼m ∈ 𝛼+2. If 𝛼 is even, then E ≡ ⋃⟮E m | m ∈ M⟯ ∈ (⋃⟮Σ𝛾 | 𝛾 ∈ 𝛼 + 1⟯)𝜎 = Σ𝛼+1 . If 𝛽 is odd, then E ≡ ⋃⟮E m | m ∈ M⟯ ∈ (⋃⟮Σ𝛾 | 𝛾 ∈ 𝛼 + 2⟯)𝜎 = Σ𝛼+2 . In both cases, E ∈ E. Consequently, E is countably additive. By the same argument, it is checked that E is countably multiplicative. Therefore, Σ𝜔1 = E𝛿 = E. Thus, Σ𝜔1 is countably additive and countably multiplicative. In the same manner, the corresponding properties for Δ-ensembles are proven. Lemma 3. Let S be an ensemble on a set T. Then, 1) the ensembles Λ 𝛼 (T, S) are perfect 𝜎-foundations for all 1 ⩽ 𝛼 ⩽ 𝜔1 ; 2) the ensembles Γ𝛼 (T, S) are co-perfect 𝛿-co-foundations for all 1 ⩽ 𝛼 ⩽ 𝜔1 . Proof. By definition, Λ 𝛼 is 𝜆-hull of some ensemble and Γ𝛼 is 𝛾-hull of some ensemble for every 𝛼 ∈ [1, 𝜔1 [. Then, by virtue of Lemma 5 (2.1.1), Λ 𝛼 is perfect 𝜎-foundation and Γ𝛼 is co-perfect 𝛿-co-foundation. Proposition 3. Let S be an ensemble on a set T. Then, 1) the ensembles Λ 𝜔1 (T, S) and Γ𝜔1 (T, S) are 𝜎-algebras; 2) Λ 𝜔1 (T, S) = ⋃⟮Λ 𝛼 (T, S) | 𝛼 ∈ 𝜔1 ⟯ and Γ𝜔1 (T, S) = ⋃⟮Γ𝛼 (T, S) | 𝛼 ∈ 𝜔1 ⟯.

24 | 2.1 Descriptive and prescriptive spaces

Proof. Consider the ensemble E = ⋃⟮Λ 𝛼 (T, S) | 𝛼 ∈ 𝜔1 ⟯. Then, Λ 𝜔1 (T, S) = E𝜆 . By Lemma 3, Λ 𝜔1 (T, S) and Γ𝜔1 (T, S) are 𝜎-foundation and 𝛿-co-foundation, respectively. Take any element E ∈ E𝜆 . By definition, E = ⋃⟮F i ∩ G i | i ∈ I⟯ for some countable set I and some collections ⟮F i ∈ E | i ∈ I⟯ and ⟮G i ∈ co-E | i ∈ I⟯. By definition, F i ∈ Λ 𝛼 for some 𝛼 ∈ 𝜔1 . Therefore, the set X i ≡ {𝛼 ∈ 𝜔1 | F i ∈ Λ 𝛼 } is non-empty. By the choice axiom from 1.1.12, there exists a mapping p : P(𝜔1 )\{⌀} → 𝜔1 such that p(P) ∈ P. Consider the element 𝛼i ≡ p(X i ) ∈ X i . Then, F i ∈ Λ 𝛼i . Similarly, G i ∈ Γ𝛼 for some 𝛼 ∈ 𝜔1 . Therefore, the set Y i ≡ {𝛼 ∈ 𝜔1 | G i ∈ co-Λ 𝛼 } is non-empty. Consider the element 𝛽i ≡ p(Y i ) ∈ Y i . Then, G i ∈ co-Λ 𝛽i . By Lemma 1 (1.3.10), 𝛼 ≡ ⋃⟮𝛼i | i ∈ I⟯ ∈ 𝜔1 , 𝛽 ≡ ⋃⟮𝛽i | i ∈ I⟯ ∈ 𝜔1 , and 𝛾 ≡ 𝛼 ∪ 𝛽 ∈ 𝜔1 . Therefore, by Corollary 1 to Lemma 3 (1.3.10) 𝛾 + 1 ∈ 𝜔1 . Besides, 𝛼i ⩽ 𝛾 and 𝛽i ⩽ 𝛾. By Lemma 1, F i ∈ Λ 𝛼i ⊂ Λ 𝛾 and G i ∈ co-Λ 𝛽i ⊂ Λ 𝛾 . Consequently, E ∈ (Λ 𝛽 )𝜆 . By virtue of Lemma 2, E ∈ Λ 𝛽+1 ⊂ E. This means that E𝜆 ⊂ E, and we conclude that Λ 𝜔1 = E𝜆 = E. Thus, we proved the first equality in (2). Take any element F ∈ Λ 𝜔1 = E. Then, F ∈ Λ 𝛼 for some 𝛼 ∈ 𝜔1 . By Corollary 1 to Lemma 2, T\F ∈ co-Λ 𝛼 ⊂ E = Λ 𝜔1 . Thus, Λ 𝜔1 is closed under the complement. It follows immediately from the properties proven above that Λ 𝜔1 is countably multiplicative. Thus, it is a 𝜎-algebra. For Γ-ensembles, the arguments are analogous. Compare now the Σ-collection for an ensemble S with the Δ-collection for its co-ensemble R and the Λ-collection for S with the Γ-collection for R. Lemma 4. Let S be an ensemble on a set T. Then, 1) co-Σ𝛼 (T, S) = Δ 𝛼 (T, R) and co-Δ 𝛼 (T, S) = Σ𝛼 (T, R) for every ordinal number 𝛼 < 𝜔1 ; 2) co-Λ 𝛼 (T, S) = Γ𝛼 (T, R) and co-Γ𝛼 (T, S) = Λ 𝛼 (T, R) for every ordinal number 𝛼 < 𝜔1 . Proof. In this proof, let us denote Σ𝛽 ≡ Σ𝛽 (T, R), Δ 𝛽 ≡ Δ 𝛽 (T, S). Take a subset B of the set 𝜔1 consisting of all ordinal numbers 𝛼 < 𝜔1 such that co-Σ𝛼 = Δ 𝛼 . It follows from the equality co-Σ0 = co-S ≡ R = Δ 0 that 0 ∈ B, i. e. B is non-empty. Let 𝛽 ∈ 𝜔1 and 𝛽 ⊂ B. If 𝛽 is odd, then co-Σ𝛽 = co- (⋃⟮Σ𝜌 | 𝜌 ∈ 𝛽⟯)𝜎 = (co- ⋃⟮Σ𝜌 | 𝜌 ∈ 𝛽⟯)𝛿 = (⋃⟮co-Σ𝜌 | 𝜌 ∈ 𝛽⟯)𝛿 = (⋃⟮Δ 𝜌 | 𝜌 ∈ 𝛽⟯)𝛿 = Δ 𝛽 . This means that 𝛽 ∈ B. If 𝛽 is even then arguing as above, we infer that also 𝛽 ∈ B. Using the principle of transfinite induction from 1.2.8, we conclude that B = 𝜔1 . All the other equalities are checked in the similar manner. Corollary 1. Let E be an ensemble on a set T with the derivative ensembles L ≡ L(T, E) and K ≡ K(T, E). Then,

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1) co-Σ𝛼 (T, K) = Δ 𝛼 (T, L) for every 𝛼 < 𝜔1 ; 2) co-Δ 𝛼 (T, L) = Σ𝛼 (T, K) for every 𝛼 < 𝜔1 . Proposition 4. Let S be a perfect ensemble on a set T. Then, 1) Σ𝛼 (T, R) ⊂ Δ 𝛼+1 (T, S) for all 𝛼 < 𝜔1 ; 2) Δ 𝛼 (T, S) ⊂ Σ𝛼+1 (T, R) for all 𝛼 < 𝜔1 . Proof. In this proof, let us denote Σ𝛽 ≡ Σ𝛽 (T, R), Δ 𝛽 ≡ Δ 𝛽 (T, S). 1. Take a subset B of a set 𝜔1 consisting of all ordinal numbers 𝛼 ∈ 𝜔1 such that Σ𝛼 ∈ Δ 𝛼+1 . For 𝛼 = 0 we have Σ0 = R ⊂ S𝛿 = Δ 1 . This means that 0 ∈ B, i. e. B is non-empty. Let 𝛽 ∈ 𝜔1 and 𝛽 ⊂ B. By assumption, Σ𝜌 ⊂ Δ 𝜌+1 for every 𝜌 ∈ B. If 𝛽 is even, then Σ𝛽 = (⋃⟮Σ𝜌 | 𝜌 ∈ 𝛽⟯)𝛿 ⊂ (⋃⟮Δ 𝜌+1 | 𝜌 ∈ 𝛽⟯)𝛿 . Since 𝜌 + 1 ⩽ 𝛽, we infer by Lemma 1 that Δ 𝜌+1 ⊂ Δ 𝛽 and Σ𝛽 ⊂ (Δ 𝛽 )𝛿 = Δ 𝛽+1 . If 𝛽 is odd, then again by the same lemma, Σ𝛽 = (⋃⟮Σ𝜌 | 𝜌 ∈ 𝛽⟯)𝜎 ⊂ (⋃⟮Δ 𝜌+1 | 𝜌 ∈ 𝛽⟯)𝜎 ⊂ (Δ 𝛽 )𝜎 = Δ 𝛽+1 . Consequently, 𝛽 ∈ B. Using the principle of transfinite induction from 1.2.8, we conclude that B = 𝜔1 . 2. For 𝛼 = 0, we have Δ 0 = S ⊂ R𝜎 = Σ1 . Further, the argument is the same as above. Corollary 1. Let E be an ensemble on a set T with the derivative ensembles L ≡ L(T, E) and K ≡ K(T, E). Then, 1) Σ𝛼 (T, K) ⊂ Δ 𝛼+1 (T, L) for every 𝛼 ∈ 𝜔1 ; 2) Δ 𝛼 (T, L) ⊂ Σ𝛼+1 (T, K) for every 𝛼 ∈ 𝜔1 . Proof. By Lemma 4 (2.1.1), L is perfect. Y and Θ collections In the sequel, for an ensemble S on a set T with the co-ensemble R, the countably additive ensembles Δ 𝛼 (T, S) with even indices 2 ⩽ 𝛼 < 𝜔1 and Σ𝛼 (T, R) with odd indices 1 ⩽ 𝛼 < 𝜔1 will be denoted also by Y𝛼 (T, S). For 𝛼 = 0, we define Y0 (T, S) ≡ S. Respectively, the countably multiplicative ensembles Δ 𝛼 (T, S) with odd indices 1 ⩽ 𝛼 < 𝜔1 and Σ𝛼 (T, R) with even indices 2 ⩽ 𝛼 < 𝜔1 will be denoted also by Θ𝛼 (T, S). For 𝛼 = 0 we define Θ0 (T, S) ≡ R. For initial indices in the additive case, we have 1) Y0 (T, S) = S, Y1 (T, S) = R𝜎 , Y2 (T, S) = S𝛿𝜎 , Y3 (T, S) = R𝜎𝛿𝜎 , Y4 (T, S) = S𝛿𝜎𝛿𝜎 , ...; 2) Λ 1 (T, S) = K𝜎 , Λ 2 (T, S) = L𝛿𝜎 , Λ 3 (T, S) = K𝜎𝛿𝜎 , Λ 4 (T, S) = L𝛿𝜎𝛿𝜎 , . . . .

26 | 2.1 Descriptive and prescriptive spaces

For initial indices in the multiplicative case, we have 1) Θ0 (T, S) = R, Θ1 (T, S) = S𝛿 , Θ2 (T, S) = R𝜎𝛿 , Θ3 (T, S) = S𝛿𝜎𝛿 , Θ4 (T, S) = R𝜎𝛿𝜎𝛿 , ...; 2) Γ1 (T, S) = L𝛿 , Γ2 (T, S) = K𝜎𝛿 , Γ3 (T, S) = L𝛿𝜎𝛿 , Γ4 (T, S) = K𝜎𝛿𝜎𝛿 , . . . . It follows from Lemma 4 that co-Y𝛼 (T, S) = Θ𝛼 (T, S) for every 𝛼 ∈ 𝜔1 . Lemma 5. Let S be an ensemble on a set T and 𝛼 ∈ 𝜔1 . Then, 1) Θ𝛼 (T, S)𝜎 = Y𝛼+1 (T, S); 2) Y𝛼 (T, S)𝛿 = Θ𝛼+1 (T, S). Proof. If 𝛼 is even, then by Lemma 1, Θ𝛼 = Σ𝛼 (T, R) implies (Θ𝛼 )𝜎 = Σ𝛼+1 (T, R) = Y𝛼+1 . If 𝛼 is odd, then Y𝛼 = Σ𝛼 (T, R) implies (Y𝛼 )𝛿 = Σ𝛼+1 (T, R) = Θ𝛼+1 . Going to the corresponding co-ensembles, we complete the proof. Lemma 6. Let S be a perfect ensemble on a set T. Then, for all 𝛼 ∈ [1, 𝜔1 [, the ensembles Y𝛼 (T, S) are perfect and the ensembles Θ𝛼 (T, S) are co-perfect. Proof. If 𝛼 is even, then Y𝛼 = Δ 𝛼 ≡ Δ 𝛼 (T, S) and Θ𝛼 = Σ𝛼 ≡ Σ𝛼 (T, S). By Lemma 1, (Θ𝛼 )𝜎 = Σ𝛼+1 . By virtue of Proposition 4, we infer that Y𝛼 = Δ 𝛼 ⊂ Σ𝛼+1 = (Θ𝛼 )𝜎 . If 𝛼 is odd, then Y𝛼 = Σ𝛼 and Θ𝛼 = Δ 𝛼 . By Lemma 1, (Θ𝛼 )𝜎 = Δ 𝛼+1 . As above, we infer that Y𝛼 = Σ𝛼 ⊂ Δ 𝛼+1 = (Θ𝛼 )𝜎 . By condition, S has the edges. Thus, R has the edges. It follows from the definition that either S ⊂ Y𝛼 or R ⊂ Y𝛼 . Consequently, Y𝛼 has the edges. By virtue of Proposition 1 Y𝛼 is latticed. Thus, Y𝛼 is perfect. Proposition 5. Let S be an ensemble on a set T such that the ensemble Y𝛼 (T, S) is perfect for some 𝛼 ∈ 𝜔1 . Then, the ensembles Y𝛽 (T, S) are perfect and latticed for all 𝛽 ∈ [𝛼 + 1, 𝜔1 [. Proof. Consider the subset B1 of the set 𝜔1 consisting of all 𝛽 ∈]𝛼, 𝜔1 [ such that Y𝛽 is perfect and latticed. By Lemma 5 (Y𝛼 )𝛿 = Θ𝛼+1 . Therefore, the condition Θ𝛼 ⊂ (Y𝛼 )𝛿 implies Y𝛼+1 = (Θ𝛼 )𝜎 ⊂ (Y𝛼 )𝛿𝜎 = (Θ𝛼+1 )𝜎 . This means that 𝛼 + 1 ∈ B1 , i. e. B1 is non-empty. Take B ≡ B1 ∪ (𝛼 + 1). Let 𝛽 ∈]𝛼 + 1, 𝜔1 [, and 𝛽 ⊂ B. At first, assume that 𝛽 is even. Then, Y𝛽 = Δ 𝛽 = E𝜎 , where E ≡ ⋃⟮Δ 𝛾 | 𝛾 ∈ 𝛽⟯. According to Lemma 1, the Δ- and Σ-collections are increasing. Hence, E ≡ ⋃⟮Δ 𝛾 | 𝛾 ∈ [𝛼 + 1, 𝛽[⟯. If 𝛾 is odd, then Δ 𝛾 = Θ𝛾 . By assumption, Y𝛾 is perfect and Θ𝛾 is co-perfect. Using this property, Lemma 5, and Lemma 1, we get Δ 𝛾 = Θ𝛾 ⊂ (Y𝛾 )𝛿 = Θ𝛾+1 = Σ𝛾+1 ⊂ Σ𝛽 . If 𝛾 is even, then by the same reason Δ 𝛾 = Y𝛾 ⊂ (Θ𝛾 )𝜎 = Y𝛾+1 = Σ𝛾+1 ⊂ Σ𝛽 . Consequently, E ⊂ Σ𝛽 . As a result, we get Y𝛽 = Esi ⊂ (Σ𝛽 )𝜎 = (Θ𝛽 )𝜎 .

2.1.2 The four transfinite collections of extensions of an ensemble |

27

If 𝛼 is odd, then Δ 𝛼 = Θ𝛼 . By condition, Θ𝛼 has the edges. Since Θ𝛼 ⊂ E ⊂ E𝜎 = Y𝛽 , we infer that Y𝛽 has the edges. If 𝛼 is even, then Δ 𝛼 = Y𝛼 ⊂ E ⊂ Y𝛽 implies the same property. Since Δ 𝛼 = Θ𝛼 or Δ 𝛼 = Y𝛼 , we conclude that the ensembles Δ 𝛾 are latticed for all 𝛾 ∈]𝛼, 𝛽[. Besides, the Δ-collection is increasing. It follows from these properties that the ensemble E is latticed. Using Lemma 1, we conclude that Y𝛽 = E𝜎 is latticed. As a result, we get that Y𝛽 is latticed and perfect. Now, assume that 𝛽 is odd. Then, Y𝛽 = Σ𝛽 = F𝜎 , where F ≡ ⋃⟮Σ𝛾 | 𝛾 ∈ 𝛽⟯. As above, F ≡ ⋃⟮Σ𝛾 | 𝛾 ∈ [𝛼+1, 𝛽[⟯. If 𝛾 is odd, then Σ𝛾 = Y𝛾 ⊂ (Θ𝛾 )𝜎 = Y𝛾+1 = Δ 𝛾+1 ⊂ Δ 𝛽 = Θ𝛽 . If 𝛾 is even, then, as above, Σ𝛾 = Θ𝛾 ⊂ (Y𝛾 )𝛿 = Θ𝛾+1 = Δ 𝛾+1 ⊂ Δ 𝛽 = Θ𝛽 . As a result, we get Y𝛽 = F𝜎 ⊂ (Θ𝛽 )𝜎 . By the same arguments as above, we check that Y𝛽 has the edges and is latticed. Thus, in this case, we also conclude that Y𝛽 is latticed and perfect. As a result, we get 𝛽 ∈ B. Using the principle of transfinite induction from 1.2.8, we conclude that B = 𝜔1 . Corollary 1. Let S be an ensemble on a set T such that the ensemble R𝜎 is perfect. Then, the ensembles Y𝛼 (T, S) are perfect and latticed for all 𝛼 ∈ [2, 𝜔1 [. If S is perfect, then we can take 𝛼 = 1. By virtue of Lemma 2, the Zakharov – Koldunov collections ⟮Λ 𝛼 (T, S) | 𝛼 ∈ 𝜔1 ⟯ and ⟮Γ𝛼 (T, S) | 𝛼 ∈ 𝜔1 ⟯ are increasing. In contrast to them the Young – Hausdorff collections ⟮Y𝛼 (T, S) | 𝛼 ∈ 𝜔1 ⟯ and ⟮Θ𝛼 (T, S) | 𝛼 ∈ 𝜔1 ⟯ have not such a property for an arbitrary ensemble S. But under some conditions, they are also increasing. Lemma 7. Let S be an ensemble on a set T with co-ensemble R such that R𝜎 is perfect. Then, 1 ⩽ 𝛼 ⩽ 𝛽 implies Y𝛼 (T, S) ⊂ Y𝛽 (T, S) and Θ𝛼 (T, S) ⊂ Θ𝛽 (T, S). If S is perfect, then we can take 0 ⩽ 𝛼 ⩽ 𝛽. Proof. In this proof, let again Σ𝛽 ≡ Σ𝛽 (T, R), Δ 𝛽 ≡ Δ 𝛽 (T, S). Let 1 ⩽ 𝛼 < 𝛽. At first assume that 𝛼 is even. According to Lemma 6, Y𝛼 is perfect for every 𝛼 ⩾ 1. Therefore, Y𝛼 ⊂ (Θ𝛼 )𝜎 = (Σ𝛼 )𝜎 , and by Lemma 1, Y𝛼 ⊂ Σ𝛼+1 . If 𝛽 is odd, then Y𝛼 ⊂ Σ𝛽 = Y𝛽 . If 𝛽 is even, then 𝛼+2 ⩽ 𝛽. As above, Y𝛼 ⊂ Σ𝛼+1 = Y𝛼+1 ⊂ (Θ𝛼+1 )𝜎 = (Δ 𝛼+1 )𝜎 = Δ 𝛼+2 ⊂ Δ 𝛽 = Y𝛽 . Now, assume that 𝛼 is odd. By the same arguments, Y𝛼 ⊂ (Θ𝛼 )𝜎 = (Δ 𝛼 )𝜎 = Δ 𝛼+1 . If 𝛽 is even, then Y𝛼 ⊂ Δ 𝛽 = Y𝛽 . If 𝛽 is odd, then 𝛼 + 2 ⩽ 𝛽. As above, Y𝛼 ⊂ Δ 𝛼+1 = Y𝛼+1 ⊂ (Θ𝛼+1 )𝜎 = (Σ𝛼+1 )𝜎 = Σ𝛼+2 ⊂ Σ𝛽 = Y𝛽 . Corollary 1. Let S be an ensemble on a set T with co-ensemble R such that R𝜎 is perfect. Then,

28 | 2.1 Descriptive and prescriptive spaces

1) Σ𝛼 (T, R) ⊂ Δ 𝛼+1 (T, S) for every 𝛼 ∈= [1, 𝜔1 [; 2) Δ 𝛼 (T, S) ⊂ Σ𝛼+1 (T, R) for every 𝛼 ∈= [1, 𝜔1 [. Lemma 8. Let S be a perfect ensemble on a set T. Then, 1) Y𝛼 (T, S) = (⋃⟮Y𝛽 (T, S) ∪ Θ𝛽 (T, S) | 𝛽 ∈ 𝛼⟯)𝜎 for every 𝛼 ∈ [1, 𝜔1 [;

2) Θ𝛼 (T, S) = (⋃⟮Y𝛽 (T, S) ∪ Θ𝛽 (T, S) | 𝛽 ∈ 𝛼⟯)𝛿 for every 𝛼 ∈ [1, 𝜔1 [. Proof. 1. First assume that 𝛼 is even. Then, Y𝛼 = Δ 𝛼 (S) = E𝜎 , where E ≡ ⋃⟮Δ 𝛽 (S) | 𝛽 ∈ 𝛼⟯. Consider F ≡ ⋃⟮Y𝛽 ∪ Θ𝛽 | 𝛽 ∈ 𝛼⟯. If 𝛽 is even, then Δ 𝛽 (S) = Y𝛽 . If 𝛽 is odd, then Δ 𝛽 (S) = Θ𝛽 . Consequently, E ⊂ F. This implies Y𝛼 = E𝜎 ⊂ F𝜎 . Check the inverse inclusion. If 𝛽 is even, then by Lemmas 7 and 1, Θ𝛽 ⊂ Θ𝛽+1 = Δ 𝛽+1 (S) ⊂ Δ 𝛼 (S) = Y𝛼 and Y𝛽 = Δ 𝛽 (S) ⊂ E ⊂ Y𝛼 . If 𝛽 is odd, then by the same reasons, Y𝛽 ⊂ Y𝛽+1 = Δ 𝛽+1 (S) ⊂ Δ 𝛼 (S) = Y𝛼 and Θ𝛽 = Δ 𝛽 (S) ⊂ E ⊂ Y𝛼 . Consequently, F ⊂ Y𝛼 . Since Y𝛼 is 𝜎-additive, it follows that F𝜎 ⊂ Y𝛼 . Thus, Y𝛼 = F𝜎 . Now, assume that 𝛼 is odd. Then, Y𝛼 = Σ𝛼 (R) = E󸀠𝜎 , where E󸀠 ≡ ⋃⟮Σ𝛽 (R) | 𝛽 ∈ 𝛼⟯. If 𝛽 is even, then Σ𝛽 (R) = Θ𝛽 . If 𝛽 is odd, then Σ𝛽 (R) = Y𝛽 . Consequently, E󸀠 ⊂ F. This implies Y𝛼 = E󸀠𝜎 ⊂ F𝜎 . Now, check the inverse inclusion. If 𝛽 is even, then by Lemmas 7 and 1, Y𝛽 ⊂ Y𝛽+1 = Σ𝛽+1 (R) ⊂ Σ𝛼 (R) = Y𝛼 and Θ𝛽 = Σ𝛽 (R) ⊂ E󸀠 ⊂ Y𝛼 . If 𝛽 is odd, then by the same reasons, Θ𝛽 ⊂ Θ𝛽+1 = Σ𝛽+1 (R) ⊂ Σ𝛼 (R) = Y𝛼 and Y𝛽 = Σ𝛽 (R) ⊂ E󸀠 ⊂ Y𝛼 . Consequently, F ⊂ Y𝛼 and F𝜎 ⊂ Y𝛼 . Equality 2 is proven in the same way. Lemma 9. Let S be a perfect 𝜎-additive ensemble on a set T. Then, 1) Y𝛼 (T, S) = Y𝛼+1 (T, R) and Θ𝛼 (T, S) = Θ𝛼+1 (T, R) for every 𝛼 ∈ 𝜔; 2) Y𝛼 (T, S) = Y𝛼 (T, R) and Θ𝛼 (T, S) = Θ𝛼 (T, R) for every 𝛼 ∈ [𝜔, 𝜔1 [. According to Lemma 2, for the Zakharov – Koldunov collections, Λ 𝛼 (T, S) = Λ 𝛼 (T, R) and Γ𝛼 (T, S) = Γ𝛼 (T, R) for all 𝛼 ∈ 𝜔1 . Proof. Note that by condition, the ensembles S and S𝜎 are perfect. Therefore, by virtue of Lemma 7, the collections ⟮Y𝜉 (T, S) | 𝜉 ∈ 𝜔1 ⟯, ⟮Θ𝜉 (T, S) | 𝜉 ∈ 𝜔1 ⟯, ⟮Y𝜉 (T, R) | 𝜉 ∈ [1, 𝜔1 ]⟯, ⟮Θ𝜉 (T, R) | 𝜉 ∈ [1, 𝜔1 ]⟯ are increasing. 1. Consider a subset N of the set 𝜔 consisting of all numbers 𝛼 ∈ 𝜔 for which property 1 is fulfilled. If n = 0, then Y0 (S) = S = S𝜎 = Y1 (R) and Θ0 (S) = R = R𝛿 = Θ1 (R). Thus, 0 ∈ N. Assume that n ∈ N. Then, by Lemma 5, Yn+1 (S) = Θn (S)𝛿 = Θn+1 (R)𝛿 = Yn+2 (R) and Θn+1 (S) = Yn (S)𝛿 = Yn+1 (R)𝛿 = Θn+2 (R). This means that n + 1 ∈ N. By the principle of natural induction, N = 𝜔. 2. First check the equalities in (2) for 𝛼 = 𝜔. By Lemma 8, we have L ≡ Y𝜔 (S) = (⋃⟮Y𝛾 (S) ∪ Θ𝛾 (S) | 𝛾 ∈ 𝜔⟯)𝜎 and M ≡ Y𝜔 (R) = (⋃⟮Y𝛾 (R) ∪ Θ𝛾 (R) | 𝛾 ∈ 𝜔⟯)𝜎 .

2.1.2 The four transfinite collections of extensions of an ensemble |

29

If E ∈ L, then E = ⋃⟮E n | n ∈ 𝜔⟯ for some sequences ⟮𝛾n ∈ 𝜔 | n ∈ 𝜔⟯ and ⟮E n | n ∈ 𝜔⟯ such that E n ∈ Y𝛾n (S) ∪ Θ𝛾n (S). According to (1), Y𝛾n (S) ∪ Θ𝛾n (S) = Y𝛾n +1 (R) ∪ Θ𝛾n +1 (R). Since 𝛾n + 1 ∈ 𝜔, we infer that E ∈ M. This means that L ⊂ M. Conversely, if F ∈ M, then as above, F = ⋃⟮F n | n ∈ 𝜔⟯ for some sequences ⟮𝛿n ∈ 𝜔 | n ∈ 𝜔⟯ and ⟮F n | n ∈ 𝜔⟯ such that F n ∈ Y𝛿n (R) ∪ Θ𝛿n (R). If 𝛿n = 0, then Y𝛿n (R) ∪ Θ𝛿n (R) = R ∪ S ⊂ R𝛿𝜎 ∪ S𝜎𝛿 = Y𝛿n +2 (R) ∪ Θ𝛿n +2 (R). If 𝛿n ⩾ 1, then by Lemma 7 Y𝛿n (R) ∪ Θ𝛿n (R) ⊂ Y𝛿n +2 (R) ∪ Θ𝛿n +2 (R). According to (1), Y𝛿n +2 (R) = Y𝛿n +1 (S) and Θ𝛿n +2 (R) = Θ𝛿n +1 (S). Since 𝛿n + 1 ∈ 𝜔, we infer that F ∈ L. This means that M ⊂ L. As a result, we get L = M. Now, consider the subset B of the set [𝜔, 𝜔1 [ consisting of all numbers 𝛼 ∈ [𝜔, 𝜔1 [ for which property 2 is fulfilled. As we checked above, 𝜔 ∈ B. Consider the set B ≡ 𝜔 ∪ B1 . Let 𝛽 ∈ B∩]𝜔0 , 𝜔1 [. By Lemma 8, L󸀠 ≡ Y𝛽 (S) = (⋃⟮Y𝛾 (S) ∪ Θ𝛾 (S) | 𝛾 ∈ 𝛽⟯)𝜎 and

M 󸀠 ≡ Y𝛽 (R) = (⋃⟮Y𝛾 (R) ∪ Θ𝛾 (R) | 𝛾 ∈ 𝛽⟯)𝜎 . If E ∈ L󸀠 , then as above, E = ⋃⟮E n | n ∈ 𝜔⟯, E n ∈ Y𝛾n (S) ∪ Θ𝛾n (S), for some sequence ⟮𝛾n ∈ 𝛽 | n ∈ 𝜔⟯ . If 𝛾n < 𝜔 , then, as above, Y𝛾n (S) ∪ Θ𝛾n (S) ⊂ Y𝛾n +2 (S) ∪ Θ𝛾n +2 (S) ⊂ Y𝜔 (S) ∪ Θ𝜔 (S). Therefore, E n ∈ E ≡ ⋃⟮Y𝛾 (S) ∪ Θ𝛾 (S) | 𝛾 ∈ [𝜔, 𝛽[⟯ for every n. Using the inductive assumption, we get E = ⋃⟮Y𝛾 (R) ∪ Θ𝛾 (R) | 𝛾 ∈ [𝜔, 𝛽[⟯ ⊂ ⋃⟮Y𝛾 (R) ∪ Θ𝛾 (R) | 𝛾 ∈ 𝛽⟯. Consequently, E ∈ M 󸀠 . This means that L󸀠 ⊂ M 󸀠 . The inverse inclusion M 󸀠 ⊂ L󸀠 is checked completely in the same way. As a result, we get L󸀠 = M 󸀠 , so 𝛽 ∈ B. Using the principle of transfinite induction, we conclude that B = 𝜔1 . Lemma 10. Let S be an ensemble on a set T. Then, for every 𝛼 ∈ [2, 𝜔1 [ and every A ∈ Y𝛼 (T, S) [A ∈ Θ𝛼 (T, S)] , there are sequences (𝛽k ∈ [1, 𝛼[| k ∈ 𝜔) and ⟮A k | k ∈ 𝜔⟯ such that A k ∈ Θ𝛽k (T, S) [A k ∈ Y𝛽k (T, S)] and A = ⋃⟮A k | k ∈ 𝜔⟯ [A = ⋂⟮A k | k ∈ 𝜔⟯]. Proof. First, assume that 𝛼 is an even number. Then, Y𝛼 = Δ 𝛼 (S) = (⋃⟮Δ 𝛾 (S) | 𝛾 ∈ 𝛼⟯)𝜎 . Therefore, we can find sequences ⟮𝛾m ∈ 𝛼 | m ∈ 𝜔⟯ and ⟮K m ∈ Δ 𝛾m (S) | m ∈ 𝜔⟯ such that A = ⋃⟮K m | m ∈ 𝜔⟯. If 𝛾m is odd, then K m ∈ Δ 𝛾m (S) = Θ𝛾m . Define new sequences ⟮K mi ∈ Θ𝛾m | i ∈ 𝜔⟯ and ⟮𝛾mi ∈ 𝛼 | i ∈ 𝜔⟯, setting K mi ≡ K m and 𝛾mi ≡ 𝛾m ⩾ 1 for every i. If 𝛾 is even and non-zero, then Y𝛾m = Δ 𝛾m (S) = (⋃⟮Δ 𝜉 (S) | 𝜉 ∈ 𝛾m ⟯)𝜎 . Therefore, there are sequences ⟮𝜉mi ∈ 𝛾m | i ∈ 𝜔⟯ and ⟮K mi ∈ Δ 𝜉mi (S) | i ∈ 𝜔⟯ such that K m = ⋃⟮K mi | i ∈ 𝜔⟯. If 𝜉mi is odd, then Δ 𝜉mi (S) = Θ𝜉mi . If 𝜉mi is even, then by Lemma 1 Δ 𝜉mi (S) ⊂ Δ 𝜉mi +1 (S) = Θ𝜉mi +1 and 𝜉mi + 1 ⩽ 𝛾m < 𝛼. Define a new sequence ⟮𝛾mi ∈ 𝛼 | i ∈ 𝜔⟯ setting 𝛾mi ≡ 𝜉mi ⩾ 1 if 𝜉mi is odd and 𝛾mi ≡ 𝜉mi + 1 ⩾ 1 if 𝜉mi is even. If 𝛾m = 0, then Y𝛾m = S ⊂ S𝛿 = Θ1 . In this case, we set K mi ≡ K m and 𝛾mi ≡ 1 for every i.

30 | 2.1 Descriptive and prescriptive spaces

We proved that in all the cases, K mi ∈ Θ𝛾mi and 𝛾mi ∈ [1, 𝛼[. From A = ⋃⟮K m | m ∈ 𝜔⟯ and K m = ⋃⟮K mi | i ∈ 𝜔⟯, we infer that A = ⋃⟮K mi | (m, i) ∈ 𝜔 × 𝜔⟯. Since there is a bijective mapping u : 𝜔 𝜔×𝜔, we define new sequences ⟮𝛽k ∈ 𝛼 | k ∈ 𝜔⟯ and ⟮A k ∈ Θ𝛽k | k ∈ 𝜔⟯ setting 𝛽k ≡ 𝛾mi and A k ≡ K mi if u(k) = (m, i). They are the necessary sequences. If 𝛼 is an odd number, then all the arguments are analogous. Lemma 11. Let T be a set and A ⊂ T. If there are sequences ⟮⟮B mn ∈ P(T) | n ∈ 𝜔⟯ | m ∈ 𝜔⟯ and ⟮⟮C mn ∈ P(T) | n ∈ 𝜔⟯ | m ∈ 𝜔⟯ such that A = ⋃⟮⋂⟮B mn | n ∈ 𝜔⟯ | m ∈ 𝜔⟯ = ⋂⟮⋃⟮C mn | n ∈ 𝜔⟯ | m ∈ 𝜔⟯ and ⋃⟮C m+1,n | n ∈ 𝜔⟯ ⊂ ⋃⟮C mn | n ∈ 𝜔⟯, then A = ⋃⟮⋂⟮A n+k | k ∈ 𝜔⟯ | n ∈ 𝜔⟯ = ⋂⟮⋃⟮A n+k | k ∈ 𝜔⟯ | n ∈ 𝜔⟯, where A n = ⋃⟮⋂⟮B ki | i ∈ n + 1⟯ ∩ ⋃⟮C ki | i ∈ n + 1⟯ | k ∈ n + 1⟯. One can see the proof of this result of W. Sierpiński in [Kuratowski, 1966, 30.IX]. However, there are two misprints there. Proof. Along with T\P, we shall write here P c . First, we shall prove the imbeddings A ⊂ ⋃⟮⋂⟮A n+k | k ∈ 𝜔⟯ | n ∈ 𝜔⟯ ≡ lim inf A n and A c ⊂ lim inf A cn . Take any t ∈ A. By condition, there is an index k such that t ∈ ⋂⟮B km | m ∈ 𝜔⟯ and there is a sequence ⟮i m ∈ 𝜔 | m ∈ 𝜔⟯ such that t ∈ ⋂⟮C mi m | m ∈ 𝜔⟯. Consider n0 = k ∨ i k . Then, for every n ⩾ n0 , we have C ki k ⊂ ⋃⟮C ki | i ∈ n + 1⟯. Consequently, t ∈ ⋂⟮B ki | i ∈ n + 1⟯ ∩ ⋃⟮C ki | i ∈ n + 1⟯ ⊂ A n . This implies the first embedding. Take any t ∈ A c . By condition, there is an index m such that t ∈ ⋂⟮C cmn | n ∈ 𝜔⟯ ⊂ ⋂⟮C ckn | n ∈ 𝜔⟯ for every k ⩾ m. Also, there is sequence ⟮i n ∈ 𝜔 | n ∈ 𝜔⟯ such that t ∈ ⋂⟮B cni n | n ∈ 𝜔⟯. Consider n0 = gr{i k | k ∈ m + 1}. Then, for every n ⩾ n0 , we have t ∈ B cki k ⊂ ⋃⟮B cki | i ∈ n + 1⟯ for every k ⩽ m and t ∈ ⋂⟮C cki | i ∈ n + 1⟯ for every k ⩾ m. Consequently, for every k ⩽ n, we have t ∈ ⋃⟮B cki | i ∈ n + 1⟯ ∪ ⋂⟮C cki | i ∈ n + 1⟯, i. e. t ∈ A cn . This implies the second imbedding. Going to the complement in this one, we get A ⊃ lim sup A n ≡ ⋂⟮⋃⟮A m+k | k ∈ 𝜔⟯ | m ∈ 𝜔⟯. Now, the necessary equality follows from the embeddings A ⊂ lim inf A n ⊂ lim sup A n ⊂ A. Using this complicated lemma, we shall derive the following result.

2.1.3 Classification of Borel sets for arbitrary and perfect ensembles | 31

Theorem 1. Let S be an ensemble on a set T such that R𝜎 is perfect. Then, for every 𝛼 ∈ [3, 𝜔1 [ and for every A ∈ Y𝛼 (T, S) ∩ Θ𝛼 (T, S), there is a sequence (A n ∈ ⋃⟮Y𝛽 ∩ Θ𝛽 | 𝛽 ∈ [1, 𝛼[⟯ | n ∈ 𝜔) such that A = ⋃⟮⋂⟮A n+k | k ∈ 𝜔⟯ | n ∈ 𝜔⟯ = ⋂⟮⋃⟮A n+k | k ∈ 𝜔⟯ | n ∈ 𝜔⟯. If S is perfect, then we can take 𝛼 ∈ [2, 𝜔1 [. Moreover, if 𝜘 is a limit number and 𝛼 = 𝜘 + 1, then we can take A n ∈ ⋃⟮Y𝛽 ∩ Θ𝛽 | 𝛽 ∈ [1, 𝜘[⟯. Proof. By Corollary 1 to Proposition 5, Y𝛽 is perfect and latticed for every 𝛽 ⩾ 2. Thus, Θ𝛽 is also latticed. By Lemma 10, we have the equalities A = ⋃⟮K m | m ∈ 𝜔⟯ and T\A = ⋃⟮M m | m ∈ 𝜔⟯ for some sequences (𝛽m ∈ [1, 𝛼[| m ∈ 𝜔), (K m ∈ Θ𝛽m | m ∈ 𝜔), (𝛿m ∈ [1, 𝛼[| m ∈ 𝜔), and (M m ∈ Θ𝛿m | m ∈ 𝜔). For every m ∈ 𝜔, consider 𝛾m ≡ gr (𝛿i + 1 | i ∈ m + 1) ∈ [2, 𝛼[ and L m ≡ ⋃⟮M i | i ∈ m + 1⟯. By Lemma 7, M i ∈ Θ𝛾m for every i ∈ m + 1. Since Θ𝛾m is latticed, it follows that L m ∈ Θ𝛾m . Besides the sequence, (L m | m ∈ 𝜔) is increasing. Now, consider 𝛼m ≡ 𝛽m ⊻ 𝛾m ⊻ 2 ∈ [2, 𝛼[. By Lemma 7, K m , L m ∈ Θ𝛼m . Again, by Lemma 10, we obtain the equalities K m = ⋂⟮B mn | n ∈ 𝜔⟯ and L m = ⋂⟮D mn | n ∈ 𝜔⟯ for some sequences (𝜇mn ∈ [1, 𝛼m [| n ∈ 𝜔), (𝜈mn ∈ [1, 𝛼m [| n ∈ 𝜔), (B mn ∈ Y𝜇mn | n ∈ 𝜔), and (D mn ∈ Y𝜈mn | n ∈ 𝜔). Denote T\D mn by C mn . Then, the sequences ((B mn | n ∈ 𝜔) | m ∈ 𝜔) and ((C mn | n ∈ 𝜔) | m ∈ 𝜔) satisfy the condition of Lemma 11. Consider the sets X kn ≡ ⋂⟮B ki | i ∈ n + 1⟯, Y kn ≡ ⋃⟮C ki | i ∈ n + 1⟯, and A n ≡ ⋃⟮X kn ∩ Y kn | k ∈ n +1⟯. By Lemma 11, the sequence (A n | n ∈ 𝜔) satisfies the necessary equalities in our theorem. Consider the numbers 𝜘n ≡ gr (𝜇ki ⊻ 𝜈ki | (k, i) ∈ (n + 1) × (n + 1)) ∈ [1, 𝛼n [. By Lemma 7, B ki ∈ Y𝜘n and C ki ∈ Θ𝜘n for all k, i ∈ n + 1. Since Y𝛽 and Θ𝛽 are latticed for 𝛽 ⩾ 2, we infer that X kn ∈ Y𝜘n ⊂ Y𝜘n +1 and Y kn ∈ Θ𝜘n ⊂ Θ𝜘n +1 . Applying Lemma 5, we get Y𝜘n ⊂ (Y𝜘n )𝛿 = Θ𝜘n +1 and Θ𝜘n ⊂ (Θ𝜘n )𝜎 = Y𝜘n +1 . Therefore, X kn ∈ Θ𝜘n +1 ∩ Y𝜘n +1 and Y kn ∈ Y𝜘n +1 ∩ Θ𝜘n +1 for every k ∈ n + 1. This implies A n ∈ Y𝜘n +1 ∩ Θ𝜘n +1 . Since 𝜘n + 1 ⩽ 𝛼n < 𝛼, we get the necessary assertion. To conclude the proof, it remains to consider the case 𝛼 = 𝜘 + 1, where 𝜘 is a limit number. Since 𝜘n < 𝛼n ⩽ 𝜘, we infer that 𝜘n + 1 < 𝜘.

2.1.3 Classification of Borel sets for arbitrary and perfect ensembles Fix a set T and an ensemble S on T. Define the Borel envelope B(T, S) of the ensemble S (the family of Borel sets with respect to the ensemble S) as the smallest 𝜎-algebra containing S.

Zakharov – Koldunov classification Theorem 1 (the first general classification theorem). Let S be an ensemble on a set T. Then,

32 | 2.1 Descriptive and prescriptive spaces

1) B(T, S) = Λ 𝜔1 (T, S) = ⋃⟮Λ 𝛼 (T, S) | 𝛼 ∈ 𝜔1 ⟯ (𝜎-additive classification); 2) B(T, S) = Γ𝜔1 (T, S) = ⋃⟮Γ𝛼 (T, S) | 𝛼 ∈ 𝜔1 ⟯ (𝛿-multiplicative classification). Proof. By Proposition 3 (2.1.2), the ensemble E ≡ ⋃⟮Λ 𝛼 | 𝛼 ∈ 𝜔1 ⟯ = Λ 𝜔1 is a 𝜎-algebra containing the ensemble S. Let A be any 𝜎-algebra containing the ensemble S. Take a subset B of the set 𝜔1 consisting of all ordinal numbers 𝛼 ∈ 𝜔1 such that Λ 𝛼 ⊂ A. From Λ 0 ≡ S, we infer that 0 ∈ B, i. e. B is non-empty. Let 𝛽 ∈ 𝜔1 and 𝛽 ⊂ B. Consider the ensemble F ≡ ⋃⟮Λ 𝛾 | 𝛾 ∈ 𝛽⟯. By assumption, Λ 𝛾 ⊂ A for every 𝛾 ∈ 𝛽. Thus, F ⊂ A and co-F ⊂ A. By construction, Λ 𝛽 = F𝜆 . Take any element H of this set. By definition, H = ⋃⟮⋂⟮F im ∩ G im | m ∈ M i ⟯ | i ∈ I⟯ for some countable set I and some finite collections ⟮F im ∈ F ∪ {T} | m ∈ M i ⟯ and ⟮G im ∈ co-F ∪ {T} | m ∈ M i ⟯. From F im ∈ A and G im ∈ A, we infer that H ∈ A. Thus, Λ 𝛽 ⊂ A. This means that 𝛽 ∈ B. Using the principle of transfinite induction from 1.2.8, we conclude that B = 𝜔1 . Therefore, E ⊂ A. This yields that E = B(T, S). For Γ-ensembles, the arguments are analogous. This result was first presented in slightly different form in papers [Zakharov and Koldunov, 1982a; 1982b] and in this form in [Zakharov, 1989b]. Young – Hausdorff classification in the general case Proposition 1. Let S be an ensemble on a set T. Then, the ensembles Σ𝜔1 (T, S) = ⋃⟮Σ𝛼 (T, S) | 𝛼 ∈ 𝜔1 ⟯ and Δ 𝜔1 (T, S) = ⋃⟮Δ 𝛼 (T, S) | 𝛼 ∈ 𝜔1 ⟯ are the smallest of all countably additive and countably multiplicative ensembles on T containing the ensemble S. Proof. Let A be such an ensemble. Consider the ensemble E ≡ ⋃⟮Σ𝛼 | 𝛼 ∈ 𝜔1 ⟯. By Proposition 2 (2.1.2), it is countably additive and countably multiplicative, S ⊂ E, and Σ𝜔1 = E. Check that E ⊂ A. Take a subset B of the set 𝜔1 consisting of all ordinal numbers 𝛼 ∈ 𝜔1 such that Σ𝛼 ⊂ A. From Σ0 ≡ S, we infer that 0 ∈ B, i. e. B is non-empty. Let 𝛽 ∈ 𝜔1 and 𝛽 ⊂ B. Consider the ensemble F ≡ ⋃⟮Σ𝛾 | 𝛾 ∈ 𝛽⟯. By assumption, Σ𝛾 ⊂ A for every 𝛾 ∈ 𝛽. Thus, F ⊂ A. If 𝛽 is odd, then Σ𝛽 = F𝜎 ⊂ A. If 𝛽 is even, then Σ𝛽 = F𝛿 ⊂ A. Consequently, 𝛽 ∈ B. Using the principle of transfinite induction, we conclude that B = 𝜔1 . Therefore, E ⊂ A. For Δ-ensembles, the arguments are analogous. Corollary 1. Let S be an ensemble on a set T. Then, Σ𝜔1 (T, S) = ⋃⟮Σ𝛼 (T, S) | 𝛼 ∈ 𝜔1 ⟯ ⊂ B(T, S) and Δ 𝜔1 (T, S) = ⋃⟮Δ 𝛼 (T, S) | 𝛼 ∈ 𝜔1 ⟯ ⊂ B(T, S). Unfortunately, the Young – Hausdorff collections of an arbitrary ensemble S do not generate its Borel envelope B(T, S) in such a manner, as the Zakharov – Koldunov

2.1.3 Classification of Borel sets for arbitrary and perfect ensembles | 33

collections make this according to Theorem 1. But using the derivative ensembles, we can prove. Theorem 2 (the second general classification theorem). Let S be an ensemble on a set T with the derivative ensembles L ≡ L(T, S) and K ≡ K(T, S). Then, 1) B(T, S) = Σ𝜔1 (T, K) = ⋃⟮Σ𝛼 (T, K) | 𝛼 ∈ 𝜔1 ⟯; 2) B(T, S) = Δ 𝜔1 (T, L) = ⋃⟮Δ 𝛼 (T, L) | 𝛼 ∈ 𝜔1 ⟯. Proof. By Proposition 2 (2.1.2), the ensemble E ≡ ⋃⟮Σ𝛼 | 𝛼 ∈ 𝜔1 ⟯ is countably additive and countably multiplicative. Take a subset B of the set 𝜔1 consisting of all ordinal numbers 𝛼 ∈ 𝜔1 such that co-Σ𝛼 ⊂ E. For 𝛼 = 0, we have co-Σ0 = co-K = L. Take any element L ∈ L. By definition, L = ⋃⟮S m ∪ R m | m ∈ M⟯ for some finite collections ⟮S m ∈ {⌀} ∪ S | m ∈ M⟯ and ⟮R m ∈ {⌀} ∪ R | m ∈ M⟯. Since {⌀} ∪ S ∪ R ⊂ K = Σ0 , we infer that L ∈ E. This means that co-Σ0 ⊂ E. Thus, 0 ∈ B, i. e, B is non-empty. Let 𝛽 ∈ 𝜔1 and 𝛽 ⊂ B. Consider the ensemble F ≡ ⋃⟮Σ𝛾 | 𝛾 ∈ 𝛽⟯. By assumption, co-Σ𝛾 ⊂ E for every 𝛾 ∈ 𝛽. If 𝛽 is odd, then Σ𝛽 = F𝜎 . Take any element F ∈ F𝜎 . By definition, F = ⋃⟮F i | i ∈ I⟯ for some countable collection (F i ∈ F | i ∈ I). By definition, F i ∈ Σ𝛾 for some 𝛾 ∈ 𝛽. Therefore, the set X i ≡ {𝛾 ∈ 𝛽 | F i ∈ Σ𝛾 } is non-empty. By the choice axiom from 1.1.12, there exists a mapping p : P(𝛽)\{⌀} → 𝛽 such that p(P) ∈ P. Consider the element 𝛾i ≡ p(X i ) ∈ X i . Then, F i ∈ Σ𝛾i . By assumption, T\F i ∈ co-Σ𝛾i ⊂ E. Therefore, T\F = ⋂⟮T\F i | i ∈ I⟯ ∈ E. If 𝛽 is even, then Σ𝛽 = F𝛿 . Take any element G ∈ F𝛿 . By definition, G = ⋂⟮G j | j ∈ J⟯ for some countable collection (G j ∈ F | j ∈ J). As above, G j ∈ Σ𝛿j for some collection (𝛿j ∈ 𝛽 | j ∈ J). By assumption, T\G j ∈ co-Σ𝛿j ⊂ E. Therefore, T\G = ⋃⟮T\G j | j ∈ J⟯ ∈ E. This shows that co-Σ𝛽 ⊂ E. Thus, 𝛽 ∈ B. Using the principle of transfinite induction from 1.2.8, we conclude that B = 𝜔1 . Therefore, co-E = ⋃⟮co-Σ𝛼 | 𝛼 ∈ 𝜔1 ⟯ ⊂ E, i. e. E is closed under the complement. Now, from S ⊂ E, we infer that B(T, S) ⊂ E. By virtue of Corollary 1 to Proposition 1, E = Σ𝜔1 (T, K) ⊂ B(T, K). From K ⊂ B(T, S), we see that B(T, K) ⊂ B(T, S). Therefore, E ⊂ B(T, S). As a result, B(T, S) = E = Σ𝜔1 (T, K). The equalities in (2) are proven in the same way. The derivative ensembles were introduced and Theorem 2 was presented in paper [Zakharov, 1991].

Perfect and topological cases Now, we shall deduce the classification theorem in the perfect case from the second general classification theorem.

34 | 2.1 Descriptive and prescriptive spaces

Proposition 2. Let S be a perfect ensemble on a set T with the derivative ensembles L ≡ L(T, S) and K ≡ K(T, S). Then, 1) Σ𝛼 (T, K) = Σ𝛼 (T, R) for every 𝛼 ∈ [1, 𝜔1 [; 2) Δ 𝛼 (T, L) = Δ 𝛼 (T, S) for every 𝛼 ∈ [1, 𝜔1 [. Proof. Take a subset B1 of the set 𝜔1 consisting of all ordinal numbers 𝛼 ∈ 𝜔1 such that Σ𝛼 (K) = Σ𝛼 (R). For 𝛼 = 1, by Corollary 2 to Lemma 4 (2.1.1), we have Σ1 (K) = K𝜎 = R𝜎 = Σ1 (R). This means that 1 ∈ B1 . Consider the set B = B1 ∪ {⌀}. Let 2 ⩽ 𝛽 ∈ 𝜔1 and 𝛽 ⊂ B. By assumption, Σ𝛾 (K) = Σ𝛾 (R) for every 𝛾 ∈ 𝛽. If 𝛽 is odd, then Σ𝛽 (K) = (⋃⟮Σ𝛾 (K) | 1 ⩽ 𝛾 < 𝛽⟯)𝜎 = (⋃⟮Σ𝛾 (R) | 1 ⩽ 𝛾 < 𝛽⟯)𝜎 = Σ𝛽 (R). If 𝛽 is even, then Σ𝛽 (K) = (⋃⟮Σ𝛾 (K) | 1 ⩽ 𝛾 < 𝛽⟯)𝛿 = (⋃⟮Σ𝛾 (R) | 1 ⩽ 𝛾 < 𝛽⟯)𝛿 = Σ𝛽 (R). This means that 𝛽 ∈ B. Using the principle of transfinite induction from 1.2.8, we conclude that B = 𝜔1 . The second assertion is checked in the same way. Corollary 1. Let S be a perfect ensemble on a set T with the derivative ensembles L ≡ L(T, S) and K ≡ K(T, S). Then, 1) Y𝛼 (T, L) = Y𝛼 (T, S) for every 𝛼 ∈ [1, 𝜔1 [; 2) Θ𝛼 (T, L) = Θ𝛼 (T, S) for every 𝛼 ∈ [1, 𝜔1 [. Proof. If 𝛼 is even, then by Proposition 2, we have Y𝛼 (L) = Δ 𝛼 (L) = Δ 𝛼 (S) = Y𝛼 (S) and Θ𝛼 (L) = Σ𝛼 (K) = Σ𝛼 (R) = Θ𝛼 (S). If 𝛼 is odd, then Y𝛼 (L) = Σ𝛼 (K) = Σ𝛼 (R) = Y𝛼 (S) and Θ𝛼 (L) = Δ 𝛼 (L) = Δ 𝛼 (S) = Θ𝛼 (S). Theorem 3 (the second Young – Hausdorff classification theorem in the perfect case). Let S be a perfect ensemble on a set T. Then, 1) B(T, S) = Σ𝜔1 (T, R) = ⋃⟮Σ𝛼 (T, R) | 𝛼 ∈ 𝜔1 ⟯ (the mixed 𝜎𝛿-classification); 2) B(T, S) = Δ 𝜔1 (T, S) = ⋃⟮Δ 𝛼 (T, S) | 𝛼 ∈ 𝜔1 ⟯ (the mixed 𝛿𝜎-classification). Proof. These equalities follows immediately from Theorem 2, Proposition 2, and Proposition 1. Theorem 4 (the first Young – Hausdorff classification theorem in the perfect case). Let S be a perfect ensemble on a set T. Then, 1) B(T, S) = ⋃⟮Y𝛼 (T, S) | 𝛼 ∈ 𝜔1 ⟯ (the 𝜎-additive classification); 2) B(T, S) = ⋃⟮Θ𝛼 (T, S) | 𝛼 ∈ 𝜔1 ⟯ (the 𝛿-multiplicative classification). Proof. Denote the right part in the first equality by E. By Theorem 3, we have B = ⋃⟮Σ𝛼 (R) | 𝛼 ∈ 𝜔1 ⟯ ≡ F. Take any E ∈ E. By definition, E ∈ Y𝛼 (S) for some 𝛼. If 𝛼 is even, then by Proposition 4 (2.1.2) E ∈ Y𝛼 (S) = Δ 𝛼 (S) ⊂ Σ𝛼+1 (R) ⊂ F. If 𝛼 is odd, then E ∈ Y𝛼 (S) = Σ𝛼 (R) ⊂ F. This means E ⊂ F.

2.1.3 Classification of Borel sets for arbitrary and perfect ensembles | 35

Now, take any F ∈ F, i. e. F ∈ Σ𝛼 (R) for some 𝛼. If 𝛼 is even, then by Lemma 1 (2.1.2), F ∈ Σ𝛼 (R) ⊂ Σ𝛼+1 (R) = Y𝛼+1 (S) ⊂ E. If 𝛼 is odd, then F ∈ Σ𝛼 (R) = Y𝛼 (S) ⊂ E. This means F ⊂ E. As a result, we get E = F. Equality 2 is checked in the same way. Now, we shall apply the classification theorems 2 and 3 to a topological space ⟮T, G⟯ with an open topology G and the ensemble of closed sets F ≡ co-G. Consider for G its derivative ensembles L ≡ L(T, S) = {G ∪ F | G ∈ G ∧ F ∈ F} and K ≡ K(T, S) = {G ∩ F | G ∈ G∧ F ∈ F} (see 2.1.1). The ensembles Δ 𝛼 (T, L) will be denoted by L𝛼 and the ensembles Σ𝛼 (T, K) will be denoted by K𝛼 . Then, Theorem 2 implies the second general classification theorem for a topological space ⟮T, G⟯: 1) B(T, G) = ⋃⟮K𝛼 | 𝛼 ∈ 𝜔1 ⟯; 2) B(T, G) = ⋃⟮L𝛼 | 𝛼 ∈ 𝜔1 ⟯. The ensembles Δ 𝛼 (T, G) are usually denoted in literature by G𝛼 and the ensembles Σ𝛼 (T, F) are denoted by F𝛼 . Then, Theorem 3 implies the Young – Hausdorff classification theorem for a perfect topological space ⟮T, G⟯ (cf. [Kuratowski, 1966, 30.II] for metric space): 1) B(T, G) = ⋃⟮F𝛼 | 𝛼 ∈ 𝜔1 ⟯; 2) B(T, G) = ⋃⟮G𝛼 | 𝛼 ∈ 𝜔1 ⟯. For a perfect space, the equalities G𝛼 = L𝛼 and F𝛼 = K𝛼 are valid for all 𝛼 ∈ 𝜔1 . For an arbitrary topological space, we have, in general, only the inclusions G𝛼 ⊂ L𝛼 and F𝛼 ⊂ K𝛼 . Comparison of Young – Hausdorff and Zakharov – Koldunov classifications Now, we have two different classifications of Borel sets of a descriptive space ⟮T, S⟯: the first one is given by Theorem 1 and the second one is given by Theorem 2. Therefore, it is necessary to compare them. The following coincidence theorem (or untwining theorem) shows that these classifications coincide: Λ-scheme and Γ-scheme give one-thread constructions for separating out the 𝜎-additive and 𝛿-multiplicative threads from the double intertwined Young – Hausdorff Σ-Δ-construction. The untwining theorem is illustrated by the following diagram, where unbroken arrows show the 𝜎-additive thread, dashed arrows show the 𝛿-multiplicative thread.

36 | 2.1 Descriptive and prescriptive spaces

Theorem 5 (the untwining theorem). Let S be an ensemble on a set T with its derivative ensembles L and K. Then, 1) Λ 𝛼 (T, S) = Δ 𝛼 (T, L) for non-zero even ordinal numbers 𝛼 ∈ 𝜔1 and Λ 𝛼 (T, S) = Σ𝛼 (T, K) for odd ordinal numbers 𝛼 ∈ 𝜔1 (the countably additive coincidence); 2) Γ𝛼 (T, S) = Δ 𝛼 (T, L) for odd ordinal numbers 𝛼 ∈ 𝜔1 and Γ𝛼 (T, S) = Σ𝛼 (T, K) for non-zero even ordinal numbers 𝛼 ∈ 𝜔1 (the countably multiplicative coincidence). Proof. 1. Take a set B1 of all non-zero ordinal numbers 𝛼 ∈ 𝜔1 such that Λ 𝛼 = Δ 𝛼 (L) for even ordinal numbers and Λ 𝛼 = Σ𝛼 (K) for odd ordinal numbers. For 𝛼 = 1 we have Λ 1 = S𝜆 = K𝜎 = Σ1 (K). This means that 1 ∈ B1 , i. e. B1 is non-empty. Consider the set B = B1 ∪ {0}. Let 1 ⩽ 𝛽 ∈ 𝜔1 and 𝛽 ⊂ B. Consider the edge ensemble E = ⋃⟮Λ 𝜁 | 𝜁 ∈ 𝛽⟯. By construction, Λ 𝛽 = E𝜆 . Take any P ∈ Λ 𝛽 . By definition, P = ⋃⟮P i | i ∈ I⟯ and P i = ⋃⟮F im ∩ G im | m ∈ M i ⟯ for some countable set I and some finite collections ⟮F im ∈ E | m ∈ M i ⟯ and ⟮G im ∈ co-E | m ∈ M i ⟯. By assumption, Λ 𝜁 = Δ 𝜁 (L) for every non-zero even 𝜁 ∈ 𝛽 and Λ 𝜁 = Σ𝜁 (K) for every odd 𝜁 ∈ 𝛽. By definition, F im ∈ Λ 𝜁 for some 𝜁 ∈ 𝛽. Therefore, the set X im ≡ {𝜁 ∈ 𝛽 | F im ∈ Λ 𝜁 } is non-empty. By the choice axiom from 1.1.12, there exists a mapping p : P(𝛽)\{⌀} → 𝛽 such that p(X) ∈ X. Consider the element 𝜁im ≡ p(X im ) ∈ X im ⊂ 𝛽. Then, F im ∈ Λ 𝜁im . Similarly, G im ∈ co-Λ 𝜂 for some 𝜂 ∈ 𝛽. Therefore, the set Y im ≡ {𝜂 ∈ 𝛽 | G im ∈ co-Λ 𝜂 } is non-empty. Consider the element 𝜂im ≡ p(Y im ) ∈ Y im ⊂ 𝛽. Then, G im ∈ co-Λ 𝜂im . Assume that 𝛽 is even. If 𝜁im = 0, then F im ∈ Λ 0 = S ⊂ Δ 𝛽 (L). If 𝜂im = 0, then G im ∈ co-Λ 0 = co-S ⊂ Δ 𝛽 (L). If 𝜁im is non-zero and even, then F im ∈ Δ 𝜁im (L) ⊂ Δ 𝛽 (L). If 𝜁im is odd, then by Corollary 1 to Proposition 4 (2.1.2), F im ∈ Σ𝜁im (K) ⊂ Δ 𝜁im +1 (L) ⊂ Δ 𝛽 (L). If 𝜂im is non-zero and even, then G im ∈ co-Δ 𝜂im (L). By Corollary 1 to Lemma 4 (2.1.2) and Corollary 1 to Proposition 4 (2.1.2), co-Δ 𝜂im (L) = Σ𝜂im (K) ⊂ Δ 𝜂im +1 (L) ⊂ Δ 𝛽 (L) If 𝜂im is odd, then by the same reason, G im ∈ co-Σ𝜂im (K) = Δ 𝜂im (L) ⊂ Δ 𝛽 (L). Since L is additive, we infer by Proposition 1 (2.1.2) that Δ 𝛽 (L) is latticed. Therefore, P i ∈ Δ 𝛽 (L). Since 𝛽 is even, Δ 𝛽 (L) is countably additive, so P ∈ Δ 𝛽 (L). This means that Λ 𝛽 ⊂ Δ 𝛽 (L). Conversely, let Q ∈ Δ 𝛽 (L) = (⋃⟮Δ 𝜉 (L) | 𝜉 ∈ 𝛽⟯)𝜎 . Then, as above, we can represent Q in the form Q = ⋃⟮H j | j ∈ J⟯ for some countable collections (𝜉j ∈ 𝛽 | j ∈ J) and (H j | j ∈ J) such that H j ∈ Δ 𝜉j (L). If 𝜉j = 0, then H j ∈ Δ 0 (L) = L ⊂ Λ 𝛽 . If 𝜉j is non-zero and even, then H j ∈ Λ 𝜉j ⊂ Λ 𝛽 . If 𝜉j is odd, then By Corollary 1 to Lemma 4 (2.1.2), H j ∈ co-Σ𝜉j (K) = co-Λ 𝜉j ⊂ co-E = E𝜆 = Λ 𝛽 . Since Λ 𝛽 is countably additive, we conclude that Q ∈ Λ 𝛽 . This means that Δ 𝛽 (L) ⊂ Λ 𝛽 . As a result, Λ 𝛽 = Δ 𝛽 (L).

2.1.3 Classification of Borel sets for arbitrary and perfect ensembles | 37

Now, assume that 𝛽 is odd. If 𝜁im = 0, then F im ∈ Λ 0 = S ⊂ Σ𝛽 (K). If 𝜂im = 0, then G im ∈ co-Λ 0 = co-S ⊂ Σ𝛽 (K). If 𝜁im is non-zero and even, then by Corollary 1 to Proposition 4 (2.1.2), F im ∈ Δ 𝜁im (L) ⊂ Σ𝜁im +1 (K) ⊂ Σ𝛽 (K). If 𝜁im is odd, then F im ∈ Σ𝜁im (K) ⊂ Σ𝛽 (K). If 𝜂im is non-zero and even, then by Corollary 1 to Proposition 4 (2.1.2), G im ∈ co-Δ 𝜂im (L) = Σ𝜂im (K) ⊂ Σ𝛽 (K). If 𝜂im is odd, then by Corollary 1 to Lemma 4 (2.1.2) and Corollary 1 to Proposition 4 (2.1.2), G im ∈ co-Σ𝜂im (K) = Δ 𝜂im (L) ⊂ Σ𝜂im +1 (K) ⊂ Σ𝛽 (K). Since K is multiplicative, we infer by Proposition 1 (2.1.2) that Σ𝛽 (K) is latticed. Therefore, P i ∈ Σ𝛽 (K). Since 𝛽 is odd, Σ𝛽 (K) is countably additive, so P ∈ Σ𝛽 (K). This means that Λ 𝛽 ⊂ Σ𝛽 (K). Conversely, let Q ∈ Σ𝛽 (K) = (⋃⟮Σ𝜉 (K) | 𝜉 ∈ 𝛽⟯)𝜎 . Then, as above, we can represent Q in the form Q = ⋃⟮H j | j ∈ J⟯ for some countable collections (𝜉j ∈ 𝛽 | j ∈ J) and (H j | j ∈ J) such that H j ∈ Σ𝜉j (K). If 𝜉j = 0, then H j ∈ Σ0 (K) = K ⊂ Λ 𝛽 . If 𝜉j is non-zero and even, then by Corollary 1 to Lemma 4 (2.1.2), H j ∈ co-Δ 𝜉j (L) = co-Λ 𝜉j ⊂ co-E = E𝜆 = Λ 𝛽 . If 𝜉j is odd, then H j ∈ Λ 𝜉j ⊂ Λ 𝛽 . Since Λ 𝛽 is countably additive, we conclude that Q ∈ Λ 𝛽 . This means that Σ𝛽 (K) ⊂ Λ 𝛽 . As a result, Λ 𝛽 = Σ𝛽 (K). Thus, we deduced that 𝛽 ∈ B. Using the principle of transfinite induction from 1.2.8, we conclude that B = 𝜔1 . Assertion 2 is proven in the same way. Corollary 1. Let S be an ensemble on a set T with its derivative ensembles L and K. Then, 1) Λ 𝛼 (T, S) = Y𝛼 (T, L) for every 𝛼 ∈ [1, 𝜔1 [; 2) Γ𝛼 (T, S) = Θ𝛼 (T, L) for every 𝛼 ∈ [1, 𝜔1 [. Proof. The equalities follows immediately from Theorem 5 and definitions of Y and Θ. Corollary 2. Let S be a perfect ensemble on a set T. Then, for all ordinal numbers 𝛼 ∈ [1, 𝜔1 [ we have 1) Λ 𝛼 (T, S) = Y𝛼 (T, S); 2) Γ𝛼 (T, S) = Θ𝛼 (T, S). Proof. The equalities follows immediately from Theorem 5 and Proposition 2. Corollary 3. Let S be an ensemble on a set T. Then, for all ordinal numbers 𝛼 ∈ [1, 𝜔1 [, we have 1) Γ𝛼 (T, S)𝜎 = Λ 𝛼+1 (T, S); 2) Λ 𝛼 (T, S)𝛿 = Γ𝛼+1 (T, S).

38 | 2.1 Descriptive and prescriptive spaces

Proof. Applying Corollary 1 to Theorem 5 and Lemma 5 (2.1.2), we get Γ𝛼 (S)𝜎 = Θ𝛼 (L)𝜎 = Y𝛼+1 (L) = Λ 𝛼+1 (S) and Λ 𝛼 (S)𝛿 = Y𝛼 (L)𝛿 = Θ𝛼+1 (L) = Γ𝛼+1 (S). Corollary 4. Let S be an ensemble on a set T and 𝛼 ∈ [1, 𝜔1 [. Then, 1) Λ 𝛼 (T, S) = (⋃⟮Λ 𝛾 (T, S) ∪ Γ𝛾 (T, S) | 𝛾 ∈ 𝛼⟯)𝜎 ; 2) Γ𝛼 (T, S)𝜎 = (⋃⟮Λ 𝛾 (T, S) ∪ Γ𝛾 (T, S) | 𝛾 ∈ 𝛼⟯)𝛿 .

Proof. The equalities follow from Lemma 8 (2.1.2), Lemma 4 (2.1.1), and Corollary 1. Lemma 1. Let S be an ensemble on a set T with the derivative ensembles L = L(T, S) and K = K(T, S). Then, 1) Y𝛼 (T, K𝜎 ) = Λ 𝛼+1 (T, S) and Θ𝛼 (T, K𝜎 ) = Γ𝛼+1 (T, S) for every 𝛼 ∈ 𝜔; 2) Y𝛼 (T, K𝜎 ) = Λ 𝛼 (T, S) and Θ𝛼 (T, K𝜎 ) = Γ𝛼 (T, S) for every 𝛼 ∈ [𝜔, 𝜔1 [. Proof. Consider the subset A of the set 𝜔 consisting of all numbers 𝛼 ∈ 𝜔 such that the necessary equalities are fulfilled. For 𝛼 = 0, we have automatically Y0 (K𝜎 ) = K𝜎 = Λ 1 (S) and Θ0 (L𝛿 ) = L𝛿 = Γ1 (S). Thus, 0 ∈ A. Let 1 ⩽ 𝛼 ∈ A. Applying Lemma 5 (2.1.2) and Corollary 3 to Theorem 5, we get Y𝛼+1 (K𝜎 ) = Θ𝛼+1 (K𝜎 )𝜎 = Γ𝛼+1 (S)𝜎 = Λ 𝛼+2 (S) and Θ𝛼+1 (K𝜎 ) = Y𝛼+1 (K𝜎 )𝛿 = Λ 𝛼+1 (S)𝛿 = Γ𝛼+2 (S). This means that 𝛼 + 1 ∈ A. By the principle of natural induction, A = 𝜔. Consider now the subset B of the set 𝜔1 consisting of all numbers 𝛼 ∈ 𝜔1 such that the necessary equalities are fulfilled. As we have proven, 𝜔 ⊂ B. Let 𝛽 ∈ [𝜔, 𝜔1 [ and 𝛽 ⊂ B. First assume that 𝛽 is a non-limit number, i. e. 𝛽 = 𝜘 + 1 for some 𝜘 ∈ 𝜔1 . Since 𝛽 ⩾ 𝜔1 , we infer that 𝜘 ⩾ 𝜔. From 𝜘 ∈ 𝛽 ⊂ B, by inductive assumption, we infer that Y𝜘 (K𝜎 ) = Λ 𝜘 (S) and Θ𝜘 (L𝛿 ) = Γ𝜘 (S). Therefore, by Lemmas 8 (2.1.2) and 7 (2.1.2), we get Y𝛽 (K𝜎 ) = (⋃⟮Y𝜉 (K𝜎 ) ∪ Θ𝜉 (K𝜎 ) | 𝜉 ∈ 𝛽⟯)𝜎 = (Y𝜘 (K𝜎 ) ∪ Θ𝜘 (K𝜎 ))𝜎 = (Λ 𝜘 (S) ∪ Γ𝜘 (S))𝜎 ≡ P. By Corollary 3 to Theorem 5, Lemmas 3 (2.1.2) and 7 (2.1.2) P ⊂ (Λ 𝜘 (S) ∪ Γ𝜘 (S)𝜎 )𝜎 = (Λ 𝜘 (S) ∪ Λ 𝛽 (S))𝜎 = Λ 𝛽 (S)𝜎 = Λ 𝛽 (S) = Λ 𝜘+1 (S) = Γ𝜘 (S)𝜎 ⊂ P. Consequently, we conclude that Y𝛽 (K𝜎 ) = Λ 𝛽 (S). In the same manner, we prove that Θ𝛽 (K𝜎 ) = Γ𝛽 (S). This means that 𝛽 ∈ B. Now, assume that 𝛽 is a limit number. Consider the sets P ≡ ⋃⟮Y𝜉 (K𝜎 ) ∪ Θ𝜉 (K𝜎 ) | 𝜉 ∈ 𝛽⟯ and Q ≡ ⋃⟮Λ 𝜉 (S) ∪ Γ𝜉 (S) | 𝜉 ∈ 𝛽⟯. Take any 𝜉 ∈ 𝛽. If 𝜉 < 𝜔, then, by inductive assumption, Y𝜉 (K𝜎 ) = Λ 𝜉+1 (S) and Θ𝜉 (K𝜎 ) = Γ𝜉+1 (S), where 𝜉 + 1 ∈ 𝛽. If 𝜉 ⩾ 𝜔, then Y𝜉 (K𝜎 ) = Λ 𝜉 (S) and Θ𝜉 (K𝜎 ) = Γ𝜉 (S). This implies the inclusion P ⊂ Q. Conversely, take any 𝜉 ∈ 𝛽. If 𝜉 < 𝜔, then, by inductive assumption, Λ 𝜉 (S) = Y𝜉−1 (K𝜎 ) and Γ𝜉 (S) = Θ𝜉−1 (K𝜎 ). If 𝜉 ⩾ 𝜔, then Λ 𝜉 (S) = Y𝜉 (K𝜎 ) and Γ𝜉 (S) = Θ𝜉 (K𝜎 ). This gives the inverse inclusion Q ⊂ P.

2.1.3 Classification of Borel sets for arbitrary and perfect ensembles | 39

It follows from Lemma 8 (2.1.2) and the proven equality P = Q that Y𝛽 (K𝜎 ) = P𝜎 = Q𝜎 . By Corollary 4 to Theorem 5, Q𝜎 = Λ 𝛽 (S). In the same manner, we prove that Θ𝛽 (K𝜎 ) = Γ𝛽 (S). This means that 𝛽 ∈ B. By the principle of transfinite induction, B = 𝜔1 .

Perfectness and separability of Λ and Γ ensembles For an arbitrary ensemble S, the Zakharov – Koldunov ensembles Λ 𝛼 (T, S) and Γ𝛼 (T, S) in contrast to the Young – Hausdorff ensembles Y𝛼 (T, S) and Θ𝛼 (T, S) have some important properties which are necessary to develop a meaningful theory of Borel-measurable functions on the descriptive space ⟮T, S⟯. According to Lemma 3 (2.1.2), all ensembles Λ 𝛼 (T, S), 𝛼 ∈ [1, 𝜔1 [, are perfect 𝜎foundations. This implies that all families M(T, Λ 𝛼 (T, S)) of Borel-measurable functions of classes 𝛼 have good algebraic properties, i. e. they are latticed R-algebras with divisions (see 2.3). But these ensembles have also such a crucial property as Λ 𝛼 (T, S) = Coz M(T, Λ 𝛼 (T, S)) (see 2.2.5 and the Lebesgue – Urysohn property in 2.3.5). But to prove this complicated property it is necessary to have first the property of separability (cf. Theorems in 2.1.1). Our aim now is to prove this property for all ensembles Λ 𝛼 (T, S). Theorem 6 (the reduction theorem). Let S be an ensemble on a set T with the coensemble R such that R𝜎 is perfect. Then, for every 𝛼 ∈ [2, 𝜔1 [ and every countable collection, (Z i ∈ Y𝛼 (T, S) | i ∈ I), there is a disjoint collection (X i ∈ Y𝛼 (T, S) | i ∈ I) such that X i ⊂ Z i and ⋃⟮Z i | i ∈ I⟯ = ⋃⟮X i | i ∈ I⟯. If S is perfect, then we can take 𝛼 ∈ [1, 𝜔1 [. Proof. For 𝛼 = 1, the assertion follows from Theorem 1 (2.1.1). Take any 𝛼 ⩾ 2. By Corollary 1 to Proposition 5 (2.1.2), Y𝛼 (T, S) is perfect for every 𝛼 ∈ [1, 𝜔1 [. At first, assume that 𝛼 is a non-limit number, i. e. 𝛼 = 𝜘 + 1 for some 𝜘 ∈ [1, 𝜔1 [. Take the ensemble S󸀠 ≡ Y𝜘 with the co-ensemble R󸀠 = Θ𝜘 . By Lemma 5 (2.1.2), R󸀠𝜎 = Y𝜘+1 = Y𝛼 . Applying now Theorem 1 (2.1.1) to S󸀠 and R󸀠 , we get the necessary assertion. Now, assume that 𝛼 is a limit number and consider the ensemble S󸀠󸀠 ≡ ⋃⟮Y𝛽 | 𝛽 ∈ 𝛼⟯. Since Y0 = S ⊂ S𝛿𝜎 = Y2 and perfect ensemble R𝜎 ⊂ S𝛿𝜎 , we infer that S󸀠󸀠 = ⋃⟮Y𝛽 | 𝛽 ∈ [2, 𝛼[⟯. Then, R󸀠󸀠 ≡ co-S󸀠󸀠 = ⋃⟮Θ𝛽 | 𝛽 ∈ [2, 𝛼[⟯. By condition, Y𝛽 ⊂ (Θ𝛽 )𝜎 for every 𝛽 ⩾ 2. Therefore, S󸀠󸀠 ⊂ ⋃⟮(Θ𝛽 )𝜎 | 𝛽 ∈ [2, 𝛼[⟯ ⊂ R󸀠󸀠 . Now, check that S󸀠󸀠 is perfect. If A, B ∈ S󸀠󸀠 , then A ∈ Y𝜇 and B ∈ Y𝜈 for some 𝜇, 𝜈 ∈ [2, 𝛼[. Take 𝜘 ≡ 𝜇 ⊻ 𝜈. By Lemma 7 (2.1.2), Y𝜇 ∪ Y𝜈 ⊂ Y𝜘 ⊂ S󸀠󸀠 . Therefore, A ∪ B belongs to S󸀠󸀠 . From Y1 ⊂ S󸀠󸀠 , we infer that S󸀠󸀠 has the edges. Thus, S󸀠󸀠 is perfect. Applying now Theorem 1 (2.1.1) to S󸀠󸀠 and R󸀠󸀠 , we get the necessary assertion.

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Corollary 1 (the general case). Let S be an ensemble on a set T. Then, for every 𝛼 ∈ [1, 𝜔1 [ and every countable collection (Z i ∈ Λ 𝛼 (T, S) | i ∈ I), there is a disjoint collection (X i ∈ Λ 𝛼 (T, S) | i ∈ I) such that X i ⊂ Z i and ⋃⟮Z i | i ∈ I⟯ = ⋃⟮X i | i ∈ I⟯. Proof. Take S󸀠 ≡ L(T, S) and R󸀠 ≡ co-S󸀠 = K(T, S). By Corollary 1 to Lemma 4 (2.1.1), R󸀠𝜎 is perfect, by Lemma 4 (2.1.1) S󸀠 is perfect. Consequently, Corollary 1 to Theorem 5 and Theorem 6 give the necessary assertion for 𝛼 ⩾ 1. Thus, the ensembles Y𝛼 (T, S), 𝛼 ∈ [2, 𝜔1 [, when R𝜎 is perfect, and Λ 𝛼 (T, S), 𝛼 ∈ [1, 𝜔1 [, in general case, are reducible. Corollary 2 (the perfect case). Let S be a perfect ensemble on a set T. Then, for every 𝛼 ∈ [1, 𝜔1 [ and every countable collection (Z i ∈ Y𝛼 (T, S) | i ∈ I), there is a disjoint collection (X i ∈ Y𝛼 (T, S) | i ∈ I) such that X i ⊂ Z i and ⋃⟮Z i | i ∈ I⟯ = ⋃⟮X i | i ∈ I⟯. Proof. If S is perfect, then by Proposition 1 (2.1.1), R𝜎 is perfect. Consequently, the assertion follows from Theorem 6. The last corollary is the result of K. Kuratowski (see [Kuratowski, 1966, 30.VII; Srivastava, 1998, 3.6.10]). Corollary 3. Under the conditions of Theorem 6 or Corollary 2, if ⋃⟮Z i | i ∈ I⟯ = T, then X i ∈ Y𝛼 (T, S) ∩ Θ𝛼 (T, S). Under the conditions of Corollary 1, if ⋃⟮Z i | i ∈ I⟯ = T, then X i ∈ Λ 𝛼 (T, S) ∩ Γ𝛼 (T, S). Proof. Since Y𝛼 is 𝜎-additive, we infer that A i ≡ ⋃⟮X j | j ∈ I\{i}⟯ ∈ Y𝛼 . Consequently, X i = T\A i ∈ Θ𝛼 . The second case is completely analogous. Theorem 7 (the separation theorem). Let S be an ensemble on a set T. If the ensemble R𝜎 is perfect, then for every 𝛼 ∈ [2, 𝜔1 [ and every countable collection (X i ∈ Θ𝛼 (T, S) | i ∈ I) such that ⋂⟮X i | i ∈ I⟯ = ⌀ there is a collection (Z i ∈ Y𝛼 (T, S) ∩ Θ𝛼 (T, S) | i ∈ I) such that X i ⊂ Z i and ⋂⟮Z i | i ∈ I⟯ = ⌀. If S is perfect, then we can take 𝛼 = 1. Proof. Consider the sets C i ≡ T\X i ∈ Y𝛼 . Then, ⋃⟮C i | i ∈ I⟯ = T. By Corollary 3 to Theorem 6, there is a collection (A i ∈ Y𝛼 ∩ Θ𝛼 | i ∈ I) such that A i ⊂ C i and ⋃⟮A i | i ∈ I⟯ = T. Consider the sets Z i ≡ T\A i . Then, the collection ⟮Z i | i ∈ I⟯ has the necessary properties. Corollary 1 (the general case). Let S be an ensemble on a set T. Then, for every 𝛼 ∈ [1, 𝜔1 [ and every countable collection (X i ∈ Γ𝛼 (T, S) | i ∈ I) such that ⋂⟮X i | i ∈ I⟯ = ⌀, there is a collection (Z i ∈ Λ 𝛼 (T, S) ∩ Γ𝛼 (T, S) | i ∈ I) such that X i ⊂ Z i and ⋂⟮Z i | i ∈ I⟯ = ⌀.

2.1.3 Classification of Borel sets for arbitrary and perfect ensembles | 41

Proof. Take S󸀠 ≡ L(T, S) and R󸀠 ≡ co-S󸀠 = K(T, S). By Corollary 1 to Lemma 4 (2.1.1), R󸀠𝜎 is perfect; by this Lemma 4 (2.1.1), S󸀠 is perfect. Applying Corollary 1 to Theorem 5 and Theorem 6, we get the necessary assertion. Thus, the ensembles Y𝛼 (T, S), 𝛼 ∈ [2, 𝜔1 [, when R𝜎 is perfect, and Λ 𝛼 (T, S), 𝛼 ∈ [1, 𝜔1 [, in general case, are separable. Corollary 2 (the perfect case). Let S be a perfect ensemble on a set T. Then, for every 𝛼 ∈ [1, 𝜔1 [ and every countable collection (X i ∈ Θ𝛼 (T, S) | i ∈ I) such that ⋂⟮X i | i ∈ I⟯ = ⌀, there is a collection (Z i ∈ Y𝛼 (T, S) ∩ Θ𝛼 (T, S) | i ∈ I) such that X i ⊂ Z i and ⋂⟮Z i | i ∈ I⟯ = ⌀. Proof. If S is perfect, then by Proposition 1 (2.1.1), R𝜎 is perfect. Consequently, the assertion follows from Theorem 6. The last corollary is the result of W. Sierpiński (see [Kuratowski, 1966, 30.VII; Srivastava, 1998, 3.6.11]). Corollary 3. In Theorem 7 and its Corollaries 1 and 2, if the set I consists of only two different elements i and j, then the collection ⟮Z i | i ∈ I⟯ can be chosen so that in addition Z j = T\Z i , i. e. ⋃⟮Z k | k ∈ I⟯ = T. Using the separation theorem, we can prove the following theorem which will be crucial for the theory of measurable functions in 2.3.5. Theorem 8 (the improvement theorem). Let S be an ensemble on a set T such that the ensemble R𝜎 is perfect. Then, for every 𝛼 ∈ [2, 𝜔1 [ the ensemble Y𝛼 (T, S) is a separable perfect 𝜎-foundation. If S is perfect, then we can take 𝛼 = 1. Proof. By Corollary 1 to Proposition 5 (2.1.2), Y𝛼 is perfect and latticed for every 𝛼 ⩾ 2. According to Lemma 7 (2.1.2), the collection ⟮Y𝛼 | 𝛼 ∈ [1, 𝜔1 [⟯ is increasing. By condition, Y1 = R𝜎 has the edges. Consequently, Y𝛼 for 𝛼 ⩾ 1 has the edges. Besides Y𝛼 is 𝜎-additive. Thus, Y𝛼 is a 𝜎-foundation. By virtue of Theorem 7 Y𝛼 is separable. If 𝛼 = 1 and S is perfect, then by Theorem 3 (2.1.1), Y1 = R𝜎 is a separable perfect 𝜎-foundation. Corollary 1 (the general case). Let S be an ensemble on a set T. Then, for every 𝛼 ∈ [1, 𝜔1 [ the ensemble Λ 𝛼 (T, S) is a separable perfect 𝜎-foundation. The proof is similar to the proof of Corollary 1 to Theorem 6. Corollary 2 (the perfect case). Let S be a perfect ensemble on a set T. Then, for every 𝛼 ∈ [1, 𝜔1 [ the ensemble Y𝛼 (T, S) is a separable perfect 𝜎-foundation. The proof is similar to the proof of Corollary 2 to Theorem 6.

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Proposition 3. Let S be an ensemble on a set T such that the ensemble R𝜎 is perfect. Then, for every ordinal number 𝛼 ∈ [2, 𝜔1 [ and for every finite disjoint collection (X i ∈ Θ𝛼 (T, S) | i ∈ I), there is a disjoint collection (Z i ∈ Y𝛼 (T, S) ∩ Θ𝛼 (T, S) | i ∈ I) such that X i ⊂ Z i and ⋃⟮Z i | i ∈ I⟯ = T. If S is perfect, then we can take 𝛼 = 1. Proof. Consider the subset N ⊂ 𝜔 consisting of all n such that the assertion is valid for every collection (X i | i ∈ I) with card I = n + 2. If n = 0, then the assertion is valid by virtue of Corollary 3 to Theorem 7. Hence, 0 ∈ N. Assume that n ∈ N. Take any finite disjoint collection (X i ∈ Θ𝛼 | i ∈ I) such that card I = n+3. Fix any different elements p, q ∈ I and consider the set K = I\{q} and the collection (Yk ∈ Θ𝛼 | k ∈ K) such that Yk ≡ X k for every k ∈ I\{p, q} and Yp ≡ X p ∪ X q . By assumption, for this collection, there is a disjoint cover (Z 󸀠k ∈ Y𝛼 ∩ Θ𝛼 | k ∈ K) of T such that Yk ⊂ Z 󸀠k . Consider the elements A p ≡ Z 󸀠p \X p and A q ≡ Z 󸀠q \X q of the ensemble Y𝛼 . By Theorem 6 for collection (A l ∈ Y𝛼 | l ∈ {p, q}), there is a disjoint collection (B l ∈ Y𝛼 | l ∈ {p, q}) such that B p ⊂ A p , B q ⊂ A q , and A p ∪ A q = B p ∪ B q . Consider the elements Z p ≡ Z 󸀠p \B p , Z q ≡ Z 󸀠p \B q , they belong to Θ𝛼 . From A p ∪ A q = Z 󸀠p , we infer that Z p ∪ Z q = Z 󸀠p and Z p ∩ Z q = ⌀. Consider the disjoint cover (Z i ∈ Θ𝛼 | i ∈ I) such that Z i ≡ Z 󸀠i for every i ∈ I\{p, q}. From Z p = T\ ⋃⟮Z i | i ∈ I\{p}⟯ ∈ Y𝛼 and Z q ∈ Y𝛼 , we conclude that the collection (Z 󸀠i ∈ Y𝛼 ∩ Θ𝛼 | i ∈ I) has all the necessary properties. This means that n + 1 ∈ N. By the principle of natural induction from 1.2.6, N = 𝜔.

2.1.4 Descriptive spaces with negligence Negligences on descriptive spaces Let ⟮T, S⟯ be a descriptive space. An ensemble N on T will be called a negligence with respect to S or a negligence on the descriptive space ⟮T, S⟯ if for every S ∈ S ∪ {T} ∪ {⌀} and every N ∈ N𝜑 the inclusion S ⊂ N implies S = ⌀. In this case, the triplet ⟮T, S, I⟯ will be called a descriptive space with negligence. An additive ensemble I on T is called an ideal ensemble or an ideal (of sets) if ∀ I ∈ I ∀ P ∈ P(T) (P ⊂ I ⇒ P ∈ I). Every ideal ensemble is a lattice with the zero element ⌀ (see 2.1.1). If an ideal ensemble is 𝜎-additive, it is called a 𝜎-additive ideal (a 𝜎-ideal). A multiplicative ensemble F is called a filter (of sets) if it follows from P ∈ P(T), F ∈ F, and P ⊃ F that P ∈ F. Any filter is a lattice with the unit T. If I is a negligence on a descriptive space ⟮T, S⟯ and I is an ideal, then it is natural to call I an ideal negligence. Note that according to the following lemma, every negligence N generates the corresponding ideal negligence I(N𝜑 ) ≡ {I ⊂ T | ∃ N ∈ N𝜑 (I ⊂ N)}

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Lemma 1. Let N be an ensemble on T. Then, I(N𝜑 ) is an ideal. If N is 𝜎-additive, then I(N𝜑 ) is a 𝜎-ideal. If ⟮T, S⟯ is a descriptive space and N is a negligence on it, then I(N𝜑 ) is an ideal negligence on ⟮T, S⟯. Proof. Let I, J ∈ I ≡ I(N𝜑 ). Then, I ⊂ M and J ⊂ N for some M, N ∈ N𝜑 . Since I ∪ J ⊂ M ∪ N ∈ N𝜑 , we get I ∪ J ∈ I. For every P ⊂ I, we have P ⊂ M. Hence, P ∈ I. Thus, I is an ideal. Let N be 𝜎-additive ensemble and (I l ∈ I | l ∈ L) be a countable collection such that I l ⊂ N l ∈ N𝜑 . Then, I ≡ ⋃⟮I l | l ∈ L⟯ ⊂ ⋃⟮N l | l ∈ L⟯ ∈ N𝜎 = N = N𝜑 . Hence, I ∈ I. Let N be a negligence on ⟮T, S⟯. Take some non-empty set S ∈ S ∪ {T}. Suppose that there exists a finite collection (I k ∈ I | k ∈ K) such that S ⊂ ⋃⟮I k | k ∈ K⟯. By definition, for every k ∈ K, there is a finite collection (N𝛼k ∈ N | 𝛼k ∈ A k ) such that I k ⊂ ⋃⟮N𝛼k | 𝛼k ∈ A k ⟯. Then, by virtue of Proposition 1 (1.1.13) and Lemma 3 (1.3.3), we get the inclusion

S ⊂ ⋃ {⋃⟮N𝛼k | 𝛼k ∈ A k ⟯ | k ∈ K } = { } = ⋃ {N(k,𝛼k ) | (k, 𝛼k ) ∈ ⋃⟮{k} × A k | k ∈ K⟯} ∈ N𝜑 , } { where N(k,𝛼k ) ≡ N𝛼k . Since N is a negligence, we get a contradiction. Let E is some fixed ensemble on T. Then, an additive ensemble I on T is called an ideal in the ensemble E if ∀ I ∈ I ∀ E ∈ E (E ⊂ I ⇒ E ∈ I). Lemma 2. Let I be an ideal in a ring R on T. Then, I is a ring. Moreover, I\R ∈ I for every I ∈ I and R ∈ R. Proof. By definition, I is additive. Let I ∈ I and R ∈ R. Then, I\R ∈ R and I\R ⊂ I. Hence, I\R ∈ I. Therefore, I is closed under the difference. Corollary 1. Let I be an ideal ensemble on T. Then, I is a ring. Present some natural ideal ensembles generated by a fixed ensemble S. It follows from Corollary 3 to Lemma 11 (2.1.1) that for every ensemble S, the 𝛿-ring H𝛿 (T, S) is the ideal 𝛿-ring generated in P(T) by the 𝛿-ring R𝛿 (T, S). Analogously, it follows from Corollary 2 to Lemma 12 (2.1.1) that the 𝜎-ring H𝜎 (T, S) is the ideal 𝜎-ring generated in P(T) by the 𝜎-ring R𝜎 (T, S). According to Corollary 4 to Proposition 4 (2.1.1), a 𝛿-ring S is an ideal in R𝜎 (T, S).

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Dense and nowhere dense sets Let S be an ensemble on a set T and S ≠ {⌀}. Let A and B be subsets of T. The set A is called S-dense in the set B if for every S ∈ S it follows from S ∩ B ≠ ⌀ that S ∩ A ≠ ⌀. If a set D ⊂ T is S-dense in the set T, then it will be called dense with respect to the ensemble S or, simply, S-dense. The ensemble of all S-dense sets D ∈ S will be denoted by D(S). A set N ⊂ T will be called nowhere dense with respect to the ensemble S if N ⊂ T\D for some D ∈ D(S). The ensemble of all such the sets will be denoted by N(S). Clearly, N(S) = I(co-D(S)). Lemma 3. Let S be a foundation on T. Then, the ensemble D(S) is multiplicative and the ensemble N(S) is an ideal. Proof. Let D1 , D2 ∈ D(S) ⊂ S. Then, D ≡ D1 ∩ D2 ≠ ⌀. Since S is multiplicative, D ∈ S. Let S ∈ S\{⌀}. It follows from D1 ∈ D(S) and D2 ∩ S ∈ S\{⌀} that D ∩ S = D1 ∩(D2 ∩ S) ≠ ⌀. Hence, D ∈ D(S). Thus, D is multiplicative, and therefore, co-D is additive. Then, Lemma 1 implies that N(S) = I(co-D) = I((co-D)𝜑 ) is an ideal. Corollary 1. Let S be a foundation on T. Then, the ensembles D(S𝜑 ), D(S𝜎 ), and D(S𝜏 ) are multiplicative and the ensembles N(S𝜑 ), N(S𝜎 ), and N(S𝜏 ) are ideal ensembles. Proof. By Lemma 1 (2.1.1), the ensembles S𝜑 , S𝜎 , and S𝜏 are also foundations. Lemma 4. Let S be a foundation on T and D ∈ S𝜎 [D ∈ S𝜑 ]. Then, the following assertions are equivalent: 1) D ∈ D(S𝜎 ) [D ∈ D(S𝜑 )]; 2) S ∩ D = / ⌀ for every S ∈ S\{⌀}. Proof. (1) ⊢ (2). This is evident. (2) ⊢ (1). By virtue of Lemma 1 (2.1.1), the ensemble S𝜎 [S𝜑 ] is a foundation. Let ⌀= / S = ⋃⟮S i ∈ S | i ∈ I⟯ ∈ S𝜎 [S𝜑 ]. Then, S i = / ⌀ for some i ∈ I. From S i ∩ D = / ⌀, we conclude that S ∩ D = ⋃⟮S i ∩ D | i ∈ I⟯ = / ⌀. Sets with the Stone property A set X ⊂ T is called a set with the Stone property with respect to an ensemble S and an ideal I if X = S ∪ N for some S ∈ S and N ∈ I. The ensemble of all such the sets will be denoted by SP(S, I). The particular case when S = G, ⟮T, G⟯ is a topological space, and I = N(S) was considered by M. H. Stone [1937] (see also [Kuratowski, 1966, 8.V] and [Semadeni, 1971, 16.1.6(B), 16.1.7]). The two other important cases are considered in 2.5.2 and 3.7.2.

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Lemma 5. Let S be a foundation on T and I be an ideal on T. Then, the ensembles SP(S, I), SP(S𝜑 , I), and SP(S𝜎 , I) are foundations. Besides, SP(S𝜑 , I) and SP(S𝜎 , I) are additive. Proof. Since ⌀ ∈ S and ⌀ ∈ I, we get ⌀ ∈ SP(S, I). It follows from T ∈ S and T = T ∪ ⌀ that T ∈ SP(S, I). Let X = P ∪ M and Y = Q ∪ N for some P, Q ∈ S and M, N ∈ I. Since S is multiplicative, we get P ∩ Q ∈ S. Since I is an ideal, we see that P ∩ N, M ∩ Y ∈ I and, therefore, (P ∩ N) ∪ (M ∩ Y) ∈ I. Hence, X ∩ Y = (P ∩ Q) ∪ ((P ∩ N) ∪ (M ∩ Y)) ∈ SP(S, I). Since, by Lemma 1, S𝜑 and S𝜎 are foundations, for SP(S𝜑 , I) and SP(S𝜎 , I), the arguments are similar. Let X = P ∪ M and Y = Q ∪ N for some P, Q ∈ S𝜑 and M, N ∈ I. Since P ∪ Q ∈ S𝜑 and M ∪ N ∈ I, we get X ∪ Y = (P ∪ Q) ∪ (M ∪ N) ∈ SP(S𝜑 , I). For SP(S𝜎 , I), the arguments are the same. Let S be an ensemble on a set T and I be an ideal on T. The order by inclusion on the full ensemble P(T) is inherited by I. According to 1.1.15, a subensemble U of I is called cofinal to I if [I, → [∩U = / ⌀ for every I ∈ I, i. e. for every I ∈ I, there exists U ∈ U such that I ⊂ U. Below, we apply this notion to U ≡ I ∩ co-S. In this case, U is cofinal to I iff V ≡ co-I ∩ S is coinitial in co-I, i. e. for every D ∈ co-I, there exists V ∈ V such that V ⊂ D. It is clear that the ensemble N(S) ∩ co-S is cofinal to the ideal N(S). Lemma 6. Let S be a foundation on T, I be an ideal on T, and I ∩ co-S is cofinal to I. If D ∈ co-I, then SP(SD , I) = SP(S, I), SP((SD )𝜑 , I) = SP(S𝜑 , I), and SP((SD )𝜎 , I) = SP(S𝜎 , I). Proof. Let X ∈ SP(SD , I), where SD is the trace of the ensemble S on the set D (see 2.1.1). Then, X = S󸀠 ∪ N, S󸀠 = S ∩ D, S ∈ S, and N ∈ I. Since I ∩ co-S is cofinal to I, there exists V ∈ co-I ∩ S such that V ⊂ D. Hence, S󸀠 = S ∩ D = S ∩ (V ∪ (D\V)) = (S ∩ V) ∪ (S ∩ (D\V)). Since S is multiplicative, S ∩ V ∈ S. Since I is an ideal, M ≡ S ∩ (D\V) ⊂ D\V ⊂ T\V ∈ I implies M ∈ I, and therefore, M ∪ N ∈ I. Finally, X = (S ∩ V) ∪ (M ∪ N) ∈ SP(S, I). Let X ∈ SP(S, I). Then, X = S ∪ N, S ∈ S, and N ∈ I. Therefore, S ∩ D ∈ SD and it follows from M ≡ S\D ⊂ T\D ∈ I that M ∈ I. Thus, X = (S ∩ D) ∪ (M ∪ N) ∈ SP(SD , I). Since (S𝜑 )D = (SD )𝜑 , (S𝜎 )D = (SD )𝜎 , and, according to Lemma 1 (2.1.1), S𝜑 and S𝜎 are foundations, the other two equalities follow from the proven one. The equivalence of sets with respect to an ideal negligence The set A 󳵻 B ≡ (A\B) ∪ (B\A) = (A ∪ B)\(A ∩ B) is called the symmetrical difference of sets A and B from T. It is clear that any ring is closed under the symmetrical difference.

46 | 2.1 Descriptive and prescriptive spaces

Lemma 7. For arbitrary elements and collections from P(T), the following statements hold: (i) A 󳵻 C ⊂ (A 󳵻 B) ∪ (B 󳵻 C); (ii) (⋃⟮A k | k ∈ K⟯) 󳵻 (⋃⟮B k | k ∈ K⟯) ⊂ ⋃⟮A k 󳵻 B k | k ∈ K⟯; (iii) (⋂⟮A k | k ∈ K⟯) 󳵻 (⋂⟮B k | k ∈ K⟯) ⊂ ⋃⟮A k 󳵻 B k | k ∈ K⟯. Proof. (i) Let t ∈ A\C. If t ∈ B, then t ∈ B\C. If t ∈ ̸ B, then t ∈ A\B. Therefore, A\C ⊂ (A\B) ∪ (B\C). Similarly, we get C\A ⊂ (B\A) ∪ (C\B). These inclusions imply (i). (ii) The inclusions ⋃ A k \ ⋃ B k ⊂ ⋃(A k \B k ) and ⋃ B k \ ⋃ A k ⊂ ⋃(B k \A k ) give the assertion. (iii) The inclusions ⋂ A k \ ⋂ B k = ⋂ {A l \ ⋂ B k | l ∈ K } = { } = ⋂ {⋃⟮A l \B k | k ∈ K⟯ | l ∈ K } ⊂ ⋃⟮A k \B k | k ∈ K⟯ } { and ⋂ B k \ ⋂ A k ⊂ ⋃(A k \B k ) give the assertion. Let ⟮T, S, I⟯ be a descriptive space with ideal negligence. Consider on S the relation 𝜃 ≡ {(A, B) ∈ S×S | A 󳵻 B ∈ I}. Obviously, it is reflexive and symmetric (see 1.1.14). The inclusion (i) from Lemma 7 implies that 𝜃 is transitive. Therefore, 𝜃 is an equivalence relation and we shall write A ∼ B mod I or simply A ∼ B. Lemma 8. Let A, B ∈ S. Then, A ∼ B mod I iff A = (B\I) ∪ J for some I, J ∈ I. Proof. Suppose A ∼ B mod I. Then, I ≡ B\A ∈ I, J ≡ A\B ∈ I, and A = (B\I) ∪ J. Conversely, if A = (B\I) ∪ J, then A\B ⊂ J ∈ I and B\A ⊂ I ∈ I, where A ∼ B. Lemma 9. The following statements hold: 1) if (A k | k ∈ K) and (B k | k ∈ K) are finite collections from S and A k ∼ B k mod I for every k ∈ K, then ⋃⟮A k | k ∈ K⟯ ∼ ⋃⟮B k | k ∈ K⟯ and ⋂⟮A k | k ∈ K⟯ ∼ ⋂⟮B k | k ∈ K⟯; 2) if A ∼ B, then T\A ∼ T\B; 3) if A ∼ B and C ∼ D, then A\C ∼ B\D. Proof. 1. This statement follows from assertions (ii) and (iii) of Lemma 7. 2. Suppose A ∼ B. Then, (T\A)\(T\B) = B\A ∈ I and (T\B)\(T\A) = A\B ∈ I implies T\A ∼ T\B. 3. Since A\C = A ∩ (T\C) and B\D = B ∩ (T\D), it follows from (1) and (2) that A\C ∼ B\D.

2.1.5 Prescriptive spaces

|

47

Lemma 10. If I is a 𝜎-ideal on T, (A k | k ∈ K) and (B k | k ∈ K) are countable collections from P(T), and A k ∼ B k mod I for every k ∈ K, then ⋃⟮A k | k ∈ K⟯ ∼ ⋃⟮B k | k ∈ K⟯ and ⋂⟮A k | k ∈ K⟯ ∼ ⋂⟮B k | k ∈ K⟯. Proof. This statement follows from assertions (ii) and (iii) of Lemma 7. For every S ∈ S, consider the class S ≡ S mod I = {S󸀠 ∈ S | S󸀠 ∼ S mod I} of equivalence of the set S with respect to I. The set {S mod I | S ∈ S} of all classes of equivalence for all sets S ∈ S is called the factor-ensemble of the ensemble S with respect to the ideal ensemble I. It will be denoted by S ≡ S/I. It follows from Lemma 9 that we can define the partial binary operations ∨, ∧, and \ and partial unary operation 󸀠 on S ≡ S/I, setting P ∨ Q ≡ P ∪ Q, P ∧ Q ≡ P ∩ Q, 󸀠 P\Q ≡ P\Q, and P = T\P. Lemma 11. The following statements hold: 1) if S is additive, then ∨ is the binary operation on S; 2) if S is multiplicative, then ∧ is the binary operation on S; 3) if S is closed under differences, then \ is the binary operation on S; 4) if S is closed under complements, then 󸀠 is the unary operation on S. Proof. All the statements are the straightforward consequences of the definitions.

2.1.5 Prescriptive spaces Let T be a set and S ⊂ T. According to 1.1.10, a collection ⟮A i ⊂ S | i ∈ I⟯ is called a cover of the set S if S = ⋃⟮A i | i ∈ I⟯. The cover ⟮A i ≡ S | i ∈ {i}⟯ will be called the onemember cover of S. A family C of covers of the set T containing the one-member cover of T will be called a covering on T. Then, the pair ⟮T, C⟯ will be called a prescriptive space or a space with a covering. A covering C on T will be called multiplicative if for every finite collection (𝜋𝛼 | 𝛼 ∈ A) of covers 𝜋𝛼 ≡ ⟮C𝛼i | i ∈ I𝛼 ⟯ ∈ C, we have ⋀ (𝜋𝛼 | 𝛼 ∈ A) ∈ C, where ⋀ (𝜋𝛼 | 𝛼 ∈ A) denotes the cover ⟮D j | j ∈ J⟯ such that J ≡ ∏⟮I𝛼 | 𝛼 ∈ A⟯ and D j ≡ ⋂⟮C𝛼j(𝛼) | 𝛼 ∈ A⟯ for all j ∈ J. To not increase levels of indexation, hereinafter, we suppose that previous indices determine the domains of subsequent ones. Taking a pair of covers {𝜘, 𝜌}, consider the collection ⟮𝜋𝛼 | 𝛼 ∈ 2⟯, where 𝜋0 ≡ 𝜘 and 𝜋1 ≡ 𝜌. Along with ⋀ (𝜋𝛼 | 𝛼 ∈ 2) we shall also write 𝜋0 ∧ 𝜋1 or 𝜘 ∧ 𝜌. For a covering C on a set T, consider its 𝜂-hull C𝜂 consisting of all collections 𝜋 ≡ ⟮P u ⊂ T | u ∈ U⟯ such that for 𝜋, there exists finite collection (𝛾i ∈ C | i ∈ I) of covers 𝛾i ≡ ⟮C im | m ∈ M i ⟯ of the set T such that U = ∏⟮M i | i ∈ I⟯ and P u = ⋂⟮C iu(i) | i ∈ I⟯ for each u ∈ U.

48 | 2.1 Descriptive and prescriptive spaces

Lemma 1. Let C be a family of finite [countable] covers of T. Then, every collection 𝜋 ∈ C𝜂 is a finite [countable] cover of T. Proof. The property of general distributivity of union with respect to intersection from Theorem 1 (1.1.13) implies the equalities ⋃ ⟮P u | u ∈ U⟯ = ⋃ ⟮⋂ ⟮C iu(i) | i ∈ I⟯ | u ∈ U⟯ = = ⋂ ⟮⋃ ⟮C im | m ∈ M i ⟯ | i ∈ I⟯ = ⋂ ⟮T | i ∈ I⟯ = T. Since every set M i is finite [countable], Lemma 5 (1.3.3) [Proposition 1 (1.3.9)] guarantees that their product U ≡ ∏⟮M i | i ∈ I⟯ is also finite [countable]. Lemma 2. Let C be a covering on T. Then, the covering C𝜂 is multiplicative. Proof. Let (𝜋𝛼 ∈ C𝜂 | 𝛼 ∈ A) be a finite collection. By definition, 𝜋𝛼 ≡ ⟮P𝛼u | u ∈ U𝛼 ⟯ for 𝛼 ∈ A and for every 𝜋𝛼 , there is finite collection (𝛾𝛼i ∈ C | i ∈ I𝛼 ) of covers such that 𝛾𝛼i ≡ ⟮C𝛼im | m ∈ M𝛼i ⟯, U𝛼 = ∏⟮M𝛼i | i ∈ I𝛼 ⟯, and P𝛼u = ⋂⟮C𝛼iu(i) | i ∈ I𝛼 ⟯ for every u ∈ U𝛼 . Consider the finite set J ≡ ⋃d ⟮I𝛼 | 𝛼 ∈ A⟯ ≡ ⋃⟮I𝛼 ∗ {𝛼} | 𝛼 ∈ A⟯. By the property of general associativity of product from Theorem 3 (1.1.12), we have the following bijections W ≡ ∏⟮U𝛼 | 𝛼 ∈ A⟯

∏ {∏⟮M𝛼i | ⟨i, 𝛼⟩ ∈ I𝛼 ∗ {𝛼}⟯ | 𝛼 ∈ A} { } ∏⟮M j | j ∈ J⟯ ≡ V ,

where M j ≡ M𝛼i for j ≡ ⟨i, 𝛼⟩ ∈ J𝛼 ≡ I𝛼 ∗ {𝛼}. Consider the finite collection (𝛿j | j ∈ J) of finite collections 𝛿j ≡ ⟮D jm | m ∈ M j ⟯ such that D jm ≡ C𝛼im for j ≡ ⟨i, 𝛼⟩ ∈ J𝛼 and m ∈ M j ≡ M𝛼i . Since 𝛿⟨i,𝛼⟩ = ⟮C𝛼im | m ∈ M𝛼i ⟯ we infer that 𝛿j ∈ C for every j ∈ J. Define the sets Q v ≡ ⋂⟮D jv(j) | j ∈ J⟯ and the collection 𝜘 ≡ ⟮Q v | v ∈ V⟯ ∈ C𝜂 . If v ∈ V is the image of w ∈ W, with respect to the mentioned bijections, then ⋂ ⟮P𝛼w(𝛼) | 𝛼 ∈ A⟯ = ⋂ ⟮⋂ ⟮C𝛼iw(𝛼)(i) | i ∈ I𝛼 ⟯ | 𝛼 ∈ A⟯ = ⋂ ⟮⋂ ⟮D jv(j) | j ∈ J𝛼 ⟯ | 𝛼 ∈ A⟯ = ⋂ ⟮D jv(j) | j ∈ J⟯ = Q v . Therefore, ⋀ (𝜋𝛼 | 𝛼 ∈ A) = 𝜘. Consequently, ⋀ (𝜋𝛼 | 𝛼 ∈ A) ∈ C𝜂 . Consider one important example of covering generated by a descriptive space ⟮T, S⟯ with the ensemble S containing T. Denote the family of all finite covers (S i ∈ S | i ∈ I) of T by Cov S. This covering is widely used in 2.4.

2.2.1 Real-valued functions and pointwise operations over them |

49

Lemma 3. Let S be a multiplicative ensemble on a set T containing T. Then, the covering Cov S is multiplicative. Proof. The assertion follows directly from the definitions of a multiplicative ensemble and a multiplicative covering.

2.2 Families of real-valued functions on a set 2.2.1 Real-valued functions and pointwise operations over them Let T be a set. A mapping f : T → R [respectively, f : T → R] is called a real-valued [extended real-valued] function on a set T. It will be also called a R-valued [R-valued] function. The set Map(T, R) of all real-valued functions on T will be denoted by F(T). We shall often say “function” instead of “real-valued function” and “number” instead of “real number”. For a function f ∈ F(T) and a set E ⊂ T, the number 𝜔( f , E) ≡ sup{|f (t) − f (s)| | t, s ∈ E} is called the oscillation of the function f on the set E. A function f on T is called a bounded function on P ⊂ T if there is a number x > 0 such that |f (t)| ⩽ x for every t ∈ P. The set of all bounded functions on T will be denoted by F b (T). Lemma 1. Let f ∈ F(T) and 𝜀 > 0. If there is a finite cover 𝜋 ≡ ⟮C i ⊂ T | i ∈ I⟯ of T such that 𝜔( f , C i ) < 𝜀 for every i ∈ I, then f ∈ F b (T). Proof. For every i ∈ I, take some point t i ∈ C i . Then, |f (t)| ⩽ |f (t i )| + 𝜀 for every t ∈ C i . Since the set I is finite, x ≡ sup (|f (t i )| + 𝜀 | i ∈ I) ∈ R. Hence, |f (t)| ⩽ x for all t ∈ T, i. e. the function f is bounded. The definitions of all operations in R using below can be founded in 1.4.3 and 1.4.6.

Lebesgue sets of a real-valued function By | we denote one of the parentheses [ or ] (see 1.1.15). For any x, y ∈ R, every set of the form f −1 [|x, y|] will be called a Lebesgue set of the function f . Note that for x > y, we have |x, y| = ⌀. Lemma 2. Let f ∈ F(T), x, y ∈ R, and x < y. Then, 1) f −1 []x, y[] = f −1 [] − ∞, y[] ∩ f −1 []x, ∞[] and f −1 [[x, y]] = f −1 [] − ∞, y]] ∩ f −1 [[x, ∞[]; 2) 3)

f −1 []x, ∞[] = ⋃ ⟮ f −1 [[x + 1/m, ∞[] | m ∈ N⟯ = ⋃ ⟮ f −1 []x, m[] | m ∈ N⟯; f −1 [] − ∞, y[] = ⋃ ⟮ f −1 [] − ∞, y − 1/n]] | n ∈ N⟯ = ⋃ ⟮ f −1 [] − n, y[] | n ∈ N⟯;

50 | 2.2 Families of real-valued functions on a set

4) f −1 [[x, ∞[] = ⋂ ⟮ f −1 []x − 1/m, ∞[] | m ∈ N⟯; 5) f −1 [] − ∞, y]] = ⋂ ⟮ f −1 [] − ∞, y + 1/n[] | n ∈ N⟯; 6) f −1 []x, y[] = ⋃ ⟮⋃ ⟮ f −1 [[x + 1/m, y − 1/n]] | m ∈ N⟯ | n ∈ N⟯ = = ⋃ ⟮ f −1 [[x + 1/m, y − 1/n]] | (m, n) ∈ N × N⟯; 7)

f −1 [[x, y]] = ⋂ ⟮⋂ ⟮ f −1 []x − 1/m, y + 1/n[] | m ∈ N⟯ | n ∈ N⟯ = = ⋂ ⟮ f −1 []x − 1/m, y + 1/n[] | (m, n) ∈ N × N⟯;

8) f −1 []x, y[] = T\(⋃ ⟮ f −1 [|x − k, x]] | k ∈ N⟯ ∪ ⋃ ⟮ f −1 [[y, y + l|] | l ∈ N⟯) = = T\(⋃ ⟮ f −1 [|x − k, x]] ∪ f −1 [[y, y + l|] | (k, l) ∈ N × N⟯); 9) f −1 [[x, y]] = T\(⋃ ⟮ f −1 [|x − k, x[] | k ∈ N⟯ ∪ ⋃ ⟮ f −1 []y, y + l|] | l ∈ N⟯) = = T\(⋃ ⟮ f −1 [|x − k, x[] ∪ f −1 []y, y + l|] | (k, l) ∈ N × N⟯). Proof. The equalities in (1) follow from the equalities ]x, y[=] − ∞, y[ ∩ ]x, ∞[ and [x, y] =] − ∞, y] ∩ [x, ∞[ and Lemma 3 (1.1.10). Assertions 2 – 5 follow from the equalities ]x, ∞[ = ⋃ ⟮ [x + 1/m, ∞[| m ∈ N⟯,

] − ∞, y[= ⋃ ⟮ ] − ∞, y − 1/n] | n ∈ N⟯,

[x, ∞[ = ⋂⟮ ]x − 1/m, ∞[| m ∈ N⟯,

] − ∞, y] = ⋂⟮ ] − ∞, y + 1/n[| n ∈ N⟯,

]x, ∞[ = ⋃ ⟮ ]x, m[| m ∈ N⟯,

] − ∞, y[= ⋃ ⟮ ] − n, y[| n ∈ N⟯,

and Lemma 3 (1.1.10). Assertions 6 and 7 follow from assertion 1, assertions 2 – 5, assertion 3 of Corollary 2 to Theorem 1 (1.1.13), and Corollary 1 to Proposition 1 (1.1.13). Assertions 8 and 9 follow from the equalities ]x, y[ = R\(⋃ ⟮|x − k, x] | k ∈ N⟯ ∪ ⋃ ⟮[y, y + l| | l ∈ N⟯), [x, y] = R\(⋃ ⟮|x − k, x[| k ∈ N⟯ ∪ ⋃ ⟮] y, y + l| | l ∈ N⟯), Lemma 4 (1.1.8), Lemma 3 (1.1.10), assertion 1 of Corollary 2 to Theorem 1 (1.1.13), and Corollary 1 to Proposition 1 (1.1.13).

Pointwise operations over real-valued functions For any finite family ( f i ∈ F(T) | i ∈ I) of functions on T, we can define the general operations of addition and multiplication setting (∑( f i | i ∈ I))(t) ≡ ∑( f i (t) | i ∈ I) and (P( f i | i ∈ I))(t) ≡ P( f i (t) | i ∈ I) for every t ∈ T. If I = k + 1, then along with ∑( f i | i ∈ k + 1) and P( f i | i ∈ k + 1) we shall also write f0 + . . . + f k and f0 ⋅ . . . ⋅ f k . Taking a pair of functions { f , g} consider the collection (h i | i ∈ 2), where h0 ≡ f and h1 ≡ g. Along with ∑(h i | i ∈ 2) and P(h i | i ∈ 2) or h0 + h1 and h0 ⋅ h1 we shall also write f + g and fg.

2.2.1 Real-valued functions and pointwise operations over them |

51

For any function f and any real number x, define the function xf setting (xf )(t) ≡ xf (t) for every t ∈ T. If m ∈ N, then mf = ∑( f i | i ∈ m) for the collection ( f i ∈ F(T) | i ∈ m), where f i ≡ f for every i. Define on T the unit function 1 setting 1(t) ≡ 1 for every t ∈ T and the zero function 0 setting 0(t) ≡ 0 for every t ∈ T. Then, 0 + f = f + 0 = f and 1f = f 1 = f for every function f . The function x1, where x ∈ R, is called the constant function on T taking the value x. For a function f , define the opposite function −f setting (−f )(t) ≡ −f (t) for every t ∈ T. Then, f − f = f + (−f ) = 0. It is clear that (−1)f = −f . For a function f : T → R\{0}, define the inverse function 1/f ≡ f −1 setting ( f −1 )(t) ≡ 1/( f (t)) for every t ∈ T. Then, ff −1 = f −1 f = 1. Along with gf −1 we shall also write g/f . It is easy to see that the mapping e : R → F(T) such that e(x) ≡ x1 is injective and preserve addition, multiplication and multiplication by reals. The following important properties of addition and multiplication of functions are simple corollaries of the definitions and corresponding properties of real numbers from 1.4.3 (Theorems 1 and 4 and their corollaries). Proposition 1. 1) Suppose that (f i | i ∈ I) is a finite collection of functions, K is a finite set, and u is a bijective mapping from K onto I; then ∑( f i | i ∈ I) = ∑( f u(k) | k ∈ K) and P( f i | i ∈ I) = P( f u(k) | k ∈ K) (the general commutativity of the sum and the product, respectively); 2) Suppose that (f i | i ∈ I) is a finite collection of functions and a collection ⟮I m ⊂ I | m ∈ M⟯ is a partition of the set I indexed by a finite non-empty set M; then ∑( f i | i ∈ I) = ∑(∑( f i | i ∈ I m ) | m ∈ M) and P( f i | i ∈ I) = P(P( f i | i ∈ I m ) | m ∈ M) (the general associativity of the sum and the product, respectively). 3) Suppose that ⟮I m | m ∈ M⟯ is a finite collection of finite non-empty sets indexed by a finite non-empty set M and (𝜘m | m ∈ M) is a finite collection of simple collections 𝜘m ≡ ( f mi ∈ F(T) | i ∈ I m ). Consider the finite set U ≡ ∏⟮I m | m ∈ M⟯. Then, P(∑( f mi | i ∈ I m ) | m ∈ M) = ∑(P( f mu(m) | m ∈ M) | u ∈ U) (the general distributivity of the product with respect to the sum). Corollary 1. Let (g j | j ∈ J) and (h k | k ∈ K) be finite simple collections of functions. Then, ∑(g j | j ∈ J) ∑(h k | k ∈ K) = ∑(g j h k | (j, k) ∈ J × K). Corollary 2. Let (( f ij ∈ F(T) | j ∈ J) | k ∈ K) be finite collection of finite collections of functions. Then, ∑ (∑ (f ij | j ∈ J) | i ∈ I) = ∑ (f ij | (i, j) ∈ I × J) and P (P (f ij | j ∈ J) | i ∈ I) = P (f ij | (i, j) ∈ I × J). Corollary 3. Let f , g, h ∈ F(T). Then, 1) f + g = g + f and fg = gf (the commutativity of the sum and of the product);

52 | 2.2 Families of real-valued functions on a set

2)

f + (g + h) = ( f + g) + h and f (gh) = ( fg)h (the associativity of the sum and of the product); 3) f (g + h) = fg + fh (the distributivity of the product with respect to the sum).

Define the operation f r of rising of a function f to a degree r setting ( f r )(t) ≡ ( f (t))r for every t ∈ T in the following four cases: i) f : T → R and r ∈ N; ii) f : T → R\{0} and r ∈ Z; iii) f : T → R+ and r ∈ Q+ ; iv) f : T → R+ \{0} and r ∈ Q. Hence, if k ∈ N, then f k = P( f i | i ∈ k) for the collection ( f i | i ∈ k), where f i ≡ f for every i ∈ k. m If f ⩾ 0 and m ∈ N, then along with f 1/m we shall also write √ f. The following properties of rising to a degree are simple corollaries of the definitions and corresponding properties of real numbers from 1.4.6 (Proposition 1). Proposition 2. a) Let ( f i : T → R\{0} | i ∈ I) and (m i ∈ Z | i ∈ I) be finite collections, f : T → R\{0}, m, n ∈ Z. Then, 1) f 0 = 1; f 1 = f ; 1m = 1; 0m = 0 for m > 0; 2) f ∑(m i |i∈I) = P( f m i | i ∈ I); 3) P( f i | i ∈ I))m = P( f im | i ∈ I); 4) f mn = ( f m )n . b) Let (f i : T → R+ \{0} | i ∈ I) and (r i ∈ Q | i ∈ I) be finite collections, f : T → R+ \{0}, r, s ∈ Q. Then, 1) 1r = 1; 0r = 0 for r > 0; 2) f ∑(r i |i∈I) = P( f r i | i ∈ I); 3) (P(f i | i ∈ I))r = P( f ir | i ∈ I); 4) f rs = ( f r )s . 2.2.2 The pointwise order between functions Define the order relation on F(T) setting f ⩽ g if f (t) ⩽ g(t) for every t ∈ T. Then, ⟮F(T), ⩽⟯ is an ordered set in sense of 1.1.15. It is clear that the mapping e : R → F(T) such that e(x) ≡ x1 preserve the order. If f ⩽ g and f = / g (i. e. f (t) < g(t) for some t ∈ T), then we shall write f < g. If f (t) < g(t) for every t ∈ T, then we shall write f ≪ g. It is easy to see that (f ∨ g)(t) = f (t)⊻ g(t) and (f ∧ g)(t) = f (t)⊼ g(t) (see 1.1.15 and 1.4.5). We also note that the inequality f ⩽ g is equivalent to f ∨ g = g and is equivalent to f ∧ g = f.

2.2.2 The pointwise order between functions

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53

Functions f : T → R+ [ f : T → R+ \{0}] are called positive [strictly positive], functions f : T → R− [f : T → R− \{0}] are called negative [strictly negative]. The sets of all positive and all negative functions will be denoted by F(T)+ and F(T)− , respectively. The following properties are direct corollaries to the definition and corresponding properties of real numbers (Proposition 4 (1.4.3)). Proposition 1. Let ( f i | i ∈ I) and (g i | i ∈ I) be finite collections of functions, f , g ∈ F(T), and r, s ∈ Q. Then, 1) if f i ⩽ g i for every i ∈ I, then ∑(f i | i ∈ I) ⩽ ∑(g i | i ∈ I); if besides f i < g i at least for one index, then ∑(f i | i ∈ I) < ∑(g i | i ∈ I); 2) f < g implies xf < xg and (−x)f > (−x)g for every x ∈ R+ ; 3) if 0 ⩽ f i ⩽ g i for every i ∈ I, then P(f i | i ∈ I) ⩽ P(g i | i ∈ I); if besides f i < g i at least for one index and g i ≫ 0 for every i ∈ I, then P(f i | i ∈ I) < P(g i | i ∈ I); 4) if 0 ≪ f < g, then f r < g r for r > 0 and f r > g r for r < 0; 5) r < s implies f r < f s for f ≫ 1 and f r > f s for 0 ≪ f ≪ 1. For a function f , define its modulus |f | setting |f |(t) ≡ |f (t)| for every t ∈ T. A collection 𝜎 ≡ (f i ∈ F b (T) | i ∈ I) will be called uniformly (order) bounded [bounded above, bounded below] if there is x ∈ R such that |f i | ⩽ x1 [f i ⩽ x1, f i ⩾ x1] for every i ∈ I. The following properties are simple corollaries to the definition and corresponding properties of real numbers (see Proposition 6 (1.4.3)). Proposition 2. Let (f i | i ∈ I) be a finite collection of functions, f , g ∈ F(T), and m ∈ Z. Then, 1) |f | = | − f |, f ⩽ |f |, and −f ⩽ |f |; 2) |P(f i | i ∈ I)| = P(|f i | | i ∈ I); 3) | ∑(f i | i ∈ I)| ⩽ ∑(|f i | | i ∈ I); 󵄨 󵄨 4) 󵄨󵄨󵄨 |f | − |g| 󵄨󵄨󵄨 ⩽ |f − g|; 5) if f : T → R\{0}, then |f m | = |f |m ; 6) if g > 0, then |f | ⩽ g is equivalent to −g ⩽ f ⩽ g, and |f | ≪ g is equivalent to −g ≪ f ≪ g. The following assertion is a corollary to Theorem 1 (1.4.5) on Dedekind completeness of the real line. Proposition 3. Let a collection 𝜎 ≡ (f i ∈ F(T) | i ∈ I) be bounded above [below]. Then, the function g ∈ F(T) such that g(t) = sup(f i (t) | i ∈ I) [g(t) = inf(f i (t) | i ∈ I)] for every t ∈ T is the supremum [infimum] of 𝜎.

54 | 2.2 Families of real-valued functions on a set

Proof. This function g is well defined By virtue of Theorem 1 (1.4.5). It is clear that g is an upper bound of 𝜎. Let h ⩾ f i for all i ∈ I. Then, h(t) ⩾ f i (t) for every t ∈ T implies h(t) ⩾ sup(f i (t) | i ∈ I) = g(t) for every t ∈ T, i. e. h ⩾ g. Therefore, g = sup 𝜎. The case of a collection bounded below is considered in the same way. The functions f+ ≡ f ∨ 0 and f− ≡ f ∧ 0 are called the positive and the negative parts of a function f with respect to the neutral element 0 (see 1.1.15). Theorem 1. Let 𝜎 ≡ (f i ∈ F(T) | i ∈ I) be a collection of functions on T and f , g ∈ F(T). Then, 1) if f = sup 𝜎, then g + f = sup(g + f i | i ∈ I); 2) if f = inf 𝜎, then g + f = inf(g + f i | i ∈ I); 3) if f = sup 𝜎, then gf = sup(gf i | i ∈ I) when g ⩾ 0 and gf = inf(gf i | i ∈ I) when g ⩽ 0; 󸀠 3 ) if f = sup 𝜎, then xf = sup(xf i | i ∈ I) for x ∈ R+ and xf = inf(xf i | i ∈ I) for x ∈ R− ; 4) if f = inf 𝜎, then gf = inf(gf i | i ∈ I) when g ⩾ 0 and gf = sup(gf i | i ∈ I) when g ⩽ 0; 4󸀠 ) if f = inf 𝜎, then xf = inf(xf i | i ∈ I) for x ∈ R+ and xf = sup(xf i | i ∈ I) for x ∈ R− ; 5) if f i ≫ 0 for every i ∈ I and f = sup 𝜎, then f −1 = inf(f i−1 | i ∈ I); 6) if f ≫ 0 and f = inf 𝜎, then f −1 = sup(f i−1 | i ∈ I); 7) if f = sup 𝜎, then g ∧ f = sup(g ∧ f i | i ∈ I); 8) if f = inf 𝜎, then g ∨ f = inf(g ∨ f i | i ∈ I). Let f , g, h ∈ F(T), then |f ∨ h − g ∨ h| + |f ∧ h − g ∧ h| = |f − g| (the Birkhoff identity), |f ∨ h − g ∨ h| ⩽ |f − g| and |f ∧ h − g ∧ h| ⩽ |f − g| (the Birkhoff inequalities). Let ⟮I m | m ∈ M⟯ be a total multivalued collection of sets indexed by a non-empty set M and U ≡ ∏⟮I m | m ∈ M⟯. Let (𝜘m | m ∈ M) be a collection of collections of functions 𝜘m ≡ (f mi | i ∈ I m ) and (h m | m ∈ M) and (g u | u ∈ U) be collections of functions. Then, 1) if h m = sup(f mi | i ∈ I m ) and g u = inf(f mu(m) | m ∈ M), then there exists f ∈ F(T) such that f = inf(h m | m ∈ M) and f = sup(g u | u ∈ U) (the general distributivity of the infimum with respect to the supremum); 2) if h m = inf(f mi | i ∈ I m ) and g u = sup(f mu(m) | m ∈ M), then there exists f ∈ F(T) such that f = sup(h m | m ∈ M) and f = inf(g u | u ∈ U) (the general distributivity of the supremum with respect to the infimum). Proof. By Proposition 3, the supremum and the infimum are pointwise operations. Therefore, all assertions of the theorem follow from the properties of real numbers collected in Lemma 4 (1.4.5), Proposition 1 (1.4.5), and Theorems 3 and 4 (1.4.5). Lemma 1. Let f ∈ F(T). Then, f = f+ + f− , |f | = f+ − f− , and |f | = f ∨ (−f ).

2.2.3 The pointwise and uniform convergences of nets and sequences of functions

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Proof. By property 4 of Lemma 5 (1.4.5) we get f (t) = f (t) + 0 = f (t) ⊻ 0 + f (t) ⊼ 0 = f+ (t) + f− (t) for every t ∈ T. Property 5) of that Lemma implies |f |(t) = |f (t) − 0| = f (t) ⊻ 0 − f (t) ⊼ 0 = f + (t) − f− (t). Property 6) of that Lemma implies |f |(t) = |f (t)| = f (t) ⊻ (−f (t)). Corollary 1. Let f , g ∈ F(T). Then, (f + g)+ ⩽ f+ + g+ and −(f + g)− ⩽ −f− − g− . Proof. By Lemma 1 and assertion 3 of Proposition 2 we get (f + g)+ = (f + g + |f + g|)/2 ⩽ (f + g + |f | + |g|)/2 = f + + g+ and −(f + g)− = (|f + g| − (f + g))/2 ⩽ (|f | + |g| − f − g)/2 = −f− + (−g− ). Remark. Using the extended structures in R defined in 1.4.3 we can introduce the pointwise structures in Map(T, ] − ∞, ∞]) and Map(T, [−∞, ∞[) extending the considered pointwise structures in F(T) ≡ Map(T, R). Then, for elements of these families an appropriate part of the assertions presented above will hold.

2.2.3 The pointwise and uniform convergences of nets and sequences of functions Let N be an upward directed preordered infinite set and (f n ∈ F(T) | n ∈ N) be a net of functions (see 1.1.15 for arbitrary nets, see 1.4.7 for nets of real numbers and their limits). If N ⊂ 𝜔, then we have a sequence (f n ∈ F(T) | n ∈ N ⊂ 𝜔) (see 1.2.6 for sequences).

Pointwise convergence A function f ∈ F(T) is called a pointwise limit of a net (f n | n ∈ N) (see 1.4.7) and this net is called pointwise convergent to f if f (t) = lim(f n (t) | n ∈ N) for every t ∈ T. By Lemma 1 (1.4.7) every net of real numbers can have only the unique limit. Therefore, a net of functions can have only the unique limit f . We shall denote it by p-lim (f n | n ∈ N). A net (f n | n ∈ N) is called increasing [strictly increasing] if f n ⩽ f n+1 [f n < f n+1 ] for every n ∈ N. Similarly, a decreasing and a strictly decreasing nets are defined. Lemma 1. Let a net (f n ∈ F(T) | n ∈ N) be increasing and bounded above or decreasing and bounded below. Then, it has the pointwise limit. Proof. By Proposition 3 (1.4.7) for every t ∈ T, the net (f n (t) | n ∈ N) of real numbers has the limit. Define the function f : T → R setting f (t) ≡ lim (f n (t) | n ∈ N). Thus, f = p-lim (f n | n ∈ N).

56 | 2.2 Families of real-valued functions on a set

The pointwise convergence in F(T) coincide with the order convergence (see 1.1.15): Lemma 2. Let f ∈ F(T) and (f n ∈ F(T) | n ∈ N) be a net in F(T). Then, the following conclusions are equivalent: 1) f = p-lim (f n | n ∈ N); 2) f = o-lim (f n | n ∈ N). Proof. (1) ⊢ (2). By the condition, f (t) = lim (f n (t) | n ∈ N) for every t ∈ T. According to Lemma 3 (1.4.7), f (t) = o-lim (f n (t) | n ∈ N). By Lemma 4 (1.1.15), f (t) = lim (f n (t) | n ∈ N) = lim (f n (t) | n ∈ N) = sup (inf (f p (t) | p ∈ N n ) | n ∈ N) = inf (sup (f p (t) | p ∈ N n ) | n ∈ N), where N n ≡ [n, →[≡ {p ∈ N | p ⩾ n} as in 1.1.15, 1.4.4, and 1.4.7. Therefore, Proposition 3 (2.2.2) implies the equalities f = sup (inf (f p | p ∈ N n ) | n ∈ N) = inf (sup (f p | p ∈ N n ) | n ∈ N). Again by Lemma 4 (1.1.15) f = o-lim (f n | n ∈ N). (2) ⊢ (1). By the condition, and the definition of order convergence there are nets (g n | n ∈ N) and (h n | n ∈ N) in F(T) such that g n ⩽ f n ⩽ h n , g n ↑ f , and h n ↓ f . According to Proposition 1 (2.2.2), g n (t) ↑ f (t), h n (t) ↓ f (t), and g n (t) ⩽ f n (t) ⩽ h n (t) for every t ∈ T. This means that f (t) = o-lim (f n (t) | n ∈ N) for every t ∈ T. According to Lemma 3 (1.4.7), f (t) = lim (f n (t) | n ∈ N), where f = p-lim (f n | n ∈ N). Proposition 1. Let (f i ∈ F(T) | i ∈ I) be a finite collection, (f n ∈ F(T) | n ∈ N) be a net, ⟮(f in ∈ F(T) | n ∈ N) | i ∈ I⟯ be a finite collection of nets, f ∈ F(T), x ∈ R, m ∈ N. If f = p-lim(f n | n ∈ N) and f i = p-lim(f in | n ∈ N) for every i ∈ I, then 1) xf = p-lim(xf n | n ∈ N); 2) ∑(f i | i ∈ I) = p-lim (∑(f in | i ∈ I) | n ∈ N); 3) P(f i | i ∈ I) = p-lim (P(f in | i ∈ I) | n ∈ N); 4) if f and f n do not take the zero value, then 1/f = p-lim(1/f n | n ∈ N); 5) f m = p-lim(f nm | n ∈ N); 6) sup(f i | i ∈ I) = p-lim(sup(f in | i ∈ I) | n ∈ N); 7) inf(f i | i ∈ I) = p-lim(inf(f in | i ∈ I) | n ∈ N); 8) f+ = p-lim((f n )+ | n ∈ N), f− = p-lim((f n )− | n ∈ N), and |f | = p-lim(|f n | | n ∈ N). Proof. Since all operations mentioned here are pointwise all assertions follow from Propositions 1 (1.4.7) and 2 (1.4.7) and their corollaries. Theorem 1. Let f ∈ F(T) and f = p-lim (f n | n ∈ 𝜔) for some sequence (f n ∈ F(T) | −1 [I n ] | k ∈ 𝜔⟯ | n ∈ N⟯ for every x ⩽ y in R, where n ∈ 𝜔). Then, f −1 [[x, y]] = ⋂ ⟮⋃ ⟮f n+k 1 1 either I n =]x − n , y + n [ or I n = [x − n1 , y + n1 ] with respect to the operations in R (see 1.4.3).

2.2.3 The pointwise and uniform convergences of nets and sequences of functions

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Proof. Denote the left and right parts of this equality by L and R, respectively. Let t ∈ L. Then, by condition, for any n and for 𝜀 ≡ 1/n there is m such that |f (t) − f p (t)| < 𝜀 for every p ⩾ m. Take p = m ⊻ n. Then, p = n + k for some k. For this p, −1 [I n ], we have x − 1/n ⩽ f (t) − 𝜀 < f p (t) < f (t) + 𝜀 ⩽ y + 1/n. This means that t ∈ f n+k consequently t ∈ R. Conversely, let t ∈ R. Take any 𝜀 > 0. Then, by condition, there is m such that |f (t)− f p (t)| < 𝜀 for every p ⩾ m. Take some n such that n ⩾ m and 1/n < 𝜀. By assumption, −1 −1 [I n ] | k ∈ 𝜔⟯. This means that t ∈ f n+k [I n ] for some k. Consider p ≡ n + k. t ∈ ⋃ ⟮f n+k Since p ⩾ m, we infer that x − 2𝜀 ⩽ x − 1/n − 𝜀 ⩽ f p (t) − 𝜀 < f (t) < f p (t) + 𝜀 ⩽ y + 1/n + 𝜀 ⩽ y + 2𝜀. Consequently, x = sup (x − 2𝜀 | 𝜀 ∈]0, ∞[) ⩽ f (t) ⩽ inf (y + 2𝜀 | 𝜀 ∈]0, ∞[) = y, i. e. t ∈ L. Thus, L = R. Corollary 1. Let f = p-lim (f n | n ∈ 𝜔) for some sequence (f n ∈ F(T) | n ∈ 𝜔) and −1 [J n ] | k ∈ 𝜔⟯ | n ∈ N⟯ ∩ some function f ∈ F(T). Then, f −1 []x, y[] = ⋃ ⟮⋂ ⟮f n+k −1 ⋃ ⟮⋂ ⟮f n+k [K n ] | k ∈ 𝜔⟯ | n ∈ N⟯ for every x < y in R, where either J n =]x + n1 , ∞[ and K n =] − ∞, y − n1 [ for all n ∈ N or J n = [x + n1 , ∞] and K n = [−∞, y − n1 ] for all n ∈ N with respect to the operations in R (see 1.4.3). Proof. Consider the closed intervals H ≡ [−∞, x] and I ≡ [y, ∞] in R. Then, ]x, y[= R\(H ∪ I) = (R\H) ∩ (R\I). By Theorem 1 we get −1 [H n ] | k ∈ 𝜔⟯ | n ∈ N⟯ and f −1 [H] = ⋂ ⟮⋃ ⟮f n+k −1 [I n ] | k ∈ 𝜔⟯ | n ∈ N⟯, f −1 [I] = ⋂ ⟮⋃ ⟮f n+k

where either H n =]−∞, x+ n1 [ and I n =]y− n1 , ∞[ or H n = [−∞, x+ n1 ] and I n = [y− n1 , ∞]. Consider the complementary sets J n ≡ R\H n = [x + n1 , ∞] and K n ≡ R\I n = [−∞, y − n1 ] in the first case and J n =]x + n1 , ∞[ and K n =] − ∞, y − n1 [ in the second case. Taking the corresponding preimages, we get f −1 []x, y[] = (f −1 [R] \f −1 [H]) ∩ (f −1 [R] \f −1 [I]) = −1 [H n ] | k ∈ 𝜔⟯ | n ∈ N⟯) ∩ = (T\ ⋂ ⟮⋃ ⟮f n+k −1 [I n ] | k ∈ 𝜔⟯ | n ∈ N⟯) = ∩ (T\ ⋂ ⟮⋃ ⟮f n+k −1 [T\H n ] | k ∈ 𝜔⟯ | n ∈ N⟯∩ = ⋃ ⟮⋂ ⟮f n+k −1 [T\I n ] | k ∈ 𝜔⟯ | n ∈ N⟯ = ∩ ⋃ ⟮⋂ ⟮f n+k −1 −1 [J n ] | k ∈ 𝜔⟯ | n ∈ N⟯ ∩ ⋃ ⟮⋂ ⟮f n+k [K n ] | k ∈ 𝜔⟯ | n ∈ N⟯. = ⋃ ⟮⋂ ⟮f n+k

58 | 2.2 Families of real-valued functions on a set

Uniform convergence A function f ∈ F(T) is called a uniform limit of a net (f n ∈ F(T) | n ∈ N) and this net is called uniformly convergent to f if for every 𝜀 > 0 there is n ∈ N such that |f p − f | ⩽ 𝜀1 for all p ⩾ n. It is clear that a uniform limit of a net is at the same time its pointwise limit. Therefore, it is unique, it will be denoted by u-lim (f n | n ∈ N). Lemma 3. Let (f n ∈ F b (T) | n ∈ N) be a net, f ∈ F(T), and f = u-lim (f n | n ∈ N). Then, f ∈ F b (T). Proof. By the definition, there is n ∈ N such that |f p − f | ⩽ 1 for all p ⩾ n. According to property 6 of Proposition 1 (2.2.2), we get f n − 1 ⩽ f ⩽ f n + 1. Since there is x > 0 such that |f n (t)| ⩽ x for all t ∈ T, we obtain −x − 1 ⩽ f (t) ⩽ x + 1, i. e. |f (t)| ⩽ x + 1. This means that f ∈ F b (T). In accordance with 2.2.2, a net (f n ∈ F b (T) | n ∈ N) is called uniformly bounded [above, below] if there is x ∈ R such that |f n | ⩽ x1 [f n ⩽ x1, f n ⩾ x1] for every n ∈ N. Lemma 4. Let (f n ∈ F(T) | n ∈ N) be a net, f = u-lim (f n | n ∈ N) and a1 ⩽ f ⩽ b1. Then, for every 𝜀 > 0 there is k ∈ N such that (a − 𝜀)1 ⩽ f n ⩽ (b + 𝜀)1 for all n ⩾ k, i. ,e. the subnet (f n | n ∈ N k ) is uniformly bounded. Proof. By the definition, we have that for every 𝜀 > 0 there is k ∈ N such that |f n − f | ⩽ 𝜀1 for all n ⩾ k. Then, property 6 of Proposition 1 (2.2.2) implies f − 𝜀1 ⩽ f n ⩽ f + 𝜀1. Since a1 ⩽ f ⩽ b1 we get (a − 𝜀)1 ⩽ f n ⩽ (b + 𝜀)1. Proposition 2. Let (f n ∈ F(T) | n ∈ N) and (g n ∈ F(T) | n ∈ N) be nets, f , g ∈ F(T), x ∈ R, m ∈ N. If f = u-lim( f n | n ∈ N) and g = u-lim(g n | n ∈ N), then 1) xf = u-lim(xf n | n ∈ N); 2) f + g = u-lim ( f n + g n | n ∈ N); 3) fg = u-lim ( f n g n | n ∈ N) for f , g ∈ F b (T); 4) if f n do not take the zero value for all n ∈ N and 0 < a1 ⩽ |f | ⩽ b1 for some a, b ∈ R, then 1/f = u-lim(1/f n | n ∈ N); 5) f m = u-lim( f nm | n ∈ N); 6) f ∨ g = u-lim( f n ∨ g n | n ∈ N); 7) f ∧ g = u-lim( f n ∧ g n | n ∈ N); 8) f+ = u-lim(( f n )+ | n ∈ N), f− = u-lim(( f n )− | n ∈ N), and |f | = u-lim(|f n | | n ∈ N). Proof. Assertions 1, 2, 5, 6, and 7 are proven similarly to corresponding assertions of Propositions 1 (1.4.7) and 2 (1.4.7) and their corollaries. By the definition, for every 𝜀 > 0 there are m(𝜀) and n(𝜀) such that |f − f p | < 𝜀1 and |g − g q | < 𝜀1 for all p ⩾ m(𝜀) and q ⩾ n(𝜀).

2.2.4 Some useful functional families |

59

3. Suppose |f | ⩽ a1, |g| ⩽ b1. By Lemma 4 there is k such that |g n | ⩽ (b + 1)1 for all n ⩾ k. Therefore, we get |fg − f p g p | ⩽ |f | |g − g p | + |g p | |f − f p | ⩽ 𝜀1 for all 𝜀 ) ⊻ n ( 2a𝜀 ). p ⩾ k ⊻ m ( 2(b+1) 4. Again by Lemma 4, there is k such that (a − a2 ) 1 ⩽ |f n | ⩽ (b − a2 ) 1 for all n ⩾ k. Therefore, we get |1/f − 1/f p | ⩽

|f −f p | |f | |f p |

2

< 𝜀 for all p ⩾ k ⊻ m ( 𝜀a2 ).

8. This follows from (6) and (7). Corollary 1. Along with assertions 2, 3, 6, and 7, more general assertions similar to assertions 2, 3, 6, and 7 of Proposition 1 are valid. A sequence (f n ∈ F(T) | n ∈ N ⊂ 𝜔) is called inner uniformly convergent (fundamental with respect to the uniform convergence or uniformly fundamental or u-fundamental) if for every 𝜀 > 0 there is n ∈ N such that |f p − f q | < 𝜀1 for all p, q ⩾ n. Lemma 5. Let s ≡ (f n ∈ F(T) | n ∈ N ⊂ 𝜔) be an infinite sequence. Then, 1) if s is uniformly inner convergent, then s is uniformly bounded; 2) if s is uniformly convergent to some g ∈ F(T), then s is uniformly inner convergent. Proof. The proof is completely similar to the proof of Lemma 2 (1.4.4) with the change 𝜀 by 𝜀1. Lemma 6. Every inner uniformly convergent sequence (f n ∈ F(T) | n ∈ N ⊂ 𝜔) is uniformly convergent to some f ∈ F(T). Proof. It is clear that for every t ∈ T the sequence (f n (t) ∈ R | n ∈ N) is inner convergent in sense of 1.4.4. So by Theorem 1 (1.4.4) there exists the set X t ⊂ R such that x = lim (f n (t) | n ∈ N) for every x ∈ X t . By Lemma 2 (1.4.4) X t = {x t }. Define the function f : T → R setting f (t) ≡ x t . Then, f (t) = lim (f n (t) | n ∈ N) for every t ∈ T. Fix 𝜀 > 0. Take n ∈ N such that |f p − f q | < 𝜀1 for all p, q ⩾ n. By property 4 of Proposition 6 (1.4.3), f q (t) − 𝜀 < f p (t) < f q (t) + 𝜀 for all p, q ⩾ n and t ∈ T. Using Corollary 3 to Proposition 1 (1.4.7), we get the inequality f q (t) − 𝜀 ⩽ f (t) ⩽ f q (t) + 𝜀 for all q ⩾ n and t ∈ T. This implies |f − f q | ⩽ 𝜀1 for all q ⩾ n, i. e. f = u-lim (f n | n ∈ N).

2.2.4 Some useful functional families Any subset A(T) of the set F(T) will be called a family of functions or functional family on T. If A(T) ⊂ F b (T), then A(T) will be called a family of bounded functions. The subfamily A b (T) ≡ A(T) ∩ F b (T) will be called the bounded part of the family A(T).

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A family A(T) is called closed under (closed with respect to) (a) multiplication by real numbers if xf ∈ A(T) for every x ∈ R and f ∈ A(T); (b) finite addition if ∑(f i | i ∈ I) ∈ A(T) for every finite collection (f i ∈ A(T) | i ∈ I); (c) finite multiplication if P(f i | i ∈ I) ∈ A(T) for every finite collection f i ∈ A(T) | i ∈ I; (d) inversion if 1/f ∈ A(T) for every f ∈ A(T) such that 0 ∈ ̸ rng f ; (d󸀠 ) bounded inversion if 1/f ∈ A(T) for every f ∈ A(T) such that 0 ∈ ̸ rng f and 1/f ∈ F b (T); (e) finite exact bounds (supremum and infimum) if sup(f i | i ∈ I) ∈ A(T) and inf(f i | i ∈ I) ∈ A(T) for every finite collection (f i ∈ A(T) | i ∈ I); (f) the Stone truncation if f ∧ (x1) ∈ A(T) for every x ∈ R and f ∈ A(T); (g) uniform convergence (of sequences) if u-lim(f n | n ∈ N) ∈ A(T) for every sequence (f n ∈ A(T) | n ∈ N) with a cofinal subset N ⊂ 𝜔 (see 1.1.15 and Lemma 1 (1.4.4)); (h) pointwise convergence (of sequences) if p-lim(f n | n ∈ N) ∈ A(T) for every sequence (f n ∈ A(T) | n ∈ N) with a cofinal subset N ⊂ 𝜔. A family A(T) is called provided with the pointwise order if the order on A(T) induced from F(T). If A(T) is provided with the pointwise order and is closed under finite exact bounds, then for every finite collection (f i ∈ A(T) | i ∈ I) its supremum and infimum in F(T) and in A(T) coincide. We shall consider the following conditions for a family A(T): 1) A(T) contains 1; 1󸀠 ) A(T) contains 0; 2) A(T) is closed under multiplication by real numbers; 2󸀠 ) A(T) is closed under multiplication by inversely natural numbers 1/n for every n ∈ N; 󸀠󸀠 2 ) A(T) is closed under multiplication by the number −1 ∈ Z; 3) A(T) is closed under finite addition; 4) A(T) is closed under finite exact bounds (supremum and infimum); 5) A(T) is closed under finite multiplication; 6) A(T) is closed under uniform convergence (of sequences); 7) A(T) is closed under inversion; 7󸀠 ) A(T) is closed under bounded inversion; 8) A(T) is closed under pointwise convergence (of sequences). It is clear that the combination of conditions 1 and 2 implies conditions 1󸀠 , 2󸀠 , and 2󸀠󸀠 , the combination of conditions 1, 2󸀠󸀠 , and 3 implies condition 1󸀠 , the combination of conditions 1, 2󸀠 , 2󸀠󸀠 , and 3 implies that A(T) is closed under multiplication by rational numbers, the combination of conditions 2󸀠󸀠 and 4 implies that A(T) is closed under modulus.

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Note that in contemporary mathematics families of functions satisfying some particular combinations of conditions 1 – 8 are usually used under the following names (see the corresponding definitions at the end of this subsection). A family A(T) ⊂ F(T) satisfying conditions 1󸀠 , 2, and 3 is a linear R-space, a family satisfying conditions 1 – 4 is a lattice-ordered linear R-space with the weak order unit 1, a family satisfying conditions 1 – 3 and 5 is a linear R-algebra with the multiplicative unit 1, a family satisfying conditions 1 – 5 is called a lattice-ordered linear R-algebra with the multiplicative and weak order unit 1. If we replace in the last three descriptions condition 1 by condition 1󸀠 , then we should omit the words “with . . . unit” in their names. In the case of A(T) ⊂ F b (T) the weak order unit 1 turns out the strong order unit. The complete list of these conditions was compiled by F. Hausdorff in book [Hausdorff, 1927]. A functional family A(T) will be called normal if it satisfies conditions 1 – 7. A functional family A(T) will be called completely normal if it satisfies conditions 1 – 8. A functional family A(T) ⊂ F b (T) will be called boundedly normal if it satisfies conditions 1 – 6 and 7󸀠 . It is clear that the whole family F(T) is a lattice-ordered linear R-algebra with the multiplicative and weak order unit 1 and it is normal and completely normal. The family F b (T) is a lattice-ordered linear R-algebra with the multiplicative and strong order unit 1 and it is boundedly normal by virtue of Lemma 3 (2.2.3). Lemma 1. For every functional family A(T) ⊂ F(T), there exist in F(T) the smallest normal family N(A(T)) and the smallest completely normal family CN(A(T)) containing A(T). For every A(T) ⊂ F b (T), there exists the smallest boundedly normal family BN(A(T)) containing A(T). Proof. The set N of all normal families B ⊂ F(T) containing A(T) is non-empty because F(T) ∈ N. It is clear that the family N(A(T)) ≡ {f ∈ F(T) | ∀ B (B ∈ N ⇒ f ∈ B)} belongs to the set N. Therefore, N(A(T)) is the necessary family. The proofs for completely normal and boundedly normal families are the same. The families N(A(T)), CN(A(T)), and BN(A(T)) from Lemma 1 will be called the normal, completely normal, and boundedly normal envelopes of the family A(T), respectively. The description of normal and completely normal families and envelopes will be given in 2.3.6 (see also the end of 2.3.7). The description of boundedly normal families and envelopes will be given in 2.4.6.

Baire (convergence) envelope In addition, we can construct new functional families from existing ones attaching limits.

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Lemma 2. 1) For every family A(T) ⊂ F(T), there are in F(T) the smallest families U(A) and B(A) containing A(T) and closed under uniform and pointwise limits of sequences in F(T), respectively. 2) For every bounded family A(T) ⊂ F b (T), there are in F b (T) the smallest families U b (A) and B b (A) containing A(T) and closed under uniform and pointwise limits of sequences in F b (T), respectively. Proof. The set U of all families containing A(T) and closed with respect to uniform limits is non-empty because F(T) ∈ U. Consider the family U(A) ≡ {f ∈ F(T) | ∀ U (U ∈ U ⇒ f ∈ U)} belonging to the set U. Therefore, U(A) is the necessary family. The proof in the case of pointwise convergence and the proof of the second assertion are similar. These smallest families U(A) and B(A) [U b (A) and B b (A)] containing A are called the uniform and the pointwise closures of A(T) in F(T) [F b (T)]. The family B(A) is also called the convergence (Baire) envelope of the family A. The smallest families U b (A b ) and B b (A b ) containing A b are called the bounded uniform and the bounded pointwise closures of A b (T) in F b (T). The family B b (A b ) is also called the bounded convergence (Baire) envelope of the bounded part of the family A. A family A(T) is called uniformly or pointwisely dense in a family E(T) ⊂ F(T) if E(T) = U(A(T)) or E(T) = B(A(T)), respectively. To describe these closures outwards, we introduce for a family A ⊂ F(T) [A ⊂ F b (T)] the hulls u-Lim A(T) [bu-Lim A(T)] consisting of all functions f ∈ F(T) [f ∈ F b (T)] such that f = u-lim (f n | n ∈ 𝜔) for some sequence (f n ∈ A(T) | n ∈ 𝜔) and p-Lim A(T) [bp-Lim A(T)] consisting of all functions f ∈ F(T) [f ∈ F b (T)] such that f = p-lim (f n | n ∈ 𝜔) for some sequence (f n ∈ A(T) | n ∈ 𝜔). Then, the uniform closure has the following simple description. Lemma 3. 1) U(A) = u-Lim A and U b (A) = bu-Lim A. 2) Every inner uniformly convergent sequence (f n ∈ U(A) | n ∈ 𝜔) has a uniform limit f in U(A). 3) If A ⊂ F b (T), then U(A) = U b (A). Proof. 1. Let (g n | n ∈ 𝜔) be a sequence from u-Lim A, g ∈ F(T), and g = u-lim (g n | n ∈ 𝜔), i. e. for any 𝜀 > 0 there is n(𝜀) such that |g − g p | < 𝜀1 for all p > n(𝜀). By the condition, g n = u-lim (f ni | i ∈ 𝜔) for a sequence (f ni ∈ A | i ∈ 𝜔), i. e. for any 𝜀 > 0 there is i n (𝜀) such that |g n − f nk | < 𝜀1 for all k > i n (𝜀). Consider the numbers k p ≡ i p (𝜀/2) + 1. Then, for every p > n(𝜀/2) we get |g − f pk p | ⩽ |g − g p | + |g p − f pk p | < (𝜀/2 + 𝜀/2)1 = 𝜀1. Hence, g = u-lim (f nk n | n ∈ 𝜔) ∈ u-Lim A. Thus, u-Lim A is closed under uniform convergence.

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Let B be closed under uniform convergence, B ⊃ A, and f ∈ u-Lim A, i. e. f = u-lim (f n | n ∈ 𝜔) for a sequence (f n ∈ A | n ∈ 𝜔). Since all f n ∈ B, we get f ∈ B. This implies u-Lim A ⊂ B. Therefore, u-Lim A is the smallest family closed under uniform convergence and containing A, i. e. u-Lim A = U(A). 2. Let (f n ∈ U(A) | n ∈ 𝜔) be an inner uniformly convergent sequence. By Lemma 6 (2.2.3) this sequence has a uniform limit f in F(T). Since U(A) is closed under uniform convergence, f ∈ U(A). 3. Let f ∈ U(A). By (1), we see that f = u-lim (f n ∈ A | n ∈ 𝜔). This implies that for 𝜀 ≡ 1 there is n ≡ n(𝜀) such that |f − f p | < 𝜀1 for all p > n. Besides, |f n+1 | ⩽ b1 for some number b. Therefore, it follows from f n+1 − 1 < f < f n+1 + 1 that |f | ⩽ (b + 1)1. Let A(T) ⊂ F(T) and D ≡ P(F(T)). Define the mapping V : ⋃ ⟮Map(𝛼, D) | 𝛼 ∈ 𝜔1 ⟯ → D setting V(0) ≡ A, V(v) ≡ p-Lim ⋃ ⟮v(𝛽) | 𝛽 ∈ 𝛼⟯ for every ordinal number 𝛼 ∈ 𝜔1 and every mapping v : 𝛼 → D. According to the scheme of construction of mappings by transfinite induction (see Theorem 1 (1.2.8)), there is a unique mapping u : 𝜔1 + 1 → D such thatu(0) = V(0) ≡ Aandu(𝛼) = V(u|𝛼)forevery𝛼 ∈ 𝜔1 +1.Denoteu(𝛼)byLim𝛼 A(T).Itis clear that Lim0 A(T) = A(T) and Lim𝛼 A(T) = p-Lim ⋃ ⟮Lim𝛽 A(T) | 𝛽 ∈ 𝛼⟯, 𝛼 ∈ [1, 𝜔1 ]. It is also clear that 𝛽 < 𝛼 implies Lim𝛽 ⊂ Lim𝛼 . The family Lim𝛼 A(T) is called the family of Baire functions of the class 𝛼 for the family A(T). The increasing transfinite collection ⟮Lim𝛼 A(T) | 𝛼 ∈ 𝜔1 + 1⟯ is called the Baire collection for the family A(T). In a similar manner for a family A(T) ⊂ F b (T), we define the families Lim0b A(T) = A(T) and Lim𝛼b A(T) = bp-Lim ⋃ ⟮Lim𝛽b A(T) | 𝛽 ∈ 𝛼⟯, 𝛼 ∈ [1, 𝜔1 ]. The family Lim𝛼b A(T) is called the family of bounded Baire functions of the class 𝛼 for the family A(T). The Baire envelope B(A) has the following transfinite description. Lemma 4. 1) If A(T) ⊂ F(T), then Lim𝜔1 A(T) = ⋃ ⟮Lim𝛽 A(T) | 𝛽 ∈ 𝜔1 ⟯; if A(T) ⊂ F b (T), then Lim𝜔b 1 A(T) = ⋃ ⟮Lim𝛽b A(T) | 𝛽 ∈ 𝜔1 ⟯. 2) If A(T) ⊂ F(T), then B(A(T)) = Lim𝜔1 A(T); if A(T) ⊂ F b (T), then B b (A(T)) = Lim𝜔b 1 A(T). Proof. Put E ≡ ⋃ ⟮Lim𝛼 A(T) | 𝛼 ∈ 𝜔1 ⟯. 1. It is easy to see that E ⊂ Lim𝜔1 A(T). Now, we shall prove the inverse inclusion. Let a function f = p-lim (f n | n ∈ 𝜔) for some sequence (f n ∈ E | n ∈ 𝜔). Check that f ∈ E. It follows from the definition that for every n ∈ 𝜔 there is an ordinal number 𝛼n ∈ 𝜔1 such that f n ∈ Lim𝛼n A(T). By Lemma 1 (1.3.10) the number 𝛼 ≡ ⋃ ⟮𝛼n | n ∈ 𝜔⟯ ∈ 𝜔1 . Further, 𝛼+1 ∈ 𝜔1 by virtue of Corollary 1 to Lemma 3 (1.3.10). Therefore, 𝛼n ∈ 𝛼+1 implies f ∈ Lim𝛼+1 A(T) ⊂ E. 2. Since A ⊂ B(A), it is easy to check by induction on 𝛼 that Lim𝛼 A(T) ⊂ B(A) for all 𝛼 ∈ 𝜔1 . Hence, E ⊂ B(A). Since B(A) is closed under pointwise convergence, we get Lim𝜔1 A(T) ⊂ B(A). In the bounded case, the arguments are similar.

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Lemma 5. Let A(T) is closed under one or some of the following operations: multiplications by real numbers, finite addition, finite exact bounds, Stone truncation, taking of positive and negative parts of a function, modulus, and finite multiplication. Then, the families U(A b ), B(A), B b (A b ), p-Lim A, bp-Lim A b , and the families Lim𝛼 A, Lim𝛼b A b for all 𝛼 ∈ 𝜔1 + 1 are also closed under the same operations. The family U(A) is closed under the same operations except finite multiplication. Proof. The assertions for U(A) and U(A b ) follow from Propositions 1 and 2 (1.4.7), its Corollaries, and Lemma 3. The assertions for B(A), B b (A b ), p-Lim A, bp-Lim A b , and Lim𝛼 A, Lim𝛼b A b , 𝛼 ∈ 𝜔1 , are derived by transfinite induction from Proposition 1 and 2 (1.4.7), its Corollaries, and Lemma 4. By Lemma 4 Lim𝜔1 A(T) = ⋃ ⟮Lim𝛽 A(T) | 𝛽 ∈ 𝜔1 ⟯. Since this Baire collection is increasing, we infer that Lim𝜔1 A(T) is also closed under the same operations. Lemma 6. Let A(T) ⊂ F(T). Then, 1) Lim𝛼 p-Lim A(T) = Lim𝛼+1 A(T) for every 𝛼 ∈ 𝜔; 2) Lim𝛼 p-Lim A(T) = Lim𝛼 A(T) for every 𝛼 ∈ [𝜔, 𝜔1 [. Proof. 1. Consider the set N ⊂ 𝜔 consisting of all numbers n such that Limn p-Lim A(T) = Limn+1 A(T). If n = 0, then by definition, we have that Lim0 p-Lim A(T) ≡ Lim1 A(T). So 0 ∈ N. Suppose that n ∈ N, then we get Limn+1 p-Lim A(T) ≡ p-Lim ⋃ ⟮ Limm p-Lim A(T) | m ∈ n + 1⟯ = = p-Lim Limn p-Lim A(T) = p-Lim Limn+1 A(T) = = p-Lim ⋃ ⟮ Limm A(T) | m ∈ n + 2⟯ ≡ Limn+2 A(T). This means that n + 1 ∈ N. By the principle of natural induction (Theorem 1 (1.2.6)) N = 𝜔. 2. If 𝛼 ≡ 𝜔, then using the equality proven above, we get Lim𝜔 p-Lim A(T) ≡ p-Lim ⋃ ⟮ Limn p-Lim A(T) | n ∈ 𝜔⟯ = = p-Lim ⋃ ⟮ Limn+1 A(T) | n ∈ 𝜔⟯ = p-Lim ⋃ ⟮ Limn A(T) | n ∈ 𝜔⟯ ≡ Lim𝜔 A(T). Consider the subset B1 of the set 𝜔1 \𝜔 consisting of all numbers 𝛼 ∈ 𝜔1 \𝜔 for which the required equality is fulfilled. As we checked above, 𝜔 ∈ B1 . Consider the set B ≡ 𝜔∪ B1 . Let 𝛽 ∈]𝜔, 𝜔1 [ and 𝛽 ⊂ B. Then, Lim𝛽 A(T) ⊂ Lim𝛽 p-Lim A(T) ≡ p-Lim ⋃ ⟮ Lim𝛾 p-Lim A(T) | 𝛾 ∈ 𝛽⟯ ⊂ ⊂ p-Lim ⋃ ⟮ Lim𝛾 p-Lim A(T) | 𝛾 ∈ [𝜔, 𝛽[⟯ = p-Lim ⋃ ⟮ Lim𝛾 A(T) | 𝛾 ∈ [𝜔, 𝛽[⟯ ⊂ ⊂ Lim𝛽 A(T).

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This means that 𝛽 ∈ B. Using the principle of transfinite induction (Proposition 1 (1.2.8)), we conclude that B = 𝜔1 . Step-functions Describe now one of the simplest ways to construct families of functions. Let R be a subset of T. The function 𝜒(R) such that 𝜒(R)(t) = 1 for all t ∈ R and 𝜒(R)(t) = 0 for all t ∈ T\R is called the characteristic function of the set R. Characteristic functions have the following properties. Lemma 7. 1) 𝜒(⌀) = 0 and 𝜒(T) = 1; 2) if R ⊂ S, then 𝜒(S) − 𝜒(R) = 𝜒(S\R). For every collection ⟮R i ⊂ T | i ∈ I⟯, we have 3) sup(𝜒(R i ) | i ∈ I) = 𝜒(⋃ ⟮R i | i ∈ I⟯) and inf(𝜒(R i ) | i ∈ I) = 𝜒(⋂ ⟮R i | i ∈ I⟯); 4) ∑(𝜒(R i ) | i ∈ I) = 𝜒(⋃ ⟮R i | i ∈ I⟯) for any finite collection of disjoint sets; 5) ∏(𝜒(R i ) | i ∈ I) = 𝜒(⋂ ⟮R i | i ∈ I⟯) for any finite collection. Proof. All statements follow directly from the definition. Let f ∈ F(T). The function 𝜒({t ∈ T | f (t) > 0})−𝜒({t ∈ T | f (t) < 0}) is called the signed function of the function f and denoted by sign f . Let S be an ensemble on T. A function f ∈ F(T) is called an S-step function or a step function with respect to S or a step function on the space ⟮T, S⟯ if there are finite collections (S i ∈ S | i ∈ I) and (x i ∈ R | i ∈ I) such that f = ∑(x i 𝜒(S i ) | i ∈ I). The family of all S-step functions on T will be denoted by St(T, S). A function f ∈ F(T) will be called a countably S-step function if there are countable collections (S i ∈ S | i ∈ I) and (x i ∈ R | i ∈ I) such that ⋂ ⟮S i | i ∈ J ⊂ I⟯ ≠ ⌀ implies J is finite and f = ∑(x i 𝜒(S i ) | i ∈ I). The family of all countably S-step functions on T will be denoted by Stc (T, S). It is clear that St(T, S) ⊂ Stc (T, S). Lemma 8. Let R be a ring on T. Then, every R-step [countably R-step] function f = ∑(x i 𝜒(R i ) | i ∈ I) can be represented in disjoint form f = ∑(y n 𝜒(S n ) | n ∈ N) for some finite [countable] disjoint collection (S n ∈ R | n ∈ N ⊂ 𝜔). Proof. Let (S n ∈ R | n ∈ N ⊂ 𝜔) be a finite [countable] disjoint collection from Lemma 8 (2.1.1). Consider the sets I n ≡ {i ∈ I | S n ⊂ R i }. Since ⋂ ⟮R i | i ∈ I n ⟯ ≠ ⌀, we see that the sets I n are finite either f is step or countably step function. Put y n ≡ ∑(x i | i ∈ I n ). Suppose t ∈ ⋃ ⟮R i | i ∈ I⟯. Then, there is the unique set S n containing t. Therefore, t ∈ R i if and only if i ∈ I n . Hence, f (t) = ∑(x i 𝜒(R i )(t) | i ∈ I) = ∑(x i 𝜒(R i )(t) | i ∈ I n ) = ∑(x i | i ∈ I n ) = y n = ∑(y n 𝜒(S n )(t) | n ∈ N).

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Proposition 1. Let R be a ring on T. Then, the family St(T, R) contains 0 and is closed under multiplication by real numbers, finite addition, finite multiplication, and finite infimum and supremum, i. e. it is a lattice-ordered linear algebra. Moreover, it is closed under inversion, rising to a degree in all the cases described in 2.2.1, positive and negative parts, modulus, and the Stone truncation. If R is a 𝜎-ring, then the family Stc (T, R) also has the mentioned properties. If R is an algebra, then St(T, R) contains 1. Proof. Let f , g ∈ A ≡ St(T, R). By Lemma 8 we get f = ∑ (x m 𝜒(R m ) | m ∈ M) and g = ∑ (y n 𝜒(S n ) | n ∈ N) for some finite disjoint collections (R m ∈ R | m ∈ M ⊂ 𝜔) and (S n ∈ R | n ∈ N ⊂ 𝜔). Since R is multiplicative, additive, and closed under the difference, the representation f + g = ∑ ((x m + y n )𝜒(R m ∩ S n ) | (m, n) ∈ M × N) + + ∑ (x m 𝜒(R m \ ⋃ ⟮S n | n ∈ N⟯) | m ∈ M) + ∑ (y n 𝜒(S n \ ⋃ ⟮R m | m ∈ M⟯) | n ∈ N) implies that f + g ∈ A. For f ∨ g and f ∧ g, we use the similar arguments. The representation fg = ∑ ((x m y n )𝜒(R m ∩ S n ) | (m, n) ∈ M × N) implies that fg ∈ A. The same assertions for finite operations are easily derived from the proven ones by induction. If f = / 0 for all t ∈ T, then ⋃ ⟮R i | i ∈ I⟯ = T and x i = / 0 for all i ∈ I. Therefore, g/f = ∑(y n /x m 𝜒(R m ∩ S n ) | (m, n) ∈ M × N) ∈ A. Further, f r = ∑ (x rm 𝜒(R m ) | m ∈ M) ∈ A, f+ = ∑ ((x m ∨ 0)𝜒(R m ) | m ∈ M) ∈ A, f− = (−f )+ ∈ A, |f | = f+ + f− ∈ A, and f ∧ (x1) = ∑ ((x m ∧ x)𝜒(R m ) | m ∈ M) ∈ A. For A ≡ Stc (T, R), we take countable collections (R m | m ∈ M) and (S n | n ∈ N) and use the 𝜎-additivity of R. If R is an algebra, then 1 = 𝜒(T) ∈ St(T, R). A function f : T → R will be called a quite S-step function [quite countably S-step function] if there are a finite [countable] partition (S i ∈ S | i ∈ I) of the set T and a collection (x i ∈ R | i ∈ I) such that f (t) = x i for any t ∈ S i . The families of all quite S-step functions and quite countably S-step functions will be denoted by qSt(T, S) and qStc (T, S), respectively. It is clear that qSt(T, S) ⊂ St(T, S) and qSt(T, S) ⊂ qStc (T, S) ⊂ Stc (T, S). Proposition 2. Let R be an algebra on T. Then, qSt(T, R) = St(T, R). If, besides, R is 𝜎-additive, then qStc (T, R) = Stc (T, R). Proof. By Lemma 8 every function f ∈ St(T, R) [f ∈ Stc (T, R)] can be represented in disjoint form f = ∑(y n 𝜒(S n ) | n ∈ N) for some finite [countable] disjoint collection (S n ∈ R | n ∈ N ⊂ 𝜔). Put S N ≡ T\ ⋃ ⟮S n | n ∈ N⟯ and y N ≡ 0. Since R is an algebra

2.2.4 Some useful functional families |

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[a 𝜎-algebra], we get S N ∈ R. Thus, f = ∑(y n 𝜒(S n ) | n ∈ N + 1) for the partition (S n ∈ R | n ∈ N + 1) of the set T, i. e. f ∈ qSt(T, R) [f ∈ qStc (T, R)]. Proposition 3. Let R be a multiplicative ensemble on T such that there is at least one finite [countable] partition of T by elements of R. Then, the family qSt(T, R) [qStc (T, R)] is non-empty, contains 0 and 1, and is closed under multiplication by real numbers, finite addition, finite multiplication, and finite infimum and supremum, i. e. it is a latticeordered linear algebra with the multiplicative and strong [weak] order unit 1. Moreover, it is closed under inversion, rising to a degree in all the cases described in 2.2.1, positive and negative parts, modulus, and the Stone truncation. Proof. It is clear that any quite R-step functions f and g can be represented in the disjoint form f = ∑ (x m 𝜒(R m ) | m ∈ M ⊂ 𝜔) and g = ∑ (y n 𝜒(S n ) | n ∈ N ⊂ 𝜔) from the proof of Proposition 1. Majorized functions and sets A function f ∈ F(T) will be called majorized by a function u ∈ F(T) if |f | ⩽ u. A set P ⊂ T will be called majorized by a function u ∈ F(T) if 𝜒(P) ⩽ u. Let E(T) and A(T) be families of functions on T. Define the subfamily E m (T, A(T)) ≡ {f ∈ E(T) | ∃ u ∈ A(T) (|f | ⩽ u)} of all functions f ∈ E(T) majorized by some functions from A(T). In a similar way for any ensemble E on T, we define its subensemble Em (A(T)) ≡ {E ∈ E | ∃ u ∈ A(T) (𝜒(E) ⩽ u)} of all sets E ∈ E majorized by some functions from A(T). The following lemma describe how conditions 1 –5 satisfied by families E(T) and/or A(T) are inherited by the family Em (T, A(T)). Lemma 9. Let E(T) and A(T) be families of functions on T and A(T) ∩ F(T)+ ≠ ⌀. Then, 1) if E(T) and A(T) satisfy conditions 1, 3, 4, or 5, then Em (T, A(T)) satisfies conditions 1, 3, 4, or 5, respectively; 󸀠 1 ) if E(T) contains 0, then Em (T, A(T)) contains 0; 2) if E(T) is closed under multiplication by real numbers and A(T) is closed under multiplication by positive real numbers or under finite addition, then Em (T, A(T)) is closed under multiplication by real numbers; 2󸀠 ) if E(T) is closed under multiplication by inversely natural numbers, then Em (T, A(T)) is closed under multiplication by inversely natural numbers. Proof. 1. If 1 ∈ E(T) and 1 ∈ A(T), then 1 ∈ Em (T, A(T)). If f , g ∈ Em (T, A(T)), then |f | ⩽ u and |g| ⩽ v for some u, v ∈ A(T). Therefore, |f + g| ⩽ u + v ∈ A(T), |f ∨ g| ⩽ |f | ∨ |g| ⩽ u ∨ v and |f ∧ g| ⩽ |f | ∧ |g| ⩽ u ∧ v, or |fg| ⩽ uv ∈ A(T), respectively. 1󸀠 . Since 0 ⩽ u for every u ∈ A(T) ∩ F(T)+ , we have 0 ∈ Em (T, A(T)).

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2. Let f ∈ Em (T, A(T)) and r ∈ R. Then, |f | ⩽ u for some u ∈ A(T). If A(T) is closed under multiplication by positive real numbers, then |rf | ⩽ |r| |f | ⩽ |r|u ∈ A(T). If A(T) is closed under addition, then the Archimedes principle (Lemma 13 (1.4.3)) guarantees that there is n ∈ N such that n > |r| and we have |rf | ⩽ |r| |f | ⩽ |r|u ⩽ nu = u + . . . + u ∈ A(T). 2󸀠 . Let f ∈ Em (T, A(T)) and n ∈ N, then |f | ⩽ u for some u ∈ A(T). Hence, |(1/n)f | ⩽ |f | ⩽ u. Corollary 1. Let E(T) and A(T) be lattice-ordered linear spaces. Then, Em (T, A(T)) is a lattice-ordered linear space. Lemma 10. Let A(T) and E(T) be families of functions on T, E be an ensemble on T. Then, Em (T, A(T)) = Em (T, Em (T, A(T))) and Em (A(T)) = Em (Em (T, A(T)). Proof. Put B(T) ≡ Em (T, A(T)) and D(T) ≡ Em (T, B(T)). If f ∈ B(T), then |f | ⩽ u ∈ A(T) ⊂ B(T) implies f ∈ D(T). Conversely, if g ∈ D(T), then |g| ⩽ v ∈ B(T), but v ⩽ w for some function w ∈ A(T). Therefore, |g| ⩽ w and g ∈ B(T). Thus, B(T) = D(T). Applying this to the family E(T) ≡ {𝜒(E) | E ∈ E}, we establish that Em (A(T)) = Em (B(T)). Let S be an ensemble on T, K ≡ K(T, S) be the corresponding ensembles of symmetrizable sets, and A ≡ A(T, S) = K𝜑 be the corresponding algebra of Alexandrov sets. Let A(T) be a family of functions on T. Consider the ensembles Km ≡ Km (A(T)) ≡ Km (T, S, A(T)) and Am ≡ Am (A(T)) ≡ Am (T, S, A(T)). It is clear that A ∈ Am if and only if A = ⋃ ⟮K i | i ∈ I⟯ for some finite collection (K i ∈ Km | i ∈ I). Consider also the family Stm (T, A, A(T)) of A-step functions majorized by functions of the family A(T). Lemma 11. Let A(T) ⊂ F(T) be closed under multiplication by real numbers. Then, Stm (T, A, A(T)) ⊂ St(T, Am (A(T))). Besides, if A(T) is closed under finite addition, then Stm (T, A, A(T)) = St(T, Am (A(T))) and is a linear space; if A(T) satisfies condition 4, then Stm (T, A, A(T)) satisfies condition 4 as well. Proof. Let f ∈ Stm , i. e. f = ∑ (x i 𝜒(A i ) | i ∈ I) for some finite collections (x i ∈ R | i ∈ I) and (A i ∈ A | i ∈ I) and |f | ⩽ u for some u ∈ A(T). Since A is an algebra, according to Proposition 2, we can suppose that all sets A i are pairwise disjoint and x i ≠ 0 for all i ∈ I. Therefore, |f | = ∑ (|x i |𝜒(A i ) | i ∈ I) ⩽ u implies 𝜒(A i ) ⩽ (1/|x i |)u ∈ A(T), where A i ∈ Am and f ∈ St(T, Am ). Let A(T) be closed under finite addition and f ∈ St(T, Am ), i. e. f = ∑(x i 𝜒(A i ) | i ∈ I) for some finite collections (x i ∈ R | i ∈ I) and (A i ∈ Am | i ∈ I). By the definition, 𝜒(A i ) ⩽ u i ∈ A(T). Therefore, |f | ⩽ ∑ (|x i |𝜒(A i ) | i ∈ I) ⩽ ∑ (|x i |u i | i ∈ I) ∈ A(T). Thus, f ∈ Stm . Using Proposition 1 and Lemma 9, we obtain the other conclusions.

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The notions of a majorized function and a majorized set will be essentially used in 3.6.2. Auxiliary definitions and statements 1∘ In the book, we repeatedly consider some concrete polygraded mathematical systems U ≡ ⟮X, s⟯, where X ≡ ⟮A0 , . . . , A m , K0 , . . . , K n−1 ⟯ is the carrier, ⟮A0 , . . . , A m ⟯ is the principle carrier, ⟮K0 , . . . , K n−1 ⟯ is the auxiliary carrier, m, n ∈ 𝜔, and s ≡ ⟮w X | w ∈ W⟯ is the general structure of U. However, in practice, it may be onerously to indicate for the system U all its auxiliary sets K0 , . . . , K n−1 (if they exist) and all its partial structures w X . Therefore, we use the informal notation U ≡ |⟮A0 , . . . , A m ⟯, w0X , . . . , w k−1 X | for some k ∈ 𝜔. Moreover, the system U is often indicated by its principal carrier ⟮A0 , . . . , A m ⟯ only. 2∘ Let A be a set. The system U ≡ |⟮A, R⟯, 0A , −A , +A , ⋅⟮A,R⟯ | with the structures 0A ∈ A, −A : A → A, +A : A ∗ A → A, and ⋅⟮A,R⟯ : R ∗ A → A is called a linear space over the system ⟮R, w r ⟯ of real numbers with the ring structure w r ≡ ⟮0R , 1R , −R , +R , ⋅R ⟯ (see 1.4.3) or a (real) (R-)linear space if the following properties hold: 1) ∀ a ∈ A (0A + a = a); 2) ∀ a ∈ A (a + (−a) = 0A ); 3) ∀ a, b, c ∈ A (a + (b + c) = (a + b) + c); 4) ∀ a, b ∈ A (a + b = b + a); 5) ∀ a ∈ A (0R a = 0A ); 6) ∀ a ∈ A (1R a = a); 7) ∀ a ∈ A (−a = (−1R )a); 8) ∀ x ∈ R ∀ a, b ∈ A (x(a + b) = xa + xb); 9) ∀ x, y ∈ R ∀ a ∈ A ((x +R y)a = xa + ya); 10) ∀ x, y ∈ R ∀ a ∈ A ((xy)a = x(ya)). Note that this list is not the shortest one among all lists determining such a system. The element 0A is called the neutral element (see 1.1.15) of the linear space U with respect to addition or the zero element. 3∘ The system U ≡ |⟮A, R⟯, 0A , −A , +A , ⋅A , ⋅⟮A,R⟯ | with the additional structure ⋅A : A ∗ A → A is called a linear algebra over the system ⟮R, w r ⟯ or a (real) (R-)linear algebra if the following properties hold: 1) the system |⟮A, R⟯, 0A , −A , +A , ⋅⟮A,R⟯ | is a real linear space; 2) ∀ a, b, c ∈ A (a(bc) = (ab)c);

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3) ∀ a, b, c ∈ A ((a(b + c) = ab + ac) ∧ ((a + b)c = ac + bc)) (the distributive laws); 4) ∀ x ∈ R ∀ a, b ∈ A (x(ab) = (xa)b = a(xb)). The system U ≡ |⟮A, R⟯, 0A , −A , +A , 1A , ⋅A , ⋅⟮A,R⟯ | with the additional structure 1A ∈ A is called a (real) (R-)linear algebra with the (multiplicative) unit 1A if: 1) the system |⟮A, R⟯, 0A , −A , +A , ⋅A , ⋅⟮A,R⟯ | is a real linear algebra; 2) ∀ a ∈ A (1A a = a1A = a). The element 1A is called also the neutral element (see 1.1.15) with respect to multiplication. 4∘ The system U ≡ |⟮A, R⟯, 0A , −A , +A , ⩽A , ⋅⟮A,R⟯ | with the additional structure ⩽A ⊂ A ∗ A is called an ordered [lattice-ordered] linear space over the system ⟮R; w r , ⩽R ⟯ of real numbers with the ring structure w r and the order ⩽R (see 1.4.3) or a (real) ordered [lattice-ordered] (R-)linear space if the following properties hold: 1) the system ⟮A, ⩽A ⟯ is an ordered [lattice-ordered] set (see 1.1.15); 2) the system |⟮A, R⟯, 0A , −A , +A , ⋅⟮A,R⟯ | is a real linear space; 3) ∀ a, b, c ∈ A (a ⩽A b ⇒ a + c ⩽A b + c); 4) ∀ a, b ∈ A (a ⩽A b ⇒ −b ⩽A −a); 5) ∀ x ∈ R+ ∀ a, b ∈ A (a ⩽A b ⇒ xa ⩽A xb); 6) ∀ x, y ∈ R ∀ a ∈ A 0 (x ⩽R y ⇒ xa ⩽A ya). Note that this list is not the shortest one among all lists determining such a system. A lattice-ordered linear space is also called a Riesz space. According to 1.1.15, for every a󸀠 , a󸀠󸀠 ∈ A, there are a󸀠 ∨ a󸀠󸀠 ≡ sup(a󸀠 , a󸀠󸀠 ) ∈ A and 󸀠 a ∧ a󸀠󸀠 ≡ inf(a󸀠 , a󸀠󸀠 ) ∈ A. The element |a| ≡ a ∨ (−a) is called the modulus of the element a. An element e ∈ A is called a (weak) order unit in the lattice-ordered linear space U if ∀ a ∈ A (|a| ∧ |e| = 0 ⇒ a = 0). An element e ∈ A is called a strong order unit in the lattice-ordered linear space U if ∀ a ∈ A ∃ x ∈ R (|a| ⩽ xe). Statement 1. Let A be an ordered linear space. Then, the following assertions hold: 1) ∀ a, b ∈ A ∀P ⊂ A (a = sup P ⇒ b + a = sup{b + c | c ∈ P}); 2) ∀ a, b ∈ A ∀P ⊂ A (a = inf P ⇒ b + a = inf{b + c | c ∈ P}); 3) ∀ a ∈ A ∀P ⊂ A (a = sup P ⇒ −a = inf{−b | b ∈ P}); 4) ∀ a ∈ A ∀P ⊂ A (a = inf P ⇒ −a = sup{−b | b ∈ P}); 5) ∀ a ∈ A ∀ x ∈ R+ ∀P ⊂ A (a = sup P ⇒ xa = sup{xb | b ∈ P}); 6) ∀ a ∈ A ∀ x ∈ R+ ∀P ⊂ A (a = inf P ⇒ xa = inf{xb | b ∈ P}). Statement 2. Let A be a lattice-ordered linear space and a ∈ A. Then, a = a + + a− and |a| = a+ − a− = a+ ∨ (−a− ).

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These two equalities are called the Riesz decomposition formulas. 5∘ The system U ≡ |⟮A, R⟯, 0A , −A , +A , ⋅A , ⩽A , ⋅⟮A,R⟯ | is called a lattice-ordered linear algebra over the system ⟮R; w r , ⩽R ⟯ or a (real) lattice-ordered (R-)linear algebra if the following properties hold: 1) the system |⟮A, R⟯, 0A , −A , +A , ⩽A , ⋅⟮A,R⟯ | is a real lattice-ordered linear space; 2) the system |⟮A, R⟯, 0A , −A , +A , ⋅A , ⋅⟮A,R⟯ | is a real linear algebra; 3) ∀ a, b ∈ A (a ⩾ 0 ∧ b ⩾ 0 ⇒ ab ⩾ 0). 2.2.5 Zero-sets and cozero-sets of functions Let f be a function on T. Then, the sets zer f ≡ {t ∈ T | f (t) = 0} and coz f ≡ {t ∈ T | f (t) = / 0} are called the zero-set and the cozero-set of the function f , respectively. Note that the set coz f is the union of the basic Lebesgue sets f −1 []0, ∞[] and f −1 [] − ∞, 0[]. Let A(T) be a family of functions on T. Then, the ensembles Zer A(T) ≡ {zer f | f ∈ A(T)} and Coz A(T) ≡ {coz f | f ∈ A(T)} are called the ensemble of all zero-sets and the ensemble of all cozero-sets of the family A(T), respectively. Lemma 1. Let x ∈ R\{0}, f , g ∈ F(T), and (f n | n ∈ 𝜔) be a sequence in F(T)+ . Then, 1) coz f = coz |f | = coz f 2 = coz(xf ); 2) coz(fg) = coz(|f | ∧ |g|) = coz f ∩ coz g; 3) coz(f 2 + g2 ) = coz(|f | + |g|) = coz(|f | ∨ |g|) = coz f ∪ coz g; 4) if f = p-lim (∑(f k | k ∈ n) | n ∈ 𝜔) or f = sup(f n | n ∈ 𝜔), then coz f = ⋃ ⟮coz f n | n ∈ 𝜔⟯. Proof. The proofs for 1, 2, and 3 are trivial. 4. If f = p-lim (∑(f k | k ∈ n) | n ∈ 𝜔), then the inclusion ⋃ ⟮coz f n | n ∈ 𝜔⟯ ⊂ coz f is fulfilled. Let t ∈ coz f . Since the sequence (∑(f k | k ∈ n) | n ∈ 𝜔) is increasing, f ⩾ ∑(f k | k ∈ n) ⩾ 0 for every n. Hence, we can take 𝜀 ≡ f (t) > 0. For this 𝜀, there is a natural number n = n(𝜀) such that 0 ⩽ f (t) − ∑(f k (t) | k ∈ n) < 𝜀. Therefore, ∑(f k (t) | k ∈ n) > 0 and t ∈ coz f k for some k ∈ n. Thus, the inverse inclusion ⋃ ⟮coz f n | n ∈ 𝜔⟯ ⊃ coz f holds and we have the equality. For the case f = sup(f n | n ∈ 𝜔), the assertion is evident. Corollary 1. Let A(T) be a family of functions on T. Then, 1) if A(T) is closed with respect to finite multiplication or finite operation inf(|f i | | i ∈ I), then the ensemble Coz A(T) is multiplicative and the ensemble Zer A(T) is additive; 2) if A(T) is closed with respect to finite operations ∑ (f i2 | i ∈ I), ∑ (|f i | | i ∈ I), or sup(|f i | | i ∈ I), then Coz A(T) is additive and Zer A(T) is multiplicative;

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3) if sup(|f n | ∧ 1 | n ∈ 𝜔) ∈ A(T) for every sequence (f n ∈ A(T) | n ∈ 𝜔), then Coz A(T) is 𝜎-additive and Zer A(T) is 𝛿-multiplicative; 4) if A(T) is closed with respect to finite sums ∑ (2−k (|f k | ∧ 1) | k ∈ n) ,

∑ (2−k (f k2 ∧ 1) | k ∈ n) ,

∑(|f k | ∧ (2−k 1) | k ∈ n), or ∑(f k2 ∧ (2−k 1) | k ∈ n), and, besides, is closed with respect to uniform convergence of sequences, then Coz A(T) is 𝜎-additive and Zer A(T) is 𝛿-multiplicative. Proof. The proofs for 1, 2, and 3 are trivial. 4. Consider a sequence (f n ∈ A(T) | n ∈ 𝜔). Take 𝜀 > 0. There is a natural number n such that 2−n+1 < 𝜀. Consider the functions g n ≡ ∑ (2−k (|f k | ∧ 1) | k ∈ n) ∈ A(T). Then, g n+m − g n ⩽ ∑(2−k 1 | k ∈ (n + m)\n) = 2−n ∑(2−i | i ∈ m)1 = 2−n (2 − 2−m+1 )1 ⩽ 2−n+1 1 < 𝜀1 for every m. This means that the sequence (g n | n ∈ 𝜔) is uniformly inner convergent. By Lemma 6 (2.2.3) there exists g ∈ F(T) such that g = u-lim (g n | n ∈ 𝜔). The condition implies that g ∈ A(T). Since g = p-lim (∑ (2−k (|f k | ∧ 1) | k ∈ n) | n ∈ 𝜔), by assertion 4 of Lemma 1 ⋃ ⟮coz f n | n ∈ 𝜔⟯ = ⋃ ⟮coz (2−n (|f n | ∧ 1)) | n ∈ 𝜔⟯ = coz g ∈ Coz A(T). For the cases of other listed sums, the arguments are the same. Corollary 2. Let A(T) be a family of functions on T. 1) If A(T) satisfies condition 1 from 2.2.4, then Coz A(T) contains T. 2) If A(T) satisfies condition 1󸀠 ), then Coz A(T) contains ⌀. 3) If A(T) satisfies conditions 2󸀠󸀠 , 3, and 4 or conditions 3 and 5, then Coz A(T) is a lattice. Proof. This follows from assertions 2 and 3 of Lemma 1 and Lemma 1 (2.2.2). Proposition 1. Let A(T) be a family of functions on T. 1) If A(T) satisfies conditions 1, 1󸀠 , 3, and 4 (or 5) ) from 2.2.4, then Coz A(T) is a 𝜑-foundation. 2) If A(T) satisfies conditions 1, 1󸀠 , 2󸀠 , 3, 4, and 6, then Coz A(T) is a 𝜎-foundation. 3) If A(T) satisfies conditions 1, 1󸀠 , 2󸀠 , 2󸀠󸀠 , 3, 5, and 8, then Coz A(T) is a 𝜎-foundation. 4) If A(T) satisfies conditions 1, 1󸀠 , 2󸀠 , 2󸀠󸀠 , 3, and 4, then Coz A(T) is a separable perfect 𝜑-foundation. 5) If A(T) satisfies conditions 1, 1󸀠 , 2󸀠 , 2󸀠󸀠 , 3, 4, and 6, then Coz A(T) is an a-foundation (i. e. a separable perfect 𝜎-foundation). 6) If A(T) satisfies conditions 1, 1󸀠 , 2󸀠 , 2󸀠󸀠 , 3, 4 (or 5) ), and 8, then Coz A(T) is a 𝜎-algebra.

2.2.5 Zero-sets and cozero-sets of functions

| 73

Proof. Assertions 1 and 2 follow from Corollary 2 to Lemma 1 and assertion 4 of Corollary 1 to it. 3. By (1), the ensemble S ≡ Coz A(T) is a 𝜑-foundation. Check that it is 𝜎-additive. Consider a sequence (f n ∈ A(T) | n ∈ 𝜔). Take 𝜀 > 0. There is a natural number n such that 2−n+1 < 𝜀. Consider the functions g n ≡ ∑ (2−k (1 − exp ∘(−f k2 )) | k ∈ n), where exp : R → R+ is the exponential function from 1.4.7. According to the definition, exp ∘(−f k2 ) = p-lim ((1 − f k2 /m)m | m ∈ N). Then, conditions 1, 2󸀠 , 2󸀠󸀠 , 3, 5, and 8 imply that g n ∈ A(T). Since −f k2 (t) ⩽ 0 for every t ∈ T and k ∈ 𝜔, Corollary 2 to Lemma 10 (1.4.7) and assertion 2 of Corollary 1 to Theorem 1 (1.4.7) imply that 0 < exp(−f k2 (t)) ⩽ 1. Consequently, for every t ∈ T the sequence (g n (t) | n ∈ 𝜔) is increasing and, owing to Lemma 3 (1.4.8), 0 ⩽ g n (t) ⩽ ∑(2−k | k ∈ n) < 2. Then, Proposition 2 (1.4.4) guarantees that (g n (t) | n ∈ 𝜔) converges for every t ∈ T. Put g ≡ p-lim (g n | n ∈ 𝜔). Condition 8) implies that g ∈ A(T). Since g = p-lim (∑ (2−k (1 − exp ∘(−f k2 )) | k ∈ n) | n ∈ 𝜔), use

of Lemma 1 yields ⋃ ⟮coz f n | n ∈ 𝜔⟯ = ⋃ ⟮coz (2−n (1 − exp ∘(−f k2 )) | n ∈ 𝜔⟯ = coz g ∈ Coz A(T). 4. By (1), the ensemble S ≡ Coz A(T) is a 𝜑-foundation. Let X, Y ∈ R ≡ co-S and X ∩ Y = ⌀. Then, X = zer f and Y = zer g for some f , g ∈ A(T). Consider the functions u = |f | − |f | ∧ |g| and v = |g| − |f | ∧ |g| belonging to A(T). If t ∈ X, then v(t) = |g(t)| ≠ 0. Therefore, X ⊂ coz v ∈ S. Similarly, Y ⊂ coz u ∈ S. Since by assertion 2 of Corollary 1 to Lemma 4 (1.4.5) u ∧ v = |f | ∧ |g| − |f | ∧ |g| = 0, we have coz u ∩ coz v = ⌀. Thus, S is separable. Finally, let X ≡ coz f and f ∈ A(T). Consider the functions g n ≡ (1/n)1 − |f | ∧ (1/n)1 ∈ A(T). Then, |f |−1 [[1/n, ∞[] = zer g n ∈ R. Hence, X = ⋃ ⟮zer g n | n ∈ N⟯. Thus, S is perfect. 5. By (4), S is perfect and separable. By (2), S is a 𝜎-foundation. 6. By (2) [or (3)], the ensemble S is a 𝜎-foundation. Let S ∈ S. Then, S = coz f for some f ∈ A(T), f ⩾ 0. Take the functions f n ≡ nf ∧ 1 ∈ A(T), n ∈ 𝜔. Since the sequence (f n | n ∈ 𝜔) is increasing and bounded above by 1, Proposition 2 (1.4.4) and condition 8) imply g ≡ 𝜒(S) = p-lim (f n | n ∈ 𝜔) ∈ A(T). Then, h ≡ −1 + g ∈ A(T). Hence, T\S = coz h ∈ S. Corollary 1. Let A(T) be a family of functions on T. 1) If A(T) is normal, then Coz A(T) is a separable perfect 𝜎-foundation. 2) If A(T) is completely normal, then Coz A(T) is a 𝜎-algebra. Lemma 2. If R is a ring on T, then Coz(St(T, R)) = R. Proof. If R ∈ R, then 𝜒(R) ∈ St(T, R) ≡ A implies R = coz 𝜒(R) ∈ Coz A. Conversely, let R = coz f for some f ∈ A. Then, by Lemma 8 (2.2.4) the function f has a representation in disjoint form f = ∑(x i 𝜒(R i ) | i ∈ I). Therefore, coz f = ⋃ ⟮R i | i ∈ I ∧ x i = / 0⟯ ∈ R.

74 | 2.2 Families of real-valued functions on a set

Lemma 3. Let f ∈ F(T) and x < y in R. Then, f −1 []x, y[] = coz g for the function g ≡ ((f − x1) ∨ 0) ∧ ((y1 − f ) ∨ 0). Proof. First, note that g ⩾ 0. Let x < f (t) < y. Then, g(t) = (f (t) − x) ⊼ (y − f (t)) > 0 implies t ∈ coz g. Conversely let t ∈ coz g. Then, f (t) − x > 0 and y − f (t) > 0 imply x < f (t) < y. Let f a function on T and n ∈ N. The set cozn f ≡ {t ∈ T | |f (t)| > 1/n} will be called the cozero-set of the level n. Let A(T) be a functional family on T. Then, the ensemble CozN A(T) ≡ {X ⊂ T | ∃f ∈ A(T) ∃n ∈ N (X = cozn f )} will be called the ensemble of all cozero-sets of all levels of the family A(T). Lemma 4. Let f , g ∈ F(T) and k, m, n ∈ N. Then, 1) cozm f ∩ cozn g = cozk ((m/k)f ∧ (n/k)g) = cozm (f ∧ (n/m)g) = cozn ((m/n)f ∧ g). 2) cozm f ∪ cozn g = cozk ((m/k)f ∨ (n/k)g) = cozm (f ∨ (n/m)g) = cozn ((m/n) f ∨ g). The proof is straightforward and is omitted. Corollary 1. If the family A(T) is closed with respect to finite infimum [supremum] and multiplication by positive rational numbers, then the ensemble CozN A(T) is multiplicative [additive]. Lemma 5. Let f ∈ F(T) and x < x󸀠 ⩽ y󸀠 < y in R. Then, f −1 [[x󸀠 , y󸀠 ]] ⊂ {t ∈ T | g(t) ⩾ (x󸀠 − x) ⊼ (y − y󸀠 )} ⊂ cozn g ⊂ coz g = f −1 []x, y[] for the function g ≡ ((f − x1) ∨ 0) ∧ ((y1 − f ) ∨ 0) and for every natural number n such that (x󸀠 − x) ⊼ (y − y󸀠 ) > 1/n. Proof. Let t ∈ f −1 [[x󸀠 , y󸀠 ]]. Then, g(t) = (f (t) − x) ∧ (y − f (t)) ⩾ (x󸀠 − x) ⊼ (y − y󸀠 ). Applying now Lemma 3 we obtain the assertion.

2.2.6 The equivalence of functions with respect to ideal ensembles Let T be a set and I be an ideal ensemble on T (see 2.1.4). Define a binary relation 𝜃I on F(T) setting f 𝜃I g if I n ≡ {t ∈ T | |f (t) − g(t)| ⩾ 1/n} ∈ I for every n ∈ N. It is evident that f 𝜃I g iff I𝜀 ≡ {t ∈ T | |f (t) − g(t)| ⩾ 𝜀} ∈ I for every real number 𝜀 > 0. Lemma 1. The relation 𝜃I is an equivalence relation on F(T). Proof. It is obvious that 𝜃I is reflexive and symmetric. Let f 𝜃I g and g 𝜃I h, i. e. I n ≡ {t ∈ T | |f (t)− g(t)| ⩾ 1/n} ∈ I and J n ≡ {t ∈ T | |g(t)− h(t)| ⩾ 1/n} ∈ I. Put K n ≡ {t ∈ T |

2.2.6 The equivalence of functions with respect to ideal ensembles |

75

|f (t) − h(t)| ⩾ 1/n}. Suppose t ∈ ̸ I2n ∪ J2n . Then, |f (t) − h(t)| ⩽ |f (t) − g(t)| + |g(t) − h(t)| < 1/n. Therefore, K n ⊂ I2n ∪ J2n ∈ I. Consequently, 𝜃I is transitive. Further, along with f 𝜃I g we shall write f ∼ g mod I or simply f ∼ g. Lemma 2. Let f , f 󸀠 , g, g󸀠 ∈ F(T) and f ∼ f 󸀠 , g ∼ g󸀠 . Then, 1) rf ∼ rf 󸀠 for every r ∈ R; 2) f + g ∼ f 󸀠 + g󸀠 ; 3) f ∨ g ∼ f 󸀠 ∨ g󸀠 and f ∧ g ∼ f 󸀠 ∧ g. Besides, if f , f 󸀠 , g, g 󸀠 ∈ F b (T), then 4) fg ∼ f 󸀠 g󸀠 ; 5) 1/f ∼ 1/f 󸀠 for every f such that 0 ∈ ̸ rng f ∪ rng f 󸀠 and 1/f , 1/f 󸀠 ∈ F b (T). Proof. By the condition, the sets I𝜀 ≡ {t ∈ T | |f (t) − f 󸀠 (t)| ⩾ 𝜀} ∈ I and J𝜁 ≡ {t ∈ T | |g(t) − g 󸀠 (t)| ⩾ 𝜁} ∈ I. 1. If r = / 0, then K𝜂 ≡ {t ∈ T | |rf (t) − rf 󸀠 (t)| ⩾ 𝜂} = I𝜀 ∈ I for 𝜀 ≡ 𝜂/|r|. 2. Put K𝜂 ≡ {t ∈ T | |(f (t) + g(t)) − (f 󸀠 (t) + g󸀠 (t))| ⩾ 𝜂}. Suppose that t ∈ ̸ I𝜀 ∪ J𝜁 for 𝜀 = 𝜁 = 𝜂/2. Then, |(f (t)+g(t))−(f 󸀠 (t)+g󸀠 (t))| ⩽ |f (t)−f 󸀠 (t)|+|g(t)−g󸀠 (t)| < 𝜂. Therefore, K𝜂 ⊂ I𝜀 ∪ J𝜁 ∈ I. 3. Put K𝜂 ≡ {t ∈ T | |f (t) ∨ g(t) − f 󸀠 (t) ∨ g󸀠 (t)| ⩾ 𝜂}. Suppose that t ∈ ̸ I𝜀 ∪ J𝜁 for 𝜀 = 𝜁 = 𝜂/2. Then, |f (t)∨ g(t)− f 󸀠 (t)∨ g󸀠 (t)| ⩽ |f (t)∨ g(t)− f 󸀠 (t)∨ g(t)|+|f 󸀠 (t)∨ g(t)− f 󸀠 (t)∨ g󸀠 (t)| ⩽ |f (t) − f 󸀠 (t)| + |g(t) − g 󸀠 (t)| < 𝜂. Therefore, K𝜂 ⊂ I𝜀 ∪ I𝜁 ∈ I. 4. By the condition, |f 󸀠 (t)| ⩽ a󸀠 and |g(t)| ⩽ b for every t ∈ T. Put K𝜂 ≡ {t ∈ T | |fg(t) − f 󸀠 g󸀠 (t)| ⩾ 𝜂}. Suppose t ∈ ̸ I𝜀 ∪ J𝜁 for 𝜀 = 𝜂/2b and 𝜁 = 𝜂/2a󸀠 . Then, |fg(t) − f 󸀠 g󸀠 (t)| ⩽ |fg(t) − f 󸀠 g(t)| + |f 󸀠 g(t) − f 󸀠 g󸀠 (t)| ⩽ |g(t)| |f (t) − f 󸀠 (t)| + |f 󸀠 (t)| |g(t) − g 󸀠 (t)| < 𝜂. Therefore, K𝜂 ⊂ I𝜀 ∪ J𝜁 ∈ J. 5. By the condition, 1/|f (t)| ⩽ a and 1/|f 󸀠 (t)| ⩽ a󸀠 for every t ∈ T. Put K𝜂 ≡ {t ∈ T | |1/f (t) − 1/f 󸀠 (t)| ⩾ 𝜂}. Suppose that t ∈ I𝜀 for 𝜀 = 𝜂/(aa󸀠 ). Then, |1/f (t) − 1/f 󸀠 (t)| = |f 󸀠 (t) − f (t)|/(|f (t)| |f (󸀠 t)|) < aa󸀠 𝜀 = 𝜂. Therefore, K𝜂 ⊂ I𝜀 ∈ I. This equivalence relation is especially simple in the following case. Lemma 3. Let I be 𝜎-ideal on T and f , g ∈ F(T). Then, f ∼ g mod I iff f |U = g|U for some set U ∈ co-I, i. e. coz(f − g) ∈ I. Proof. It is evident that the second condition implies the first one. Conversely, if I n ≡ {t ∈ T | |f (t) − g(t)| ⩾ 1/n} ∈ I for every n ∈ N, then I ≡ {t ∈ T | f (t) = / g(t)} = ⋃ ⟮I n | n ∈ N⟯ ∈ I, where U ≡ T\I ∈ co-I. Let A(T) ⊂ F(T). Consider on A(T) the equivalence relation 𝜃I,A(T) ≡ 𝜃I ∩ (A(T) × A(T)). For every function f ∈ A(T), consider the class f ≡ f mod 𝜃I,A(T) ≡ {f 󸀠 ∈ A(T) | f 󸀠 𝜃I,A(T) f } of equivalence of the function f with respect to 𝜃I,A(T) . Consider also the factor-family A(T, I) ≡ A(T)/𝜃I,A(T) of the family A(T) with respect to 𝜃I,A(T) . Let

76 | 2.2 Families of real-valued functions on a set

p : A(T) A(T, I) be the corresponding factor-mapping (see 1.1.14). Along with the notation A(T)/𝜃I,A(T) , we shall use the simple notation A(T)/I. Lemma 4. The equivalence relation 𝜃I,A(T) is a congruence. Proof. Let f , f 󸀠 , f 󸀠󸀠 ∈ A(T) and f ⩽ f 󸀠 ⩽ f 󸀠󸀠 . Suppose that f 𝜃f 󸀠󸀠 . Then, J n ≡ {t ∈ T | |f (t) − f 󸀠󸀠 (t)| ⩾ 1/n} ∈ I for every n ∈ N. Consider the sets I n ≡ {t ∈ T | |f (t) − f 󸀠 (t)| ⩾ 1/n}. If t ∈ I n , then f (t) − f 󸀠 (t) ⩽ 0 implies f (t) − f 󸀠 (t) ⩽ −1/n. Hence, f (t) − f 󸀠󸀠 (t) ⩽ −1/n, i. e. t ∈ J n . This means that I n ⊂ J n . Therefore, I n ∈ I, i. e. f 𝜃f 󸀠 . If the family A(T) contains the zero function 0, then we can define the equivalence relation 𝜃I,A(T) by another way. Consider the neutral subfamily A0 (T, I) ≡ {f ∈ A(T) | f 𝜃I,A(T) 0} of A(T) with respect to the neutral element 0 and the equivalence relation 𝜃I,A(T) (see 1.1.14). Define on A(T) containing 0 the binary relation 𝜃A0 (T,I) setting f 𝜃A0 (T,I) g if f − g ∈ A0 (T, I) for f , g ∈ A(T). Consider the factor-family A(T)/𝜃 of the family A(T) with respect to 𝜃A0 (T,I) . Along with the notation A(T)/𝜃A0 (T,I) we shall use the simple notation A(T)/A0 (T, I). It is clear that if 0 ∈ A(T), then 𝜃I,A(T) = 𝜃A0 (T,I) . Hence, A(T, I) ≡ A(T)/I = A(T)/A0 (T, I). If A(T) is a linear space [linear algebra with the unit 1, lattice-ordered linear space, lattice-ordered linear algebra with the unit 1], then A0 (T, I) is a linear subspace [ideal, l-ideal, l-ideal] of A(T). Therefore, the factor-family A(T, I) ≡ A(T)/I = A(T)/A 0 (T, I) has the natural general structure of linear space [linear algebra with the unit 1, lattice-ordered linear space, lattice-ordered linear algebra with the unit 1] (see 2.2.4). In particular, the relation ⩽A ≡ {(f1̄ , f2̄ ) | ∀ f1󸀠 ∈ f1̄ ∃ f2󸀠 ∈ f2̄ (f1󸀠 ⩽ f2󸀠 )} is the order relation. By virtue of Lemma 4 and Statement 1. Taking the whole family F(T) and its subfamily F b (T) we obtain the lattice-ordered linear algebras with the unit F(T, I) ≡ F(T)/I = F(T)/F 0 (T) and F b (T, I) ≡ F b (T)/I = F b (T)/F 0b (T). Moreover, the multiplicative unit 1 is a weak order unit in F(T, I) and a strong order unit in F b (T, I).

Auxiliary definitions and statements 1∘ Let ⟮A, ⩽⟯ be an ordered set and 𝜀 be an equivalence relation on A. Consider the relation 𝜃 on A ≡ A/𝜀 setting 𝜃 ≡ {(ā 1 , ā 2 ) | ∀ a󸀠1 ∈ ā 1 ∃ a󸀠2 ∈ ā 2 (a󸀠1 ⩽ a󸀠2 )}. The equivalence relation 𝜀 is called a congruence if a ⩽ a 󸀠 ⩽ a󸀠󸀠 and a𝜀a󸀠󸀠 implies a𝜀a󸀠 . Statement 1. If 𝜀 is a congruence, then 𝜃 is an order. In this case, it will be denoted by ⩽A .

2.2.7 Seminorms and norms of the uniform convergence on families and factor-families | 77

2∘ Let U ≡ |⟮A, R⟯, s ls | be a (real) linear space. A real linear space V ≡ |⟮B, R⟯, t ls | is called a linear subspace of the linear space U if the following properties hold: 1) B ⊂ A; 2) 0B = 0A ; 3) −B = −A |B ∗ B; 4) +B = +A |(B ∗ B) ∗ B; 5) ⋅⟮B,R⟯ = ⋅⟮A,R⟯ |(R ∗ B) ∗ B. In a similar way, we can define the notion of a linear subalgebra of a (real) linear algebra. 3∘ Let U ≡ |⟮A, R⟯, s lols | be a (real) lattice-ordered linear space. A (real) lattice-ordered linear space V ≡ |⟮B, R⟯, t lols | is called an l-ideal of the lattice-ordered linear space U if the following properties hold: 1) |⟮B, R⟯, t ls | is a linear subspace of the linear space |⟮A, R⟯, t ls |; 2) |B, ⩽B | is an l-ideal of the lattice-ordered set ⟮A, ⩽A ⟯. Let U ≡ |⟮A, R⟯, s la | be a (real) linear algebra. A (real) linear algebra V ≡ |⟮B, R⟯, t la | is called an ideal of the linear algebra U if the following properties hold: 1) |⟮B, R⟯, t ls | is a linear subspace of the linear space |⟮A, R⟯, t ls |; 2) ⋅B = ⋅A |(B ∗ B) ∗ B (i. e. V is a linear subalgebra of U); 3) ∀ a ∈ A ∀ b ∈ B (a ⋅A b ∈ B). Let U ≡ |⟮A, R⟯, s lola | be a (real) lattice-ordered linear algebra. A (real) lattice-ordered linear algebra V ≡ |⟮B, R⟯, t lola | is called an l-ideal of the lattice-ordered linear algebra U if the following properties hold: 1) |⟮B, R⟯, t lols | is a l-ideal of the lattice-ordered linear space |⟮A, R⟯, t lols |; 2) |⟮B, R⟯, t la | is an ideal of the linear algebra |⟮A, R⟯, t la |. 2.2.7 Seminorms and norms of the uniform convergence on families and factor-families of functions Norm of uniform convergence On the lattice-ordered linear algebra F b (T) of all bounded real-valued functions on a fixed set T we can consider the functional r : F b (T) → R+ such that rf ≡ sup(|f (t)| | t ∈ T). Lemma 1. The functional r is a modulusly monotone unitary submultiplicative norm on the lattice-ordered linear algebra F b (T) with the unit 1.

78 | 2.2 Families of real-valued functions on a set

Proof. It is evident that rf = 0 implies f = 0. Besides, r1 = 1. Let f , g ∈ F b (T) and x ∈ R. Then, by Proposition 6 (1.4.3) and Lemma 4 (1.4.5) r(f + g) ⩽ sup(|f (t)| + |g(t)| | t ∈ T) ⩽ rf + rg, r(fg) ⩽ sup(|f (t)||g(t)| | t ∈ T) ⩽ r(f )r(g), and r(xf ) = sup(|x||f (t)| | t ∈ T) = |x|rf . Let |f | ⩽ |g|. Then, |f (t)| ⩽ rg for every t ∈ T implies rf ⩽ rg. We shall denote further the norm r on F b (T) by ‖ ⋅ ‖u . Corollary 1. ⟮F b (T), ‖⋅‖u ⟯ is a normed lattice-ordered linear algebra with the multiplicative and strong order unit 1 such that ‖ ⋅ ‖u = ‖ ⋅ ‖1 . Proof. According to the definition at the end of this subsection ‖f ‖1 ≡ inf{x ∈ R+ | |f | ⩽ x1}. From |f | ⩽ ‖f ‖u 1 we conclude that ‖f ‖1 ⩽ ‖f ‖u . Fix 𝜀 > 0. Then, there is a number x such that |f | ⩽ x1 and x < ‖f ‖1 + 𝜀. Therefore, ‖f ‖u ⩽ x < ‖f ‖1 + 𝜀. Since 𝜀 is arbitrary, we conclude that ‖f ‖u ⩽ ‖f ‖1 . Lemma 2. Let (f m ∈ F b (T) | m ∈ M ⊂ 𝜔) be a sequence for some cofinal subset M of 𝜔 and f ∈ F b (T). Then, the following assertions are equivalent: 1) f = u-lim (f m | m ∈ M) (see 2.2.4); 2) 0 = lim (‖f m − f ‖u | m ∈ M) in R; 3) f = lim (f m | m ∈ M) in the topological space ⟮F b (T), G(‖ ⋅ ‖u )⟯. Proof. The equivalence of (1) and (2) is evident. The equivalence of (2) and (3) follows from Corollary 1 to Statement 6. Corollary 1. ⟮F b (T), ‖ ⋅ ‖u ⟯ is a Banach lattice-ordered algebra with the unit 1. Proof. By Corollary 1 to Lemma 1 ⟮F b (T), ‖ ⋅ ‖u ⟯ is a normed lattice linear algebra with the unit 1. Let (f m | m ∈ 𝜔) be a fundamental sequence with respect to the normed metric 𝜌‖⋅‖u , i. e. for every 𝜀 > 0 there is n ∈ 𝜔 such that ‖f p − f q ‖u < 𝜀 for all p, q ⩾ n (see the definitions in 2∘ and 6∘ at the end of this subsection). It is clear that |f p − f q | < 𝜀1, i. e. this sequence is uniformly inner convergent. By Lemma 5 (2.2.3) there is a function f ∈ F b (T) such that f = u-lim (f m | m ∈ 𝜔). By Lemma 2 lim (𝜌‖⋅‖u (f m , f ) | m ∈ 𝜔) = 0. By Statement 2 we obtain f = lim (f m | m ∈ 𝜔) with respect to the metric topology G(𝜌‖⋅‖u ). Then, the metric space ⟮F b (T), 𝜌‖⋅‖u ⟯ is complete, and therefore, ⟮F b (T), ‖ ⋅ ‖u ⟯ is a Banach space. Corollary 2. ⟮F b (T), ‖ ⋅ ‖u ⟯ is an M-algebra. Proof. The statement follows from Corollary 1, Corollary 1 to Lemma 1, and Statement 14.

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The norm ‖ ⋅ ‖u is called the norm of uniform convergence on F b (T). For A(T) ⊂ F b (T), consider the functional ‖ ⋅ ‖u,A(T) ≡ ‖ ⋅ ‖u | A(T) on A(T). It follows from Lemma 1 and its corollary 1 that if A(T) is a linear space [latticeordered linear space, linear algebra, lattice-ordered linear algebra], then ⟮A(T), ‖ ⋅ ‖u,A(T) ⟯ is a normed linear space [lattice-ordered linear space, linear algebra, latticeordered linear algebra]. The norm ‖ ⋅ ‖u,A(T) is called the norm of uniform convergence on A(T). Lemma 3. Let A(T) be a linear subspace [linear subalgebra] of F b (T) closed with respect to the uniform convergence of sequences in F b (T) (see property g) in 2.2.4). Then, ⟮A(T), ‖ ⋅ ‖u,A(T) ⟯ is a Banach space [Banach algebra]. If, besides, A(T) is a lattice-ordered linear subspace [lattice-ordered linear subalgebra] of F b (T), containing the unit function 1, then ⟮A(T), ‖ ⋅ ‖u,A(T) ⟯ is a Banach lattice-ordered space [Banach lattice-ordered algebra with the unit 1] with the strong order unit 1 such that ‖ ⋅ ‖u,A(T) = ‖ ⋅ ‖1 . Proof. Let (f m ∈ A(T) | m ∈ M ⊂ 𝜔) be a fundamental sequence with respect to the normed metric 𝜌‖⋅‖u . By Corollary 1 to Lemma 2 there is f ∈ F b (T) such that lim (‖f m − f ‖u | m ∈ M) = 0. By Lemma 2 f = u-lim (f m | m ∈ M). Hence, by the condition, f ∈ A(T). If A(T) is a lattice-ordered linear subspace with 1, then by Corollary 1 to Lemma 1 ‖f ‖u,A(T) = ‖f ‖1 for every f ∈ A(T). Let I be an ideal ensemble on T (see 2.1.4). Consider the equivalence relation 𝜃I,A(T) from 2.2.6 and the factor-family A(T, I) ≡ A(T)/𝜃I,A(T) ≡ A(T)/I. Lemma 4. Let I be an ideal ensemble on T and A(T) ⊂ F b (T). Then, for every f ∈ A(T) its class of equivalence f is a closed set in the topological space ⟮A(T), G (‖ ⋅ ‖u,A(T) )⟯. Proof. Let g ∈ A(T), (g m ∈ f | m ∈ M ⊂ 𝜔) be a infinite sequence and g = lim (g m | m ∈ M) in the mentioned topological space. Fix 𝜀 > 0 and take 𝜀1 ≡ 𝜀/2. Then, by virtue of Corollary 1 to Statement 6 there is a number m ∈ M such that |g(t) − g m (t)| < 𝜀1 for every t ∈ T. By the definition, P ≡ {t ∈ T | |g(t)− g m (t)| < 𝜀1 } ∈ co-I. Therefore, for every t ∈ P, we have |g(t) − f (t)| ⩽ |g(t) − g m (t)| + |g m (t) − f (t)| < 2𝜀1 = 𝜀. This means that g ∈ f . By Statement 7 this means that f ̄ is closed. Corollary 1. Let I be an ideal ensemble on T and 0 ∈ A(T) ⊂ F b (T). Then, the subset A0 (T, I) ≡ {f ∈ A(T) | f ∼ 0 mod I} is closed in ⟮A(T), G(‖ ⋅ ‖u,A(T) )⟯. Proof. A0 (T, I) = 0 mod 𝜃I,A(T) .

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Define on A(T) ⊂ F(T) containing 0 the binary relation 𝜃A0 (T,I) setting f 𝜃A0 (T,I) g if f − g ∈ A0 (T, I) for f , g ∈ A(T). Consider the factor-family A(T)/𝜃 of the family A(T) with respect to 𝜃A0 (T,I) . Along with the notation A(T)/𝜃A0 (T,I) we shall use the simple notation A(T)/A0 (T, I). It is clear that 𝜃I,A(T) = 𝜃A0 (T,I) . Therefore, A(T, I) ≡ A(T)/I = A(T)/𝜃I,A(T) = A(T)/𝜃A0 (T,I) = A(T)/A0 (T, I). If A(T) is a linear space [lattice-ordered linear space, linear algebra, latticeordered linear algebra], then A0 (T, I) is a linear subspace [l-ideal, ideal, l-ideal] of A(T). Therefore, the factor-family A(T, I) has the natural general structure of linear space [lattice-ordered linear space, linear algebra, lattice-ordered linear algebra]. If ⟮A(T), ‖ ⋅ ‖u,A(T) ⟯ is a seminormed linear space [lattice-ordered linear space, linear algebra, lattice-ordered linear algebra], then we can consider the seminormed linear space [lattice-ordered linear space, linear algebra, lattice-ordered linear algebra] ⟮A(T, I), ‖ ⋅ ‖u,A(T,I) ⟯ (see Statement 8 below). Corollary 2. Let I be an ideal ensemble on T and A(T) be a linear subspace of F b (T). Then, ⟮A(T, I), ‖ ⋅ ‖u,A(T,I) ⟯ is a normed linear space. If A(T) is closed with respect to the uniform convergence of sequences in F b (T), then ⟮A(T, I), ‖⋅‖u,A(T,I) ⟯ is a Banach space. Proof. By Statement 9 ‖⋅‖u,A(T,I) is a norm on the linear space A(T, I). If A(T) is closed with respect to the uniform convergence, then By virtue of Corollary 1 to Lemma 4 and Statement 10 ⟮A(T, I), ‖ ⋅ ‖u,A(T,I) ⟯ is a Banach space. The norm ‖ ⋅ ‖u,A(T,I) is called the norm of the uniform convergence on A(T, I). Corollary 3. Let I be an ideal ensemble on T and A(T) be a linear subalgebra of F b (T) containing the unit function 1. Then, ⟮A(T, I), ‖ ⋅ ‖u,A(T,I) ⟯ is a normed linear algebra with the unit 1.̄ If, besides, A(T) is closed with respect to the uniform convergence of sequences in F b (T), then ⟮A(T, I), ‖ ⋅ ‖u,A(T,I) ⟯ is a Banach algebra with the unit 1.̄ Lemma 5. Let A(T) be a lattice-ordered linear subspace [lattice-ordered linear subalgebra] of F b (T) containing the unit function 1. Then, A(T, I) is a lattice-ordered linear space [lattice-ordered linear algebra] with the strong order unit 1̄ and ‖f ̄‖u,A(T,I) = ‖f ̄‖1̄ for every f ̄ ∈ A(T, I). Proof. The equality of these norms follows from Statement 15. The norm ‖ ⋅ ‖1̄ = ‖ ⋅ ‖u,A(T,I) is usually denoted by ‖ ⋅ ‖∞ . Corollary 1. Let A(T) be a lattice-ordered linear subspace [lattice-ordered linear subalgebra] of F b (T) containing the unit function 1. Then, (A(T, I), ‖ ⋅ ‖∞ ) is a normed latticē orderedlinearspace[lattice-orderedlinearalgebrawiththeunit 1]suchthat‖ 1‖̄ ∞ = 1and ‖f ̄ ∨ g‖̄ ∞ = ‖f ̄‖∞ ⊻‖g‖̄ ∞ whenever f ̄ , ḡ ∈ A(T, I)+ , i. e. it is a pre-M-space [pre-M-algebra].

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Proof. It follows from Lemma 5 and Statements 13, 14, and 15. Corollary 2. Let A(T) be a lattice-ordered linear subspace [lattice-ordered linear subalgebra] of F b (T) containing the unit function 1 and closed with respect to the uniform convergence of sequences in F b (T). Then, (A(T, I), ‖ ⋅ ‖∞ ) is a Banach lattice-ordered ̄ Moreover, it is an M-space space [Banach lattice-ordered algebra with the unit 1]. [M-algebra]. Proof. By Corollary 2 [Corollary 3] to Lemma 4 (A(T, I), ‖ ⋅ ‖∞ ) is a Banach space [algebra]. Using Corollary 1 we conclude that it a Banach lattice-ordered space [algebra] and it is an M-space [M-algebra].

Seminorm of essentially uniform convergence With respect to the equivalence relation 𝜃I on the family F(T) its subfamily F b (T) has the following shortcoming. If g ∈ F b (T) and f ∼ g mod I, then not necessary f ∈ F b (T). Therefore, we shall consider the family ̂ F b (T, I) ≡ {f ∈ F(T) | ∃ g ∈ F b (T) (f ∼ g mod I)}. The first task is to give some its internal description. A function f ∈ F(T) is called essentially bounded with respect to the ideal ensemble I if there is x ∈ R+ such that C x f ≡ {t ∈ T | |f (t)| > x} ∈ I, i. e. |f (t)| ⩽ x for every t ∈ U for some U ∈ co-I. The family of all essentially bounded functions is denoted by F∞ (T, I). It is evident that F b (T) ⊂ F∞ (T, I). If I = {⌀}, then F∞ (T, I) = F b (T). On the family F∞ (T, I) we can consider the functional r : F∞ (T, I) → R+ such that rf ≡ inf{x ∈ R+ | C x f ∈ I}. Lemma 6. Let I be an ideal ensemble on T. Then, 1) F∞ (T, I) is a lattice-ordered linear subalgebra of the lattice-ordered linear algebra F(T) and 1 ∈ F∞ (T, I); 2) F∞ (T, I) is an l-ideal of the lattice-ordered linear space F(T) and 1 is a week order unit in F∞ (T, I); 3) the functional r is a modulusly monotone unitary submultiplicative seminorm on F∞ (T, I). Proof. It is clear that r0 = 0 and r1 = 1. Let f , g ∈ F∞ (T, I) and z ∈ R\{0}. Then, for every numbers x, y ∈ R+ such that rf < x and rg < y, we have C x f ∈ I and C y g ∈ I. Therefore, {t ∈ T | |f (t) + g(t)| > x + y} ⊂ C x f ∪ C y g ∈ I implies f + g ∈ F∞ (T, I) and r(f + g) ⩽ x + y, where r(f + g) ⩽ rf + rg. Besides, {t ∈ T | |zf (t)| > |z|x} = C x f implies zf ∈ F∞ (T, I) and r(zf ) ⩽ |z|x, where r(zf ) ⩽ |z|rf . Consequently, r((1/z)g) ⩽ (1/|z|)rg. Taking in this inequality g = zf we get rf ⩽ (1/|z|)r(zf ), where |z|rf ⩽ r(zf ). As a result, r(zf ) = |z|rf . Further, {t ∈ T | |f (t)g(t)| > xy} ⊂ C x f ∪ C y g ∈ I implies fg ∈ F∞ (T, I) and r(fg) ⩽ xy, where r(fg) ⩽ rf rg. Finally, if f ∈ F(T), g ∈ F∞ (T, I), and |f | ⩽ |g|, then C y f ⊂ C y g

82 | 2.2 Families of real-valued functions on a set

implies f ∈ F∞ (T, I) and rf ⩽ y, where rf ⩽ rg. This means that F∞ (T, I) is an l-ideal in the lattice linear space F(T). Consequently, F∞ (T, I) is a lattice-ordered linear space itself and 1 is a week order unit in F∞ (T, I). We shall denote further the seminorm r on F∞ (T, I) by ‖ ⋅ ‖eu . It is called the seminorm of the essentially uniform convergence on F∞ (T, I). If I = {⌀}, then F∞ (T, I) = F b (T) and ‖ ⋅ ‖eu = ‖ ⋅ ‖u . Lemma 7. Let I be an ideal ensemble on T and f , g ∈ F∞ (T, I). Then, the seminorm ‖ ⋅ ‖eu on F∞ (T, I) has the following properties: 1) {t ∈ T | |f (t)| > ‖f ‖eu + 𝜀} ∈ I for every 𝜀 > 0; 2) if f ∼ g mod I, then ‖f ‖eu = ‖g‖eu ; 3) f ∼ 0 mod I iff ‖f ‖eu = 0. Moreover, if I is a 𝜎-ideal ensemble, then 4) {t ∈ T | |f (t)| > ‖f ‖eu } ∈ I; 5) coz f ∈ I iff ‖f ‖eu = 0. Proof. We shall write rf for this seminorm. 1) Supposing that the set in (1) does not belong to I we get rf ⩾ rf + 𝜀 but this is impossible. 2) By the condition, I n ≡ {t ∈ T | |f (t) − g(t)| ⩾ 1/n} ∈ I for every n ∈ N. Then, for every x ∈ R+ such that rf < x, we have C x f ∈ I. Since {t ∈ T | |g(t)| > x + 1/n} ⊂ C x f ∪ I n ∈ I we conclude that rg ⩽ x + 1/n for every n. Hence, rg ⩽ rf . Analogously, rg ⩽ rf . As a result, rg = rf . 3) If f ∼ 0, then by (2), we have rf = r0 = 0. If rf = 0, then {t ∈ T | |f (t)| > 1/n} ∈ I for every n. Hence, f ∼ 0. 4) The statement follows from (1) and the equality {t ∈ T | |f (t)| > rf } = ⋃⟮{t ∈ T | |f (t)| > rf + 1/n} | n ∈ N⟯. 5) The statement follows from (3) and Lemma 3 (2.2.6). If A(T) ⊂ F(T), then the subfamily A∞ (T, I) ≡ A(T) ∩ F∞ (T, I) will be called the essentially bounded part of A(T) with respect to I. 󵄨 Let A(T) ⊂ F∞ (T, I). Consider the functional ‖ ⋅ ‖eu,A(T) ≡ ‖ ⋅ ‖eu 󵄨󵄨󵄨A(T) on A(T). If A(T) is a linear subspace [lattice linear subspace, linear subalgebra, lattice linear subalgebra] of F∞ (T, I), then ⟮A(T), ‖ ⋅ ‖eu,A(T) ⟯ is a seminormed linear space [lattice linear space, linear algebra, lattice linear algebra]. The seminorm ‖ ⋅ ‖eu,A(T) is called the seminorm of the essentially uniform convergence on A(T). Lemma 8. Let I be an ideal ensemble on T and f ∈ F(T). Then, 1) f ∈ F∞ (T, I) iff there is g ∈ F b (T) such that f ∼ g mod I;

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2) if f ∈ F b (T), then ‖f ‖eu ⩽ ‖f ‖u ; 3) if f ∈ F∞ (T, I), then ‖f ‖eu = inf(‖g‖u | g ∈ F b (T) ∧ f ∼ g mod I). Moreover, if I is a 𝜎-ideal ensemble, then 4) for f ∈ F∞ (T, I) there is g ∈ F b (T) such that f ∼ g mod I and ‖f ‖eu = ‖g‖u . Proof. 1) Let f ∈ F∞ (T, I), i. e. C x f ∈ I for some x ∈ R+ . Consider the function g ≡ f 𝜒(T\C x f ) ∈ F b (T). Then, I n ≡ {t ∈ T | |f (t) − g(t)| ⩾ 1/n} ⊂ C x f for every n ∈ N implies f ∼ g mod I. Conversely, if f ∼ g mod I, then by Lemma 7 ‖f ‖eu = ‖g‖eu < ∞. Hence, f ∈ F∞ (T, I). 2) If f ∈ F b (T), then f (t) ⩽ ‖f ‖u for every t ∈ T implies ‖f ‖eu ⩽ ‖f ‖u . 3) For f ∈ F∞ (T, I), consider x ≡ ‖f ‖eu and x m = x + m1 for m ∈ N. By Lemma 7 the set E m ≡ {t ∈ T | |f (t)| > x m } ∈ I. Consider the functions g m ≡ f 𝜒(T\E m ) ∈ F b (T). Then, I mn ≡ {t ∈ T | |f (t) − g m (t)| ⩾ 1/n} ⊂ E m implies f ∼ g m mod I. By Lemma 7 x = ‖g m ‖eu ⩽ ‖g m ‖u ⩽ x m . By virtue of Lemma 1 (1.4.5), we get the necessary equality in (3). 4) For f ∈ F∞ (T, I), consider x ≡ ‖f ‖eu . By Lemma 7 C x f ∈ I. Consider the function g m ≡ f 𝜒(T\C x f ) ∈ F b (T). Then, I n ≡ {t ∈ T | |f (t) − g(t)| ⩾ 1/n} ⊂ C x f implies f ∼ g mod I. By Lemma 7 x = ‖g‖eu ⩽ ‖g‖u ⩽ x. This lemma shows that F∞ (T, I) = ̂ F b (T, I). Proposition 1. Let I be a 𝜎-ideal ensemble on T. Then, (F∞ (T, I), ‖ ⋅ ‖eu ) is a complete seminormed lattice-ordered linear algebra with the unit 1. Proof. Let (f n | n ∈ 𝜔) be a inner convergent sequence with respect to the seminormed pseudometric 𝜌‖⋅‖eu , i. e. for every 𝜀 > 0 there is n ∈ 𝜔 such that ‖f p − f q ‖eu < 𝜀 for every p, q ⩾ n. Then, for every k ∈ N there is n k such that ‖f p − f q ‖eu < 1/k for all p, q ⩾ n k . Therefore, there exist sets E pq ∈ I such that |f p (t) − f q (t)| ⩽ 1/k for all t ∈ ̸ E pq . Consider the set E ≡ ⋃ ⟮⋃ ⟮E pq | p, q ⩾ n k ⟯ | k ∈ N⟯ ∈ I and the functions g n ≡ f n 𝜒(T\E). Since |f p (t) − f q (t)| ⩽ 1/k for all t ∈ ̸ E and all p, q ⩾ n k , the sequence (g n | n ∈ 𝜔) is inner uniformly convergent in F(T) (see 2.2.3). Therefore, by Lemma 6 (2.2.3) there is the unique function f ∈ F(T) such that f = u-lim (g n | n ∈ 𝜔). Applying the triangle inequality to the seminorm u ≡ ‖ ⋅ ‖eu : F∞ (T, I) → R (see 4∘ , 6∘ ) we get for the numbers x n ≡ ‖f n ‖eu the inequality |x p −x q | ⩽ ‖f p −f q ‖eu < 1/k for all p, q ⩾ n k . Therefore, by Theorem 1 (1.4.4) there is the unique number x ⩾ 0 such that x = lim (x n | n ∈ 𝜔). By Lemma 7 F n ≡ {t ∈ T | |f n (t)| > x n } ∈ I. So F ≡ ⋃ ⟮F n | n ∈ 𝜔⟯ ∈ I and |f n (t)| ⩽ x n for every t ∈ ̸ F. By Corollary 3 to Proposition 1 (1.4.7) |f (t)| ⩽ x for every t ∈ ̸ E ∪ F ∈ I. This means that f ∈ F∞ (T, I). By the definition, of the uniform convergence from 2.2.3 for every 𝜀 > 0 there is m = m(𝜀) such that |g n − f |u < 𝜀 for all n ⩾ m. By virtue of Lemmas 7 and 8, f n ∼ g n mod I implies ‖f n − f ‖eu = ‖g n − f ‖eu = ‖g n − f ‖u < 𝜀 for all n ⩾ m. Thus, f = lim (f n | n ∈ 𝜔) with respect to 𝜌‖⋅‖eu .

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Consider now the equivalence relation 𝜃I,A(T) on A(T) ⊂ F∞ (T, I) and the factorfamily A(T, I) ≡ A(T)/𝜃I,A(T) = A(T)/I. Lemma 9. Let I be an ideal ensemble on T and A(T) ⊂ F∞ (T, I). Then, for every f ∈ A(T) its class of equivalence f ̄ in A(T, I) is a closed set in the topological space ⟮A(T), G(‖ ⋅ ‖eu,A(T) )⟯. Proof. Let g ∈ A(T), (g m ∈ f ̄ | m ∈ M ⊂ 𝜔) be a sequence for some cofinal subset M of 𝜔 and g = lim (g m | m ∈ M) in the mentioned topological space. Fix 𝜀 > 0 and take 𝜀1 ≡ 𝜀/2. Then, by virtue of Corollary 1 to Statement 6 there is a number m ∈ M such that ‖g − g m ‖eu < 𝜀1 . Therefore, there is a set E ∈ I such that |g(t) − g m (t)| < 𝜀1 for every t ∈ ̸ E. By the definition, P ≡ {t ∈ T | |g m (t) − f (t)| < 𝜀1 } ∈ co-I. Therefore, for every t ∈ T\E ∈ co-I, we have |g(t) − f (t)| ⩽ |g(t) − g m (t)| + |g m (t) − g(t)| < 2𝜀1 = 𝜀. This means that g ∈ f ̄. By Statement 7 this means that f ̄ is closed. Corollary 1. Let I be an ideal ensemble on T and 0 ∈ A(T) ⊂ F∞ (T, I). Then, the subset A0 (T, I) ≡ {f ∈ A(T) | f ∼ 0 mod I} is closed in ⟮A(T), G(‖ ⋅ ‖eu,A(T) )⟯. Proof. A0 (T, I) = 0 mod 𝜃I,A(T) . If ⟮A(T), ‖ ⋅ ‖eu,A(T) ⟯ is a seminormed linear space [lattice-ordered linear space, linear algebra, lattice-ordered linear algebra], then we can consider the seminormed linear space [lattice-ordered linear space, linear algebra, lattice-ordered linear algebra] ⟮A(T, I), ‖ ⋅ ‖eu,A(T,I) ⟯. By virtue of the definition of the factor-seminorm and the assertion 2 of Lemma 7, we have ‖f ̄‖eu,A(T,I) ≡ inf{‖f 󸀠 ‖eu | f 󸀠 ∈ f ̄} = ‖f 󸀠 ‖eu for every f 󸀠 ∈ f ̄. Corollary 2. Let I be an ideal ensemble on T and A(T) be a linear subspace of F∞ (T, I). Then, (A(T, I), ‖ ⋅ ‖eu,A(T,I) ) is a normed linear space. Proof. By Statement 9 ‖ ⋅ ‖u,A(T,I) is a norm on the linear space A(T, I). Corollary 3. Let I be an ideal ensemble on T and A(T) be a linear subalgebra of F∞ (T, I) containing the unit function 1. Then, (A(T, I), ‖ ⋅ ‖eu,A(T,I) ) is a normed linear algebra with the unit 1.̄ Proposition 2. Let I be an ideal ensemble on T and A(T) be a lattice-ordered linear subspace [lattice-ordered linear subalgebra] of F∞ (T, I) containing the unit function 1. Then, the normed lattice-ordered linear spaces [lattice-ordered linear algebras] ⟮A(T, I), ‖ ⋅ ‖eu,A(T,I) ⟯ and ⟮A b (T, I), ‖ ⋅ ‖eu,A b (T,I) ⟯ are naturally isomorphic with respect to the (bijective isometric) isomorphism u : A(T, I) → A b (T, I) defined as u f ̄ = f ̄ ∩ A b (T).

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Proof. We shall denote in the proof the norms ‖ ⋅ ‖eu,A(T,I) and ‖ ⋅ ‖eu,A b (T,I) by ‖ ⋅ ‖󸀠 and ‖ ⋅ ‖󸀠󸀠 , respectively. Also we shall denote the classes f ̄ mod 𝜃I,A(T) for f ∈ A(T) ḡ mod 𝜃I,A b (T) for g ∈ A b (T) by f ̄ and g,̄̄ respectively. Let f ̄ ∈ A(T, I). Since f ∈ F (T, I) there is x > 0 such that C f ∈ I. By the condi∞

x

tion, the function g ≡ (−x)1 ∨ f ∧ x1 belongs to A b (T). Then, I n ≡ {t ∈ T | |f (t) − g(t)| ⩾ 1/n} ⊂ C x f for every n ∈ N implies f ∼ g mod I. So g ∈ f ̄ ∩ A b (T). By Lemma 7 ‖f ‖eu = ‖g‖eu . Therefore, we can define correctly a mapping u : A(T, I)(T) → A b (T, I) setting u f ̄ = f ̄ ∩ A b (T). Then, ‖u f ̄‖󸀠󸀠 = inf{‖g‖u | g ∈ f ̄ ∩ A b (T)}. Consider the numbers y ≡ ‖f ‖eu and y m = y + m1 for m ∈ N. By Lemma 7 the set E m ≡ {t ∈ T | |f (t)| > y m } ∈ I. It was checked above that g m ≡ (−y m )1 ∨ f ∧ y m 1 ∈ f ̄ ∩ A b (T). By Lemmas 7 and 8 y = ‖g m ‖eu ⩽ ‖g m ‖u ⩽ y m . By virtue of Lemma 1 (1.4.5), we get ‖u f ̄‖󸀠󸀠 = y = ‖f ̄‖󸀠 , i. e. u is isometric. Let g1 ∈ f1̄ ∩ A b (T) and g2 ∈ f2̄ ∩ A b (T). Then, g1 + g2 ∈ (f1̄ + f2̄ ) ∩ A b (T). By the definition, u f1̄ = ḡ̄ 1 , u f2̄ = ḡ̄ 2 , and u(f1̄ + f2̄ ) = ḡ̄ 1 + ḡ̄ 2 = u f1̄ + u f2̄ . It is checked in the same way that u is the corresponding homomorphism. Since u is isometric, we deduce that u is injective. For any g ∈ A b (T), consider the classes ḡ ∈ A(T) and ḡ̄ ∈ A b (T). It is clear that u ḡ = ḡ ∩ A b (T) = g.̄̄ So u is bijective. According to Statement 5 below, u is an isomorphism. By virtue of this proposition, we may denote the coinciding norms ‖ ⋅ ‖eu,A(T,I) and ‖ ⋅ ‖eu,A b (T) by the same symbol ‖ ⋅ ‖∞ . Corollary 1. Let I be an ideal ensemble on T. Then, the normed lattice-ordered linear algebras (F∞ (T), ‖ ⋅ ‖∞ ) and (F b (T), ‖ ⋅ ‖∞ ) are naturally isomorphic with respect to the isomorphism u : F∞ (T) → F b (T) such that u f ̄ = f ̄ ∩ F b (T). Corollary 2. Let I be an ideal ensemble on T and A(T) be a lattice linear subspace [lattice linear subalgebra] of F∞ (T, I, I) containing the unit function 1. Then, (A∞ (T), ‖ ⋅ ‖∞ ) is a pre-M-space [pre-M-algebra]. Moreover, if A b (T) is closed with respect to the uniform convergence of sequences in F b (T), then (A(T, I), ‖ ⋅ ‖∞ ) is an M-space [M-algebra]. Proof. The first assertion follows from Proposition 2 and Corollary 1 to Lemma 5. If A(T) is closed with respect to the uniform convergence in F b (T), then the second assertion follows from Corollary 2 to Lemma 5. Auxiliary definitions and statements 1∘ Let A be a set. The system U ≡ |A, 𝜌⟮A,R⟯ | with the structure 𝜌⟮A,R⟯ : A×A → R+ is called a metric space if the following properties hold: 1) 𝜌(a󸀠 , a󸀠󸀠 ) = 0 iff a󸀠 = a󸀠󸀠 ;

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2) ∀ a󸀠 , a󸀠󸀠 ∈ A 𝜌(a󸀠 , a󸀠󸀠 ) = 𝜌(a󸀠󸀠 , a󸀠 ); 3) ∀ a󸀠 , a󸀠󸀠 , a󸀠󸀠󸀠 ∈ A 𝜌(a󸀠 , a󸀠󸀠 ) + 𝜌(a󸀠󸀠 , a󸀠󸀠󸀠 ) ⩾ 𝜌(a󸀠 , a󸀠󸀠󸀠 ) (the triangle inequality). The mapping 𝜌 is called a metric on A and the number 𝜌(a 󸀠 , a󸀠󸀠 ) is called a distance between the elements (points) a󸀠 and a󸀠󸀠 . If 𝜌 satisfies conditions 2, 3, and 1󸀠 ) ∀ a ∈ A 𝜌(a, a) = 0, then it is called a pseudometric on A and U is called a pseudometric space. Now, let |A, 𝜌| be a [pseudo]metric space. For every a ∈ A and r ∈ R+ , the set B(a, r) ≡ {b ∈ A | 𝜌(a, b) < r} is called an open ball in A. A set G ⊂ A is called open in A if there is a collection ⟮B i | i ∈ I⟯ of open balls such that G = ⋃ ⟮B i | i ∈ I⟯. The open topology (see 2.1.1) consisting of all open sets is called the [pseudo]metric topology on the set A and is denoted by G(𝜌). Statement 1. Let |A, 𝜌| be a pseudometric space. Then, the topological space ⟮A, G(𝜌)⟯ is perfect and normal. 2∘ Let ⟮A, G⟯ be a topological space (see 2.1.1). A net s ≡ (a n | n ∈ N) is called convergent to an element [a point] a ∈ A and the element a is called a limit of the net s if for every G ∈ G such that a ∈ G there is n ∈ N such that a p ∈ G for all p ∈ N such that p ⩾ n. Statement 2. Let |A, 𝜌| be a pseudometric space. Then, a sequence (a n ∈ A | n ∈ N ⊂ 𝜔) converges to an element a ∈ A with respect to the pseudometric topology G(𝜌) iff the corresponding sequence (𝜌(a n , a) | n ∈ N) converges to the zero element 0 in R. Let |A, 𝜌| be a pseudometric space. A net s ≡ (a n | n ∈ N) is called inner convergent (fundamental) if for every 𝜀 > 0 there is n ∈ N such that 𝜌(a p , a q ) < 𝜀 for all p, q ∈ N such that p, q ⩾ n. The pseudometric space |A, 𝜌| is called complete if each inner convergent sequence has a limit with respect to the pseudometric topology G(𝜌). 3∘ Let U ≡ |⟮A, R⟯, s ls | and V ≡ |⟮B, R⟯, t ls | be linear spaces with the principal carriers A and B and the structures s ls and t ls , respectively (see 2.2.4). Let u : A → B be a mapping. If ((u ∗m u) ∗m u)[+A ] ⊂ +B , i. e. u(a󸀠 + a󸀠󸀠 ) = ua󸀠 + ua󸀠󸀠 for every a󸀠 , a󸀠󸀠 ∈ A, then u is called additive. If ((idR ∗m u) ∗m u)[⋅⟮A,R⟯ ] ⊂ ⋅⟮B,R⟯ , i. e. u(xa) = xua for every x ∈ R and a ∈ A, then u is called homogeneous. An additive and homogeneous mapping is called linear. It is called also a linear operator.

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A mapping u : A → B will be called a homomorphism from the linear space U to the linear space V if the following properties hold: 1) u[{0A }] ⊂ {0B }, 2) (u ∗m u)[−A ] ⊂ −B , 3) ((u ∗m u) ∗m u)[+A ] ⊂ +B , 4) ((idR ∗m u) ∗m u)[⋅⟮A,R⟯ ] ⊂ ⋅⟮B,R⟯ , where, for example, the notation u ∗m u : A ∗ A → B ∗ B means that (u ∗m u)⟨a󸀠 , a󸀠󸀠 ⟩ = ⟨ua󸀠 , ua󸀠󸀠 ⟩ and the notation (u ∗m u) ∗m u : (A ∗ A) ∗ A → (B ∗ B) ∗ B means that ((u ∗m u) ∗m u)⟨⟨a󸀠 , a󸀠󸀠 ⟩, a󸀠󸀠󸀠 ⟩ = ⟨⟨ua󸀠 , ua󸀠󸀠 ⟩, ua󸀠󸀠󸀠 ⟩ (see 1.1.8). This means that u preserves all the indicated structures. Statement 3. Let U and V be linear spaces. Then, every linear mapping u : A → B is a homomorphism from U into V. 4∘ Let U ≡ |⟮A, R⟯, s ols | and V ≡ |⟮B, R⟯, t ols | be ordered linear spaces with the principal carriers A and B and the structures s ols and t ols , respectively (see 2.2.4). A mapping u : A → B is called positive if a ⩾ 0 implies ua ⩾ 0. A mapping u : A → B is called bipositive (≡ order regular) if u = u 1 − u2 for some positive mappings u1 : A → B and u2 : A → B. A mapping u : A → B is called subadditive if u(a 󸀠 + a󸀠󸀠 ) ⩽ ua󸀠 + ua󸀠󸀠 for every 󸀠 a , a󸀠󸀠 ∈ A (the triangle inequality). A mapping u : A → B is called positively homogeneous if u(xa) = xua for every x ∈ R+ and a ∈ A; u is called absolutely homogeneous if u(xa) = |x|ua for every x ∈ R and a ∈ A. A subadditive and positively homogeneous mapping is called sublinear or convex. Let U ≡ |⟮A, R⟯, s la | and V ≡ |⟮B, R⟯, t la | be real linear algebras. A mapping u : A → B is called multiplicative if u(a󸀠 a󸀠󸀠 ) = u(a󸀠 )u(a󸀠󸀠 ) for every a󸀠 , a󸀠󸀠 ∈ A. Let U ≡ |⟮A, R⟯, s la , 1A | and V ≡ |⟮B, R⟯, t la , 1B | be real linear algebras with the units 1A and 1B , respectively. A mapping u : A → B is called unitary if u1A = 1B . Let U ≡ |⟮A, R⟯, s la | be a real linear algebra and V ≡ |⟮B, R⟯, t lola | be a real latticeordered linear algebra. A mapping u : A → B is called submultiplicative if u(a 󸀠 a󸀠󸀠 ) ⩽ u(a󸀠 )u(a󸀠󸀠 ) for every a󸀠 , a󸀠󸀠 ∈ A. Let U ≡ |⟮A, R⟯, s lola | and V ≡ |⟮B, R⟯, t lola | be real lattice-ordered linear algebras. A mapping u : A → B is called modulusly monotone if ∀ a 󸀠 , a󸀠󸀠 ∈ A (|a󸀠 | ⩽ |a󸀠󸀠 | ⇒ ua󸀠 ⩽ ua󸀠󸀠 ). Statement 4. Let U ≡ |⟮A, R⟯, s lols | and V ≡ |⟮B, R⟯, t lols | be lattice-ordered linear spaces, V be Dedekind complete (see 1.1.15), and u : A → B be a linear mapping. Then, the following conclusions are equivalent: 1) u is bipositive; 2) u is (order) bounded in the sense of 1.1.15.

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5∘ Let U ≡ |⟮A, R⟯, s lols | and V ≡ |⟮B, R⟯, t lols | be lattice-ordered linear spaces with derivative binary operations ∨A : A ∗ A → A, ∧A : A ∗ A → A and ∨B : B ∗ B → B, ∧B : B ∗ B → B, respectively (see 2.2.4). A mapping u : A → B will be called an isomorphism from the lattice-ordered linear space U onto the lattice-ordered linear space V if the following properties hold: 1) u is a bijection, 2) u[{0A }] = {0B }, 3) (u ∗m u)[−A ] = −B , 4) ((u ∗m u) ∗m u)[+A ] = +B , 5) (u ∗m u)[⩽A ] = ⩽B , 6) ((u ∗m u) ∗m u)[∨A ] = ∨B , 7) ((u ∗m u) ∗m u)[∧A ] = ∧B , 8) ((idR ∗m u) ∗m u)[⋅⟮A,R⟯ ] = ⋅⟮B,R⟯ . This means that the bijection u quite preserves all the indicated structures. If we omit property 1 and replace the equality symbol “=” by the inclusion symbol “⊂” in properties 2 – 8, then we get the definition of the homomorphism (≡ operator) u from the system U into the system V. In other words, this means that the mapping u preserves all the indicated structures. If U ≡ |⟮A, R⟯, s lola | and V ≡ |⟮B, R⟯, t lola | are lattice-ordered linear algebras, then a mapping u : A → B will be called an isomorphism from the lattice-ordered linear algebra U onto the lattice-ordered linear algebra V if in addition to properties 1 – 8 the following property holds: 9) ((u ∗m u) ∗m u)[⋅A ] = ⋅B . Statement 5. A bijective homomorphism of lattice-ordered linear spaces [algebras] is an isomorphism of these spaces [algebras]. 6∘ Let U ≡ |⟮A, R⟯, s ls | be a linear space and V ≡ ⟮R; w r , ⩽R ⟯ be the system of real numbers with the ring structure w r and the order ⩽R (see 1.4.3 and 2.2.4). A mapping u : A → R is called also a functional from U into V or simply a functional on U. The set of all linear functionals 𝜑 : A → R will be denoted by U × or simply A× . A sublinear absolutely homogeneous functional p on U is called a seminorm on U or simply a seminorm on A. For a seminorm p, we have pa = 2p(a/2) = p(a/2) + p(a/2) = p(a/2) + p(−a/2) ⩾ p(a/2 − a/2) = 0 for every a ∈ A. Therefore, a seminorm is also denoted by ‖ ⋅ ‖ : A → R+ . A seminorm p is called a norm if pa > 0 for every non-zero a ∈ A. Every (semi)norm ‖ ⋅ ‖ generates the (semi)normed pseudometric 𝜌‖⋅‖ in the following way 𝜌‖⋅‖ (a󸀠 , a󸀠󸀠 ) ≡ ‖a󸀠 − a󸀠󸀠 ‖. If ‖ ⋅ ‖ is a norm, then 𝜌‖⋅‖ is a metric.

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Let |⟮A, R⟯, s ls | be a linear space. The system U ≡ |⟮A, R⟯, s ls , ‖ ⋅ ‖⟮A,R⟯ | with the additional structure ‖⋅‖⟮A,R⟯ : A × A → R+ is called a seminormed [normed] linear space if ‖ ⋅ ‖ is a seminorm [norm]. A seminormed [normed] linear space |⟮A, R⟯, s ls , ‖ ⋅ ‖| is called complete if the pseudometric [metric] space |A, 𝜌‖⋅‖ | is complete. A complete normed linear space is called also a Banach space. Let U ≡ |⟮A, R⟯, s ls , ‖ ⋅ ‖A | and V ≡ |⟮B, R⟯, t ls , ‖ ⋅ ‖B | be seminormed linear spaces. A mapping u : A → B will be called an (isometric) isomorphism from the seminormed linear space U onto the seminormed linear space V if in addition to properties 1 – 4 and 8 from 5∘ the following property holds: 10) (u ∗m idR )[‖ ⋅ ‖A ] = ‖ ⋅ ‖B , i. e. ‖ua‖B = ‖a‖A for every a ∈ A. In the similar manner, an (isometric) isomorphism from the seminormed latticeordered linear space [algebra] U onto the seminormed lattice-ordered linear space [algebra] V can be defined. 7∘ Let |⟮A, R⟯, s ls , ‖ ⋅ ‖| be a seminormed linear space. The set D(‖ ⋅ ‖) ≡ {a ∈ A | ‖a‖ < 1} is called the open unit ball. Consider the ensemble U󸀠0 ≡ {U ⊂ A | ∃ x ∈ R+ U = {a ∈ A | ∃ b ∈ D(‖ ⋅ ‖) (a = xb)}} and the ensembles U󸀠a ≡ {V ⊂ A | ∃ U ∈ U󸀠0 (V = {b ∈ A | ∃ u ∈ U (b = a + u)})} for every a ∈ A. The set G ⊂ A is called open in A if for every point a ∈ G there exists a set U ∈ U󸀠a such that U ⊂ G. The open topology (see 2.1.1) consisting of all open sets is called the seminormed topology on the set A and is denoted by G(‖ ⋅ ‖). Statement 6. Let |⟮A, R⟯, s ls , ‖⋅‖| be a seminormed linear space. Then, G(‖⋅‖) = G(𝜌‖⋅‖ ). Corollary 1. A sequence (a n ∈ A | n ∈ N ⊂ 𝜔) converges to an element a ∈ A with respect to the seminormed topology G(‖ ⋅ ‖) iff the corresponding sequence (‖a n − a‖ | n ∈ N) converges to the zero element 0 in R. Statement 7. Let |⟮A, R⟯, s ls , ‖ ⋅ ‖| be a seminormed linear space and B ⊂ A. Then, the following conclusions are equivalent: 1) B is closed in the topological space ⟮A, G(‖ ⋅ ‖)⟯, i. e. B ∈ co-G(‖ ⋅ ‖); 2) for every sequence (b n ∈ B | n ∈ N ⊂ 𝜔) and for every element a ∈ A such that a = lim (b n | n ∈ N), we have a ∈ B. Let |⟮A, R⟯, s ls , ‖ ⋅ ‖| be a seminormed linear space, I be a linear subspace of A, and A ≡ A/I be a linear factor-space of A. Define the functional r : A → R setting ̄ r ā ≡ inf{‖a 󸀠 ‖A | a󸀠 ∈ a}.

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Statement 8. The functional r is a seminorm on A. This seminorm will be denoted by ‖ ⋅ ‖A . Statement 9. The following conclusions are equivalent: 1) ‖ ⋅ ‖A is a norm; 2) the set I is closed in the topological space ⟮A, G(‖ ⋅ ‖A )⟯. Statement 10. Let |⟮A, R⟯, s ls , ‖ ⋅ ‖| be a complete seminormed linear space and the set I be closed in the topological space ⟮A, G(‖ ⋅ ‖A )⟯. Then, the seminormed linear factor-space ⟮A, ‖ ⋅ ‖A ⟯ is a Banach space. Statement 11. Let |⟮A, R⟯, s ls , ‖ ⋅ ‖| be a seminormed linear space. Then, the seminorm ‖ ⋅ ‖A is modulusly monotone. Let |⟮A, R⟯, s ls , ‖ ⋅ ‖A | and |⟮B, R⟯, t ls , ‖ ⋅ ‖B | be seminormed (real) linear spaces. A mapping u : A → B is called (seminorm) bounded if there is some real number r > 0 such that ‖ua‖B ⩽ r‖a‖A for every a ∈ A. Statement 12. Let ⟮A, ‖ ⋅ ‖A ⟯ and ⟮B, ‖ ⋅ ‖B ⟯ be seminormed linear spaces and u : A → B be a linear operator. Then, the following conclusions are equivalent: 1) u : ⟮A, G(‖ ⋅ ‖A )⟯ → ⟮B, G(‖ ⋅ ‖B )⟯ is continuous; 2) u : ⟮A, ‖ ⋅ ‖A ⟯ → ⟮B, ‖ ⋅ ‖B ⟯ is (seminorm) bounded. Let U ≡ |⟮A, R⟯, s ls , ‖ ⋅ ‖A | be a (semi)normed linear space and V ≡ ⟮R; w r , ⩽R ⟯ be the system of real numbers (see 6∘ ). The set of all (seminorm) bounded linear functionals 𝜑 : A → R will be denoted by U 󸀠 , ⟮A, ‖ ⋅ ‖A ⟯󸀠 , or simply by A󸀠 . It is called the (seminorm) dual (≡ conjugate, adjoint) to the system U. Define on the set A󸀠 the structures of a normed linear space setting (r𝜑)a ≡ r(𝜑a), (𝜑 + 𝜓)a ≡ 𝜑a + 𝜓a for every a ∈ A and r ∈ R, and ‖𝜑‖󸀠 ≡ sup{|𝜑a| | ‖a‖A ⩽ 1} = sup{|𝜑a| | ‖a‖A < 1} = sup{|𝜑a| | ‖a‖A = 1}. 8∘ A system U ≡ |⟮A, R⟯, s lols , ‖ ⋅ ‖| is called a seminormed lattice-ordered [real] linear space if: 1) |⟮A, R⟯, s lols | is a seminormed (real) linear space; 2) |⟮A, R⟯, s ls , ‖ ⋅ ‖| is a seminormed (real) linear space; 3) ∀ a󸀠 , a󸀠󸀠 ∈ A (|a󸀠 | ⩽ |a󸀠󸀠 | ⇒ ‖a󸀠 ‖ ⩽ ‖a󸀠󸀠 ‖) (i. e. the seminorm ‖ ⋅ ‖ is modulusly monotone in the sense of 4∘ ). If ‖ ⋅ ‖ is a norm, then U is called a normed lattice-ordered linear space.

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The system U is called a Banach lattice-ordered space if U is a normed latticeordered linear space and |⟮A, R⟯, s ls , ‖ ⋅ ‖| is a Banach space. A system U ≡ |⟮A, R⟯, s la , ‖ ⋅ ‖| is called a seminormed [real] linear algebra if: 1) |⟮A, R⟯, s la | is a (real) linear algebra; 2) |⟮A, R⟯, s ls , ‖ ⋅ ‖| is a seminormed (real) linear space; 3) ∀ a󸀠 , a󸀠󸀠 ∈ A (‖a 󸀠 a󸀠󸀠 ‖ ⩽ ‖a󸀠 ‖‖a󸀠󸀠 ‖) (i. e. the seminorm ‖ ⋅ ‖ is submultiplicative in the sense of 4∘ ). If ‖ ⋅ ‖ is a norm, then U is called a normed linear algebra. A system U ≡ |⟮A, R⟯, s la , 1A , ‖ ⋅ ‖| is called a seminormed [real] linear algebra with the unit 1A if: 1) |⟮A, R⟯, s la , 1A | is a (real) linear algebra with the unit 1A ; 2) |⟮A, R⟯, s la , ‖ ⋅ ‖| is a seminormed (real) linear algebra; 3) ‖1A ‖ = 1, i. e. the seminorm ‖ ⋅ ‖ is unitary in the sense of 4∘ . If ‖ ⋅ ‖ is a norm, then U is called a normed linear algebra with the unit 1A . The system U is called a Banach algebra if U is a normed linear algebra with the unit 1A and |⟮A, R⟯, s ls , ‖ ⋅ ‖| is a Banach space. A system U ≡ |⟮A, R⟯, s lola , ‖ ⋅ ‖| is called a seminormed [real] lattice-ordered linear algebra if: 1) |⟮A, R⟯, s lola | is a (real) lattice-ordered linear algebra; 2) |⟮A, R⟯, s la , ‖ ⋅ ‖| is a seminormed (real) linear algebra; 3) |⟮A, R⟯, s lols , ‖ ⋅ ‖| is a seminormed (real) lattice-ordered linear space. If ‖ ⋅ ‖ is a norm, then U is called a normed lattice-ordered linear algebra. A system U ≡ |⟮A, R⟯, s lola , 1A , ‖ ⋅ ‖| is called a Banach lattice-ordered algebra if: 1) |⟮A, R⟯, s lola , ‖ ⋅ ‖| is a normed lattice-ordered linear algebra; 2) |⟮A, R⟯, s la , 1A , ‖ ⋅ ‖| is a Banach algebra. Let U ≡ |⟮A, R⟯, s lols , ‖ ⋅ ‖| be a seminormed lattice-ordered (real) linear space. It will be called a pre-M-space if ‖a󸀠 ∨ a󸀠󸀠 ‖ = ‖a󸀠 ‖ ⊻ ‖a󸀠󸀠 ‖ for all a󸀠 , a󸀠󸀠 ∈ A+ . It will be called a pre-L-space if ‖a󸀠 + a󸀠󸀠 ‖ = ‖a󸀠 ‖ + ‖a󸀠󸀠 ‖ for all a󸀠 , a󸀠󸀠 ∈ A+ . If, moreover, U is a Banach lattice-ordered space, then it is called an M-space or an L-space, respectively. A seminormed lattice-ordered linear algebra U with the unit is called a pre-Malgebra if ‖a󸀠 ∨ a󸀠󸀠 ‖ = ‖a󸀠 ‖ ⊻ ‖a󸀠󸀠 ‖ for all a󸀠 , a󸀠󸀠 ∈ A+ . If, moreover, U is a Banach lattice-ordered algebra, then it is called an M-algebra. 9∘ An ordered linear space A is called Archimedean if na ⩽ b for all n ∈ Z implies a = 0. If A is a lattice-ordered linear space, then A is Archimedean iff a ⩾ 0 and na ⩽ b for all n ∈ N implies a = 0.

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Let U ≡ |⟮A, R⟯, s lols , e| be a lattice-ordered linear space with the strong order unit e (see 2.2.4). Consider the functional ‖ ⋅ ‖e : A → R+ setting ‖a‖e ≡ inf{x ∈ R+ | |a| ⩽ xe}. Statement 13. Let |⟮A, R⟯, s lols , e| be a lattice-ordered linear space with the strong order unit e. Then, the functional ‖ ⋅ ‖e is a seminorm such that |a| ⩽ ‖a‖e e for every a ∈ A. If A is Archimedean, then ‖ ⋅ ‖e is a norm. Statement 14. The system |⟮A, R⟯, s lols , e, ‖ ⋅ ‖e | is a seminormed lattice-ordered linear space such that ‖e‖e = 1 and ‖a󸀠 ∨ a󸀠󸀠 ‖e = ‖a󸀠 ‖e ⊻ ‖a󸀠󸀠 ‖e for all a󸀠 , a󸀠󸀠 ∈ A, i. e. it is a pre-M-space. Statement 15. Let |⟮A, R⟯, s lols , e| be a lattice-ordered linear space with the strong order unit e and I be an l-ideal of A. Then, the seminorms ‖ ⋅ ‖e,A and ‖ ⋅ ‖ē on A are equal. If the set I is closed in the topological space ⟮A, G(‖ ⋅ ‖e )⟯, then these seminorms are norms and the lattice-ordered linear space A is Archimedean.

2.2.8 Pointwise continuous linear functionals on lattice-ordered linear spaces of functions Let T be a fixed set and A(T) be a lattice-ordered linear space of functions on T (see 2.2.4). We shall consider A(T) with the pointwise convergence of nets (see 2.2.3). According to Lemma 2 (2.2.3), this convergence in F(T) coincides with the order convergence. But for A(T) it is not valid in general. A mapping 𝜑 : A(T) → R will be called pointwise continuous [pointwise 𝜎-continuous] if f = p-lim(f m | m ∈ M) implies 𝜑f = lim(𝜑f m | m ∈ M) for every increasing net (f m ∈ A(T) | m ∈ M) [respectively, sequence (f m ∈ A(T) | m ∈ M ⊂ 𝜔)] and every function f ∈ A(T). A linear functional 𝜑 : A(T) → R will be called pointwise bicontinuous [pointwise 𝜎-bicontinuous] if 𝜑 = 𝜑1 − 𝜑2 for some pointwise continuous [𝜎-continuous] positive linear functionals 𝜑1 and 𝜑2 . Proposition 1 below shows that every continuous [𝜎-continuous] functional is bicontinuous [𝜎-bicontinuous] and vice versa. Lemma 1. Let A(T) be a lattice-ordered linear space of functions on a set T, 𝜑 be a positive linear functional on A(T). Then, the following conditions are equivalent: 1) 𝜑 is pointwise continuous [𝜎-continuous]; 2) for every net (f m ∈ A(T)+ | m ∈ M) [sequence (f m ∈ A(T)+ | m ∈ M ⊂ 𝜔)] the condition (f m | m ∈ M) ↓ 0 in F(T) implies (𝜑f m | m ∈ M) ↓ 0; 3) for every net (f m ∈ A(T)+ | m ∈ M) [sequence (f m ∈ A(T)+ | m ∈ M ⊂ 𝜔)] and function f ∈ A(T) the condition (f m | m ∈ M) ↑ f in F(T) implies (𝜑f m | m ∈ M) ↑ 𝜑f .

2.2.8 Pointwise continuous linear functionals on spaces of functions

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Proof. (1) ⊢ (2). Suppose (f m | m ∈ M) ↓ 0. Then, (−f m | m ∈ M) ↑ 0. Hence, lim (𝜑f m | m ∈ M) = − lim (𝜑(−f m ) | m ∈ M) = 0. (2) ⊢ (1). Let (f m ∈ A(T) | m ∈ M) be an increasing sequence, f ∈ A(T) and f = p-lim (f m | m ∈ M). If (f m | m ∈ M) ↑ f , then (f − f m | m ∈ M) ↓ 0 implies (𝜑f − 𝜑f m | m ∈ M) ↓ 0, i. e. (𝜑f m | m ∈ M) ↑ 𝜑f . (2) ⊢ (3). Suppose (f m ∈ A(T) | m ∈ M) ↑ f ∈ A(T). Then, (f − f m ) ↓ 0 in F(T) implies (𝜑f − 𝜑f m ) ↓ 0, where 𝜑f m ↑ 𝜑f . (3) ⊢ (2). Let (f m ∈ A(T)+ | m ∈ M) ↓ 0 in F(T). Fix some n ∈ M, then (f n − f m | m ∈ M) ↑ f n . Consider the set L ≡ {l ∈ M | l ⩾ n}. For every m ∈ M, there is l ∈ L such that l ⩾ m. Therefore, f l ⩽ f m implies f n ⩾ f n −f l ⩾ f n −f m ; hence ((f n −f l ) ∈ A(T)+ | l ∈ L) ↑ f n in F(T). By the condition, we have (𝜑f n −𝜑f l | l ∈ L) ↑ 𝜑f n . This implies (𝜑f l | l ∈ L) ↓ 0, and, finally, (𝜑f m | m ∈ M) ↓ 0. Lemma 2. Let A(T) be a lattice-ordered linear space of functions on a set T and 𝜑 be a pointwise 𝜎-continuous linear functional on A(T). Then 𝜑 is bounded. Proof. Suppose 𝜑 is not (order) bounded. Then, there are a function g ∈ A(T)+ and a sequence (f n ∈ A(T) | n ∈ N) such that |f n | ⩽ g and 𝜑f n > n for every n ∈ N. Fix any point t ∈ T. Suppose that x ≡ inf (g(t)/n | n ∈ N) > 0. Then, nx ⩽ g(t) for every n ∈ N. This contradicts the Archimedean principle (Lemma 13 (1.4.3)). So x = 0. By Lemma 7 (1.4.7) 0 = lim (g(t)/n | n ∈ N). Then, by Lemma 6 (1.4.7) 0 = lim (|f n (t)|/n | n ∈ N). This means that 0 = lim (f n (t)/n | n ∈ N). Thus, p-lim ((1/n)f n | n ∈ N) = 0. Hence, by the condition, lim ((1/n)𝜑f n | n ∈ N) = 0. But (1/n)𝜑f n > 1 for every n. It follows from this contradiction that 𝜑 is (order) bounded. Proposition 1. Let A(T) be a lattice-ordered linear space of functions on a set T and 𝜑 be a linear functional on A(T). Then, the following conclusions are equivalent: 1) 𝜑 is pointwise 𝜎-bicontinuous [bicontinuous]; 2) 𝜑 is pointwise 𝜎-continuous [continuous]. Proof. (1) ⊢ (2). By virtue of Proposition 1 (1.4.7), 𝜑 is pointwise continuous [𝜎-continuous]. (2) ⊢ (1). By Lemma 2 the functional 𝜑 is (order) bounded. By Statement 4 (2.2.7), 𝜑 ∈ A(T)∼ . By virtue of Corollary 3 to Statement 5 , 𝜑 = 𝜑+ +𝜑− . Now, we shall prove that 𝜑+ and −𝜑− are pointwise 𝜎-continuous. Let (f m ∈ A(T)+ | m ∈ M ⊂ 𝜔) ↓ 0 in F(T) and 𝜀 > 0. By assertion 2 of Corollary 3 to Statement 5 for every m ∈ M, the set G m ≡ {g ∈ A(T)+ | g m ⩽ f m ∧ 𝜑+ f m − 𝜀/2 ⩽ 𝜑g m ⩽ 𝜑+ f m } is not empty. By the axiom of choice, there is a mapping c : P(A(T)+ )\{⌀} → A(T)+ such that cP ∈ P. Define the sequence (g m ∈ A(T)+ | m ∈ M) setting g m ≡ c(G m ). Then, by Lemmas 6 and 7 (1.4.7) 0 = p-lim (g m | m ∈ M), and so

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0 = lim (𝜑g m | m ∈ M). By the definition, there is k ∈ M such that |𝜑g m | < 𝜀/2 for every m ⩾ k. Consequently, 0 ⩽ 𝜑+ f m ⩽ |𝜑g m | + 𝜀/2 < 𝜀 for every m ⩾ k. This means that (𝜑+ f m | m ∈ M) ↓ 0. By Lemma 1, 𝜑+ is pointwise 𝜎-continuous. The functional −𝜑 is also pointwise 𝜎-continuous. Therefore, −𝜑− = (−𝜑)+ is pointwise 𝜎-continuous. Thus, 𝜑 = 𝜑+ − (−𝜑− ) is pointwise 𝜎-continuous. In the second case, the proof is completely analogous. The set of all pointwise (bi)continuous [𝜎-(bi)continuous] linear functionals will be denoted by A(T)∨ [A(T)∧ ]. Its subset of all positive pointwise continuous [𝜎-continuous] functionals will be denoted by (A(T)∨ )0 [(A(T)∧ )0 ]. It is clear that A(T)∨ ⊂ A(T)∧ ⊂ A(T)∼ . Theorem 1. Let A(T) be a lattice-ordered linear space of functions on a set T. Then, A(T)∧ and A(T)∨ are completely (order) closed l-ideals in the Dedekind complete latticeordered linear space A(T)∼ . Besides, (A(T)∧ )+ = (A(T)∧ )0 and (A(T)∨ )+ = (A(T)∨ )0 . Proof. Denote the set (A(T)∧ )0 of all positive pointwise 𝜎-continuous linear functionals on A(T) by P. Let 𝜑, 𝜓 ∈ P and x, y ∈ R+ . It is clear that x𝜑+b𝜓 is linear and positive. By virtue of Proposition 1 (1.4.7), x𝜑 + b𝜓 ∈ P. Consequently, A(T)∧ = P − P is a linear subspace of the linear space A(T)∼ . Let 𝜑 ∈ A(T)∼ , 𝜓 ∈ P, and 0 ⩽ 𝜑 ⩽ 𝜓. If f = p-lim (f m | m ∈ M) in A(T), then 0 = p-lim (|f m − f | | m ∈ M). By virtue of Corollary 3 to Statement 5, 0 ⩽ |𝜑f m −𝜑f | ⩽ 𝜑(|f m − f |) ⩽ 𝜓(|f m − f |) → 0, where 𝜑f = lim (𝜑f m | m ∈ M). Thus, 𝜑 ∈ P. Let 𝜑 ∈ (A(T)∧ )+ . Then, by the definition 𝜑 = 𝜑1 − 𝜑2 for some 𝜑1 , 𝜑2 ∈ P. From 0 ⩽ 𝜑 ⩽ 𝜑1 by the property proven above we infer that 𝜑 ∈ P. This means that (A(T)∧ )+ = P. Now, let 𝜑 ∈ A(T)∼ , 𝜓 ∈ A(T)∧ , and |𝜑| ⩽ |𝜓|. By the definition, 𝜓 = 𝜓1 − 𝜓2 for some 𝜓1 , 𝜓2 ∈ P. Therefore, by the property proven above the conditions 0 ⩽ 𝜑+ ⩽ |𝜑| ⩽ 𝜓1 + 𝜓2 ∈ P imply 𝜑+ ∈ P. Analogously, −𝜑− ∈ P. Hence, 𝜑 = 𝜑+ − (−𝜑− ) ∈ A(T)∧ . Then, Statement 2 (see below) implies that A(T)∧ is an l-ideal in A(T)∼ . In particular, A(T)∧ is a lattice-ordered linear subspace of A(T)∼ . Finally, suppose that Φ ⊂ A(T)∧ , Φ ≠ ⌀, 𝜃 ∈ A(T)∼ , and 𝜃 = sup Φ. Fix 𝜑0 ∈ Φ and consider the set Ψ ≡ {sup(X ∪ {𝜑0 }) | X ⊂ Φ ∧ P(X) ∈ 𝜔}. By proven above Ψ ⊂ A(T)∧ , Ψ is bounded below by 𝜑0 , is bounded above by 𝜃 = sup Φ, and Ψ is upward directed. We know that 𝜃−𝜑0 is positive and 𝜓−𝜑0 ∈ A(T)∧ for every 𝜓 ∈ Ψ. Let f ∈ A(T)+ and (f m ∈ A(T)+ | m ∈ M ⊂ 𝜔) ↑ f in F(T). Then, (𝜓 − 𝜑0 )f = sup ((𝜓 − 𝜑0 )f m | m ∈ M). By Corollary 1 to Statement 5 𝜃f = sup (𝜓f | 𝜓 ∈ Ψ) and 𝜃f m = sup (𝜓f m | 𝜓 ∈ Ψ) for every m ∈ M. Therefore, by Theorem 1 (2.2.2) and Corollary 1 to Proposition 2 (1.1.15) (𝜃 − 𝜑0 )f = sup ((𝜓 − 𝜑0 )f | 𝜓 ∈ Ψ) = = sup (sup ((𝜓 − 𝜑0 )f m | m ∈ M) | 𝜓 ∈ Ψ) = sup ((𝜓 − 𝜑0 )f m | (m, 𝜓) ∈ M × Ψ) .

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Similarly, sup ((𝜃 − 𝜑0 )f m | m ∈ M) = sup (sup ((𝜓 − 𝜑0 )f m | 𝜓 ∈ Ψ) | m ∈ M) = = sup ((𝜓 − 𝜑0 )f m | (m, 𝜓) ∈ M × Ψ) . Thus, we have (𝜃 − 𝜑0 )f = sup ((𝜃 − 𝜑0 )f m | m ∈ M). This means by Lemma 1 that 𝜃 − 𝜑0 is pointwise 𝜎-continuous. Consequently, 𝜃 = (𝜃 − 𝜑0 ) + 𝜑0 ∈ A(T)∧ . In this manner, we deduce that ⌀ ≠ Φ ⊂ A(T)∧ , 𝜃 ∈ A(T)∼ , and 𝜃 = inf Φ imply 𝜃 ∈ A(T)∧ . This means that A(T)∧ is completely (order) closed in A(T)∼ . For A(T)∨ , the proof is completely analogous. Corollary 1. Let A(T) be a lattice-ordered linear space of functions on a set T. Then, the sets A(T)∧ and A(T)∨ are Dedekind complete lattice-ordered linear spaces, the smallest upper [the greatest lower] bounds of bounded above [below] sets in the ordered sets A(T)∼ and A(T)∧ (and in the ordered sets A(T)∼ and A(T)∨ , respectively) coincide, and they are expressed by the formulas from Statement 5 and its Corollaries. Proof. The assertion follows from Theorem 1 and Statements 3 and 1 below. Consider now some important subfamilies of A(T)∧ and A(T)∨ . A functional 𝜑 on A(T) will be called uniformly (order) bounded if the set 𝜑[E] is bounded in R for every uniformly (order) bounded subset E of A b (T) (see 2.2.2). The set of all uniformly (order) bounded linear functionals 𝜑 : A(T) → R will be denoted by A(T) ⃝ . The subfamilies in A(T)∧ , A(T)∨ , A(T)† , and A(T)∼ consisting of all uniformly (order) bounded functionals will be denoted by A(T) ∧ , A(T) ∨ , A(T) † , and A(T) ∼ , respectively. Lemma 3. Let A(T) be a linear subspace of F b (T). Then, A(T) (2.2.7)).

⃝

= ⟮A(T), ‖ ⋅ ‖u ⟯󸀠 (see 7∘

Proof. Let 𝜑 ∈ A(T) ⃝ . Take the unit ball B ≡ {g ∈ A(T) | ‖g‖u ⩽ 1} = {g ∈ A(T) | |g| ⩽ 1}. By the definition, there is s ∈ R such that |𝜑g| ⩽ s for every g ∈ B. If f ∈ A(T) and f ≠ 0, then xf ∈ B for x ≡ 1/‖f ‖u . So |𝜑(xf )| ⩽ s implies |𝜑f | ⩽ s‖f ‖u . This means that 𝜑 ∈ A(T)󸀠 . Conversely, let 𝜑 ∈ A(T)󸀠 , i. e. |𝜑f | ⩽ s‖f ‖u for some s ∈ R. Take some E ⊂ A(T) such that |e| ⩽ x1 for every e ∈ E and some x ∈ R. Then, ‖e‖u ⩽ x implies |𝜑e| ⩽ s‖e‖u ⩽ sx, i. e. 𝜑[E] is bounded in R. This means that 𝜑 ∈ A(T) ⃝ . Corollary 1. Let A(T) be a lattice-ordered linear subspace of F b (T). Then, ⟮A(T), ‖⋅‖u ⟯󸀠 = A(T) ⃝ = A(T) † = A(T) ∼ ⊂ A(T)∼ = A(T)† . If, besides, 1 ∈ A(T), then ⟮A(T), ‖ ⋅ ‖u ⟯󸀠 = A(T) ⃝ = A(T) † = A(T) ∼ = A(T)∼ = A(T)† .

96 | 2.2 Families of real-valued functions on a set

Proof. By Lemma 3 ⟮A(T), ‖ ⋅ ‖u ⟯󸀠 = A(T) ⃝ . According to Statement 4, A(T)† = A(T)∼ . Hence, A(T) † = A(T) ∼ . Since every bounded set in F b (T) is uniformly bounded, it follows directly from the definitions of bounded and uniformly bounded functionals that A(T) ⃝ ⊂ A(T)† . Thus, A(T) ⃝ = A(T) ⃝ ∩ A(T)† ≡ A(T) † . If 1 ∈ A(T), then every uniformly bounded set in F b (T) is bounded, and therefore, we have A(T)† ⊂ A(T) ⃝ . If A(T) is a linear subspace of F b (T), then Lemma 3 guarantees that A(T) ∧ = A(T)∧ ∩ ⟮A(T), ‖ ⋅ ‖u ⟯󸀠 and A(T) ∨ = A(T)∨ ∩ ⟮A(T), ‖ ⋅ ‖u ⟯󸀠 . Lemma 4. Let A(T) be a lattice-ordered linear space of functions on a set T and 𝜑 ∈ A(T)∼ . Then, 𝜑 is uniformly (order) bounded iff |𝜑| has the same property. Proof. Let F ⊂ A(T) and |f | ⩽ x1 for some x ∈ R+ and every f ∈ F. Let 𝜑 is uniformly bounded. By Corollary 3 to Statement 5 |𝜑||f | = sup{|𝜑g| | g ∈ A(T) ∨ |g| ⩽ |f |}. Consider the set G ≡ {g ∈ A(T) | ∃ f ∈ F(|g| ⩽ |f |)}. It is clear that |g| ⩽ x1 for every g ∈ G. Hence G is uniformly bounded in F(T). By the condition, there is y ∈ R+ such that 󵄨 󵄨 |𝜑g| ⩽ y for every g ∈ G. Therefore, 󵄨󵄨󵄨𝜑|f |󵄨󵄨󵄨 ⩽ |𝜑||f | ⩽ sup{|𝜑g| | g ∈ G} ⩽ y. This means that |𝜑| is uniformly bounded. Now, let |𝜑| is uniformly bounded. Then, for the set E ≡ {|f | | f ∈ F} there is a 󵄨 󵄨 number z such that 󵄨󵄨󵄨𝜑|e|󵄨󵄨󵄨 ⩽ z for every e ∈ E. Therefore, by Corollary 3 to Statement 5 |𝜑f | ⩽ |𝜑||f | ⩽ z for every f ∈ F. This means that |𝜑| is uniformly bounded. Corollary 1. Let A(T) be a lattice-ordered linear space of functions on a set T. Then, A(T) ∧ and A(T) ∨ are l-ideals in the Dedekind complete lattice-ordered linear spaces A(T)∧ and A(T)∨ , respectively. Proof. Let 𝜑 ∈ A(T)∧ , 𝜓 ∈ A(T) ∧ and |𝜑| ⩽ |𝜓|. By Lemma 4 |𝜓| ∈ A(T) ∧ . Let F ⊂ A(T) and |f | ⩽ x1 for some x ∈ R+ and every f ∈ F. Consider the set E ≡ {|f | | 󵄨 󵄨 f ∈ F}. Then, there is y ∈ R+ such that 󵄨󵄨󵄨𝜓|e|󵄨󵄨󵄨 ⩽ y for every e ∈ E. Therefore, by Corollary 3 to Statement 5 |𝜑f | ⩽ |𝜑||f | ⩽ |𝜓||f | ⩽ y for every f ∈ F. Consequently, 𝜑 is uniformly bounded. Finally, Statement 2 guarantees that A(T) ∧ is an l-ideal in A(T)∧ . Corollary 2. Let A(T) be a lattice-ordered linear space of functions on a set T. Then, A(T) ∧ and A(T) ∨ are Dedekind complete lattice-ordered linear spaces, the smallest upper [the greatest lower] bounds of bounded above [below] sets in the ordered sets A(T)∧ and A(T) ∧ (and A(T)∨ and A(T) ∨ , respectively) coincide and they are expressed by the formulas from Statement 5 and its Corollaries. Proof. The assertion follows from Corollary 1 and Statements 3 and 1 below.

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Auxiliary definitions and statements 1∘ Let ⟮A, ⩽⟯ be an ordered class and 𝛼 be a cardinal number. This class will be called upward [downward] 𝛼-Dedekind complete if for every bounded above [below] collection (a i ∈ A | i ∈ I) such that card I ⩽ 𝛼 there is a ∈ A such that a = sup (a i | i ∈ I) [a = inf (a i | i ∈ I)]. It is called 𝛼-Dedekind complete if it is upward and downward 𝛼-Dedekind complete. If V is 𝛼-Dedekind complete for every 𝛼, then V is Dedekind complete in the sense of 1.1.15. Let U ≡ ⟮A, ⩽A ⟯ be an ordered class. An ordered class V ≡ ⟮B, ⩽B ⟯ will be called upward [downward] (order) 𝛼-closed in the ordered class U if the following properties hold: 1) B ⊂ A; 2) ⩽B =⩽A |B ∗ B; 3) if a ∈ A and a = supA (b i | i ∈ I) [a = inf A (b i | i ∈ I)] for some simple collection (b i ∈ B | i ∈ I) such that card I ⩽ 𝛼, then a ∈ B. It is called (order) 𝛼-closed in U if it is upward and downward 𝛼-closed. If V is (order) 𝛼-closed for every 𝛼, then V is completely (order) closed in the sense of 1.1.15. Statement 1. Let A be a Dedekind complete [𝛼-Dedekind complete] lattice-ordered linear space and B be an l-ideal in A. Then, B is a Dedekind complete [𝛼-Dedekind complete] lattice-ordered linear space as well. Statement 2. Let A be a lattice-ordered linear space and B be a linear subspace of A. Then, the following conclusions are equivalent: 1) B is an l-ideal (in the sense of 1.1.15); 2) ∀ a ∈ A ∀ b ∈ B (|a| ⩽ |b| ⇒ a ∈ B). Statement 3. Let A be a lattice-ordered linear space and B be an l-ideal in A. Then, for any simple collection u ≡ (b i ∈ B | i ∈ I) and for any elements a ∈ A and b ∈ B such that a = supA u [a = inf A u] and b = supB u [b = inf B u], we have a = b. 2∘ Let U ≡ |⟮A, R⟯, s ols | be an ordered linear space and V ≡ ⟮R; w r , ⩽R ⟯ be the system of real numbers with the ring structure w r and the order ⩽R . The set of all (order) bounded linear functionals 𝜑 : A → R (see 1.1.15 and 3∘ , 6∘ (2.2.7)) will be denoted by U † or simply by A† . The set of all bipositive linear functionals 𝜑 : A → R (see 4∘ (2.2.7)) will be denoted by U ∼ or simply by A∼ . It is called the (order) dual (≡ conjugate, adjoint) to the system U. Statement 4 (2.2.7) implies the following statement.

98 | 2.2 Families of real-valued functions on a set

Statement 4. Let U ≡ |⟮A, R⟯, s lols | be a lattice-ordered linear space. Then, A∼ = A† . Define on the set A∼ the structures of an ordered linear space setting (r𝜑)a ≡ r(𝜑a), (𝜑 + 𝜓)a ≡ 𝜑a + 𝜓a for every a ∈ A and r ∈ R, and 𝜓 ⩾ 𝜑 iff 𝜓a ⩾ 𝜑a for every a ∈ A+ . It is remarkable that the ordered linear space A∼ is lattice-ordered. This follows from the following famous theorem. Statement 5 (the Riesz – Kantorovich theorem). Let U ≡ |⟮A, R⟯, s lols | be a latticeordered linear space, a be an element of A + , and Φ be a subset of A∼ . Then, 1) the ordered linear space A∼ is lattice-ordered and Dedekind complete; 2) if Φ is bounded above, then (sup Φ)a = sup {∑(𝜑i a i | i ∈ I) | (a i | i ∈ I) ∈ Map(I, A+ ) ∧ ∧ (𝜑i | i ∈ I) ∈ Map(I, Φ) ∧ P(I) ∈ 𝜔 ∧ a = ∑(a i | i ∈ I)} (see the convenient designations in 1.1.5); 3) if Φ is bounded below, then (inf Φ)a = inf {∑(𝜑i a i | i ∈ I) | (a i | i ∈ I) ∈ Map(I, A+ ) ∧ ∧ (𝜑i | i ∈ I) ∈ Map(I, Φ) ∧ P(I) ∈ 𝜔 ∧ a = ∑(a i | i ∈ I)} . Corollary 1. In the conditions of Statement 5, we have: 1) if Φ is bounded above and upward directed, then (sup Φ)a = sup(𝜑a | 𝜑 ∈ Φ); 2) if Φ is bounded below and downward directed, then (inf Φ)a = inf(𝜑a | 𝜑 ∈ Φ). Corollary 2. Let 𝜑, 𝜓 ∈ A ∼ and a ∈ A+ . Then, 1) (𝜑 ∨ 𝜓)a = sup{𝜑e + 𝜓f | (e, f ) ∈ A + × A+ ∧ a = e + f }; 2) (𝜑 ∧ 𝜓)a = inf{𝜑e + 𝜓f | (e, f ) ∈ A + × A+ ∧ a = e + f }. Corollary 3. Let 𝜑 ∈ A∼ and a ∈ A+ . Then, 1) 𝜑+ a = sup{𝜑b | b ∈ A+ ∧ b ⩽ a}; 2) 𝜑− a = inf{𝜑b | b ∈ A+ ∧ b ⩽ a}; 3) |𝜑|a = sup{𝜑b | b ∈ A + ∧ |b| ⩽ a} = sup{|𝜑b| | b ∈ A + ∧ |b| ⩽ a}; 4) 𝜑 = 𝜑+ + 𝜑− and |𝜑| = 𝜑+ − 𝜑− . The equality 𝜑 = 𝜑+ + 𝜑− is called the Riesz decomposition of 𝜑. The functionals 𝜑+ and 𝜑− are called the positive part and the negative part of the functional 𝜑, respectively. 2.2.9 Truncatable lattice-ordered linear spaces of functions In this subsection, we consider a class of families of functions important for integration theory.

2.2.9 Truncatable lattice-ordered linear spaces of functions

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Let A(T) be a function family on a set T. It is called truncatable or with the Stone property if f ∧ 1 ∈ A(T) for every f ∈ A(T). We will say that a family A(T) envelopes [𝜎-envelopes] from above a function h ∈ F(T) if there is a net (f m ∈ A(T) | m ∈ M) [a sequence (f m ∈ A(T) | m ∈ M ⊂ 𝜔)] such that (f m (t) | m ∈ M) ↓ h(t) in each point t ∈ T. In the same way, we will say that A(T) envelope [𝜎-envelope] from below a function g ∈ F(T) if (f m (t) | m ∈ M) ↑ g(t) in each point t ∈ T. A set K ⊂ T will be called inscribed [𝜎-inscribed] in a family A(T) if A(T) envelopes [𝜎-envelopes] from above the function 𝜒(K). The ensembles of all subsets of T inscribed [𝜎-inscribed] in A(T) will be denoted by I(A(T)) [I𝜎 (A(T))]. Lemma 1. Let A(T) be a truncatable lattice-ordered linear space of functions on a set T and f ∈ A(T). Then, 1) f ∧ x1 ∈ A(T) for every x > 0; 2) Sf ≡ {t ∈ T | f (t) ⩾ 1} ∈ I𝜎 (A(T)). Proof. 1. It is evident that f ∧ x1 = x(x−1 f ∧ 1) ∈ A(T). 2. Take f n ≡ 2n (f ∧1− f ∧(1−2−n )1) ∈ A(T). Then, f n (t) = 1 for every t ∈ Sf and Sf ⊂ coz f n+1 ⊂ f n . Let t ∈ coz f n+1 \Sf . Then, f (t) < 1 implies f n+1 (t) = 2n+1 (f (t) − f (t) ∧ (1 − 2−n−1 )) = 2n (2f (t)−2f (t) ∧ (2−2−n )) = 2n (−0 ∧ (2−2−n −2f (t))) = 2n (0 ∨ (2f (t)+2−n −2)) ⩽ 2n (0 ∨ (2f (t) + 2−n − 1)) = 2n (−0 ∧ (1 − 2−n − f (t))) = 2n (f (t) − f (t) ∧ (1 − 2−n )) = f n (t). Consequently, f n+1 ⩽ f n . Let t ∈ ̸ Sf . Then, by Corollary 2 to Proposition 5 (1.4.3) there is m such that 2−m < 1 − f (t). Therefore, f m (t) = 2m (f (t) − f (t)) = 0. Thus, we obtain that f n ↓ 𝜒(Sf ) in F(T). Proposition 1. Let A(T) be a truncatable lattice-ordered linear space of functions on a set T. Then, the ensemble I𝜎 (A(T)) is additive and 𝛿-multiplicative and the ensemble I(A(T)) is additive and completely multiplicative. Proof. Let K ≡ I𝜎 (A(T)). Suppose K, L ∈ K, (f m󸀠 | m ∈ M ⊂ 𝜔) ↓ 𝜒(K), and (g󸀠n | n ∈ N ⊂ 𝜔) ↓ 𝜒(L). By Theorem 1 (1.3.7) there are the unique isotone bijections u : 𝜔 M and v : 𝜔 N. Therefore, we may consider the new sequences (f k | k ∈ 𝜔) ↓ 󸀠 𝜒(K)and (g k | k ∈ 𝜔) ↓ 𝜒(L) such that f k = f u(k) and g k = g󸀠v(k) . Consider the functions h k = f k ∨ g k . If x > 1 and t ∈ K ∪ L, then there are k and l such that f k (t) < x and g l (t) < x. Let m = k ∨ l. Then, h m (t) ⩽ f k (t) ∨ g l (t) < x, where 1 ⩽ y ≡ inf (h m (t) | m ∈ 𝜔) < x. Since x is arbitrary, we conclude that y = 1. If x > 0 and t ∈ ̸ K ∪ L, then by the same reason, 0 ⩽ z ≡ inf (h m (t) | m ∈ 𝜔) < x. Since x is arbitrary, we conclude that z = 0. Thus, (h k | k ∈ 𝜔) ↓ 𝜒(K ∪ L) in F(T), where K ∪ L ∈ K, and K is additive. Let (K m ∈ K | m ∈ 𝜔) be a sequence. Then, (f mn ∈ A(T) | n ∈ 𝜔) ↓ 𝜒(K m ) for some sequences. Consider the set K ≡ ⋂ ⟮K m | m ∈ 𝜔⟯ and the functions

100 | 2.3 Families of measurable and distributable functions on a descriptive space

g n ≡ inf (f ij | (i, j) ∈ (n + 1) × (n + 1)) ∈ A(T). By Corollary 1 to Proposition 2 (1.1.15) we have 𝜒(K) = inf (𝜒(K m ) | m ∈ 𝜔) = inf ((f mn ∈ A(T) | n ∈ 𝜔) | m ∈ 𝜔) = inf(f mn | (m, n) ∈ 𝜔 × 𝜔) = inf (g n | n ∈ 𝜔) in F(T). Hence, K ∈ K. Thus, K is 𝛿-multiplicative. Now, let K ≡ I(A(T)). Suppose K, L ∈ K, (f 𝛾 | 𝛾 ∈ Γ) ↓ 𝜒(K), and (g𝛿 | 𝛿 ∈ Δ) ↓ 𝜒(L). Consider the upward directed set Z ≡ Γ × Δ (with the order (𝛾󸀠 , 𝛿󸀠 ) ⩽ (𝛾󸀠󸀠 , 𝛿󸀠󸀠 ) ⇔ 𝛾󸀠 ⩽ 𝛾󸀠󸀠 and 𝛿󸀠 ⩽ 𝛿󸀠󸀠 ) and the net (h𝜁 | 𝜁 ∈ Z) ↓ such that h𝜁 ≡ f𝛾 ∨ g𝛿 for every 𝜁 ≡ (𝛾, 𝛿). If x > 1 and t ∈ K ∪ L, then there is 𝜁 ≡ (𝛾, 𝛿) such that f𝛾 (t) < x and g𝛿 (t) < x. Then, h𝜁 (t) < x, where 1 ⩽ y ≡ inf (h𝜁 (t) | 𝜁 ∈ Z) < x. Since x is arbitrary, we conclude that y = 1. If x > 0 and t ∈ ̸ K ∪ L, then by the same reason, 0 ⩽ z ≡ inf (h𝜁 (t) | 𝜁 ∈ Z) < x. Since x is arbitrary, we conclude that z = 0. Thus, (h𝜁 | 𝜁 ∈ Z) ↓ 𝜒(K∪L) in F(T), where K ∪ L ∈ K, and K is additive. Let (K i ∈ K | i ∈ I) be a collection. Then, (f i𝛾 ∈ A(T) | 𝛾 ∈ Γi ) ↓ 𝜒(K i ) for some nets. Consider the sets K ≡ ⋂ ⟮K i | i ∈ I⟯ and S ≡ ⋃ ⟮{f i𝛾 | 𝛾 ∈ Γi } | i ∈ I⟯. Let Z be the set of all finite subsets 𝜁 of S. Consider on Z the order by inclusion (𝜁󸀠 ⩽ 𝜁󸀠󸀠 ⇔ 𝜁󸀠 ⊂ 𝜁󸀠󸀠 ). Define a decreasing net (h𝜁 ∈ A(T)+ | 𝜁 ∈ Z) setting h𝜁 ≡ inf 𝜁. It is clear that h𝜁 ⩾ 𝜒(K) for every 𝜁 ∈ Z. Let t ∈ K and x > 1. Fix some i. Then, there is 𝛾 ∈ Γi such that f i𝛾 (t) < x. Take 𝜁 ≡ {f i𝛾 }. Then, h𝜁 (t) = f i𝛾 (t) < x. Therefore, y ≡ inf (h𝜁 (t) | 𝜁 ∈ Z) < x. Since x is arbitrary, we conclude that y = 1. If x > 0 and t ∈ ̸ K, then by the same reason, 0 ⩽ z ≡ inf (h𝜁 (t) | 𝜁 ∈ Z) < x. Since x is arbitrary, we conclude that z = 0. As a result, (h𝜁 | 𝜁 ∈ Z) ↓ 𝜒(K) in F(T), where K ∈ K. Thus, K is completely multiplicative.

2.3 Families of measurable and distributable functions on a descriptive space The notion of a measurable function took its shape in the works of E. Borel, R. Baire, H. Lebesgue, et al. [Borel, 1898; Baire, 1899; 1905; Borel, 1905; Lebesgue, 1902; 1903; 1904; Young, 1905; 1911]. It became one of the most important notions in contemporary mathematics. 2.3.1 Measurable and distributable functions Let T and T 󸀠 be sets and S and S󸀠 be ensembles (non-empty families of subsets) on T and T 󸀠 , respectively (see 2.1.1). A mapping 𝜏 : T → T 󸀠 is called measurable with respect to S and S󸀠 or (S, S󸀠 )-measurable if 𝜏−1 [S󸀠 ] ∈ S for every S󸀠 ∈ S󸀠 . If T 󸀠 = R, then the ensemble S󸀠 = Oint ≡ {]x, y[| x, y ∈ R} of all open intervals in R will be considered. Thus, a function f : T → R is called measurable on the descriptive space ⟮T, S⟯ or measurable with respect to the ensemble S or S-measurable if it is (S, Oint )-measurable. The family of all S-measurable real-valued functions on T is

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denoted by M(T, S). Its subfamily of all bounded S-measurable functions on T is denoted by M b (T, S). In the classical case of a topological space ⟮T, G⟯, the term “continuous” is used along with the term “measurable” and the notation C(T, G) is also used. This is explained by the following lemma. Lemma 1. Let S be a 𝜎-additive ensemble on a set T and f ∈ F(T). Then, the following conclusions are equivalent: 1) f ∈ M(T, S); 2) f −1 []x, y[] ∈ S for every interval ]x, y[⊂ R; 3) f −1 [G] ∈ S for every G ∈ Ost , where Ost is the usual (standard) topology on R. Proof. (1) ⊢ (2). Suppose x = −∞, y ∈ R. Then, ] − ∞, y[= ⋃⟮]n, y[| n ∈ N⟯. Therefore, f −1 [] − ∞, y[] = f −1 [ ⋃⟮]n, y[| n ∈ N⟯] = ⋃⟮f −1 []n, y[] | n ∈ N⟯. Since S n ≡ f −1 []n, y[] ∈ S for every n ∈ N and S is 𝜎-additive, we have f −1 [] − ∞, y[] = ⋃⟮S n | n ∈ N⟯ ∈ S. Thus, f −1 [] − ∞, y[] ∈ S. The cases of x ∈ R, y = ∞ and x = −∞, y = ∞ are considered in the completely similar way. (2) ⊢ (3). Statement 1 provides Ost = (Oint )𝜎 . Then, for every G ∈ Ost , we have G = ⋃⟮]x i , y i [| i ∈ I⟯ for some countable set I. By Lemma 3 (1.1.10) this implies f −1 [G] = ⋃⟮f −1 []x i , y i [] | i ∈ I⟯ ∈ S. (3) ⊢ (1). This is evident. Corollary 1. Let S be a 𝜎-additive ensemble on T and f ∈ M(T, S). Then, f −1 [[x, y]] ∈ S𝛿 ∩ co-S. Proof. The assertion follows from Lemma 1 and assertions 7 and 9 of Lemma 2 (2.2.1).

Corollary 2. Let S be a 𝜎-algebra on T and f ∈ F(T). Then, the following conclusions are equivalent: 1) f ∈ M(T, S); 2) f −1 [[x, y]] ∈ S for every interval [x, y] ⊂ R. Proof. (1) ⊢ (2). It follows from Lemma 1, 𝛿-multiplicativity of S, and assertions 4, 5, and 7 of Lemma 2 (2.2.1). (2) ⊢ (1). It follows from assertion 6 of Lemma 2 (2.2.1). Recall that a collection ⟮A i ⊂ T | i ∈ I⟯ is called a cover of the set T if T = ⋃⟮A i | i ∈ I⟯ (see 1.1.10), the number 𝜔( f , E) ≡ sup{|f (t) − f (s)| | t, s ∈ E} is called the oscillation of the function f on the set E (see 2.2.1). If 𝜋 ≡ ⟮C i | i ∈ I⟯ is a cover of the set T, then

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the number 𝜔( f , 𝜋) ≡ sup⟮𝜔( f , C i ) | i ∈ I⟯ is called the oscillation of the function f on the cover 𝜋 [Zakharov, 1993b]. The S-measurable functions for a 𝜎-additive S may be characterized by means of their oscillations on countable or finite covers in the following way. Proposition 1. Let S be an ensemble on T. Then, the following assertions are equivalent: 1) f ∈ M(T, S𝜎 ); 2) for every 𝜀 > 0 there is a countable cover 𝜋 ≡ (S i ∈ S | i ∈ I) of T such that 𝜔(f , 𝜋) < 𝜀. Proof. (1) ⊢ (2). Let f ∈ M(T, S𝜎 ). Take 𝜀 > 0 and n ∈ N such that 2/n < 𝜀. For each m ∈ Z, consider the sets X m ≡ f −1 [](m − 1)/n, (m + 1)/n[] ∈ S𝜎 . For every X m , there exists a countable collection (Q mp ∈ S | p ∈ P m ) such that X m = ⋃⟮Q mp | p ∈ P m ⟯. Consider the set I ≡ ⋃⟮{m} × P m | m ∈ Z⟯. By Theorem 1 (1.3.9) the set I is countable. Put S i ≡ Q mp for every i = (m, p) ∈ I. Then, 𝜔( f , S i ) ⩽ 𝜔( f , X m ) < 2/n < 𝜀. Besides, using Proposition 1 (1.1.13), we get ⋃⟮S i | i ∈ I⟯ = ⋃⟮Q mp | (m, p) ∈ I⟯ = ⋃⟮⋃⟮Q mp | p ∈ P m ⟯ | m ∈ Z⟯ = ⋃⟮X m | m ∈ Z⟯ = T, and therefore, the collection (S i | i ∈ I) is a countable cover of T. (2) ⊢ (1). For every n ∈ N, consider a countable cover 𝜋n ≡ (S ni ∈ S | i ∈ I n ) such that 𝜔( f , 𝜋n ) < 1/n. Take any real numbers x < y and consider the sets E ≡ f −1 []x, y[], E n ≡ f −1 [[x + 1/n, y − 1/n]], J n ≡ {i ∈ I n | S ni ∩ E n ≠ ⌀}, and F n ≡ ⋃⟮S ni | i ∈ J n ⟯ ∈ S𝜎 . Since 𝜋n is a cover of T, we have E n ⊂ F n . If t ∈ F n , then t ∈ S ni for some i ∈ J n . Hence, there is s ∈ S ni ∩ E n . Consequently, f (t) < f (s) + 1/n < y and f (t) > f (s) − 1/n > x, i. e. t ∈ E. This means that F n ⊂ E. Thus, E ⊂ ⋃⟮E n | n ∈ N⟯ ⊂ ⋃⟮F n | n ∈ N⟯ ⊂ E. As a result, we get E = ⋃⟮F n | n ∈ N⟯ ∈ S𝜎 . This means that f ∈ M(T, S𝜎 ). Corollary 1. Let S be a 𝜎-additive ensemble on T. Then, the following assertions are equivalent: 1) f ∈ M(T, S) [ f ∈ M b (T, S)]; 2) for every 𝜀 > 0 there is a countable [finite] cover 𝜋 ≡ (S i ∈ S | i ∈ I) of T such that 𝜔( f , 𝜋) < 𝜀. Proof. (1) ⊢ (2). If f ∈ M(T, S), then it follows directly from Proposition 1. Suppose f ∈ M b (T, S); then, |f | ⩽ a1 for some a > 0. Take 𝜀 > 0 and n ∈ N such that 2/n < 𝜀. Therefore, the set I ≡ {i ∈ Z | |i| ⩽ na} is finite. Define S i ≡ f −1 [](i − 1)/n, (i + 1)/n[] ∈ S. Then, ⋃⟮S i | i ∈ I⟯ = T and 𝜔( f , S i ) < 2/n < 𝜀. (2) ⊢ (1). By Proposition 1 f ∈ M(T, S). By Lemma 1 (2.2.1) the function f is bounded. Using the cover language in Proposition 1 gives the basis to introduce a new class of functions possessing good properties for any foundation S, unlike the class of

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measurable functions possessing these properties only for a 𝜎-foundation S (cf. Corollary 2 to Proposition 2 (2.3.4) and Corollary 2 to Proposition 3 (2.3.4) below). A function f ∈ F(T) will be called distributable with respect to an ensemble S or Sdistributable if for every 𝜀 > 0 there is a countable cover 𝜋 ≡ (S i ∈ S | i ∈ I) such that 𝜔( f , 𝜋) < 𝜀. The set of all such functions will be denoted by D(T, S). Proposition 1 means that D(T, S) = M(T, S𝜎 ) ⊃ M(T, S) for every ensemble S and D(T, S) = M(T, S) for a 𝜎-additive S. This shows that the notion of a distributable function uses the other language in comparison with the classical preimage language but does not generate any new essence different from the notion of measurable function. However, the situation changes cardinally for quasidistributable functions considered in 2.5.2. Lemma 2. Let S be an additive [𝜎-additive] ensemble on a set T. Then, qSt(T, S) ⊂ M b (T, S) ⊂ D b (T, S) [qStc (T, S) ⊂ M(T, S) = D(T, S)]. Proof. Let f ∈ qSt(T, S) [ f ∈ qStc (T, S)], i. e. there are a finite [countable] partition (S i ∈ S | i ∈ I) of T and a collection (x i ∈ R | i ∈ I) such that f (t) = x i for every t ∈ S i . For any ]a, b[⊂ R, we get f −1 []a, b[] = ⋃⟮S i | i ∈ I a,b ⟯ ∈ S for the finite [countable] set I a,b ≡ {i ∈ I | x i ∈]a, b[}. This means that f ∈ M b (T, S) [ f ∈ M(T, S)]. Corollary 1. Let S be an algebra on T. Then, St(T, S) ⊂ M b (T, S). If, besides, S is 𝜎-additive, then Stc (T, S) ⊂ M(T, S). Proof. The assertion follows from Lemma 2 and Proposition 2 (2.2.4). Let |T, 𝜌| be a metric space with the metric topology G ≡ G(𝜌) (see 1∘ (2.2.7)). A function f ∈ C(T, G) ≡ M(T, G) is called uniformly continuous if for every 𝜀 > 0 there is 𝛿 > 0 such that for every t󸀠 , t󸀠󸀠 ∈ T the inequality 𝜌(t󸀠 , t󸀠󸀠 ) < 𝛿 implies |f (t󸀠 )− f (t󸀠󸀠 )| < 𝜀. Proposition 2. Let |T, 𝜌| be a compact metric space with the metric topology G ≡ G(𝜌), i. e. T is a G-compact set (see 2.1.1). Then, every continuous function f is uniformly continuous. Proof. Since G is 𝜎-additive, Corollary 1 to Proposition 1 guarantees that for every 𝜀 > 0 there is a countable cover (G m ∈ G | m ∈ M) such that 𝜔(f , G m ) < 𝜀/3 for every m ∈ M. By the definition, of the metric topology for every m ∈ M there is a collection (B(t ma , r ma ) | a ∈ A m ) of open balls B ma ≡ B(t ma , r ma ) with centra t ma ∈ T and radii r ma > 0 such that G m = ⋃⟮B ma | a ∈ A m ⟯. For every t ∈ T, consider non-empty sets M t ≡ {m ∈ M | t ∈ G m } and A mt ≡ {a ∈ A m | t ∈ B ma } for m ∈ M t . For every m ∈ M t and a ∈ A mt , put r mat ≡ (r ma ⊼ (r ma − 𝜌(t, t ma )))/3. Then, B mat ≡ B(t, r mat ) ⊂ B(t ma , r ma ) ⊂ G m , and therefore, 𝜔(f , B mat ) ⩽ 𝜔(f , G m ) < 𝜀/3.

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Consider the sets N t ≡ ⋃⟮{m} × A mt | m ∈ M t ⟯ and L ≡ ⋃⟮{t} × N t | t ∈ T⟯. For every l ≡ (t, n) ∈ L, where n ≡ (m, a) ∈ N t , put B l ≡ B mat and r l ≡ r mat . Using Proposition 1 (1.1.13), we get T ⊃ ⋃⟮B l | l ∈ L⟯ = ⋃⟮⋃⟮B(t,n) | n ∈ N t ⟯ | t ∈ T⟯ = ⋃⟮⋃⟮⋃⟮B mat | a ∈ A mt ⟯ | m ∈ M t ⟯ | t ∈ T⟯ ⊃ ⋃⟮{t} | t ∈ T⟯ = T. Thus, ⟮B l | l ∈ L⟯ is a cover of T. Hence, by definition of a G-compact set, there is a finite set K ⊂ L such that ⋃⟮B k | k ∈ K⟯ = T. Let t󸀠 , t󸀠󸀠 ∈ T and 𝜌(t󸀠 , t󸀠󸀠 ) < 𝛿 ≡ sm (r k | k ∈ K). Then, there are k󸀠 , k󸀠󸀠 ∈ K such that t󸀠 ∈ B k󸀠 ≡ B m󸀠 a󸀠 s󸀠 ≡ B(s󸀠 , r m󸀠 a󸀠 s󸀠 ) and t󸀠󸀠 ∈ B k󸀠󸀠 ≡ B m󸀠󸀠 a󸀠󸀠 s󸀠󸀠 ≡ B(s󸀠󸀠 , r m󸀠󸀠 a󸀠󸀠 s󸀠󸀠 ). Hence, |f (t󸀠 ) − f (s󸀠 )| ⩽ 𝜔( f , B m󸀠 a󸀠 s󸀠 ) < 𝜀/3 and |f (t󸀠󸀠 ) − f (s󸀠󸀠 )| ⩽ 𝜔( f , B m󸀠󸀠 a󸀠󸀠 s󸀠󸀠 ) < 𝜀/3. Besides, 𝜌(s󸀠 , s󸀠󸀠 ) ⩽ 𝜌(s󸀠 , t󸀠 ) + 𝜌(t󸀠 , t󸀠󸀠 ) + 𝜌(t󸀠󸀠 , s󸀠󸀠 ) < r k󸀠 + 𝛿 + r k󸀠󸀠 ⩽ r m󸀠 a󸀠 /3 + 𝛿 + r m󸀠󸀠 a󸀠󸀠 /3 ≡ x. If r m󸀠 a󸀠 ⩾ r m󸀠󸀠 a󸀠󸀠 , then x ⩽ r m󸀠 a󸀠 /3 + r m󸀠 a󸀠 /3 + r m󸀠 a󸀠 /3 = r m󸀠 a󸀠 . Therefore, s󸀠󸀠 ∈ B m󸀠 a󸀠 s󸀠 and |f (s󸀠 ) − f (s󸀠󸀠 )| ⩽ 𝜔( f , B m󸀠 a󸀠 s󸀠 ) < 𝜀/3. Similarly, if r m󸀠 a󸀠 < r m󸀠󸀠 a󸀠󸀠 , then x < r m󸀠󸀠 a󸀠󸀠 /3 + r m󸀠󸀠 a󸀠󸀠 /3 + r m󸀠󸀠 a󸀠󸀠 /3 = r m󸀠󸀠 a󸀠󸀠 .Therefore,s󸀠 ∈ B m󸀠󸀠 a󸀠󸀠 s󸀠󸀠 and|f (s󸀠 ) − f (s󸀠󸀠 )| ⩽ 𝜔(f , B m󸀠󸀠 a󸀠󸀠 s󸀠󸀠 ) < 𝜀/3. Finally, |f (t󸀠 ) − f (t󸀠󸀠 )| ⩽ |f (t󸀠 ) − f (s󸀠 )| + |f (s󸀠 ) − f (s󸀠󸀠 )| + |f (t󸀠󸀠 ) − f (s󸀠󸀠 )| < 𝜀. Auxiliary definitions and statements 1∘ The most important example of a topological space for our book is the space ⟮R, Ost ⟯, where R is the set of all real numbers (see 1.4.3) and the ensemble Ost is the standard topology on R consisting of all sets G ⊂ R such that for every x ∈ G there is an open interval ]y, z[ such that x ∈]y, z[⊂ G. It is clear that all open intervals ]y, z[, ]y, → [=]y, ∞[, and ] ←, z[=] − ∞, z[ are open sets. Denote by Oint the ensemble of all open intervals of R. Statement 1. For every open set G ∈ Ost , there exists a countable collection (O i ∈ Oint | i ∈ I) of pairwise disjoint open intervals such that G = ⋃⟮O i | i ∈ I⟯. For every i ∈ I, the end-points of the interval O i are not in G.

2.3.2 Pointwise operations over measurable and distributable functions The class of 𝜎-foundations S seems to be the widest class of ensembles such that the families of S-measurable functions are closed under pointwise operations. Theorem 1. Let S be a 𝜎-foundation on T, f , g ∈ M(T, S), r ∈ R, and m ∈ N. Then, 1) 1 ∈ M(T, S); 2) rf ∈ M(T, S); 3) f + g ∈ M(T, S); 4) fg ∈ M(T, S); 5) 1/f ∈ M(T, S) for every function f : T → R \ {0}; m 6) √ f ∈ M(T, S) for every f ⩾ 0.

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Proof. 1. If 1 ∈]x, y[, then 1−1 []x, y[] = T ∈ S. If 1 ∈]x, ̸ y[, then 1−1 []x, y[] = ⌀ ∈ S. −1 −1 2. If r > 0, then (rf ) []x, y[] = f []x/r, y/r[] ∈ S. If r < 0, then (rf )−1 []x, y[] = f −1 []y/r, x/r[] ∈ S. If r = 0, then (rf )−1 []x, y[] equals to T ∈ S or to ⌀ ∋ S. 3. Suppose t ∈ (f + g)−1 []x, y[], i. e. x − g(t) < f (t) < y − g(t). By Lemma 14 (1.4.3) there are rational numbers p and q such that x − g(t) < p < f (t) < q < y − g(t). Therefore, t ∈ f −1 []p, q[] ∩ g−1 []x − p, y − q[] ≡ T pq . Then, ( f + g)−1 []x, y[] ⊂ ⋃⟮T pq | p, q ∈ Q (p < q)⟯ ∈ S. Conversely, if t belongs to the last set, then there are p, q ∈ Q such that p < f (t) < q and x − p < g(t) < y − q. Hence, t ∈ ( f + g)−1 []x, y[]. Thus, f + g ∈ M(T, S). 4. First, we prove that f 2 ∈ M(T, S). Let 0 ⩽ x < y. If t ∈ (f 2 )−1 []x, y[] ≡ R, then by Proposition 2 (1.4.6) 0 ⩽ √x < √f 2 (t) < √y, i. e. 0 ⩽ √x < |f (t)| < √y. If f (t) > 0, then 0 ⩽ √x < f (t) < √y, i. e. t ∈ f −1 []√x, √y[]. If f (t) < 0, then 0 ⩽ √x < −f (t) < √y, i. e. t ∈ f −1 [] − √y, −√x[]. Thus, R ⊂ f −1 []√x, √y[] ∪ f −1 [] − √y, −√x[] ≡ S. Conversely, let t ∈ S. If f (t) > 0, then √x < f (t) < √y. By assertion 4 of Proposition 4 (1.4.3) x < f 2 (t) < y, i. e. t ∈ R. If f (t) < 0, then −√y < f (t) < −√x. By assertions 3 and 4 of Proposition 4 (1.4.3), x < f 2 (t) < y, i. e. t ∈ R. This means that R = S ∈ S. If x < y ⩽ 0, then (f 2 )−1 []x, y[] = ⌀ ∈ S. Let x < 0 < y. If t ∈ (f 2 )−1 []x, y[] ≡ R, then x < 0 ⩽ f 2 (t) < y. By the previous arguments 0 ⩽ √f 2 (t) < √y, i. e. 0 ⩽ |f (t)| < √y. If f (t) ⩾ 0, then t ∈ f −1 [[0, √y[] ≡ S1 . If f (t) < 0, then 0 < −f (t) < √y implies t ∈ f −1 [] − √y, 0[] ≡ S2 . In both cases, t ∈ S1 ∪

S2 = f −1 [] − √y, √y[] ≡ S. Conversely, let t ∈ S, i. e. −√y < f (t) < √y. If f (t) ⩾ 0, then 0 ⩽ f (t) < √ y implies x < 0 ⩽ f 2 (t) < y, i. e. t ∈ R. If f (t) < 0, then −√y < f (t) < 0. By the previous arguments x < 0 < f 2 (t) < y, i. e. t ∈ R. This means that R = S ∈ S. Suppose now that f , g ∈ M(T, S). Then, using the equality fg = ((f + g)2 − (f − g)2 )/4 and proven assertions 2 and 3, we immediately get fg ∈ M(T, S). 5. We check, as above, that (1/f )−1 []x, y[] = (f −1 []0, ∞[] ∩ (xf )−1 [] − ∞, 1[] ∩ (yf )−1 []1, ∞[]) ∪ ∪ (f −1 [] − ∞, 0[] ∩ (xf )−1 []1, ∞[] ∩ (yf )−1 [] − ∞, 1[]) ∈ S. Hence, 1/f ∈ M(T, S). −1 m f ) []x, y[]. If y ⩽ 0, then X = ⌀ ∈ S. If x < 0 < y, then 6. Let X ≡ ( √ X = f −1 [[0, y m [] = f −1 [] − y m , y m [] ∈ S. If x ⩾ 0, then X = f −1 []x m , y m [] ∈ S. Corollary 1. Let S be a 𝜎-foundation on T and (f i ∈ M(T, S) | i ∈ I) be a finite collection. Then, ∑ (f i | i ∈ I) ∈ M(T, S) and P (f i | i ∈ I) ∈ M(T, S). The proof is by induction.

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Corollary 2. Let S be a 𝜎-foundation on T. Then, the family M(T, S) [M b (T, S)] satisfies conditions 1, 2, 3, 5, and 7 [7󸀠 )] from 2.2.4. In contrast to the families of S-measurable functions, the families of S-distributable functions are closed under the pointwise operations listed in Theorem 1 for any foundation S. Moreover, for distributable functions the analogue of Theorem 1 can be proven by essentially different method, not using the set Q of rational numbers but using only the definition of these functions. This method will be crucial in 2.4 and 2.5. The notion of S-distributable function gives another method of reasoning but, due to Proposition 1 (2.3.1), it does not give any new essence in comparison with the notion of S-measurable function. However, the situation is changed cardinally under transition from countable covers to finite ones in 2.4 and 2.5. Theorem 2. Let S be a foundation on T, f , g ∈ D(T, S), r ∈ R, and m ∈ N. Then, 1) 1 ∈ D(T, S); 2) rf ∈ D(T, S); 3) f + g ∈ D(T, S); 4) fg ∈ D(T, S); 5) 1/f ∈ D(T, S) for every function f : T → R \ {0}; m 6) √ f ∈ D(T, S) for every f ⩾ 0. Proof. Denote D(T, S) by A(T). Let 𝜀 > 0. 1. For any cover 𝜋 ≡ (S i ∈ S | i ∈ I), we have 𝜔(1, 𝜋) = 0. 2. Let r ∈ R\{0}. For the number 𝜀1 ≡ 𝜀/|r|, there exists a countable cover 𝜋 ≡ (S i ∈ S | i ∈ I) such that 𝜔(f , 𝜋) < 𝜀1 . If s, t ∈ S i , then |(rf )(s) − (rf )(t)| = |r| |f (s) − f (t)| ⩽ |r| 𝜔(f , 𝜋). Therefore, 𝜔(rf , 𝜋) ⩽ |r| 𝜔(f , 𝜋) < 𝜀 and rf ∈ A(T). For r = 0, we have 𝜔(rf , 𝜋) = 𝜔(0, 𝜋) = 0 for every cover 𝜋 ≡ (S i ∈ S | i ∈ I). 3. For the numbers 𝜀1 ≡ 𝜀/2 and 𝜀2 ≡ 𝜀/2, there are countable covers 𝜋 ≡ (C i ∈ S | i ∈ I) and 𝜘 ≡ (D j ∈ S | j ∈ J) such that 𝜔(f , 𝜋) < 𝜀1 and 𝜔(g, 𝜘) < 𝜀2 . Consider the cover 𝜌 ≡ (E k | k ∈ K), where K ≡ I × J and E k ≡ C i ∩ D j for every k = (i, j) ∈ K. By virtue of Theorem 1 (1.3.9), the set K is countable. Since S is multiplicative, we infer that E k ∈ S. If s, t ∈ E k , then |(f + g)(s) − (f + g)(t)| ⩽ |f (s) − f (t)| + |g(s) − g(t)| ⩽ 𝜔(f , 𝜋) + 𝜔(g, 𝜘). Therefore, 𝜔(f + g, 𝜌) ⩽ 𝜔(f , 𝜋) + 𝜔(g, 𝜘) < 𝜀. Thus, f + g ∈ A(T). 4. Take m ∈ N such that 1/m < 𝜀. There are countable covers 𝜘m ≡ (G mi ∈ S | i ∈ I m ) and 𝜆 m ≡ (H mj ∈ S | j ∈ J m ) such that 𝜔(f , 𝜘m ) < 1/m and 𝜔(g, 𝜆 m ) < 1/m. Define the sets A k ≡ f −1 [[−k, k]], B l ≡ g−1 [[−l, l]], I mk ≡ {i ∈ I m | G mi ∩ A k = / ⌀}, and J ml ≡ {j ∈ J m | H mj ∩ B l = / ⌀} for every k, l ∈ N. Consider the countable collections 𝜘mk ≡ (G mi | i ∈ I mk ) and 𝜆 ml ≡ (H mj | j ∈ J ml ) covering the sets rng 𝜘mk = ⋃⟮G i | i ∈ I mk ⟯ ⊃ A k and rng 𝜆 ml = ⋃⟮H j | j ∈ J ml ⟯ ⊃ B l , respectively. It is clear that 𝜔(f , 𝜘mk ) < 1/m and 𝜔(g, 𝜆 ml ) < 1/m.

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Obviously, sup{|f (t)| | t ∈ rng 𝜘mk } ⩽ k+1/m ≡ a mk and sup{|g(t)| | t ∈ rng 𝜆 ml } ⩽ l + 𝜀 ≡ b ml . Take the natural numbers p ml ≡ 2ml + 2 and q mk ≡ 2mk + 2. For every ordered pair z ≡ (k, l) ∈ N2 , define the set W mz ≡ I mk × J ml × I p ml × J q mk . For every suite w ≡ (i1 , j1 , i2 , j2 ) ∈ W mz , consider the set Q mzw ≡ G mi1 ∩ H mj1 ∩ G p ml i2 ∩ H q mk j2 ∈ S. If s, t ∈ Q mzw , then |(fg)(t) − (fg)(s)| ⩽ |f (t)| |g(t) − g(s)| + |g(s)| |f (t) − f (s)| ⩽ a mk /q mk + b ml /p ml ⩽ 1/m. Consequently, 𝜔(fg, Q mzw ) ⩽ 1/m. Define the set C m ≡ ⋃⟮{z} × W mz | z ∈ N2 ⟯, it is countable By virtue of Theorem 1 (1.3.9). Then, the countable collection 𝜇m ≡ (S mc ∈ S | c ∈ C m ), where S mc ≡ Q mzw for c ≡ (z, w), covers the set T and we have 𝜔(fg, 𝜇m ) ⩽ 1/m < 𝜀. 5. For every m ∈ N, there is a countable cover 𝜋m ≡ (S mi m ∈ S | i m ∈ I m ) such that 𝜔(f , 𝜋m ) < 1/m. Consider the sets X a ≡ {t ∈ T | |f (t)| > 1/a} for all a ∈ N. It is clear that ⋃⟮X a | a ∈ N⟯ = T. Take a number p ∈ N such that 1/p < 𝜀. For every a ∈ N, put m ap ≡ 9pa2 . Then, m ap > 2a and 4a2 /m ap < 1/2p. Consider the set J m ap ≡ {i m ap ∈ I m ap | S m ap i m ∩ X a ≠ ⌀} and the collection 𝜘m ap ≡ ap ⟮S m ap i m | i m ap ∈ J m ap ⟯. It is clear that X a ⊂ rng 𝜘m ap . ap

Let i m ap ∈ I m ap . Take s󸀠 , t󸀠 ∈ S m ap i m . Since S m ap i m ∩ X a ≠ ⌀, there exists s, t ∈ ap ap S m ap i m ∩ X a . ap

If f (s) > 0, then f (s) = |f (s)| > 1/a. It follows from |f (s)−f (s󸀠 )| < 1/m ap < 1/2a that f (s󸀠 ) > f (s) − 1/2a > 1/a − 1/2a = 1/2a > 0. Hence, |f (s󸀠 )| = f (s󸀠 ) > 1/2a. If f (s) < 0, then −f (s) = |f (s)| > 1/a implies f (s󸀠 ) < f (s) + 1/2a < −1/a + 1/2a = −1/2a < 0. Hence, |f (s󸀠 )| = −f (s󸀠 ) > 1/2a. Thus, |f (s󸀠 )| > 1/2a. Similarly, |f (t󸀠 )| > 1/2a. Hence, |1/f (s󸀠 ) − 1/f (t󸀠 )| = |f (s󸀠 ) − f (t󸀠 )|/(|f (s󸀠 )| |f (t󸀠 )|) < 2a ⋅ 2a/m ap < 1/2p. Consider the set J p ≡ ⋃⟮{a}×J m ap | a ∈ N⟯. It is countable by Theorem 1 (1.3.9). If j ∈ J p , then j = (a, i m ap ) for some a ∈ N and some i m ap ∈ J m ap . Denote S m ap i m by S j . Thus, ap

we obtain the countable collection 𝜌p ≡ (S j ∈ S | j ∈ J p ) such that 𝜔(1/f , 𝜌p ) < 1/p. Since, according to Proposition 1 (1.1.13), we have T = ⋃⟮X a | a ∈ N⟯ ⊂ ⋃⟮⋃⟮S m ap i m | ap

i m ap ∈ J m ap ⟯ | a ∈ N⟯ = (S j | j ∈ J p ) , the collection 𝜌p covers the set T. 6. According to the Archimedes principle (Lemma 13 (1.4.3)), there is n ∈ N such that 1/n < 𝜀m /2. Divide the set R+ = [0, ∞[ by points x k = k/n, k ∈ 𝜔. Then, 0 = x0 < x1 < . . . and x k+1 − x k = 1/n. Take a countable cover 𝜋 ≡ (S i ∈ S | i ∈ I) such that 𝜔(f , 𝜋) < 1/n. Consider the / ⌀} is nonsets X k ≡ f −1 [[x k , x k+1 ]]. For every i ∈ I, the set K i ≡ {k ∈ 𝜔 | S i ∩ X k = empty. According to Theorem 2 (1.2.6) and Proposition 2 (1.2.2) the set K i has the unique smallest element k i ≡ sm K i ∈ K i . Suppose s ∈ S i and take a point t ∈ S i ∩ X k i . Then, f (s) = f (s) − f (t) + f (t) < 𝜔(f , S i ) + x k i +1 and f (s) ⩾ −𝜔(f , S i ) + x k i . Hence, m

m

m m x k i < √f (s) < √x √ k i +1 + 1/n for every s ∈ S i . This implies the inequality 𝜔( √ f , S i ) < m m √x k i +1 + 1/n − √ x k i . If k i ⩾ 2, then using Corollary 1 to Lemma 1 (1.4.6), we get m

m m m m m m m 𝜔(√f , S i ) < √x k i + 1/n − √ x k i < √ x k i − √x k i − 1/n = √x k i −1 + 1/n − √ x k i −2 < √ x k i −1 −

108 | 2.3 Families of measurable and distributable functions on a descriptive space

m

m m m m m m √𝜀m = 𝜀. If k i = 0, √x k i −2 − 1/n = √x k i −2 + 1/n − √ x k i −3 < . . . < √ x 2 − √ x 0 = √2/n < m m m m m m m x0 = √ x1 − √ x0 < √ x2 − √ x0 < 𝜀. If k i = 1, then then 𝜔(√ f , S i ) < √x0 + 1/n − √ m m m m m m m 𝜔(√ f , S i ) < √x1 + 1/n − √x1 = √x2 − √x1 < √x2 − √x0 < 𝜀. Thus, we obtain that m 𝜔(√ f , 𝜋) < 𝜀.

Corollary 1. Let S be a foundation on T. Then, the family D(T, S) satisfies conditions 1, 2, 3, 5, and 7 from 2.2.4.

2.3.3 The pointwise order between measurable and distributable functions Define the relations f ⩽ g, f < g, and f ≪ g for f , g ∈ M(T, S) and for f , g ∈ D(T, S) as pointwise relations induced from F(T). Proposition 1. Let S be a 𝜎-foundation on T, f ∈ M(T, S), and (f i ∈ M(T, S) | i ∈ I) be a finite collection. Then, 1) if g = sup (f i | i ∈ I) in F(T), then g ∈ M(T, S); 2) if g = inf (f i | i ∈ I) in F(T), then g ∈ M(T, S); 3) f+ , f− , |f | ∈ M(T, S). If S is, besides, 𝛿-multiplicative, then we can take countable collections (f i | i ∈ I). Proof. 1. Consider the function g ∈ M(T, S) such that g = sup (f i | i ∈ I) in F(T). Let t ∈ R ≡ g −1 []x, y[]. Then, x < sup (f i (t) | i ∈ I) < y − 1/n < y for some n ∈ N. Hence, f i (t) < y − 1/n for all i ∈ I and there is i ∈ I such that f i (t) > x. This means that t ∈ ⋃⟮ f i−1 []x, ∞[] | i ∈ I⟯ ∩ ⋃⟮⋂⟮ f i−1 [] − ∞, y − 1/n[] | i ∈ I⟯ | n ∈ N⟯ ≡ S. Therefore, R ⊂ S. Conversely, if t ∈ S, then f i (t) > x. Besides, there is n ∈ N such that f i (t) < y − 1/n for all i ∈ I. Therefore, g(t) > x and g(t) < y − 1/n, i. e. t ∈ R. Thus, R = S ∈ S, what implies g ∈ M(T, S). 2. Since inf (f i | i ∈ I) = − sup (−f i | i ∈ I), assertion 2 follows from (1) and assertion 1 of Theorem 1 (2.3.2). 3. This assertion follows from formulae f+ = f ∨ 0, f− = f ∧ 0, |f | = f+ − f− , the previous assertions and Theorem 1 (2.3.2). Corollary 1. Let S be a 𝜎-foundation on T. Then, M(T, S) is a lattice-ordered linear space. Corollary 2. Let S be a 𝜎-foundation on T. Then, the family M(T, S) satisfies condition 4 from 2.2.4.

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Corollary 3. Let S be a 𝜎-foundation on T. Then, finite exact bounds in F(T) and M(T, S) coincides. If S is, besides, 𝛿-multiplicative, then countable exact bounds in F(T) and M(T, S) coincides. Corollary 4. Let S be a 𝛿-multiplicative 𝜎-foundation on T, f ∈ M(T, S), and (f n ∈ M(T, S) | n ∈ 𝜔) is a sequence. Then, the following conclusions are equivalent: 1) f = o-lim (f n | n ∈ 𝜔) in F(T); 2) f = o-lim (f n | n ∈ 𝜔) in M(T, S). Proof. (1) ⊢ (2). Suppose f = o-lim (f n | n ∈ 𝜔) in F(T). Then, by Lemma 4 (1.1.15) we have f = sup (inf (f p | p > n) | n ∈ 𝜔) = inf (sup (f p | p > n) | n ∈ 𝜔) in F(T). By virtue of Proposition 1, these equalities are also valid in M(T, S). Then, again by Lemma 4 (1.1.15) f = o-lim (f n | n ∈ 𝜔) in M(T, S). (2) ⊢ (1). The proof is completely similar. Proposition 2. Let S be a foundation on T, f , g ∈ D(T, S). Then, f ∨ g, f ∧ g ∈ D(T, S) and f+ , f− , |f | ∈ D(T, S). Proof. Let 𝜀 > 0. There are countable covers 𝜋 ≡ (C i ∈ S | i ∈ I) and 𝜘 ≡ (D j ∈ S | j ∈ J) such that 𝜔(f , 𝜋) < 𝜀/2 and 𝜔(g, 𝜘) < 𝜀/2. Consider the cover 𝜌 ≡ (E k | k ∈ K), where K ≡ I × J and E k ≡ C i ∩ D j for every k = (i, j) ∈ K. By virtue of Theorem 1 (1.3.9), the set K is countable. Since S is multiplicative, we infer that E k ∈ S. If s, t ∈ C i ∩ D j , then by Birkhoff inequality (Corollary 1 to Theorem 4 (1.4.5)) |(f ∨ g)(s) − (f ∨ g)(t)| ⩽ |f (s) ⊻ g(s) − f (s) ⊻ g(t)| + |f (s) ⊻ g(t) − f (t) ⊻ g(t)| ⩽ |g(s) − g(t)| + |f (s) − f (t)| ⩽ 𝜔(g, 𝜘) + 𝜔(f , 𝜋) < 𝜀. Thus, we get f ∨ g ∈ D(T, S). Completely in the same way, we check that f ∧ g ∈ D(T, S). By the definition, f + ≡ f ∨ 0, f− ≡ f ∧ 0. Therefore, these functions belong to D(T, S). By Lemma 1 (2.2.2) |f | = f+ − f− , so that |f | ∈ D(T, S). Corollary 1. Let S be a foundation on a set T and (f i ∈ D(T, S) | i ∈ I) be a finite collection. Then, sup (f i | i ∈ I) ∈ D(T, S) and inf (f i | i ∈ I) ∈ D(T, S). The proof is by induction. Corollary 2. Let S be a foundation on T. Then, the family D(T, S) satisfies condition 4 from 2.2.4. 2.3.4 The pointwise and uniform convergences of sequences of measurable and distributable functions Here we consider the pointwise and uniform convergences introduced in 2.2.3 for sequences of measurable and distributable functions.

110 | 2.3 Families of measurable and distributable functions on a descriptive space

Pointwise convergence Lemma 1. Let S be a 𝛿-multiplicative 𝜎-foundation on T, f ∈ M(T, S), and (f n ∈ M(T, S) | n ∈ 𝜔) is a sequence. Then, the following conclusions are equivalent: 1) f = p-lim (f n | n ∈ 𝜔); 2) f = o-lim (f n | n ∈ 𝜔) in M(T, S). Proof. By Lemma 2 (2.2.3) f = p-lim (f n | n ∈ 𝜔) iff f = o-lim (f n | n ∈ 𝜔) in F(T). By Corollary 4 to Proposition 1 (2.3.3) this is equivalent to the equality f = o-lim (f n | n ∈ 𝜔) in M(T, S). Lemma 2. Let S be an ensemble on a set T with the co-ensemble R ≡ co-S, f ∈ F(T), and f = p-lim (f n | n ∈ 𝜔) for some sequence (f n ∈ M(T, S) | n ∈ 𝜔). Then, 1) f −1 [[x, y]] ∈ S𝜎𝛿 ∩ R𝜎𝛿 for every x ⩽ y in R; 2) f −1 []x, y[] ∈ S𝛿𝜎 ∩ R𝛿𝜎 for every x < y in R. Proof. 1. By Theorem 1 (2.2.3), −1 [I n ] | k ∈ 𝜔⟯ | n ∈ N⟯, E ≡ f −1 [[x, y]] = ⋂⟮⋃⟮f n+k −1 [I n ] belongs where either I n =]x − n1 , y + n1 [ or I n = [x − n1 , y + n1 ]. By the condition, f n+k either to S or to R. Consequently, E ∈ S𝜎𝛿 ∩ R𝜎𝛿 . 2. By virtue of Corollary 1 to Theorem 1 (2.2.3), we have the equality

F ≡ f −1 []x, y[] = −1 −1 [J n ] | k ∈ 𝜔⟯ | n ∈ N⟯ ∩ ⋃⟮⋂⟮f n+k [K n ] | k ∈ 𝜔⟯ | n ∈ N⟯, = ⋃⟮⋂⟮f n+k where either J n =]x + n1 , ∞[ and K n =] − ∞, y − n1 [ or J n = [x + n1 , ∞] and K n = [−∞, −1 −1 [J n ] and f n+k [K n ] belong either to S or to R. Consey − n1 ]. By the condition, f n+k quently, F = G ∩ H for some sets G and H belonging to S𝛿𝜎 or to R𝛿𝜎 . By Lemma 1 (2.1.1) these ensembles are multiplicative. Therefore, F ∈ S𝛿𝜎 ∩ R𝛿𝜎 . Corollary 1. In the conditions of Lemma 2, if the ensemble S is 𝜎-additive, then 1) f −1 [[x, y]] ∈ S𝛿 ∩ R𝜎𝛿 for every x ⩽ y in R; 2)

f −1 []x, y[] ∈ S𝛿𝜎 ∩ R𝜎 for every x < y in R.

Corollary 2. In the conditions of Lemma 2, if the ensemble S is 𝜎-additive, then f ∈ M(T, S𝛿𝜎 ) ∩ M(T, R𝜎 ). Proof. The assertion follows from Corollary 1 and Lemma 1 (2.3.1).

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Corollary 3. Let S is a 𝛿-multiplicative 𝜎-foundation (in particular, a 𝜎-algebra), f ∈ F(T), and f = p-lim (f n | n ∈ 𝜔) for some sequence (f n ∈ M(T, S) | n ∈ 𝜔). Then, f ∈ M(T, S). Proposition 1. Let S is a 𝛿-multiplicative 𝜎-foundation (in particular, a 𝜎-algebra) on a set T and (f i ∈ M(T, S) | i ∈ I) be a denumerable collection bounded above. Then, there is a function f ∈ M(T, S) such that f (t) = sup (f i (t) | i ∈ I), i. e. f = sup (f i | i ∈ I) in the ordered set ⟮M(T, S), ⩽⟯. The similar assertion is valid for collections bounded below and their infima. Proof. Take some bijection u : 𝜔 I and for every n ∈ 𝜔 define the function g n ≡ sup (f u(m) | m ∈ n + 1) ∈ M(T, S). It is clear that (g n (t) | n ∈ 𝜔) ↑ for every t ∈ T. By Proposition 2 (1.4.4) there is a function f ∈ F(T) such that f (t) = lim (g n (t) | n ∈ 𝜔). By virtue of Corollary 3 to Lemma 2, f ∈ M(T, S). By Lemma 7 (1.4.7) f (t) = sup (g n (t) | n ∈ 𝜔). Evidently, f ⩾ f i for every i ∈ I. Consider the sets I n ≡ u[n+1]. Then, I = ⋃⟮I n | n ∈ 𝜔⟯. Using Proposition 1 (1.1.15), we infer that f (t) = sup (f i (t) | i ∈ I) for every t ∈ T. According to Proposition 1 (2.3.3), f = sup (f i | i ∈ I) in M(T, S). Proposition 1 means that in the case of a 𝛿-multiplicative 𝜎-foundation S the ordered set ⟮M(T, S), ⩽⟯ is 𝜔-Dedekind complete in the sense of 1∘ (2.2.6). Uniform convergence Proposition 2. Let S be a 𝜎-additive ensemble on T, (f n ∈ M(T, S) | n ∈ N) be a net, f ∈ F(T), and f = u-lim (f n | n ∈ N). Then, f ∈ M(T, S). If, besides, all f n ∈ M b (T, S), then f ∈ M b (T, S). Proof. Let l ∈ N, x, y ∈ R, and x < y. By the condition, there is m such that |f − f p | < (1/l)1 for all p ⩾ m. Consider the sets S ≡ f −1 []x, y[] and S l ≡ f −1 [[x + 3/l, y − 3/l]]. For every k ∈ N, put r k ≡ k/(2l) and R k ≡ f m−1 []r k−1 , r k+1 []. It is clear that the sets R k belong to S and cover the set T. Consider the sets Q l ≡ ⋃⟮R k | k ∈ Z ∧ R k ∩S l ≠ ⌀⟯ ∈ S. It is easy to see that S l ⊂ Q l . Suppose t ∈ Q l . Hence, there are k ∈ Z and s ∈ T such that t ∈ R k and s ∈ R k ∩S l . Then, f (t) < f m (t)+1/l < f m (s)+2/l < f m (s)+3/l ⩽ y−3/l+3/l = y and f (t) > f m (t) − 1/l > f m (s) − 2/l > f m (s) − 3/l ⩾ x + 3/l − 3/l = x guarantee that t ∈ S. Therefore, Q l ⊂ S and S = ⋃⟮S l | l ∈ N⟯ ⊂ ⋃⟮Q l | l ∈ N⟯ ⊂ S. Thus, S = ⋃⟮Q l | l ∈ N⟯ ∈ S and f ∈ M(T, S). In the case of bounded functions, f n the function f is bounded owing to Lemma 3 (2.2.3). Corollary 1. Let S be a 𝜎-additive ensemble on T. Then, the families M(T, S) and M b (T, S) satisfy condition 6 from 2.2.4.

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Corollary 2. Let S be a 𝜎-foundation on T. Then, the family M(T, S) is normal and the family M b (T, S) is boundedly normal (in the sense of 2.2.4). Proof. The assertion follows from Corollary 2 to Theorem 1 (2.3.2), Corollary 2 to Proposition 1 (2.3.3), and Corollary 1. Theorem 1 (the Dini theorem). Let S be an ensemble on a set T, P be an S-compact subset of T, and (f n ∈ M(T, S) | n ∈ N) be a decreasing net such that p-lim (f n | n ∈ N) = 0. Then, u-lim (f n |P | n ∈ N) = 0|P. Proof. For 𝜀 > 0, consider the increasing net 𝜋 ≡ (S n ∈ S | n ∈ N) such that S n ≡ {t ∈ T | 0 ⩽ f n (t) < 𝜀} = {t ∈ T | −𝜀 < f n (t) < 𝜀} = f n−1 []x, y[] ∈ S. It is clear that 𝜋 is a cover of T. Since P is S-compact, there is a finite subset M ⊂ N such that P ⊂ ⋃⟮S m | m ∈ M⟯. Since 𝜋 is increasing, we can choose some k ∈ N such that k ⩾ m for every m ∈ M. It 󵄨 󵄨 is clear that P ⊂ S k . Since (f n | n ∈ N) is decreasing, we conclude that 󵄨󵄨󵄨f l |P󵄨󵄨󵄨 ⩽ 𝜀1|P for all l ⩾ k. Corollary 1. Let S be a 𝜎-foundation on a set T, P be an S-compact subset of T, f ∈ M(T, S), and (f n ∈ M(T, S) | n ∈ N) be an increasing net such that p-lim (f n | n ∈ N) = f . Then, u-lim (f n |P | n ∈ N) = f |P. Proof. By Theorem 1 (2.3.2), we have g n ≡ f − f n ∈ M(T, S). Since (g n | n ∈ N) ↓ 0, we can apply Theorem 1. Proposition 3. Let S be an ensemble on T, f ∈ F(T), and f = u-lim (f n | n ∈ N) for some net (f n ∈ D(T, S) | n ∈ N). Then, f ∈ D(T, S). Proof. For any 𝜀 > 0, take m ∈ N such that |f − f n | < (𝜀/3)1 for every n ⩾ m. Consider a countable cover 𝜋 ≡ ⟮S i ∈ S | i ∈ I⟯ such that 𝜔(f m , 𝜋) < 𝜀/3. Then, |f (s) − f (t)| = |f (s) − f m (s) + f m (s) − f m (t) + f m (t) − f (t)| ⩽ |f (s) − f m (s)| + |f m (s) − f m (t)| + |f m (t) − f (t)| < 𝜀 for all s, t ∈ S i . Therefore, 𝜔(f , 𝜋) < 𝜀 and f ∈ D(T, S). Corollary 1. Let S be an ensemble on T. Then, the family D(T, S) satisfies condition 6 from 2.2.4. Corollary 2. Let S be a foundation on T. Then, the family D(T, S) is normal (in the sense of 2.2.4). Proof. The assertion follows from Corollary 1 to Theorem 2 (2.3.2), Corollary 2 to Proposition 2 (2.3.3), and Corollary 1. Lemma 3. Let S be a 𝜎-additive ensemble on T. Then, every inner uniformly convergent sequence in M(T, S) [M b (T, S)] has a uniform limit in M(T, S) [M b (T, S)].

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Proof. Let (f n ∈ M(T, S) [M b (T, S)] | n ∈ 𝜔) be inner uniformly convergent. By Lemma 6 (2.2.3) there is f ∈ F(T) such that f = u-lim (f n | n ∈ 𝜔). By Proposition 2 f ∈ M(T, S) [M b (T, S)]. Corollary 1. Let S be a 𝜎-additive ensemble on T. Then, every inner uniformly convergent sequence in qSt(T, S) [qStc (T, S)] has a uniform limit in M b (T, S) [M(T, S)]. Proof. By Lemma 2 (2.3.1) qSt(T, S) ⊂ M b (T, S) and qStc (T, S) ⊂ M(T, S). Corollary 2. Let S be a 𝜎-algebra on T. Then, every inner uniformly convergent sequence in St(T, S) [Stc (T, S)] has a uniform limit in M b (T, S) [M(T, S)]. Proof. By Corollary 1 to Lemma 2 (2.3.1) St(T, S) ⊂ M b (T, S) and Stc (T, S) ⊂ M(T, S).

Measurable functions as limits of sequences of step functions Lemma 4. Let S be a 𝛿-multiplicative ensemble on a set T, f ∈ M(T, S), and rng f is countable. Then, there is a sequence (f n ∈ St(T, S) | n ∈ 𝜔) (see 2.2.4) such that f = p-lim (f n | n ∈ 𝜔). / x j for i = / j. Take the sets S i ≡ f −1 [{x i }] ∈ S Proof. Let rng f = {x i | i ∈ 𝜔} and x i = and the step functions f n such that f n (t) ≡ x i for t ∈ S i and i ∈ n and f n (t) ≡ 0 for t ∈ T\ ⋃⟮S i | i ∈ n⟯. Proposition 4. Let S be a ring [𝜎-ring] on T. Then, for every function f ∈ M(T, S𝜎 ) [f ∈ M b (T, S)] there is a sequence (f n ∈ qStc (T, S) | n ∈ 𝜔) [(f n ∈ qSt(T, S) | n ∈ 𝜔)] such that f = u-lim (f n | n ∈ 𝜔) and |f n | ⩽ |f | for every n ∈ 𝜔. Proof. By virtue of Proposition 1 (2.3.1) [its Corollary 1] for every n ∈ N, there is a countable [finite] cover (R ni ∈ S | i ∈ I n ) of T such that 𝜔(f , R ni ) < 1/(2n + 2) for every i ∈ I n . Use of Lemma 8 (2.1.1) yields that there is a countable [finite] partition (S nk ∈ S | k ∈ K n ⊂ 𝜔) such that S nk ⊂ R ni for some i. Then, 𝜔(f , S nk ) < 1/(2n + 2) for every k ∈ K n . Consider the numbers x nk ≡ inf(f (t) | t ∈ S nk ) and y nk ≡ inf(|f (t)| | t ∈ S nk ). Put f0 ≡ 0 and for every n ∈ N define the quite S-step function f n setting f n (t) ≡ x nk y nk /|x nk | for every t ∈ S nk and every k ∈ K n such that x nk ≠ 0 and f n (t) ≡ 0 for every t ∈ S nk and every k ∈ K n such that x nk = 0. Then, we have |f n | ⩽ |f | for every n. Take any n ∈ N and t ∈ T. Then, t ∈ S nk for some k ∈ K n . If x nk = 0, then |f (t) − f n (t)| = |f (t) − x nk | ⩽ 𝜔(f , S nk ) < 1/(n + 1). If x nk ≠ 0, then |f (t) − f n (t)| ⩽ |f (t) − x nk | + |x nk − x nk y nk /|x nk || ⩽ 𝜔(f , S nk ) + ||x nk | − y nk | < 1/(2n + 2) + 𝜔(f , S nk ) < 1/(n + 1). This means that f = u-lim (f n | n ∈ 𝜔). Corollary 1. Let S be a 𝜎-ring on T. Then, M(T, S) [M b (T, S)] is a uniform closure of q c St (T, S) [qSt(T, S)] in F(T).

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Proof. It follows from Proposition 4 and Corollary 1 to Lemma 3. Corollary 2. Let S be a 𝜎-ring on T, f ∈ M(T, S). Then, there is a sequence (f n ∈ M(T, S) | n ∈ 𝜔) such that f = u-lim (f n | n ∈ 𝜔) and rng f n is countable and |f n | ⩽ |f | for every n ∈ 𝜔. Proof. By Lemma 2 (2.3.1) qStc (T, S) ⊂ M(T, S). Corollary 3. Let S be a 𝜎-algebra on T. Then, for every function f ∈ M b (T, S) there is a sequence of S-measurable S-step functions uniformly convergent to f . This Corollary means that for a 𝜎-algebra S the family St(T, S) is uniformly dense in M b (T, S) in the sense of 2.2.4. Theorem 2. Let S be a 𝜎-algebra on T and f ∈ M(T, S). Then, there exists a sequence (f n ∈ St(T, S) ⊂ M(T, S) | n ∈ 𝜔) such that (|f n | | n ∈ 𝜔) ↑ |f | and f = p-lim (f n | n ∈ 𝜔). If f is bounded, then f n can be chosen such that f = u-lim (f n | n ∈ 𝜔). If f ⩾ 0, then in both cases, f n can be chosen such that 0 ⩽ f n ⩽ f n+1 ⩽ f . Proof. Put f0 ≡ 0. First, consider the case f ⩾ 0. For each n ∈ N and k ∈ n2n + 1, take the sets R nk ≡ f −1 [[k/2n , (k + 1)/2n [] and S n ≡ f −1 [[n, ∞[]. Since S is 𝜎-additive and closed under the complement, according to Lemma 1 (2.3.1), we have R nk ∈ S and S n ∈ S. Consider the functions f n ≡ ∑ (k/2n 𝜒(R nk ) + n𝜒(S n ) | k ∈ n2n + 1) ∈ St(T, S). It is easy to see that 0 ⩽ f n ⩽ f n+1 ⩽ f , |f n | ⩽ n1, and |f (t) − f n (t)| < 1/2n for all t ∈ T \ S n . This implies f = p-lim (f n | n ∈ 𝜔). If f is bounded, i. e. there exists b ∈ R such that |f | ⩽ b1, then |f (t) − f n (t)| < 1/2n for all n ⩾ b and t ∈ T. Therefore, f = u-lim (f n | n ∈ 𝜔). Now, consider the general case. For f+ ≡ f ∨ 0 and f− ≡ f ∧ 0, we have sequences (g n ∈ St(T, S)+ | n ∈ 𝜔) and (h n ∈ St(T, S)+ | n ∈ 𝜔) such that (g n | n ∈ 𝜔) ↑, (h n | n ∈ 𝜔) ↑, g n ⩽ f+ , h n ⩽ −f− , f+ = p-lim (g n | n ∈ 𝜔), and, finally, −f− = p-lim (h n | n ∈ 𝜔). Take the functions f n ≡ g n − h n . They belong to St(T, S). By virtue of Proposition 1 (2.2.4). According to Proposition 1 (2.2.3) f = f+ + f− = f+ − (−f− ) = p-lim (f n | n ∈ 𝜔) and |f | = p-lim (|f n | | n ∈ 𝜔). It follows from coz g n ∩coz h n ⊂ coz f+ ∩ coz(−f− ) = ⌀ that g n ∧ h n = 0. Applying assertion 5 of Lemma 5 (1.4.5) and assertion 1 of Lemma 6 (1.4.5), we get |f n (t)| = |g n (t) − h n (t)| = g n (t) ⊻ h n (t) − g n (t) ⊼ h n (t) = g n (t) ⊻ h n (t) ⩽ g n (t) ⊻ h n+1 (t) ⩽ g n+1 (t) ⊻ h n+1 (t) = |f n+1 (t)| in each point t ∈ T. Theorem 3. Let S be a perfect ensemble on a set T with the co-ensemble R. Then, for every function f ∈ M(T, R𝜎 ) there is a sequence (f n ∈ qStc (T, R𝜎 ) | n ∈ 𝜔) such that

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f = u-lim (f n | n ∈ 𝜔). Moreover, if f is bounded, then we can reckon that every f n ∈ St(T, R𝜎 ).

q

Proof. Consider the rational numbers r ni ≡ i/(n + 1) for all n ∈ 𝜔 and i ∈ 𝜔. By the condition, K ni ≡ f −1 []r n,i−1 , r n,i+1 [] ∈ R𝜎 . Since the collection (K ni ∈ R𝜎 | i ∈ Z) covers the set T, by Theorem 1 (2.1.1) there is a disjoint partition (H ni ∈ R𝜎 | i ∈ Z) of T such that H ni ⊂ K ni for every i. Therefore, we can define correctly a function f n on T setting f n (t) ≡ r n,i−1 for every t ∈ H ni . This means that f n ∈ qStc (T, R𝜎 ). If t ∈ T, then t ∈ H ni for some i. This implies 2 0 ⩽ f (t) − f n (t) ⩽ r n,i+1 − r n,i−1 = n+1 for all t ∈ T. Thus, f = u-lim (f n | n ∈ 𝜔). If f is bounded, then the set J n = {j ∈ Z | K nj ≠ ⌀} is finite. Therefore, the function f n takes its values in the finite set {r n,i−1 | i ∈ J n }. Hence, f ∈ qSt(T, R𝜎 ).

Essentiality of the 𝜎-additivity of S for good properties of M(T , S) The following lemma shows that the condition of 𝜎-additivity of S in Theorem 1 (2.3.2) and in Proposition 2 is crucially essential, i. e. for some non-𝜎-additive foundations S the family M(T, S) may not be closed with respect at least to addition and uniform convergence, and therefore, M(T, S) [M b (T, S)] may not be normal [boundedly normal]. Lemma 5. There is a foundation K on T ≡ [0, 1[⊂ R, functions h󸀠 , h󸀠󸀠 ∈ M b (T, K) and f , g ∈ F b (T), and a sequence (f n ∈ M b (T, K) | n ∈ N) such that g = h󸀠 − h󸀠󸀠 , f = u-lim (f n | n ∈ N), and f , g ∈ ̸ M(T, K). Proof. Define the topological space ⟮T, G⟯ setting T ≡ [0, 1[⊂ R and G ≡ {[0, b[ | 0 < b ⩽ 1} ∪ {⌀}. Then, F ≡ co-G = {[a, 1[ | 0 ⩽ a < 1}. By Corollary 1 to Lemma 2 (2.1.1) K ≡ K(T, G) = {G ∩ F | G ∈ G ∧ F ∈ F} = {[a, b[ | 0 ⩽ a < b ⩽ 1} ∪ {⌀}. It is easy to see that K is a foundation. For every n ∈ 𝜔, put Δ n ≡ [2−(n+1) , 2−n [. For every n ∈ N, define the function f n : T → [0, 1] setting f n (t) ≡ 2−k for all t ∈ Δ k , k ∈ n, and f n (t) ≡ 0 otherwise. Take n ∈ N. If x ⩾ 1 or y ⩽ 0, then f n−1 []x, y[] = ⌀ ∈ K. If x < 1 and y > 0, then f n−1 []x, y[] = [a, b[ ∈ K or f n−1 []x, y[] = ⌀ ∈ K. Hence, f n ∈ M b (T, K). Consider the functions h󸀠 ≡ f1 , h󸀠󸀠 ≡ −f2 , and g ≡ h󸀠 + h󸀠󸀠 . Since g(t) = −1/2 for every t ∈ Δ 1 and g(t) = 0 otherwise, we see that g−1 [] − 1/3, 1/3[] = T\Δ 1 = [0, 1/4[∪[1/2, 1[ ∈ ̸ K, and, therefore, g ∈ ̸ M(T, K). Define the function f : T → [0, 1] setting f (t) ≡ 2−n for all t ∈ Δ n , n ∈ 𝜔, and f (0) ≡ 0. Since f −1 []0, 2[] =]0, 1[∈ ̸ K, we see that f ∈ ̸ M(T, K). It is easy to see that f = u-lim (f n | n ∈ N). Corollary 1. There is a foundation K on T ≡ [0, 1[⊂ R such that M(T, K) ≠ D(T, K).

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2.3.5 Separability of sets by measurable and distributable functions Adding the condition of separability to the 𝜎-foundations S considered above provides the family M(T, S) by a row of new crucial properties. Lemma 1. Let S be an ensemble on a set T. Then, 1) if S is a 𝜎-foundation, then Coz M(T, S) ⊂ S; 2) if S is a 𝜎-algebra, then Coz M(T, S) = S. Proof. 1. If f ∈ M(T, S), then coz f = ⋃⟮f −1 [] − n, −1/n[] | n ∈ N⟯ ∪ ⋃⟮f −1 []1/n, n[] | n ∈ N⟯ ∈ S. 2. If S ∈ S, then T \ S ∈ S. Therefore, 𝜒(S) ∈ M(T, S). This implies S = coz 𝜒(S) ∈ Coz M(T, S). Proposition 1. Let S be a 𝜎-foundation. Then, the ensemble Coz M(T, S) is a separable perfect 𝜎-foundation. Proof. Denote Coz M(T, S) by E and M(T, S) by A. By Corollary 2 to Theorem 1 (2.3.2) the family A satisfies conditions 1, 1󸀠 , 2, 3, and 5 from 2.2.4. By Corollary 1 to Proposition 2 (2.3.4) it also satisfies condition 6. Then, according to Proposition 1 (2.2.5), we conclude that E is a 𝜎-foundation. Let X and Y are disjoint sets from Zer A = co-E, that is, X ≡ zer f and Y ≡ zer g for some functions f , g ∈ A and X ∩ Y = ⌀. By virtue of Lemma 1 (2.2.5), coz(|f | + |g|) = coz f ∪ coz g = T, where we can define the function h ≡ |f |/(|f | + |g|). By Proposition 1 (2.3.3) |f |, |g| ∈ A, then according to Theorem 1 (2.3.2), h ∈ A. By the definition, h(t) = 0 for every t ∈ X and h(t) = 1 for every t ∈ Y. Consider the functions v ≡ (h − (1/2)1)+ and u ≡ (h − (1/2)1)− . They belong to A, hence coz u, coz v ∈ E. Besides, X ⊂ coz u, Y ⊂ coz v, and coz u ∩ coz v = ⌀. Thus, E is separable. Finally, suppose X ∈ E, i. e. X = coz f , f ∈ A. Consider the functions g n ≡ (1/n)1 − |f | ∧ (1/n)1, n ∈ N. They belong to A. Therefore, |f |−1 [[1/n, ∞[] = zer g n ∈ Zer A = co-E. Consequently, X = ⋃⟮zer g n | n ∈ N⟯ ∈ (co-E)𝜎 . Thus, E is perfect. Lemma 2. Let S be a 𝜎-foundation. Then, M(T, S) = M(T, Coz M(T, S)). Proof. By Lemma 1 M(T, Coz M(T, S)) ⊂ M(T, S). Conversely, if f ∈ M(T, S), then by Lemma 3 (2.2.5) f −1 []x, y[] = coz g for the function g ≡ ((f − x1) ∨ 0) ∧ ((y1 − f ) ∨ 0) ∈ M(T, S). Consequently, f ∈ M(T, Coz M(T, S)). The following theorem generalizes Urysohn’s lemma on normal topological spaces [Kuratowski, 1966, 14.IV], which originates in Lebesgue’s construction of the

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Cantor staircase (the Lebesgue singular measure) on a segment (see, e. g., [Semadeni, 1971, Ch. II, 8.3.2]). Theorem 1 (the Lebesgue – Urysohn theorem on the Cantor staircase). Let S be a separable foundation on a set T. Then, for every non-empty disjoint sets A, B ∈ R there is a function f ∈ D b (T, S) such that 0 ⩽ f ⩽ 1, f [A] = {0}, and f [B] = {1}. Proof. 1. Consider the set K n ≡ 2n , i. e. K n = {0, 1, . . . , 2n − 1}. Define by induction two sequences (𝛾n | n ∈ 𝜔) and (𝛿n | n ∈ 𝜔) of finite collections 𝛾n ≡ (G nk ∈ S | k ∈ K n ) and 𝛿n ≡ (F nk ∈ R | k ∈ K n ) with the following properties: 1) A ⊂ F n,k−1 ⊂ G nk ⊂ F nk ⊂ G n,k+1 ⊂ B󸀠 ≡ T \ B for every k = 1, . . . , 2n − 2; 2) G n+1,2k = G nk and F n+1,2k = F nk for every k ∈ K n . By Lemma 7 (2.1.1) for every sets P ∈ R and Q ∈ S the set U(P, Q) ≡ {(X, Y) | X ∈ S ∧ Y ∈ R ∧ (P ⊂ X ⊂ Y ⊂ Q)} is non-empty. By the choice axiom from 1.1.12 there is a mapping p : P(S × R) \ {⌀} → S × R such that p(E) ∈ E. For n = 0, consider the pair (X0 , Y0 ) ≡ p(U(A, B󸀠 )) and the collections 𝛾0 ≡ (G0k ∈ S | k ∈ 1) and 𝛿0 ≡ (F0k ∈ R | k ∈ 1) such that G00 ≡ X0 and F00 ≡ Y0 . Then, A ⊂ G00 ⊂ F00 ⊂ B󸀠 . Assume now that such collections are constructed for n. Define some collections 𝛾n+1 ≡ (G n+1,k | k ∈ K n+1 ) and 𝛿n+1 ≡ (F n+1,k | k ∈ K n+1 ) in the following way. If k = 2m, then take G n+1,2m ≡ G nm and F n+1,2m ≡ F nm for every m ∈ K n . Then, G n+1,2m ⊂ F n+1,2m . If k = 2m + 1, then consider the pair (X k , Y k ) ≡ p(U(F n+1,2m , G n+1,2m+2 )) and take G n+1,2m+1 ≡ X k and F n+1,2m+1 ≡ Y k . Then, F n+1,2m ⊂ G n+1,2m+1 ⊂ F n+1,2m+1 ⊂ G n+1,2m+2 for every m = 0, . . . , 2n − 1. It is clear that 𝛾n and 𝛿n are increasing. It can be checked by induction that G mj = G m+i,k and F mj = F m+i,k for k = j2i . It follows from this fact that if r ≡ j/2m = k/2n , then p ≡ j2n = k2m ≡ q implies G mj = G m+n,p = G n+m,q = G nk . Therefore, we can correctly define for the set D ≡ {k/2n | n ∈ 𝜔 ∧ k ∈ 2n } a collection (G r ∈ S | r ∈ D) setting G r ≡ G nk for every r ≡ k/2n . By the same way we can define a collection (F r ∈ R | r ∈ D) setting F r ≡ F nk . Then, G r ⊂ F r for every r. Let r, s ∈ D and r < s. By definition, r = j/2m , s = k/2n . Therefore, p ≡ j2n < k2m ≡ q implies G r ≡ G mj = G m+n,p ⊂ G m+n,q = G nk ≡ G s . In the same way, F r ⊂ F s . This means that these collections are increasing. 2. Now, define a function f : T → [0, 1] setting f (t) ≡ 0 for every t ∈ G0 = G n0 , f (t) ≡ 1 for every t ∈ B, and f (t) ≡ sup{r ∈ D | t ∈ ̸ G r } for the other t. Fix a number n and consider the set S k ≡ G n,k+1 \ F n,k−1 for k = 1, . . . , 2n − 2. Since S is multiplicative, S k ∈ S. Let t ∈ S k . Then, t ∈ ̸ F n,k−1 implies t ∈ ̸ G n,k−1 . Therefore, f (t) ⩾ (k − 1)/2n . Assume that f (t) > (k + 1)/2n . Then, there is r ∈ D such that f (t) ⩾ r > (k + 1)/2n and t ∈ ̸ G r , but t ∈ G n,k+1 ⊂ G r . It follows from this contradiction that f (t) ⩽ (k + 1)/2n . This implies 𝜔(f , S k ) ⩽ 2/2n .

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Take any t ∈ B󸀠 . From the chain A ⊂ G n0 ⊂ F n0 ⊂ G n1 ⊂ F n1 ⊂ G n2 ⊂ F n2 ⊂ . . . ⊂ F n,k−2 ⊂ G n,k−1 ⊂ F n,k−1 ⊂ G nk ⊂ F nk ⊂ G n,k+1 ⊂ . . . ⊂ G n𝜘 ⊂ F n𝜘 ⊂ B󸀠 , where 𝜘 ≡ 2n − 1, we infer that either t ∈ S𝜘 ≡ B󸀠 \F n,𝜘 , or t ∈ S0 ≡ G n1 , or t ∈ F nk \G nk ⊂ S k , or t ∈ G n,k+1 \F nk ⊂ S k for some k = 0, . . . , 2n − 2. In the first case, t ∈ ̸ G n𝜘 implies f (t) ⩾ 𝜘/2n . Therefore, 𝜔(f , S𝜘 ) ⩽ 1/2n and S𝜘 ∈ S. In the second case, f (t) ⩽ 1/2n . In fact, assume that f (t) > 1/2n . Then, there is r ∈ D such that f (t) ⩾ r > 1/2n and t ∈ ̸ G r , but t ∈ G n1 ⊂ G r . Thus, we come to the contradiction. Therefore, 𝜔(f , S0 ) ⩽ 1/2n . In the other cases, t ∈ S k for some k = 1, . . . , 2n − 2. As a result, we obtain that B󸀠 ⊂ ⋃⟮S k | k ∈ K n ⟯ and 𝜔(f , S k ) ⩽ 2/2n . Finally, if t ∈ S𝜘+1 ≡ T\F n𝜘 , then t ∈ ̸ G n𝜘 implies f (t) ⩾ 𝜘/2n . Therefore, 𝜔(f , S𝜘+1 ) ⩽ 1/2n and S𝜘+1 ∈ S. Since B ⊂ S𝜘+1 , we get the cover (S k ∈ S | k ∈ 2n + 1) of T with the oscillations 𝜔(f , S k ) ⩽ 2/2n for all k. Thus, f ∈ D b (T, S). Corollary 1. Let S be a separable 𝜎-foundation on a set T. Then, for every non-empty disjoint sets A, B ∈ R there is a function f ∈ M b (T, S) such that 0 ⩽ f ⩽ 1, f [A] = {0}, and f [B] = {1}. Proof. Theorem 1 gives a function f ∈ D b (T, S). Using Corollary 1 to Proposition 1 (2.3.1), we conclude that f ∈ M b (T, S). Corollary 2. Let S be a separable 𝜎-foundation on a set T. Then, for every non-empty disjoint sets A, B ∈ R and for every real numbers a ⩽ b there is a function f ∈ M b (T, S) such that rng f ⊂ [a, b], f [A] = {a}, and f [B] = {b}. Proof. Take the function f from Corollary 1. Then, the function f 󸀠 ≡ a1 + (b − a)f has the necessary properties. Now, we shall prove that all a-foundations (i. e. separable perfect 𝜎-foundations) can be represented in the form Coz M(T, S) from Proposition 1. The following result was obtained by A. D. Alexandrov [Alexandrov, 1941]. Theorem 2 (the Alexandrov theorem). Let S be an a-foundation on a set T. Then, for every set S ∈ S there is a positive function f ∈ M b (T, S) such that S = coz f . Proof. By the condition, S = ⋃⟮R n | n ∈ 𝜔⟯ for some sequence (R n ∈ R | n ∈ 𝜔), where R ≡ co-S. By Corollary 1 to Theorem 1 the set U n ≡ {f ∈ M b (T, S) | 0 ⩽ f ⩽ 1 ∧ (f [T\S] = {0}) ∧ (f [R n ] = {1})} is non-empty for every n ∈ 𝜔. By the choice axiom from 1.1.12 there is a mapping p : P( Map(T, [0, 1]))\{⌀} → Map(T, [0, 1]) such that p(E) ∈ E. For every n, take the function f n ≡ p(U n ). Consider the sequence (g n ∈ M b (T, S) | n ∈ 𝜔) such that g n ≡ ∑ ((1/2)m f m | m ∈ n). Then, for every t ∈ T, we have 0 ⩽ g n+1 (t) − g n (t) ⩽ (1/2)n . Therefore, by Lemma 4 (1.4.8) the sequence

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(g n | n ∈ 𝜔) is inner uniformly convergent. By Lemma 6 (2.2.3) it is uniformly convergent to some f ∈ F(T). By virtue of Proposition 2 (2.3.4), f ∈ M b (T, S). It is clear that S = coz f . The following result gives a characterization of a-foundations in the class of all 𝜎-foundations. Corollary 1. Let S be a 𝜎-foundation on T. Then, the following conclusions are equivalent: 1) S is an a-foundation; 2) S = Coz M b (T, S) = Coz M(T, S). Proof. (1) ⊢ (2). It is clear that Coz M b (T, S) ⊂ Coz M(T, S). The inclusion S ⊂ Coz M b (T, S) follows Theorem 2. The inclusion Coz M(T, S) ⊂ S follows from Lemma 1. (2) ⊢ (1). It follows from Proposition 1.

2.3.6 Description of normal and completely normal families and envelopes. Naturalness of the family of measurable functions In this subsection, we shall show that the notion of a measurable function is very natural. The first reason for this conclusion is that E. Borel, H. Lebesgue, and F. Hausdorff proved that the class of all families of measurable functions on descriptive spaces ⟮T, S⟯ with 𝜎-foundations S coincides with the class of all normal families of functions [Borel, 1905; Lebesgue, 1905; Hausdorff, 1927]. The second reason is that G. Regoli proved that the class of all families of measurable functions on descriptive spaces ⟮T, S⟯ with 𝜎-algebras S coincides with the class of all completely normal families of functions [Regoli, 1977]. Proposition 1. Let A(T) be a normal family of functions on a set T. Then, A(T) = M(T, Coz A(T)). Proof. Since A(T) is normal, by Corollary 1 to Proposition 1 (2.2.5), S ≡ Coz A(T) is a 𝜎foundation. Consequently, the family B(T) ≡ M(T, S) is normal by virtue of Corollary 2 to Proposition 2 (2.3.4). Let f ∈ B(T)+ . Consider the function g ≡ u ∘ f ≡ f /(1 + f ) ∈ B(T)+ , where u : R ] − 1, 1[ is the mapping such that u(x) ≡ x/(1 + |x|) (see Lemma 1 (1.4.9)). It is clear that 0 ⩽ g(t) < 1 for every t ∈ T. Take the points x ni ≡ i/n for all i ∈ [−1, n + 1] ∩ Z and n ∈ N and consider the sets S ni ≡ g−1 []x n,i−1 , x n,i+1 [] ∈ S. Then, S ni = coz a ni for some functions a ni ∈ A(T)+ . Since (S ni | i ∈ n + 1) is a cover of T we can consider the

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functions g ni ≡ a ni / ∑ (a ni | i ∈ n + 1) ∈ A(T)+ . Let t ∈ T. Suppose that g ni (t) < 1/(n+1) for every i ∈ n + 1; then, 1 = ∑ (g ni (t) | i ∈ n + 1) < n/(n + 1) < 1, but it is impossible. Therefore, g ni (t) ⩾ 1/(n + 1) for some i and (R ni | i ∈ n + 1), where R ni ≡ cozn+2 g ni , is a cover of T. Consider the functions h ni ≡ (x ni−1 (n + 2) g ni ) ∧ (x ni−1 1) and h n ≡ sup(h ni | i ∈ n + 1) in A(T)+ . If t ∈ S ni , then h ni (t) ⩽ x n,i−1 ⩽ g(t). If t ∈ ̸ S ni , then h ni (t) = 0 ⩽ g(t). Therefore, h ni ⩽ g for all indices. If t ∈ R ni , then h ni (t) = x n,i−1 . Hence, 0 ⩽ g(t) − h n (t) ⩽ g(t) − h ni (t) < x n,i+1 − x n,i−1 = 2/n for every t ∈ R ni . This means that g = u-lim (h n | n ∈ N). Consequently, g ∈ A(T). Applying the inverse mapping v ≡ u −1 : ] − 1, 1[ R such that v(y) ≡ y/(1 − |y|) (see Lemma 1 (1.4.9)), we get f = v ∘ g = g/(1 − g) ∈ A(T). If f ∈ B(T), then f = f + − (−f− ) ∈ A(T). Thus, we proved that B(T) ⊂ A(T). Conversely, if f ∈ A(T), then by Lemma 3 (2.2.5), f −1 []x, y[] ∈ S for every ]x, y[⊂ R. Therefore, f ∈ B(T) by virtue of Lemma 1 (2.3.1). This guarantees that B(T) = A(T). For any family A(T) ⊂ F(T), consider the associated family A(T)∗ ≡ {((u − x1) ∨ 0) ∧ ((y1 − u) ∨ 0) | u ∈ A(T) ∧ x, y ∈ R ∧ x < y} (see Lemma 3 (2.2.5)). Lemma 1. Let A(T) ⊂ F(T) satisfy conditions 1 – 4 from 2.2.4 (i. e. A(T) be a latticeordered linear space with the unit 1). Then, A(T)∗ = A b (T)+ . Proof. The conditions imply that f ≡ ((u − x1) ∨ 0) ∧ ((y1 − u) ∨ 0) ∈ A(T). From 0 ⩽ f ⩽ (y − x)1 we infer A(T)∗ ⊂ A b (T)+ . Conversely, let u ∈ A b (T)+ . Take x ≡ 0 and y ∈ R such that y1 ⩾ 2u. Consider the function f = ((u − x1) ∨ 0) ∧ ((y1 − u) ∨ 0) ∈ A(T)∗ . Since (y1 − u) ∨ 0 ⩾ u, we have f = u. Thus, u ∈ A(T)∗ . Corollary 1. Let A(T) be a normal family on T. Then, A(T) = M(T, Coz A(T)∗ ). Proof. Lemma 1 implies Coz A(T)∗ = Coz A b (T)+ . Obviously, Coz A b (T)+ ⊂ Coz A(T). Since for every f ∈ A(T) we get coz f = coz g, where g ≡ |f | ∧ 1 = (f+ − f− ) ∧ 1 = (f ∨ 0 − f ∧ 0) ∧ 1 ∈ A b (T)+ we also obtain Coz A b (T)+ ⊃ Coz A(T). Thus, Coz A(T) = Coz A(T)∗ and we can apply Proposition 1. The following theorem gives us a characterization of normal families A(T) ⊂ F(T). Theorem 1 (the Borel – Lebesgue – Hausdorff theorem on normal families). Let A(T) be a family of functions on T. Then, the following conclusions are equivalent: 1) the family A(T) is normal; 2) A(T) = M(T, S) for some 𝜎-foundation S; 3) Coz A(T) is a 𝜎-foundation and A(T) = M(T, Coz A(T)); 4) A(T) = M(T, {⌀, T} ∪ (Coz A(T))𝜂𝜎 ) (for the operations 𝜂 and 𝜎 we refer to 2.1.1); 5) A(T) = M(T, {⌀, T} ∪ (Coz A(T)∗ )𝜂𝜎 ).

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Proof. (1) ⊢ (3). According to Corollary 1 to Proposition 1 (2.2.5) the ensemble Coz A(T) is a 𝜎-foundation. By Proposition 1 A(T) = M(T, Coz A(T)). (3) ⊢ (4). Since Coz A(T) is a 𝜎-foundation, {⌀, T} ∪ (Coz A(T))𝜂𝜎 = Coz A(T). (4) ⊢ (2). We need to check that S ≡ {⌀, T} ∪ (Coz A(T))𝜂𝜎 is a 𝜎-foundation. By virtue of Lemma 1 (2.1.1), S is 𝜎-additive, (Coz A(T))𝜂 is multiplicative, and therefore, S is also multiplicative. (2) ⊢ (1). This follows from Corollary 2 to Proposition 2 (2.3.4). (1) ⊢ (5). By virtue of Corollary 1 to Lemma 1, Lemma 1 and proven deduction (1) ⊢ (3), we get A(T) = M(T, Coz A(T)∗ ) ⊂ M(T, {⌀, T} ∪ (Coz A(T)∗ )𝜂𝜎 ) ⊂ M(T, {⌀, T} ∪ (Coz A(T))𝜂𝜎 ) = A(T). (5) ⊢ (2). As in the proof of deduction (3) ⊢ (2) we infer from Lemma 1 (2.1.1) that S ≡ {⌀, T} ∪ (Coz A(T)∗ )𝜂𝜎 is a 𝜎-foundation. The equivalence of assertions 1 and 2 of this Theorem shows that two essentially different ways (the descriptive way through the measurability and the operational way through normality) lead to one and the same remarkable class of functions. Theorem 1 gives the opportunity to describe the normal envelope N(A(T)) of an arbitrary family of functions A(T) ⊂ F(T), i. e. the smallest normal family containing the family A(T). Theorem 2 (the Hausdorff theorem on the normal envelope). Let A(T) be a family of functions on T. Then, the family M(T, {⌀, T} ∪ (Coz A(T)∗ )𝜂𝜎 ) is the normal envelope N(A(T)) of the family A(T). Proof. By Lemma 1 (2.1.1) S ≡ {⌀, T} ∪ (Coz A(T)∗ )𝜂𝜎 is a 𝜎-foundation. Then, by Corollary 2 to Proposition 2 (2.3.4) E(T) ≡ M(T, S) is a normal family. If f ∈ A(T), then according to Lemma 3 (2.2.5), f −1 []x, y[] = coz g for the function g ≡ ((f − x1) ∨ 0) ∧ ((y1 − f ) ∨ 0) ∈ A(T)∗ . Therefore, f ∈ E(T) by virtue of Lemma 1 (2.3.1). Thus, A(T) ⊂ E(T). Let B(T) be a normal family and A(T) ⊂ B(T). Then, Coz A(T)∗ ⊂ Coz B(T)∗ implies E(T) ⊂ M(T, {⌀, T} ∪ (Coz B(T)∗ )𝜂𝜎 ). Applying Theorem 1, we get E(T) ⊂ B(T). Proposition 2. Let A(T) ⊂ F(T) satisfy conditions 1, 2󸀠 , 2󸀠󸀠 , 3, and 4 from 2.2.4. Then, N(A(T)) = M(T, (Coz A(T))𝜎 ). Proof. By Proposition 1 (2.2.5) S ≡ Coz A(T) is a 𝜑-foundation. Therefore, S𝜎 is a 𝜎-foundation and by Corollary 2 to Proposition 2 (2.3.4) M(T, S𝜎 ) is a normal family. Let f ∈ A(T), x < y in R, and L ≡ f −1 []x, y[]. By Corollary 1 to Lemma 3 (1.4.5) there are sequences 𝛼 ≡ (r n ∈ Q | n ∈ 𝜔) ↓ and 𝛽 ≡ (s n ∈ Q | n ∈ 𝜔) ↑ such that x = inf 𝛼 and y = sup 𝛽. For each n ∈ N, consider the set K n ≡ f −1 []r n , s n []. By Lemma 3 (2.2.5) K n = coz g for the function g ≡ ((f − r n 1) ∨ 0) ∧ ((s n 1 − f ) ∨ 0). Since A(T) is

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closed under addition, multiplication by −1 ∈ Z, and by the numbers of the form 1/k, k ∈ N, we conclude that A(T) is closed under multiplication by rational numbers and contains 0. Besides, A(T) contains 1 and is closed under finite exact bounds. Hence, g ∈ A(T), and therefore, K n ∈ S. It is clear that ]x, y[= ⋃⟮]r n , s n [| n ∈ N⟯. Then, according to Lemma 3 (1.1.10), L = ⋃⟮K n | n ∈ 𝜔⟯ ∈ S𝜎 , and therefore, f ∈ M(T, S𝜎 ). Thus, A(T) ⊂ M(T, S𝜎 ). Let B(T) be a normal family such that A(T) ⊂ B(T). Corollary 1 to Proposition 1 (2.2.5) guarantees that Coz B(T) is a 𝜎-foundation, and therefore, S𝜎 ⊂ Coz B(T). Using Proposition 1, we infer that M(T, S𝜎 ) ⊂ M(T, Coz B(T)) = B(T). Thus, M(T, S𝜎 ) = N(A(T)). The following theorem gives us a characterization of completely normal families A(T) ⊂ F(T). Note that the proof of the equivalence 1 ∼ 3 (see 1.1.3) in this theorem unexpectedly requires some portion of hard analysis, namely, the exponential function exp : R → R from 1.4.7. Theorem 3 (the Regoli theorem on completely normal families). Let A(T) be a family of functions on T. Then, the following conclusions are equivalent: 1) the family A(T) is completely normal; 2) the family A(T) satisfies conditions 1 – 4 and 8 from 2.2.4; 3) the family A(T) satisfies conditions 1 – 3, 5, and 8 from 2.2.4; 4) A(T) = M(T, S) for some 𝜎-algebra S; 5) Coz A(T) is a 𝜎-algebra and A(T) = M(T, Coz A(T)); 6) A(T) = M(T, B(T, Coz A(T))). Proof. (1) ⊢ (5). By Theorem 1 A(T) = M(T, {⌀, T} ∪ (Coz A(T))𝜂𝜎 ). By virtue of Corollary 1 to Proposition 1 (2.2.5) Coz A(T) is a 𝜎-algebra, and therefore, {⌀, T} ∪ (Coz A(T))𝜂𝜎 = Coz A(T). Hence, A(T) = M(T, Coz A(T)). (5) ⊢ (6). Since Coz A(T) is a 𝜎-algebra, we have B(T, Coz A(T)) = Coz A(T). (6) ⊢ (4). It is evident. (4) ⊢ (1). By Theorem 1, the family A(T) is normal. By virtue of Corollary 3 to Lemma 2 (2.3.4), A(T) is closed under pointwise convergence. (1) ⊢ (2). It is evident. (2) ⊢ (5). By Proposition 1 (2.2.5) S ≡ Coz A(T) is a 𝜎-algebra. First, check that A(T) ⊂ M(T, S). Let f ∈ A(T). According to Lemma 3 (2.2.5) S ≡ f −1 []x, y[] = coz g for the function g ≡ ((f − x1) ∨ 0) ∧ ((y1 − f ) ∨ 0) ∈ A(T). Hence, S ∈ S. Therefore, by Lemma 1 (2.3.1) f ∈ M(T, S). Now, check the inverse inclusion. Take some set S ∈ S and the function g ≡ 𝜒(S). By the definition, S = coz a for some a ∈ A(T)+ . Then, g = p-lim (na ∧ 1 | n ∈ 𝜔) ∈ A(T). Consequently, St(T, S) ⊂ A(T). Let f ∈ M(T, S). By virtue of Theorem 2 (2.3.4), f = p-lim (f n | n ∈ 𝜔) for some sequence (f n ∈ St(T, S) | n ∈ 𝜔). By condition 8, f ∈ A(T).

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(1) ⊢ (3). It is evident. (3) ⊢ (5). By Proposition 1 (2.2.5) S ≡ Coz A(T) is a 𝜎-foundation. First, check that M(T, S) ⊂ A(T). Take some set S ∈ S. By the definition, S = coz a for some a ∈ A(T). Consider the function g ≡ exp ∘(−a2 ) (see 1.4.7). By conditions 1, 2, 3, 5, and 8 g ∈ A(T). Since −a2 (t) < 0 for every t ∈ S and −a2 (t) = 0 for every t ∈ T\S, Corollary 2 to Lemma 10 (1.4.7) and assertion 2 of Corollary 1 to Theorem 1 (1.4.7) imply that 0 < g(t) < 1 for every t ∈ S and g(t) = 1 for every t ∈ T \ S. Consider the function h ≡ p-lim (g n | n ∈ 𝜔) ∈ A(T). According to Lemma 7 (1.4.4) h(t) = 0 for every t ∈ S and h(t) = 1 for every t ∈ T\S, i. e. h = 𝜒(T\S). Thus, T\S ∈ S, and therefore, S is a 𝜎-algebra. Besides, 𝜒(S) = 1 − h ∈ A(T). Consequently, St(T, S) ⊂ A(T). Let f ∈ M(T, S). By virtue of Theorem 2 (2.3.4), f = p-lim (f n | n ∈ 𝜔) for some sequence (f n ∈ St(T, S) | n ∈ 𝜔). By condition 8 f ∈ A(T). Now, check the inverse inclusion. Let f ∈ A(T) and x, y ∈ R, x < y. Put S ≡ f −1 []x, y[] and S󸀠 ≡ f −1 [[x, y]]. Then, S = {t ∈ T | x < f (t) < y} = {t ∈ T | (f (t) − x)(y − f (t)) > 0} and S󸀠 \S = {t ∈ T | f (t) = x ∨ f (t) = y} = {t ∈ T | (f (t) − x)(y − f (t)) = 0}. Conditions 1, 2, 3, 5, and 8 imply that g ≡ exp ∘((f − x1)(y1 − f )) ∈ A(T). According to Corollary 2 to Lemma 10 (1.4.7) and assertion 2 of Corollary 1 to Theorem 1 (1.4.7) g(t) > 1 for t ∈ S, g(t) = 1 for t ∈ S󸀠 \ S, and 0 < g(t) < 1 for t ∈ T\S󸀠 . For every t ∈ T, denote the sequence (g n (t) | n ∈ 𝜔) by z t . Then, Lemma 7 (1.4.4) implies that lim z t = ∞ for t ∈ S, lim z t = 1 for t ∈ S󸀠 \S, and lim z t = 0 for t ∈ T\S󸀠 . Define the function h ≡ p-lim (exp(−g n ) | n ∈ 𝜔) ∈ A(T). By Theorem 2 (1.4.7), we conclude that h(t) = 0 for t ∈ S, h(t) = exp(−1) for t ∈ S 󸀠 \S, and h(t) = 1 for t ∈ T\S󸀠 . Thus, T\S = coz h ∈ S. As was shown above, S is a 𝜎-algebra. Hence, S ∈ S. Therefore, by Lemma 1 (2.3.1) f ∈ M(T, S). This theorem except conclusion 6 was proven in [Regoli, 1977]. Theorem 3 gives the opportunity to describe the completely normal envelope CN(A(T)) of an arbitrary family of functions A(T) ⊂ F(T), i. e. the smallest completely normal family containing the family A(T). Theorem 4 (the Zakharov – Rodionov theorem on the completely normal envelope in the descriptive form). Let A(T) be a family of functions on T. Then, the family M(T, B(T, Coz A(T)∗ )) is the completely normal envelope CN(A(T)) of the family A(T). Proof. By virtue of Theorem 2, A(T) ⊂ N(A(T)) = M(T, {⌀, T} ∪ (Coz A(T)∗ )𝜂𝜎 ). Since {⌀, T} ∪ (Coz A(T)∗ )𝜂𝜎 ⊂ B(T, Coz A(T)∗ )), we obtain the inclusion A(T) ⊂ M(T, B(T, Coz A(T)∗ )) ≡ E(T). According to Theorem 3, the family E(T) is completely normal. Let B(T) be a completely normal family and A(T) ⊂ B(T). Then, Coz A(T)∗ ⊂ Coz B(T)∗ implies E(T) ⊂ M(T, B(T, Coz B(T)∗ )). Using Lemma 1 and Theorem 3, we get E(T) ⊂ M(T, B(T, Coz B(T))) = B(T).

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This result was obtained in [Zakharov and Rodionov, 2014b]. Corollary 1. Let A(T) ⊂ F(T) satisfy conditions 1 – 4 from 2.2.4. Then, the equality CN(A(T)) = M(T, B(T, Coz A(T))) holds. This result was obtained in [Regoli, 1977]. Proposition 3. Let A(T) ⊂ F(T) satisfy conditions 1 – 4 from 2.2.4. Then, the equality CN(A(T)) = Lim𝜔1 M(T, (Coz A(T))𝜎 ) holds. Proof. By Lemma 5 (2.2.4) the family B(T) ≡ Lim𝜔1 N(A(T)) satisfies conditions 1 – 4 from 2.2.4. Proposition 2 implies that B(T) = Lim𝜔1 M(T, S𝜎 ), where S ≡ Coz A(T). According to Lemma 4 (2.2.4), B(T) is closed under pointwise convergence. Therefore, B(T) is completely normal By virtue of Theorem 3. Take some completely normal family E(T) ⊃ A(T) and the ensemble E ≡ Coz E(T). By Theorem 3 E(T) = M(T, B(T, E)). It is clear that S ⊂ E. Consequently, S𝜎 ⊂ B(T, E). This implies the inclusion M(T, S𝜎 ) ⊂ E(T). Since E(T) satisfies condition 8, we conclude that B(T) ⊂ E(T). This means that B(T) = CN(A(T)). Conclusion 6 of Theorem 3 allows us to give other descriptions of completely normal families and envelopes at the end of 2.3.7 and at the end of 2.4.7 using the Baire collection (see 2.2.4).

2.3.7 Correlations between Baire’s and Borel’s functional collections Let S be an ensemble on a set T with the co-ensemble R. Consider the Borel envelope B(T, S) of the ensemble S and the corresponding separable perfect 𝜎-foundations Λ 𝛼 (T, S) of all classes 𝛼 ∈ [1, 𝜔1 [ (see 2.1.1 – 2.1.3). Recall that By virtue of Corollary 2 to Theorem 5 (2.1.3) Λ 𝛼 (T, S) = Y𝛼 (T, S) for all 𝛼 ∈ [1, 𝜔1 [. The 𝜎-algebra B(T, S) assigns the family BM(T, S) ≡ M(T, B(T, S)) of all Borel (measurable) functions. Every 𝜎-foundation Λ 𝛼 (T, S) assigns the family M(T, Λ 𝛼 (T, S)) of all Borel (measurable) functions of the class 𝛼. Thus, we have the Borel collection ⟮M(T, Λ 𝛼 (T, S)) | 𝛼 ∈ 𝜔1 ⟯. By Theorem 1 (2.1.3) we get B(T, S) = ⋃⟮Λ 𝛼 (T, S) | 𝛼 ∈ 𝜔1 ⟯. Therefore, BM(T, S) = ⋃⟮M(T, Λ 𝛼 (T, S)) | 𝛼 ∈ 𝜔1 ⟯. The Baire – Borel correlations between this Borel collection and the Baire collection ⟮Lim𝛼 A(T) | 𝛼 ∈ 𝜔1 + 1⟯ (see 2.2.4) for various families A(T) ⊂ F(T) and various descriptive spaces ⟮T, S⟯ were investigated by many authors. For the first time, H. Lebesgue and F. Hausdorff proved that in the case of a metric space, ⟮T, G𝜌 ⟯ some Baire – Borel correlation exists and gave its exact description (see [Lebesgue, 1905; Hausdorff, 1927] and also [Kuratowski, 1966, 31.IX]). The same correlation holds for a perfectly normal space.

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S. Banach considered the case of a perfect topological space (see [Banach, 1931] and [Kuratowski, 1966, 31.IX]). The cases of an arbitrary topological and descriptive spaces were considered in [Zakharov, 2002a; Zakharov and Rodionov, 2008a]. This subsection is devoted to some detailed description of all these Baire – Borel correlations for the general, the perfect, and the quasiclassical cases of a descriptive space ⟮T, S⟯ with an arbitrary, a perfect, and a separable perfect ensemble S, respectively. First, we investigate properties of the families M(T, Λ 𝛼 (T, S)) relative to the pointwise convergence of sequences in F(T) (see 2.3.4, 2.2.4, 2.2.3). One-step pointwise limits Consider now p-Lim-hulls of families M(T, Y𝛼 (T, S)) and M(T, Λ 𝛼 (T, S)), 𝛼 ∈ 𝜔1 . Lemma 1. Let S be a 𝜎-additive ensemble on a set T. Then, p-Lim M(T, S) ⊂ M(T, R𝜎 ). Proof. The assertion follows immediately from Corollary 2 to Lemma 2 (2.3.4). Now, we shall prove the inverse result under some conditions. Proposition 1. Let S be a separable 𝜎-foundation on a set T. Let f ∈ M b (T, R𝜎 ) and f takes only finite set of values rng f = {a i | i ∈ I}. Then, there is a sequence (f m ∈ M b (T, S) | m ∈ 𝜔) such that f = p-lim (f m | m ∈ 𝜔) and rng f m ⊂ [a, b] for every m, where a = sm (a i | i ∈ I) and b = gr (a i | i ∈ I). Proof. Consider the sets A i ≡ f −1 [{a i }] ∈ R𝜎 . Then, the collection (A i | i ∈ I) is disjoint. Since S is multiplicative, R is additive. Therefore, A i = ⋃⟮R im | m ∈ 𝜔⟯ for some increasing sequences (R im ∈ R | m ∈ 𝜔). It is clear that R im ∩ R jn = ⌀ for every i ≠ j and every m, n. Consider the sets R󸀠im ≡ ⋃⟮R jm | j ∈ I \ {i}⟯ ∈ R. By Corollary 2 to Theorem 1 (2.3.5) for disjoint sets R󸀠im , R im ∈ R and for numbers a ⩽ a i there is a function f im ∈ M(T, S) such that rng f im ⊂ [a, a i ], f im [R󸀠im ] = {a}, and f im [R im ] = {a i }. Consider the function f m ≡ sup (f jm | j ∈ I) ∈ M(T, S). If t ∈ R im and j ≠ i, then t ∈ 󸀠 R jm , f im (t) = a i , and f jm (t) = a. Therefore, f m (t) = a i , i. e. f m [R im ] = {a i } for every i ∈ I. If t ∈ T, then t ∈ R im ⊂ R in for some indices i and m and every n ⩾ m. Consequently, f (t) = a i = f m (t) = f n (t) for all n ⩾ m. This means that f = p-lim (f m | m ∈ 𝜔). Besides, rng f m ⊂ [a, b]. Theorem 1. Let S be a separable perfect 𝜎-foundation on a set T. Then, te equality p-Lim M(T, S) = M(T, R𝜎 ) holds. Proof. 1. Take any function f ∈ M b (T, R𝜎 ). By Theorem 3 (2.3.4) for f there is a sequence (f k ∈ M b (T, R𝜎 ) | k ∈ 𝜔) such that f = u-lim (f k | k ∈ 𝜔) and every f k takes a finite set of values. For 𝜀m = (1/2)m , there is a number k m such that

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‖f − f p ‖u ⩽ 𝜀m for every p ⩾ k m . Consider a new sequence (g m ∈ M b (T, R𝜎 ) | m ∈ 𝜔) such that g0 ≡ 0 and g m ≡ f k m for m ⩾ 1. Then, ‖g m − g m−1 ‖u ⩽ 2𝜀m . By Proposition 1 for every function g m − g m−1 there is a sequence (g mn ∈ M b (T, S) | n ∈ 𝜔) such that g m − g m−1 = p-lim (g mn | n ∈ 𝜔) and ‖g mn ‖u ⩽ ‖g m − g m−1 ‖u ⩽ 𝜀m−1 . Since g m = ∑ (g i − g i−1 | i ∈ (m + 1)\1), it is naturally to consider the functions h mn = ∑ (g in | i ∈ (m + 1)\1). They functions belong to M b (T, S) by virtue of Theorem 1 (2.3.2). Proposition 1 (2.2.3) implies g m = ∑ (g i − g i−1 | i ∈ (m + 1)\1) = p-lim (h mn | n ∈ 𝜔). Take any t ∈ T and 𝜀 > 0. Then, there is m such that 𝜀m−1 < 𝜀 and |f (t) − g m (t)| < 𝜀. For every m, there is n m > m such that |g m (t) − h mn (t)| < 𝜀 for every n > n m . Therefore, for n > n m using Lemma 3 (1.4.8), we get |f (t)− h nn (t)| ⩽ |f (t)− g m (t)|+|g m (t)− h mn (t)|+ |h mn (t) − h m,n+1 (t)| + . . . + |h n−1,n (t) − h nn (t)| < 𝜀 + 𝜀 + |g m+1,n (t)| + . . . + |g nn (t)| < 2𝜀 + 𝜀m + . . . + 𝜀n−1 < 2𝜀 + 𝜀m−1 < 3𝜀. Hence, f (t) = lim (h nn (t) | n ∈ 𝜔). As a result, we obtain the inclusion M b (T, R𝜎 ) ⊂ bp-Lim M b (T, S). 2. Denote now M(T, S) by A and M(T, R𝜎 ) by B. Consider the mappings u : R → ]− 1, 1[ and v : ]−1, 1[→ R from 1.4.9 such that u(x) = x/(1+|x|) and v(y) = y/(1−|y|). Take any function f ∈ B + and consider the function g ≡ u ∘ f = f /(1 + f ). By virtue of Theorem 1 (2.3.2), g ∈ B b . Therefore, by (1), there is a sequence (g n ∈ A b | n ∈ 𝜔) such that g = p-lim (g n | n ∈ 𝜔). Define the functions h n ≡ (1/n)1 ∨ g n ∧ (1 − (1/n))1. By virtue of Proposition 1 (2.3.3), h n ∈ A b . Since rng h n ⊂ ]0, 1[ , we can consider the functions f n (t) ≡ v ∘ h n = h n /(1 − h n ) ∈ A+ . Using Proposition 2 (1.4.7), we infer that g(t) = 0 ∨ g(t) ∧ 1 = lim (h n (t) | n ∈ 𝜔). Then, by Lemma 4 (1.4.9), we get v(g(t)) = lim (v(h n (t)) | n ∈ 𝜔) = lim (f n (t) | n ∈ 𝜔). But by Lemma 1 (1.4.9), v(g(t)) = (v ∘ u)(f (t)) = f (t). Consequently, f = p-lim (f n | n ∈ 𝜔). If f ∈ B, then f = f+ + f− and f+ , −f− ∈ B+ . As proven above, f+ = p-lim⟮f n󸀠 | n ∈ 𝜔⟯ and −f− = p-lim⟮f n󸀠󸀠 | n ∈ 𝜔⟯ for some sequences (f n󸀠 ∈ A+ | n ∈ 𝜔) and (f n󸀠󸀠 ∈ A+ | n ∈ 𝜔). Consider the functions f n ≡ f n󸀠 − f n󸀠󸀠 ∈ A. Then, we obtain f = p-lim⟮f n | n ∈ 𝜔⟯. This means that B ⊂ p-Lim A. Using Lemma 1, we get the necessary equality. Corollary 1. Let S be an ensemble on a set T such that the ensemble S𝜎 is a separable perfect 𝜎-foundation. Then, p-Lim M(T, S𝜎 ) = M(T, R𝛿𝜎 ). Proof. Taking in Theorem 1 S󸀠 ≡ S𝜎 and R󸀠 ≡ co-S󸀠 = R𝛿 , we get the necessary equality. Corollary 1 might be better understood if we shall use the classification of Borel sets developed in 2.1.3. In terms of Y-ensembles, the conclusion of Corollary 1 takes the form p-Lim M(T, Y1 (T, R)) = M(T, Y2 (T, R)). This form gives the opportunity to generalize this equality to all consequent ordinal numbers. The following results will be called the theorems about one-step pointwise limits.

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Theorem 2. Let S be a perfect ensemble on a set T. Then, p-Lim M(T, Y𝛼 (T, S)) = M(T, Y𝛼+1 (T, S)) for every 𝛼 ∈ [1, 𝜔1 [. Proof. Take S󸀠 ≡ Y𝛼 and R󸀠 ≡ co-S󸀠 = Θ𝛼 . Applying Lemma 5 (2.1.2), we obtain R󸀠 = (Θ𝛼 )𝜎 = Y𝛼+1 . By Corollary 2 to Theorem 8 (2.1.3) (the improvement theorem in the perfect case) S󸀠 = Y𝛼 is a separable perfect 𝜎-foundation for 𝛼 ⩾ 1. Applying now Theorem 1 to S󸀠 and R󸀠 , we complete the proof. Corollary 1 (the general case). Let S be an ensemble on a set T. Then, we have p-Lim M(T, Λ 𝛼 (T, S)) = M(T, Λ 𝛼+1 (T, S)) for every 𝛼 ∈ [1, 𝜔1 [. Proof. Take S󸀠 ≡ L(T, S). By Lemma 4 (2.1.1), S󸀠 is perfect. Besides, by Corollary 1 to Theorem 5 (2.1.3), Y𝛼 (T, S󸀠 ) = Λ 𝛼 (T, S). Applying now Theorem 2 to S󸀠 , we get the necessary equality. Corollary 2 (the quasiclassical case). Let S be a separable perfect 𝜎-foundation on a set T. Then, we have p-Lim M(T, Y𝛼 (T, S)) = M(T, Y𝛼+1 (T, S)) for every 𝛼 ∈ 𝜔1 . Proof. The assertion follows from Theorem 1 and Theorem 2. Pointwise limits for unions of the classes Here we consider p-Lim-hulls of the families ⋃⟮M(T, Y𝛽 ) | 𝛽 ∈ 𝛼⟯ and ⋃⟮M(T, Λ 𝛽 ) | 𝛽 ∈ 𝛼⟯, 𝛼 ∈ [1, 𝜔1 [. Lemma 2. Let S be a perfect ensemble on a set T. Then, for every number 𝛼 ∈ [2, 𝜔1 [ we have p-Lim ⋃⟮M(T, Y𝛽 (T, S)) | 𝛽 ∈ 𝛼⟯ ⊂ M(T, Y𝛼+1 (T, S)). Proof. It follows from Lemma 7 (2.1.2) that Y𝛽 (T, S) ⊂ Y𝛼 (T, S) for any 𝛽 ∈ 𝛼. Then, we get the inclusion ⋃⟮M(T, Y𝛽 (T, S)) | 𝛽 ∈ 𝛼⟯ ⊂ M(T, Y𝛼 (T, S)). Finally, owing to Theorem 2, we have p-Lim ⋃⟮M(T, Y𝛽 (T, S)) | 𝛽 ∈ 𝛼⟯ ⊂ p-Lim M(T, Y𝛼 (T, S)) = M(T, Y𝛼+1 (T, S)). For non-limits ordinal numbers 𝛼, we obtain more precise result. Proposition 2. Let S be a perfect ensemble on a set T. Then, we have the equality p-Lim ⋃⟮M(T, Y𝛽 (T, S)) | 𝛽 ∈ 𝛼⟯ = M(T, Y𝛼 (T, S)) for every non-limit number 𝛼 ∈ [2, 𝜔1 [. Proof. Denote M(T, Y𝛼 (T, S)) by A and ⋃⟮M(T, Y𝛽 (T, S)) | 𝛽 ∈ 𝛼⟯ by B. Since 𝛼 is a non-limit number, there is 𝜘 ∈ [1, 𝛼[ such that 𝛼 = 𝜘 + 1. Consequently, M(T, Y𝜘 ) ⊂ B. By Proposition 1 (1.2.3), we have 𝛽 ⩽ 𝜘 for all 𝛽 ∈ 𝛼. Since S is perfect, according to Lemma 7 (2.1.2), we have that Y𝛽 ⊂ Y𝜘 for every 𝛽 ∈ 𝛼. Then,

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B ⊂ M(T, Y𝜘 ). Thus, B = M(T, Y𝜘 ). It follows from Theorem 2 that A = p-Lim M(T, Y𝜘 ) = p-Lim B. Proposition 3. Let S be a perfect ensemble on a set T and 𝛼 ∈ [2, 𝜔1 [. Suppose that f ∈ M b (T, Y𝛼 (T, S)) and f takes only finite set of values rng f = {a i | i ∈ I}; then, there is a sequence (f m ∈ ⋃⟮M b (T, Y𝛽 (T, S)) | 𝛽 ∈ [1, 𝛼[⟯ | m ∈ 𝜔) such that f = p-lim (f m | m ∈ 𝜔) and rng f m ⊂ rng f for every m ∈ 𝜔. Moreover, if 𝜘 is a limit number and 𝛼 = 𝜘 + 1, then we can take f m ∈ ⋃⟮M b (T, Y𝛽 (T, S)) | 𝛽 ∈ [1, 𝜘[⟯. Proof. Consider the sets A i ≡ f −1 [{a i }] ∈ Y𝛼 . By Corollary 1 to Proposition 5 (2.1.2) Y𝛼 and Θ𝛼 are perfect and latticed. Hence, A i = T\ ⋃⟮A j | j ∈ I\{i}⟯ ∈ Θ𝛼 . By Theorem 1 (2.1.2) for every i there is a sequence (A in ∈ E | n ∈ 𝜔) such that A i = ⋃⟮⋂⟮A i,n+k | k ∈ 𝜔⟯ | n ∈ 𝜔⟯ = ⋂⟮⋃⟮A i,n+k | k ∈ 𝜔⟯ | n ∈ 𝜔⟯, where E ≡ ⋃⟮Y𝛽 ∩ Θ𝛽 | 𝛽 ∈ [1, 𝛼[⟯. We can assume without loss of generality that I ∈ N. Consider the new disjoint sets F0n ≡ A0n and F in ≡ A in \ ⋃⟮A jn | j ∈ i⟯ for 1 ⩽ i < I. Using the choice axiom from 1.1.12, we can find a collection (𝛽in ∈ [1, 𝛼[ | (i, n) ∈ I × 𝜔) such that A in ∈ Y𝛽in ∩Θ𝛽in . Consider the number 𝛽n ≡ max⟮𝛽in | i ∈ I⟯ ∈ [1, 𝛼[.By virtue of Lemma 7 (2.1.2), A in ∈ Y𝛽n ∩Θ𝛽n ≡ An for every i. Therefore, B in ≡ ⋃⟮A jn | j ∈ i⟯ ∈ An . This implies F in = A in ∩ (T \ B in ) ∈ An for every i ∈ I. According to Proposition 3 (2.1.3), for every n, there is a disjoint collection (E ni ∈ An | i ∈ I) such that F in ⊂ E in and ⋃⟮E ni | i ∈ I⟯ = T. Define a function f n : T → R setting f n (t) ≡ a i for every t ∈ E ni . Since Y𝛽n and Θ𝛽n are perfect and latticed, it follows that An is latticed foundation and, by definition of measurable function, f n ∈ M b (T, An ). Take any t ∈ T. Then, t ∈ A i for some i. Therefore, t ∈ ⋂⟮A i,p+k | k ∈ 𝜔⟯ for some p ∈ 𝜔, i. e. t ∈ A i,p+k for every k ∈ 𝜔. Because t ∈ ̸ A j for every j ∈ I \ {i}, we deduce that t ∈ ̸ ⋃⟮A j,q j +k | k ∈ 𝜔⟯ for some q j ∈ 𝜔. Consequently, t ∈ ̸ A j,q j +l for every l ∈ 𝜔. Consider the number r i ≡ p ∨ gr (q j | j ∈ I \ {i}). For every n > r i , consider k ≡ n − p and l j ≡ n − q j . Then, p + k = n = q j + l j for every j ≠ i. This implies t ∈ A in and t ∈ ̸ A jn for every j ≠ i. It easily follows from this property that t ∈ F in ⊂ E in . As a result, we obtain f n (t) = a i = f (t) for every n > r, where r ≡ gr (r i | i ∈ I). This implies f = p-lim (f n | n ∈ 𝜔). If 𝛼 = 𝜘 + 1, then by Theorem 1 (2.1.2) we can take A in ∈ ⋃⟮Y𝛽 ∩ Θ𝛽 | 𝛽 ∈ [1, 𝜘[⟯, so that 𝛽n ∈ [1, 𝜘[. Theorem 3. Let S be a perfect ensemble on a set T. Then, we have the equality p-Lim ⋃⟮M(T, Y𝛽 (T, S)) | 𝛽 ∈ 𝛼⟯ = M(T, Y𝛼+1 (T, S)) for every limit number 𝛼 ∈ [𝜔, 𝜔1 [. Proof. 1. Take any f ∈ M b (T, Y𝛼+1 ). Applying Theorem 3 (2.3.4) to the ensemble Y𝛼 we see that for f there is a sequence (f k ∈ M b (T, Y𝛼+1 ) | k ∈ 𝜔) such that

2.3.7 Correlations between Baire’s and Borel’s functional collections | 129

f = u-lim (f k | k ∈ 𝜔) and every f k takes a finite set of values. For 𝜀m = 2−m , there is a number k m such that ‖f − f p ‖u ⩽ 𝜀m for every p ⩾ k m . Consider a new sequence ⟮g m ∈ M b (T, Y𝛼+1 ) | m ∈ 𝜔⟯ such that g0 ≡ 0 and g m ≡ f k m , m ⩾ 1. Then, ‖g m − g m−1 ‖u ⩽ 2𝜀m . By Proposition 3 for every function g m − g m−1 there is a sequence (g mn ∈ ⋃⟮M b (T, Y𝛽 ) | 𝛽 ∈ [1, 𝛼[ ⟯ | n ∈ 𝜔) such that g m − g m−1 = p-lim (g mn | n ∈ 𝜔) and ‖g mn ‖u ⩽ ‖g m − g m−1 ‖u ⩽ 𝜀m−1 . Since g m = ∑ (g i − g i−1 | i ∈ (m + 1) \ 1), it is naturally to consider the functions h mn = ∑⟮g in | i ∈ (m + 1) \ 1⟯. It follows from Proposition 1 (2.2.3) that g m = p-lim (h mn | n ∈ 𝜔). Using the choice axiom from 1.1.12, we can find the collections (𝛽mn ∈ [1, 𝛼[| n ∈ 𝜔) such that g mn ∈ M b (T, Y𝛽mn ). Consider the number 𝛾mn ≡ gr (𝛽in | i ∈ (m + 1) \ 1) ∈ [1, 𝛼[. It follows from Lemma 7 (2.1.2) that g in ∈ M b (T, Y𝛾mn ) for every i ∈ (m + 1) \ 1. By virtue of Corollary 1 to Proposition 5 (2.1.2), 𝜎-additive (by the definition) ensemble Y𝛾mn is perfect and latticed, hence it is a 𝜎-foundation. Therefore, by Theorem 1 (2.3.2), h mn ∈ M b (T, Y𝛾mn ) for every m and n. Hence, h mn ∈ ⋃⟮M b (T, Y𝛽 ) | 𝛽 ∈ [1, 𝛼[⟯. Completely in the same way as in the proof of Theorem 1, it is checked that f = p-lim (h nn | n ∈ 𝜔). As a result, we obtain the inclusion M b (T, Y𝛼+1 (T, S)) ⊂ bp-Lim ⋃⟮M b (T, Y𝛽 (T, S)) | 𝛽 ∈ 𝛼⟯. 2. Denote now M(T, Y𝛽 ) by A𝛽 and M(T, Y𝛼+1 ) by B. Consider the mappings u : R → ] − 1, 1[ and v : ] − 1, 1[→ R from 1.4.9 such that u(x) = x/(1 + |x|) and v(y) = y/(1 − |y|0. Take any function f ∈ B + and consider the function g ≡ u ∘ f = f /(1 + f ). By virtue of Theorem 1 (2.3.2) g ∈ B b . Therefore, by (1), there are sequences (g n ∈ ⋃⟮(A𝛽 )b | 𝛽 ∈ [1, 𝛼[⟯ | n ∈ 𝜔) and (𝛽n ∈ [1, 𝛼[| n ∈ 𝜔) such that g = p-lim (g n | n ∈ 𝜔) and g n ∈ (A𝛽n )b for every n ∈ 𝜔. Define the functions h n ≡ (1/n)1 ∨ g n ∧ (1 − (1/n))1. By virtue of Proposition 1 (2.3.3), h n ∈ (A𝛽n )b . Since rng h n ⊂ ]0, 1[ , we can consider the functions f n (t) ≡ v ∘ h n = h n /(1 − h n ) ∈ (A𝛽n )+ . Using the same arguments as in the proof of the second part of Theorem 1, we check that f = p-lim (f n | n ∈ 𝜔). If f ∈ B, then f = f+ + f− and f+ , −f− ∈ B+ . As proven above, f+ = p-lim⟮f n󸀠 | n ∈ 𝜔⟯ and −f− = p-lim⟮f n󸀠󸀠 | n ∈ 𝜔⟯ for some sequences (f n󸀠 | n ∈ 𝜔), (f n󸀠󸀠 | n ∈ 𝜔), (𝛽n ∈ [1, 𝛼[| n ∈ 𝜔), and (𝛾n ∈ [1, 𝛼[| n ∈ 𝜔) such that f n󸀠 ∈ (A𝛽n )+ and f n󸀠󸀠 ∈ (A𝛾n )+ for every n ∈ 𝜔. Consider the numbers 𝛿n ≡ 𝛽n ⊻ 𝛾n . By Lemma 7 (2.1.2) A𝛽n ∪ A𝛾n ⊂ A𝛿n . Consequently, f n ≡ f n󸀠 − f n󸀠󸀠 ∈ A𝛿n . By Proposition 1 (2.2.3), f = p-lim⟮f n | n ∈ 𝜔⟯. This means that B ⊂ p-Lim ⋃⟮A 𝛽 | 𝛽 ∈ 𝛼⟯. Using Lemma 2, we get the necessary equality. Corollary 1 (the general case). Let S be an ensemble on a set T. Then, 1) p-Lim ⋃⟮M(T, Λ 𝛽 (T, S)) | 𝛽 ∈ 𝛼⟯ = M(T, Λ 𝛼 (T, S)) for every non-limit number 𝛼 ∈ [2, 𝜔1 [; 2) p-Lim ⋃⟮M(T, Λ 𝛽 (T, S)) | 𝛽 ∈ 𝛼⟯ = M(T, Λ 𝛼+1 (T, S)) for every limit number 𝛼 ∈ [𝜔, 𝜔1 [.

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Proof. Take S󸀠 ≡ L ≡ L(T, S). By Lemma 4 (2.1.1) S󸀠 is perfect. Besides, by Corollary 1 to Theorem 5 (2.1.3) Y𝛾 (T, S󸀠 ) = Λ 𝛾 (T, S) for every 𝛾 ⩾ 1. Applying now Proposition 2 and Theorem 3 to S󸀠 , we get the necessary equalities. Corollary 2 (the quasiclassical case). Let S be a separable perfect 𝜎-foundation on a set T. Then, 1) p-Lim ⋃⟮M(T, Y𝛽 (T, S)) | 𝛽 ∈ 𝛼⟯ = M(T, Y𝛼 (T, S)) for every non-limit number 𝛼 ∈ [1, 𝜔1 [; 2) p-Lim ⋃⟮M(T, Y𝛽 (T, S)) | 𝛽 ∈ 𝛼⟯ = M(T, Y𝛼+1 (T, S)) for every limit number 𝛼 ∈ [𝜔, 𝜔1 [. Proof. For 𝛼 ⩾ 2, the assertions follow from Proposition 2 and Theorem 3. For 𝛼 = 1, equality 1 follows from Theorem 1.

Convergence classification theorems Consider now the Baire collection ⟮Lim𝛼 M(T, S) | 𝛼 ∈ 𝜔1 + 1⟯ for the family of S-measurable functions M(T, S), where Lim0 M(T, S) = M(T, S) and Lim𝛼 M(T, S) = p-Lim ⋃⟮Lim𝛽 M(T, S) | 𝛽 ∈ 𝛼⟯, 𝛼 ∈ [1, 𝜔1 ] (see 2.2.4). Having all the previous results, we can finally obtain the desired Baire – Borel correlations. We establish the local classification theorems, i. e. the equalities between Borel and Baire families of functions of the same or different classes, and, the (full) classification theorems, i. e. the representation of the family of all Borel-measurable functions as the last step in the Baire transfinite process. Lemma 3. Let S be a separable perfect 𝜎-foundation on a set T. Then, 1) Lim𝛼 M(T, S) = M(T, Y𝛼 (T, S)) for every 𝛼 ∈ 𝜔; 2) Lim𝛼 M(T, S) ⊂ M(T, Y𝛼+1 (T, S)) for every 𝛼 ∈ [𝜔, 𝜔1 [. Proof. Denote Lim𝛼 M(T, S) by L𝛼 . Consider the subset A of the set 𝜔 consisting of all numbers 𝛼 ∈ 𝜔 such that the necessary equality is fulfilled. For 𝛼 = 0, we have automatically L0 = M(T, S) = M(T, Y0 (T, S)). For 𝛼 = 1 by Theorem 1 L1 ≡ Lim1 L0 = p-Lim L0 = p-Lim M(T, S) = M(T, R𝜎 ) = M(T, Y1 (T, S)). This means that 0, 1 ∈ A. Let 2 ⩽ 𝛼 ∈ A. Then, L𝛼+1 ≡ p-Lim ⋃⟮L𝛽 | 𝛽 ∈ 𝛼 + 1⟯ = p-Lim L𝛼 = p-Lim M(T, Y𝛼 (T, S)) = M(T, Y𝛼+1 (T, S)). By virtue of Theorem 2. So 𝛼 + 1 ∈ A. By the principle of natural induction A = 𝜔. Consider now the subset B of the set 𝜔1 consisting of all numbers 𝛼 ∈ 𝜔1 such that the necessary inclusion is fulfilled. As we proved, 𝜔 ⊂ B. Let 𝛽 ∈ [𝜔, 𝜔1 [ and 𝛽 ⊂ B. Consider the set M ≡ ⋃⟮L𝛾 | 𝛾 ∈ 𝛽⟯. By the definition, L𝛽 = p-Lim M. Let f ∈ L𝛽 . Using the choice axiom from 1.1.12, we can find the sequences (𝛾n ∈ 𝛽 | n ∈ 𝜔) and (f n ∈ M | n ∈ 𝜔) such that f n ∈ L𝛾n and f = p-lim (f n | n ∈ 𝜔). By Lemma 7 (2.1.2) the collection (Y𝛼 | 𝛼 ∈ 𝜔1 ) is increasing. Therefore, 𝛾n < 𝛽 implies

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131

f n ∈ L𝛾n ⊂ M(T, Y𝛾n +1 ) ⊂ M(T, Y𝛽 ). Consequently, f ∈ p-Lim M(T, Y𝛽 ). By virtue of Theorem 2, f ∈ p-Lim M(T, Y𝛽+1 ). This gives the inclusion L𝛽 ⊂ M(T, Y𝛽+1 ), i. e. 𝛽 ∈ B. Using the principle of transfinite induction, we conclude that B = 𝜔1 . Theorem 4 (the Lebesgue – Hausdorff local convergence classification theorem in quasiclassical case). Let S be a separable perfect 𝜎-foundation on a set T. Then, 1) Lim𝛼 M(T, S) = M(T, Y𝛼 (T, S)) for every 𝛼 ∈ 𝜔; 2) Lim𝛼 M(T, S) = M(T, Y𝛼+1 (T, S)) for every 𝛼 ∈ [𝜔, 𝜔1 [. Proof. Denote Lim𝛼 M(T, S) by L𝛼 . Consider the subset B of the set 𝜔1 consisting of all numbers 𝛼 ∈ 𝜔1 such that L𝛼 = M(T, Y𝛼 ). By Lemma 3 𝜔 ⊂ B. Let 𝛽 ∈ [𝜔, 𝜔1 [ and 𝛽 ⊂ B. First, assume that 𝛽 is a non-limit number, i. e. 𝛽 = 𝜘 + 1, 𝜘 ∈ 𝜔1 . Take any f ∈ M(T, Y𝛽+1 ). According to Proposition 2, f = p-lim (f n | n ∈ 𝜔) for some sequences (𝛾n ∈ 𝛽 + 1 | n ∈ 𝜔) and (f n ∈ M(T, Y𝛾n ) | n ∈ 𝜔). By Lemma 7 (2.1.2) Y𝛾n ⊂ Y𝛽 , then f n ∈ M(T, Y𝛽 ) for every n. From 𝜘 ∈ 𝛽, by inductive assumption, we infer that M(T, Y𝛽 ) = L𝜘 . This implies f n ∈ L𝜘 ⊂ ⋃⟮L𝛾 | 𝛾 ∈ 𝛽⟯ for all n, then, by definition, f ∈ L𝛽 . Thus, we proved the inclusion M(T, Y𝛽+1 ) ⊂ L𝛽 . Using Lemma 3, we get the exact equality. Consequently, 𝛽 ∈ B. Now, assume that 𝛽 is a limit number. Take any f ∈ M(T, Y𝛽+1 ). According to Theorem 3, f = p-lim (f n | n ∈ 𝜔) for some sequences (𝛾n ∈ 𝛽 | n ∈ 𝜔) and (f n ∈ M(T, Y𝛾n ) | n ∈ 𝜔). Since 𝛽 is a limit number, we infer 𝛿n ≡ 𝛾n + 2 ∈ 𝛽. By Lemma 7 (2.1.2), Y𝛾n ⊂ Y𝛿n . Hence, f n ∈ M(T, Y𝛿n ). From 𝛿n ∈ 𝛽, by inductive assumption, we infer that M(T, Y𝛿n ) = L𝛾n +1 . This implies f n ∈ ⋃⟮L𝛾 | 𝛾 ∈ 𝛽⟯ for every n and, by definition, f ∈ L 𝛽 . Thus, we again get the inclusion M(T, Y𝛽+1 ) ⊂ L𝛽 . Using Lemma 3, we obtain the exact equality. Consequently, 𝛽 ∈ B. Using the principle of transfinite induction, we conclude that B = 𝜔1 . For metric spaces, this is a result of H. Lebesgue and F. Hausdorff (see [Hausdorff, 1927] and [Kuratowski, 1966, 31.IX]). Corollary 1 (the local convergence classification theorem in the general case). Let S be an ensemble on a set T with the derivative ensembles L ≡ L(T, S) and K ≡ K(T, S) (see 2.1.1). Then, Lim𝛼 M(T, K𝜎 ) = M(T, Λ 𝛼+1 (T, S)) for every 𝛼 ∈ 𝜔1 . Proof. By Corollary 1 to Theorem 3 (2.1.1) the ensemble S󸀠 ≡ K𝜎 is a separable perfect 𝜎-foundation. Besides, by Lemma 1 (2.1.3), Y𝛼 (T, S󸀠 ) = Λ 𝛼+1 (T, S) for every 𝛼 ∈ 𝜔 and Y𝛼 (T, S󸀠 ) = Λ 𝛼 (T, S) for every 𝛼 ∈ [𝜔, 𝜔1 [. Applying now Theorem 4 to S󸀠 , we get the necessary equality. Corollary 2 (the Banach local convergence classification theorem in the perfect case). Let S be a perfect ensemble on a set T. Then, Lim𝛼 M(T, R𝜎 ) = M(T, Y𝛼+1 (T, S)) for every 𝛼 ∈ 𝜔1 .

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Proof. Using Corollary 2 to Lemma 4 (2.1.1) (K𝜎 = R𝜎 for perfect S), and Corollary 2 to Theorem 5 (2.1.3) (Λ 𝛼 = Y𝛼 for perfect S), we derive the assertion from the previous Corollary. For perfect topological spaces, this is a result of S. Banach (see [Banach, 1931] and [Kuratowski, 1966, 31.IX]). Combining these local classification theorems with the classification theorems for Borel sets in 2.1.3, we come to the full convergence classification theorems for Borel measurable functions [Zakharov and Rodionov, 2008a]. Theorem 5 (the main convergence classification theorem). Let S be an ensemble on a set T and E is a separable perfect 𝜎-foundation such that S ⊂ E ⊂ B(T, S). Then, BM(T, S) ≡ M(T, B(T, S)) = Lim𝜔1 M(T, E) = ⋃⟮Lim𝛼 M(T, E) | 𝛼 ∈ 𝜔1 ⟯. Proof. It is clear that E ⊂ B(T, S) ⊂ B(T, E). Since B(T, S) is a 𝜎-algebra and B(T, E) is the smallest 𝜎-algebra containing E, we conclude that B(T, E) = B(T, S). By Theorem 4 (2.1.3) the Borel envelope B(T, E) = ⋃⟮Y𝛼 (T, E) | 𝛼 ∈ 𝜔1 ⟯, then M(T, B(T, S)) = ⋃⟮M(T, Y𝛼 (T, E)) | 𝛼 ∈ 𝜔1 ⟯. Since E is perfect, it follows from Lemma 7 (2.1.2) that Y𝛼 (T, E) ⊂ Y𝛼+1 (T, E). Taking into account Theorem 4, we get M(T, Y𝛼 (T, E)) ⊂ M(T, Y𝛼+1 (T, E)) = Lim𝛼 M(T, E) for 𝛼 ∈ [𝜔, 𝜔1 [ and M(T, Y𝛼 (T, E)) = Lim𝛼 M(T, E) for 𝛼 ∈ 𝜔. This implies M(T, B(T, S)) = ⋃⟮Lim𝛼 M(T, E) | 𝛼 ∈ 𝜔1 ⟯ ≡ L. According to Lemma 4 (2.2.4) the family L is closed with respect to the pointwise convergence of sequences. Consequently, L = Lim𝜔1 M(T, E). Corollary 1 (the convergence classification theorem in the general case). Let S be an ensemble on a set T. Then, BM(T, S) = Lim𝜔1 M(T, K𝜎 ) = ⋃⟮Lim𝛼 M(T, K𝜎 ) | 𝛼 ∈ 𝜔1 ⟯. Proof. By Corollary 1 to Theorem 3 (2.1.1) the ensemble E ≡ K𝜎 is a separable perfect 𝜎-foundation. Besides, B(T, E) = B(T, S), as in the beginning of the proof of Theorem 5. Applying now Theorem 5 to S and E, we get the necessary equalities. Corollary 2 (the Banach convergence classification theorem in the perfect case). Let S be a perfect ensemble on a set T. Then, BM(T, S) = Lim𝜔1 M(T, R𝜎 ) = ⋃⟮Lim𝛼 M(T, R𝜎 ) | 𝛼 ∈ 𝜔1 ⟯. Proof. By Theorem 3 (2.1.1) the ensemble E ≡ R𝜎 is a separable perfect 𝜎-foundation. Applying now Theorem 5 to S and E, we get the necessary assertion.

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Corollary 3 (the Lebesgue – Hausdorff convergence classification theorem in quasiclassical case). Let S be a separable perfect 𝜎-foundation on a set T. Then, BM(T, S) = Lim𝜔1 M(T, S) = ⋃⟮Lim𝛼 M(T, S) | 𝛼 ∈ 𝜔1 ⟯. Proof. In the case of E = S, Theorem 5 implies immediately the necessary assertion. The boundedness of the Lebesgue – Hausdorff and the Banach correlations The following assertions shows that for general topological spaces ⟮T, G⟯ the Lebesgue – Hausdorff and the Banach correlations are not valid because for some spaces the family BM(T, G) of Borel functions is rich while the initial family C(T, G) ≡ M(T, G) of continuous functions and the initial family M(T, F𝜎 ) of F𝜎 -measurable functions are trivial. Proposition 4. There is a topological space ⟮T, G⟯ with T = [0, 1[⊂ R such that C(T, G) = M(T, F𝜎 ) = {x1 | x ∈ R} and BM(T, G) = BM(T, O) where O ≡ Ost |[0, 1[, and Ost is the standard topology on R. Proof. Consider the topological space ⟮T, G⟯ from Lemma 5 (2.3.4) taking T ≡ [0, 1[⊂ R and G ≡ {[0, b[ | 0 < b ⩽ 1} ∪ {⌀}. Then, F ≡ co-G = {[a, 1[ | 0 ⩽ a < 1} and K = {G ∩ F | G ∈ G ∧ F ∈ F} = {[a, b[ | 0 ⩽ a < b ⩽ 1} ∪ {⌀}. 1. Take a function f ∈ C(T, G) and suppose that it takes at least two different values x, y ∈ R. Put 𝜀 ≡ |x − y|/3. Consider the sets X ≡ f −1 []x − 𝜀, x + 𝜀[] ∈ G and Y ≡ f −1 []y − 𝜀, y + 𝜀[] ∈ G. On the one hand, X = [0, a[ and Y = [0, b[ with a > 0, b > 0. Hence, 0 ∈ X ∩ Y. On the other hand, X ∩ Y = f −1 []x − 𝜀, x + 𝜀[∩]y − 𝜀, y + 𝜀[] = ⌀. This contradiction proves that f can take only one value, i. e. it is a constant function. 2. It is easy to see that F𝜎 = F ∪ {]a, 1[ | 0 ⩽ a < 1}. Take a function f ∈ M(T, F𝜎 ) and suppose that it takes at least two different values x, y ∈ R. Put 𝜀 ≡ |x − y|/3. Consider the sets X ≡ f −1 []x − 𝜀, x + 𝜀[] ∈ F𝜎 and Y ≡ f −1 []y − 𝜀, y + 𝜀[] ∈ F𝜎 . On the one hand, X = [a, 1[ or X =]a, 1[ and Y = [b, 1[ or Y =]b, 1[ with 0 < a < 1, 0 < b < 1. Hence, ⌀ =]a ̸ ∨ b, 1[⊂ X ∩ Y. On the other hand, X ∩ Y = −1 f []x − 𝜀, x + 𝜀[∩]y − 𝜀, y + 𝜀[] = ⌀. This contradiction proves that f can take only one value, i. e. it is a constant function. 3. It follows from K = {[a, b[ | 0 ⩽ a < b ⩽ 1}∪{⌀} that every interval ]c, d[ ⊂ [0, 1[ belongs to K𝜎 , where Oint |[0, 1[⊂ K𝜎 . According to Statement 1 (2.3.1), this implies that O ≡ Ost |[0, 1[= (Oint |[0, 1[)𝜎 ⊂ K𝜎 ⊂ B(T, G). Since G ⊂ O and B(T, O) is the smallest 𝜎-algebra containing O, we obtain B(T, G) ⊂ B(T, O) ⊂ B(T, G). Constructive descriptions of completely normal families and envelopes The convergence classification theorem in the general case allows us to derive some constructive descriptions of completely normal families and envelopes from descriptive ones given in 2.3.6 (see Theorems 3 and 4 (2.3.6)).

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Theorem 6 (the Zakharov – Rodionov theorem on completely normal families). Let A(T) be a family of functions on T. Then, the following conclusions are equivalent: 1) the family A(T) is completely normal; 2) A(T) = Lim𝜔1 M(T, K𝜎 (T, Coz A(T))). Proof. (1) ⊢ (2). Let S ≡ Coz A(T), E ≡ K𝜎 (T, S). By Theorem 3 (2.3.6) A(T) = M(T, B(T, S)) and by Corollary 1 to Theorem 5 M(T, B(T, S)) = Lim𝜔1 M(T, E). (2) ⊢ (1). By Corollary 1 to Theorem 5 M(T, B(T, S)) = Lim𝜔1 M(T, E) = A(T). Then, by Theorem 3 (2.3.6) the family A(T) is completely normal. Theorem 7 (the Zakharov – Rodionov theorem on the completely normal envelope in the constructive form). Let A(T) ⊂ F(T). Then, the family Lim𝜔1 M(T, K𝜎 (T, Coz A(T)∗ )) is the completely normal envelope CN(A(T)) of the family A(T). Proof. Denote Coz A(T)∗ by S. According to Theorem 4 (2.3.6), CN(A(T)) = M(T, B(T, S)). Then, by Corollary 1 to Theorem 5, we get the equality M(T, B(T, S)) = Lim𝜔1 M(T, K𝜎 (T, S)). 2.3.8 Families of semimeasurable functions on a space with an ensemble Here we shall consider some families of functions that are more extensive than the families of measurable functions. Let S be an ensemble on a fixed set T. A function f : T → R will be called lower S-semimeasurable or lower semimeasurable with respect to the ensemble S if f −1 []x, ∞[] ∈ S for every real number x; similarly, a function f will be called upper S-semimeasurable or upper semimeasurable with respect to the ensemble S if f −1 [] − ∞, x[] ∈ S for every real number x. The sets of all such functions will be denoted by SM l (T, S) and SM u (T, S), respectively. The subsets of all bounded semimeasurable functions will be denoted by SM bl (T, S) and SM bu (T, S). In the classical case of a topological space, the term “semicontinuous” is used along with the term “semimeasurable” and the notation SC is also used. The following lemma shows correlations between the families of Ssemimeasurable and S-measurable functions. Lemma 1. Let S be an ensemble on a set T. Then, 1) if S is 𝜎-additive, then M(T, S) ⊂ SM l (T, S) ∩ SM u (T, S); 2) if S is multiplicative, then SM l (T, S) ∩ SM u (T, S) ⊂ M(T, S); 3) if S is a 𝜎-foundation, then SM l (T, S) ∩ SM u (T, S) = M(T, S); 4) if S is a 𝜎-algebra, then M(T, S) = SM l (T, S) = SM u (T, S); 5) if S is a 𝜎-foundation, but is not closed under the complement, then M(T, S) ⫋ SM l (T, S) and M(T, S) ⫋ SM u (T, S).

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Proof. 1. The inclusion follows from assertions 2 and 3 of Lemma 2 (2.2.1). 2. The inclusion follows from assertion 1 of Lemma 2 (2.2.1). 3. This follows from (1) and (2). 4. Let f ∈ SM l (T, S). Then, X n ≡ f −1 []x − n1 , ∞[] ∈ S for every x ∈ R and n ∈ N. Since S is an algebra, X −n ≡ f −1 [] − ∞, x − n1 ]] = T\X n ∈ S. Assertion 4

of Lemma 2 (2.2.1) implies that f −1 [] − ∞, x[] = ⋃⟮X −n | n ∈ N⟯ ∈ S. Hence, f ∈ SM u (T, S). Thus, SM l (T, S) ⊂ SM u (T, S). In a similar way, we conclude that SM u (T, S) ⊂ SM l (T, S) using assertion 5 of Lemma 2 (2.2.1). Finally, by virtue of assertion 3 proven above we get SM l (T, S) = SM u (T, S) = SM l (T, S) ∩ SM u (T, S) = M(T, S). 5. If S is not closed under the complement, then there is a set S ∈ S such that T\S ∈ ̸ S. Therefore, 𝜒(S) ∈ SM l (T, S), −𝜒(S) ∈ SM u (T, S) but 𝜒(S), −𝜒(S) ∈ ̸ M(T, S). Proposition 1. Let S be a 𝜎-foundation on T, I be a countable set, f , g ∈ SM l (T, S), u ≡ (f i ∈ SM l (T, S) | i ∈ I), and r ∈ R. Then, 1) rf ∈ SM l (T, S) for r ⩾ 0 and rf ∈ SM u (T, S) for r ⩽ 0; 2) f + g ∈ SM l (T, S); 3) fg ∈ SM l (T, S) for f ⩾ 0, g ⩾ 0; 4) fg ∈ SM u (T, S) for f ⩽ 0, g ⩽ 0; 5) 1/f ∈ SM u (T, S) for every function f : T → R+ \{0}; 6) f ∧ g ∈ SM l (T, S); 7) sup u ∈ SM l (T, S); if, moreover, S is a 𝜏-foundation, then sup u ∈ SM l (T, S) for an arbitrary set I; 8) f+ , f− ∈ SM l (T, S). Proof. 1. If r > 0, then (rf )−1 []x, ∞[] = f −1 []x/r, ∞[] ∈ S. If r < 0, then (rf )−1 [] − ∞, x[2] = f −1 []x/r, ∞[] ∈ S. 2. Suppose t ∈ (f + g)−1 []x, ∞[] ≡ S, i. e. x < f (t) + g(t). By Lemma 14 (1.4.3) there is a rational number p such that x − g(t) < p < f (t). Then, t ∈ f −1 []p, ∞[] ∩ g−1 []x − p, ∞[] ≡ T pq implies S ⊂ ⋃⟮T pq | p ∈ Q⟯ ≡ R. Conversely, if t ∈ R, then there is p ∈ Q such that p < f (t) and x − p < g(t). Adding these inequalities we get t ∈ S. Hence, S = R ∈ S. 3. Let x ⩾ 0 and t ∈ (fg)−1 []x, ∞[] ≡ S. Since g(t) > 0, we have 0 ⩽ x/g(t) < f (t). Take p ∈ Q+ such that 0 ⩽ x/g(t) < p < f (t). Then, t ∈ ⋃⟮f −1 []p, ∞[]∩g−1 []x/p, ∞[] | p ∈ Q+ \ {0}⟯ ≡ R. Conversely, if t ∈ R, then there is a rational number p > 0 such that p < f (t) and x/p < g(t). Multiplying these inequalities we get t ∈ S. Hence, S = R ∈ S. If x < 0, then (fg)−1 []x, ∞[] = (fg)−1 [[0, ∞[] = f −1 [[0, ∞[] ∩ g−1 [[0, ∞[] = −1 f []x, ∞[] ∩ g−1 [[x, ∞[] ∈ S. 4. Let x ⩾ 0 and t ∈ (fg)−1 [] − ∞, x[] ≡ S, i. e. 0 ⩽ f (t)g(t) < x. First, assume that f (t) ≠ 0. Then, g(t) > x/f (t) implies g(t) > p > x/f (t) for some rational number p < 0. Therefore, pf (t) < x implies f (t) > x/p. This means that t ∈ f −1 []x/p, ∞[] ∩

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g−1 []p, ∞[] ∈ S. If g(t) ≠ 0, then as above t ∈ g−1 []x/p, ∞[] ∩ f −1 []p, ∞[] ∈ S. Finally, assume that f (t) = 0 and g(t) = 0. Then, xf (t) > −x and g(t) = 0 > −x imply t ∈ (xf )−1 [] − x, ∞[] ∩ g−1 [] − x, ∞[] ∈ S. In all three cases, t ∈ ⋃ ⟮f −1 []x/p, ∞[] ∩ g−1 []p, ∞[] | p ∈ Q− \ {0}⟯ ∪ ∪ ⋃ ⟮g−1 []x/p, ∞[] ∩ f −1 []p, ∞[] | p ∈ Q− \ {0}⟯ ∪ ∪ ((xf )−1 [] − x, ∞[] ∩ g−1 [] − x, ∞[]) ≡ R. Conversely, if t ∈ R, then it can be checked that t ∈ S. Hence, S = R ∈ S. If x ⩽ 0, then (fg)−1 [] − ∞, x[] = ⌀ ∈ S. 5. If x > 0, then (1/f )−1 [] − ∞, x[] = (1/f )−1 []0, x[] = f −1 []1/x, ∞[] ∈ S. If x ⩽ 0, then (1/f )−1 [] − ∞, x[] = ⌀ ∈ S. 6. Let t ∈ (f ∧ g)−1 []x, ∞[] ≡ S, i. e. x < f (t) ∧ g(t). Then, t ∈ f −1 []x, ∞[] ∩ g−1 []x, ∞[] ≡ R. Conversely, if t ∈ R, then x < f (t) and x < g(t) imply x < f (t) ∧ g(t). Therefore, t ∈ S. As a result, we obtain S = R ∈ S. 7. Consider the function g ∈ F(T) such that g = sup u in F(T). Let t ∈ g −1 []x, ∞[] ≡ S. Then, x < sup (f i (t) | i ∈ I). Consequently, x < f i (t) for some i ∈ I. This means that t ∈ ⋃ ⟮f i−1 []x, ∞[] | i ∈ I⟯ ≡ R. Conversely, if t ∈ R, then f i (t) > x for some i ∈ I, where g(t) > x, i. e. t ∈ S. As a result, we obtain S = R. Since I is countable, we infer that R ∈ S. If S is completely additive, then R ∈ S for any set I. 8. This assertion follows from (6) and (7). Proposition 2. Let S be a 𝜎-foundation on T such that S ⊂ (Coz M(T, S))𝜏 . Let f ∈ SM l (T, S), g ∈ M(T, S), and g ⩽ f . Then, there exists a collection u ≡ (f i ∈ M(T, S) | i ∈ I) for some set I such that f = sup u in F(T). If S ⊂ Coz M(T, S), then the set I can be assumed to be countable. Proof. Consider the numbers x nk ≡ k/n for all n ∈ N and k ∈ Z and the sets R nk ≡ f −1 []x nk , ∞[] ∈ S. Consider also the sets K nk ≡ {Q ∈ Coz M(T, S) | Q ⊂ R nk }. Using the choice axiom, we can take collections 𝜘nk ≡ ⟮Q nk𝛼nk ∈ K nk | 𝛼nk ∈ A nk ⟯ such that R nk = ⋃⟮Q nk𝛼nk | 𝛼nk ∈ A nk ⟯. Similarly, we can take collections 𝜑nk ≡ (f nk𝛼nk ∈ M(T, S) | 𝛼nk ∈ A nk ) such that Q nk𝛼nk = coz f nk𝛼nk . Consider the functions f nk𝛼nk m ≡ (g + m|f nk𝛼nk |) ∧ x nk 1 ∈ M(T, S) for all m ∈ 𝜔. For every t ∈ Q nk𝛼nk ⊂ R nk , we get f nk𝛼nk m (t) ⩽ x nk < f (t), and for every t ∈ ̸ Q nk𝛼nk , we get f nk𝛼nk m (t) ⩽ g(t) ⩽ f (t). This means that f nk𝛼nk m ⩽ f . Consider the set I ≡ (∏⟮A nk | (n, k) ∈ N × Z⟯) ∗ 𝜔. In this set, take the element i ≡ ⟨(𝛼nk | (n, k) ∈ N × Z) , m⟩. Then, the functions f nk𝛼nk m may be denoted also by f i . Consider the collection u ≡ (f i | i ∈ I) and define the function h ≡ sup u in F(T). Assume that h(t) < f (t) for some t ∈ T. Then, there is some number x nk such that

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h(t) < x nk < f (t). Therefore, t ∈ R nk . Hence, t ∈ Q nk𝛼nk for some 𝛼nk . Consequently, there is m such that f nk𝛼nk m (t) = x nk . As a result, we obtain the inequality h(t) < f nk𝛼nk m (t), which is impossible. Thus, h = f . If S ⊂ Coz M(T, S), then we can take A nk consisting of a single element. Hence, ∏⟮A nk | (n, k) ∈ N × Z⟯ = ∏⟮{a nk } | (n, k) ∈ N × Z⟯ = {(a nk | (n, k) ∈ N × Z)}. Then, card ∏⟮A nk | (n, k) ∈ N × Z⟯ = 1, and therefore, card I = 𝜔. The following theorem gives some sufficient condition for a function f ∈ SM l (T, S) to be bounded below by some function g ∈ M(T, S). Theorem 1. Let S be a 𝜎-foundation on T such that S ⊂ Coz M(T, S). Then, every function f ∈ SM l (T, S) is bounded below by some function g ∈ M(T, S). Proof. First, suppose that f : T →] − 1, 1[. Consider the points y k ≡ −1 + 1/2k and the sets R k ≡ f −1 []y k , ∞[] ∈ S for all k ∈ 𝜔. By the condition, there are functions f k ∈ M(T, S)+ such that R k = coz f k . We can assume that f k ⩽ (1/2k+1 )1. Consider the functions h m ≡ ∑ (f k | k ∈ m + 1). Using Lemma 3 (1.4.8), for every p ⩾ m we obtain h p (t) − h m (t) = ∑ (f k (t) | m + 1 ⩽ k ⩽ p) = ∑ (1/2k+1 | m + 1 ⩽ k ⩽ p) ⩽ (1/2m+2 ) ∑ (1/2k | k ∈ p − m − 1) < 2/2m+2 = 1/2m+1 . This implies that the sequence (h m | m ∈ 𝜔) is inner uniformly convergent. By Lemma 3 (2.3.4) there is h ∈ M(T, S) such that h = u-lim (h m | m ∈ 𝜔). Clearly, coz h = ⋃⟮coz h m | m ∈ 𝜔⟯ = ⋃⟮coz f k | k ∈ 𝜔⟯ = T. Consider the function g ≡ −1 + h ≫ −1. Take t ∈ T. If f (t) > 0, then h(t) = lim (h m | m ∈ 𝜔) ⩽ lim (∑ (1/2k+1 | k ∈ m + 1) | m ∈ 𝜔) ⩽ 1 implies g(t) ⩽ 0 < f (t). If f (t) ⩽ 0, then y p+1 < f (t) ⩽ y p for some p ∈ 𝜔. Therefore, t ∈ R p+1 \R p ⊂ R p+1 \R q for every q ∈ p + 1. Thus, h(t) = lim (∑ (f k (t) | p + 1 ⩽ k ⩽ m) | m ∈ 𝜔) ⩽ lim (∑ (1/2k+1 | p + 1 ⩽ k ⩽ m) | m ∈ 𝜔) ⩽ (1/2p+2 ) lim (∑ (1/2k | k ∈ m − p) | m ∈ 𝜔) ⩽ 1/2p+1 implies g(t) ⩽ −1 + 1/2p+1 < f (t). Now, let 𝜑 ∈ SM l (T, S). Use the mapping u : R ] − 1, 1[ from Lemma 1 (1.4.9), i. e. consider the function f ≡ u ∘ 𝜑 = 𝜑/(1 + |𝜑|). Take any number x ∈] − 1, 1[ and 𝜉 ≡ u−1 (x) = x/(1 − |x|). Since u maps bijectively and strong monotonically the interval ]𝜉, ∞[ onto the interval ]x, 1[, the conditions t ∈ f −1 []x, 1[] and t ∈ 𝜑−1 []𝜉, ∞[] are equivalent. Consequently, f −1 []x, 1[] = 𝜑−1 []𝜉, ∞[] ∈ S. This means that f ∈ SM l (T, S). As was shown above, there is a function g ∈ M(T, S) such that −1 ≪ g ⩽ f . Apply the inverse mapping u−1 :] − 1, 1[ s R and consider the function 𝜓 ≡ u−1 ∘ g ≡ g/(1 − g). Take any real numbers 𝜉 < 𝜂 and consider the corresponding numbers

138 | 2.3 Families of measurable and distributable functions on a descriptive space

x ≡ u(𝜉) = 𝜉/(1 + |𝜉|) and y ≡ u(𝜂) = 𝜂/(1 + |𝜂|). Since u maps bijectively and strong monotonically the interval ]𝜉, 𝜂[ onto the interval ]x, y[, we infer that 𝜓−1 []𝜉, 𝜂[] = g −1 []x, y[]. Hence, 𝜓 ∈ M(T, S). Besides, 𝜓 ⩽ u−1 ∘ f = 𝜑. Corollary 1. Let S be a 𝜎-foundation on T such that S ⊂ Coz M(T, S). Then, for every function f ∈ SM l (T, S), there is a countable collection u ≡ (f i ∈ M(T, S) | i ∈ I) such that f = sup u in F(T). Proof. The assertion directly follows from Proposition 2 and Theorem 1. Corollary 2. Let S be a separable perfect 𝜎-foundation on T. Then, for every function f ∈ SM l (T, S) there is a countable collection u ≡ (f i ∈ M(T, S) | i ∈ I) such that f = sup u in F(T). Proof. Corollary 1 to Theorem 2 (2.3.5) provides that S = Coz M(T, S). Now, the assertion follows from the previous Corollary. Proposition 1 and Theorem 1 with its Corollaries will be used in 2.5.2 and then in 3.7 for some new characterizations of Riemann integrable functions different from the famous Lebesgue characterization. Since the families SM l (T, S), SM u (T, S), SM bl (T, S), and SM bu (T, S) are not linear spaces, it is natural to consider some their linear extensions. Consider the families BSM(T, S) ≡ {f1 − f2 | f1 , f2 ∈ SM l (T, S)+ } = {f1 − f2 | f1 , f2 ∈ SM u (T, S)+ } of all binary-semimeasurable functions and BSM b (T, S) ≡ {f1 − f2 | f1 , f2 ∈ SM bl (T, S)+ } = {f1 − f2 | f1 , f2 ∈ SM bu (T, S)+ } of all boundedly binary-semimeasurable functions on the space ⟮T, S⟯. It is clear that BSM b (T, S) = {f1 − f2 | f1 , f2 ∈ SM bl (T, S)} = {f1 − f2 | f1 , f2 ∈ SM bu (T, S)}. Lemma 2. Let S be a 𝜎-foundation on T. If f , g ∈ BSM(T, S) and x ∈ R, then xf , f + g, f ∧ g, f ∨ g, and fg belong to BSM(T, S). The similar assertions are valid for BSM b (T, S). Proof. By the definition, f = f1 − f2 and g = g1 − g2 for some f1 , f2 , g1 , g2 ∈ SM l (T, S). If x > 0, then by assertion 1 of Proposition 1 xf = xf1 − xf2 ∈ BSM. If x < 0, then xf = (−x)f2 − (−x)f1 ∈ BSM. By assertion 2 of Proposition 1 f + g = (f1 + g1 ) − (f2 + g2 ) ∈ BSM. By assertions 2 and 6 of Proposition 1 and Corollary 1 to Lemma 4 (1.4.5) f ∧ g = (f1 − f2 ) ∧ (g1 − g2 ) + f2 − f2 = f1 ∧ (g1 + f2 − g2 ) − f2 = f1 ∧ (g1 + f2 − g2 ) + g2 − g2 − f2 = (f1 + g2 ) ∧ (g1 + f2 ) − (f2 + g2 ) ∈ BSM. In a similar way, f ∨ g = (f1 + g2 ) ∨ (g1 + f2 )− (f2 + g2 ) ∈ BSM.

2.3.8 Families of semimeasurable functions on a space with an ensemble

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Let f , g ∈ BSM(T, S). Then, f = f1 − f2 and g = g1 − g2 for some f1 , f2 , g1 , g2 ∈ SM l (T, S)+ . By assertions 2 and 3 of Proposition 1, fg = (f1 g1 +f2 g2 )−(f1 g2 +f2 g1 ) ∈ BSM. For BSM b , the arguments are the same. Corollary 1. Let S be a 𝜎-foundation on T. Then, the families BSM(T, S) and BSM b (T, S) satisfy the conditions 1 – 5 from 2.2.4. The following result shows that BSM(T, S) contains S-step functions. Moreover, this family contains all K𝜑 -step functions, where K ≡ K(T, S) is the multiplicative derivative ensemble for S. Lemma 3. Let S be a 𝜎-foundation on a set T. Then, St(T, K𝜑 ) ⊂ BSM b (T, S). Proof. Since S is additive and multiplicative, Lemma 3 (2.1.1) implies that K𝜑 = K. Let A ∈ K𝜑 = K. Corollary 1 to Lemma 2 (2.1.1) guarantees that A = S ∩ R for some S ∈ S and R ∈ co-S. It is clear that 𝜒(S) ∈ SM bl (T, S) and 𝜒(R) ∈ SM bu (T, S). Consequently, 𝜒(A) = 𝜒(S)𝜒(R) ∈ BSM b (T, S) due to Lemma 7 (2.2.4) and Lemma 2. By the definition, if f ∈ St(T, K𝜑 ), then f = ∑(x i 𝜒(A i ) | i ∈ I) for some finite collections (A i ∈ K𝜑 | i ∈ I) and (x i ∈ R | i ∈ I). Now, again by Lemma 2 f ∈ BSM b (T, S). The following theorem shows that the “good” (i. e. normal) family M(T, K𝜎 ) of K𝜎 measurable functions is the normal envelope of the “bad” families SM l (T, S) and SM u (T, S) and the “semigood” family BSM(T, S). Theorem 2. Let S be a 𝜎-foundation on a set T. Then, N(SM l (T, S)) = N(SM u (T, S)) = N(BSM(T, S)) = M(T, K𝜎 ). Proof. By Corollary 2 to Proposition 2 (2.3.4) the family A(T) ≡ M(T, K𝜎 ) is normal. Let f ∈ SM l (T, S) and x, y ∈ R, x < y. Consider the sets X ≡ f −1 []x, y[], R ≡ f −1 []x, ∞[] ∈ S, and R n ≡ f −1 []y − 1/n, ∞[] ∈ S. By assertion 4 of Lemma 2 (2.2.1) f −1 [[y, ∞[] = ⋂⟮R n | n ∈ N⟯. This implies that X = R \ ⋂⟮R n | n ∈ N⟯ = ⋃⟮R ∩ (T \ R n ) | n ∈ N⟯ ∈ K𝜎 . Hence, f ∈ M(T, K𝜎 ). Thus, SM l (T, S) ⊂ A(T). Let B(T) be a normal family such that SM l (T, S) ⊂ B(T). Then, BSM(T, S) ⊂ B(T). By Proposition 1 (2.3.6) B(T) = M(T, E), where E ≡ Coz B(T). According to Proposition 3 (2.1.1), the ensemble K𝜑 is an algebra; therefore, it is a ring. By virtue of Lemma 2 (2.2.5), this implies the equality K𝜑 = Coz St(T, K𝜑 ). Then, using Lemma 3, we obtain K ⊂ Coz St(T, K𝜑 ) ⊂ Coz BSM(T, S) ⊂ E. By Corollary 1 to Proposition 1 (2.2.5), E is a 𝜎-foundation, so that K𝜎 ⊂ E. As a result, A(T) ⊂ B(T). For the upper semimeasurable functions, the arguments are the same. Corollary 1. Let S be a perfect 𝜎-foundation on a set T. Then, N(SM l (T, S)) = N(SM u (T, S)) = N(BSM(T, S)) = M(T, R𝜎 ), where R = co-S.

140 | 2.4 Families of uniform functions on a prescriptive space

Proof. Recall that K𝜎 = R𝜎 by virtue of Corollary 2 to Lemma 4 (2.1.1). The description of the boundedly normal envelope of the family BSM b (T, S) will be given in Theorem 1 (2.4.5) by means of the new family U(T, K) of K-uniform functions.

2.4 Families of uniform functions on a prescriptive space Some new class of functions has been introduced by V. K. Zakharov in papers [1981, 1985, 1989a, 1991, 1993b, 2006a]. This class extends the class of bounded measurable functions and plays an important role in modern analysis. These functions were introduced in the mentioned papers under the names of almost semicontinuous, metasemicontinuous, small oscillated, and uniform functions. The families M(T, S) and M b (T, S) of S-measurable functions have good properties only in the case when S is a 𝜎-foundation (see 2.3.2 – 2.3.4). But sometimes, the property of 𝜎-additivity is too much restrictive. In a more general case when S is only a foundation, the family of S-uniform functions has the same good properties as the families M b (T, S) for 𝜎-foundations S (see 2.4.2 and 2.4.3 below). 2.4.1 Uniform functions and their properties Recall that for a function f ∈ F(T) and a set E ⊂ T the number 𝜔(f , E) ≡ sup{|f (t) − f (s)| | t, s ∈ E} is called the oscillation of the function f on the set E (see 2.2.1). According to 2.2.3, if 𝜋 ≡ ⟮C i | i ∈ I⟯ is a cover of the set T, then the number 𝜔(f , 𝜋) ≡ sup⟮𝜔(f , C i ) | i ∈ I⟯ is called the oscillation of the function f on the cover 𝜋. Let ⟮T, C⟯ be a prescriptive space. A function f is called uniform on the prescriptive space ⟮T, C⟯ or uniform with respect to the covering C if for every 𝜀 > 0 there is a finite cover 𝜋 ∈ C such that 𝜔(f , 𝜋) < 𝜀. The family of all such functions will be denoted by U(T, C). By Lemma 1 (2.2.1), U(T, C) ⊂ F b (T). Let ⟮T, S⟯ be a descriptive space and T ∈ S. A function f is called uniform on a descriptive space ⟮T, S⟯ or uniform with respect to the ensemble S if it is uniform with respect to the covering Cov S. The family U(T, Cov S) of all such functions will be denoted also by U(T, S). It is clear that U(T, S) ⊂ D(T, S). The necessity to consider along with the class of families U(T, S) the wider class of families U(T, C) will be clarified in 2.4.6. Lemma 1. Let S be an ensemble on a set T. Then, U(T, S) = U(T, S𝜑 ). Proof. The assertion follows directly from the definition. Lemma 2. Let ⟮T, S⟯ be a compact descriptive space. Then, U(T, S) = D(T, S) = D b (T, S) = M b (T, S𝜎 ) = M(T, S𝜎 ).

2.4.1 Uniform functions and their properties |

141

Proof. Since the set T is S-compact, we get U(T, S) = D(T, S), and therefore, U b (T, S) = D b (T, S). By Lemma 1 (2.2.1) U(T, S) = U b (T, S). The definition of a distributable function and Proposition 1 (2.3.1) imply that D(T, S) = M(T, S𝜎 ), and therefore, D b (T, S) = M b (T, S𝜎 ). In particular, this assertion generalizes the following classical result. Corollary 1. Let ⟮T, G⟯ be a compact topological space. Then, C(T, G) = C b (T, G). Prove now two simple lemmas on relations between uniform and measurable functions. Lemma 3. Let S be a 𝜎-additive ensemble on a set T. Then, U(T, S) = M b (T, S). Proof. The assertion follows directly from Corollary 1 to Proposition 1 (2.3.1). Corollary 1. Let S be a 𝜎-additive ensemble on a set T. Then, U(T, S) = D b (T, S). Lemma 4. Let S be an ensemble on a set T. Then, M b (T, S) ⊂ U(T, S) ⊂ M b (T, S𝜎 ). Proof. Let f ∈ M b (T, S). Then, f [T] ⊂] − a, a [for some a > 0. Take any 𝜀 > 0 and positive natural number n such that 2a/n < 𝜀. Consider the sets C i ≡ f −1 []a(i − 1)/n, a(i + 1)/n[] ∈ S, |i| ∈ n. It is clear that ⋃ ⟮C i | |i| ∈ n⟯ = T and 𝜔(f , C i ) ⩽ 2a/n < 𝜀. Thus, f ∈ U(T, S). The second inclusion follows from the trivial inclusion U(T, S) ⊂ U(T, S𝜎 ) and Lemma 3. For non-𝜎-additive S, both the inclusions in Lemma 4 may be strict. We tell more about it in 2.4.5. Corollary 1 to Proposition 1 (2.3.1) characterizes the families M(T, S) and M b (T, S) of S-measurable functions for a 𝜎-additive ensemble S by means of their oscillations on countable or finite covers, i. e. in the “uniform” way. It is also possible to characterize uniform functions in the “measurable” way, i. e. by means of preimages. In this sense, the following result has a remote likeness with Proposition 1 (2.3.1). Proposition 1. Let S be a foundation on T and f ∈ F b (T). Then, the following assertions are equivalent: 1) f ∈ U(T, S); 2) if [x󸀠 , y󸀠 ] ⊂]x, y[, then there is a set X ∈ S𝜑 such that f −1 [[x󸀠 , y󸀠 ]] ⊂ X ⊂ f −1 []x, y[]. Proof. (1) ⊢ (2). Let [x󸀠 , y󸀠 ] ⊂]x, y[. Take 𝜀 ≡ sm(y − y󸀠 , x󸀠 − x)/2. Then, there is a finite cover 𝜋 ≡ (S i ∈ S | i ∈ I) such that 𝜔(f , 𝜋) < 𝜀. Consider the sets Y ≡ f −1 [[x󸀠 , y󸀠 ]], Z ≡ f −1 []x, y[], J ≡ {i ∈ I | S i ∩ Y ≠ ⌀}, and X ≡ ⋃ ⟮S j | j ∈ J⟯ ∈ S𝜑 . Clearly, Y ⊂ X. If t ∈ X, then t ∈ S j for some j ∈ J. Take a point s ∈ S j ∩Y. By construction, |f (t)−f (s)| < 𝜀.

142 | 2.4 Families of uniform functions on a prescriptive space

Therefore, f (t) ⩽ f (s) + 𝜀 ⩽ y󸀠 + 𝜀 ⩽ y󸀠 + (y − y󸀠 )/2 < y and f (t) ⩾ f (s) − 𝜀 ⩾ x󸀠 − 𝜀 ⩽ x󸀠 − (x󸀠 − x)/2 > x. Hence, X ⊂ Z. (2) ⊢ (1). Let 𝜀 > 0. By condition, there are a, b ∈ R such that rng f ⊂ [a, b]. By the Archimedes principle (Lemma 13 (1.4.3)), there is a number n ∈ N such that n > (b−a)/𝜀. Take the points x i ≡ a +(b − a)i/(3n) for i ∈ 3n +2 and x−1 ≡ a −(b − a)/(3n). Consider the sets Y i ≡ f −1 [[x i , x i+1 ]] and Z i ≡ f −1 []x i−1 , x i+2 [] for i ∈ 3n. Since [a, b] = ⋃ ⟮[x i , x i+1 ] | i ∈ 3n⟯, the collection ⟮Y i | i ∈ 3n⟯ is a cover of T. By condition, Xi ≡ {X ∈ S𝜑 | Y i ⊂ X ⊂ Z i } ≠ ⌀ for i ∈ 3n. Put X ≡ ⋃ ⟮Xi | i ∈ 3n⟯. According to the choice axiom from 1.1.12, there is a function p : P(X) \ {⌀} → X such that p(A) ∈ A. Consider the sets X i ≡ p(Xi ) ∈ Xi . Since Y i ⊂ X i ⊂ Z i , the collection 𝜉 ≡ (X i ∈ S𝜑 | i ∈ 3n) is a cover of T. From 𝜔(f , Z i ) < x i+2 − x i−1 < 𝜀, we infer that 𝜔(f , 𝜉) < 𝜀, and so f ∈ U(T, S𝜑 ) = U(T, S). Lemma 2 (2.3.1) guarantees that quite S-step functions are S-measurable for an additive ensemble S. The following lemma shows that these functions are S-uniform for an arbitrary ensemble S. Lemma 5. Let S be an ensemble on T. Then, qSt(T, S) ⊂ U(T, S). Proof. By definition of quite step functions, there are a finite partition 𝜋 ≡ ⟮S i ∈ S | i ∈ I⟯ of the set T and a collection (x i ∈ R | i ∈ I) such that f (t) = x i for any t ∈ S i . This partition is a cover of T and 𝜔(f , 𝜋) = 0. Corollary 1. Let S be an algebra on T. Then, St(T, S) ⊂ U(T, S). Proof. The assertion follows from Lemma 5 and Proposition 2 (2.2.4). Corollary 2. Let S be an algebra on T. Then, S ⊂ Coz(U(T, S)). Proof. Take any set S ∈ S and the function 𝜒(S) ∈ St(T, S). Then, by the previous corollary, 𝜒(S) ∈ U(T, S).

2.4.2 Pointwise operations over uniform functions The following theorems show that U(T, C) and U(T, S) have the same good properties as M b (T, S) (see 2.3.2 and 2.3.3). Theorem 1. Let C be a multiplicative covering on a set T. If f , g ∈ U(T, C) and r ∈ R, then 1) 1 ∈ U(T, C); 2) rf ∈ U(T, C); 3) f + g ∈ U(T, C);

2.4.2 Pointwise operations over uniform functions

4) 5) 6) 7) 8)

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fg ∈ U(T, C); 1/f ∈ U(T, C) for every function f : T → R \ {0} such that 1/f ∈ F b (T); f ∨ g, f ∧ g ∈ U(T, C); f + , f− , |f | ∈ U(T, C); m √ f ∈ U(T, C) for every f ⩾ 0.

Proof. Denote U(T, C) by A(T). Let 𝜀 > 0. Take positive numbers a and b such that |f | ⩽ a1 and |g| ⩽ b1. 1. For any cover 𝜋 ∈ C, we have 𝜔(1, 𝜋) = 0. 2. Let r ∈ R\{0}. For the number 𝜀1 ≡ 𝜀/|r|, there exists a finite cover 𝜋 ≡ ⟮C i | i ∈ I⟯ ∈ C such that 𝜔(f , 𝜋) < 𝜀1 . If s, t ∈ C i , then |(rf )(s) − (rf )(t)| = |r| |f (s) − f (t)| ⩽ |r| 𝜔(f , 𝜋). Therefore, 𝜔(rf , 𝜋) ⩽ |r| 𝜔(f , 𝜋) < 𝜀 and rf ∈ A(T). For r = 0, we have 𝜔(rf , 𝜋) = 𝜔(0, 𝜋) = 0 for every cover 𝜋 ∈ C. 3. For the numbers 𝜀1 ≡ 𝜀/2 and 𝜀2 ≡ 𝜀/2, there are finite covers 𝜋 ≡ ⟮C i | i ∈ I⟯ ∈ C and 𝜘 ≡ ⟮D j | j ∈ J⟯ ∈ C such that 𝜔(f , 𝜋) < 𝜀1 and 𝜔(g, 𝜘) < 𝜀2 . If s, t ∈ C i ∩ D j , then |(f + g)(s) − (f + g)(t)| ⩽ |f (s) − f (t)| + |g(s) − g(t)| ⩽ 𝜔(f , 𝜋) + 𝜔(g, 𝜘). Therefore, 𝜔(f + g, 𝜋 ∧ 𝜘) ⩽ 𝜔(f , 𝜋) + 𝜔(g, 𝜘) < 𝜀. Since C is multiplicative, we infer that 𝜋 ∧ 𝜘 ∈ C. Thus, f + g ∈ A(T). 4. Take the numbers 𝜀1 ≡ 𝜀/(2b) and 𝜀2 ≡ 𝜀/(2a) and consider finite covers 𝜋 ≡ ⟮C i | i ∈ I⟯ ∈ C and 𝜘 ≡ ⟮D j | j ∈ J⟯ ∈ C such that 𝜔(f , 𝜋) < 𝜀1 and 𝜔(g, 𝜘) < 𝜀2 . If s, t ∈ C i ∩D j , then |(fg)(s)−(fg)(t)| = |(f (s) g(s)− f (t) g(s)|+|(f (t) g(s)− f (t) g(t)| ⩽ |g(s)| |f (s)− f (t)|+ |f (t)| |g(s)−g(t)| ⩽ b𝜔(f , 𝜋)+a𝜔(g, 𝜘). Therefore, 𝜔(fg, 𝜋∧𝜘) ⩽ b𝜔(f , 𝜋)+a𝜔(g, 𝜘) < 𝜀. This means that fg ∈ A(T). 5. Suppose that f (t) = / 0 for all t ∈ T and 1/|f | ⩽ c1 for some positive number c. For 𝜀1 ≡ 𝜀/c2 , there exists a finite cover 𝜋 ≡ ⟮C i | i ∈ I⟯ ∈ C such that 𝜔(f , 𝜋) < 𝜀1 . If s, t ∈ C i , then |(1/f )(s) − (1/f )(t)| = |f (s) − f (t)|/(|f (s)| |f (t)|) ⩽ c2 𝜔(f , 𝜋). Hence, 𝜔(1/f , 𝜋) ⩽ c2 𝜔(f , 𝜋) < 𝜀. So that, 1/f ∈ A(T). 6. For the numbers 𝜀1 ≡ 𝜀/2 and 𝜀2 ≡ 𝜀/2, there are finite covers 𝜋 ≡ ⟮C i | i ∈ I⟯ ∈ C and 𝜘 ≡ ⟮D j | j ∈ J⟯ ∈ C such that 𝜔(f , 𝜋) < 𝜀1 and 𝜔(g, 𝜘) < 𝜀2 . If s, t ∈ C i ∩ D j , then by Birkhoff inequality (Corollary 1 to Theorem 4 (1.4.5)) |(f ∨ g)(s) − (f ∨ g)(t)| ⩽ |f (s) ⊻ g(s) − f (s) ⊻ g(t)| + |f (s) ⊻ g(t) − f (t) ⊻ g(t)| ⩽ |g(s) − g(t)| + |f (s) − f (t)| ⩽ 𝜔(g, 𝜘) + 𝜔(f , 𝜋). Hence, 𝜔(f ∨ g, 𝜋 ∧ 𝜘) ⩽ 𝜔(g, 𝜘) + 𝜔(f , 𝜋) < 𝜀, and we get f ∨ g ∈ A(T). Completely in the same way, we check that f ∧ g ∈ A(T). 7. By the definition f+ ≡ f ∨ 0, f− ≡ f ∧ 0. Therefore, these functions belong to A(T). By Lemma 1 (2.2.2) |f | = f+ − f− , so that |f | ∈ A(T). 8. According to the Archimedes principle (Lemma 13 (1.4.3)), there is n ∈ N such that a/n < 𝜀m /2. Divide the interval [0, a] by points x k = ak/n, k ∈ n + 1. Then, 0 = x0 < . . . < x n = a and x k+1 − x k = a/n. Take 𝜋 ≡ ⟮C i | i ∈ I⟯ ∈ C such that 𝜔(f , 𝜋) < a/n. Consider the sets X k ≡ −1 f [[x k , x k+1 ]]. For every i ∈ I, the set K i ≡ {k ∈ n | C i ∩ X k = / ⌀} is non-empty and finite. Put k i ≡ sm K i . Suppose s ∈ C i and take a point t ∈ C i ∩ X k i . Then, m x f (s) = f (s) − f (t) + f (t) < 𝜔(f , C i ) + x k i +1 and f (s) ⩾ −𝜔(f , C i ) + x k i . Hence, √ ki
0, there is m ∈ N such that |f (t) − f n (t)| < 𝜀/3 for every n ⩾ m. Consider a cover finite 𝜋 ≡ ⟮C i | i ∈ I⟯ ∈ C such that 𝜔(f m , 𝜋) < 𝜀/3. Then, |f (s) − f (t)| = |f (s) − f m (s) + f m (s) − f m (t) + f m (t) − f (t)| ⩽ |f (s) − f m (s)| + |f m (s) − f m (t)| + |f m (t) − f (t)| < 𝜀 for all s, t ∈ C i . Therefore, 𝜔(f , 𝜋) < 𝜀 and f ∈ U(T, C). Corollary 1. Let C be a covering on a set T. Then, the family U(T, C) satisfies condition 6 from 2.2.4. Corollary 2. Let C be a multiplicative covering on a set T. Then, the family U(T, C) is boundedly normal (in the sense of 2.2.4). Proof. The assertion follows from Corollary 2 to Theorem 1 (2.4.2) and the previous Corollary. Corollary 3. Let S be an ensemble on a set T containing T, f ∈ F(T), and f = u-lim (f n | n ∈ N) for some net (f n ∈ U(T, S) | n ∈ N). Then, f ∈ U(T, S). Proof. The assertion follows directly from Proposition 1. Corollary 4. Let S be a foundation on T. Then, the family U(T, S) is boundedly normal (in the sense of 2.2.4). Proof. The assertion follows from Corollary 5 to Theorem 1 (2.4.2) and the previous Corollary. Corollary 3 shows that the family of S-uniform functions is closed under uniform convergence for every foundation S without the condition of 𝜎-additivity of S. In contrast, Lemma 5 (2.3.4) shows that for the family of S-measurable functions this condition is essential. Lemma 1. Let S be a foundation on T. Then, every inner uniformly convergent sequence in U(T, S) has a uniform limit in U(T, S). Proof. Let (f n ∈ U(T, S) | n ∈ 𝜔) be inner uniformly convergent. By Lemma 6 (2.2.3), there is f ∈ F(T) such that f = u-lim (f n | n ∈ 𝜔). By Corollary 3 to Proposition 1, f ∈ U(T, S). Corollary 1. Let S be a foundation on T. Then, every inner uniformly convergent sequence in qSt(T, S) has a uniform limit in U(T, S). Proof. By Lemma 5 (2.4.1), qSt(T, S) ⊂ U(T, S). Corollary 2. Let S be an algebra on T. Then, every inner uniformly convergent sequence in St(T, S) has a uniform limit in U(T, S).

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Proof. By Corollary 1 to Lemma 5 (2.4.1), we have St(T, S) ⊂ U(T, S). Proposition 2. Let R be a ring on T. Then, for every function f ∈ U(T, R), there is an increasing sequence (f n ∈ qSt(T, R) | n ∈ 𝜔) such that f = u-lim (f n | n ∈ 𝜔). Proof. Consider finite covers (R ni ∈ R | i ∈ I n ) of T such that 𝜔(f , R ni ) < 1/n for every i ∈ I n . Use of Lemma 8 (2.1.1) yields that there are finite partitions (S nk ∈ R | k ∈ K ⊂ 𝜔) such that S nk ⊂ R ni for some i. Consider the numbers x nk ≡ inf(f (t) | t ∈ S nk ). Put g0 ≡ 0 and define the quite R-step functions g n setting g n (t) ≡ x nk for every t ∈ S nk . Take any n ∈ N and t ∈ T. Then, t ∈ S nk for some k ∈ K. Therefore, using Lemma 4 (1.4.5), we get 0 ⩽ f (t) − g n (t) = f (t) − inf(f (s) | s ∈ S nk ) = inf(f (t) − f (s) | s ∈ S nk ) ⩽ 𝜔(f , R ni ) < 1/n. This means that f = u-lim (g n | n ∈ 𝜔). Setting f n ≡ sup (g k | k ∈ n + 1) for every n ∈ 𝜔, we obtain the increasing sequence (f n | n ∈ 𝜔). By Proposition 3 (2.2.4), all functions f n also belong to qSt(T, R). Take any 𝜀 > 0 and p ∈ N such that |f − g n | < 𝜀1 for all n ⩾ p. Then, for any n ⩾ p, we have 0 ⩽ f (t) − f n (t) ⩽ f (t) − f p (t) ⩽ f (t) − g p (t) < 𝜀 for all t ∈ T. This means that f = u-lim (f n | n ∈ 𝜔). Corollary 1. Let R be a ring on T. Then, U(T, R) is the uniform closure of qSt(T, R) in F(T). Proof. It follows from Proposition 2 and Corollary 1 to Lemma 1. 2.4.4 Separability of sets by uniform functions Here, analogously to 2.3.5, we shall see that adding the condition of separability to the foundations S considered above provides the family U(T, S) by a row of new crucial properties. Proposition 1. Let C be a multiplicative covering on a set T. Then, the ensemble Coz U(T, C) is an a-foundation (i. e. a separable perfect 𝜎-foundation). Proof. By Corollary 2 to Theorem 1 (2.4.2) and Corollary 1 to Proposition 1 (2.4.3), the family U(T, C) satisfies conditions 1, 2, 3, 4, and 6 from 2.2.4. Therefore, Proposition 1 (2.2.5) implies that Coz U(T, C) is an a-foundation. Corollary 1. Let S be a foundation. Then, the ensemble Coz U(T, S) is an a-foundation. Proof. This follows from Lemma 3 (2.1.5) and Proposition 1. Now we present the Lebesgue – Urysohn – Alexandrov theorem (Theorem 1 (2.3.5)) in a new formulation.

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Theorem 1. Let S be a separable foundation on a set T. Then, for any two non-empty disjoint sets A, B ∈ co-S, there is a function f ∈ U(T, S) such that 0 ⩽ f ⩽ 1, f [A] = {0}, and f [B] = {1}. Proof. Note that in the proof of Theorem 1 (2.3.5), we have constructed the finite cover of T, and therefore, we have already obtained f ∈ U(T, S). Corollary 1. Let S be a separable foundation on a set T. Then, for every non-empty disjoint sets A, B ∈ co-S and for every real number a ⩽ b, there is a function f ∈ U(T, S) such that rng f ⊂ [a, b], f [A] = {a}, and f [B] = {b}. Proof. Take the function f from Theorem 1. Then, the function f 󸀠 ≡ a1 + (b − a)f has the necessary properties. Theorem 2. Let S be a separable perfect foundation on T. Then, for every set S ∈ S there is a positive function f ∈ U(T, S) such that S = coz f . Proof. Since S is perfect, S = ⋃ ⟮R n | n ∈ 𝜔⟯ for some sequence (R n ∈ R | n ∈ 𝜔), where R ≡ co-S. By Theorem 1, the set V n ≡ {f ∈ U(T, S) | 0 ⩽ f ⩽ 1 ∧ (f [T \ S] = {0}) ∧ (f [R n ] = {1})} is non-empty for every n ∈ 𝜔. By the choice axiom from 1.1.12, there is a mapping p : P( Map(T, [0, 1])) \ {⌀} → Map(T, [0, 1]) such that p(E) ∈ E. For every n, take the function f n ≡ p(V n ). Define the sequence (g n | n ∈ 𝜔) setting g n ≡ ∑ ((1/2)m f m | m ∈ n). By Corollary 3 to Theorem 1 (2.4.2), g n ∈ U(T, S). For every t ∈ T, we have 0 ⩽ g n+1 (t) − g n (t) ⩽ (1/2)n . Therefore, by Lemma 4 (1.4.8), the sequence (g n | n ∈ 𝜔) is inner uniformly convergent. By Lemma 6 (2.2.3), it is uniformly convergent to some f ∈ F(T). By virtue of Corollary 3 to Proposition 1 (2.4.3), f ∈ U(T, S). It is clear that S = coz f . The following result gives a characterization of a-foundations in the class of all foundations (cf. Corollary 1 to Theorem 2 (2.3.5)). Corollary 1. Let S be a foundation on T. Then, the following conclusions are equivalent: 1) S is an a-foundation (i. e. a separable perfect 𝜎-foundation); 2) S = Coz U(T, S). Proof. (1) ⊢ (2). The inclusion S ⊂ Coz U(T, S) follows from Theorem 2. Since S is a 𝜎-foundation, Lemma 3 (2.4.1) and Lemma 1 (2.3.5) imply Coz U(T, S) = Coz M b (T, S) ⊂ Coz M(T, S) ⊂ S. (2) ⊢ (1). It follows from Proposition 1. Clarify the relation between an arbitrary foundation S and the corresponding a-foundation Coz U(T, S) in one partial case.

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Lemma 1. Let ⟮S𝛼 | 𝛼 ∈ A⟯ be a collection of separable perfect foundations on T. Then, Coz U(T, S), where S ≡ ⋃ ⟮S𝛼 | 𝛼 ∈ A⟯, is the smallest a-foundation on T containing every S𝛼 . Proof. Let S ∈ S𝛼 for some 𝛼 ∈ A. Since S𝛼 is a separable perfect foundation, Theorem 2 implies that there is a function f ∈ U(T, S𝛼 ) ⊂ U(T, S) such that S = coz f . Hence, S ∈ Coz U(T, S). Thus, S𝛼 ⊂ Coz U(T, S). If E is an a-foundation containing all S𝛼 , then U(T, S) ⊂ U(T, E). Using Corollary 1 to Theorem 2, we get Coz U(T, S) ⊂ Coz U(T, E) = E. Corollary 1. Let S be a separable perfect foundation on T. Then, Coz U(T, S) is the smallest a-foundation on T containing S, i. e. the a-envelope A(S) of S. Unfortunately, the description of the a-envelope A(S) in the form Coz U(T, S) is not internal. By this reason, we present below some internal description of A(S). Let C be a multiplicative family of finite covers of T. Consider the family U(T, C) of all C-uniform functions on T and the a-foundation Coz U(T, C) of all cozero-sets of the family U(T, C) (see Proposition 1). Recall that according to the definition in 1.1.10, rng 𝜘 = ⋃ ⟮Q i | i ∈ I⟯ for any collection 𝜘 ≡ ⟮Q i ⊂ T | i ∈ I⟯. A finite cover 𝜘 ≡ ⟮Q i ⊂ T | i ∈ I⟯ of the set T will be called a cover with the graduation (𝜘(k), 𝜘(k) | k ∈ l + 1) if 𝜘(k) and 𝜘(k) 𝜘(k) ≡ ⟮Q i ⊂ T | i ∈ I(k) ⊂ I⟯ and 𝜘(k) ≡ ⟮Q i ⊂ T | i ∈ I(k) ⊂ I⟯ are subcollections of the collection 𝜘 such that 1) ⌀ = I(0) ⊂ I(1) ⊂ . . . ⊂ I(l) ⊂ I, I ⊃ I(0) ⊃ I(1) ⊃ . . . ⊃ I(l) = ⌀; 2) rng 𝜘(k) ∩ rng 𝜘(k) = ⌀ for every k ∈ l; 3) rng 𝜘(k + 1) ∪ rng 𝜘(k) = T for every k ∈ l. The following theorem gives an internal description of the a-foundation Coz U(T, C) generated by the multiplicative covering C. Theorem 3. Let C be a multiplicative covering on a set T and G ⊂ T. Then, the following conclusions are equivalent: 1) G ∈ Coz U(T, C); 2) there exists a sequence of finite covers (𝜘n ∈ C | n ∈ N) with the graduations (𝜘n (k), 𝜘n (k) | k ∈ 2n + 1) such that a) G = ⋃ ⟮rng 𝜘n (2) | n ∈ N⟯; b) rng 𝜘n+1 (2k) ⊃ rng 𝜘n (k) and rng 𝜘n+1 (2k) ⊃ rng 𝜘n (k) for every k ∈ 2n + 1; c) rng 𝜘n+1 (2k + 1) ∪ rng 𝜘n (k) = T and rng 𝜘n+1 (2k + 1) ∪ rng 𝜘n (k + 1) = T for every k ∈ 2n−1 + 1. Proof. (1) ⊢ (2). Let G = coz f for some function f ∈ U(T, C). Without loss of generality, we can assume that 0 ⩽ f ⩽ 1. Consider the sets X n (k) ≡ f −1 [[0, k/2n − 1/2n+2 ]] and

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Y n (k) ≡ f −1 [[k/2n + 1/2n+2 , 1]]. Obviously, X n (k) ⊂ X n (k+1) and Y n (k) ⊃ Y n (k+1). Using the choice axiom we can take for every n such cover 𝜘n ≡ ⟮Q ni ⊂ T | i ∈ I n ⟯ ∈ C that 𝜔(f , 𝜘n ) < 1/2n+3 . Consider the sets I n (k) ≡ {i ∈ I n | Q ni ∩ X n (k) ≠ ⌀} and I n (k) ≡ {i ∈ I n | Q ni ∩ Y n (k) ≠ ⌀} and the corresponding collections 𝜘n (k) ≡ ⟮Q ni | i ∈ I n (k)⟯ and 𝜘n (k) ≡ ⟮Q ni | i ∈ I n (k)⟯. It is clear that I n (k) ⊂ I n (k + 1) and I n (k) ⊃ I n (k + 1). Suppose s ∈ Q ni for some i ∈ I n (k); then, there is a point r ∈ Q ni ∩ X n (k). Hence, f (s) < f (r) + 1/2n+3 ⩽ k/2n −1/2n+3 . Similarly, f (s) > k/2n +1/2n+3 for every s ∈ Q ni and every i ∈ I n (k). Therefore, rng 𝜘n (k)∩rng 𝜘n (k) = ⌀ for every k ∈ 2n . Since X n (k+1)∪Y n (k) = T, we have I n (k + 1) ∪ I n (k) = I n for every k ∈ 2n . Besides, X n (0) = ⌀ implies I n (0) = ⌀ and Y n (2n ) = ⌀ implies I n (2n ) = ⌀. Thus, 𝜘n is a cover with the graduation (𝜘(k), 𝜘(k) | k ∈ 2n + 1). It is clear that G = ⋃ ⟮rng 𝜘n (k) | n ∈ N⟯ for every k. If s ∈ Q ni for some i ∈ I n (k), then f (s) < 2k/2n+1 −1/2n+3 means that s ∈ X n+1 (2k). Hence, rng 𝜘n (k) ⊂ rng 𝜘n+1 (2k). If s ∈ Q ni for some i ∈ I n (k), then f (s) < 2k/2n+1 − 1/2n+3 means that s ∈ Y n+1 (2k). Hence, rng 𝜘n (k) ⊂ rng 𝜘n+1 (2k). Since Y n (k) = f −1 [[2k/2n+1 + 1/2n+2 , 1]] and X n+1 (2k + 1) = f −1 [[0, (2k + 1)/2n+1 − 1/2n+3 ]], we have X n (2k + 1) ∪ Y n (k) = T. Therefore, rng 𝜘n+1

(2k + 1) ∪ rng 𝜘n (k) = T. Finally, X n (k + 1) = f −1 [[0, (2k + 2)/2n+1 − 1/2n+2 ]] and Y n+1 (2k+1) = f −1 [[(2k + 1)/2n+1 + 1/2n+3 , 1]] imply X n (k+1)∪Y n+1 (2k+1) = T, and therefore, rng 𝜘n+1 (2k + 1) ∪ rng 𝜘n (k + 1) = T. (2) ⊢ (1). For every n ∈ N and k ∈ 2n + 1, define the sets E n (k) ≡ rng 𝜘n (k), G n (k) ≡ rng 𝜘n (k), and Z n (k) ≡ T \ G n (k). According to the definition of graduation, E n (k) ⊂ E n (k + 1), G n (k) ⊃ G n (k + 1), and E n (k) ∩ G n (k) = ⌀, where E n (k) ⊂ Z n (k) and Z n (k) ⊂ Z n (k+1). Define the function f : T → R setting f (s) ≡ sup{k/2n | s ∈ ̸ E n (k+1)} for every s ∈ T. Therefore, 0 ⩽ f ⩽ 1. Consider the sets D n (k) ≡ E n (k+1)\Z n (k−1) for k ∈ 2n \1. Let s ∈ D n (k). Then, s ∈ ̸ Z n (k − 1) implies s ∈ ̸ E n (k − 1). Hence, f (s) ⩾ (k − 2)/2n . Prove that f (s) ⩽ (k+1)/2n . Assume the converse. Since s ∈ E n (k+1) and f (s) > (k+1)/2n , there is a dyadic rational number l/2m such that l/2m > (k+1)/2n and s ∈ ̸ E m (l+1). If we combine this with condition (b) and the inequality 2n 2l/2m+n+1 > 2m+1 (k + 1)/2m+n+1 , then we get s ∈ E n (k + 1) ⊂ E m+n+1 (2m+1 (k + 1)) ⊂ E m+n+1 (2n 2l) ⊂ Z m+n+1 (2n 2l) ⊂ Z m+n+1 (2n (2l + 1)). On the other hand, by condition (c), it follows from s ∈ ̸ E m (l + 1) that s ∈ G m+1 (2l + 1), i. e. s ∈ ̸ Z m+1 (2l + 1) ⊃ Z m+n+1 (2n (2l + 1)). This contradiction proves the inequality f (s) ⩽ (k+1)/2n . Therefore, (k−2)/2n ⩽ f (s) ⩽ (k+1)/2n for s ∈ D n (k), k ∈ 2n \1, and f (s) ⩽ 1/2n for s ∈ E n (1). Besides, we establish that f (s) ⩾ (2n − 2)/2n for s ∈ G n (2n − 1) = T \ Z n (2n − 1). Thus, 𝜔(f , D n (k)) ⩽ 3/2n , 𝜔(f , E n (1)) ⩽ 1/2n , and 𝜔(f , G n (2n − 1)) ⩽ 2/2n . Let Q ni be a member of the collection 𝜘n . If i ∈ I n (1), then Q ni ⊂ E n (1). If i ∈ I n \ I n (2n ), then I n = I n (2n )∪I n (2n −1) implies i ∈ I n (2n −1), and therefore, Q ni ⊂ G n (2n −1). If the previous two cases are not realized, then i ∈ I n (k+1)\I n (k) for some k ∈ 2n \1. Then, it follows from I n = I n (k)∪I n (k−1) that i ∈ I n (k−1). Therefore, Q ni ⊂ E n (k+1)∩G n (k−1). Consequently, 𝜔(f , Q ni ) < 4/2n . This means that f ∈ U(T, C). Let s ∈ G. By condition (a), s ∈ ̸ Z n (2) for some n ∈ N. This implies f (s) ⩾ n 1/2 > 0. Let s ∈ ̸ G. Again, by condition (a) s ∈ ⋂ ⟮E n (2) | n ∈ N⟯. Hence, f (s) ⩽ inf (2/2n | n ∈ N), i. e. f (s) = 0. Consequently, G = coz f .

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Corollary 1. Let S be a foundation on T and G ⊂ T. Then, the following conclusions are equivalent: 1) G ∈ Coz U(T, S); 2) there is a sequence of finite covers (𝜘n ∈ Cov S | n ∈ N) with the graduations (𝜘n (k), 𝜘n (k) | k ∈ 2n + 1) possessing properties (a) – (c) from Theorem 3. Proof. The family C ≡ Cov S is multiplicative by virtue of Lemma 3 (2.1.5). Corollary 2. Let S be a 𝜎-foundation on T and G ⊂ T. Then, the following conclusions are equivalent: 1) G ∈ Coz M b (T, S); 2) there is a sequence of finite covers (𝜘n ∈ Cov S | n ∈ N) with the graduations (𝜘n (k), 𝜘n (k) | k ∈ 2n + 1) possessing properties (a) – (c) from Theorem 3. Proof. By Lemma 3 (2.4.1), U(T, S) = M b (T, S). Corollary 3. Let S be a separable perfect foundation on T and G ⊂ T. Then, the following conclusions are equivalent: 1) G ∈ A(S); 2) there is a sequence of finite covers (𝜘n ∈ Cov S | n ∈ N) with the graduations (𝜘n (k), 𝜘n (k) | k ∈ 2n + 1) possessing properties (a) – (c) from Theorem 3. Proof. By Corollary 1 to Lemma 1 A(S) = Coz U(T, S).

2.4.5 Symmetrizable functions on a space with an ensemble Now we consider one important subclass of the class of uniform functions. This class is crucial for solving the Riesz – Radon – Fréchet problem of characterization of Radon integrals as linear functionals and will be widely used in 3.6. Consider for an ensemble S on a set T the co-ensemble R ≡ co-S and the derivative ensemble K ≡ K(T, S) from 2.1.1. The elements of the ensemble K will be called symmetrizable sets. Uniform functions on the set T with respect to the ensemble K will be called symmetrizable on the descriptive space ⟮T, S⟯. The family U(T, K) of all symmetrizable functions will be denoted by S(T, S). By Lemma 1 (2.4.1), S(T, S) = U(T, K𝜑 ). Lemma 4 (2.4.1) implies that M b (T, S) ⊂ M b (T, K) ⊂ S(T, S) ⊂ M b (T, K𝜎 ). In the particular case of a 𝜎-algebra S, we have K = S = S𝜎 = K𝜎 , and therefore, S(T, S) = M b (T, S). The following result shows that not only bounded K-measurable functions are symmetrizable but also all bounded S-semimeasurable (and binary-semimeasurable) functions.

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Lemma 1. Let S be an ensemble on T. Then, BSM b (T, S) ⊂ S(T, S). Proof. Let f ∈ SM bl (T, S). Take any 𝜀 > 0 and divide an interval containing rng f by points a k such that a k+1 − a k ⩽ 𝜀/3. Consider the sets A k ≡ f −1 []a k−1 , ∞[\]a k+1 , ∞[] = f −1 []a k−1 , ∞[]\f −1 []a k+1 , ∞[]. Since K𝜑 is an algebra containing S, we infer that A k ∈ K𝜑 . It follows from f −1 [[a k , a k+1 ]] ⊂ A k that the sets A k cover T. Besides, 𝜔(f , A k ) ⩽ a k+1 − a k−1 < 𝜀. Hence, f ∈ U(T, K𝜑 ) = S(T, S). If f ∈ SM bu (T, S), then by Proposition 1 (2.3.8), −f ∈ SM bl (T, S). Using assertion 2 of Corollary 3 to Theorem 1 (2.4.2), we obtain that f = −(−f ) ∈ S(T, S). Therefore, using assertions 2 and 3 of this Theorem, we can conclude that BSM b (T, S) ⊂ S(T, S).

Consider the algebra A(T, S) generated by the ensemble S. According to Proposition 3 (2.1.1), the algebra A(T, S) consists of unions ⋃ ⟮K i | i ∈ I⟯ of finite collections (K i ∈ K | i ∈ I) of symmetrizable sets, i. e. A = K𝜑 . Lemma 2. Let S be an ensemble on T and f ∈ S(T, S). Then, there is an increasing sequence (f n ∈ qSt(T, A) | n ∈ 𝜔) such that f = u-lim (f n | n ∈ 𝜔). Proof. Since S(T, S) ≡ U(T, K) = U(T, K𝜑 ) and A = K𝜑 is an algebra, we can apply Proposition 2 (2.4.3) to R = A. Theorem 1. Let S be a ensemble on T. Then, the family S(T, S) is the uniform closure of the family St(T, K𝜑 ). Besides, if S is a 𝜎-foundation, then the family S(T, S) is the uniform closure and also the boundedly normal envelope of the family BSM b (T, S). Proof. Put A(T) ≡ BSM b (T, S) and B(T) ≡ S(T, S) ≡ U(T, K). By Corollary 1 to Proposition 2 (2.4.3), the family B(T) ≡ U(T, K) = U(T, K𝜑 ) is a uniform closure of q St(T, K𝜑 ). For a 𝜎-foundation S, we have St(T, K𝜑 ) ⊂ A(T) by Lemma 3 (2.3.8). Then, Lemma 1 implies B(T) = U(St(T, K𝜑 )) ⊂ U(A(T)) ⊂ U(B(T)) = B(T). By Corollary 4 to Proposition 1 (2.4.3), the family B(T) is boundedly normal. Moreover, by Lemma 1 A(T) ⊂ B(T). Let E(T) be some boundedly normal family such that A(T) ⊂ E(T). Then, B(T) = U(A(T)) ⊂ E(T). This means that B(T) = BN(A(T)). Lemma 3. Let S be an ensemble on T, Q be a 𝜎-foundation on T, and S(T, S) = M b (T, Q). Then, S(T, S) = M b (T, A(T, S)𝜎 ). Proof. By Theorem 1 (2.3.6), the family M(T, Q) is normal, and therefore, by Proposition 1 (2.2.5), Coz M(T, Q) is a 𝜎-foundation. Then, Proposition 1 (2.3.6) guarantees the equality M(T, Q) = M(T, Coz M(T, Q)). Hence, S(T, S) = M b (T, Q) = M b (T, Coz M(T, Q)) = = M b (T, Coz M b (T, Q)) = M b (T, (Coz M b (T, Q))𝜎 ) = M b (T, (Coz S(T, S))𝜎 ).

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By Proposition 3 (2.1.1), K𝜑 = A(T, S) is an algebra, by Lemma 1 (2.4.1) S(T, S) ≡ U(T, K) = U(T, K𝜑 ). Then, Corollary 2 to Lemma 5 (2.4.1) implies A ⊂ Coz S(T, S). Therefore, A𝜎 ⊂ (Coz S(T, S))𝜎 and M b (T, A𝜎 ) ⊂ M b (T, (Coz S(T, S))𝜎 ) = S(T, S). But S(T, S) ⊂ M b (T, S𝜎 ) ⊂ M b (T, A𝜎 ) by Lemma 4 (2.4.1). Thus, S(T, S) = M b (T, A𝜎 ). Proposition 1. There is a topological space ⟮T, G⟯ with T = [0, 1[⊂ R such that M b (T, K(T, G)) ⫋ S(T, G) ⫋ M b (T, K(T, G)𝜎 ). Proof. Consider the topological space ⟮T, G⟯ from Lemma 5 (2.3.4) taking T ≡ [0, 1[⊂ R and G ≡ {[0, b[ | 0 < b ⩽ 1}∪{⌀}. Then, F ≡ co-G = {[a, 1[ | 0 ⩽ a < 1} and K = {G∩F | G ∈ G ∧ F ∈ F} = {[a, b[ | 0 ⩽ a < b ⩽ 1} ∪ {⌀}. The inclusions hold by virtue of Lemma 4 (2.4.1). Further, prove that S(T, G) ≡ U(T, K(T, G)) ⊄ M b (T, K). Take the function g : T → [0, 1] defined in the proof of Lemma 5 (2.3.4). Recall that g(t) ≡ 2−n for all t ∈ Δ n ≡ [2−(n+1) , 2−n [, n ∈ 𝜔, and g(0) ≡ 0. There we proved that g ∈ ̸ M(T, K) because g −1 []0, 2[] =]0, 1[∈ ̸ K. Check that g ∈ U(T, K). For 𝜀 > 1, we have 𝜔(g, 𝜋0 ) = 1 < 𝜀 for the cover 𝜋0 ≡ {T}. For any 𝜀 ∈]0, 1], take j0 ∈ N such that 2−j0 < 𝜀 ⩽ 2−j0 +1 . Put K j ≡ Δ j ∈ K for j ∈ j0 , K j0 ≡ [0, 2−j0 [∈ G ⊂ K. Therefore, 𝜔(g, K j ) = 0 for j ∈ j0 and 𝜔(g, K j0 ) = 2−j0 . Thus, 𝜔(g, 𝜋) = 2−j0 < 𝜀 for the cover 𝜋 ≡ ⟮K j | j ∈ j0 + 1⟯. Finally, prove that M b (T, K𝜎 ) ⊄ U(T, K). Define the sets Γn ≡ [1 − 2−n , 1 − 2−(n+1) [, n ∈ 𝜔, and the function h : T → [0, 1], setting h(t) ≡ 1 − 2−2k for t ∈ Γ2k , k ∈ 𝜔, and h(t) ≡ 0 otherwise. If x ⩾ 1 or y ⩽ 0, then h−1 []x, y[] = ⌀; if x < 0 < 1 ⩽ y, then h−1 []x, y[] = T; if 0 ⩽ x < 1 ⩽ y, then h−1 []x, y[] = ⋃ ⟮Γ2n | n > n0 ⟯; if 0 ⩽ x < y < 1, then h−1 []x, y[] is a finite union ⋃ ⟮Γi | i ∈ I⟯ or is empty; if x < 0 < y < 1, then h−1 []x, y[] is a countable union ⋃ ⟮Γm | m ∈ M⟯. Since Γn ∈ K, n ∈ 𝜔, we infer that h−1 []x, y[] ∈ K𝜎 in all the cases above. Thus, h ∈ M b (T, K𝜎 ). Check now that h ∈ ̸ U(T, K). Take 𝜀 ≡ 1/2 and a finite cover 𝜋󸀠 ≡ ⟮K j ∈ K | j ∈ J⟯ of T. Since all non-empty elements of K have the form [a, b[ with 0 ⩽ a < b ⩽ 1, there is j1 ∈ J such that K j1 = [a, 1[. This implies that 1 − 2−n ∈ K j1 for every n ⩾ n1 ∈ 𝜔, where 2−n1 ⩽ 1 − a < 2−n1 +1 . Therefore, ⋃ ⟮Γn | n ⩾ n1 ⟯ ⊂ K j1 . Hence, 󵄨 󵄨 𝜔(h, 𝜋󸀠 ) ⩾ 𝜔(h, K j1 ) ⩾ 𝜔(h, Γn1 +1 ∪ Γn1 +2 ) = 󵄨󵄨󵄨h(1 − 2−(n1 +1) ) − h(1 − 2−(n1 +2) )󵄨󵄨󵄨 ⩾ 1 − 2−(n1 +1) ⩾ 1/2 ≡ 𝜀. Corollary 1. There is a topological space ⟮T, G⟯ such that S(T, G) ≡ U(T, K(T, G)) = / M b (T, Q) for every 𝜎-foundation Q on T. Proof. Take the topological space ⟮T, G⟯ from Proposition 1. Then, S(T, G) ⫋ M b (T, K(T, G)𝜎 ) ⊂ M b (T, A(T, G)𝜎 ). Hence, Lemma 3 implies that S(T, G) = / M b (T, Q).

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This corollary shows that, generally speaking, it is impossible to describe the family of uniform functions as a family of bounded measurable functions. Therefore, the class of families of uniform functions is really wider than the accustomed class of families of measurable functions. Remark. In addition to Proposition 1, we can prove that even in the classical case of the unit interval T ≡ [0, 1[ with the usual topology O ≡ Ost |T (see 2.3.1), the imbedding M b (T, K(T, O)) ⫋ S(T, O) is strict. Since for every x, y ∈ R, x ⩽ y, we have [x, y] = ⋂ ⟮]x − 1/n, y + 1/n[| n ∈ N⟯, we infer that Oint ⊂ (co-Ost )𝜎 . Then, Statement 1 (2.3.1) provides that Ost = (Oint )𝜎 ⊂ (co-Ost )𝜎 . Thus, Ost is a perfect ensemble. Using Corollary 2 to Lemma 4 (2.1.1), we conclude that K𝜎 = F𝜎 . It follows from Corollary 1 to Lemma 2 (2.1.1) that K = {G ∩ F | G ∈ O ∧ F ∈ F}, where F = co-O. Consider the sets D k ≡ {p2−k | p = 1, 3, 5, . . . , 2k − 1} for every k ∈ N and the set D of all dyadic-rational numbers in ]0, 1[. Then, D = ⋃ ⟮D k | k ∈ N⟯. Define the function h : T → [0, 1] setting h(t) ≡ 2−k if x ∈ D k and h(t) ≡ 0 if t ∈ T \D. Prove that h ∈ ̸ M(T, K). For the interval ]0, 3/4[, we have h−1 []0, 3/4[] = D. Suppose that D ∈ K, i. e. D = G ∩ F for some G ∈ O and F ∈ F. Note that F =]0, ̸ 1[ because {0} ∈ ̸ O. Hence, if F ≠ [0, 1[, then there is t 0 ∈]0, 1[ such that t0 ∈ ̸ F, i. e. t0 ∈ T \F ∈ O. This implies that there are x0 , y0 ∈]0, 1[ such that t0 ∈]x0 , y0 [⊂ T \ F. According to Lemma 17 (1.4.3), there is a point t1 ∈ D such that t1 ∈]x0 , y0 [. Therefore, t1 ∈ T \ F. We have a contradiction with t1 ∈ D ⊂ F. This means that F = [0, 1[. Consequently, D = G ∈ O. But there is no interval ]x, y[ contained in D. Hence, we get a contradiction. Thus, D ∈ ̸ K, and therefore, h ∈ ̸ M(T, K). Now prove that h ∈ U(T, K). Take 𝜀 > 0. If 𝜀 > 1/2 , then 𝜔(h, 𝜘0 ) = 1/2 < 𝜀 for the cover 𝜘0 ≡ {T}. Let 𝜀 ∈]2−(k+1) , 2−k ] for some k ∈ N. Put A j ≡ {j2−k } and B j ≡ ](j − 1)2−k , j2−k [ for j = 1, . . . , 2k , A0 ≡ {0}. We see that A j ∈ F ⊂ K, B j ∈ O ⊂ K, and 𝜔(h, A j ) = 0. Besides, it follows from B j ∩ D i = ⌀ for all j = 1, . . . , 2k and i = 1, . . . , k that 0 ⩽ h(t) ⩽ 2−(k+1) for any t ∈ B j . Therefore, 𝜔(h, B j ) < 𝜀. Thus, for the cover 𝜘k ≡ (C n | n = 0, 1, . . . , 2k+1 ), where C n ≡ A n for n = 0, . . . , 2k and C n ≡ B n−2k for n = 2k + 1, . . . , 2k+1 we have that 𝜔(h, 𝜘k ) < 𝜀. Therefore, h ∈ U(T, K).

2.4.6 Descriptions of boundedly normal families and envelopes. Naturalness of the family of uniform functions In this subsection, we shall show that the notion of a uniform function is as well natural in F b (T) as the notion of a measurable function in F(T) because all boundedly normal families A(T) ⊂ F b (T) (see 2.2.4) are represented as families of uniform functions. We describe boundedly normal families and envelopes following the papers [Zakharov, 2008b; Zakharov and Rodionov, 2008b].

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Lemma 1. Let A(T) ⊂ F b (T). 1) If A(T) is boundedly normal (or even satisfies only conditions 1, 1󸀠 , 2󸀠 , 3, and 4 from 2.2.4), then A(T) ⊂ U(T, Coz A(T)) and the family U(T, Coz A(T)) is also boundedly normal. 2) There exists a boundedly normal family A(T) such that A(T) ≠ U(T, Coz A(T)) = M b (T, Coz A(T)). Proof. 1. Let f ∈ A(T) and 𝜀 > 0. Since f ∈ F b (T), there exists N ∈ N such that |f | < N1. Take n ∈ N such that 2N/n < 𝜀 and consider the rational numbers r nm = Nm/n, m ∈ 𝜔, and the sets S nm = f −1 []r n,m−1 , r n,m+1 [], |m| ∈ n. The collection 𝜋n ≡ (S nm | |m| ∈ n) is a cover of T such that 𝜔(f , 𝜋n ) ⩽ 2N/n < 𝜀. By Lemma 3 (2.2.5), S nm = coz g nm for the function g nm ≡ ((f − r n,m−1 1) ∨ 0) ∧ ((r n,m+1 1 − f ) ∨ 0). Since A(T) is closed under addition and multiplication by the numbers of the form 1/k, k ∈ N, we conclude that A(T) is closed under multiplication by rational numbers. Besides, A(T) contains 1 and is closed under finite exact bounds. Hence, g ∈ A(T), and therefore, S nm ∈ S ≡ Coz A(T). Thus, f ∈ U(T, S) and therefore, A(T) ⊂ U(T, S). By Proposition 1 (2.2.5), S is a foundation. Then, by Corollary 4 to Proposition 1 (2.4.3), U(T, S) is a boundedly normal family. 2. Consider the topological space ⟮T, G⟯ from Proposition 1 (2.4.5) and put A(T) ≡ S(T, G). By Corollary 4 to Proposition 1 (2.4.3), the family A(T) is boundedly normal. By Proposition 1 (2.2.5), Coz A(T) is a 𝜎-foundation. Therefore, Corollary 1 to Proposition 1 (2.4.5) and Lemma 3 (2.4.1) imply that A(T) ≠ U(T, Coz A(T)) = M b (T, Coz A(T)). This lemma shows that for the description of the boundedly normal families the sampled boundedly normal families U(T, Coz A(T)) and M b (T, Coz A(T)) are not appropriate. We need a smaller boundedly normal family. For this purpose, consider the covering Cov A(T) consisting of all finite covers 𝜘 ≡ ⟮C i | i ∈ I⟯ of a set T such that 𝜘 is either the one-member cover ⟮C i | i ∈ {i}⟯ with C i ≡ T or for 𝜘 there exist a collection (f i ∈ A(T)+ | i ∈ I) and a number l ∈ N such that C i = coz f i and (cozl f i | i ∈ I) is also a cover of T, where cozl f ≡ {t ∈ T | |f (t)| > 1/l} (see 2.2.5). First, we have the desired inclusion U(T, Cov A(T)) ⊂ U(T, Coz A(T)). Lemma 2. Let A(T) ⊂ F b (T) be closed under finite infimum. Then, the covering Cov A(T) is multiplicative and the family of functions U(T, Cov A(T)) is boundedly normal. Proof. Let 𝜘, 𝜌 ∈ C ≡ Cov A(T), 𝜘 ≡ ⟮C i | i ∈ I⟯, C i = coz f i , ⋃ ⟮cozl f i | i ∈ I⟯ = T, 𝜌 ≡ ⟮D j | j ∈ J⟯, D j = coz g j , and ⋃ ⟮cozm g j | j ∈ J⟯ = T. For K ≡ I × J, k ≡ (i, j) ∈ K, E k ≡ C i ∩ D j , and h k ≡ f i ∧ g j ∈ A(T), we have E k = coz h k . Similarly, for n ≡ l ⊻ m, we obtain cozn h k = cozn f i ∩ cozn g j ⊃ cozl f i ∩ cozm g j . Therefore, ⋃ ⟮cozn h k | k ∈ K⟯ = T. This means that 𝜘 ∧ 𝜌 ≡ (E k | k ∈ K) ∈ C, i. e. C is binary multiplicative. It can be proven by induction that C is (finitely) multiplicative.

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By Corollary 2 to Proposition 1 (2.4.3), the family U(T, Cov A(T)) is boundedly normal. For the definition of the associated family A(T)∗ , we refer to 2.3.6. Lemma 3. Let A(T) ⊂ F b (T). Then, A(T) ⊂ U(T, Cov A(T)∗ ). Proof. Suppose u ∈ A(T) and u(t) ∈ [−z, z] for all t ∈ T. Take 𝜀 > 0 and n ∈ N such that 1/n < 𝜀. Consider the numbers x i ≡ −z + 2zi/(3n) for i ∈ I ≡ 3n + 1. Consider also the functions f i ≡ ((u − x i−1 1) ∨ 0) ∧ ((x i+2 1 − u) ∨ 0) ∈ A(T)∗ and a collection 𝜘 ≡ (C i | i ∈ I) such that C i ≡ coz f i . Suppose t ∈ T. Then, u(t) ∈ [x i , x i+1 ] for some i. Hence, f i (t) ⩾ (x i − x i−1 ) ∧ (x i+2 − x i+1 ) ⩾ 1/3n > 1/4n, i. e. t ∈ coz4n f i . This implies that (coz4n f i | i ∈ I) is a cover of T. Therefore, 𝜘 ∈ Cov A(T)∗ . If s, t ∈ C i , then ((u(s) − x i−1 ) ∨ 0) ∧ ((x i+2 − u(s)) ∨ 0) = f i (s) > 0 yields that x i−1 < u(s) < x i+2 . Similarly, x i−1 < u(t) < x i+2 . Consequently, |u(s)− u(t)| < x i+2 − x i−1 = 1/n and 𝜔(u, C i ) ⩽ 1/n < 𝜀. This means that A(T) ⊂ U(T, Cov A(T)∗ ). Proposition 1. Let A(T) ⊂ F b (T) be a boundedly normal family. Then, A(T) = U(T, Cov A(T)). Proof. By Lemma 2, the family C ≡ Cov A(T) is multiplicative. Then, by Corollary 2 to Proposition 1 (2.4.3), the family B(T) ≡ U(T, C) is boundedly normal. It follows from Lemma 1 (2.3.6) that Cov A(T)∗ = Cov A b (T)+ = Cov A(T). Then, Lemma 3 implies A(T) ⊂ U(T, Cov A(T)∗ ) = B(T). Conversely, suppose f ∈ B(T)+ . Take 𝜀 > 0. By definition, 𝜔(f , 𝜘) < 𝜀 for a finite collection (f i ∈ A(T) | i ∈ I) and a number l ∈ N such that 𝜘 = (C i | i ∈ I), C i = coz f i , and ⋃ ⟮cozl f i | i ∈ I⟯ = T. Consider the numbers r i ≡ inf(f (t) | t ∈ C i ) and the functions g i ≡ (r i lf i ) ∧ (r i 1) and g ≡ sup(g i | i ∈ I) from A(T). If t ∈ D i ≡ cozl f i , then g i (t) = r i . If t ∈ C i , then g i (t) ⩽ r i ⩽ f (t). If t ∈ ̸ C i , then g i (t) = 0 ⩽ f (t). Therefore, g i ⩽ f and we get 0 ⩽ f (t)− g(t) ⩽ f (t)− g i (t) = f (t)− r i < 𝜀 for every t ∈ D i . This implies that using the choice axiom from 1.1.12, we can construct a sequence of functions (g n ∈ A(T) | n ∈ N) such that f = u-lim (g n | n ∈ N). Since A(T) is uniformly closed, we conclude that f ∈ A(T). If f is an arbitrary function from B(T), then f = f+ +f− , where f+ ≡ f ∨0 ∈ B(T)+ and −f− ≡ −(f ∧ 0) ∈ B(T)+ . By the above, f+ ∈ A(T) and −f− ∈ A(T). Thus, f = f+ +f− ∈ A(T) and therefore, B(T) ⊂ A(T). Corollary 1. Let A(T) ⊂ F b ((T) be a boundedly normal family. Then, A(T) = U(T, Cov A(T)∗ ). Proof. It follows from Lemma 1 (2.3.6) that Cov A(T)∗ = Cov A b (T)+ = Cov A(T). Further, we apply Proposition 1.

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The following theorem gives us a characterization of boundedly normal families A(T) ⊂ F b (T). Theorem 1 (the Zakharov theorem on boundedly normal families). Let A(T) ⊂ F b (T). Then, the following assertions are equivalent: 1) the family A(T) is boundedly normal; 2) A(T) = U(T, C) for some multiplicative covering C containing the one-member cover; 3) A(T) = U(T, (Cov A(T))𝜂 ); 4) A(T) = U(T, (Cov A(T)∗ )𝜂 ). Proof. (1) ⊢ (3). By Lemma 2, the covering C ≡ Cov A(T) is multiplicative. Therefore, by Proposition 1, we get A(T) = U(T, C) = U(T, C𝜂 ). (3) ⊢ (2). By Lemma 2 (2.1.5), the covering (Cov A(T))𝜂 is multiplicative. Obviously, (Cov A(T))𝜂 contains one-member cover. (2) ⊢ (1). It follows from Corollary 2 to Proposition 1 (2.4.3). (3) ⊢ (4). It was already prove that A(T) is boundedly normal. Therefore, Lemma 1 (2.3.6) implies that Cov A(T)∗ = Cov A b (T)+ = Cov A(T). (4) ⊢ (2). By Lemma 2 (2.1.5), the covering (Cov A(T)∗ )𝜂 is multiplicative. Theorem 1 gives us the opportunity to describe the boundedly normal envelope BN(A(T)) of an arbitrary family of bounded functions A(T) ⊂ F b (T), i. e. the smallest boundedly normal family B(T) containing the family A(T) (see 2.2.4). Theorem 2 (the Zakharov – Rodionov theorem on the boundedly normal envelope). Let A(T) ⊂ F b (T). Then, the family U(T, (Cov A(T)∗ )𝜂 ) is the boundedly normal envelope BN(A(T)) of the family A(T). Proof. By Lemma 2 (2.1.5), the covering C ≡ (Cov A(T)∗ )𝜂 is multiplicative. Therefore, by Corollary 2 to Proposition 1 (2.4.3), the family E(T) ≡ U(T, C) is boundedly normal. By Lemma 3, we have A(T) ⊂ U(T, Cov A(T)∗ ). Hence, A(T) ⊂ E(T). Finally, suppose B(T) is a boundedly normal family and A(T) ⊂ B(T). Then, by Theorem 1, we have B(T) = U(T, (Cov B(T)∗ )𝜂 ). Since Cov A(T)∗ ⊂ Cov B(T)∗ , we obtain E(T) ⊂ U(T, (Cov B(T)∗ )𝜂 ) = B(T). Further, we shall give another description of some boundedly normal families ascending to the alternative description of the family of uniform functions presented in Proposition 1 (2.4.1). Lemma 4. Let A(T) ⊂ F b (T) be closed with respect to finite supremum and multiplication by natural numbers. Then, for every function f ∈ A(T), the following conclusions are equivalent:

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for any real number x < x󸀠 ⩽ y󸀠 < y, there are a function g ∈ A(T)+ and a natural number n such that f −1 [[x󸀠 , y󸀠 ]] ⊂ cozn g ⊂ coz g ⊂ f −1 []x, y[]; 2) for every 𝜀 > 0 there are finite collections (g i ∈ A(T)+ | i ∈ I) and (n i ∈ N | i ∈ I) such that ⋃ ⟮cozn i g i | i ∈ I⟯ = T and 𝜔(f , coz g i ) < 𝜀 for every i ∈ I.

1)

Proof. (1) ⊢ (2). Take a function f ∈ A(T) and a number 𝜀 > 0. Divide an interval containing rng f by some points a i such that a i+1 − a i < 𝜀/3. By (1), there are functions g i ∈ A(T)󸀠+ and numbers n i such that f −1 [[a i , a i+1 ]] ⊂ cozn i g i ⊂ coz g i ⊂ f −1 []a i−1 , a i+2 []. Then, ⋃ ⟮cozn i g i | i ∈ I⟯ = T and 𝜔(f , coz g i ) ⩽ a i+2 − a i−1 < 𝜀. (2) ⊢ (1). Take a function f ∈ A(T), any real number x < x󸀠 ⩽ y󸀠 < y, and a number 𝜀 ≡ sm(x󸀠 − x, y − y󸀠 )/2. By (2), there are (g i ∈ A(T)+ | i ∈ I) and (n i ∈ N | i ∈ I) such that ⋃ ⟮cozn i g i | i ∈ I⟯ = T and 𝜔(f , coz g i ) < 𝜀. Consider the sets S ≡ f −1 []x, y[],

S󸀠 ≡ f −1 [[x󸀠 , y󸀠 ]], and J ≡ {i ∈ I | cozn i g i ∩ S󸀠 ≠ ⌀}. It is clear that S󸀠 ⊂ ⋃ ⟮cozn i g i | i ∈ J⟯. Take the function g ≡ sup (n i g i | i ∈ I) ∈ A(T)+ . It follows from Lemma 4 (2.2.5) that ⋃ ⟮cozn i g i | i ∈ J⟯ = coz1 g. Lemma 1 (2.2.5) implies that ⋃ ⟮coz g i | i ∈ J⟯ = coz g. If t ∈ coz g, then t ∈ coz g i for some i ∈ J. Take a point s ∈ coz g i ∩S󸀠 . Then, f (s) ∈ [x󸀠 , y󸀠 ]. Hence, f (t) < f (s) + 𝜀 < y󸀠 + y − y󸀠 = y and f (t) > f (s) − 𝜀 > x󸀠 − x󸀠 + x = x. This means that coz g ⊂ S.

Proposition 2. Let A(T) ⊂ F b (T). Then, the following conclusions are equivalent: 1) A(T) is a boundedly normal family; 2) A(T) is closed with respect to finite supremum and infimum and to multiplication by natural numbers and A(T) consists of all functions f ∈ F b (T) such that for any x < x󸀠 ⩽ y󸀠 < y there are g ∈ A(T)+ and n ∈ N such that f −1 [[x󸀠 , y󸀠 ]] ⊂ cozn g ⊂ coz g ⊂ f −1 []x, y[]. Proof. (1) ⊢ (2). Take a boundedly normal family A(T) and denote the family with the property mentioned in (2) by B(T). It follows from Lemma 5 (2.2.5) that A(T) ⊂ B(T). Take any f ∈ B(T) such that |f | ⩽ 1. Fix a number n ∈ N. Take a number m ∈ N such that 1/m < 1/(3n) and consider the numbers x ni ≡ i/m for all |i| ∈ m + 2. By the condition, there are functions g ni ∈ A(T)+ and numbers l ni such that X ni ≡ f −1 [[x n,i−1 , x n,i+1 ]] ⊂ cozl ni g ni ⊂ coz g ni ⊂ f −1 []x n,i−2 , x n,i+2 [] ≡ Y ni . It is clear that ⋃ ⟮X ni | |i| ∈ m + 2⟯ = T. Consider the functions h ni ≡ −1 − 1/n + ((m + i − 1)l/m)(g ni ∧ (1/l)1) ∈ A(T) and the function h n ≡ sup (h ni | |i| ∈ m + 2) ∈ A(T). If t ∈ Y ni , then h ni (t) ⩽ −1 − 1/n+ (m + i − 1)/m = (i − 1)/m − 1/n = (i − 2)/m + 1/m − 1/n < f (t). If t ∈ ̸ Y ni , then h ni (t) = −1 − 1/n < f (t). Consequently, h n ⩽ f . If t ∈ T, then t ∈ X ni for some i. Therefore, 0 ⩽ f (t) − h n (t) ⩽ f (t) − h ni (t) ⩽ x n,i+2 + 1 + 1/n − (m + i − 1)/m = (i + 2)/m + 1/n− (i − 1)/m = 3/m + 1/n < 2/n. This means that 0 ⩽ f − h n ⩽ (2/n)1, where f = u-lim (h n | n ∈ N). Since A(T) is uniformly closed, we infer that f ∈ A(T).

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Now if f is any function in B(T), then |xf | ⩽ 1 for some number x > 0. Therefore, xf ∈ A(T) by the arguments above, where f ∈ A(T). Thus, B(T) ⊂ A(T). (2) ⊢ (1). Let f , g ∈ A(T) and 𝜀, 𝜀1 , 𝜀2 > 0. By Lemma 4, there are finite collections (g j ∈ A(T)+ | j ∈ J), (m j ∈ N | j ∈ J), (h k ∈ A(T)󸀠+ | k ∈ K), and (n k ∈ N | k ∈ K) such that ⋃ ⟮cozm j g j | j ∈ J⟯ = T, 𝜔(f , coz g j ) < 𝜀1 , ⋃ ⟮cozn k h k | k ∈ K⟯ = T, and 𝜔(g, coz h k ) < 𝜀2 . By Lemma 4 (2.2.5), cozm j g j ∩ cozn k h k = cozl i f i for i = (j, k), some p j , q k , l i ∈ N, and f i ≡ p j g j ∧ q k h k ∈ A(T)+ . Lemma 1 (2.2.5) implies that coz f i = coz g j ∩ coz h k . It is clear that ⋃ ⟮cozl i f i | i ∈ J × K⟯ = T. Take 𝜀1 ≡ 𝜀2 ≡ 𝜀/2. If s, t ∈ coz f i , then |(f (s) + g(s)) − (f (t) + g(t))| ⩽ |f (s) − f (t)| + |g(s) − g(t)| ⩽ 𝜔(f , coz g j ) + 𝜔(g, coz h k ) < 𝜀, where 𝜔(f + g, coz f i ) < 𝜀. By Corollary 1 to Theorem 4 (1.4.5), |f (s) ∨ g(s) − f (t) ∨ g(t)| ⩽ |f (s) ∨ g(s) − f (s) ∨ g(t)| + |f (s) ∨ g(t) − f (t) ∨ g(t)| ⩽ |g(s)−g(t)|+|f (s)−f (t)| ⩽ 𝜔(g, coz g j )+𝜔(f , coz h k ) < 𝜀, where 𝜔(f ∨ g, coz f i ) < 𝜀. Similarly, 𝜔(f ∧ g, coz f i ) < 𝜀. Then, by Lemma 4, we get f + g, f ∨ g, f ∧ g ∈ A(T). Take now 𝜀1 ≡ 𝜀/‖f ‖, 𝜀2 ≡ 𝜀/‖g‖, where ‖f ‖ ≡ ‖f ‖u ≡ sup (|f (t)| | t ∈ T), ‖g‖ ≡ ‖g‖u ≡ sup (|g(t)| | t ∈ T). Then, |f (s)g(s)−f (t)g(t)| ⩽ |f (s)g(s)−f (s)g(t)|+|f (s)g(t)−f (t)g(t)| < ‖f ‖𝜀1 + ‖g‖𝜀2 = 𝜀, where 𝜔(fg, coz f i ) < 𝜀 and fg ∈ A(T). Further, let 1/f ∈ F b (T), i. e. |1/f | ⩽ z1 for some z ∈ R+ . Take 𝜀1 ≡ 𝜀/z2 . Then, |1/f (s) − 1/f (t)| ⩽ |f (s) − f (t)|/(|f (s)| |f (t)|) < 𝜀1 z2 = 𝜀, where 𝜔(1/f , coz f i ) < 𝜀 and 1/f ∈ A(T). Finally, take a sequence (f n ∈ A(T) | n ∈ 𝜔) and a function f ∈ F b (T) such that f = u-lim (f n | n ∈ 𝜔). Fix 𝜀 > 0 and take some m ∈ 𝜔 such that |f − f m | < (𝜀/3)1. By Lemma 4, there are finite collections (g i ∈ A(T)+ | i ∈ I) and (n i ∈ N | i ∈ I) such that ⋃ ⟮cozn i g i | i ∈ I⟯ = T and 𝜔(f m , coz g i ) < 𝜀. If s, t ∈ coz g i , then |f (s) − f (t)| ⩽ |f (s) − f m (s)| + |f m (s) − f m (t)| + |f m (t) − f (t)| < 𝜀, where f ∈ A(T) by Lemma 4. Thus, the family A(T) is boundedly normal. Using this description, we can characterize some boundedly normal envelopes. Proposition 3. Let A(T) ⊂ F b (T) satisfy conditions 1 – 4 from 2.2.4. Then, BN(A(T)) consists of all functions f ∈ F b (T) such that for any real number x < x󸀠 ⩽ y󸀠 < y, there are g ∈ A(T)+ and n ∈ N such that f −1 [[x󸀠 , y󸀠 ]] ⊂ cozn g ⊂ coz g ⊂ f −1 []x, y[]. Proof. Denote the mentioned family of functions by B(T). If f ∈ A(T), then f ∈ B(T) by virtue of Lemma 5 (2.2.5). Hence, A(T) ⊂ B(T). By Proposition 2, the family B(T) is boundedly normal. Let E(T) be a boundedly normal family containing A(T). By Proposition 2, E(T) has property 2 from it. Therefore, B(T) ⊂ E(T). This means that B(T) is the boundedly normal envelope of A(T). 2.4.7 Fine correlations between Baire’s and Borel’s functional collections In 2.3.7, the general Baire – Borel correlation was investigated. In the capacity of the initial family, the family M(T, K𝜎 ) of measurable functions was used.

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The notion of uniform function gives the a good opportunity to find another Baire – Borel correlation with considerably more narrow initial family (fine correlation). Fine convergence classification theorems Proposition 4 (2.3.4) and Theorem 3 (2.3.4) shows that one can uniformly approximate measurable functions by some complicated step functions. The next result shows the opportunity of pointwise approximation, but by more simple step functions. Proposition 1. Let S be an ensemble on a set T and f ∈ M b (T, K𝜎 ). Then, there exist sequences (g n ∈ qSt(T, K𝜑 ) | n ∈ 𝜔) ↑ and (h n ∈ qSt(T, K𝜑 ) | n ∈ 𝜔) ↓ such that f = sup (g n | n ∈ 𝜔) = inf (h n | n ∈ 𝜔) in F(T). Proof. Let f : T →] − a, a[. Consider the numbers x ni ≡ ai/(2n) for all n ∈ N and |i| ∈ 2n + 1. Consider also the corresponding sets X ni ≡ f −1 []x n,i−1 , x n,i+1 [] ∈ K𝜎 . By definition, X ni = ⋃ ⟮K nim | m ∈ 𝜔⟯ for some sequence (K nim ∈ K | i ∈ 𝜔). Define functions g nim ⩽ f and h nim ⩾ f setting g nim (t) ≡ −a and h nim (t) ≡ a for all t ∈ ̸ K nim and g nim (t) ≡ x n,i−1 and h nim (t) ≡ x n,i+1 for all t ∈ K nim . Since K ⊂ K𝜑 and by Proposition 3 (2.1.1), K𝜑 is an algebra, we infer that these functions belong to qSt(T, K𝜑 ). Let t ∈ T. Then, t ∈ K nim for every n ∈ N and some i and m. Therefore, 0 ⩽ f − g nim (t) < x n,i+1 − x n,i−1 = 1/n and similarly 0 ⩽ h nim (t) − f (t) < 1/n. By Lemma 1 (1.4.5), f (t) = sup (g nim (t) | m ∈ 𝜔, |i| ⩽ 2n + 1, n ∈ N) and f (t) = inf (h nim (t) | m ∈ 𝜔, |i| ⩽ 2n + 1, n ∈ N). By Lemma 3 (1.3.9), the set J ≡ 𝜔×]−2a, 2a[×N of indices j ≡ (n, i, m) is countable. Consider some bijective mapping u : 𝜔 J and the functions g n ≡ sup (g u(i) | i ∈ n) and h n ≡ inf (g u(i) | i ∈ n). Then, g n ↑, h n ↓, and f = sup (g n | n ∈ 𝜔) = inf (h n | n ∈ 𝜔). Corollary 1. Let S be an ensemble on a set T. Then, for every function f ∈ M b (T, K𝜎 ), there are sequences (g n ∈ U(T, K) | n ∈ 𝜔) ↑ and (h n ∈ U(T, K) | n ∈ 𝜔) ↓ such that f = sup (g n | n ∈ 𝜔) = inf (h n | n ∈ 𝜔) in F(T). Proof. Lemma 5 (2.4.1) provides the inclusion qSt(T, K𝜑 ) ⊂ U(T, K𝜑 ) = U(T, K). Proposition 2. Let S be an ensemble on a set T and f ∈ M(T, K𝜎 )+ . Then, there exists a sequence (f n ∈ U(T, K)+ | n ∈ 𝜔) ↑ such that f = sup (f n | n ∈ 𝜔) in F(T). Proof. Consider the functions h k ≡ f ∧ (k1). By Corollary 1 to Proposition 1, there are sequences (g kn ∈ U(T, K) | n ∈ 𝜔) ↑ such that h k = sup (g kn | n ∈ 𝜔). By Theorem 1 (1.3.8), the set 𝜔 × 𝜔 is denumerable. Therefore, there is a bijective mapping u:𝜔×𝜔 𝜔. Denote the function g kn such that u(k, n) = p by g p . It is clear that f = sup (g p | p ∈ 𝜔). Define new functions f q ≡ sup (g p | p ∈ q) for q ∈ N and f0 ≡ f1 . Then, (f q | q ∈ 𝜔) ↑ and f = sup (f q | q ∈ 𝜔).

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Corollary 1. Let S be an ensemble on a set T. Then, M(T, K𝜎 ) ⊂ p-Lim U(T, K). Under some additional assumptions, we can replace the family U(T, K) in this corollary by the more narrow family U(T, S). Proposition 3. Let S be a separable foundation on a set T. Let f ∈ M b (T, R𝜎 ) and f takes only finite set of values rng f = {a i | i ∈ I}. Then, there is a sequence (f m ∈ U(T, S) | m ∈ 𝜔) such that f = p-lim⟮f m | m ∈ 𝜔⟯ and rng f m ⊂ [a, b] for every m, where a = sm (a i | i ∈ I) and b = gr (a i | i ∈ I). Proof. All the arguments are the same as the arguments in the proof of Proposition 1 (2.3.7); only, instead of Theorem 1 (2.3.5), we should use Theorem 1 (2.4.4). Proposition 4. Let S be a separable perfect foundation on a set T. Then, M b (T, K𝜎 ) = M b (T, R𝜎 ) ⊂ bp-Lim U(T, S). Proof. Since S is perfect, Corollary 2 to Lemma 4 (2.1.1) guarantees that K𝜎 = R𝜎 . All the further arguments are the same as the arguments in the first part of the proof of Theorem 1 (2.3.7); only, instead of Proposition 1 (2.3.7), we should use Proposition 3, and instead of Theorem 1 (2.3.2), we should use Corollary 3 to Theorem 1 (2.4.2). The following result was obtained in [Zakharov and Rodionov, 2014a]. Theorem 1 (the main fine convergence classification theorem). Let S be an ensemble on a set T and E be a separable perfect foundation on T such that S ⊂ E ⊂ K𝜎 . Then, 1) M(T, Λ 𝛼 (T, S)) ⊂ Lim𝛼 U(T, E) ⊂ M(T, Λ 𝛼+1 (T, S)) for every 𝛼 ∈ [1, 𝜔[; 2) Lim𝛼 U(T, E) = M(T, Λ 𝛼+1 (T, S)) for every 𝛼 ∈ [𝜔, 𝜔1 [; 3) BM(T, S) ≡ M(T, B(T, S)) = Lim𝜔1 U(T, E). Proof. 1. Using Corollary 1 to Theorem 4 (2.3.7), Proposition 4, the first equality from Lemma 6 (2.2.4), and Lemma 4 (2.4.1), we get M(T, Λ n (T, S)) = Limn−1 M(T, K𝜎 ) ⊂ ⊂ Limn−1 M(T, K(T, E)𝜎 ) ⊂ Limn−1 p-Lim U(T, E) = Limn U(T, E) ⊂ ⊂ Limn M(T, E𝜎 ) ⊂ Limn M(T, K𝜎 ) = M(T, Λ n+1 (T, S)). for every n ∈ [1, 𝜔[. 2. Using Proposition 4, the second equality from Lemma 6 (2.2.4), and Lemma 4 (2.4.1), we get Lim𝛼 M(T, K𝜎 ) ⊂ Lim𝛼 M(T, K(T, E)𝜎 ) ⊂ Lim𝛼 p-Lim U(T, E) = = Lim𝛼 U(T, E) ⊂ Lim𝛼 M(T, E𝜎 ) ⊂ Lim𝛼 M(T, K𝜎 ).

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for every 𝛼 ∈ [𝜔, 𝜔1 [. Consequently, Lim𝛼 U(T, E) = Lim𝛼 M(T, K𝜎 ). Now, by virtue of Corollary 1 to Theorem 4 (2.3.7), we get the required equality. 3. From the inclusions E ⊂ K𝜎 ⊂ B(T, S) ⊂ B(T, E), we see that B(T, S) = B(T, E). By Theorem 1 (2.1.3), the Borel envelope B(T, S) = ⋃ ⟮Λ 𝛼 (T, S) | 𝛼 ∈ 𝜔1 ⟯. Then, M(T, B(T, S)) = ⋃ ⟮M(T, Λ 𝛼 (T, S)) | 𝛼 ∈ 𝜔1 ⟯. Since M(T, Λ 0 (T, S)) ⊂ M(T, Λ 1 (T, S)), M(T, Λ 𝛼 (T, S)) ⊂ Lim𝛼 U(T, E) for 𝛼 ∈ [1, 𝜔[ by assertion 1, and also M(T, Λ 𝛼 (T, S)) ⊂ M(T, Λ 𝛼+1 (T, S)) = Lim𝛼 U(T, E) for 𝛼 ∈ [𝜔, 𝜔1 [by assertion 2, we infer that M(T, B(T, S)) ⊂ ⋃ ⟮Lim𝛼 U(T, E) | 𝛼 ∈ 𝜔1 ⟯ ⊂ ⊂ ⋃ ⟮M(T, Λ 𝛼+1 (T, S) | 𝛼 ∈ 𝜔1 ⟯ ⊂ M(T, B(T, S)). Thus, we get BM(T, S) ≡ M(T, B(T, S)) = ⋃ ⟮Lim𝛼 U(T, E) | 𝛼 ∈ 𝜔1 ⟯ ≡ L. According to Lemma 4 (2.2.4), we obtain that L = Lim𝜔1 U(T, E). Corollary 1 (the general case). Let S be an ensemble on a set T. Then, 1) M(T, Λ 𝛼 (T, S)) ⊂ Lim𝛼 U(T, K) ⊂ M(T, Λ 𝛼+1 (T, S)) for every 𝛼 ∈ [1, 𝜔[; 2) Lim𝛼 U(T, K) = M(T, Λ 𝛼+1 (T, S)) for every 𝛼 ∈ [𝜔, 𝜔1 [; 3) BM(T, S) ≡ M(T, B(T, S)) = Lim𝜔1 U(T, K). Proof. By Proposition 3 (2.1.1), the ensemble K𝜑 is a perfect foundation and an algebra. Then, it follows immediately from the definition of a separable ensemble that K𝜑 is separable. It is clear from the definition of uniform function that U(T, K𝜑 ) = U(T, K). Applying now Theorem 1 to E ≡ K𝜑 , we get the required assertions. Under additional assumptions, the family U(T, K) can be replaced by more narrow family U(T, S). Corollary 2 (the quasiclassical case). Let S be a separable perfect foundation on a set T. Then, 1) M(T, Y𝛼 (T, S)) ⊂ Lim𝛼 U(T, S) ⊂ M(T, Y𝛼+1 (T, S)) for every 𝛼 ∈ [1, 𝜔[; 2) Lim𝛼 U(T, S) = M(T, Y𝛼+1 (T, S)) for every 𝛼 ∈ [𝜔, 𝜔1 [; 3) BM(T, S) ≡ M(T, B(T, S)) = Lim𝜔1 U(T, S). Proof. For perfect S, we have that Λ 𝛼 (T, S) = Y𝛼 (T, S) for all 𝛼 ∈ [1, 𝜔1 [ by virtue of Corollary 2 to Theorem 5 (2.1.3). Now, apply Theorem 1 to E ≡ S. Theorem 1, with its corollaries, constitutes the fine correlations between the Baire functional hierarchy and the Borel functional hierarchy. These fine correlations allow us to obtain fine constructive descriptions of completely normal families and envelopes (cf. Theorems 6 and 7 (2.3.7)) and therefore to take off the operation of countable union K 󳨃→ K𝜎 .

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Fine constructive descriptions of completely normal families and envelopes Theorem 2. Let A(T) ⊂ F(T). Then, the following conclusions are equivalent: 1) the family A(T) is completely normal; 2) A(T) = Lim𝜔1 U(T, K(T, Coz A(T))). Proof. (1) ⊢ (2). Let S ≡ Coz A(T), K ≡ K(T, S). By Theorem 3 (2.3.6), A(T) = M(T, B(T, S)) and by Corollary 1 to Theorem 1 M(T, B(T, S)) = Lim𝜔1 U(T, K). (2) ⊢ (1). By Corollary 1 to Theorem 1, M(T, B(T, S)) = Lim𝜔1 U(T, K) = A(T). Then, by Theorem 3 (2.3.6), the family A(T) is completely normal. Theorem 3. Let A(T) ⊂ F(T). Then, CN(A(T)) = Lim𝜔1 U(T, K(T, Coz A(T)∗ )). Proof. Denote Coz A(T)∗ by S. According to Theorem 4 (2.3.6), CN(A(T)) = M(T, B(T, S)). Further, by Corollary 1 to Theorem 1, we obtain the equality M(T, B(T, S)) = Lim𝜔1 U(T, K(T, S)). Corollary 2 to Theorem 1 allows us to obtain a simpler fine constructive description of the completely normal envelope under additional assumptions on A(T). Theorem 4. Let A(T) ⊂ F(T) satisfy conditions 1 – 4 from 2.2.4. Then, CN(A(T)) = Lim𝜔1 U(T, Coz A(T)). Proof. Denote Coz A(T) by S. According to Corollary 1 to Theorem 4 (2.3.6), CN(A(T)) = M(T, B(T, S)). By virtue of Proposition 1 (2.2.5), ensemble S is a separable perfect foundation. Hence, by Corollary 2 to Theorem 1, M(T, B(T, S)) = Lim𝜔1 U(T, S). Note that the description of completely normal envelope in Theorem 4 is finer than the description in Proposition 3 (2.3.6). This became possible because the family U(T, S) of uniform function is comparison with the family M(T, S) of measurable functions has the same good properties even without the necessity for S to be 𝜎-additive.

2.5 Families of functions on a descriptive space with negligence If ⟮T, S, I⟯ is a descriptive space with negligence, then it determines some more extensive families of functions (in comparison with the families proper for descriptive spaces) having the considered properties of measurability, distributability, and uniformity everywhere except negligible sets. 2.5.1 Almost measurable, almost distributable, and almost uniform functions Let ⟮T, S, I⟯ be a descriptive space with negligence. A function f ∈ F(T) is called almost measurable on ⟮T, S, I⟯ if there is E ∈ co-I such that f |E ∈ M(E, SE ). The family of all almost measurable functions on ⟮T, S, I⟯ is denoted by AM(T, S, I).

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A function f ∈ F(T) is called almost distributable [almost uniform] on ⟮T, S, I⟯ if for some set E ∈ co-I and every 𝜀 > 0, there is a countable [finite] cover 𝜋 ≡ (S i ∈ SE | i ∈ I) of E such that 𝜔(f , 𝜋) < 𝜀. The family of all almost distributable [almost uniform] functions on ⟮T, S, I⟯ is denoted by AD(T, S, I) [AU(T, S, I)]. The following two assertions are direct corollaries of the definitions. Lemma 1. Let ⟮T, S, I⟯ be a descriptive space with negligence. Then, f ∈ AD(T, S, I) [f ∈ AU(T, S, I)] iff there is E ∈ co-I such that f |E ∈ D(E, SE ) [f |E ∈ U(E, SE )]. Lemma 2. Let ⟮T, S, I⟯ be a descriptive space with negligence. Then AU(T, S, I) ⊂ AM(T, S, I) ⊂ AM(T, S𝜎 , I) = AD(T, S, I). The families AM(T, S, N(S𝜎 )), AD(T, S, N(S𝜎 )), and AU(T, S, N(S𝜑 )) are denoted also by AM(T, S), AD(T, S), and AU(T, S), respectively. Their elements are also called almost measurable, almost distributable, and almost uniform functions with respect to the ensemble S. These families are the extensions of the classical family of functions with the Baire property (see e. g. [Kuratowski, 1966, 32] and [Semadeni, 1971, 15.6]) from the narrow class of Baire (topological) spaces ⟮T, G⟯ to the wide class of descriptive spaces ⟮T, S⟯; they are also the extensions of the family of almost continuous functions (i. e. continuous on dense open sets) firstly used in [Fine et al., 1965]. Lemma 3. Let S be a foundation on T and E ⊂ T. If f ∈ M(T, S), f ∈ D(T, S), or f ∈ U(T, S), then f |E ∈ M(E, SE ), f |E ∈ D(E, SE ), or f |E ∈ U(E, SE ), respectively. Proof. Let f ∈ M(T, S) and ]x, y[⊂ R. Then, S ≡ f −1 []x, y[] ∈ S. Therefore, (f |E)−1 []x, y[] = S ∩ E ∈ SE . Hence, f |E ∈ M(E, SE ). Now, let f ∈ D(T, S) [f ∈ U(T, S)] and 𝜀 > 0. Then, there is a countable [finite] cover 𝜋 ≡ (S i ∈ S | i ∈ I) of T such that 𝜔(f , 𝜋) < 𝜀. Consider the countable [finite] cover 𝜘 ≡ ⟮S i ∩ E | i ∈ I⟯ of E. Since S i ∩ E ∈ SE and 𝜔(f |E, 𝜘) < 𝜀, we conclude that f |E ∈ D(E, SE ) [f |E ∈ U(E, SE )]. Corollary 1. Let S be a foundation on a set T and I be an ensemble on T. Then, M(T, S) ⊂ AM(T, S, I), D(T, S) ⊂ AD(T, S, I), and U(T, S) ⊂ AU(T, S, I). The requirement that the triplet ⟮T, S, I⟯ is a descriptive space with negligence guarantees us the important property of bijectivity of the factor-mapping f → f (mod I) considered on the families M(T, S), D(T, S), and U(T, S) (see 2.2.6). Namely, consider the factor-family AD(T, S, I) ≡ AD(T, S, I)/𝜃I,AD(T,S,I) and the AD(T, S, I). factor-mapping p : AD(T, S, I) Proposition 1. Let ⟮T, S, I⟯ be a descriptive space with negligence. Then, the restriction p|D(T, S) is injective.

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Proof. Take some f1 , f2 ∈ D(T, S) such that f1 ≠ f2 . Then, there is t ∈ T such that |f1 (t) − f2 (t)| > a > 0 for some a ∈ R. By definition, there is some countable cover 𝜋 ≡ (S i ∈ S | i ∈ I) such that 𝜔(f1 − f2 , 𝜋) < a/3. Take some i ∈ I such that t ∈ S i . Then, for every s ∈ S i , we have the inequalities |f1 (s) − f2 (s)| ⩾ f1 (s) − f1 (t) + f1 (t) − f2 (t) + f2 (t) − f2 (s) > −a/3 + a − a/3 = a/3 or |f1 (s) − f2 (s)| ⩾ f2 (s) − f2 (t) + f2 (t) − f1 (t) + f1 (t) − f1 (s) > −a/3 + a − a/3 = a/3. Since S i ≠ ⌀, we conclude that S ∈ ̸ I. This means that f1 is not equivalent to f2 with respect to I. Hence, pf1 ≠ pf2 . Lemma 4. Let S be a 𝜎-foundation and I be an ensemble on a set T. Then, AD(T, S, I) = AM(T, S, I) and AU b (T, S, I) = AM b (T, S, I). Proof. For every E ⊂ T, the foundation SE is 𝜎-additive. Therefore, Lemma 3 (2.4.1) implies that M b (E, SE ) = U(E, SE ). By Proposition 1 (2.3.1), M(E, SE ) = D(E, SE ). Proposition 2. Let S be a foundation on a set T and I be an additive ensemble on T. Then, the family AD(T, S, I) [AU(T, S, I)] satisfies conditions 1, 2, 3, 4, 5, and 7 [7󸀠 )] from 2.2.4. If, besides, S is 𝜎-additive, then the family AM(T, S, I) satisfies the conditions 1, 2, 3, 4, 5, and 7. Moreover, if I is 𝜎-additive, then AD(T, S, I), AU(T, S, I), and AM(T, S, I) satisfy condition 6. Proof. First, it follows directly from the definition that 1 ∈ AU(T, S, I). Let f , g ∈ AU(T, S, I). Then, f |D ∈ U(D, SD ) and g|E ∈ U(E, SE ) for some D, E ∈ co-I. Since I is additive, co-I is multiplicative. Hence, H ≡ D ∩ E ∈ co-I. By Lemma 3, f |H ∈ U(H, SH ) and g|H ∈ U(H, SH ). By Corollary 4 to Proposition 1 (2.4.3), the family U(H, SH ) is boundedly normal. Therefore, the functions rf |H (for r ∈ R), (f + g)|H, (fg)|H, (f ∨ g)|H, (f ∧ g)|H belong to U(H, SH ). Consequently, the functions rf , f + g, fg, f ∨ g, f ∧ g belong to AU(T, S, I). Let f ∈ AU(T, S, I) and 1/f ∈ F b (T). Then, there is D ∈ co-I such that f |D ∈ U(D, SD ). By Corollary 3 to Theorem 1 (2.4.2), (1/f )|D ∈ U(D, SD ), where 1/f ∈ AU(T, S, I). Let the ensemble I is 𝜎-additive and f = u-lim (f n | n ∈ 𝜔) for some sequence (f n ∈ AU(T, S, I) | n ∈ 𝜔). Then, f n |D n ∈ U(D n , SD n ) for some D n ∈ co-I. Since I is 𝜎-additive, co-I is 𝛿-multiplicative. Hence, H ≡ ⋂⟮D n | n ∈ 𝜔⟯ ∈ co-I. By Lemma 3, f n |H ∈ U(H, SH ) for every n ∈ 𝜔. Corollary 3 to Proposition 1 (2.4.3) implies that f ∈ U(H, SH ). In the cases of AD(T, S, I) and AM(T, S, I), the arguments are analogous. Corollary 1. Let S be a foundation on a set T and I be an additive ensemble on T. Then, the ensembles Coz AD(T, S, I) and Coz AU(T, S, I) are separable perfect 𝜑-foundations. If, besides, S is 𝜎-additive, then Coz AM(T, S, I) is a separable perfect 𝜑-foundation too.

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Proof. The assertion follows from Propositions 2 and 1 (2.2.5). Corollary 2. Let S be a foundation on a set T. Then, the family AD(T, S) [AU(T, S)] satisfies conditions 1 – 5 and 7 [7󸀠 ] from 2.2.4. If, besides, S is 𝜎-additive, then the family AM(T, S) satisfies conditions 1 – 5 and 7. Proof. By Corollary 1 to Lemma 3 (2.1.4), the ensembles N(S𝜑 ) and N(S𝜎 ) are additive. This proposition inspires the natural question about some functional description of the uniform closure of the considered families AM(T, S, I), AD(T, S, I), and AU(T, S, I). It will be talked in the next subsection.

2.5.2 Quasimeasurable, quasidistributable, and quasiuniform functions Here we shall consider some more extensive classes of functions than the classes of almost measurable, almost distributable, and almost uniform functions.

Definitions and properties Let ⟮T, S, I⟯ be a descriptive space with negligence. A function f ∈ F(T) will be called quasimeasurable on ⟮T, S, I⟯ if for every 𝜀 > 0, there are D ∈ co-I and g ∈ M(D, SD ) such that |f |D−g| < 𝜀1|D. The family of all quasimeasurable functions on ⟮T, S, I⟯ will be denoted by QM(T, S, I). It follows directly from the definitions that AM(T, S, I) ⊂ QM(T, S, I). A function f ∈ F(T) will be called quasidistributable [quasiuniform] on ⟮T, S, I⟯ if for every 𝜀 > 0, there are D ∈ co-I and a countable [finite] cover 𝜋 ≡ (S i ∈ SD | i ∈ I) of D such that 𝜔(f , 𝜋) < 𝜀. The family of all quasidistributable [quasiuniform] functions on ⟮T, S, I⟯ will be denoted by QD(T, S, I) [QM(T, S, I)]. It is clear that QU(T, S, I) ⊂ QD(T, S, I). It follows directly from the definitions that AU(T, S, I) ⊂ QU(T, S, I) and AD(T, S, I) ⊂ QD(T, S, I). The families QM(T, S, N(S𝜎 )), QD(T, S, N(S𝜎 )), and QU(T, S, N(S𝜑 )) are denoted also by QM(T, S), QD(T, S), and QU(T, S), respectively. Their elements are also called quasimeasurable, quasidistributable, and quasiuniform functions with respect to the ensemble S. These functions were introduced in [Zakharov, 1980; 1984] to replace the classical family of functions with the Baire property (see e. g. [Kuratowski, 1966, 32] and [Semadeni, 1971, 15.6]) which has good properties only for the narrow class of Baire (topological) spaces. These families allowed to solve the Fine – Gillman – Lambek problem of a functional description of the uniform completion of the bounded part of the classical ring of quotients and the rationally complete ring of quotients of the ring of continuous functions and the Nakano – Shimogaki problem of a functional description of the Dedekind and the Cantor completions of the lattice linear space of continuous functions not only for Baire spaces but

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for general Tychonoff spaces [Zakharov, 1984; 1987a; 1987b; 1995a; 1995b; 1995d; 1996]. Lemma 1. Let ⟮T, S, I⟯ be a descriptive space with negligence. Then, QM(T, S, I) ⊂ QD(T, S, I). Proof. For every 𝜀 > 0, take E ∈ co-I and g ∈ M(E, SE ) such that |f |E − g| < (𝜀/2)1|E. By Proposition 1 (2.3.1), there is a countable cover 𝜋 ≡ (S i ∈ SE | i ∈ I) of E such that 𝜔(g, 𝜋) < 𝜀/2. Therefore, |f (t) − f (s)| ⩽ |f (t) − g(t)| + |g(t) − g(s)| < 𝜀/2 + 𝜀/2 = 𝜀 for all s, t ∈ S i ⊂ E and every i ∈ I. Thus, 𝜔(f , 𝜋) < 𝜀. Proposition 1. Let S be a 𝜎-foundation on T and I be an ensemble on T. Then, QD b (T, S, I) = QU b (T, S, I). Proof. We must only prove the inclusion QD b (T, S, I) ⊂ QU b (T, S, I). Let f ∈ QD b (T, S, I) and rng f ⊂ [−z, z]. Then, for every 𝜀 > 0, there is n ∈ N such that 1/n < 𝜀 and there are a set D n ∈ co-I and a countable cover 𝜋n ≡ (R ni ∈ SD n | i ∈ I n ) of D n such that 𝜔(f , 𝜋n ) < 1/n < 𝜀. By the Archimedes principle (Lemma 13 (1.4.3))m there is k n ∈ N such that k n > 6nz. Consider the points x kn ≡ −z + k/(3n) ∈ R for all k ∈ k n . Then, ⋃⟮[x nk , x n,k+1 ] | k ∈ k n ⟯ ⊃ [−z, z]. Define the sets A nk ≡ f −1 [[x nk , x n,k+1 ]], J nk ≡ {i ∈ I n | R3n,i ∩ A nk ≠ ⌀}, and E nk ≡ ⋃⟮R3n,i | i ∈ J nk ⟯ ∈ SD n . Then, the collection 𝜆 n ≡ (E nk | k ∈ k n ) is a finite cover of D n . Let s, t ∈ E nk . Then, s ∈ R3n,i and t ∈ R3n,j for some i and j such that X ≡ R3n,i ∩ A nk = / ⌀ and Y ≡ R3n,j ∩ A nk = / ⌀. Take some points s i ∈ X and s j ∈ Y. Then, we get |f (s) − f (t)| ⩽ |f (s) − f (s i )| + |f (s i ) − f (s j )| + |f (s j ) − f (t)| < 1/n. If s ∈ E nk and t ∈ N nk , then s ∈ R3n,i implies |f (s)− f (t)| ⩽ |f (s)− f (s i )|+|f (s i )− f (t)| < 2/(3n). Thus, 𝜔(f , 𝜆 n ) < 1/n < 𝜀. Hence, f ∈ QU b (T, S, I). Corollary 1. Let S be a 𝜎-foundation on T. Then, QD b (T, S) = QU b (T, S). Proof. Since S𝜑 = S = S𝜎 , we get the equalities QD b (T, S) ≡ QD b (T, S, N(S𝜎 )) = QD b (T, S, N(S𝜑 )) = QU b (T, S, N(S𝜑 )) ≡ QU b (T, S). Proposition 2. Let S be a foundation on a set T and I be an additive ensemble on T. Then, the families QD(T, S, I) and QU(T, S, I) satisfy conditions 1 – 4, 6, and 7󸀠 from 2.2.4. If, besides, S is 𝜎-additive, then the family QM(T, S, I) satisfies the same conditions. Proof. First consider the case of QD(T, S, I). Denote A(T) ≡ QD(T, S, I). Taking any D ∈ I and the one-element cover ⟮D i ≡ D | i ∈ {i}⟯ of D ⊂ T, we see that 1 ∈ A(T).

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Let f , g ∈ A(T). Then, for every 𝜀 > 0, there is n ∈ N such that 1/n < 𝜀 and there are D n , E n ∈ co-I and countable covers 𝜘n ≡ (G ni ∈ SD n | i ∈ I n ) of D n and 𝜆 n ≡ (H nj ∈ SE n | j ∈ J n ) of E n such that 𝜔(f , 𝜘n ) < 1/n and 𝜔(g, 𝜆 n ) < 1/n. For every r ∈ R, take p(r, n) ≡ sm{l ∈ N | |r|/l < 1/n} and 𝜋n ≡ 𝜘p(r,n) . Then, 𝜔(rf , 𝜋n ) < 1/n, and therefore, rf ∈ A(T). Define the sets A nk ≡ f −1 [[−k, k]] ∩ D n , B nl ≡ g−1 [[−l, l]] ∩ E n , I nk ≡ {i ∈ I n | G ni ∩ A nk = / ⌀}, and J nl ≡ {j ∈ J n | H nj ∩ B nl = / ⌀} for every k, l ∈ N. Consider the collection 𝜈n ≡ (G2n,i ∩ H2n,j ∈ SD2n ∩E2n | (i, j) ∈ I2n × J2n ). It is countable by virtue of Lemma 3 (1.3.9). This collection is a cover of D2n ∩ E2n ∈ co-I and 𝜔(f + g, 𝜈n ) ⩽ 𝜔(f , 𝜈n ) + 𝜔(g, 𝜈n ) < 1/n. Hence, f + g ∈ A(T). If t, s ∈ G2n,i ∩ H2n,j , then |f (t) ∨ g(t) − f (s) ∨ g(s)| ⩽ |f (t) ∨ g(t) − f (t) ∨ g(s)| + |f (t) ∨ g(s) − f (s) ∨ g(s)| ⩽ |g(t) − g(s)| + |f (t) − f (s)| < 1/n, and therefore, 𝜔(f ∨ g, 𝜈n ) < 1/n. This implies f ∨ g ∈ A(T). Similarly, f ∧ g ∈ A(T). Let f ∈ F(T) and f = u-lim (f q | q ∈ N) for some sequence (f q ∈ A(T) | q ∈ N). Take m ∈ N such that |f (t) − f q (t)| < 1/(3n) for all t ∈ T and q ⩾ m. For the function h ≡ f m , there is V n ∈ co-I and 𝜇n ≡ (S ni ∈ SV n | i ∈ I n ) covering V n such that 𝜔(h, 𝜇n ) < 1/(3n). Then, |f (t) − f (s)| ⩽ |f (t) − h(t)| + |h(t) − h(s)| + |h(s) − f (s)| < 1/n for all s, t ∈ S ni . Therefore, 𝜔(f , 𝜇n ) < 1/n. Consequently, f ∈ A(T). Suppose that f ∈ A(T), f (t) = / 0 for all t ∈ T, and 1/|f | ⩽ c1 for some positive number c. For every 𝜀 > 0, there exist D ∈ co-I and a countable cover 𝜋 ≡ (S i ∈ SD | i ∈ I) of D such that 𝜔(f , 𝜋) < 𝜀/c2 . If s, t ∈ S i , then |(1/f )(s) − (1/f )(t)| = |f (s) − f (t)|/(|f (s)| |f (t)|) ⩽ c2 𝜔(f , 𝜋). Hence, 𝜔(1/f , 𝜋) ⩽ c2 𝜔(f , 𝜋) < 𝜀. Thus, 1/f ∈ A(T). In the case of QU(T, S, I), the arguments are analogous. Now, let us consider the case of 𝜎-additive S. Since 0|D, 1|D ∈ M(D, SD ) for every D ⊂ T, we have 0, 1 ∈ B(T) ≡ QM(T, S, I). Let f ∈ B(T), 𝜀 > 0, and r ∈ R\{0}. Then, there are D ∈ co-I and g ∈ M(D, SD ) such that |f |D − g| < (𝜀/|r|)1|D. Since rg ∈ M(D, SD ), we have |(rf )|D − (rg)| < 𝜀1|D, i. e. rf ∈ B(T). For f1 , f2 ∈ B(T) and 𝜀 > 0 take D1 , D2 ∈ co-I, g1 ∈ M(D1 , SD1 ), and g2 ∈ M(D2 , SD2 ) such that |f1 |D1 − g1 | < (𝜀/2)1|D1 and |f2 |D2 − g2 | < (𝜀/2)1|D2 . Consider the set D ≡ D1 ∩ D2 ∈ co-I and the functions f ≡ (f1 + f2 )|D, g ≡ (g1 + g2 )|D, f ̃ ≡ (f1 ∨ f2 )|D, and g̃ ≡ (g1 ∨ g2 )|D. By Lemma 3 (2.5.1), g1 |D, g2 |D ∈ M(D, SD ). Theorem 1 (2.3.2) 󵄨 󵄨 󵄨 󵄨 󵄨 implies that g ∈ M(D, SD ). Then, it follows from 󵄨󵄨󵄨f − g󵄨󵄨󵄨 ⩽ 󵄨󵄨󵄨f1 |D − g1 |D󵄨󵄨󵄨 + 󵄨󵄨󵄨f2 |D − 󵄨 g2 |D󵄨󵄨󵄨 < (𝜀/2)1|D + (𝜀/2)1|D = 𝜀1|D that f ∈ B(T). Proposition 1 (2.3.3) guarantees that g̃ ∈ M(T, SD ). The Birkhoff inequality from Theorem 1 (2.2.2) implies that 󵄨󵄨 ̃ ̃ 󵄨󵄨 󵄨󵄨 󵄨 󵄨 󵄨 󵄨 󵄨 󵄨 󵄨 󵄨󵄨f − g󵄨󵄨 ⩽ 󵄨󵄨(f1 ∨ f2 )|D − (f1 ∨ g2 )|D󵄨󵄨󵄨 + 󵄨󵄨󵄨(f1 ∨ g2 )|D − (g1 ∨ g2 )|D󵄨󵄨󵄨 ⩽ 󵄨󵄨󵄨f2 |D − g2 |D󵄨󵄨󵄨 + 󵄨󵄨󵄨f1 |D − g1 |D󵄨󵄨󵄨 < (𝜀/2)1|D + (𝜀/2)1|D = 𝜀1|D and we conclude that f ̃ ∈ B(T). The arguments for f1 ∧ f2 are the same. Let f ∈ F(T) and f = u-lim (f q | q ∈ N) for some sequence (f q ∈ B(T) | q ∈ N). Take m ∈ N such that |f (t) − f q (t)| < 𝜀/2 for all t ∈ T and q ⩾ m. Take D ∈ co-I and

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󵄨 󵄨 󵄨 󵄨 󵄨 󵄨 󵄨 󵄨 g ∈ M(D, SD ) such that 󵄨󵄨󵄨f q |D − g󵄨󵄨󵄨 < (𝜀/2)1|D. Then, 󵄨󵄨󵄨f |D − g󵄨󵄨󵄨 ⩽ 󵄨󵄨󵄨f |D − f q |D󵄨󵄨󵄨 + 󵄨󵄨󵄨f q |D − g󵄨󵄨󵄨 < 𝜀1|D. Thus, f ∈ B(T). Suppose that f ∈ B(T), f (t) = / 0 for all t ∈ T, and 1/|f | ⩽ c1 for some positive number c. Let 𝜀 > 0 and put 𝛽 ≡ (𝜀/(2c2 )) ⊼ (1/2c). There exist D ∈ co-I and g ∈ 󵄨 󵄨 M(D, SD ) such that 󵄨󵄨󵄨f |D − g󵄨󵄨󵄨 < 𝛽1|D. Hence, |g| > (1/c − 𝛽)1|D. It follows from 󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨(1/f )|D−1/g󵄨󵄨 = 󵄨󵄨f |D−g󵄨󵄨/(|f |D| |g|) < (c𝛽/(1/c−𝛽))1|D ⩽ (𝜀/(2−2c𝛽))1|D ⩽ 𝜀1|D that 1/ f ∈ B(T). Corollary 1. Let S be a foundation on a set T and I be an additive ensemble on T. Then, the ensembles Coz QD(T, S, I) and Coz QU(T, S, I) are a-foundations. If, besides, S is 𝜎-additive, then Coz QM(T, S, I) is an a-foundation too. Proof. The assertion follows from Proposition 2 and Proposition 1 (2.2.5). Corollary 2. Let S be a foundation on a set T. Then, the families QD(T, S) and QU(T, S) satisfy conditions 1 – 4, 6, and 7󸀠 from 2.2.4. If, besides, S is 𝜎-additive, then the family QM(T, S, I) satisfies the same conditions. Lemma 2. Let S be a 𝜎-foundation on a set T and I be an additive ensemble on T. Then, the family QM b (T, S, I) satisfies condition 5 from 2.2.4. Proof. Let f1 , f2 ∈ QM b (T, S, I) and 𝜀 > 0. Take c > 0 such that |f1 | ⩽ c1 and |f2 | ⩽ c1. Put 𝛽 ≡ (𝜀/3c) ⊼ (√𝜀/2). There are D1 , D2 ∈ co-I, g1 ∈ M(D1 , SD1 ), and g2 ∈ M(D2 , SD2 ) such that |f1 |D1 − g1 | < 𝛽1|D1 and |f2 |D2 − g2 | < 𝛽1|D2 . Consider the set D ≡ D1 ∩ D2 ∈ co-I and the functions f ≡ (f1 f2 )|D and g ≡ (g1 g2 )|D. By Lemma 3 (2.5.1), g1 |D, g2 |D ∈ M(D, SD ). Theorem 1 (2.3.2) implies that g ∈ M(D, SD ). It is obvious that |g2 | < (c + 󵄨 󵄨 󵄨 󵄨 󵄨 󵄨 𝛽)1|D. Then, it follows from 󵄨󵄨󵄨f − g󵄨󵄨󵄨 ⩽ 󵄨󵄨󵄨(f1 f2 )|D − (f1 g2 )|D󵄨󵄨󵄨 + 󵄨󵄨󵄨(f1 g2 )|D − (g1 g2 )|D󵄨󵄨󵄨 < 󵄨 󵄨 󵄨 󵄨 c󵄨󵄨󵄨f2 |D − g2 |D󵄨󵄨󵄨 + (c + 𝛽)󵄨󵄨󵄨f2 |D − g2 |D󵄨󵄨󵄨 < (2c𝛽 + 𝛽2 )1|D ⩽ (2𝜀/3 + 𝜀/4)1|D < 𝜀1|D that f ∈ QM b (T, S, I). Corollary 1. Let S be a 𝜎-foundation on a set T and I be an additive ensemble on T. Then, the family QM b (T, S, I) is boundedly normal. Proposition 3. Let S be a foundation on a set T and I be an additive ensemble on T. Then, the families QD b (T, S, I) and QU(T, S, I) satisfy condition 5 from 2.2.4. Proof. Let f , g ∈ QD b (T, S, I). Take the positive numbers c and d such that |f | ⩽ c1 and |g| ⩽ d1. Then, for every 𝜀 > 0, there is n ∈ N such that (c + d)/n < 𝜀/2 and there are D n , E n ∈ co-I and countable covers 𝜘n ≡ (G ni ∈ SD n | i ∈ I n ) of D n and 𝜆 n ≡ (H nj ∈ SE n | j ∈ J n ) of E n such that 𝜔(f , 𝜘n ) < 1/n and 𝜔(g, 𝜆 n ) < 1/n.

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The set W n ≡ I n × J n is countable by Lemma 3 (1.3.9). For every pair w ≡ (i, j) ∈ W n consider the set Q nw ≡ G ni ∩ H nj ∈ SD n ∩E n . If s, t ∈ Q nw , then |(fg)(t) − (fg)(s)| ⩽ |f (t)| |g(t) − g(s)| + |g(s)| |f (t) − f (s)| ⩽ c/n + d/n < 𝜀/2. Consequently, 𝜔(fg, Q nw ) ⩽ 𝜀/2. Consider the set F n ≡ ⋃⟮Q nw | w ∈ W n ⟯ and its countable cover 𝜇n ≡ (Q nw ∈ SD n ∩E n | w ∈ W n ). We have 𝜔(fg, 𝜇n ) ⩽ 𝜀/2 < 𝜀. Using assertion 3 of Corollary 2 to Theorem 1 (1.1.13) we get F n ≡ ⋃⟮Q nw | w ∈ W n ⟯ = ⋃⟮G ni ∩ H nj | (i, j) ∈ I n × J n ⟯ = = ⋃⟮G ni | i ∈ I n ⟯ ∩ ⋃⟮H nj | j ∈ J n ⟯ = D n ∩ E n . Hence, T\F n = (T\D n ) ∪ (T\E n ). Since I is additive, we get T\F n ∈ I. Thus, fg ∈ QD b (T, S, I). Let f , g ∈ QU(T, S, I). Then, for every 𝜀 > 0, there is n ∈ N such that 1/n < 𝜀 and there are D n , E n ∈ co-I and finite covers 𝜘n ≡ (G ni ∈ S | i ∈ I n ) of D n and 𝜆 n ≡ (H nj ∈ S | j ∈ J n ) of E n such that 𝜔(f , 𝜘n ) < 1/n and 𝜔(g, 𝜆 n ) < 1/n. By Lemma 1 (2.2.1), the functions f |D n and f |E n are bounded. Put a n ≡ sup{|f (t)| | t ∈ D n } and b n ≡ sup{|g(t)| | t ∈ E n }. By the Archimedes principle (Lemma 13 (1.4.3)), there are p n , q n ∈ N such that p n > 2nb n and q n > 2na n . Consider the set W n ≡ I n × J n × I p n × J q n , it is finite by Lemma 4 (1.3.3). For every suite w ≡ (i1 , j1 , i2 , j2 ) ∈ W n , consider the set Q nw ≡ G ni1 ∩H nj1 ∩G p n i2 ∩H q n j2 ∈ SD n ∩E n ∩D p ∩E q . n n If s, t ∈ Q nw , then |(fg)(t) − (fg)(s)| ⩽ |f (t)| |g(t) − g(s)| + |g(s)| |f (t) − f (s)| ⩽ a n /q n + b n /p n < 1/n. Consequently, 𝜔(fg, Q nw ) ⩽ 1/n. Consider the finite collection 𝜇n ≡ ⟮Q nw | w ∈ W n ⟯ covering the set F n ≡ rng 𝜇n ≡ ⋃⟮Q nw | w ∈ W n ⟯. Then, 𝜔(fg, 𝜇n ) ⩽ 1/n < 𝜀. Using assertion 3 of Corollary 2 to Theorem 1 (1.1.13), we get F n ≡ ⋃⟮Q nw | w ∈ W n ⟯ = = ⋃⟮G ni1 ∩ H nj1 ∩ G p n i2 ∩ H q n j2 | (i1 , j1 , i2 , j2 ) ∈ I n × J n × I p n × J q n ⟯ = = ⋃⟮G ni1 | i1 ∈ I n ⟯ ∩ ⋃⟮H nj1 | j1 ∈ J n ⟯ ∩ ⋃⟮G p n i2 | i2 ∈ I p n ⟯ ∩ ⋃⟮H q n j2 | j2 ∈ J q n ⟯ = = Dn ∩ E n ∩ Dpn ∩ E qn . Hence, T\F n = T\(D n ∩ E n ∩ D p n ∩ E q n ) = (T\D n ) ∪ (T\E n ) ∪ (T\D p n ) ∪ (T\E q n ). Since I is additive, we get T\F n ∈ I. Thus, fg ∈ QU(T, S, I). Corollary 1. Let S be a foundation on a set T. Then, the families QD b (T, S) and QU(T, S) satisfy condition 5 from 2.2.4. Proof. By Corollary 1 to Lemma 3 (2.1.4), the ensembles N(S𝜎 ) and N(S𝜑 ) are ideals.

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Corollary 2. Let S be a foundation on a set T and I be an additive ensemble on T. Then, the families QD b (T, S, I) and QU b (T, S, I) are boundedly normal. The following proposition is a module variant of generalizing of Proposition 3 for quasidistributable functions. Proposition 4. Let S be a foundation on a set T and I be an ideal ensemble on T. Then, for any f ∈ QD(T, S, I) and g ∈ AD b (T, S, I), we have fg ∈ QD(T, S, I). Proof. Take E ∈ co-I such that there for every n ∈ N, there is a countable cover 𝜆 n ≡ (H nj ∈ SE | j ∈ J n ) of E such that 𝜔(g, 𝜆 n ) < 1/n. Take a positive number b such that |g| ⩽ b1. Let 𝜀 > 0. Fix a number n ∈ N such that (b + 1)/n < 𝜀. There are D n ∈ co-I and a countable cover 𝜘n ≡ (G ni ∈ SD n | i ∈ I n ) of D n such that 𝜔(f , 𝜘n ) < 1/n. / ⌀} for Define the sets A nk ≡ f −1 [[−k, k]] ∩ D n and I nk ≡ {i ∈ I n | G ni ∩ A nk = every k ∈ N. Consider the countable collections 𝜘nk ≡ (G ni | i ∈ I nk ) covering the sets rng 𝜘nk ≡ ⋃⟮G ni | i ∈ I nk ⟯ ⊃ A nk . It is clear that 𝜔(f , 𝜘nk ) < 1/n. It is also clear that sup{|f (t)| | t ∈ rng 𝜘nk } ⩽ k + 1/n ≡ a nk . Take the natural numbers q nk ≡ 2nk + 2. For every k ∈ N, define the set W nk ≡ I nk × J q nk . For every pair w ≡ (i, j) ∈ W nk , consider the set Q nkw ≡ G ni ∩ H q nk j ∈ SD n ∩E . If s, t ∈ Q nkw , then |(fg)(t) − (fg)(s)| ⩽ |f (t)| |g(t) − g(s)| + |g(s)| |f (t) − f (s)| ⩽ a nk /q nk + b/n = 1/(2n) + b/n < (b + 1)/n. Consequently, 𝜔(fg, Q nzw ) ⩽ (b + 1)/n. Define the set C n ≡ ⋃⟮{k}×W nk | k ∈ N⟯, it is countable by virtue of Theorem 1 (1.3.9). Consider the set F n ≡ ⋃⟮Q nkw | (k, w) ∈ C n ⟯ and its countable cover 𝜇n ≡ (S nc ∈ S | c ∈ C n ), where S nc ≡ Q nkw for c ≡ (k, w). We have 𝜔(fg, 𝜇n ) ⩽ (b + 1)/n < 𝜀. Using Proposition 1 (1.1.13) and assertion 3 of Corollary 2 to Theorem 1 (1.1.13), we get F n ≡ ⋃⟮Q nkw | (k, w) ∈ C n ⟯ = ⋃⟮⋃⟮Q nkw | w ∈ W nk ⟯ | k ∈ N⟯ = = ⋃⟮⋃⟮G ni ∩ H q nk j | (i, j) ∈ I nk × J q nk ⟯ | k ∈ N⟯ = = ⋃⟮⋃⟮G ni | i ∈ I nk ⟯ ∩ ⋃⟮H q nk j | j ∈ J q nk ⟯ | k ∈ N⟯ ⊃ ⊃ ⋃⟮A nk ∩ E | k ∈ N⟯ = ⋃⟮A nk | k ∈ N⟯ ∩ E = D n ∩ E. Hence, T\F n ⊂ (T\D n ) ∪ (T\E). Since I is an ideal ensemble, we see that T\F n ∈ I. Thus, fg ∈ QD(T, S, I). Some connections of “quasi” functions with uniform, semimeasurable, and measurable functions It is remarkable that boundedly normal families QD b (T, S, I) and QU b (T, S, I) of bounded quasidistributable and quasiuniform functions can be represented as

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families of functions uniform with respect to some derivative foundations. This underlines once again the principle importance of the notion of a uniform function. Lemma 3. Let S be a foundation on T, I be an ideal on T, I ∩ co-S is cofinal to I (see 2.1.4), and ]x, y[⊂ R. If f ∈ QD(T, S, I), then f −1 []x, y[] ∈ (SP(S𝜎 , I))𝜎 . If f ∈ QU(T, S, I), then f −1 []x, y[] ∈ (SP(S𝜑 , I))𝜎 .

Proof. Let f ∈ QD(T, S, I). Then, for every n ∈ N, there is D n ∈ co-I and a countable cover 𝜋n ≡ (S ni ∈ SD n | i ∈ I n ) of D n such that 𝜔(f , 𝜋n ) < 1/n. Consider the sets X ≡ f −1 []x, y[], X n ≡ f −1 [[x + 1/n, y − 1/n]], J n ≡ {i ∈ I n | S ni ∩ X n = / ⌀}, and G n ≡ ⋃⟮S ni | i ∈ J n ⟯ ∈ (SD n )𝜎 . It is clear that X n ∩ D n ⊂ G n . It follows from N n ≡ X\D n ⊂ T\D n ∈ I that N n ∈ I. Let s ∈ G n . Then, there are i ∈ J n and t ∈ T such that s ∈ S ni and t ∈ S ni ∩ X n . Therefore, x ⩽ f (t) − 1/n < f (s) < f (t) + 1/n ⩽ y implies G n ⊂ X. Consequently, X = ⋃⟮X n | n ∈ N⟯ = ⋃⟮(X n ∩ D n ) ∪ (X n \D n ) | n ∈ N⟯ ⊂ ⋃⟮G n ∪ N n | n ∈ N⟯ ⊂ X implies X = ⋃⟮G n ∪ N n | n ∈ N⟯ ∈ (SP((SD n )𝜎 , I))𝜎 . Then, Lemma 6 (2.1.4) provides that X ∈ (SP(S𝜎 , I))𝜎 . The proof for f ∈ QU(T, S, I) is analogous. Corollary 1. Let S be a foundation on T, I be an ideal on T, and I ∩ co-S is cofinal to I. Then, we have QD(T, S, I) ⊂ M(T, (SP(S𝜎 , I))𝜎 ) = D(T, SP(S𝜎 , I)) and QU(T, S, I) ⊂ M(T, (SP(S𝜑 , I))𝜎 ) = D(T, SP(S𝜑 , I)). Proposition 5. Let S be a foundation on T, I be an ideal on T, and I ∩ co-S is cofinal to I. Then, QD b (T, S, I) = U(T, SP(S𝜎 , I)) and QU b (T, S, I) = U(T, SP (S𝜑 , I)). Proof. Let f ∈ QD b (T, S, I) and rng f ⊂ [−z, z]. Then, for every 𝜀 > 0, there is n ∈ N such that 1/n < 𝜀 and there are a set D n ∈ co-I and a countable cover 𝜋n ≡ (R ni ∈ SD n | i ∈ I n ) of D n such that 𝜔(f , 𝜋n ) < 1/n < 𝜀. By the Archimedes principle (Lemma 13 (1.4.3)), there is k n ∈ N such that k n > 6nz. Consider the points x kn ≡ −z + k/(3n) ∈ R for all k ∈ k n . Then, ⋃⟮[x nk , x n,k+1 ] | k ∈ k n ⟯ ⊃ [−z, z]. Define the sets A nk ≡ f −1 [[x nk , x n,k+1 ]], J nk ≡ {i ∈ I n | R3n,i ∩ A nk ≠ ⌀}, E nk ≡ ⋃⟮R3n,i | i ∈ J nk ⟯ ∈ (SD n )𝜎 , and N nk ≡ A nk \D n . Since I is an ideal and N nk ⊂ T\D n ∈ I, we get N nk ∈ I. Therefore, S nk ≡ E nk ∪ N nk ∈ SP((SD n )𝜎 ), I). Then, Lemma 6 (2.1.4) implies that S nk ∈ SP(S𝜎 , I). Besides, we see that the collection 𝜘n ≡ (S nk | k ∈ k n ) is a finite cover of T. Let s, t ∈ S nk . If s, t ∈ E nk , then s ∈ R3n,i and t ∈ R3n,j for some i and j such that / ⌀ and Y ≡ R3n,j ∩ A nk = / ⌀. Take some points s i ∈ X and s j ∈ Y. X ≡ R3n,i ∩ A nk = Then, we get |f (s) − f (t)| ⩽ |f (s) − f (s i )| + |f (s i ) − f (s j )| + |f (s j ) − f (t)| < 1/n. If s ∈ E nk and t ∈ N nk , then s ∈ R3n,i implies |f (s) − f (t)| ⩽ |f (s) − f (s i )| + |f (s i ) − f (t)| < 2/(3n). Finally, if s, t ∈ N nk , then |f (s)−f (t)| < 1/3n. Thus, 𝜔(f , S nk ) < 1/n < 𝜀. This means that f ∈ U(T, SP(S𝜎 , I)).

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Let f ∈ U(T, SP(S𝜎 , I)). Then, for every 𝜀 > 0, there is a finite cover 𝜘 ≡ (X i ∈ SP(S𝜎 , I) | i ∈ I) of T such that 𝜔(f , X i ) < 𝜀. By definition, X i = S i ∪ N i , where S i ∈ S𝜎 and N i ∈ I. Consider the sets N ≡ ⋃⟮N i | i ∈ I⟯ ∈ I and D ≡ T\N. By definition, S i = ⋃⟮S ij | j ∈ J i ⟯ for some countable collections (S ij ∈ S | j ∈ J i ). Consider the set K ≡ ⋃⟮{i} × J i | i ∈ I⟯. It is countable by virtue of Theorem 1 (1.3.9). The countable collection (R k ∈ SD | k ∈ K), where R k ≡ S ij ∩D for k = (i, j) ∈ K forms a cover of the set D such that 𝜔(f , R k ) ⩽ 𝜔(f , X i ) < 𝜀. Therefore, f ∈ QD(T, S, I). According to Lemma 4 (2.4.1), every uniform function is bounded, where f ∈ QD b (T, S, I). The arguments for the equality QU b (T, S, I) = U(T, SP(S𝜑 , I)) are quite similar. One should only replace S𝜎 by S𝜑 , QD b by QU b , “countable” by “finite” and the reference to Theorem 1 (1.3.9) by the reference to Lemma 3 (1.3.3). Corollary 1. Let S be a foundation. Then, QD b (T, S, N(S𝜎 )) = U(T, SP(S𝜎 , N(S𝜎 ))) and QU b (T, S, N(S𝜑 )) = U(T, SP(S𝜑 , N(S𝜑 ))). Proof. Take some N ∈ N(S𝜎 ). By definition, there is D ∈ D(S𝜎 ) ⊂ D(S) ⊂ S such that N ⊂ T\D ∈ N(S𝜎 ). Hence, N(S𝜎 ) ∩ co-S is cofinal to N(S𝜎 ). The arguments for the second equality are quite similar. The notion of equivalence of functions with respect to an ideal ensemble allows to connect quasiuniform functions with semimeasurable functions. Theorem 1. Let S be a 𝜎-foundation on T, I be an ideal on T, and f ∈ F b (T). Then, the following assertions are equivalent: 1) f ∈ QU b (T, S, I); 2) there are functions g ∈ SM bl (T, S) and h ∈ SM bu (T, S) such that g ⩽ f ⩽ h and g ∼ h mod I; 3) there are functions g ∈ SM bl (T, S) ∩ QU b (T, S, I) and h ∈ SM bu (T, S) ∩ QU b (T, S, I) such that g ⩽ f ⩽ h and g ∼ h mod I. Proof. (1) ⊢ (2). Suppose rng f ⊂ [−z, z]; then, by definition, for every n ∈ N, there are D n ∈ co-I and a finite cover 𝜋n ≡ (S ni ∈ SD n | i ∈ I n ) of D n such that 𝜔(f , 𝜋n ) < 1/n. Consider the numbers x ni ≡ inf (f (t) | t ∈ S ni ) and y ni ≡ sup (f (t) | t ∈ S ni ). Define the functions g ni setting g ni (t) ≡ −z for t ∈ ̸ S ni and g ni (t) ≡ x ni for t ∈ S ni and the functions h ni setting h ni (t) ≡ z for t ∈ ̸ S ni and h ni (t) ≡ y ni for t ∈ S ni . It is clear that g ni ∈ SM bl (T, S) and h ni ∈ SM bu (T, S). Consider the functions g n ≡ sup (g ni | i ∈ I n ), h n ≡ inf (h ni | i ∈ I n ), g ≡ sup (g n | n ∈ N), and h ≡ inf (h n | n ∈ N). By Proposition 1 (2.3.8), we obtain that g n , g ∈ SM bl (T, S) and h n , h ∈ SM bu (T, S). Evidently, g ⩽ f ⩽ h. We also have 0 ⩽ h(t) − g(t) ⩽ h n (t) − g n (t) ⩽ 𝜔(f , 𝜋n ) < 1/n for every t ∈ D n . This means that g ∼ h mod I. (2) ⊢ (3). It follows from g ∼ h mod I that for every n ∈ N, there is a set D n ∈ co-I such that 0 ⩽ h(t) − g(t) ⩽ 1/(2n) for all t ∈ D n . Consider the sets S nk ≡

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g −1 [](k − 1)/(2n), ∞[] ∩ h−1 [] − ∞, (k + 1)/(2n)[] ∩ D n . Since g and h are lower and upper S-semimeasurable, respectively, we see that S nk ∈ SD n . If t ∈ D n , then (k − 1)/(2n) < g(t) ⩽ k/(2n) for some k ∈ Z, and therefore, (k − 1)/(2n) < h(t) ⩽ (k + 1)/(2n). Thus, t ∈ S nk . Consequently, D n = ⋃⟮S nk | k ∈ Z⟯. Further, for s, t ∈ S nk , we get g(s)−g(t) ⩽ h(s)−g(t) < (k+1)/(2n)−(k−1)/(2n) = 1/n and g(s) − g(t) ⩽ g(s) − h(t) > (k − 1)/(2n) − (k + 1)/(2n) = −1/n. This means that 𝜔(g, S nk ) < 1/n. Similarly, 𝜔(h, S nk ) < 1/n. Since the functions g and h are bounded, this entails g, h ∈ QD b (T, S, I). Then, Proposition 1 implies that g, h ∈ QU b (T, S, I). (3) ⊢ (1). Let 𝜀 > 0. Lemma 4 (2.2.6) guarantees that f ∼ g mod I. Therefore, I ≡ {t ∈ T | |f (t) − g(t)| ⩾ 𝜀/2} ∈ I. By definition, there are D ∈ co-I and its finite cover 𝜋 ≡ (S k ∈ SD | k ∈ K) such that 𝜔(g, 𝜋) < 𝜀/2. Consider the set E ≡ D ∩ (T\I) ∈ co-I. Then, the collection 𝜘 ≡ (S k ∩ E ∈ SE | k ∈ K) is a finite cover of E and 𝜔(f , 𝜘) ⩽ 𝜔(g, 𝜋) + 𝜀/2 < 𝜀. Hence, f ∈ QU b (T, S, I). Theorem 2. Let S be a separable perfect 𝜎-foundation on T, I be an ideal on T, and f ∈ F b (T). Then, the following assertions are equivalent: 1) f ∈ QU b (T, S, I); 2) there exist countable collections of functions u ≡ (g i ∈ M b (T, S) | i ∈ I) and v ≡ (h j ∈ M b (T, S) | j ∈ J) such that g i ⩽ f ⩽ h j for all i ∈ I and j ∈ J and g ∼ h mod I, where g ≡ sup u and h ≡ inf v in F(T). Proof. (1) ⊢ (2). By Theorem 1, there exist functions g ∈ SM bl (T, S) and h ∈ SM bu (T, S) such that g ⩽ f ⩽ h and g ∼ h mod I. Since g and h are bounded, there is z ∈ R+ such that rng g ∪ rng h ⊂ [−z, z]. Corollary 2 to Theorem 1 (2.3.8) implies that there is a countable collection u 󸀠 ≡ 󸀠 (g i ∈ M(T, S) | i ∈ I) such that g = sup u󸀠 in F(T). Consider the functions g i ≡ g󸀠i ∨ (−z)1 ∈ M(T, S). It follows from (−z)1 ⩽ g i ⩽ g ⩽ z1 that g i ∈ M b (T, S). According to Corollary 1 to Proposition 2 (1.1.15), we get g = sup u for u ≡ (g i | i ∈ I). Similarly, h = inf v for v ≡ (h j ∈ M b (T, S) | j ∈ J). Clearly, g i ⩽ g ⩽ f ⩽ h ⩽ h j for all i ∈ I and j ∈ J. (2) ⊢ (1). By Proposition 1 (2.3.8), we obtain g ∈ SM bl (T, S) and h ∈ SM bu (T, S). Since f is an upper bound of u and a lower bound of v, we get g ⩽ f ⩽ h. Then, Theorem 1 implies that f ∈ QU b (T, S, I). Theorems 1 and 2 will be essentially used in 3.7 for some new characterizations of Riemann integrable functions different from the famous Lebesgue characterization. Some important classes of quasiuniform functions on topological spaces Let ⟮T, G⟯ be a topological space. For the ensemble G of all open sets and the ideal N(G) of all nowhere dense sets of ⟮T, G⟯ (see 2.1.4), consider the family QU b (T, G, N(G)) of all bounded quasiuniform functions on the space ⟮T, G⟯. In the same manner, for the ensemble G0 ≡ Coz C b (T, G) (see 2.3.1) of all cozero-sets of

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⟮T, G⟯ and the ideal N(G0 ) of all nowhere dense sets of the descriptive space ⟮T, G0 ⟯, consider the family QU b (T, G0 , N(G0 )) of all bounded quasiuniform functions on the descriptive cozero space ⟮T, G⟯ (see [Alexandrov, 1940; 1941; 1943] and [Gordon, 1971]). These functional families were considered in papers [Fine et al., 1965; Dashiell et al., 1980; Zakharov, 1995a; 1995b; 2005c; 2006a]. It is clear that QU b (T, G0 , N(G0 )) ⊂ QU b (T, G, N(G)). Now, consider the important intermediate functional family. Define the ensemble O0 of main open sets of the space ⟮T, G⟯ by the following transfinite procedure written in a non-formal manner: O00 ≡ G0 , O0𝛼 ≡ {⋃⟮O n ∩ int(T\O󸀠n ) | n ∈ 𝜔⟯ | ⟮O n ∈ ⋃⟮O0𝛽 | 𝛽 ∈ 𝛼⟯ | n ∈ 𝜔⟯, ⟮O󸀠n ∈ ⋃⟮O0𝛽 | 𝛽 ∈ 𝛼⟯ | n ∈ 𝜔⟯} , and O0 ≡ ⋃⟮O0𝛼 | 𝛼 ∈ 𝜔1 ⟯ (the formal manner is presented in 2.1.2). The same procedure applied to the family G gives nothing but G itself. Consider the ideal N(O0 ) of all nowhere dense sets of the descriptive space ⟮T, O0 ⟯. As above, consider the family QU b (T, O0 , N(O0 )) of all bounded quasiuniform functions on the descriptive space ⟮T, O0 ⟯. This family was considered in papers [Zakharov, 1994; 1995b; 2005c; 2006b]. According to Corollary 1 to Proposition 5, we get the equalities QU b (T, G, N(G)) = U(T, SP(G, N(G))), QU b (T, G0 , N(G0 )) = U(T, SP(G0 , N(G0 ))), and QU b (T, O0 , N(O0 )) = U(T, SP(O0 , N(O0 ))). The ideals N(G), N(G0 ), and N(O0 ) generate the corresponding equivalence relations on these families of uniform functions (see 2.2.6). As a result, for a Tychonoff space ⟮T, G⟯ we obtain the following factor-families Z ≡ U(T, SP(G, N(G))), Z 0 ≡ U(T, SP(G0 , N(G0 ))), and J0 ≡ U(T, SP(O0 , N(O0 ))) and the corresponding extensions C b Z, C b Z 0 , and C b J 0 of the initial family C b ≡ C b (T, G). These extensions were characterized in terms of divisibility and order completeness in the papers [Zakharov, 1994; 1995a; 1995b; 2005c; 2006a; 2006b]. More precisely, they are the divisible envelopes of the family C b with respect to the properties of total, classical, and countable divisibility, and the Dedekind envelopes of the family C b with respect to properties of total, complimentary countable, and countable order completeness, respectively. It is remarkable that the intermediate extension C b J 0 is the divisible envelope and at the same time the Dedekind envelope of the family C b of the same type as the Baire extension C b B b ≡ Lim𝜔b 1 C b considered in 2.2.4 (see Lemmas 2 and 4

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there). They were distinguished only by choosing some different refinements on the family C b . If ⟮T, G, M, 𝜇⟯ is a Tychonoff space with a positive bounded Radon measure, 𝜇 : M → R+ and N𝜇 is an ideal of subsets of all 𝜇-negligible zero-sets considered in 3.7.2, then we obtain the factor-family R𝜇 ≡ RI(T, G, M, 𝜇) with respect to this ideal and the corresponding 𝜇-Riemann extension C b R𝜇 . 0 0 Z and the 𝜇-Riemann extension C b Surprisingly, the Z -extension C b R𝜇 turned out to be the divisible envelopes and the Dedekind envelopes of the same type but with respect to some different refinements on C b (see [Zakharov, 1995a; 1996]). Some connections between “quasi” and “almost” functions Lemma 4. Let S be a 𝜎-foundation on a set T, I be an ensemble on T, and f ∈ F(T). Then, the following assertions are equivalent: 1) f ∈ QD(T, S, I); 2) for every n ∈ N, there are D n ∈ co-I and a countable cover 𝜘n ≡ (S nk ∈ S | k ∈ Z) of D n such that S nk ∩ S nl ≠ ⌀ implies |k − l| ⩽ 2 and 𝜔(f , 𝜘n ) < 1/n. Proof. (1) ⊢ (2). Let f ∈ QD(T, S, I). By definition, for every n ∈ N, there is E n ∈ co-I and a countable cover 𝜋n ≡ (R nk ∈ S | k ∈ I n ) of the set E n such that 𝜔(f , 𝜋n ) < 1/n. Consider the points a kn ≡ k/(3n) for all n ∈ N and k ∈ Z. Define the sets A nk ≡ f −1 [[a nk , a n,k+1 ]], J nk ≡ {i ∈ I n | R3n,i ∩ A nk ≠ ⌀}, and S nk ≡ ⋃⟮R3n,i | i ∈ J nk ⟯ ∈ S. Then, for every n ∈ N the collection 𝜘n ≡ (S nk | k ∈ Z) is a cover of the set D n ≡ E3n and 𝜔(f , S nk ) ⩽ a n,k+1 − a nk < 1/n. Let s ∈ S nk ∩ S nl . Then, f (s) ∈]k/(3n) − 1/(3n), (k + 1)/(3n) + 1/(3n)[ and f (s) ∈ ]l/(3n) − 1/(3n), (l + 1)/(3n) + 1/(3n)[ imply either (k − 1)/(3n) < f (s) < (k + 2)/(3n) or (l − 1)/(3n) < f (s) < (l + 2)/(3n). Consequently, either l − k < 3 or k − l < 3. Thus, |k − l| ⩽ 2. (2) ⊢ (1). It follows directly from the definition. Theorem 3. Let S be an a-foundation on a set T and I be an additive ensemble on T. Then, the family QD(T, S, I) is the uniform closure in F(T) of the family AD(T, S, I) = AM(T, S, I). Proof. Let f ∈ QD(T, S, I)+ . Take the sets D n and the covers 𝜘n ≡ (S nk ∈ S | k ∈ Z) from Lemma 4. Fix a number n ∈ N. By Theorem 2 (2.3.5) for every k ∈ Z, there is a positive function 𝜑k ∈ M b (T, S) such that coz 𝜑k = S nk . Define the function 𝜓 : T →]0, ∞] setting 𝜓(t) ≡ ∑net (𝜑k | k ∈ Z) (see 1.4.8). Let t ∈ S nk . If t ∈ S nl , then |k−l| ⩽ 2. Therefore, 𝜓(t) = ∑ (𝜑l (t) | |k − l| ⩽ 2) ∈ ]0, ∞[ for every point t ∈ S nk . Consider the functions 𝜓k ≡ ∑ (𝜑l | |k − l| ⩽ 2) ∈ M b (T, S). Since 𝜓|S nk = 𝜓k |S nk , we conclude that 𝜓|D n ∈ M(D n , SD n ). In fact, if x < y,

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then 𝜓−1 []x, y[] ∩ D n = ⋃⟮𝜓−1 []x, y[] ∩ S nk | k ∈ Z⟯ = ⋃⟮𝜓k−1 []x, y[] ∩ S nk | k ∈ Z⟯ ∈ SD n . Define the functions g nk on T setting g nk (t) ≡ 𝜑k (t)/𝜓(t) for t ∈ D n and g nk (t) ≡ 0 for t ∈ ̸ D n . By Theorem 1 (2.3.2), g nk |D n ∈ M(D n , SD n ). Let t ∈ D n . Assume that g nk (t) < 1/5 for every k ∈ Z. There is k ∈ Z such that t ∈ S nk . If t ∈ S nl , then |k − l| ⩽ 2. If t ∈ ̸ R nl , then g nl (t) = 0. Therefore, 1 = ∑net (g nl (t) | l ∈ Z) = ∑ (g nl (t) | |k − l| ⩽ 2) < 5/5 = 1, but this is impossible. Consequently, g nk (t) ⩾ 1/5 for some k ∈ Z, where t ∈ G nk ≡ {t ∈ D n | g nk (t) > 1/6}. Thus, 𝜆 n ≡ (G nk | k ∈ Z) is a cover of D n . Let x nk ≡ inf{f (t) | t ∈ S nk }. Consider the functions h nk ≡ (6x nk g nk ) ∧ (x nk 1). By Proposition 1 (2.3.3), h nk |D n ∈ M(D n , SD n ). Define the function f n : T → [0, ∞] setting f n (t) ≡ sup (h nk (t) | k ∈ Z) for t ∈ D n and f n (t) ≡ f (t) for t ∈ ̸ D n . Let t ∈ S nk . If t ∈ S nl , then |k − l| ⩽ 2. Therefore, f n (t) = sup (h nl (t) | |k − l| ⩽ 2). Consider the functions p nk ≡ sup (h nl | |k − l| ⩽ 2). By Proposition 1 (2.3.3), g nk |D n ∈ M(D n , SD n ). Since f n |S nk = p nk |S nk , we conclude as above that f n |D n ∈ M(D n , SD n ). Hence, f n ∈ AD(T, S, I). If t ∈ S nk , then h nk (t) ⩽ x nk ⩽ f (t). If t ∈ ̸ S nk , then h nk (t) = 0 ⩽ f (t). Consequently, h nk ⩽ f for every k ∈ Z. This implies f n ⩽ f . If t ∈ G nk , then h nk (t) = x nk . Therefore, 0 ⩽ f (t) − f n (t) ⩽ f (t) − h nk (t) ⩽ x nk + 1/n − x nk = 1/n. Since 𝜆 n is a cover of D n , it follows that 0 ⩽ f (t) − f n (t) ⩽ 1/n for every t ∈ D n . Thus, 0 ⩽ f (t) − f n (t) ⩽ 1/n for every t ∈ T. If f ∈ QD(T, S, I), then f = f+ + f− . As was shown above, there are g n , h n ∈ AD(T, S, I) such that 0 ⩽ f+ (t) − g n (t) ⩽ 1/n and 0 ⩽ −f− (t) − h n (t) ⩽ 1/n for every t ∈ T. Take f n ≡ g n − h n . It belongs to AD(T, S, I) by virtue of Proposition 2 (2.5.1). Then, |f (t) − f n (t)| ⩽ |f+ (t) − g n (t)| + |f− (t) + h n (t)| ⩽ 2/n. Thus, AD(T, S, I) is uniformly dense in QD(T, S, I). According to Proposition 2 the family QD(T, S, I) is uniformly closed. Corollary 1. Let S be an a-foundation on T and I be an additive ensemble on T. Then, the family QD b (T, S, I) is the uniform closure in F b (T) of the family AD b (T, S, I). Corollary 2. Let S be an a-foundation on T. Then, the family QD(T, S) is the uniform closure in F(T) of the family AD(T, S) = AM(T, S). Corollary 3. Let S be an a-foundation on T and I be an additive ensemble on T. Then, QD(T, S, I) = QM(T, S, I). Proof. The inclusion QM(T, S, I) ⊂ QD(T, S, I) holds by virtue of Lemma 1. Check the inverse inclusion. Let f ∈ QD(T, S, I) and 𝜀 > 0. By Theorem 3, there is g ∈ AM(T, S, I) such that |f − g| < 𝜀1. By definition, there is D ∈ co-I such that g|D ∈ M(T, SD ). Hence, |f |D − g|D| < 𝜀1|D. Thus, f ∈ QM(T, S, I). Corollary 4. Let S be an a-foundation on T. Then, QM b (T, S) = U(T, SP(S), N(S)).

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Proof. It follows from Corollary 1 to Proposition 5 and Corollary 3. Theorem 4. Let S be a separable perfect foundation on a set T and I be an additive ensemble on T. Then, the family QU(T, S, I) is the uniform closure in F(T) of the family AU(T, S, I). Proof. Let f ∈ QU(T, S, I)+ . Fix a number n ∈ N. Take the set D n ∈ co-I and its finite cover 𝜘n ≡ (S nk ∈ S | k ∈ K n ) such that 𝜔(f , 𝜘n ) < 1/n. By Theorem 2 (2.4.4), for every k ∈ K n , there is a positive function 𝜑k ∈ U(T, S) such that coz 𝜑k = S nk . Define the function 𝜓 ≡ ∑ (𝜑k | k ∈ K n ). Lemma 3 (2.5.1) and Corollary 3 to Theorem 1 (2.4.2) guarantee that 𝜓|D n ∈ U(D n , SD n ). Define the functions g nk on T setting g nk (t) ≡ 𝜑k (t)/𝜓(t) for t ∈ D n and g nk (t) ≡ 0 for t ∈ ̸ D n . By Corollary 3 to Theorem 1 (2.4.2), g nk |D n ∈ U(D n , SD n ). Let t ∈ D n . Put m n ≡ card K n . Assume that g nk (t) < 1/m n for every k ∈ K n . Therefore, 1 = ∑ (g nk (t) | k ∈ K n ) < m n /m n = 1, but this is impossible. Consequently, g nk (t) ⩾ 1/m n for some k ∈ K n , where t ∈ G nk ≡ {t ∈ D n | g nk (t) > 1/(m n + 1)}. Thus, 𝜆 n ≡ (G nk | k ∈ K n ) is a cover of D n . Let x nk ≡ inf{f (t) | t ∈ S nk }. Consider the functions h nk ≡ ((m n + 1)x nk g nk ) ∧ (x nk 1). By Corollary 3 to Theorem 1 (2.4.2) h nk |D n ∈ U(D n , SD n ). Define the function f n : T → [0, ∞[ setting f n (t) ≡ sup (h nk (t) | k ∈ K n ) for t ∈ D n and f n (t) ≡ f (t) for t ∈ ̸ D n . By Corollary 4 to Theorem 1 (2.4.2), f n |D n ∈ U(D n , SD n ). Hence, f n ∈ AU(T, S, I). If t ∈ S nk , then h nk (t) ⩽ x nk ⩽ f (t). If t ∈ ̸ S nk , then h nk (t) = 0 ⩽ f (t). Consequently, h nk ⩽ f for every k ∈ K n . This implies f n ⩽ f . If t ∈ G nk , then h nk (t) = x nk . Therefore, 0 ⩽ f (t) − f n (t) ⩽ f (t) − h nk (t) ⩽ x nk + 1/n − x nk = 1/n. Since 𝜆 n is a cover of D n , it follows that 0 ⩽ f (t) − f n (t) ⩽ 1/n for every t ∈ D n . Thus, 0 ⩽ f (t) − f n (t) ⩽ 1/n for every t ∈ T. If f ∈ QU(T, S, I), then f = f+ + f− . As was shown above, there are g n , h n ∈ AU(T, S, I) such that 0 ⩽ f+ (t) − g n (t) ⩽ 1/n and 0 ⩽ −f− (t) − h n (t) ⩽ 1/n for every t ∈ T. Take f n ≡ g n − h n . It belongs to AU(T, S, I) by virtue of Proposition 2 (2.5.1). Then, |f (t) − f n (t)| ⩽ |f+ (t) − g n (t)| + |f− (t) + h n (t)| ⩽ 2/n. Thus, AU(T, S, I) is uniformly dense in QU(T, S, I). According to Proposition 2, the family QU(T, S, I) is uniformly closed. Corollary 1. Let S be a separable perfect foundation on a set T and I be an additive ensemble on T. Then, the family QU b (T, S, I) is the uniform closure in F b (T) of the family AU b (T, S, I). Corollary 2. Let S be a separable perfect foundation on a set T and I be an additive ensemble on T. Then, the family QU(T, S) is the uniform closure in F(T) of the family AU(T, S). We conclude this section by describing the normal envelopes of the considered families.

178 | 2.5 Families of functions on a descriptive space with negligence

Proposition 6. Let S be a foundation on T and I be an additive ensemble on T. Then, 1) N(QU(T, S, I)) = M(T, Coz QU(T, S, I)); 2) N(QD(T, S, I)) = M(T, Coz QD(T, S, I)); 3) N(AU(T, S, I)) = M(T, (Coz AU(T, S, I))𝜎 ); 4) if, besides, S is 𝜎-additive, then N(AD(T, S, I)) = M(T, (Coz AD(T, S, I))𝜎 ); 5) if S is perfect and separable, then N(AU(T, S, I)) = N(QU(T, S, I)); 6) if S is an a-foundation, then N(AD(T, S, I)) = N(QD(T, S, I)). Proof. 1) By Proposition 2, the family QU(T, S, I) satisfies conditions 1 – 4 from 2.2.4. By its Corollary 1, the ensemble Coz QU(T, S, I) is 𝜎-additive. Then, it follows from Proposition 2 (2.3.6) that N(QU(T, S, I)) = M(T, Coz QU(T, S, I)). 2) The arguments are the same. 3) By Proposition 2 (2.5.1) AU(T, S, I) satisfies conditions 1 – 4 from 2.2.4, and therefore, N(AU(T, S, I)) = M(T, (Coz AU(T, S, I))𝜎 ) by virtue of Proposition 2 (2.3.6). 4) The arguments are the same. 5) Since S is a separable perfect foundation, by Theorem 4, the uniform closure of AU(T, S, I) is QU(T, S, I). Therefore, QU(T, S, I) ⊂ N(AU(T, S, I)). Then, using AU(T, S, I) ⊂ QU(T, S, I), we get N(AU(T, S, I)) ⊂ N(QU(T, S, I)) ⊂ N(AU(T, S, I)). 6) The arguments are similar, one should only use Theorem 3 instead of Theorem 4.

3 Fundamentals of the measure theory Introduction This chapter is devoted to the general theory of measure and integral. It is based on the two preceding chapters and especially on the theory of series in the extended real line. This allows us to consider measures taking their values in [−∞, ∞] at very beginning. The usage of these general measures gives the opportunity to bring together two classical lines of the measure theory: the line of bounded (signed) measures 𝜇 : M → [a, b] ⊂ R and the line of (unbounded) positive measures 𝜇 : M → R+ ≡ [0, ∞]. Such the general measure theory is expounded in 3.1 and 3.2. Similarly, in the integral theory, two parallel points of view were initially formed: to consider the integral as special structure over a descriptive space ⟮T, M, 𝜇⟯ with the 𝜎-algebra M and the measure 𝜇 : M → R and to consider the integral as a superstructure over the functional system ⟮A(T), 𝜑⟯ with the linear space of functions A(T) and the linear functional 𝜑 : A(T) → R. The first point of view was developed by H. Lebesgue [1902, 1904], J. Radon [1913], M. Fréchet [1915a], P. Halmos [1950], and others. The second point of view was developed by W. H. Young [1905, 1911, 1914], P. Daniell [1918], S. Banach [1937], N. Bourbaki [1956, 1959, 1965, 1969], and others. The parallelism of these points of view was observed by M. Stone [1948a, 1948b, 1948c, 1949], R. Arens [1950, 1963] et al. But this parallelism was neither strictly formulated nor strictly proven. Two mentioned conceptual approaches to the integral are expounded in sections 3.3 and 3.4. Besides, in 3.3.6, the supposed parallelism is strictly formulated and in 3.4.2, it is strictly proven in the form of the isomorphism between the Lebesgue integrals i𝜇 and some linear functionals 𝜑 : A(T) → R (Theorem 3 (3.4.2)). In a more popular topological case ⟮T, G, M, 𝜇⟯ with the open topology G ⊂ M and the Radon measure 𝜇, this supposed parallelism is known as the Riesz – Radon – Fréchet problem of characterization of Radon integrals as linear functionals (see 3.5.3). Partial solution of this problem were given by J. Hadamard [1903], M. Frechét [1904], F. Riesz [1909, 1914], J. Radon [1913], S. Saks [1938], S. Kakutani [1941], P. Halmos [1950], E. Hewitt [1952], R. E. Edwards [1953], N.Bourbaki [1969], H. König [2000, 2008], and others. Its complete solution is expounded in 3.6.4. It is remarkable that after the appearance of the Lebesgue integral, the celebrated Riemann integral did not lose his value and continued developing. Therefore, the last section of this chapter is devoted to the Riemann integral in the very general form invented by N. Bourbaki [1965, 5, ex.∘ 17] for a topological measurable space ⟮T, G, M, 𝜇⟯ with the Tychonoff topology G and the bounded positive Radon measure 𝜇. The most conceptual and technical achievement of this chapter is Theorem 3 (3.7.2) giving some new characterization of Riemann 𝜇-integrable functions different from the famous Lebesgue one. https://doi.org/10.1515/9783110550962-002

180 | 3.1 Spaces with semimeasures and measures

3.1 Spaces with semimeasures and measures 3.1.1 Spaces with evaluations, semimeasures and measures Let T be a set and S be an ensemble on T (see 2.1.1). Every mapping 𝜀 : S → R is called an evaluation (≡ a set function) on the descriptive space ⟮T, S⟯ or simply on the ensemble S. A descriptive space ⟮T, S⟯ with an evaluation 𝜀 will be called a space with an evaluation and will be denoted by ⟮T, S, 𝜀⟯. The elements of the ensemble dom 𝜀 ≡ S are called 𝜀-measurable sets. The evaluation 𝜀 is called bounded, if rng 𝜀 ⊂ [a, b] ⊂ R for some a, b ∈ R. An evaluation 𝜀 is called increasing [decreasing] (see 1.1.15), if 𝜀R ⩽ 𝜀S [𝜀R ⩾ 𝜀S] for every R, S ∈ S such that R ⊂ S. It is called positive if rng 𝜀 ⊂ R+ ≡ [0, ∞] and R+ ∩ rng 𝜀 = / ⌀. It is called negative if rng 𝜀 ⊂ R− ≡ [−∞, 0] and R− ∩ rng 𝜀 = / ⌀. It is called finite if rng 𝜀 ⊂ R. We shall use often the following subsets of S: – the subset S0 (𝜀) ≡ {S ∈ S | 𝜀S = 0} of 𝜀-negligible sets (≡ 𝜀-null sets); – the subset domf 𝜀 ≡ Sf (𝜀) ≡ {S ∈ S | 𝜀S ∈ R} of sets of finite value of 𝜀 (≡ 𝜀squarable sets). For any ensemble X on T, we also denote Xf (𝜀) ≡ X ∩ Sf (𝜀) and X0 (𝜀) ≡ X ∩ S0 (𝜀). An increasing [decreasing] evaluation 𝜀 will be called 𝜏-finite on a set S ∈ S if there is an arbitrary collection 𝜋 ≡ (S i ∈ Sf (𝜀) | i ∈ I) such that S = ⋃⟮S i | i ∈ I⟯ and 𝜀S = sup(𝜀S i | i ∈ I) [𝜀S = inf(𝜀S i | i ∈ I)]. If the collection 𝜋 is countable, then 𝜀 will be called 𝜎-finite. For the increasing or decreasing evaluation 𝜀 on S, consider also the following subsets of S: – the subset S𝜎f (𝜀) ≡ {S ∈ S | 𝜀 is 𝜎-finite on S} of sets of 𝜎-finite value of 𝜀; – the subset S𝜏f (𝜀) ≡ {S ∈ S | 𝜀 is 𝜏-finite on S} of sets of 𝜏-finite value of 𝜀. The increasing or decreasing evaluation 𝜀 on an ensemble S is called 𝜎-finite [𝜏-finite], if S = S𝜎f (𝜀) [S = S𝜏f (𝜀)]. An evaluation 𝜀 : S → R will be called natural, if 1. S contains ⌀; 2. 𝜀(⌀) = 0; and 3. either rng 𝜀 ⊂ {−∞} ∪ R = [−∞, ∞[ or rng 𝜀 ⊂ R ∪ {∞} =] − ∞, ∞]. A space with an evaluation ⟮T, S, 𝜀⟯ will be called 𝜎-finite, if T = ⋃⟮T i | i ∈ I⟯ for some countable collection (T i ∈ Sf (𝜀) | i ∈ I). An evaluation 𝜀 : S → R will be called internally finite, if for every S ∈ S such that 𝜀S = ∞ [𝜀S = −∞] there exists R ∈ S such that R ⊂ S and 0 < 𝜀R < ∞ [−∞ < 𝜀R < 0]. For a justification of this term, see Theorem 5 (3.1.4). The following group of properties of evaluations is connected with the property of additivity. A natural evaluation 𝜀 is called (binary) additive, if 𝜀(Q ∪ R) = 𝜀Q + 𝜀R for every pair of disjoint sets Q, R ∈ S such that Q ∪ R ∈ S. A natural evaluation 𝜀 will be called finitely [countably, completely] additive, if 𝜀(⋃⟮S i | i ∈ I⟯) = ∑net (𝜀(S i ) |

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i ∈ I) (see 1.4.8) for every finite [countable, arbitrary] pairwise disjoint collection (S i ∈ S | i ∈ I) (see 1.1.10) such that ⋃⟮S i | i ∈ I⟯ ∈ S. It is evident that every subsequent kind of additivity implies all preceding ones. Lemma 1. Let 𝜀 be a binary additive natural evaluation on a disjointly additive ensemble S. Then, 𝜀 is finitely additive. Proof. By the remark to Proposition 1 (2.1.1), S is finitely disjointly additive. Consider the set X of all n ∈ 𝜔 such that 𝜀(⋃⟮S i | i ∈ I⟯) = ∑(𝜀S i | i ∈ I) for every collection (S i ∈ S | i ∈ I) of pairwise disjoint sets such that ⋃⟮S i | i ∈ I⟯ ∈ S and card I = 2 + n. Let card I = 2, i. e. I = {j, k} and j = / k. Then, ⋃⟮S i | i ∈ I⟯ = S j ∪ S k implies 𝜀(⋃⟮S i | i ∈ I⟯) = 𝜀S j +𝜀S k . According to Lemma 4 (1.4.3), the last number is equal to the number ∑(𝜀S i | i ∈ I). Consequently, 0 ∈ X. Let n ∈ X and card I = n + 3. Then, I = {j} ∪ K for some j ∈ I and K ≡ I\{j}. Since P(K) = 2 + n, then 𝜀(⋃⟮S k | k ∈ K⟯) = ∑(𝜀S k | k ∈ K). By Corollary 2 to Proposition 1 (1.1.10), the sets R ≡ ⋃⟮S k | k ∈ K⟯ and S j do not intersect. According to this Corollary, Theorem 1 (1.4.3), and Lemma 4 (1.4.3), we have 𝜀(⋃⟮S i | i ∈ I⟯) = 𝜀(S j ∪ R) = 𝜀S j + 𝜀R = 𝜀S j + ∑(𝜀S k | k ∈ K) = ∑(𝜀S i | i ∈ I). Hence, n + 1 ∈ X. By the principle of natural induction from 1.2.6 (Theorem 1), we get X = 𝜔. Lemma 2. Let 𝜀 be a binary additive natural evaluation on an ensemble S. If Q, R ∈ S, Q ⊂ R, R\Q ∈ S and |𝜀Q| < ∞, then 𝜀(R\Q) = 𝜀R − 𝜀Q. Proof. We have R = (R\Q)∪Q and (R\Q)∩Q = ⌀. Therefore, 𝜀R = 𝜀(R\Q)+𝜀(Q) implies the desired equality. A natural evaluation 𝜀 is called (binary) subadditive, if 𝜀(Q ∪ R) ⩽ 𝜀Q + 𝜀R for every pair of disjoint sets Q, R ∈ S such that Q ∪ R ∈ S. A natural evaluation 𝜀 is called finitely [countably, completely] subadditive, if 𝜀S ⩽ ∑net (𝜀S i | i ∈ I) for every finite [countable, arbitrary] collection (S i ∈ S | i ∈ I) of pairwise disjoint sets such that S ≡ ⋃⟮S i | i ∈ I⟯ ∈ S. A natural evaluation 𝜀 is called (binary) superadditive, if 𝜀(Q ∪ R) ⩾ 𝜀Q + 𝜀R for every pair of disjoint sets Q, R ∈ S such that Q ∪ R ∈ S. A natural evaluation 𝜀 is called finitely [countably, completely] superadditive, if 𝜀S ⩾ ∑net (𝜀S i | i ∈ I) for every finite [countable, arbitrary] collection (S i ∈ S | i ∈ I) of pairwise disjoint sets such that S ≡ ⋃⟮S i | i ∈ I⟯ ∈ S. By the same induction as in the proof of Lemma 1, it is deduced that if 𝜀 is subadditive [superadditive] and S is disjointly additive, then 𝜀 is finitely subadditive [superadditive]. Lemma 3. Let 𝜀 be an increasing natural subadditive [countably subadditive] evaluation on a ring R. Then, 𝜀R ⩽ ∑(𝜀R i | i ∈ I) for every finite [countable] collection (R i ∈ R | i ∈ I) and every R ∈ R such that R ⊂ ⋃⟮R i | i ∈ I⟯.

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Proof. According to the definition of finite [countable] sets in 1.2.6, Proposition 1 (1.1.10), and Theorem 1 (1.4.3) [Theorem 1 (1.4.8)], we may assume without loss of generality that I = n ∈ N [I = 𝜔]. Consider the disjoint sets S i ≡ R i \ ⋃⟮R j | j ∈ i⟯. Since R ⊂ ⋃⟮R i | i ∈ I⟯ = ⋃⟮S i | i ∈ I⟯, we get 𝜀R ⩽ 𝜀(⋃⟮S i | i ∈ I⟯) ⩽ ∑net (𝜀S i | i ∈ I) ⩽ ∑ (𝜀R i | i ∈ I). A natural finitely additive evaluation 𝜇 defined on a ring R (see 2.1.1) will be called a semimeasure on the space ⟮T, R⟯ or simply on the ring R. A space with an evaluation ⟮T, R, 𝜇⟯, where the evaluation 𝜇 is a semimeasure, is called a space with a semimeasure or a semimeasurable space. Lemma 4. Let 𝜀 be a positive semimeasure on a ring R. Then, 𝜀 is increasing and completely superadditive. Proof. Let P ⊂ Q in R. Since Q = P ∪ (Q\P), we have 𝜀P ⩽ 𝜀P + 𝜀(Q\P) = 𝜀Q. Let (R i ∈ R | i ∈ I) be a pairwise disjoint collection such that R ≡ ⋃⟮R i | i ∈ I⟯ ∈ R. For every finite set J ⊂ I, we have ∑(𝜀R i | i ∈ J) = ∑(⋃⟮R i | i ∈ J⟯) ⩽ 𝜀R. Therefore, using the definition of a netful series from 1.4.8, Theorem 2 (1.4.5), Lemma 7 (1.4.7), and Corollary 3 to Proposition 1 (1.4.7) we conclude that ∑net (𝜀R i | i ∈ I) ⩽ 𝜀R. A natural evaluation 𝜀 will be called sequentially additive or 𝜎-additive, if 𝜀S = ∑(𝜀S n | n ∈ N) (see 1.4.8) for every cofinal subset N of 𝜔 and every pairwise disjoint sequence (S n ∈ S | n ∈ N) such that S ≡ ⋃⟮S n | n ∈ N⟯ ∈ S. It is evident that a sequentially additive evaluation is finitely additive and additive. A natural 𝜎-additive evaluation 𝜇 : R → R defined on a ring R will be called a measure on the space ⟮T, R⟯ or simply on the ring R. A space with an evaluation ⟮T, R, 𝜇⟯, where the evaluation 𝜇 is a measure, is called a space with a measure or a measurable space. A measure defined on 𝛿-ring [𝜎-algebra] will be called narrow [wide]. There are some equivalent reformulations of the definition of a measure below. Lemma 5. Let 𝜀 be a semimeasure on a ring R. Then, the following conclusions are equivalent: 1) 𝜀(⋃⟮R m | m ∈ M⟯) = lim(𝜀R m | m ∈ M) for every increasing sequence (R m ∈ R | m ∈ M ⊂ 𝜔) such that ⋃⟮R m | m ∈ M⟯ ∈ R; 2) 𝜀 is a measure. Proof. (1) ⊢ (2). Let (S m ∈ R | m ∈ M) be a pairwise disjoint sequence such that S ≡ ⋃⟮S m | m ∈ M⟯ ∈ R. Consider the sequence (R m | m ∈ 𝜔) ↑ such that R m ≡ ⋃⟮S i | i ∈ M ∩ (m + 1)⟯. It follows from (1) that 𝜀S = 𝜀(⋃⟮R m | m ∈ 𝜔⟯) = lim(𝜀R m | m ∈ 𝜔) = lim (∑(𝜀S i ) | i ∈ M ∩ (m + 1) | m ∈ 𝜔) = ∑(𝜀S m | m ∈ M). (2) ⊢ (1). Let (R m ∈ R | m ∈ M) ↑ be a sequence such that R ≡ ⋃⟮R m | m ∈ 𝜔⟯ ∈ R. Consider the pairwise disjoint sequence (S m | m ∈ M) such that S m ≡ R m \ ⋃⟮R i | i ∈

3.1.1 Spaces with evaluations, semimeasures and measures | 183

M ∩ m⟯. Since R = ⋃⟮S m | m ∈ M⟯, we have 𝜀R = lim(∑(𝜀S i | i ∈ M ∩ (m + 1)) | m ∈ 𝜔) = lim (𝜀(⋃⟮S i | i ∈ M ∩ (m + 1)⟯) | m ∈ M) = lim(𝜀R m | m ∈ M). Lemma 6. Let 𝜇 be a measure on a ring R. Then, 𝜇(⋂⟮R m | m ∈ 𝜔⟯) = lim(𝜇R m | m ∈ 𝜔) for every decreasing sequence (R m ∈ R | m ∈ 𝜔) such that ⋂⟮R m | m ∈ 𝜔⟯ ∈ R, |𝜇(⋂⟮R m | m ∈ 𝜔⟯)| < ∞ and |𝜇R m | < ∞ for every m. Proof. The sequence S m ≡ R0 \R m is increasing and ⋃⟮S m | m ∈ 𝜔⟯ = R0 \ ⋂⟮R m | m ∈ 𝜔⟯ ∈ R. Therefore, by Lemma 2, Lemma 5 and Proposition 1 (1.4.7) we obtain 𝜇(⋃⟮S m | m ∈ 𝜔⟯) = lim(𝜇S m | m ∈ 𝜔) = lim(𝜇R0 − 𝜇R m | m ∈ 𝜔) = 𝜇R0 − lim(𝜇R m | m ∈ 𝜔). On the other hand 𝜇(⋃⟮S m | m ∈ 𝜔⟯) = 𝜇R0 − 𝜇(⋂⟮R m | m ∈ 𝜔⟯). Since all the considered numbers are real, we get 𝜇(⋂⟮R m | m ∈ 𝜔⟯) = lim(𝜇R m | m ∈ 𝜔). Lemma 7. Let 𝜀 be a finite semimeasure on a ring R. If lim(𝜀R m | m ∈ 𝜔) = 0 for every decreasing sequence (R m ∈ R | m ∈ 𝜔) such that ⋂⟮R m | m ∈ 𝜔⟯ = ⌀, then 𝜀 is a measure. Proof. Let (S i ∈ R | i ∈ M ⊂ 𝜔) be a pairwise disjoint sequence such that S ≡ ⋃⟮S i | i ∈ 𝜔⟯ ∈ R. Consider the sequence (R m | m ∈ 𝜔) ↓ ⌀ such that R m ≡ S\ ⋃⟮S i | i ∈ M ∩ (m + 1)⟯. Then, lim(𝜀R m | m ∈ 𝜔) = 0. Since 𝜀S = 𝜀R m + ∑(𝜀S i | i ∈ M ∩ (m + 1)), we get ∑(𝜀S i | i ∈ M ∩ (m + 1)) = 𝜀S − 𝜀R m . By Corollary 1 to Lemma 1 (1.4.8), ∑(𝜀S i | i ∈ M) = 𝜀S. Lemma 8. Let 𝜀 be a positive countably subadditive semimeasure on a ring R. Then, 𝜀 is a measure. Proof. Let (R i ∈ R | i ∈ 𝜔) be a sequence of pairwise disjoint sets with R ≡ ⋃⟮R i | i ∈ 𝜔⟯ ∈ R. Since by Lemma 4, 𝜀 is completely superadditive, we have 𝜀R ⩾ ∑net (𝜀R i | i ∈ 𝜔). Since 𝜀 is countably subadditive, we have 𝜀R ⩽ ∑net (𝜀R i | i ∈ 𝜔). Thus, 𝜀R = ∑net (𝜀R i | i ∈ 𝜔). According to Corollary 1 to Lemma 1 (1.4.8) the last sum is equal to ∑(𝜀R i | i ∈ 𝜔). Proposition 1. Let 𝜀 be a positive evaluation on a ring R. Then, the following conclusions are equivalent: 1) 𝜀 is 𝜎-additive, i. e. 𝜀 is a measure; 2) 𝜀 is countably additive. Proof. (1) ⊢ (2). Let (R i ∈ R | i ∈ I) be a countable pairwise disjoint collection such that R ≡ ⋃⟮R i | i ∈ I⟯ ∈ R. We can assume that I = 𝜔. Since 𝜀 is natural, finitely additive, and positive, Lemma 4 implies that 𝜀R ⩾ ∑net (𝜀R i | i ∈ I). Besides, 𝜀R = lim (∑ (𝜀R i | i ∈ n + 1) | n ∈ 𝜔) ⩽ ∑net (𝜀R i | i ∈ I). As a result, we get 𝜀R = ∑net (𝜀R i | i ∈ I).

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(2) ⊢ (1). Since 𝜀 is finitely additive and countably subadditive, it follows from Lemma 8 that 𝜀 is a measure. Lemma 9. Let 𝜀 be an increasing subadditive positive evaluation on a ring R. Then, the following statements are fulfilled: 1) the ensembles R0 (𝜀), Rf (𝜀), and R𝜎f (𝜀) are rings and ideals in R (see 2.1.4); 2) if R is a 𝛿-ring, then R0 (𝜀), Rf (𝜀) and R𝜎f (𝜀) are 𝛿-rings; 3) if R is a 𝜎-ring and 𝜀 is countably subadditive, then R0 (𝜀) and R𝜎f (𝜀) are 𝜎-rings. Proof. 1. Let (R m ∈ Rf | m ∈ M) be a finite collection. Then, by Lemma 3 for R ≡ ⋃⟮R m | m ∈ M⟯ ∈ R, we have 0 ⩽ 𝜀R ⩽ ∑ (𝜀R m | m ∈ M) < ∞. Hence, R ∈ Rf , i. e. Rf is finitely additive. The same is valid for R0 . Further, let R ∈ Rf and S ∈ R. Then, 0 ⩽ 𝜀(R ∩ S) ⩽ 𝜀R < ∞ implies R ∩ S ∈ Rf . Therefore, Rf is an ideal in R. The same is valid for R0 . Then, by Lemma 2 (2.1.4), Rf and R0 are rings. Let (R m ∈ R𝜎f | m ∈ M) be a finite collection. Then, R m = ⋃⟮R mi | i ∈ I m ⟯ for some countable collection (R mi ∈ Rf | i ∈ I m ). By Proposition 1 (1.1.13), R ≡ ⋃⟮R m | m ∈ M⟯ = ⋃⟮⋃⟮R mi |∈ I m ⟯ | m ∈ M⟯ = ⋃⟮R p | p ∈ P⟯, where p ≡ (m, i) ∈ P ≡ ⋃⟮{m} × I m | m ∈ M⟯. By Theorem 1 (1.3.9) the set P is countable. Therefore, R ∈ R𝜎f . Let R ∈ R𝜎f and S ∈ R. Then, R = ⋃⟮R i | i ∈ I⟯ for some countable collection (R i ∈ Rf | i ∈ I). The equalities R ∩ S = ⋃⟮R i ∩ S | i ∈ I⟯ and R i ∩ S ∈ Rf imply R ∩ S ∈ R𝜎f . Therefore, R𝜎f is an ideal in R. Then, by Lemma 2 (2.1.4), R𝜎f is a ring. 2. Let (R m ∈ Rf | m ∈ M) be a countable collection such that S ≡ ⋃⟮R m | m ∈ M⟯ ⊂ R ∈ Rf . By Lemma 9 (2.1.1), S ∈ R. Since according to (1), the ensemble Rf is an ideal in R, we get S ∈ Rf . Then, again by Lemma 9 (2.1.1), Rf is a 𝛿-ring. The same is true for R0 and R𝜎f . 3. Let (R m ∈ R0 | m ∈ M) be a countable collection. Then, for R ≡ ⋃⟮R m | m ∈ M⟯ ∈ R by Proposition 1 and Lemma 3 we have 0 ⩽ 𝜀R ⩽ ∑(𝜀R m | m ∈ M) = 0, where R ∈ R0 . Thus, R0 is a 𝜎-ring. Let (R m ∈ R𝜎f | m ∈ M) be a countable collection. Then, R m = ⋃⟮R mi | i ∈ I m ⟯ for some countable collection (R mi ∈ Rf | i ∈ I). Repeating the argument from the proof of assertion 1 we deduce that R ≡ ⋃⟮R m | m ∈ M⟯ = ⋃⟮R mi | p ∈ P⟯ ∈ R𝜎f , where p ≡ (m, i) ∈ P ≡ ⋃⟮{m} × I m | m ∈ M⟯. Thus, R𝜎f is a 𝜎-ring. Now, we shall prove some curious property of ensemble of pairwise disjoint sets. Lemma 10. Let 𝜀 be a bounded positive finitely superadditive evaluation on a finitely additive ensemble S and L ⊂ S consists of pairwise disjoint sets such that 𝜀L > 0. Then, L is countable. Proof. Let rng 𝜀 ⊂ [0, a], a ∈ N. Take the sets Ln ≡ {L ∈ L | 𝜀L ⩾ 1/n}. Then, L = ⋃⟮Ln | n ∈ N⟯. Suppose that there exists a number n such that the set Ln is infinite.

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By Corollary 4 to Proposition 1 (1.3.2), there is an injective mapping u : 𝜔 → Ln . Let L i ≡ u(i) and S ≡ ⋃⟮L i | i ∈ (a + 1)n⟯. Then, 𝜀S ⩾ ∑(𝜀L i | i ∈ (a + 1)n) ⩾ (1/n)(a + 1)n = a + 1 > a. On the other hand, 𝜀S ⩽ a. It follows from this contradiction that all sets Ln are finite. By Theorem 1 (1.3.9), the set L is countable. At the end of this subsection, we shall give some simple condition for a positive measure to be 𝜎-finite. Lemma 11. Let 𝜇 be a positive measure on a ring R such that the space ⟮T, R, 𝜇⟯ is 𝜎-finite. Then, 𝜇 is 𝜎-finite. Proof. By the condition, we have T = ⋃⟮T i | i ∈ 𝜔⟯ for some countable collection I. For every (T i ∈ Sf (𝜀) | i ∈ I). Since I is countable, there is a bijection u : 𝜔 f k ∈ 𝜔, consider the set S k ≡ ⋃⟮T u(j) | j ∈ k + 1⟯ ∈ R (𝜇). Then, (S k | k ∈ 𝜔) ↑ T. If M ∈ R, then (M k | k ∈ 𝜔) ↑ M, where M k ≡ M ∩ S k ∈ Rf (𝜇). By virtue of Lemma 5 and Lemma 7 (1.4.7), 𝜇M = lim(𝜇M k | k ∈ 𝜔) = sup (𝜇M k | k ∈ 𝜔). Corollary 1. A positive measure 𝜇 on an algebra A is 𝜎-finite iff the space ⟮T, A, 𝜇⟯ is 𝜎-finite.

3.1.2 Families of evaluations, semimeasures, and measures on a descriptive space Let ⟮T, S⟯ be a fixed descriptive space and ⌀ ∈ S. The family of all natural evaluations on ⟮T, S⟯ will be denoted by Eval(T, S). Its subfamilies of all natural evaluations 𝜀 : S →] − ∞, ∞] and 𝜀 : S → [−∞, ∞[ will be denoted by Eval(T, S, ] − ∞, ∞]) and Eval(T, S, [−∞, ∞[), respectively. The subfamilies of all finite [bounded] natural evaluations will be denoted by Evalf (T, S) [Evalb (T, S)]. It is clear that Eval(T, S) = Eval(T, S, ] − ∞, ∞]) ∪ Eval(T, S, [−∞, ∞[) and Evalb (T, S) ⊂ Evalf (T, S) = Eval(T, S, ] − ∞, ∞]) ∩ Eval(T, S, [−∞, ∞[). The families Evalf (T, S) and Evalb (T, S) are the subfamilies of the family F(S) ≡ Map(S, R) of all real-valued functions on S (see 2.2.1). The zero element 0, the multiplication by real numbers, the addition, the order relation, and so on in F(S) were defined in 2.2.1 and 2.2.2. These pointwise structures are inherited by Evalf (T, S) and Evalb (T, S) but in this case they are called setwise. Note that the families Evalf (T, S) and Evalb (T, S) are closed with respect to the mentioned operations. Hence, the analogues of appropriate assertions of Proposition 1 (2.2.1), its Corollaries, Proposition 1 (2.2.2), and Theorem 1 (2.2.2) are valid. In particular, these families are linear spaces. The whole family Eval(T, S) is a subfamily of the family Map(S, R). Therefore, the extended pointwise structures in the family Map(S, R) (see Remark at the end

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of 2.2.2) are inherited by its subfamilies Eval(T, S), Eval(T, S, ] − ∞, ∞]), and Eval(T, S, [−∞, ∞[). Note that the families Eval(T, S, ] − ∞, ∞]) and Eval(T, S, [−∞, ∞[) are closed with respect to addition and the multiplication by positive and negative real numbers, respectively. The whole family Eval(T, S) is closed under the multiplication by real numbers. Let T be a fixed set and R be a fixed ring on T. The families of all semimeasures and measures on R will be denoted by SMeas(T, R) and Meas(T, R), respectively. The subfamily of SMeas(T, R) of all semimeasures 𝜇 : R →] − ∞, ∞] [𝜇 : R → [−∞, ∞[ ] will be denoted by SMeas(T, R, ] − ∞, ∞]) [SMeas(T, R, [−∞, ∞[) ]. In the similar way, the subfamilies Meas(T, R, ] − ∞, ∞]) and Meas(T, R, [−∞, ∞[) of Meas(T, R) are defined. The subfamilies of SMeas(T, R) of all finite [bounded] semimeasures will be denoted by SMeasf (T, R) [SMeasb (T, R)]. The corresponding subfamilies for measures will be denoted by Measf (T, R) and Measb (T, R), respectively. The orders and the operations on the subfamilies of SMeas(T, R) and Meas(T, R) are induced by the order and the operations on the corresponding subfamilies of Eval(T, R) considered above. If we consider the families Eval(T, R), SMeas(T, R), and Meas(T, R) together with the mentioned orders on them and we fix the neutral element 0, then according to 1.1.15 we can single out their main parts Eval(T, R)0 , SMeas(T, R)0 , and Meas(T, R)0 , respectively. For a fixed set T, we can consider the extensive families M(T) ≡ ⋃⟮Meas(T, R) | R is a ring ⟯, Mn (T) ≡ ⋃⟮Meas(T, R) | R is a 𝛿-ring ⟯, and Mw (T) ≡ ⋃⟮Meas(T, R) | R is a 𝜎-algebra ⟯ of all measures on T, of all narrow measures on T, and of all wide measures on T, respectively. Lemma 1. Let R be a ring on T. Then, the families SMeas(T, R, ] − ∞, ∞]) and Meas(T, R, ]−∞, ∞]) [ SMeas(T, R, [−∞, ∞[) and Meas(T, R, [−∞, ∞[) ] are closed with respect to addition and multiplication on positive [negative] real numbers. Proof. Let 𝜇, 𝜈 ∈ SMeas(T, R, ] − ∞, ∞]) and x, y ∈]0, ∞[. Let (R i ∈ R | i ∈ I) be a finite collection of pairwise disjoint sets and R ≡ ⋃⟮R i | i ∈ I⟯. If 𝜇R i and 𝜈R i are finite for all i, then by Corollary 1 to Lemma 4 (1.4.3), (x𝜇 + y𝜈)R ≡ x𝜇R + y𝜈R = ∑(x𝜇R i | i ∈ I) + ∑(y𝜈R i | i ∈ I) = ∑((x𝜇 + y𝜈)R i | i ∈ I). If 𝜇R i are finite for all i and 𝜈R j = ∞ for some j ∈ I, then x𝜇R ∈ R, y𝜈R = ∞, (x𝜇 + y𝜈)R j = ∞ and ∑((x𝜇 + y𝜈)R i | i ∈ I) = ∞. Therefore, (x𝜇 + y𝜈)R ≡ x𝜇R + y𝜈R = ∞ = ∑((x𝜇 + y𝜈)R i | i ∈ I). If 𝜇R j = ∞

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and 𝜈R k = ∞ for some j, k ∈ I, then the arguments are the same. This means that x𝜇 + y𝜈 ∈ SMeas(T, R, ] − ∞, ∞]). Suppose now that 𝜇, 𝜈 ∈ Meas(T, R, ] − ∞, ∞]). Take any increasing infinite sequence (R m ∈ R | m ∈ M ⊂ 𝜔) such that R ≡ ⋃⟮R m | m ∈ M⟯ ∈ R. By Lemma 5 (3.1.1), 𝜇R = lim(𝜇R m | m ∈ M) and 𝜈R = lim(𝜈R m | m ∈ M). At first suppose that 𝜇R and 𝜈R are finite. Then, for any 𝜀 > 0 there are p and q in M such that |𝜇R − 𝜇R m | < 𝜀/2x for every m ⩾ p and |𝜈R − 𝜈R m | < 𝜀/2y for every m ⩾ q. So 𝜇R m and 𝜈R m are finite for all m ⩾ gr{p, q} and |(x𝜇 + y𝜈)R − (x𝜇 + y𝜈)R m | ⩽ x|𝜇R − 𝜇R m | + y|𝜈R − 𝜈R m | < 𝜀. This means that (x𝜇 + y𝜈)R = lim((x𝜇 + y𝜈)R m | m ∈ M). Now, suppose that 𝜇R < ∞ and 𝜈R = ∞. Then, for any 𝛿 > 0 there are p and q in M such that |𝜇R − 𝜇R m | < 𝛿/x for every m ⩾ p and 𝜈R m > (2𝛿 + x|𝜇R|)/y for every m ⩾ q. So 𝜇R m > −𝛿/x + 𝜇R implies (x𝜇 + y𝜈)R m > −𝛿 + x𝜇R + 2𝛿 + x|𝜇R| ⩾ 𝛿 for all m ⩾ gr{p, q}. This means that (x𝜇 + y𝜈)R = ∞ = lim((x𝜇 + y𝜈)R m | m ∈ M). Finally, suppose that 𝜇R = ∞ and 𝜈R = ∞. Then, for any 𝛿 > 0 there are p and q in M such that 𝜇R m > 𝛿/x for every m ⩾ p and 𝜈R m > 𝛿/y for every m ⩾ q. Therefore, for every m ⩾ gr{p, q} we have (x𝜇+ y𝜈)R m > 2𝛿 > 𝛿. This means that (x𝜇+ y𝜈)R = ∞ = lim((x𝜇 + y𝜈)R m | m ∈ M). By virtue of Lemma 5 (3.1.1) we conclude that x𝜇 + y𝜈 ∈ Meas(T, R, ] − ∞, ∞]). For the second case, the arguments are analogous. Corollary 1. The families SMeasf (T, R) and Measf (T, R) are closed with respect to addition and multiplication on real numbers. Corollary 2. The families SMeasf (T, R) and Measf (T, R) are ordered linear spaces. Proof. Since F(R) is an ordered linear space (see 2.2.1 and 2.2.2) and the subfamily SMeasf (T, R) ⊂ F(R) is closed under addition and multiplication on real numbers, SMeasf (T, R) is also an ordered linear space. Generally speaking the families SMeasf (T, R) and Measf (T, R) are not closed with respect to the lattice operations in F(R). Hence, we need some “good” subfamilies of these families. They will be considered in 3.2.1. 3.1.3 The total variation of a natural evaluation Let T be a set, S be an ensemble on T, and E ⊂ T. Denote by Parf (S, E) [Par𝜎 (S, E)] the set of all finite [countable] partitions (S i ∈ S | i ∈ I) of E. Denote by Parf (S, E) [Par𝜎 (S, E)] the set of all finite [countable] pairwise disjoint collections (S i ∈ S | i ∈ I) such that ⋃⟮S i | i ∈ I⟯ ⊂ E. It is clear that Parf (S, E) ⊂ Par𝜎 (S, E), Parf (S, E) ⊂ Par𝜎 (S, E), Parf (S, E) ⊂ Parf (S, E), and Par𝜎 (S, E) ⊂ Par𝜎 (S, E). Similarly, denote by 𝜎 Par (S, E) the set of all countable collections (S i ∈ S | i ∈ I) such that E ⊂ ⋃⟮S i | i ∈ I⟯.

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Let 𝜀 : S → R be an evaluation and 𝜋 ≡ (S i ∈ S | i ∈ I) be a finite [countable] collection. Put v(𝜀, 𝜋) ≡ ∑ (|𝜀(S i )| | i ∈ I) [v(𝜀, 𝜋) ≡ ∑net (|𝜀(S i )| | i ∈ I)]. According to the definitions of finite and countable sets in 1.2.6 and Theorem 1 (1.4.8), we may assume without loss of generality that I ⊂ 𝜔. Then, Proposition 1 (1.4.8) implies that v(𝜀, 𝜋) = ∑ (|𝜀(S i )| | i ∈ I) either I is finite or not. If 𝜀 is positive, then v(𝜀, 𝜋) = ∑ (𝜀(S i ) | i ∈ I). For a natural evaluation 𝜀 : S → R, define the evaluation 𝜀̄̄ : P(T) → R+ setting ̄𝜀(E) ̄ ≡ sup{v(𝜀, 𝜋) | 𝜋 ∈ Parf (S, E)} for every subset E of T. This evaluation is called the total variation of 𝜀. Along with the notation 𝜀̄̄ we shall use also the notation var 𝜀. Thus, we have the mapping var : Eval(T, S) → Eval(T, P(T)) such that var(𝜀) ≡ 𝜀.̄̄ ̄̄ ∈ Eval(T, S) is called the variation of the evaluation 𝜀 and will be The evaluation 𝜀|S denoted by v(𝜀). ̄̄ Lemma 1. Let 𝜀 be a natural evaluation on an ensemble S. Then, 𝜀(E) = sup{v(𝜀, 𝜋) | 𝜋 ∈ Par𝜎 (S, E)} for every E ⊂ T. Proof. Denote the right part of the desired equality by x. It follows from the definition ̄̄ of 𝜀̄̄ that 𝜀(E) ⩽ x. Take a collection 𝜋 ≡ (S i | i ∈ I ⊂ 𝜔) ∈ Par𝜎 (S, E). According to Proposition 1 (1.4.8) v(𝜀, 𝜋) ≡ ∑ (|𝜀(S i )| | i ∈ I) = sup (∑ (|𝜀(S i )| | i ∈ I ∩ (n + 1)) | n ∈ 𝜔) ⩽ ̄̄ ⩽ sup{v(𝜀, 𝜌) | 𝜌 ∈ Parf (S, E)} = 𝜀(E). ̄̄ Therefore, x ⩽ 𝜀(E). Lemma 2. Let 𝜀 be a natural evaluation on a ring R and R ∈ R. Then, v(𝜀)(R) = sup{v(𝜀, 𝜋) | 𝜋 ∈ Parf (R, R)}. ̄̄ Proof. Denote the right part of the desired equality by y. Then, y ⩽ 𝜀(R). On the other f hand, if (R i | i ∈ I) ∈ Par (R, R), then there is a bijection u : n I for some n ∈ 𝜔. Consider the pairwise disjoint collection (S k ∈ R | k ∈ n + 1) such that S0 ≡ R u(0) , S k ≡ R u(k) \ ⋃⟮R u(m) | m ∈ k⟯ for k ∈ n ∩ N, and S n ≡ R\ ⋃⟮R i | i ∈ I⟯. Then, ⋃⟮S k | k ∈ ̄̄ n + 1⟯ = R implies ∑ (|𝜀R i | | i ∈ I) ⩽ ∑ (|𝜀S k | | k ∈ n + 1) ⩽ y, where 𝜀(R) ⩽ y. Lemma 3. Let 𝜀 be a natural evaluation on a 𝜎-ring R and R ∈ R. Then, v(𝜀)(R) = sup{v(𝜀, 𝜋) | 𝜋 ∈ Par𝜎 (R, R)}. ̄̄ Proof. Denote the right part of the desired equality by z. Then, by Lemma 1, z ⩽ 𝜀(R). 𝜎 On the other hand, if (R k | k ∈ 𝜔) ∈ Par (R, R), then ⋃⟮R k | k ∈ 𝜔⟯ ∈ R. By virtue of Lemma 9 (2.1.1). Therefore, we can consider the sequence (S i ∈ R | i ∈ 𝜔) such that S0 ≡ R\ ⋃⟮R k | k ∈ 𝜔⟯ and S i+1 ≡ R i for i ∈ 𝜔. Then, ⋃⟮S i | i ∈ 𝜔⟯ = R implies ∑(|𝜀R k | | ̄̄ k ∈ 𝜔) ⩽ ∑ (|𝜀S i | | i ∈ 𝜔) ⩽ z, where, by Lemma 1, 𝜀(R) ⩽ z.

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Lemma 4. Let 𝜀 be a natural evaluation on an ensemble S. Then, 𝜀̄̄ has the following properties: ̄̄ for every S ∈ S; 1) |𝜀S| ⩽ 𝜀S ̄ 2) 𝜀̄ is increasing; ̄̄ = 0 iff 𝜀S = 0 for every S ∈ S such that S ⊂ E. 3) 𝜀E Proof. 1. In fact, we can take (S i ∈ S | i ∈ 1) such that S0 ≡ S. 2. Let F ⊂ E ⊂ T. If (S i ∈ S | i ∈ I) is a disjoint collection such that X ≡ ⋃⟮S i | i ∈ I⟯ ⊂ F, then X ⊂ E. Hence, Parf (S, F) ⊂ Parf (S, E). ̄̄ = 0, then |𝜀(S)| ⩽ 𝜀(S) ̄̄ ̄̄ 3. If 𝜀E ⩽ 𝜀(E) = 0 for every S ∈ S such that S ⊂ E by assertions 1 and 2. If 𝜀(S) = 0 for every S ∈ S such that S ⊂ E, then v(𝜀, 𝜋) = 0 for every 𝜋 ∈ Parf (S, E). Proposition 1. Let S be an ensemble on a set T, 𝜀 be a natural evaluation on S, and 𝜀̄̄ be its total variation. Then, 𝜀̄̄ is completely superadditive. ̄̄ Proof. Take any two disjoint subsets E, F and two numbers x, y ∈ R+ such that x < 𝜀E, ̄̄ For them, there are collections (Q i | i ∈ I) ∈ Parf (S, E) and (R j ∈ S | j ∈ J) ∈ y < 𝜀F. Parf (S, F) such that x < ∑(|𝜀Q i | | i ∈ I) and y < ∑(|𝜀R j | | j ∈ J). Consider the set K ≡ I ∪d J ≡ ⋃⟮A m × {m} | m ∈ 2⟯ from 1.2.2, where A0 ≡ I and A1 ≡ J. Define a collection (S k ∈ S | k ∈ K) setting S k ≡ Q i for k = (i, 0) and S k ≡ R j for k = (j, 1). Then, according to Lemma 4 (1.4.3) and Theorem 1 (1.4.3), ⋃⟮S k | k ∈ K⟯ = ⋃⟮Q i | i ∈ I⟯ ∪ ⋃⟮R j | j ∈ J⟯ ⊂ E∪F implies that x+y < ∑(|𝜀S k | | k ∈ A0 ×{0})+∑(|𝜀S k | | k ∈ A1 ×{1}) = ∑(∑(|𝜀S k | | k ∈ ̄̄ ∪ F). From x < 𝜀(E ̄̄ ∪ F) − y, it follows that A m × {m}) | m ∈ 2) = ∑(|𝜀S k | | k ∈ K) ⩽ 𝜀(E ̄̄ ⩽ 𝜀(E∪F)−y. ̄̄ ̄̄ < ∞. Then, y ⩽ 𝜀(E∪F)− ̄̄ ̄̄ implies 𝜀F ̄̄ ⩽ 𝜀(E∪F)− ̄̄ ̄̄ 𝜀E Suppose that 𝜀E 𝜀E 𝜀E, ̄̄ + 𝜀F ̄̄ ⩽ 𝜀(E ̄̄ ∪ F). If 𝜀E ̄̄ = ∞, then by Lemma 4, 𝜀(E ̄̄ ∪ F) = ∞ as well. Therefore, where 𝜀E ̄̄ + 𝜀F ̄̄ ⩽ 𝜀(E ̄̄ ∪ F). As a result, 𝜀E ̄̄ + 𝜀F ̄̄ ⩽ 𝜀(E ̄̄ ∪ F). in this case, we have also 𝜀E ̄̄ = ∞ Let ⟮E i ⊂ T | i ∈ I⟯ be a pairwise disjoint collection and E ≡ ⋃⟮E i | i ∈ I⟯. If 𝜀E ̄̄ i ⩽ 𝜀E, ̄̄ where 𝜀E ̄̄ = ∞ = ∑(𝜀E ̄̄ i | for some i, then by Lemma 4, E i ⊂ E implies ∞ = 𝜀E ̄̄ i < ∞ for every i ∈ I. i ∈ I). Therefore, we shall further suppose that 𝜀E ̄̄ i | i ∈ I) ⩽ Consider the subset X of 𝜔 consisting of all numbers n such that ∑(𝜀E ̄𝜀(⋃⟮E ̄ i | i ∈ I⟯) for all finite pairwise disjoint collections ⟮E i ⊂ T | i ∈ I⟯ such that card I = n + 2. Let I = {j, k} and j = / k. Then, according to Lemma 4 (1.4.3) and the ̄̄ i | i ∈ I) = 𝜀E ̄̄ j + 𝜀E ̄̄ k ⩽ property proven above, ⋃⟮E i | i ∈ I⟯ = E j ∪ E k implies that ∑(𝜀E ̄𝜀(⋃⟮E ̄ i | i ∈ I⟯). This means that 0 ∈ X. Let n ∈ X and card I = n + 3. Take some j ∈ I and consider the sets K ≡ I\{j} and ̄̄ k | k ∈ K) ⩽ 𝜀E. ̄̄ By Corollary 2 F ≡ ⋃⟮E k | k ∈ K⟯. Since card K = n+2, we have x ≡ ∑(𝜀E ̄̄ i | to Proposition 1 (1.1.10), E ≡ ⋃⟮E i | i ∈ I⟯ = E j ∪ F and by Theorem 1 (1.4.3), y ≡ ∑(𝜀E ̄ ̄ ̄ ̄ ̄ ̄ j +x. By the property proven above, we have y = 𝜀E ̄ j +x ⩽ 𝜀E ̄ j + 𝜀F ̄ ⩽ 𝜀(E ̄ j ∪F) = i ∈ I) = 𝜀E ̄̄ Consequently, n + 1 ∈ X. 𝜀E. By the principle of natural induction (Theorem 1 (1.2.6)), X = 𝜔.

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Finally, let (E i ⊂ T | i ∈ I) be an arbitrary collection. Then, for every finite subset ̄̄ ̄̄ because 𝜀̄̄ is increasing. As a ̄̄ i | i ∈ J) ⩽ 𝜀(⋃⟮E J ⊂ I, we have ∑(𝜀E i | i ∈ J⟯) ⩽ 𝜀E consequence, we get the necessary inequality. Corollary 1. The total variation 𝜀̄̄ is a positive increasing completely superadditive natural evaluation on P(T). Corollary 2. The total variation 𝜀̄̄ is the smallest of all increasing completely superadditive evaluations 𝛿 : P(T) → R+ satisfying the inequality |𝜀S| ⩽ 𝛿S for every S ∈ S. Proof. Let (S i | i ∈ I) ∈ Parf (S, E). Then, ∑ (|𝜀S i | | i ∈ I) ⩽ ∑ (𝛿S i | i ∈ I) ⩽ 𝛿(⋃⟮S i | ̄̄ ⩽ 𝛿E. i ∈ I⟯) ⩽ 𝛿E implies 𝜀E Proposition 2. Let S be an ensemble on a set T, 𝜀 be a natural evaluation on S. Then, ̄̄ 1) if 𝜀 is finitely additive, then 𝜀|U(T, S) is finitely additive, where U(T, S) is the ̄̄ = ∑ (𝜀E ̄̄ i | i ∈ I) for any finite pairwise saturation of S (see 2.1.1); moreover, 𝜀E disjoint collection (E i ∈ U(T, S) | i ∈ I); ̄̄ ̄̄ = ∑ (𝜀E ̄̄ i | i ∈ 𝜔) for any 2) if 𝜀 is 𝜎-additive, then 𝜀|U(T, S) is 𝜎-additive; moreover, 𝜀E countable pairwise disjoint collection (E i ∈ U(T, S) | i ∈ 𝜔). Proof. We may and shall suppose that rng 𝜀 ⊂] − ∞, ∞]. 1. Let (E i ∈ U(T, S) | i ∈ I) be a finite pairwise disjoint collection and E ≡ ⋃⟮E i | i ∈ I⟯. Note that E does not necessary belong to U(T, S). Take a collection (S j ∈ S | j ∈ J) ∈ Parf (S, E). We have E i ∩ S j ∈ S and ⋃⟮E i ∩ S j | i ∈ I⟯ = S j . Since 𝜀 is finitely additive we get 𝜀S j = ∑(𝜀(E i ∩ S j ) | i ∈ I). Therefore, using Corollary 1 to Proposition 1 (1.4.3), we obtain ∑(|𝜀S j | | j ∈ J) ⩽ ∑(∑(|𝜀(E i ∩ S j )| | i ∈ I) | j ∈ J) = ̄̄ i | i ∈ I). Consequently, 𝜀E ̄̄ ⩽ ∑(𝜀E ̄̄ i | i ∈ I). ∑(∑(|𝜀(E i ∩ S j )| | j ∈ J) | i ∈ I) ⩽ ∑(𝜀E ̄ ̄ ̄ = ∑(𝜀E ̄ i | i ∈ I). The inverse inequality follows from Proposition 1. Thus, 𝜀E 2. Let (E i ∈ U(T, S) | i ∈ 𝜔) be a sequence of pairwise disjoint elements. Take E and (S j | j ∈ J) as above. Put p ≡ card J. Consider the sequences (x jn | n ∈ 𝜔) such that x jn ≡ ∑(𝜀(E i ∩ S j ) | i ∈ n +1). Since ⋃⟮E i ∩ S j | i ∈ 𝜔⟯ = S j and 𝜀 is 𝜎-additive, we get 𝜀S j = lim (x jn | n ∈ 𝜔). By Corollary 2 to Proposition 1 (1.4.7), |𝜀S j | = lim (|x jn | | n ∈ 𝜔). Suppose that 𝜀S j < ∞ for every j ∈ J. Take an arbitrary number 𝛿 > 0 and put x ≡ 𝛿/p. Then, there is m j ∈ 𝜔 such that |𝜀S j | − x < |x jn | for every n ⩾ m j . Take m = gr(m j | j ∈ J). Then, for every n ⩾ m, we get |𝜀S j | − x < |x jn |. Therefore, we have ∑(|𝜀S j | | j ∈ J) − 𝛿 = ∑(|𝜀S j | − x | j ∈ J) < ∑(|x jn | | j ∈ J) ⩽ ∑(∑(|𝜀(E i ∩ S j )| | i ∈ n + 1) | ̄̄ i | i ∈ n + 1) | n ∈ 𝜔). Since 𝛿 is arbitrary, we ̄̄ i | i ∈ n + 1) ⩽ sup (∑(𝜀E j ∈ J) ⩽ ∑(𝜀E ̄̄ i | i ∈ n + 1) | n ∈ 𝜔). This implies that 𝜀E ̄̄ ⩽ conclude that ∑(|𝜀S j | | j ∈ J) ⩽ sup (∑(𝜀E ̄̄ i | i ∈ n + 1) | n ∈ 𝜔). The inverse inequality follows from Proposition 1. sup (∑(𝜀E ̄ ̄̄ i | i ∈ 𝜔). ̄ = ∑(𝜀E Thus, 𝜀E

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̄̄ = ∞. Consider K ≡ {k ∈ J | Now, suppose that 𝜀S k = ∞ for some k ∈ J. Then, 𝜀E 𝜀S k = ∞}. Fix some numbers y > 0 and z > 0. If j ∈ J\K, then there is n(j, y) such that |x jp − 𝜀S j | < y for every p ⩾ n(j, y). Consider the number a ≡ sm (𝜀S j − y | j ∈ J\K). If k ∈ K, then there is n(k, z) such that x kp > z + |a| for every p ⩾ n(k, z). Put n(y, z) ≡ gr (n(j, y) | j ∈ J\K) ⊻ gr (n(k, z) | k ∈ K). Then, for every p ⩾ n we have ∑(x jp | j ∈ J) = ∑(x kp | k ∈ K) + ∑(x jp | j ∈ J\K) ⩾ (z + |a|) card K + a card(J\K) ⩾ z. This implies z ⩽ ∑(x jp | j ∈ J) = | ∑(x jp | j ∈ J)| ⩽ ∑(∑(|𝜀(E i ∩S j )| | i ∈ p+1) | j ∈ J) ≡ c. If |𝜀(E i ∩S j )| < ∞ ̄̄ i | i ∈ for all i ∈ p + 1 and j ∈ J, then z ⩽ c = ∑(∑(|𝜀(E i ∩ S j )| | j ∈ J) | i ∈ p + 1) ⩽ ∑(𝜀E ̄ ̄ i | i ∈ 𝜔) whereas ⋃⟮E i ∩ S j | j ∈ J⟯ ⊂ E i . If |𝜀(E i ∩ S j )| = ∞ for some i ∈ p +1 p +1) ⩽ ∑(𝜀E ̄̄ i = ∞ implies ∑(𝜀E ̄̄ i | i ∈ 𝜔) = ∞, where z ⩽ c = ∞ ⩽ ∞ = ∑(𝜀E ̄̄ i | and j ∈ J, then 𝜀E i ∈ 𝜔). In both cases, we have the same inequality. Since z is arbitrary, we conclude ̄̄ i | i ∈ 𝜔) = ∞ = 𝜀E. ̄̄ Thus, we get the necessary inequality again. that ∑(𝜀E Corollary 1. Let R be a ring on a set T. Then, 1) if 𝜇 is a semimeasure on R, then v(𝜇) is a semimeasure as well; 2) if 𝜇 is a measure on R, then v(𝜇) is a measure as well. Proof. By virtue of Corollary 1 to Lemma 15 (2.1.1), R ⊂ U(T, R). ̄̄ ≡ var(𝜀X ). For an evaluation 𝜀 on an ensemble S and ⌀ ∈ X ⊂ S, put 𝜀X ≡ 𝜀|X and 𝜀X Lemma 5. Let 𝜀 be a natural evaluation on S and ⌀ ∈ X ⊂ S. Then, ̄̄ ⩽ 𝜀;̄̄ 1) 𝜀X 2) if for every set S ∈ S there exists a collection (X i | i ∈ I) ∈ Parf (X, S) [(X i | i ∈ I) ∈ ̄̄ = 𝜀.̄̄ Par𝜎 (X, S)] and 𝜀 is finitely [countably] additive, then 𝜀X Proof. 1. For every E ⊂ T and (Y j | j ∈ J) ∈ Parf (X, E) ⊂ Parf (S, E), we get ̄̄ Hence, 𝜀X ̄̄ (E) ⩽ 𝜀(E). ̄̄ ∑ (|𝜀X (Y j )| | j ∈ J) = ∑ (|𝜀(Y j )| | j ∈ J) ⩽ 𝜀E. 2. For every S ∈ S, we have |𝜀S| = | ∑(𝜀X i | i ∈ I)| ⩽ ∑(|𝜀X i | | i ∈ I) = ∑(|𝜀X X i | | ̄̄ (S), where in the case of card I = 𝜔 we use Lemma 1. Then, Corollary 1 i ∈ I) ⩽ 𝜀X ̄̄ (S k ) | k ∈ K) ⩽ 𝜀X ̄̄ (E) for every to Proposition 1 implies that ∑ (|𝜀S k | | k ∈ K) ⩽ ∑ (𝜀X f ̄̄ ̄̄ (E). E ⊂ T and every finite collection (S k | k ∈ K) ∈ Par (S, E). Hence, 𝜀(E) ⩽ 𝜀X Corollary 1. Let S be a semiring and 𝜀 be an additive natural evaluation on the ring ̄̄ . R(T, S) generated by S (see 2.1.1). Then, 𝜀̄̄ = 𝜀S Proof. It follows from Theorem 4 (2.1.1) and Lemma 5. Corollary 2. Let S be a 𝛿-ring and 𝜀 a countably additive evaluation on the 𝜎-ring ̄̄ . R𝜎 (T, S) generated by S (see 2.1.1). Then, 𝜀̄̄ = 𝜀S Proof. It follows from Corollary 1 to Proposition 4 (2.1.1) and Lemma 5.

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Lemma 6. Let 𝜀 be a natural evaluation on S and ⌀ ∈ X ⊂ S. If for every set S ∈ S we ̄̄ . have |𝜀S| ⩽ sup{|𝜀X| | X ∈ X ∧ X ⊂ S}, then 𝜀̄̄ = 𝜀X ̄̄ ̄̄ (E) ⩽ 𝜀(E) for every E ⊂ T. Proof. It is clear that 𝜀X Let E ⊂ T and (S i ∈ S | i ∈ I) ∈ Parf (S, E). Put p ≡ card I and take x > 0. For each i ∈ I, there is a set X i ⊂ S i such that |𝜀S i | ⩽ |𝜀X i | + x/p. Then, ∑(|𝜀S i | | i ∈ I) ⩽ ̄̄ E + x. Therefore, 𝜀E ̄̄ ⩽ 𝜀X ̄̄ E + x implies 𝜀̄̄ ⩽ 𝜀X ̄̄ . ∑(|𝜀X X i | | i ∈ I) + x ⩽ 𝜀X Lemma 7. Let 𝜀 be a positive finitely superadditive natural evaluation on a disjointly ̄̄ = sup{𝜀S | S ∈ S ∧ S ⊂ E} = sup{𝜀(E ̄̄ ∩ S) | S ∈ S} for additive ensemble S. Then, 𝜀E every E ⊂ T. Proof. By Remark to Proposition 1 (2.1.1), S is finitely disjointly additive. Let (S i ∈ S | i ∈ I) ∈ Parf (S, E) and S ≡ ⋃⟮S i | i ∈ I⟯. Then, S ∈ S and ∑ |𝜀S i | ⩽ 𝜀S ⩽ sup{𝜀S | ̄̄ ⩽ x ⩽ sup{𝜀S ̄̄ | S ∈ S ∧ S ⊂ E} ⩽ S ∈ S ∧ S ⊂ E} ≡ x. Using Lemma 4, we obtain 𝜀E ̄ ̄ ̄ ∩ S) | S ∈ S} ⩽ 𝜀E. ̄ sup{𝜀(E ̄̄ = 𝜀S for every S ∈ S. Corollary 1. If, moreover, 𝜀 is increasing, then 𝜀S ̄̄ = sup{𝜀R | R ∈ S ∧ R ⊂ S} ⩽ 𝜀S. Proof. According to Lemma 4, we get 𝜀S ⩽ 𝜀S ̄̄ Corollary 2. Let 𝜇 be a positive semimeasure on a ring R. Then, v(𝜇) = 𝜇 and 𝜇(E) = ̄̄ ∩ R) | R ∈ R} for every E ⊂ T. sup{𝜇R | R ∈ R ∧ R ⊂ E} = sup{𝜇(E Proof. It follows from Lemma 4 (3.1.1) that 𝜇 is increasing and completely superadditive. Since every ring is additive, Corollary 1 implies that v(𝜇) ≡ 𝜇̄̄ = 𝜇. By Lemma 7, we have two other equalities. Lemma 8. Let 𝜀 be an additive natural evaluation on S and S be disjointly additive. ̄̄ ⩽ 2 sup{|𝜀S| | S ∈ S ∧ S ⊂ E} for every E ⊂ T. Then, sup{|𝜀S| | S ∈ S ∧ S ⊂ E} ⩽ 𝜀E Proof. The first inequality follows from Lemma 4. Let (S i ∈ S | i ∈ I) ∈ Parf (S, E). Then, ⋃⟮S i | i ∈ I⟯ ∈ S. Consider the sets I 󸀠 ≡ {i ∈ I | 𝜀S i > 0}, I 󸀠󸀠 ≡ {i ∈ I | 𝜀S i ⩽ 0}, S󸀠 ≡ ⋃⟮S i | i ∈ I 󸀠 ⟯, and S󸀠󸀠 ≡ ⋃⟮S i | i ∈ I 󸀠󸀠 ⟯. Then, ∑(|𝜀S i | | i ∈ I) = ∑(𝜀S i | i ∈ I 󸀠 ) − ∑(𝜀S i | i ∈ I 󸀠󸀠 ) = 𝜀S󸀠 − 𝜀S󸀠󸀠 = |𝜀S󸀠 | + |𝜀S󸀠󸀠 | ⩽ 2 sup{|𝜀S| | ̄̄ ⩽ x. S ∈ S ∧ S ⊂ E} ≡ x, where 𝜀E Corollary 1. Let 𝜀 be an additive natural evaluation on a disjointly additive ensemble S. ̄̄ Then, 𝜀 is bounded iff 𝜀(T) is finite. ̄̄ Proof. Applying Lemma 8 to E = T, we get sup{|𝜀S| | S ∈ S} ⩽ 𝜀(T) ⩽ 2 sup{|𝜀S| | S ∈ S}.

3.1.4 Some extensions of additive evaluations defined on semirings and rings

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Corollary 2. Let 𝜇 be a semimeasure on a ring R. Then, sup{|𝜇S| | S ∈ R ∧ S ⊂ R} ⩽ v(𝜇)(R) ⩽ 2 sup{|𝜇S| | S ∈ R ∧ S ⊂ R} for every R ∈ R. Lemma 9. Let 𝜀 be a natural evaluation on a disjointly additive ensemble S. Then, ̄̄ var(𝜀|S) = 𝜀.̄̄ ̄̄ by 𝛿. By virtue of Lemma 4 and Corollary 1 to Proposition 1, 𝛿 is Proof. Denote 𝜀|S natural, positive, increasing and completely superadditive; besides, |𝜀S| ⩽ 𝛿S for every S ∈ S. ̄̄ Let E ⊂ T and (S i | i ∈ I) ∈ Parf (S, E). Then, ∑(|𝜀S i | | i ∈ I) ⩽ ∑(𝛿S i | i ∈ I) ⩽ 𝛿E. ̄̄ Conversely, if S ∈ S and S ⊂ E, then 𝛿S = 𝜀S ̄̄ ⩽ 𝛿E. ̄̄ ⩽ 𝜀E. ̄̄ Therefore, by Hence, 𝜀E ̄ ̄ ̄ ̄ ̄ ̄ ̄ Thus, 𝜀E ̄ = 𝛿E, where 𝜀̄̄ = 𝛿.̄̄ Lemma 7, 𝛿E = sup{𝛿S | S ∈ S ∧ S ⊂ E} ⩽ 𝜀E. Lemma 10. Let S be an ensemble on a set T containing the empty subset ⌀. Then, 1) if 𝛿, 𝜀 ∈ Eval(T, S), then var(𝛿 + 𝜀) ⩽ 𝛿̄̄ + 𝜀;̄̄ 2) if 𝜀 ∈ Evalf (T, S) and x ∈ R, then var(x𝜀) = |x|𝜀;̄̄ 3) if 𝜀 ∈ Eval(T, S), then 𝜀 = 0 iff 𝜀̄̄ = 0; 4) if 0 ⩽ 𝛿 ⩽ 𝜀 in Eval(T, S), then 𝛿̄̄ ⩽ 𝜀.̄̄ Proof. 1. Let E ⊂ T and (S i | i ∈ I) ∈ Parf (S, E). Then, ∑(|(𝛿 + 𝜀)(S i )| | i ∈ I) ⩽ ∑(|𝛿S i | | ̄̄ + 𝜀E. ̄̄ ̄̄ This implies var(𝛿 + 𝜀)E ⩽ (𝛿̄̄ + 𝜀)E. i ∈ I) + ∑(|𝜀S i | | i ∈ I) ⩽ 𝛿E 2. Similarly, ∑(|x𝜀(S i )| | i ∈ I) = |x| ∑(|𝜀S i | | i ∈ I) implies var(x𝜀) = |x|𝜀.̄̄ By virtue of Lemma 4 (1.4.5). 3. This assertion follows from Lemma 4. 4. Let E ⊂ T, (S i | i ∈ I) ∈ Parf (S, E). Then, ∑(|𝛿(S i )| | i ∈ I) ⩽ ∑(|𝜀(S i )| | i ∈ I) ⩽ ̄̄ ⩽ 𝜀E. ̄̄ ̄̄ implies 𝛿E 𝜀E Corollary 1. Let S be a disjointly additive ensemble on T containing ⌀. Then, the map̄̄ ping 𝜀 󳨃→ 𝜀(T) is a norm ‖ ⋅ ‖ on the linear space Evalb (T, S). Proof. The assertion follows from Lemma 10, Lemma 9, and Lemma 4.

3.1.4 Some extensions of additive evaluations defined on semirings and rings Extension of a finitely additive evaluation defined on a semiring The following result shows that it is natural to introduce the notion of a semimeasure and a measure only for rings but not for semirings. Theorem 1. Let 𝜀 be a finitely additive natural evaluation defined on a semiring S. Then, 𝜀 can be uniquely extended to semimeasure 𝛿 defined on the ring R(T, S) generated by S. If, in addition, 𝜀 is 𝜎-additive, then 𝛿 is a measure.

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Proof. We shall use the description of R(T, S) from Theorem 4 (2.1.1). Let R ∈ R(T, S) and R = ⋃⟮X i | i ∈ I⟯ = ⋃⟮Y j | j ∈ J⟯ for some finite pairwise disjoint collections (X i ∈ S | i ∈ I) and (Y j ∈ S | j ∈ J). Then, 𝜀X i = ∑(𝜀(X i ∩ Y j ) | j ∈ J) implies x ≡ ∑(𝜀X i | i ∈ I) = ∑(∑(𝜀(X i ∩ Y j ) | j ∈ J) | i ∈ I). In the same way we deduce that y ≡ ∑(𝜀Y j | j ∈ J) = ∑(∑(𝜀(X i ∩ Y j ) | i ∈ I) | j ∈ J). By Corollary 1 to Proposition 1 (1.4.3), x = y. This means that we can define correctly a mapping 𝛿 : R(T, S) → R setting 𝛿R ≡ x and 𝛿 is a natural evaluation. For S ∈ S, we have 𝛿S = 𝜀S. If 𝜀 is finite, then 𝛿 is finite. If 𝜀 is positive, then 𝛿 is also positive. Let P, Q ∈ R(T, S), P = ⋃⟮X i | i ∈ I⟯, Q = ⋃⟮Y j | j ∈ J⟯ and P ∩ Q = ⌀. Consider the set K ≡ I ∪d J ≡ ⋃⟮A m × {m} | m ∈ 2⟯ from 1.2.2, where A0 ≡ I and A1 ≡ J. Define a collection (Z k ∈ S | k ∈ K) setting Z k ≡ X i for k = (i, 0) and Z k ≡ Y j for k = (j, 1). Then, P ∪ Q = ⋃⟮Z k | k ∈ K⟯ implies 𝛿(P ∪ Q) = ∑(𝜀Z k | k ∈ K) and, according to Lemma 4 (1.4.3) and Theorem 1 (1.4.3), we get 𝛿(P ∪ Q) = ∑(∑(𝜀Z k | k ∈ A m × {m}) | m ∈ 2) = ∑(𝜀Z k | k ∈ A0 × {0}) + ∑(𝜀Z k | k ∈ A1 × {1}) = 𝛿P + 𝛿Q. Thus, 𝛿 is binary additive. By Lemma 1 (3.1.1), 𝛿 is finitely additive. Let 𝛾 : R(T, S) → R be a finitely additive evaluation extending 𝜀. Then, for R = ⋃⟮X i | i ∈ I⟯, we have 𝛾R = ∑ 𝛾X i = ∑ 𝜀X i = 𝛿R. Thus, 𝛿 is unique. Now, let 𝜀 be 𝜎-additive and (R k ∈ R(T, S) | k ∈ 𝜔) be a sequence of pairwise disjoint sets such that R ≡ ⋃⟮R k | k ∈ 𝜔⟯ ∈ R(T, S). Every R k has the form R k = ⋃⟮X ki | i ∈ I k ⟯ for some finite pairwise disjoint collection (X ki ∈ S | i ∈ I k ). Therefore, 𝛿R k = ∑(𝜀X ki | i ∈ I k ). We have R = ⋃⟮⋃⟮X ki | i ∈ I k ⟯ | k ∈ 𝜔⟯. Consider at first the case of R ∈ S. To use the 𝜎-additivity of 𝜀 we need to have a sequence, but we have only a collection ((X ki | i ∈ I k ) | k ∈ 𝜔) of collections. Since every I k is finite we can assume that I k = [1, m k ] ⊂ 𝜔. Then, we can define for the set P ≡ ⋃⟮{k} × I k | k ∈ 𝜔⟯ the surjective mapping u : P N setting u(k, i) ≡ ∑(m j | j ∈ k) + i. Consider the lexicographic linear order on P setting p ≡ (k, i) ⩽ q ≡ (l, j) if either k < l or k = l and i ⩽ j. Since 1 ⩽ i ⩽ m k the mapping u is strictly increasing. Therefore, u is bijective. Take v ≡ u−1 : N P. Now, we can define a sequence (Y𝜈 ∈ S | 𝜈 ∈ N) setting Y𝜈 ≡ X v(𝜈) . It is clear that R = ⋃⟮Y𝜈 | 𝜈 ∈ N⟯. As 𝜀 is 𝜎-additive, we get 𝜀R = lim(∑(𝜀Y𝜈 | 𝜈 ∈ [1, n] ⊂ 𝜔) | n ∈ N). Consider at first the case where 𝜀 is positive. Suppose that 𝜀X ki = ∞ for some k and i. By the construction 𝛿 is positive. Therefore, by Lemma 4 (3.1.1), 𝛿R ⩾ 𝛿X ki = ∞. Besides, 𝛿R k ⩾ 𝛿X ki = ∞ implies ∑(𝛿R k | k ∈ 𝜔) = ∞. Thus, 𝛿R = ∑(𝛿R k | k ∈ 𝜔). Now, suppose that 𝜀X ki < ∞ for all k and i. For 𝜈 = n, consider (k n , i n ) ≡ v(n). Since 𝛿 is increasing, we get x n ≡ ∑(𝜀Y𝜈 | 𝜈 ∈ [1, n]) = ∑(𝜀X ki | (k, i) ⩽ (k n , i n )) = ∑(𝜀X ki | k < k n ∨ (k = k n ∧ i ⩽ i n )) = ∑(𝜀X ki | i ∈ I k ∧ k < k n ) + ∑(𝜀X k n i | 1 ⩽ i ⩽ i n ) = ∑(∑(𝜀X ki | i ∈ I k ) | k < k n ) + ∑(𝜀X k n i | 1 ⩽ i ⩽ i n ) = ∑(𝛿R k | k < k n ) + ∑(𝛿X k n i | 1 ⩽ i ⩽ i n ) ⩽ ∑(𝛿R k | k ⩽ k n ) ⩽ ∑(𝛿R k | k ∈ 𝜔). Therefore, 𝛿R = sup (x n | n ∈ N) ⩽ ∑(𝛿R k | k ∈ 𝜔). On the other hand, ∑(𝛿R k | k ∈ 𝜔) = ∑(𝛿R k | k ∈ 𝜔) ⩽ 𝛿R. Now, let 𝜀 be finite. Consider the mapping w : 𝜔 → N written as w ≡ (n l ∈ N | l ∈ 𝜔) such that n l ≡ u(l, m l ) = ∑(m j | j ⩽ l). It is clear that rng w is cofinal in N. Besides, w is increasing. According to 1.2.7 the sequence (x n l | l ∈ 𝜔) is a subsequence

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of the sequence (x n | n ∈ N). By Lemma 5 (1.4.4), 𝜀R = lim(x n | n ∈ N) implies 𝛿R = 𝜀R = lim(x n l | l ∈ 𝜔). It follows from the above that x n l = ∑(𝜀Y𝜈 | 𝜈 ∈ [1, n l ]) = ∑(𝜀X ki | (k, i) ⩽ (l, m l )) = ∑(𝜀X ki | i ∈ I k ∧ k ⩽ l) + ∑((∑ 𝜀X ki | i ∈ I k ) | k ⩽ l) = ∑(𝛿R k | k ⩽ l). Thus, 𝛿R = lim(∑(𝛿R k | k ⩽ l) | l ∈ 𝜔) = ∑(𝛿R k | k ∈ 𝜔). Finally, if R ∉ S, then R = ⋃⟮Y l | l ∈ L⟯ for some finite pairwise disjoint collection (Y l ∈ S | l ∈ L). Therefore, 𝛿R = ∑(𝛿Y l | l ∈ L). For each l, we have Y l = ⋃⟮R k ∩ Y l | k ∈ 𝜔⟯ ∈ S, where (R k ∩Y l ∈ R(T, S) | k ∈ 𝜔) is a pairwise disjoint collection. The property proven above guarantees the equality 𝛿Y l = ∑(𝛿(R k ∩Y l ) | k ∈ 𝜔). Similarly, for each k, we have R k = ⋃⟮R k ∩Y l | l ∈ L⟯. Since 𝛿 is finitely additive, we have 𝛿R k = ∑(𝛿(R k ∩Y l ) | l ∈ L). Using Proposition 2 (1.4.7) and Corollary 1 to Proposition 1 (1.4.3), we get 𝛿R = ∑(lim(∑(𝛿(R k ∩ Y l ) | k ∈ n + 1) | n ∈ 𝜔) | l ∈ L) = lim(∑(∑(𝛿(R k ∩ Y l ) | k ∈ n + 1) | l ∈ L) | n ∈ 𝜔) = lim(∑(∑(𝛿(R k ∩ Y l ) | l ∈ L) | k ∈ n + 1) | n ∈ 𝜔) = lim(∑(𝛿R k | k ∈ n + 1) | n ∈ 𝜔) = ∑(𝛿R k | k ∈ 𝜔). The initial Borel – Lebesgue measure on ℝn Consider the most known example of the constructed extension. Example 1 (the initial Borel – Lebesgue measure on Rn ). Consider the ensemble Ppar of all the parallelepipeds P|s, t| ≡ ∏⟮|a i , b i | | i ∈ n⟯ and the semiring Spar of all the half-open parallelepipeds P[s, t[= ∏⟮[a i , b i [| i ∈ n⟯ from Example 2 (2.1.1). Define an evaluation v : Ppar → R+ setting v(P[s, t[) ≡ ∏(b i − a i | i ∈ n), i. e. taking the volume of parallelepipeds. By Theorem 2 (2.1.1), the ring R(Rn , Spar ) consists of all subsets R of Rn , which can be represented in the form R = ⋃⟮P i | i ∈ I⟯ for some finite collection (P i ∈ Spar | i ∈ I) of pairwise disjoint parallelepipeds. Define the initial Borel – Lebesgue evaluation 𝜆 : R(Rn , Spar ) → R+ setting 𝜆(⋃⟮P i | i ∈ I⟯) ≡ ∑(v(P i ) | i ∈ I). Theorem 2. The volume v is finitely additive on Spar . Proof. Let P ≡ P[r, s[ be a parallelepiped and (P𝛼 | 𝛼 ∈ A) be a finite collection of pairwise disjoint parallelepipeds P𝛼 ≡ P[r𝛼 , s𝛼 [ such that P = ⋃⟮P𝛼 | 𝛼 ∈ A⟯. Then, P[r, s[= ∏⟮[a i , b i [| i ∈ n⟯ and P[r𝛼 , s𝛼 [= ∏⟮[a i𝛼 , b i𝛼 [| i ∈ n⟯. Consider the set C i ≡ {a i𝛼 | 𝛼 ∈ A} ∪ {b i𝛼 | 𝛼 ∈ A}. The set C i assigns a partition of [a i , b i [. Enumerate points of C i such that C i = {c ik | k ∈ K i ∈ 𝜔} and c ik < c i(k+1) . Consider the set U ≡ ∏⟮K i | i ∈ n⟯. We obtain the set T ≡ ∏⟮C i | i ∈ n⟯ of points t u ≡ (c iu(i) ∈ C i | i ∈ n) ∈ P for all u ∈ U. It is clear that r, s ∈ T and r𝛼 , s𝛼 ∈ T for every 𝛼 ∈ A. Since b i − a i = ∑(c i,k+1 − c ik | k ∈ K i ), using Theorem 4 (1.4.3), we get v(P) = ∏(b i − a i | i ∈ n) = ∏(∑(c i,k+1 − c ik | k ∈ K i ) | i ∈ n) = ∑(∏(c i,u(i)+1 − c iu(i) | i ∈ n) | u ∈ U). Consider the sets T𝛼 ≡ T ∩ P𝛼 . Then, (T𝛼 | 𝛼 ∈ A) is a partition of T. Define the bijective mapping 𝛽 : U → T setting 𝛽(u) ≡ t u . This mapping assigns a partition (U𝛼 ≡ 𝛽−1 [T𝛼 ] | 𝛼 ∈ A) of the parallelepiped U = 𝜔n ∩ P[0, p[ in the set 𝜔n , which defines by

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the corner points 0 ≡ (00 , . . . , 0n−1 ) and p ≡ (K0 , . . . , K n−1 ) in 𝜔n . Consider the points v𝛼 ≡ 𝛽−1 (r𝛼 ) and w𝛼 ≡ 𝛽−1 (s𝛼 ) from U. Take any point t u ≡ (c iu(i) | i ∈ n) for u ∈ U. If u ∈ U𝛼 , then t u ∈ T𝛼 implies c iu(i) ∈ [a i𝛼 , b i𝛼 [ for every i. Therefore, c iv𝛼 (i) = a i𝛼 ⩽ c iu(i) < b i𝛼 = c iw𝛼 (i) . This gives v𝛼 (i) ⩽ u(i) < w𝛼 (i) for every i, where u ∈ 𝜔n ∩ P[v𝛼 , w𝛼 [ ≡ N𝛼 . Therefore, U𝛼 ⊂ N𝛼 . Conversely, if u ∈ N𝛼 , then u(i) ∈ [v𝛼 (i), w𝛼 (i)[ implies a i𝛼 = c iv𝛼 (i) ⩽ c iu(i) < c iw𝛼 (i) = b i𝛼 for every i, where t u ∈ T𝛼 . Hence, u ∈ U𝛼 . Thus, U𝛼 is equal to the parallelepiped N𝛼 in 𝜔n . Since b i𝛼 − a i𝛼 = ∑(c ik+1 − c ik | k ∈ [v𝛼 (i), w𝛼 (i)[), we get v(P𝛼 ) = ∏(b i𝛼 − a i𝛼 | i ∈ n) = ∏(∑(c ik+1 − c ik | k ∈ [v𝛼 (i), w𝛼 (i)[) | i ∈ n) = ∑(∏(c iu(i)+1 − c iu(i) | i ∈ n) | u ∈ U𝛼 ). Finally, using Theorem 1 (1.4.3), we get ∑(v(P𝛼 ) | 𝛼 ∈ A) = ∑(∏(c iu(i)+1 − c iu(i) | i ∈ n) | u ∈ U) = v(P). Corollary 1. The initial Borel – Lebesgue evaluation 𝜆 is a semimeasure. Proof. Since v is finitely additive, the assertion follows from Theorem 1. Let P ≡ P[r, s[≡ ∏⟮[a i , b i [| i ∈ n⟯ be a half-open parallelepiped. For every number, l ∈ N such that 1/l < sm (b i − a i | i ∈ n), consider the closed parallelepiped F l ≡ ∏⟮[a i , b i − 1/l] | i ∈ n⟯ ⊂ P, the open parallelepiped G l ≡ ∏⟮]a i − 1/l, b i [| i ∈ n⟯ ⊃ P and the half-open parallelepipeds P󸀠l ≡ ∏⟮[a i , b i − 1/l[| i ∈ n⟯ and P󸀠󸀠l ≡ ∏⟮[a i − 1/l, b i [| i ∈ n⟯. Then, we have P󸀠l ⊂ F l ⊂ P ⊂ G l ⊂ P󸀠󸀠l , F l ⊂ F l+1 , and G l ⊃ G l+1 . Lemma 1. For every 𝜀 > 0, there is l such that v(P󸀠l ) = v(F l ) = ∏(b i − 1/l − a i | i ∈ n) > v(P) − 𝜀 and v(P󸀠󸀠l ) = v(G l ) = ∏(b i − a i + 1/l | i ∈ n) < v(P) + 𝜀. Proof. We have v(F l ) = v(P) − f (l), where f (l) ≡ A0 /l + A 1 /l2 + . . . + A n−1 /l n for some real numbers A1 , . . . , A n such that f (l) > 0 for every l ∈ N. Let A ≡ gr (|A i | | i ∈ n) and l > 1. Then, f (l) ⩽ A ∑ (1/l i | i ∈ n). Applying Lemma 3 (1.4.8) to the infinite geometric progression with the base 1/l, we obtain that f (l) < A(∑ (1/l i | i ∈ 𝜔) − 1) = A(1/(1 − 1/l) − 1) = A/(l − 1). By the Archimedes principle (Lemma 13 (1.4.3)) for every 𝜀 > 0, there is l ∈ N such that (l − 1)𝜀 > A. Hence, f (l) < 𝜀. Similarly, v(G l ) = v(P) + g(l), where g(l) ≡ B0 /l + B1 /l2 + . . . + B n−1 /l n for some real numbers B1 , . . . , B n such that g(l) > 0 for every l ∈ N. We can prove, as above, that there is l such that g(l) < 𝜀. Theorem 3. The initial Borel – Lebesgue semimeasure 𝜆 is a measure. Proof. Let (R𝛼 ∈ R(Rn , Spar ) | 𝛼 ∈ 𝜔) be a pairwise disjoint countable sequence such that R ≡ ⋃⟮R𝛼 | 𝛼 ∈ 𝜔⟯ ∈ R(Rn , Spar ). By Theorem 4 (2.1.1), R = ⋃⟮P k | k ∈ m ∈ N⟯ and

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R𝛼 = ⋃⟮P𝛼k | k ∈ m𝛼 ∈ N⟯ for some finite sequences (P k ∈ Spar | k ∈ m) and (P𝛼k ∈ Spar | k ∈ m𝛼 ) of pairwise disjoint parallelepipeds P k ≡ P[r k , s k [= ∏⟮[a ik , b ik [| i ∈ n⟯ and P𝛼k ≡ P[r𝛼k , s𝛼k [= ∏⟮[a i𝛼k , b i𝛼k [| i ∈ n⟯. Take some 𝜀 > 0. According to Lemma 1, for every k ∈ m, there is the closed parallelepiped F k ⊂ P k such that v(F k ) > v(P k )−𝜀/2 and for every k ∈ m𝛼 , there is the open parallelepiped G𝛼k ⊃ P𝛼k such that v(G𝛼k ) < v(P𝛼k ) + 𝜀/2𝛼+k . Then, using the inequalities ∑(1/2k+2 | k ∈ m) < ∑(1/2k+2 | k ∈ 𝜔) = 1/2 provided by Lemma 3 (1.4.8), we get x ≡ ∑(v(F k ) | k ∈ m) > 𝜆R − 𝜀/2 and y𝛼 ≡ ∑(v(G𝛼k ) | k ∈ m𝛼 ) < 𝜆R𝛼 + 𝜀/2𝛼+2 . Consider the bounded closed set F ≡ ⋃⟮F k | k ∈ m⟯ and the open sets G𝛼 ≡ ⋃⟮G𝛼k | k ∈ m𝛼 ⟯. Since F ⊂ ⋃⟮G𝛼 | 𝛼 ∈ 𝜔⟯, Statement 1 guarantees that there is a finite subcollection (G𝛼 | 𝛼 ∈ B ⊂ 𝜔) such that F ⊂ ⋃⟮G𝛼 | 𝛼 ∈ B⟯. Take some half-open parallelepipeds P󸀠k ⊂ F k and P󸀠𝛼k ⊃ G𝛼k such that v(P󸀠k ) = v(F k ) and v(P󸀠𝛼k ) = v(G𝛼k ). Since ⋃⟮P󸀠k | k ∈ m⟯ ⊂ F ⊂ ⋃⟮G𝛼 | 𝛼 ∈ B⟯ ⊂ ⋃⟮⋃⟮P󸀠𝛼k | k ∈ m𝛼 ⟯ | 𝛼 ∈ B⟯, using Lemmas 4 and 3 (3.1.1), we obtain x = 𝜆(⋃⟮P󸀠k | k ∈ m⟯) ⩽ ∑(∑(𝜆(P󸀠𝛼k ) | k ∈ m𝛼 ) | 𝛼 ∈ B). Finally, we get 𝜆R < x + 𝜀 ⩽ ∑(y𝛼 | 𝛼 ∈ B) + 𝜀/2 < ∑(𝜆R𝛼 + 𝜀/2𝛼+2 | 𝛼 ∈ B) + 𝜀/2 < ∑(𝜆R𝛼 | 𝛼 ∈ 𝜔) + 𝜀 ∑(1/2𝛼+2 | 𝛼 ∈ 𝜔) + 𝜀/2 = ∑(𝜆R𝛼 | 𝛼 ∈ 𝜔) + 𝜀. Since 𝜀 is arbitrary, we get 𝜆R ⩽ ∑(𝜆R𝛼 | 𝛼 ∈ 𝜔) = ∑(𝜆R𝛼 | 𝛼 ∈ 𝜔). This means that 𝜆 is countably subadditive. Consequently, by Lemma 8 (3.1.1), 𝜆 is a measure. The extension of 𝜆 to some 𝜎-algebra in Rn will be considered in 3.1.5. The completion of a semimeasure Let S be an ensemble on T and 𝜀 be an evaluation on S. Consider the ensemble N0 (T, S, 𝜀) ≡ {N0 ∈ P(T) | ∀S ∈ S (𝜀(N0 ∩ S) = 0)}. It is clear that N0 (T, S, 𝜀) = {N0 ∈ P(T) | ∀S ∈ S (N0 ∩ S ∈ S0 (𝜀))} ⊂ U(T, S), where U(T, S) is the saturation of the ensemble S (see 2.1.1). Consider also the wider ensemble N(T, S, 𝜀) ≡ {N ∈ P(T) | ∃N0 ∈ N0 (T, S, 𝜀) (N ⊂ N0 )}. The evaluation 𝜀 will be called complete if N(T, S, 𝜀) ⊂ S. Lemma 2. Let 𝜇 be a semimeasure on a ring R. Then, N0 (T, R, 𝜇) is a ring and an ideal in U(T, R). Moreover, if 𝜇 is a measure, then N0 (T, R, 𝜇) is a 𝜎-ring. Proof. We shall denote N0 (T, R, 𝜇) simply by N0 . Let M ∈ N0 , V ∈ U(T, R), and R ∈ R. Then, V ∩ R ∈ R, and therefore, (M ∩ V) ∩ R = M ∩ (V ∩ R) ∈ R0 (𝜀). Therefore, M ∩ V ∈ N0 . By Corollary 1 to Proposition 6 (2.1.1) the saturation U(T, R) is an algebra containing R. Then, T\V ∈ U(T, R), where M\V = M ∩ (T\V) ∈ N0 . In particular, it follows that for M, N ∈ N0 , we have M\N ∈ N0 and M ∩ N ∈ N0 . If M ∩ N = ⌀ and R ∈ R, then R ∩ M, R ∩ N ∈ R and 𝜇(R ∩ M) = 𝜇(R ∩ N) = 0 imply 𝜇(R ∩ (M ∪ N)) = 𝜇(R ∩ M) + 𝜇(R ∩ N) = 0. Therefore, M ∪ N ∈ N0 . For arbitrary M, N ∈ N0 , we have M ∪ N = (M\N) ∪ (M ∩ N) ∪ (N\M) ∈ N0 , i. e. N0 is additive. Thus, N0 is a ring and an ideal in U(T, R).

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Now, let 𝜇 be a measure and (N i ∈ N0 | i ∈ 𝜔) be a pairwise disjoint sequence. Take any R ∈ R. Then, N i ∩ R ∈ R and 𝜇(N i ∩ R) = 0. Since N ∩ R = ⋃⟮N i ∩ R | i ∈ 𝜔⟯, we get 𝜇(N ∩ R) = ∑(𝜇(N i ∩ R) | i ∈ 𝜔) = 0. Therefore, ⋃⟮N i | i ∈ 𝜔⟯ ∈ N0 , i. e. N0 is countably disjointly additive. Corollary 1 to Lemma 8 (2.1.1) guarantees that the ring N0 is countably additive. Corollary 1. If 𝜇 is a semimeasure on a ring R, then R\N0 ∈ R for every R ∈ R and N0 ∈ N0 (T, R, 𝜇). Proof. By Corollary 1 to Proposition 6 (2.1.1), the saturation U(T, R) is an algebra containing R. Therefore, R\N0 ∈ U(T, R). This implies that R\N0 = (R\N0 )∩ R ∈ R. Corollary 2. If 𝜇 is semimeasure on a ring R, then N(T, R, 𝜇) is an ideal and a ring. Proof. Since N0 (T, S, 𝜇) is a ring, Lemma 1 (2.1.4) implies that N(T, R, 𝜇) = I(N0 (T, R, 𝜇)) is an ideal. Then, by Corollary 1 to Lemma 2 (2.1.4) N is a ring. Corollary 3. If 𝜇 is a measure on a ring R, then N(T, R, 𝜇) is an ideal and a 𝜎-ring on T. Consider the ensemble K(T, S, 𝜀) ≡ {S ∪ N | S ∈ S ∧ N ∈ N(T, S, 𝜀)}. If 𝜀 is complete and S is additive then K(T, S, 𝜀) = S. Theorem 4. Let 𝜇 be a semimeasure on a ring R. Then, K(T, R, 𝜇) is a ring containing R and N(T, R, 𝜇). The semimeasure 𝜇 can be extended uniquely to a complete semimeasure 𝜇̌ defined on the ring K(T, R, 𝜇) vanishing on N(T, R, 𝜇) and such that ̌ ∪ N) = 𝜇R for every R ∈ R and N ∈ N. 𝜇(R If 𝜇 is a measure, then 𝜇̌ is a measure. Moreover, N0 (T, K(T, R, 𝜇), 𝜇)̌ = N0 (T, R, 𝜇), N(T, K(T, R, 𝜇), 𝜇)̌ = N(T, R, 𝜇) and K(T, K(T, R, 𝜇), 𝜇)̌ = K(T, R, 𝜇). Proof. We shall denote N0 (T, R, 𝜇), N(T, R, 𝜇), and K(T, R, 𝜇) simply by N0 , N and K. Let P, Q ∈ K. Then, P = R ∪ M and Q = S ∪ N for some R, S ∈ R and M, N ∈ N. Then, by Corollary 2 to Lemma 2 we obtain P ∪ Q = (R ∪ S) ∪ (M ∪ N) ∈ K. Since N is an ideal, we get M\Q ∈ N. Take a set N0 ∈ N0 such that N ⊂ N0 . Then, R\Q = (R\S)\N = ((R\S)\N 0 ) ∪ ((R\S) ∩ (N0 \N)). Since R\S ∈ R, by Corollary 1 to Lemma 2, we get (R\S)\N0 ∈ R. Since (R\S) ∩ (N0 \N) ⊂ N0 , we have (R\S) ∩ (N0 \N) ∈ N. Hence, R\Q ∈ K. Thus, P\Q = (R\Q) ∪ (M\Q) ∈ K, and therefore, K is a ring. We shall prove now that P = Q implies 𝜀R = 𝜀S. In fact, M ⊂ M 0 ∈ N0 and N ⊂ N0 ∈ N0 . Then, P = Q implies R\S ⊂ P\S = Q\S ⊂ N ⊂ N0 . Since R\S ∈ R,

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we conclude 𝜀(R\S) = 𝜀((R\S) ∩ N0 ) = 0, where 𝜀R = 𝜀(R ∩ S). Similarly, 𝜀S = 𝜀(R ∩ S) = 𝜀R. ̌ ≡ 𝜀R for Consequently, we can define correctly a mapping 𝜇̌ : K → R setting 𝜇P ̌ P = R ∪ M. It is clear that 𝜇̌ extends 𝜀 and 𝜇[N] = {0}. Let P = R ∪ M, Q = S ∪ N and P ∩ Q = ⌀. Then, R ∩ S = ⌀ and M ∪ N ∈ N imply ̌ ∪ Q) = 𝜀(R ∪ S) = 𝜀R + 𝜀S = 𝜇P ̌ + 𝜇Q. ̌ Thus, 𝜇̌ is binary additive. By Lemma 1 (3.1.1) 𝜇(P the evaluation 𝜇̌ is finitely additive. Let 𝜈 be a semimeasure on K extending 𝜇 and vanishing on N. If P = R ∪ M ∈ K, then P = R∪(M\R) and M\R ∈ N by virtue to Corollary 2 of Lemma 2. So 𝜈P = 𝜈R = 𝜇P. Thus, 𝜇 is unique. ̌ Then, for any R ∈ R ⊂ K we have 𝜀(R ∩ X0 ) = 𝜇(R ̌ ∩ X0 ) = 0, Let X0 ∈ N0 (T, K, 𝜇). where X0 ∈ N0 . Conversely, let N0 ∈ N0 . Then, for any K ∈ K, we have K∩N0 ∈ N and ̌ = 0. This means N0 ∈ N0 (T, K, 𝜇). ̌ Finally, N0 (T, K, 𝜇)̌ = N0 . so 𝜇K Consequently, N(T, K, 𝜇)̌ = N ⊂ K, i. e. 𝜇̌ is complete and also K(T, K, 𝜇)̌ = K. Suppose now that 𝜀 is a measure. Let (P i ∈ K | i ∈ 𝜔) be a pairwise disjoint sequence such that P ≡ ⋃⟮P i | i ∈ 𝜔⟯ ∈ K. Then, P i = R i ∪ M i and P = R ∪ M for P i , P ∈ S and M i , M ∈ N. Besides, P = ⋃⟮R i | i ∈ 𝜔⟯ ∪ ⋃⟮M i | i ∈ 𝜔⟯. By Corollary 3 to Lemma 2 ⋃⟮M i | i ∈ 𝜔⟯ ∈ N and N ≡ M ∪ ⋃⟮M i | i ∈ 𝜔⟯ ∈ N. Therefore, N ⊂ N0 ∈ N0 . Now, we get R\N0 = P\N0 = ⋃⟮P i \N0 | i ∈ 𝜔⟯ = ⋃⟮R i \N0 | i ∈ 𝜔⟯. By Corollary 1 to ̌ = 𝜇R ̌ = 𝜇(R\N ̌ Lemma 2, R\N0 ∈ R and R i \N0 ∈ R. Consequently, 𝜇P 0 ) = 𝜀(R\N 0 ) = ̌ i \N0 ) = ∑ 𝜇R ̌ i = ∑ 𝜇P ̌ i . Thus, 𝜇̌ is a measure. ∑ 𝜀(R i \N0 ) = ∑ 𝜇(R The semimeasurable [measurable] space ⟮T, K(T, R, 𝜇), 𝜇⟯̌ is called the completion of the semimeasurable [measurable] space ⟮T, R, 𝜇⟯ and 𝜇̌ is called the completion of 𝜇. Corollary 1. Let 𝜇 be a measure on a 𝛿-ring R. Then, K(T, R, 𝜇) is a 𝛿-ring. Proof. Let (K i ∈ K | i ∈ I) be a countable collection such that L ≡ ⋃⟮K i | i ∈ I⟯ ⊂ K ∈ K. Then, K i = R i ∪ M i and K = R ∪ M for some R i , R ∈ R and M i , M ∈ N. Consider the sets N i󸀠 ≡ R i ∩ M, N i󸀠󸀠 ≡ R ∩ M i , N i ≡ N i󸀠 ∪ N i󸀠󸀠 from N and S i ≡ R i ∩ R from R. Then, K i = K i ∩ K = (K i ∩ R) ∪ N i󸀠 = S i ∪ N i󸀠󸀠 ∪ N i󸀠 = S i ∪ N i . From S ≡ ⋃⟮S i | i ∈ I⟯ ⊂ R ∩ L, we infer by Lemma 9 (2.1.1) that S ∈ R. Furthermore, by Corollary 3 to Lemma 2, N ≡ L\S = ⋃⟮K i \S | i ∈ I⟯ ⊂ ⋃⟮K i \S i | i ∈ I⟯ ⊂ ⋃⟮N i | i ∈ I⟯ ∈ N implies N ∈ N. As a result, L = S ∪ N ∈ K. By Lemma 9 (2.1.1) this means that K is a 𝛿-ring. Corollary 2. Let 𝜇 be an internally finite measure on a 𝛿-ring R. Then, 𝜇̌ is an internally finite measure on a 𝛿-ring K(T, R, 𝜇). ̌ = ∞, then 𝜇R = 𝜇K ̌ = ∞ implies that there Proof. Let K ∈ K and K = R ∪ N. If 𝜇K ̌ = 𝜇S < ∞. If 𝜇K ̌ = −∞, then the arguis S ∈ R ⊂ K such that S ⊂ R ⊂ K and 0 < 𝜇S ments are the same. Thus, 𝜇̌ is internally finite.

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The local completion of a positive semimeasure Now, we shall consider the other variant of completeness. Instead of the ensemble N0 (T, S, 𝜀) we shall take the ensemble S0 (𝜀) and instead of the ensemble N(T, S, 𝜀) we shall take the ensemble Nl (T, S, 𝜀) ≡ {N ⊂ T | ∃S0 ∈ S0 (𝜀) (N ⊂ S0 )}. The evaluation 𝜀 will be called locally complete if Nl (T, S, 𝜀) ⊂ S. If 𝜇 is a positive semimeasure on a ring R, then by virtue of Lemma 9 (3.1.1), R0 (𝜇) is a subring of the ring R and an ideal in the lattice R. Therefore, by Lemma 1 (2.1.4) and Corollary 1 to Lemma 2 (2.1.4), Nl (T, R, 𝜇) is an ideal and a ring on T. If 𝜇 is a measure on the 𝜎-ring R, then R0 (𝜇) is a 𝜎-ring and Nl (T, R, 𝜇) is an ideal and a 𝜎-ring on T. Lemma 3. Let 𝜇 be a positive semimeasure on a ring R. Then, R0 (𝜇) ⊂ N0 (T, R, 𝜇) and Nl (T, R, 𝜇) ⊂ N(T, R, 𝜇). Proof. By Corollary 1 to Lemma 15 (2.1.1) R ⊂ U(T, R). Let N0 ∈ R0 (𝜇). Then, N0 ∈ U(T, R) and 𝜇(N0 ∩ R) ⩽ 𝜇N0 = 0 for every R ∈ R. Therefore, 𝜇(N0 ∩ R) = 0 implies N0 ∈ N0 (T, R, 𝜇). Then, the second inclusion is obvious. Corollary 1. Let 𝜇 be a positive semimeasure on a ring R. If 𝜇 is complete, then 𝜇 is locally complete. Consider now the ensemble Kl (T, S, 𝜀) ≡ {S ∪ N | S ∈ S ∧ N ∈ Nl (T, S, 𝜀)}. If 𝜀 is locally complete and S is additive then Kl (T, S, 𝜀) = S. Proposition 1. Let 𝜇 be a positive semimeasure on a ring R. Then, Kl (T, R, 𝜇) is a subring of the ring K(T, R, 𝜇) containing R and Nl (T, R, 𝜇), the semimeasure 𝜇 can be extended uniquely to a locally complete positive semimeasure 𝜇̄ on the ring ̄ ∪ N) = 𝜇R. Kl (T, R, 𝜇) vanishing on Nl (T, R, 𝜇) and such that 𝜇(R If 𝜇 is a measure, then 𝜇̄ is a measure. Moreover, Nl (T, Kl (T, R, 𝜇), 𝜇)̄ = Nl (T, R, 𝜇) = (Kl (T, R, 𝜇))0 (𝜇)̄ and Kl (T, Kl (T, R, 𝜇), 𝜇)̄ = Kl (T, R, 𝜇). Proof. We shall denote R0 (𝜇), Nl (T, R, 𝜇) and Kl (T, R, 𝜇) simply by R0 , Nl and Kl . It is checked as in the proof of Theorem 4 that Kl is a subring ̌ l . By virtue of Theoof the ring K(T, R, 𝜇). Consider the evaluation 𝜇̄ ≡ 𝜇|K rem 4, 𝜇̄ is a semimeasure. If 𝜇 is a measure, then 𝜇̄ is a measure as well. If R ∪ N ∈ Kl , where R ∈ R and N ∈ Nl , then by virtue of Lemma 3 and Theō ∪ N) = 𝜇R. Now, the uniqueness of 𝜇̄ is checked as in the proof of rem 4, 𝜇(R Theorem 4. ̄ Then, X ⊂ X0 ∈ K0l (𝜇). ̄ We have X0 ∈ Kl and 𝜇X ̄ 0 = 0. Let X ∈ Nl (T, Kl , 𝜇). ̄ 0 = 0 we conclude Therefore, X0 = R ∪ M for some R ∈ R and M ∈ Nl . From 𝜇R = 𝜇X ̄ Conversely that R ∈ R0 (𝜇). Thus, X0 ∈ Nl . So that X ∈ Nl ⊂ Kl , where X ∈ K0l (𝜇).

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̄ As a result, Nl (T, Kl , 𝜇)̄ = Nl = K0l (𝜇). ̄ Thus, Nl (T, Kl , 𝜇)̄ = K0l (𝜇) ⊂ Nl (T, Kl , 𝜇). Nl ⊂ Kl means that the measure 𝜇̄ is locally complete. ̄ then E = K ∪ N for some K ∈ Kl and N ∈ Finally, if E ∈ Kl (T, Kl , 𝜇), Nl (T, K, 𝜇)̄ = Nl . From K = R ∪ M for some R ∈ R and M ∈ Nl , we infer that E = R ∪ (M ∪ N) ∈ Kl . Thus, Kl (T, Kl , 𝜇)̄ = Kl . Corollary 1. Let 𝜇 be a positive 𝜎-finite semimeasure on a ring R. Then, 𝜇̄ is 𝜎-finite as well. Proof. Let K = R ∪ N ∈ Kl . By the condition, there is a countable collection (R i ∈ Rf (𝜇) | i ∈ I) such that R = ⋃⟮R i | i ∈ I⟯ and 𝜇R = sup (𝜇R i | i ∈ I). Consider the sets ̄ Then, K = ⋃⟮𝜇R i | i ∈ I⟯ and 𝜇K ̄ = 𝜇R = sup (𝜇R i | i ∈ I) = K i ≡ R i ∪ N ∈ Kfl (𝜇). ̄ i | i ∈ I). sup (𝜇K The semimeasurable [measurable] space ⟮T, Kl (T, R, 𝜇), 𝜇⟯̄ is called the local completion of the semimeasurable [measurable] space ⟮T, R, 𝜇⟯ and 𝜇̄ is called the local completion of 𝜇. The completions coincide in the following case. Lemma 4. Let 𝜇 be a positive semimeasure on an algebra A. Then, 1) N0 (T, A, 𝜇) = A0 (𝜇); 2) N(T, A, 𝜇) = Nl (T, A, 𝜇); 3) K(T, A, 𝜇) = Kl (T, A, 𝜇) and 𝜇̌ = 𝜇;̄ 4) K(T, A, 𝜇) is an algebra; 5) if, moreover, A is a 𝜎-algebra and 𝜇 is a measure, then K(T, A, 𝜇) is a 𝜎-algebra as well. Proof. By Corollary 3 to Lemma 15 (2.1.1), U(R) = R. Therefore, N0 (T, A, 𝜇) = {N0 ∈ A | ∀A ∈ A (A ⊂ N0 ⇒ 𝜇A = 0)} = A0 (𝜇), where N(T, A, 𝜇) = {N ⊂ T | ∃N0 ∈ A0 (𝜇) (N ⊂ N0 )} = Nl (T, A, 𝜇). Assertion 3 follows from assertions 1 and 2. By Theorem 4, T ∈ A ⊂ K(T, A, 𝜇) and the last ensemble is a ring. Hence, it is an algebra. Now, let A be a 𝜎-algebra. Take any countable collection (K i ∈ K(T, A, 𝜇) | i ∈ I). The definition and assertion 2 imply K i = A i ∪N i and N i ⊂ M i for some A i ∈ A and M i ∈ A0 (𝜇). By Lemma 9 (3.1.1), A0 (𝜇) is a 𝜎-ring. Therefore, M ≡ ⋃⟮M i | i ∈ I⟯ ∈ A0 (𝜇). Besides, A ≡ ⋃⟮A i | i ∈ I⟯ ∈ A and N ≡ ⋃⟮N i | i ∈ I⟯ ⊂ M. As a result, ⋃⟮K i | i ∈ I⟯ = A ∪ N ∈ K(T, A, 𝜇), i. e. this ensemble is 𝜎-additive. As a result, it is a 𝜎-algebra. Corollary 1. Let 𝜇 be a locally complete positive semimeasure on an algebra A. Then, N(T, A, 𝜇) = Nl (T, A, 𝜇) = A0 (𝜇), K(T, A, 𝜇) = Kl (T, A, 𝜇) = A, and A0 (𝜇) is an ideal and a ring.

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Proof. By Lemma 4 (3.1.1), 𝜇 is increasing. Therefore, N ∈ N(T, A, 𝜇) ⊂ A and N ⊂ N0 ∈ A0 (𝜇) imply N ∈ A0 (𝜇) ⊂ N(T, A, 𝜇). By Corollary 2 to Lemma 2 N(T, A, 𝜇) is an ideal and a ring. Corollary 2. Let 𝜇 be a positive semimeasure on an algebra A. Then, 𝜇 is complete iff 𝜇 is locally complete. Lemma 5. Let 𝜇 be a positive semimeasure on an algebra M, A be a subalgebra of M, and 𝜇̃ ≡ 𝜇|A. Then, 1) A0 (𝜇)̃ ⊂ M0 (𝜇); 2) N(T, A, 𝜇)̃ ⊂ N(T, M, 𝜇); ̌̃ ̌ for every K ∈ K(T, A, 𝜇), ̃ i. e. 𝜇̌ is 3) A ⊂ K(T, A, 𝜇)̃ ⊂ K(T, M, 𝜇) and 𝜇(K) = 𝜇K ̌ an extension of 𝜇;̃ ̃ i. e. 4) if, moreover, 𝜇 is complete, then A ⊂ K(T, A, 𝜇)̃ ⊂ M and 𝜇̌̃ = 𝜇|K(T, A, 𝜇), 𝜇|K(T, A, 𝜇)̃ is the completion of 𝜇.̃ ̃ = 0} = {A ∈ A | 𝜇A = 0} ⊂ M0 (𝜇). Proof. It is evident that A0 (𝜇)̃ ≡ {A ∈ A | 𝜇A Therefore, by Lemma 4 for N ∈ N(T, A, 𝜇)̃ we conclude that N ⊂ N0 ∈ A0 (𝜇)̃ ⊂ M0 (𝜇) implies N ∈ N(T, M, 𝜇). ̃ then K = A ∪ N for some A ∈ A ⊂ M and As a result, if K ∈ K(T, A, 𝜇), ̌̃ = 𝜇A ̃ = N ∈ N(T, A, 𝜇)̃ ⊂ N(T, M, 𝜇) implies K ∈ K(T, M, 𝜇). By Theorem 4, 𝜇K ̌ 𝜇A = 𝜇K. Finally, if 𝜇 is complete, then by Corollary 1 to Lemma 3 𝜇 is locally complete. Then, by Corollary 1 to Lemma 4, K(T, M, 𝜇) = M and N(T, M, 𝜇) = M0 (𝜇). Hence, ̌̃ = 𝜇A = 𝜇K. 𝜇K

The saturation and the strong saturation of a positive semimeasure Let 𝜀 be an evaluation on an ensemble S and E ⊂ S. The evaluation 𝜀 will be called E-saturated if S is E-saturated, i. e. S = UE (T, S) (see 2.1.1). It will be simply called saturated if S is saturated, i. e. S = U(T, S). Now, we shall consider the extension of a non-saturated semimeasure and a measure on the saturation U(T, R) of a ring R. Recall that, according to Corollary 1 to Proposition 6 (2.1.1), for every ring R its saturation U(T, R) is an algebra. Note that, according to Corollary 3 to Lemma 15 (2.1.1), if R is an algebra, then 𝜇 is saturated. Proposition 2. Let 𝜇 be a positive semimeasure on a ring R, U(T, R) be the saturation of R, and 𝜇̄̄ : P(T) → R+ be the variation of 𝜇. Then, ̄̄ 1) the evaluation 𝜇󸀠 ≡ 𝜇|U(T, R) : U(T, R) → R+ is a saturated semimeasure extending the semimeasure 𝜇; 2) 𝜇󸀠 E = sup{𝜇R | R ∈ R ∧ R ⊂ E} = sup{𝜇(E ∩ R) | R ∈ R} for every E ∈ U(T, R);

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3) the evaluation 𝜇󸀠 is the smallest of all increasing evaluations 𝜈 : U(T, R) → R+ extending the semimeasure 𝜇; 4) if 𝜇 is a measure on a 𝛿-ring R, then 𝜇󸀠 is a measure on the 𝜎-algebra U(T, R). ̄̄ ̄̄ = 𝜇󸀠 |R, assertions 1 and 2 follow from Proof. Since 𝜇󸀠 ≡ 𝜇|U(T, R) and v(𝜇) = 𝜇|R Proposition 2 (3.1.3), Corollary 2 to Lemma 7 (3.1.3), and Corollary 1 to Lemma 18 (2.1.1). Let 𝜈 : U(T, R) → R+ be some increasing evaluation such that 𝜈|R = 𝜇. Then, for every E ∈ U(T, R) and R ∈ R, we have 𝜈E ⩾ 𝜈(E ∩ R) = 𝜇(E ∩ R). Applying (2), we get 𝜈E ⩾ 𝜇󸀠 E. If R is a 𝛿-ring, then by Corollary 1 to Proposition 7 (2.1.1), U(T, R) is a 𝜎-algebra. By Proposition 2 (3.1.3), 𝜇󸀠 is 𝜎-additive. The semimeasurable [measurable] space ⟮T, U(T, R), 𝜇⟯ will be called the saturation of the semimeasurable [measurable] space ⟮T, R, 𝜇⟯ and 𝜇󸀠 will be called the saturation of 𝜇. Finally, we shall consider the important case of strong saturation of an internally finite (see 3.1.1) semimeasure. Let S be an ensemble on T and 𝜀 be an evaluation on S. Consider the Sf (𝜀)saturation USf (𝜀) (T, S) of S (see 2.1.1 and 3.1.1). It will be called also the strong saturation of S with respect to 𝜀 and denoted by Us (T, S, 𝜀). An evaluation 𝜀 on S will be called strongly saturated if S = Us (T, S, 𝜀). It is clear that U(T, S) ⊂ Us (T, S, 𝜀). If 𝜀 is finite, then U(T, S) = Us (T, S, 𝜀). Lemma 6. Let 𝜀 be a strongly saturated evaluation on a multiplicative ensemble S. Then, 𝜀 is saturated. Proof. According to Lemma 15 (2.1.1), S ⊂ U(T, S) ⊂ Us (T, S, 𝜀) = S. Lemma 7. Let 𝜀 be an increasing evaluation on an ensemble S and rng 𝜀 ⊂] − ∞, ∞]. Then, Us (T, S, 𝜀) ⊂ USf (𝜀) (T, Sf (𝜀)). Proof. For every U ∈ Us (T, S, 𝜀) ≡ USf (𝜀) (T, S) and E ∈ Sf (𝜀), we get U ∩ E ∈ S. Then, −∞ < 𝜀(U ∩ E) ⩽ 𝜀E < ∞. Hence, U ∩ E ∈ Sf (𝜀), i. e. U ∈ USf (𝜀) (T, Sf (𝜀)). Lemma 8. Let 𝜀 be an evaluation on an ensemble S. Then, the ensemble Us (T, S, 𝜀) has the following properties: 1) T ∈ Us (T, S, 𝜀); if ⌀ ∈ S, then ⌀ ∈ U(T, S); 2) if rng 𝜀 ⊂] − ∞, ∞] and 𝜀 is increasing, then Us (T, S, 𝜀) is multiplicative; 3) if U ∈ Us (T, S, 𝜀) and U ⊂ S ∈ Sf (𝜀), then U ∈ S; 4) if T ∈ Sf (𝜀), then Us (T, S, 𝜀) ⊂ S; 5) if S is multiplicative, then S ⊂ Us (T, S, 𝜀); 6) if S is additive, then Us (T, S, 𝜀) is additive as well.

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Proof. 2. By Lemma 7 Us (T, S, 𝜀) ⊂ USf (𝜀) (T, Sf (𝜀)). By Lemma 15 (2.1.1) this implies that Us (T, S, 𝜀) is multiplicative. All the other properties follow directly from the corresponding assertions of Lemma 15 (2.1.1). Corollary 1. If S is multiplicative and T ∈ Sf (𝜀), then Us (T, S, 𝜀) = S, i. e. 𝜀 is strongly saturated. Proposition 3. Let 𝜀 be increasing evaluation on a ring R such that rng 𝜀 ⊂] − ∞, ∞]. Then, Us (T, R, 𝜀) is an algebra containing R. Proof. The assertion follows from Lemma 7 and Proposition 6 (2.1.1). Proposition 4. Let 𝜀 be an increasing evaluation on 𝛿-ring R such that rng 𝜀 ⊂]−∞, ∞]. Then, Us (T, R, 𝜀) is a 𝜎-algebra containing R. Proof. The assertion follows from Lemma 7 and Proposition 7 (2.1.1). Lemma 9. Let 𝜀 be an additive positive evaluation on a disjointly additive ensemble S such that 𝜀S ⩽ m for some natural number m and every S ∈ Sf (𝜀). Then, every subensemble E of Sf (𝜀) consisting of pairwise disjoint sets E with 𝜀E > 0 is countable. Proof. Consider the subensembles En ≡ {E ∈ E | 𝜀E ⩾ 1/n}. Then, E = ⋃⟮En | n ∈ N⟯. Suppose that there exists n ∈ N such that the set E is infinite. By Corollary 4 to Proposition 1 (1.3.2), there is an injective mapping u : 𝜔 → En . Let E i ≡ u(i) and E ≡ ⋃⟮E i | i ∈ (m + 1)n⟯ ∈ Sf (𝜀). Then, 𝜀E = ∑(𝜀E i | i ∈ (m + 1)n) ⩾ (1/n)(m + 1)n > m. On the other hand, 𝜀E ⩽ m. It follows from this contradiction that all the sets En are finite. By Theorem 1 (1.3.9), the set E is countable. Theorem 5. Let 𝜇 be an internally finite positive measure on a 𝜎-ring R, R ∈ R, and 𝜇R = ∞. Then, 𝜇R = sup{𝜇S | S ∈ Rf (𝜇) ∧ S ⊂ R}. Proof. Consider the ensemble S ≡ {S ∈ Rf (𝜇) | S ⊂ R, 𝜇S > 0}. Since 𝜇 is internally finite it is non-empty. Therefore, the set D of all subensembles E of S, consisting of pairwise disjoint sets is non-empty as well. Consider on D the order by inclusion E ⊂ F. By the Hausdorff maximality principle from Theorem 1 (1.2.11) the set D contains a maximal chain C. Consider the ensemble A ≡ ⋃⟮C | C ∈ C⟯ ∈ D. It is clear that A ⩾ C for every C ∈ C. Suppose that 𝜇S ⩽ m for some natural number m and every S ∈ Sf (𝜇) = S. Then, by Lemma 9 the ensemble A is countable. If A is finite, then A ≡ ⋃⟮A󸀠 | A󸀠 ∈ A⟯ ∈ Rf (𝜇) and A ⊂ R. Therefore, A ∈ S, where 𝜇A ⩽ m. If A is denumerable, then there is a bijective mapping u : 𝜔 → A. Let A i ≡ u(i) and A ≡ ⋃⟮A i | i ∈ 𝜔⟯ =

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⋃⟮A󸀠 | A󸀠 ∈ A⟯ ∈ R. Consider the sequence (A i | i ∈ 𝜔). By Proposition 1 (3.1.1), 𝜇 is countably additive. Therefore, 𝜇A = ∑net (𝜇A i | i ∈ 𝜔) ≡ sup{∑(𝜇A i | i ∈ J) | J ∈ Pf (𝜔)} (see 1.4.8). If J ∈ Pf (𝜔), then S J ≡ ⋃⟮A i | i ∈ J⟯ ∈ Rf (𝜇) and S J ⊂ R. Therefore, S J ∈ S, where 𝜇S J ⩽ m. Since A consists of pairwise disjoint elements we have ∑(𝜇A i | i ∈ J) = 𝜇S J ⩽ m. As a result, 𝜇A ⩽ m. Thus, in both the cases we have 𝜇A ⩽ m and A ⊂ R. Consider the set B ≡ R\A ∈ R. Since ∞ = 𝜇R = 𝜇A + 𝜇B, we infer that 𝜇B = ∞. Since 𝜇 is internally finite, there exists a set C ∈ Rf (𝜇) such that C ⊂ B and 𝜇C > 0. Since C ∈ S we may consider the ensemble B ≡ A∪{C} ∈ D and the chain C󸀠 ≡ C∪{B}. Since B > A ⩾ C for every C ∈ C we conclude that B ∉ C. As a result, C󸀠 > C, but this contradicts the maximality of the chain C. It follows from the obtained contradiction that there is no such a number m, i. e. for every m ∈ N there exists a set S ∈ Sf (𝜇) such that 𝜇S > m. Consequently, we get the equality sup{𝜇S | S ∈ Rf (𝜇) ∧ S ⊂ R} = ∞. Define for an evaluation 𝜀 on an ensemble S an evaluation 𝜀󸀠󸀠 : Us (T, S, 𝜀) → R setting 𝜀󸀠󸀠 U ≡ sup{𝜀(U ∩ S) | S ∈ Sf (𝜀)} for every U ∈ Us (T, S, 𝜀). Theorem 6. Let 𝜇 be a internally finite positive semimeasure on a ring R. Then: 1) the evaluation 𝜇󸀠󸀠 is a strongly saturated internally finite positive semimeasure on the algebra Us (T, R, 𝜇) extending the semimeasure 𝜇; 2) the evaluation 𝜇󸀠󸀠 is the smallest of all increasing evaluations 𝜈 : Us (T, R, 𝜇) → R+ extending the semimeasure 𝜇; 3) if 𝜇 is a measure on a 𝛿-ring R, then 𝜇󸀠󸀠 is a measure on the 𝜎-algebra Us (T, R, 𝜇). Proof. By Proposition 3, Us ≡ Us (T, R, 𝜇) is an algebra containing R. We shall denote Rf (𝜇) simply by Rf . Let R ∈ R. Then, 𝜇󸀠󸀠 R = sup{𝜇(R ∩ S) | S ∈ Rf } ⩽ 𝜇R. If R ∈ Rf then 𝜇R ⩽ 𝜇󸀠󸀠 R. If 𝜇R = ∞, then by Theorem 5, 𝜇R = sup{𝜇S | S ∈ Rf ∧ S ⊂ R} ⩽ 𝜇󸀠󸀠 R. Thus, in both cases, 𝜇󸀠󸀠 R = 𝜇R, i. e. 𝜇󸀠󸀠 extends 𝜇. Let U ∈ Us and 𝜇󸀠󸀠 U = ∞. From the equality 𝜇󸀠󸀠 U = sup{𝜇(U ∩ R) | R ∈ R f } we infer, that there exists R ∈ Rf such that 0 < 𝜇(U ∩ R) < ∞. Hence, 𝜇󸀠󸀠 is internally finite. Let U, V ∈ Us and U ∩ V = ⌀. If R ∈ Rf then 𝜇((U ∪ V) ∩ S) = 𝜇(U ∩ S) + 𝜇(V ∩ S) ⩽ 󸀠󸀠 𝜇 U + 𝜇󸀠󸀠 V. Therefore, 𝜇󸀠󸀠 (U ∪ V) ⩽ 𝜇󸀠󸀠 U + 𝜇󸀠󸀠 V. Take any two real numbers x, y such that x < 𝜇󸀠󸀠 U and y < 𝜇󸀠󸀠 V. For them, there are sets R, S ∈ Rf such that x < 𝜇(U ∩ R) and y < 𝜇(V ∩ S). Therefore, for P ≡ R ∪ S ∈ Rf we have 𝜇󸀠󸀠 (U ∪ V) ⩾ 𝜇((U ∪ V) ∩ P) = 𝜇(U ∩ P) + 𝜇(V ∩ P) ⩾ 𝜇(U ∩ R) + 𝜇(V ∩ S) > x + y. From x < 𝜇󸀠󸀠 (U ∪ V) − y it follows that 𝜇󸀠󸀠 U ⩽ 𝜇󸀠󸀠 (U ∪ V) − y. If 𝜇󸀠󸀠 U < ∞, then y ⩽ 𝜇󸀠󸀠 (U ∪ V) − 𝜇󸀠󸀠 U implies 𝜇󸀠󸀠 V ⩽ 𝜇󸀠󸀠 (U ∪ V) − 𝜇󸀠󸀠 U, where 𝜇󸀠󸀠 U + 𝜇󸀠󸀠 V ⩽ 𝜇󸀠󸀠 (U ∪ V). If 𝜇󸀠󸀠 U = ∞, then ∞ = 𝜇󸀠󸀠 U ⩽ 𝜇󸀠󸀠 (U ∪ V) ⩽ ∞, whereas 𝜇 is increasing. Therefore, in this case we have also 𝜇󸀠󸀠 U + 𝜇󸀠󸀠 V ⩽ 𝜇󸀠󸀠 (U ∪ V). This means that 𝜇󸀠󸀠 is additive. By Lemma 1 (3.1.1), 𝜇󸀠󸀠 is finitely additive. Hence, it is a positive semimeasure.

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Let now 𝜇 be a measure on a 𝛿-ring R and (U i ∈ Us | i ∈ 𝜔) be a pairwise disjoint sequence. By Proposition 4, Us is a 𝜎-algebra containing R. Therefore, U ≡ ⋃⟮U i | i ∈ 𝜔⟯ ∈ Us . Take any R ∈ Rf . Consider the sequence (x n | n ∈ 𝜔) such that x n ≡ ∑(𝜇(U i ∩ R) | i ∈ n + 1). Since ⋃⟮U i ∩ R | i ∈ 𝜔⟯ = U ∩ R and 𝜇 is 𝜎-additive, we get 𝜇(U ∩ R) = lim(x n | n ∈ 𝜔). We have 𝜇(U ∩ R) < ∞. Take any number x > 0. Then, there is m ∈ 𝜔 such that 𝜇(U ∩ R) − x < x n for every n ⩾ m. Therefore, 𝜇(U ∩ R) − x < ∑(𝜇󸀠󸀠 U i | i ∈ n +1) ⩽ sup{∑(𝜇󸀠󸀠 U i | i ∈ n +1) | n ∈ 𝜔} = ∑net (𝜇󸀠󸀠 U i | i ∈ 𝜔) ≡ y (see 1.4.8). Since x can be taken arbitrarily small we conclude that 𝜇(U ∩ R) ⩽ y. This implies 𝜇󸀠󸀠 U ⩽ y, i. e. 𝜇󸀠󸀠 is countably subadditive. By virtue of Lemma 8 (3.1.1), 𝜇󸀠󸀠 is a measure. Finally, we shall check that 𝜇󸀠󸀠 is strongly saturated. Consider the ensemble E ≡ Us (T, Us , 𝜇󸀠󸀠 ). Let E ∈ E. Then, for every U ∈ Ufs (𝜇󸀠󸀠 ) ≡ Ufs we have E ∩ U ∈ Us . If R ∈ Rf then R ∈ Ufs whereas 𝜇󸀠󸀠 extends 𝜇. Therefore, E ∩ R ∈ Us implies E ∩ R = (E ∩ R) ∩ R ∈ R. This means that E ∈ Us . Thus, E ⊂ Us . On the other hand, Lemma 8 guarantees that Us ⊂ E. Let 𝜈 : Us → R+ be some increasing evaluation such that 𝜈|R = 𝜇. Then, for every U ∈ Us and R ∈ Rf , we have U ⩾ 𝜈(U ∩ R) = 𝜇(U ∩ R), where 𝜈U ⩾ 𝜇󸀠󸀠 U. For a semimeasure 𝜇 on a ring R, consider its restriction 𝜇f ≡ 𝜇|Rf (𝜇). Corollary 1. Let 𝜇 be an internally finite positive semimeasure on a ring R. Then, 𝜇󸀠󸀠 = (𝜇f )󸀠 . Proof. If U ∈ Us (T, R, 𝜇) then U ∩ R ∈ R for every R ∈ Rf (𝜇). Since 𝜇 is increasing, we get U ∩ R ∈ Rf (𝜇), where U ∈ U(T, Rf (𝜇)). Thus, Us (T, R, 𝜇) ⊂ U(T, Rf (𝜇)). The converse inclusion is evident. Therefore, for every U ∈ Us (T, R, 𝜇) = U(T, Rf (𝜇)) we have (𝜇f )󸀠 U = sup (𝜇f (U ∩ R) | R ∈ Rf ) = sup (𝜇(U ∩ R) | R ∈ Rf ) = 𝜇󸀠󸀠 U. The semimeasurable [measurable] space ⟮T, Us (T, R, 𝜇), 𝜇󸀠󸀠 ⟯ will be called the strong saturation of the semimeasurable [measurable] space ⟮T, R, 𝜇⟯ with internally finite positive 𝜇 and 𝜇󸀠󸀠 will be called the strong saturation of 𝜇. Note that if 𝜇 is finite and positive, then the strong saturation 𝜇󸀠󸀠 and the saturation 𝜇󸀠 coincide. Lemma 10. Let 𝜇 be a 𝜎-finite positive measure on a 𝜎-algebra A. Then, 𝜇 is strongly saturated, and therefore, saturated. Proof. By the conditions T = ⋃⟮A i | i ∈ I⟯ for some countable collection (A i ∈ Af (𝜇) | i ∈ I). Let U ∈ Us (T, A, 𝜇). Then, U ∩ A i ∈ A implies U = ⋃⟮U ∩ A i | i ∈ I⟯ ∈ A. Thus, Us (T, A, 𝜇) ⊂ A. The inverse inclusion follows from assertions 4 and 5 of Lemma 8. In conclusion, we shall prove that the saturation and the strong saturation preserve or strengthen the properties of the completeness.

3.1.4 Some extensions of additive evaluations defined on semirings and rings

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Lemma 11. Let 𝜇 be a positive locally complete (in particular, complete) semimeasure on a ring R. Then, 1) the saturation 𝜇󸀠 of 𝜇 is complete; 2) if, moreover, 𝜇 is internally finite, then the strong saturation 𝜇󸀠󸀠 of 𝜇 is complete. Proof. 1. We shall denote U(T, R) by V. By Lemma 18 (2.1.1), U(T, V) = V. Consequently, N0 (T, V, 𝜇󸀠 ) = {N0 ∈ U(T, V) | ∀E ∈ V (𝜇󸀠 (N0 ∩ E) = 0)} = {N0 ∈ V | ∀E ∈ V (𝜇󸀠 (N0 ∩ E) = 0)} = V0 (𝜇󸀠 ). Let N ∈ N(T, V, 𝜇󸀠 ). Then, there is N0 ∈ N0 (T, V, 𝜇󸀠 ) such that N ⊂ N0 . Then, for every R ∈ R we have N ∩ R ⊂ N0 ∩ R ∈ R0 (𝜇). Since 𝜇 is locally complete we infer that N ∩ R ∈ R. This means that N ∈ V. Thus, 𝜇 is complete. 2. Now, we shall denote Us (T, R, 𝜇) by V. By Theorem 6, 𝜇󸀠󸀠 is strongly saturated. By Lemma 6, 𝜇󸀠󸀠 is saturated, i. e. U(T, V) = V. The further proof is completely the same. Lemma 11 shows that we can consider the following important extensions of a measure 𝜇 : R → R+ on a 𝛿-ring R. The saturation 𝜇󸀠̌ : U(T, K(T, R, 𝜇)) → R+ of the completion 𝜇̌ : K(T, R, 𝜇) → R+ of 𝜇 will be called the complete saturated extension of the positive measure 𝜇 on the 𝛿-ring R. It will be denoted by 𝜇=/ : M=/ (T, R, 𝜇) → R+ . According Corollary 1 to Theorem 4, 𝜇̌ is a measure on the 𝛿-ring K(T, R, 𝜇). Therefore, by Proposition 2, 𝜇=/ is a positive measure on the 𝜎-algebra M=/ (T, R, 𝜇) ≡ U(T, K(T, R, 𝜇)). If 𝜇 is internally finite then by Corollary 2 to Theorem 4, 𝜇̌ is also internally finite measure on a 𝛿-ring. Therefore, for an internally finite measure 𝜇 the strong saturation 𝜇󸀠󸀠̌ : Us (T, K(T, R, 𝜇), 𝜇)̌ → R+ of the completion 𝜇̌ of 𝜇 will be called the complete strongly saturated extension of the internally finite positive measure 𝜇 on the 𝛿-ring R. It will be denoted by 𝜇#: M#(T, R, 𝜇) → R+ . By Theorem 6, 𝜇# is a positive measure ̌ on the 𝜎-algebra M#(T, R, 𝜇) ≡ Us (T, K(T, R, 𝜇), 𝜇). If 𝜇 is finite, then these extensions coincide. Lemma 12. Let 𝜇 be a positive, internally finite, and 𝜎-finite measure on a 𝜎-algebra M. Then, 𝜇# = 𝜇̌ = 𝜇̄ and 𝜇# is 𝜎-finite as well. Proof. By Corollary 1 of Proposition 1, 𝜇̄ is 𝜎-finite. By virtue of Lemma 4, 𝜇̌ = 𝜇̄ and K(T, M, 𝜇) is a 𝜎-algebra. Therefore, by Lemma 10, 𝜇̌ is strongly saturated, and so 𝜇# = 𝜇̌ = 𝜇.̄ Auxiliary definitions and statements 1∘ A set E ⊂ Rn is called (topologically) bounded if there is a parallelepiped P[s, t] such that E ⊂ P[s, t]. Statement 1. Let F be a bounded closed set in Rn . Then, F is Ost -compact (see 2.1.1).

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3.1.5 Extension of a positive measure to a wide complete saturated measure There are some methods of extending of every positive measure to a wide complete measure. Historically the first method was introduced by H. Lebesgue in the beginning of the XX century. The other one was offered by C. Caratheodory. We shall consider a very general modern method described in the book [Dinculeanu, 1967]. We shall realize the extension of positive measure 𝜇 : R → R+ defined on a ring R in two steps. First, we extend 𝜇 to an outer evaluation 𝜇∗ on the ideal 𝜎-ring H𝜎 (T, R) (see 2.1.1 and 2.1.4) and we delimitate the smaller 𝜎-ring L(T, R, 𝜇) containing R on which 𝜇× ≡ 𝜇∗ |L(T, R, 𝜇) is countably additive. Second, we take the saturation (𝜇× )󸀠 : U(T, L(T, R, 𝜇)) → R+ of the measure 𝜇× : L(T, R, 𝜇) → R+ (see 3.1.4). This two-step approach will play the key role in sections 3.4 – 3.6 for solving the Riesz – Radon – Fréchet problem.

The outer evaluation induced by a positive measure Define an evaluation 𝜇∗ : H𝜎 (T, R) → R+ setting 𝜇∗ (H) ≡ inf{v(𝜇, 𝜋) | 𝜋 ∈ 𝜎 Par (R, H)} (see 3.1.3). The evaluation 𝜇∗ is called the outer (exterior) evaluation induced by 𝜇. For the ensemble H𝜎 (T, R) and the set function 𝜇∗ , consider the subensembles: – H𝜎0 (T, R, 𝜇) ≡ H𝜎 (T, R)0 (𝜇∗ ) = {H ∈ H𝜎 (T, R) | 𝜇∗ H = 0}, – H𝜎f (T, R, 𝜇) ≡ H𝜎 (T, R)f (𝜇∗ ) = {H ∈ H𝜎 (T, R) | 𝜇∗ H < ∞}, and – H𝜎𝜎f (T, R, 𝜇) ≡ H𝜎 (T, R)𝜎f (𝜇∗ ) = {H ∈ H𝜎 (T, R) | 𝜇∗ is 𝜎-finite on H}. For the ensemble R𝜎 (T, R) ⊂ H𝜎 (T, R), we consider the subensembles R0𝜎 (T, R, 𝜇), R𝜎f (T, R, 𝜇), and R𝜎f 𝜎 (T, R, 𝜇) defined in the same way. Lemma 1. The evaluation 𝜇∗ has the following properties: 1) 𝜇∗ is increasing and natural; 2) 𝜇∗ is countably subadditive; 3) 𝜇∗ R = 𝜇R for every R ∈ R. Proof. We shall denote H𝜎 (T, R) simply by H. 1. It is evident. 2. Take a pairwise disjoint countable collection (H k ∈ H | k ∈ K) such that H ≡ ⋃⟮H k | k ∈ K⟯ ∈ H. We can assume that K = 𝜔. Let 𝜀 > 0. For each k, there is a countable collection (R ki ∈ R | i ∈ I k ) such that H k ⊂ ⋃⟮R ki | i ∈ I k ⟯ and ∑(𝜇R ki | i ∈ I k ) ⩽ 𝜇∗ H k + 𝜀/2k . Consider the set P ≡ ⋃⟮{k} × I k | k ∈ 𝜔⟯. By Theorem 1 (1.3.9), P is countable. Since H ⊂ ⋃⟮R ki | (k, i) ∈ P⟯, we have 𝜇∗ H ⩽ ∑(𝜇R ki | (k, i) ∈ P) ≡ x. By Proposition 1 (1.4.3), x = ∑(∑(𝜇R ki | i ∈ I k ) | k ∈ K) ⩽ ∑(𝜇∗ H k + 𝜀/2k | k ∈ 𝜔) = ∑(𝜇∗ H k + 𝜀/2k | k ∈ 𝜔) ≡ y. The definitions in 1.4.8 and Proposition 2 (1.4.7) imply y = ∑(𝜇∗ H k |

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k ∈ 𝜔) + 𝜀 ∑(1/2k | k ∈ 𝜔). According to Lemma 3 (1.4.8), ∑(1/2k | k ∈ 𝜔) = 2. Thus, 𝜇∗ H ⩽ ∑(𝜇∗ H k | k ∈ 𝜔) + 2𝜀. Since 𝜀 is arbitrary, we get 𝜇∗ H ⩽ ∑ 𝜇∗ H k . 3. If R ∈ R and (R i ∈ R | i ∈ I) is a countable collection such that R ⊂ ⋃⟮R i | i ∈ I⟯, then by Lemma 3 (3.1.1), 𝜇R ⩽ ∑ 𝜇R i , where 𝜇R ⩽ 𝜇∗ R. On the other hand, it is evident that 𝜇∗ R ⩽ 𝜇R. Corollary 1. If (H i ∈ H𝜎 (T, R) | i ∈ I) is a countable collection, H ∈ H𝜎 (T, R) and H ⊂ ⋃⟮H i | i ∈ I⟯, then 𝜇∗ H ⩽ ∑(𝜇∗ H i | i ∈ I). Proof. It follows from Lemma 1 and Lemma 3 (3.1.1). Corollary 2. H𝜎0 (T, R, 𝜇) is an ideal and a 𝜎-ring. Proof. Lemma 1 and Lemma 9 (3.1.1) imply that H𝜎0 (T, R, 𝜇) is an ideal in H𝜎 (T, R) and a 𝜎-ring. Since H𝜎 (T, R) is an ideal, H𝜎0 (T, R, 𝜇) is an ideal too. ̄̄ ⩽ 𝜇∗ H for every H ∈ H𝜎 (T, R). Lemma 2. Let 𝜇̄̄ be the total variation of 𝜇. Then, 𝜇H 𝜎

Proof. Let (S i ∈ R | i ∈ I) ∈ Par (R, H). ̄̄ = ∞, then for any x ∈ R+ there is a finite pairwise disjoint collection If 𝜇H (R k ∈ R | k ∈ K) such that R ≡ ⋃⟮R k | k ∈ K⟯ ⊂ H and ∑ (𝜇R k | k ∈ K) ⩾ x. Since R k = ⋃⟮R k ∩ S i | i ∈ I⟯ we have by Lemma 3 (3.1.1) that 𝜇R k ⩽ ∑(𝜇(R k ∩ S i ) | i ∈ I). Therefore, using Corollary 1 to Proposition 1 (1.4.3), we get x ⩽ ∑(∑(𝜇(R k ∩ S i | i ∈ I) | k ∈ K)) = ∑(∑(𝜇(R k ∩ S i ) | k ∈ K) | i ∈ I) = ∑(𝜇(⋃⟮R k ∩ S i | k ∈ K⟯) | i ∈ I) ⩽ ∑(𝜇S i | i ∈ I). ̄̄ ⩽ 𝜇∗ H. Consequently, x ⩽ 𝜇∗ H, where 𝜇∗ H = ∞. Thus, 𝜇H ̄̄ < ∞. Then, for any 𝜀 > 0 there is a finite pairwise disjoint Now, suppose that 𝜇H ̄̄ ⩽ ∑ 𝜇R k + 𝜀. By collection (R k ∈ R | k ∈ K) such that R ≡ ⋃⟮R k | k ∈ K⟯ ⊂ H and 𝜇H ∗ ∗ ̄ ̄ ̄ − 𝜀 ⩽ 𝜇 H, where 𝜇H ̄ ⩽ 𝜇 H. the above, we obtain 𝜇H Lemma 3. 𝜇∗ H = inf{𝜇∗ E | E ∈ R𝜎 (T, R) ∧ H ⊂ E} for every H ∈ H𝜎 (T, R). Proof. Denote the right part of this equality by x. As 𝜇∗ is increasing we have 𝜇∗ H ⩽ x. If 𝜇∗ H = ∞ then we get the necessary equality. Suppose 𝜇∗ H < ∞; then, for every 𝜀 > 0, there is a countable collection (R i ∈ R | i ∈ I) such that H ⊂ ⋃⟮R i | i ∈ I⟯ ≡ E ∈ R𝜎 (T, R) and ∑ 𝜇R i ⩽ 𝜇∗ H + 𝜀. By Corollary 1 to Lemma 1, 𝜇∗ E ⩽ ∑ 𝜇∗ R i ⩽ 𝜇∗ H + 𝜀, where x ⩽ 𝜇∗ H + 𝜀. Finally, x ⩽ 𝜇∗ H. Corollary 1. If R is a 𝜎-ring then 𝜇∗ H = inf{𝜇R | R ∈ R ∧ H ⊂ R} for every H ∈ H𝜎 (T, R). Lemma 4. For every H ∈ H𝜎 (T, R), there is E ∈ R𝜎 (T, R) such that H ⊂ E and 𝜇∗ H = 𝜇∗ E.

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Proof. If 𝜇∗ H < ∞, then by Lemma 3 for each n ∈ N there is E n ∈ R𝜎 (T, R) such that H ⊂ E n and 𝜇∗ E n ⩽ 𝜇∗ H +1/n. Then, H ⊂ E ≡ ⋂⟮E n | n ∈ N⟯ ∈ R𝜎 (T, R) implies 𝜇∗ H ⩽ 𝜇∗ E ⩽ 𝜇∗ E n ⩽ 𝜇∗ H + 1/n, where 𝜇∗ H = 𝜇∗ E. If 𝜇∗ H = ∞, then for every E ∈ R𝜎 (T, R) such that H ⊂ E we have 𝜇∗ E = ∞. Corollary 1. Every set H ∈ H𝜎0 (T, R, 𝜇∗ ) is contained in a set E ∈ R0𝜎 (T, R, 𝜇∗ ). The Lebesgue – Caratheodory extension of a positive measure Consider the ensemble L(T, R, 𝜇) of all L ∈ H𝜎 (T, R) such that 𝜇∗ H = 𝜇∗ (H ∩ L) + 𝜇∗ (H\L) for every H ∈ H𝜎 (T, R). The restriction 𝜇∗ |L(T, R, 𝜇) will be denoted by 𝜇× . By Lemma 1, we have always 𝜇∗ H ⩽ 𝜇∗ (H ∩ L) + 𝜇∗ (H\L) for H, L ∈ H𝜎 (T, R). Therefore, L(T, R, 𝜇) consists of all L ∈ H𝜎 (T, R) such that 𝜇∗ H ⩾ 𝜇∗ (H ∩ L)+𝜇∗ (H\L) for every H ∈ H𝜎 (T, R). Theorem 1. Let 𝜇 be a positive measure on a ring R. Then, the ensemble L(T, R, 𝜇) is a 𝜎-ring containing the ring R and 𝜇× : L(T, R, 𝜇) → R+ is a measure extending the measure 𝜇. Proof. We shall denote H𝜎 (T, R) and L(T, R, 𝜇) simply H and L. Let E, F ∈ L. Fix a set H ∈ H. We have successively 𝜇∗ H = 𝜇∗ (H ∩ E ∩ F) + 𝜇∗ ((H ∩ E)\F) + 𝜇∗ ((H\E) ∩ F) + 𝜇∗ ((H\E)\F). Writing this equality for the set H ∩ (E ∪ F) instead of H, we obtain 𝜇∗ (H ∩ (E ∪ F)) = 𝜇∗ (H ∩ E ∩ F) + 𝜇∗ ((H ∩ E)\F) + 𝜇∗ ((H\E) ∩ F). From these two equalities we deduce 𝜇∗ H = 𝜇∗ (H ∩ (E ∪ F)) + 𝜇∗ (H\(E ∪ F)). This means that E ∪ F ∈ L. Writing now the first equality for the set H\(E\F) instead of H, we receive 𝜇∗ (H\(E\F)) = 𝜇∗ (H ∩ E ∩ F) + 𝜇∗ ((H\E) ∩ F) + 𝜇∗ ((H\E)\F) because ((H\(E\F)) ∩ E)\ F = ⌀. From the first and the last equalities we obtain 𝜇∗ H = 𝜇∗ (H\(E\F)) + 𝜇∗ (H ∩ (E\F)). This means that E\F ∈ L. Thus, we proved that L is a ring. Consider the subset X of 𝜔 of all n such that 𝜇∗ (H ∩ (⋃⟮L i | i ∈ I⟯)) = ∑(𝜇∗ (H ∩ L i ) | i ∈ I) for every H ∈ H and for every finite pairwise disjoint collection (L i ∈ L | i ∈ I) / k. Using such that card I = n + 2 and 𝜇∗ (H ∩ (⋃⟮L i | i ∈ I⟯)) < ∞. Let I = {j, k} and j = the second equality for the disjoint sets, E ≡ L i and F ≡ L k we obtain 𝜇∗ (H ∩(⋃⟮L i | i ∈ I⟯)) = ∑(𝜇∗ (H ∩ L i ) | i ∈ I), where we used the Corollary 3 to Proposition 1 (1.1.10), and Lemma 4 (1.4.3). Therefore, 0 ∈ X. Suppose n ∈ X and take any (L i | i ∈ I) such that card I = n + 3. Take some j ∈ I and consider the set K ≡ I\{j} such that card K = n + 2. By the same reasons as above, we have 𝜇∗ (H ∩ (⋃⟮L i | i ∈ I⟯)) = 𝜇∗ (H ∩ (L j ∪ ⋃⟮L k | k ∈ K⟯)) = 𝜇∗ (H ∩ L j ) + 𝜇∗ (H ∩ (⋃⟮L k | k ∈ K⟯)) = 𝜇∗ (H ∩ L j ) + ∑(𝜇∗ (H ∩ L k ) | k ∈ K) = ∑(𝜇∗ (H ∩ L i ) | i ∈ I), where we used the Corollary 3 to Proposition 1 (1.1.10), Theorem 1 (1.4.3) and Lemma 4 (1.4.3). Consequently, n + 1 ∈ X. By the principle of natural induction from Theorem 1 (1.2.6), X = 𝜔.

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If (L i | i ∈ I) is a collection such that 𝜇∗ (H ∩ (⋃⟮L i | i ∈ I⟯)) = ∞, then by Corollary 1 to Lemma 1 we obtain ∞ ⩽ ∑(𝜇∗ (H ∩ L i ) | i ∈ I). Thus, in this case we also have the same equality as above. Let now (L i ∈ L | i ∈ I) be any pairwise disjoint countable collection and M ≡ ⋃⟮L i | i ∈ I⟯. Then, for every finite subset J of I, we obtain 𝜇∗ H = 𝜇∗ (H ∩ (⋃⟮L i | i ∈ J⟯)) + 𝜇∗ (H\ ⋃⟮L i | i ∈ J⟯) ⩾ ∑(𝜇∗ (H ∩ L i ) | i ∈ J) + 𝜇∗ (H\M). If 𝜇∗ (H\M) = ∞, then 𝜇∗ H = ∞ and we have 𝜇∗ H = 𝜇∗ (H ∩ M) + 𝜇∗ (H\M). If 𝜇∗ (H\M) < ∞, then 𝜇∗ H ⩾ ∑(𝜇∗ (H ∩L i ) | i ∈ I)+𝜇∗ (H\M). By Corollary 1 to Lemma 1, 𝜇∗ H ⩾ 𝜇∗ (H ∩M)+𝜇∗ (H\M). Consequently, M ∈ L, and therefore, L is countably disjointly additive. Then, by Corollary 1 to Lemma 8 (2.1.1), L is countably additive, i. e. L is a 𝜎-ring. Since it is always fulfilled 𝜇∗ H ⩽ 𝜇∗ (H ∩ M) + 𝜇∗ (H\M), we conclude that ∑(𝜇∗ (H ∩ L i ) | i ∈ I) + 𝜇∗ (H\M) = 𝜇∗ (H ∩ M) + 𝜇∗ (H\M). Writing this equality for H = M, we obtain ∑(𝜇∗ L i | i ∈ I) = 𝜇∗ M. This means that 𝜇∗ is countably additive on L. By Proposition 1 (3.1.1), 𝜇∗ is 𝜎-additive. It remains to prove that R ⊂ L. Let R ∈ R. Suppose that 𝜇∗ H = ∞. Then, ∑(𝜇R i | i ∈ I) = ∞ for every countable collection (R i ∈ R | i ∈ I) such that H ⊂ ⋃⟮R i | i ∈ I⟯. Therefore, 𝜇∗ H = ∑(𝜇R i | i ∈ I) = ∑(𝜇(R i ∩ R) + 𝜇(R i \R) | i ∈ I) = ∑(𝜇(R i ∩ R) | i ∈) + ∑(𝜇(R i \R) | i ∈ I) ⩾ 𝜇∗ (H ∩ R) + 𝜇∗ (H\R) because H ∩ R ⊂ ⋃⟮R i ∩ R | i ∈ I⟯ and H\R ⊂ ⋃⟮R i \R | i ∈ I⟯. Suppose now that 𝜇∗ H < ∞. Then, for any 𝜀 > 0, there is a countable collection (R i ∈ R | i ∈ I) such that H ⊂ ⋃⟮R i | i ∈ I⟯ and 𝜇∗ H + 𝜀 ⩾ ∑(𝜇R i | i ∈ I). Therefore, as above, 𝜇∗ H +𝜀 ⩾ 𝜇∗ (H ∩R)+𝜇∗ (H\R), where 𝜇∗ H ⩾ 𝜇∗ (H ∩R)+𝜇∗ (H\R). Consequently, R ∈ L. The measure 𝜇× : L(R, R, 𝜇) → R+ is called the Lebesgue – Caratheodory extension of the measure 𝜇 : R → R+ . For the ensemble L(T, R, 𝜇) and the measure 𝜇× , consider the following subensembles: – L0 (T, R, 𝜇) ≡ L(T, R, 𝜇)0 (𝜇× ) = {L ∈ L(T, R, 𝜇) | 𝜇× L = 0}, – Lf (T, R, 𝜇) ≡ L(T, R, 𝜇)f (𝜇× ) = {L ∈ L(T, R, 𝜇) | 𝜇× L < ∞}, and – L𝜎f (T, R, 𝜇) ≡ L(T, R, 𝜇)𝜎f (𝜇× ) = {L ∈ L(T, R, 𝜇) | 𝜇× is 𝜎-finite on L}. Corollary 1. R𝜎 (T, R) ⊂ L(T, R, 𝜇). Corollary 2. L0 (T, R, 𝜇) = H𝜎0 (T, R, 𝜇), L0 (T, R, 𝜇) is an ideal and a 𝜎-ring, and the measure 𝜇× is locally complete. Proof. Let E ∈ H𝜎0 (T, R, 𝜇) and H ∈ H𝜎 (T, R). Then, 0 ⩽ 𝜇∗ (H ∩ E) ⩽ 𝜇∗ E = 0 implies 𝜇∗ H ⩾ 𝜇∗ (H\E) = 𝜇∗ (H ∩ E) + 𝜇∗ (H\E). This means that E ∈ L0 (T, R, 𝜇). Thus, L0 (T, R, 𝜇) = H𝜎0 (T, R, 𝜇). By Corollary 2 to Lemma 1, H𝜎0 (T, R, 𝜇) is an ideal and a 𝜎-ring. Hence, 𝜇× is locally complete (see 3.1.4).

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Corollary 3. Let L ⊂ T. Then, L ∈ Lf (T, R, 𝜇) iff L = F\K for some F ∈ R𝜎f (T, R, 𝜇) and K ∈ L0 (T, R, 𝜇). Proof. By Lemma 4, there is F ∈ R𝜎 (T, R) such that L ⊂ F and 𝜇∗ L = 𝜇∗ F. Hence, F ∈ R𝜎f (T, R, 𝜇). Consider the set K ≡ F\L ∈ L(T, R, 𝜇). By Theorem 1, 𝜇∗ F = 𝜇∗ L + 𝜇∗ K. Therefore, K ∈ L0 (T, R, 𝜇). Finally, L = F\K. Corollary 4. Let L ⊂ T. Then, L ∈ Lf (T, R, 𝜇) iff L = E ∪ N for some E ∈ R𝜎f (T, R, 𝜇) and N ∈ L0 (T, R, 𝜇). Proof. By Corollary 3, L = F\K, for some F ∈ R𝜎f (T, R, 𝜇) and K ∈ L0 (T, R, 𝜇). By Corollary 1 to Lemma 4, there is M ∈ R0𝜎 (T, R, 𝜇) such that K ⊂ M. Consider the sets E ≡ F\M ∈ R𝜎f (T, R, 𝜇) and N ≡ L\E ∈ L(T, R, 𝜇). Then, 𝜇∗ N = 𝜇∗ (L ∩ M) = 0 and L = E ∪ N. Corollary 5. Let L ⊂ T. Then, L ∈ Lf (T, R, 𝜇) iff E ⊂ L ⊂ F for some E, F ∈ R𝜎f (T, R, 𝜇) such that F\E ∈ L0 (T, R, 𝜇). Proof. By Corollaries 3 and 4, L = F\K and L = E ∪ N, where F, E ∈ R𝜎f (T, R, 𝜇) and K, N ∈ L0 (T, R, 𝜇). Then, E ⊂ L ⊂ F and 0 ⩽ 𝜇∗ (F\E) = 𝜇∗ (F\L) + 𝜇∗ (L\E) ⩽ 𝜇∗ K + 𝜇∗ N = 0 because 𝜇∗ is additive and increasing. Corollary 6. Let L ⊂ T. Then, L ∈ L𝜎f (T, R, 𝜇) iff L = F\K for some F ∈ R𝜎f 𝜎 (T, R, 𝜇) 0 and K ∈ L (T, R, 𝜇). Proof. If L ∈ L𝜎f (T, R, 𝜇), then L = ⋃⟮L i | i ∈ I⟯ for some countable collection (L i ∈ Lf (T, R, 𝜇) | i ∈ I). By Corollary 3, L i = F i \K i for some F i ∈ R𝜎f (T, R, 𝜇) and K i ∈ L0 (T, R, 𝜇). Then, F ≡ ⋃⟮F i | i ∈ I⟯ ∈ R𝜎f 𝜎 (T, R, 𝜇). Since K ≡ F\L = ⋃⟮F i \ ⋃⟮L j | j ∈ I⟯ | i ∈ I⟯ ⊂ ⋃⟮K i | i ∈ I⟯, using Corollary 2, we get K ∈ L0 (T, R, 𝜇). Finally, L = F\K. Corollary 7. Let L ⊂ T. Then, L ∈ L𝜎f (T, R, 𝜇) iff L = E ∪ N for some E ∈ R𝜎f 𝜎 (T, R, 𝜇) and N ∈ L0 (T, R, 𝜇). Proof. If L ∈ L𝜎f (T, R, 𝜇), then L = ⋃⟮L i | i ∈ I⟯ for some countable collection (L i ∈ Lf (T, R, 𝜇) | i ∈ I). By Corollary 4, L i = E i ∪ N i for some E i ∈ R𝜎f (T, R, 𝜇) and N i ∈ L0 (T, R, 𝜇). Then, E ≡ ⋃⟮E i | i ∈ I⟯ ∈ R𝜎f 𝜎 (T, R, 𝜇). By Corollary 2, N ≡ ⋃⟮N i | i ∈ I⟯ ∈ L0 (T, R, 𝜇). Finally, L = E ∪ N. Corollary 8. Let L ⊂ T. Then, L ∈ L𝜎f (T, R, 𝜇) iff E ⊂ L ⊂ F for some E, F ∈ 0 R𝜎f 𝜎 (T, R, 𝜇) such that F\E ∈ L (T, R, 𝜇).

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The proof is completely the same as for Corollary 5. Proposition 1. Let 𝜇 be 𝜎-finite. Then, L(T, R, 𝜇) ⊂ H𝜎 (T, Rf (𝜇)) and 𝜇× is 𝜎-finite as well. Proof. We shall denote L(T, R, 𝜇), Lf (T, R, 𝜇), L𝜎f (T, R, 𝜇) simply by L, Lf , and L𝜎f . Let L ∈ L. There is a countable collection (R m ∈ R | m ∈ M) such that L ⊂ ⋃⟮R m | m ∈ M⟯. By the condition, R m = ⋃⟮S mi | i ∈ I m ⟯ for some countable collection (S mi ∈ Rf (𝜇) | i ∈ I m ). So by Proposition 1 (1.1.10) L ⊂ ⋃ {⋃⟮S mi | i ∈ I m ⟯ | m ∈ M } = ⋃⟮S mi | p ≡ (m, i) ∈ P⟯, { } where P ≡ ⋃⟮{m}×I m | m ∈ M⟯. By Theorem 1 (1.3.9), the set P is countable. This means that L ⊂ H𝜎 (T, Rf ). Finally, L = ⋃⟮L ∩ S p | p ∈ P⟯ and L ∩ S p ∈ Lf . Thus, L ∈ L𝜎f . Hence, L = L𝜎f , i. e. 𝜇× is 𝜎-finite. Theorem 2. Let 𝜇 be a 𝜎-finite positive measure on a ring R. Then, 𝜇̃ ≡ 𝜇∗ |R𝜎 (T, R) is the unique positive 𝜎-additive extension of 𝜇 from R to R𝜎 (T, R). Proof. We shall denote R𝛿 (T, R) and R𝜎 (T, R) simply by R𝛿 and R𝜎 . Let 𝜆 and 𝜈 be positive countably additive evaluations defined on R𝜎 and extending 𝜇. At first, suppose that 𝜆 and 𝜈 are finite on R. Let E ∈ R𝛿 . Then, by Corollary 1 to Lemma 11 (2.1.1), E ⊂ ⋃⟮R i | i ∈ I⟯ for some finite collection (R i ∈ R | i ∈ I). Therefore, by Lemma 3 (3.1.1), 𝜈E ⩽ ∑(𝜈R i | i ∈ I) < ∞. Similarly, 𝜆E < ∞. This means that 𝜈 and 𝜆 are finite on R𝛿 . Denote by X the ensemble of all the sets X ∈ R𝛿 such that 𝜈X = 𝜆X. Then, R ⊂ X. Let (X n ∈ X | n ∈ 𝜔) ↑ E ∈ R𝛿 . Then, by Lemma 5 (3.1.1), 𝜆E = lim(𝜆X n | n ∈ 𝜔) = lim(𝜈X n | n ∈ 𝜔) = 𝜈E. So E ∈ X. Let (X n ∈ X | n ∈ 𝜔) ↓ E ∈ R𝛿 . Then, by Lemma 6 (3.1.1), 𝜆E = 𝜈E implies E ∈ X. This means that X is closed in R𝛿 with respect to limits (unions and intersections) of monotone (increasing or decreasing) sequences. Consider the non-empty set A of all ensembles Y ⊂ R𝛿 such that R ⊂ Y and Y is closed in R𝛿 with respect to limits of monotone sequences. Then, the ensemble S ≡ ⋂⟮Y | Y ∈ A⟯ = {Y | ∀ Y (Y ∈ A ⇒ Y ∈ Y)} is also closed in R𝛿 with respect to monotone sequences and contains R. Since X ∈ A, we have S ⊂ X. Now, we shall prove that S = R𝛿 . For every set F ∈ R𝛿 , we consider an ensemble S(F) ≡ {E ∈ R𝛿 | E\F, F\E, E ∪ F ∈ S}. We see that E ∈ S(F) iff F ∈ S(E). Let (E n ∈ S(F) | n ∈ 𝜔) ↑ E ∈ R𝛿 . This means that E n \F, F\E n , E n ∪ F ∈ X. Since (E n \F ∈ S | n ∈ 𝜔) ↑ E\F ∈ R𝛿 , we have E\F ∈ S. Similarly, F\E, E ∪ F ∈ S. Therefore, E ∈ S(F). Analogously, if (E n ∈ S(F) | n ∈ 𝜔) ↓ E ∈ R𝛿 , then E ∈ S(F). This means that S(F) is closed in R𝛿 with respect to limits of monotone sequences.

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If E, F ∈ R, then E\F, F\E, E ∪ F ∈ R ⊂ S implies E ∈ S(F), i. e. R ⊂ S(F) for every F ∈ R. Therefore, S ⊂ S(F) for every F ∈ R. Consequently, if E ∈ S and F ∈ R, then E ∈ S(F), and so F ∈ S(E). This implies R ⊂ S(E), where S ⊂ S(E) for every E ∈ S. We conclude that if F ∈ S ⊂ S(E), then F\E, E\F, E ∪ F ∈ S. This means that S is a ring. Let (S i ∈ S | i ∈ 𝜔) be a sequence and H ≡ ⋂⟮S i | i ∈ 𝜔⟯ ∈ R𝛿 . Consider the sets R i ≡ ⋂⟮S i | i ∈ n + 1⟯ ∈ S. Then, (R i | i ∈ 𝜔) ↓ H implies H ∈ S. Consequently, S is a 𝛿-ring. Finally, S = R𝛿 , where X = R𝛿 . Now, let E ∈ R𝜎 . By Corollary 2 to Proposition 4 (2.1.1), E = ⋃⟮R i | i ∈ I⟯ for some countable pairwise disjoint collection (R i ∈ R𝛿 | i ∈ I). Therefore, 𝜆E = ∑(𝜆R i | i ∈ I) = ∑(𝜈R i | i ∈ I) = 𝜈E. Suppose now that 𝜈 and 𝜆 are 𝜎-finite on R and consider the ring R0 ⊂ R on which 𝜈 and 𝜆 are finite. It was proven above that 𝜈 coincides with 𝜆 on R𝜎 (T, R0 ). Let R ∈ R. Then, R = ⋃⟮R i | i ∈ I⟯ for some countable collection (R i ∈ R | i ∈ I) such that 𝜈R i < ∞. Similarly, R = ⋃⟮S j | j ∈ J⟯ for some countable collection (S j ∈ R | j ∈ J) such that 𝜆S j < ∞. Then, R = ⋃⟮R i ∩ S j | (i, j) ∈ I × J⟯ and R i ∩ S j ∈ R0 imply R ∈ R𝜎 (T, R0 ). Therefore, R ⊂ R𝜎 (T, R), where R𝜎 (T, R) = R𝜎 (T, R0 ). Thus, 𝜈 = 𝜆. Corollary 1. Let 𝜇 be 𝜎-finite. Then, 𝜇× is the unique positive countably additive extension of 𝜇 from R to L(T, R, 𝜇). Proof. We shall use the notations from the proof of Theorem 2. Let 𝜈 be such an extension. Then, by Theorem 2, 𝜈|R𝜎 = 𝜇∗ |R𝜎 . If N ∈ L0 (T, R, 𝜇), then by Corollary 1 to Lemma 4, N ⊂ E ∈ R𝜎 and 𝜇∗ E = 0 imply 𝜈N ⩽ 𝜈E = 0. If L ∈ L(R, 𝜇), then Proposition 1 and Corollary 7 to Theorem 1 imply L = E ∪ N for some E ∈ R𝜎 and N ∈ L0 (R, 𝜇). Hence, 𝜈L = 𝜇× L.

The large complete saturated extension of a positive measure ̂ Consider now the saturation M(T, R, 𝜇) ≡ U(T, L(T, R, 𝜇)) of the 𝜎-ring L(T, R, 𝜇) × 󸀠 and the saturation 𝜇̂ ≡ (𝜇 ) of the measure 𝜇× . Lemma 5. Let 𝜇 be a positive measure on a ring R. Then, ̂ 1) M(T, R, 𝜇) is a 𝜎-algebra; ̂ 2) 𝜇̂ : M(T, R, 𝜇) → R+ is a wide positive saturated measure extending the measure 𝜇× ; ̂ ̂ = sup{𝜇× L | L ∈ L(T, R, 𝜇) ∧ L ⊂ M} = sup{𝜇× 3) if M ∈ M(T, R, 𝜇), then 𝜇M (M ∩ L) | L ∈ L(T, R, 𝜇)}; ̂ 4) M ∈ M(T, R, 𝜇) iff M ∩ R ∈ L(T, R, 𝜇) for every R ∈ R.

3.1.5 Extension of a positive measure to a wide complete saturated measure | 215

Proof. Statement 1 follows from Theorem 1 and Corollary 1 to Proposition 7 (2.1.1). Statements 2 and 3 follow from Proposition 2 (3.1.4). Statement 4 follows from Lemma 17 (2.1.1). ̂ Corollary 1. B(T, R) ⊂ M(T, R, 𝜇). Proof. It follows from the definition of the Borel 𝜎-algebra B(T, R) generated by R as the smallest 𝜎-algebra on T containing R (see 2.1.1 and 2.1.3). ̂ The measure 𝜇̂ : M(T, R, 𝜇) → R+ will be called the large complete saturated extension of the measure 𝜇 : R → R+ . ̂ For the ensemble M(T, R, 𝜇) and the measure 𝜇,̂ consider the following subensembles: ̂ 0 (T, R, 𝜇) ≡ M(T, ̂ ̂ ̂ = 0}, M R, 𝜇)0 (𝜇)̂ = {M ∈ M(T, R, 𝜇) | 𝜇M ̂ ̂ ̂ f (T, R, 𝜇) ≡ M(T, ̂ < ∞}, and R, 𝜇)f (𝜇)̂ = {M ∈ M(T, R, 𝜇) | 𝜇M M ̂ 𝜎f (T, R, 𝜇) ≡ M(T, ̂ ̂ M R, 𝜇)𝜎f (𝜇)̂ = {M ∈ M(T, R, 𝜇) | 𝜇̂ is 𝜎-finite on M}. ̂ ̂ f (T, R, 𝜇), and M ̂ 0 (T, R, 𝜇) According to 3.1.1, the elements of M(T, R, 𝜇), M ̂ ̂ ̂ ̂ will be called 𝜇-measurable, 𝜇-squarable (≡ 𝜇-integrable), and 𝜇-negligible, respectively, or measurable, squarable (≡ integrable), and negligible with respect to 𝜇.̂ ̂ 0 (T, R, 𝜇) (see 2.1.1) will be called 𝜇-conegligible. ̂ The elements of co-M ̂ Lemma 6. Let 𝜇 be 𝜎-finite and M ⊂ T. Then, M ∈ M(T, R, 𝜇) iff M ∩ R ∈ L(T, R, 𝜇) for every R ∈ Rf (𝜇). Proof. By Proposition 1, Rf (𝜇) ⊂ L(T, R, 𝜇) ⊂ H𝜎 (T, Rf ). If M ∩ R ∈ L(T, R, 𝜇) for ̂ R, 𝜇). The inverse statement every R ∈ Rf (𝜇), then by Lemma 17 (2.1.1), M ∈ M(T, follows from the definition of M(T, R, 𝜇) = U(T, L(T, R, 𝜇)). ̂ Corollary 1. Let 𝜇 be 𝜎-finite and M ⊂ T. Then, M ∈ M(T, R, 𝜇) iff M ∩ L ∈ L(T, R, 𝜇) f for every L ∈ L (T, R, 𝜇). ̂ ̂ = sup{𝜇× (M ∩ R) | R ∈ R} = sup{𝜇× (M ∩ E) | Lemma 7. If M ∈ M(T, R, 𝜇), then 𝜇M × E ∈ R𝛿 (T, R)} = sup{𝜇 (M ∩ F) | F ∈ R𝜎 (T, R)} = sup{𝜇× (M ∩ L) | L ∈ L(T, R, 𝜇)}. Proof. We shall denote L(T, R, 𝜇), R𝜎 (T, R), R𝛿 (T, R) simply by L, R𝜎 , R𝛿 . Since R ⊂ R𝛿 ⊂ R𝜎 ⊂ L we have sup{𝜇∗ (M ∩ R) | R ∈ R} ⩽ sup{𝜇∗ (M ∩ E) | E ∈ R𝛿 } ⩽ ̂ sup{𝜇∗ (M ∩ F) | F ∈ R𝜎 } ⩽ sup{𝜇∗ (M ∩ L) | L ∈ L} ⩽ 𝜇M. ̂ ⩽ sup{𝜇∗ (M ∩ R) | R ∈ R} ≡ x. Let L ∈ L and L ⊂ It remains to prove that 𝜇M M. Then, L ⊂ ⋃⟮R n | n ∈ 𝜔⟯ ≡ F ∈ R𝜎 for some sequence (R n ∈ R | n ∈ 𝜔) ↑. Since

216 | 3.1 Spaces with semimeasures and measures

(M ∩ R n | n ∈ 𝜔) ↑ (M ∩ F), then by Lemma 5 (3.1.1) and Lemma 7 (1.4.7) we have 𝜇∗ L ⩽ 𝜇∗ (M ∩ F) = lim(𝜇∗ (M ∩ R n ) | n ∈ 𝜔) = sup{𝜇∗ (M ∩ R n ) | n ∈ 𝜔} ⩽ x. Consequently, ̂ ⩽ x. 𝜇M ̂ ̂ = sup{𝜇× L | L ∈ Lemma 8. Let 𝜇 be 𝜎-finite on R. If M ∈ M(T, R, 𝜇) then 𝜇M f × f L (T, R, 𝜇) ∧ L ⊂ M} = sup{𝜇 (M ∩ L) | L ∈ L (T, R, 𝜇)}. Proof. We shall denote L(T, R, 𝜇) and Lf (T, R, 𝜇) simply by L and Lf . Denote the second and third numbers of this equality by x and y, respectively. We have x ⩽ y. If L ∈ Lf , then by Lemma 6, L󸀠 ≡ M ∩ L ∈ Lf and L󸀠 ⊂ M. This means that y ⩽ x. It is ̂ evident that x ⩽ 𝜇M. Let now L ∈ L and L ⊂ M. Then, by Lemma 6, there is a sequence (R n ∈ Rf | n ∈ 𝜔) ↑ such that L ⊂ ⋃⟮R n | n ∈ 𝜔⟯ ≡ F ∈ R𝜎 . Since L n ≡ R n ∩ L ∈ Lf , L n ⊂ M and (L n | n ∈ 𝜔) ↑ L, by Lemma 5 (3.1.1), we obtain 𝜇∗ L = lim(𝜇∗ L n | n ∈ 𝜔) = sup{𝜇∗ L n | ̂ ⩽ x. n ∈ 𝜔} ⩽ x. Finally, by Lemma 5, 𝜇M Lemma 9. ̂ f (T, R, 𝜇) is a 𝛿-ring and M ̂ 𝜎f (T, R, 𝜇) is a 𝜎-ring. 1) M ̂ ̂ 0 (T, R, 𝜇) iff L ∈ L0 (T, R, 𝜇) for every L ∈ 2) Let M ∈ M(T, R, 𝜇). Then, M ∈ M L(T, R, 𝜇) such that L ⊂ M and iff M ∩ R ∈ L0 (T, R, 𝜇) for every R ∈ R. ̂ 0 (T, R, 𝜇) is an ideal and a 𝜎-ring and L0 (T, R, 𝜇) = H0 (T, R, 𝜇) ⊂ 3) M 𝜎 ̂ 0 (T, R, 𝜇). M Proof. 1. This assertion follows from Lemma 9 (3.1.1). 2. This assertion follows from Lemmas 6 and 7. 3. By Lemma 9 (3.1.1), M0 (T, R, 𝜇) is a 𝜎-ring. Let E ⊂ M ∈ M0 (T, R, 𝜇). If R ∈ R, then E ∩ R ⊂ M ∩ R ∈ L0 (T, R, 𝜇) implies E ∩ R ∈ L0 (T, R, 𝜇) according to Corollary 2 to Theorem 1. By (2), E ∈ M0 (T, R, 𝜇). Thus, M0 (R, 𝜇) is an ideal and a 𝜎-ring. By Corollary 2 to Theorem 1, we get the last statement of (3). Lemma 10. The measure 𝜇̂ is complete. Proof. The statement follows from Corollary 2 to Theorem 1 and Lemma 11 (3.1.4). Now, we shall clarify a structure of elements of Mf (T, R, 𝜇) and M𝜎f (T, R, 𝜇). ̂ f (T, R, 𝜇) iff M = E ∪ N for some E ∈ Rf (T, R, 𝜇) Lemma 11. Let M ⊂ T. Then, M ∈ M 𝜎 0 ̂ (T, R, 𝜇). and N ∈ M ̂ ̂ L. etc. without Proof. We shall denote M(T, R, 𝜇), L(T, R, 𝜇), etc. simply by M, f ̂ ̂ < ∞ using Lemma 5 we their arguments in round brackets. Let M ∈ M . From 𝜇M

3.1.5 Extension of a positive measure to a wide complete saturated measure | 217

̂ = sup{𝜇× L n | n ∈ 𝜔} for some sequence (L n ∈ Lf | n ∈ 𝜔) ↑ such that deduce that 𝜇M L n ⊂ M. By Corollary 4 to Theorem 1, L n = E n ∪ N n for some E n ∈ R𝜎f and N n ∈ L0 . Consider the sets E ≡ ⋃⟮E n | n ∈ 𝜔⟯ ∈ R𝜎 and K ≡ ⋃⟮N n | n ∈ 𝜔⟯. From E ⊂ M we deduce that E ∈ R𝜎f . By Corollary 2 to Theorem 1, K ∈ L0 . Consider the set F ≡ ̂ where ⋃⟮L n | n ∈ 𝜔⟯ = E ∪ K ∈ L. By Lemma 5 (3.1.1), 𝜇× F = sup{𝜇× L n | n ∈ 𝜔} = 𝜇M, ̂ 0 , we obtain M = E ∪ N. ̂ 𝜇(M\F) = 0. Setting N ≡ K ∪ (M\F) ∈ M ̂ 𝜎f (T, R, 𝜇) iff M = E ∪ N for some E ∈ Corollary 1. Let M ⊂ T. Then, M ∈ M 𝜎f 0 ̂ R𝜎 (T, R, 𝜇) and N ∈ M (T, R, 𝜇). ̂ 𝜎f , then M = ⋃⟮M | i ∈ I⟯ for some countable collection (M ∈ M ̂f | Proof. If M ∈ M i i i ∈ I). By Lemma 11, M i = E i ∪ N i for some E i ∈ R𝜎f and N i ∈ M0 . Then, E ≡ ⋃⟮E i | ̂0 i ∈ I⟯ ∈ R𝜎f 𝜎 . By Lemma 9 (3.1.1), N ≡ ⋃⟮N i | i ∈ I⟯ ∈ M . Finally, M = E ∪ N. ̂ f (T, R, 𝜇). Then, for every 𝜀 > 0, there is a set F ∈ Rf (T, R, 𝜇) Corollary 2. Let M ∈ M 𝛿 ̂ such that F ⊂ M and 𝜇(M\F) < 𝜀. ̂ 0 (T, R, 𝜇). Proof. By Lemma 11, M = E ∪ N for some E ∈ R𝜎f (T, R, 𝜇) and N ∈ M By Corollary 2 to Proposition 4 (2.1.1), E = ⋃⟮F i | i ∈ I⟯ for some countable pairwise disjoint collection (F i ∈ R𝛿 (T, R) | i ∈ I). It follows from 𝜇E = ∑net (𝜇F i | i ∈ I) that 𝜇E < 𝜇(⋃⟮F i | i ∈ J⟯) + 𝜀 for some finite subset J ⊂ I. Take F ≡ ⋃⟮F i | i ∈ J⟯ ∈ R𝛿 (T, R). ̂ f (T, R, 𝜇), [L ∈ Lf (T, R, 𝜇)]. Then, for every 𝜀 > 0 there is Proposition 2. Let M ∈ M f ̂ 󳵻 R) < 𝜀 [𝜇× (L 󳵻 R) < 𝜀]. a set R ∈ R (𝜇) such that 𝜇(M ̂0 Proof. By Lemma 11, M = E ∪ N for some E ∈ R𝜎f ⊂ H𝜎f (T, R, 𝜇) and N ∈ M (T, R, 𝜇). By the definitions of 𝜇∗ and 𝜇× , there is a sequence (R i ∈ R | i ∈ 𝜔) such that E ⊂ ⋃⟮R i | i ∈ 𝜔⟯ ≡ L and x ≡ ∑net (𝜇R i | i ∈ 𝜔) < 𝜇× E + 𝜀/2. According to Proposition 1 (1.4.8), x = ∑(𝜇R i | i ∈ 𝜔). Therefore, there is a number n such that y ≡ ∑(𝜇R i | i ∈ 𝜔\n) < 𝜀/2. Consider the set R ≡ ⋃⟮R i | i ∈ n⟯ ∈ R. Then, by Lemma 3 (3.1.1), ̂ ̂ 𝜇(M\R) = 𝜇(E\R) ⩽ 𝜇× (L\R) ⩽ 𝜇× (⋃⟮R i | i ∈ 𝜔\n⟯) ⩽ y < 𝜀/2, whereas 𝜇̂ extends 𝜇× . ̂ ̂ Similarly, by Lemma 2 (3.1.1) and Lemma 3 (3.1.1), 𝜇(R\M) = 𝜇(R\E) ⩽ 𝜇× (L\E) = 𝜇× L − × × f ̂ 󳵻 R) < 𝜀. For L , the proof is the same. 𝜇 E ⩽ x − 𝜇 E < 𝜀/2. Hence, 𝜇(M Now, we shall consider the most important cases of the large saturated extension. Lemma 12. Let 𝜇 be a positive measure on a ring R. Then, the Lebesgue – Caratheodory extension 𝜇× : L(T, R, 𝜇) → R+ of the measure 𝜇 coincides with the local completion 𝜇̄̃ : Kl (T, R𝜎 (T, R), 𝜇)̃ → R+ of the restriction 𝜇̃ ≡ 𝜇× |R𝜎 (T, R) of the measure 𝜇× on the 𝜎-ring R𝜎 (T, R) generated by the ring R.

218 | 3.1 Spaces with semimeasures and measures

Proof. We shall denote R𝜎 (T, R) and L(T, R, 𝜇) simply by R𝜎 and L. By Corollary 1 to Theorem 1, R𝜎 ⊂ L. Consequently, R0𝜎 (𝜇)̃ ⊂ L0 (𝜇× ). By virtue of Corollary 2 to Theorem 1, Nl (T, R𝜎 , 𝜇)̃ ⊂ L0 (𝜇× ). As a result, Kl (T, R𝜎 , 𝜇)̃ ⊂ L. Let L ∈ L. By Lemma 4, there is E ∈ R𝜎 such that L ⊂ E and 𝜇× L = 𝜇× E. Consider the set M ≡ E\L ∈ L. Then, 𝜇× N = 0, where N ∈ L0 (𝜇× ). Now, by Corollary 1 to Lemma 4, there is N0 ∈ R0𝜎 (𝜇)̃ such that M ⊂ N0 . Consider ̃ Then, L = (L\N 0 ) ∪ (L ∩ N0 ) = the sets H ≡ E\N0 ∈ R𝜎 and N ≡ L ∩ N0 ∈ Nl (T, R𝜎 , 𝜇). ̃ Thus, L ⊂ Kl (T, R𝜎 , 𝜇). ̃ ((E\M)\N0 ) ∪ N = H ∪ N ∈ Kl (T, R𝜎 , 𝜇). Corollary 1. Let 𝜇 be a positive measure on a 𝜎-ring R. Then, the Lebesgue – Caratheodory extension 𝜇× : L(T, R, 𝜇) → R+ of the measure 𝜇 coincides with the local completion 𝜇̄ : Kl (T, R, 𝜇) → R+ of 𝜇. Corollary 2. Let 𝜇 be a locally complete positive measure on a 𝜎-ring R. Then, ̂ L(T, R, 𝜇) = R, and so the large saturated extension 𝜇̂ : M(T, R, 𝜇) → R+ of the mea󸀠 sure 𝜇 coincides with the saturation 𝜇 : U(T, R) → R+ of 𝜇. Lemma 13. Let 𝜇 be a positive measure on a ring R such that T = ⋃⟮R i | i ∈ I⟯ for some ̂ countable collection (R i ∈ R | i ∈ I). Then, M(T, R, 𝜇) = L(T, R, 𝜇) and the large ̂ complete saturated extension 𝜇̂ : M(T, R, 𝜇) → R+ of the measure 𝜇 coincides with the Lebesgue – Caratheodory extension 𝜇× : L(T, R, 𝜇) → R+ of 𝜇. Proof. It is clear that R𝜎 (T, R) contains T, and therefore, it is a 𝜎-algebra. Hence, L(T, R, 𝜇) is a 𝜎-algebra as well. Then, Corollary 3 to Lemma 15 (2.1.1) implies ̂ that M(T, R, 𝜇) ≡ U(T, L(T, R, 𝜇)) = L(T, R, 𝜇). This means that the measure 𝜇× is saturated, i. e. 𝜇̂ = (𝜇× )󸀠 = 𝜇× . Corollary 1. Let A be a 𝜎-algebra and 𝜇 a positive measure on A. Then, the large sat̂ urated extension 𝜇̂ : M(T, A, 𝜇) → R+ of the measure 𝜇 coincides with the Lebesgue – Caratheodory extension 𝜇× : L(T, A, 𝜇) → R+ of 𝜇 and with the local completion 𝜇̄ : Kl (T, A, 𝜇) → R+ of 𝜇. Proof. It follows from Lemma 13 and Corollary 1 to Lemma 12. Corollary 2. Let A be a 𝜎-algebra and 𝜇 a locally complete positive measure on A. ̂ Then, M(T, A, 𝜇) = A and 𝜇̂ = 𝜇. ̂ ̂ Corollary 3. Let 𝜇 be a positive measure on a ring R. Then, M(T, M(T, R, 𝜇), 𝜇)̂ = ̂ M(T, R, 𝜇). ̂ Proof. Applying Corollaries 1 and 2 to the measure 𝜇̂ on the 𝜎-algebra M(T, R, 𝜇), we obtain the necessary equality.

3.1.5 Extension of a positive measure to a wide complete saturated measure | 219

Thus, if we repeat the method of the extension of a positive measure 𝜇 starting from the large complete saturated extension 𝜇,̂ we obtain no new measure.

The large complete strongly saturated extension of a positive measure If the Lebesgue – Caratheodory extension 𝜇× : L(T, R, 𝜇) → R+ of the measure 𝜇 : R → R+ is internally finite, then we can consider some larger then 𝜇̂ extension ̂ ̂ of 𝜇 taking the strong saturation M(T, R, 𝜇) ≡ Us (T, L(T, R, 𝜇), 𝜇× ) of the 𝜎-ring L(T, R, 𝜇) with respect to the measure 𝜇× and the strong saturation 𝜇̂̂ ≡ (𝜇× )󸀠󸀠 of the measure 𝜇× . For instance, if 𝜇 is 𝜎-finite, then by Proposition 1, 𝜇× is 𝜎-finite, and therefore, internally finite.

Proposition 3. Let 𝜇 be a positive measure on a ring R such that its Lebesgue – Caratheodory extension 𝜇× : L(T, R, 𝜇) → R+ is internally finite. Then, ̂ ̂ 1) M(T, R, 𝜇) is a 𝜎-algebra; ̂ ̂ 2) 𝜇̂̂ : M(T, R, 𝜇) → R+ is a wide positive complete strongly saturated measure extending the measure 𝜇× ; ̂ ̂ ̂̂ = sup{𝜇× (M ∩ L) | L ∈ Lf (T, R, 𝜇)} = sup{𝜇× L | L ∈ 3) if M ∈ M(T, R, 𝜇), then 𝜇M Lf (T, R, 𝜇) ∧ L ⊂ M}; ̂ ̂ 4) M ∈ M(T, R, 𝜇) iff M ∩ R ∈ L(T, R, 𝜇) for every R ∈ Rf (𝜇). Proof. Assertions 1 and 2 and the first equality in (3) follow from Theorem 1, its Corollary 2, Theorem 6 (3.1.4), Lemma 11 (3.1.4), and the definition of the strong saturation of a measure. Check the second equality in (3). It is clear that x ≡ sup{𝜇× L | L ∈ Lf ∧ L ⊂ M} = ̂̂ sup{𝜇× (M ∩ L) | L ∈ Lf ∧ L ⊂ M} ⩽ 𝜇M. Further, if L ∈ Lf , then M ∩ L ∈ Lf and × ̂ ̂ ⩽ x. M ∩ L ⊂ M imply 𝜇 (M ∩ L) ⩽ x, where 𝜇M 4. Let M ∩ R ∈ L(T, R, 𝜇) for every R ∈ Rf (𝜇). Take any L ∈ Lf (T, R, 𝜇). By the definition, of 𝜇∗ and 𝜇× for the set L there is a countable collection (R i ∈ R | i ∈ I) such that L ⊂ ⋃⟮R i | i ∈ I⟯ and 𝜇× L ⩽ ∑net (𝜇R i | i ∈ I) < ∞. Hence, R i ∈ Rf (𝜇) for every i ∈ I. By virtue of Theorem 1, M ∩ L = (⋃⟮M ∩ R i | i ∈ I⟯) ∩ L ∈ L(T, R, 𝜇). ̂ ̂ This means that M ∈ M(T, R, 𝜇). The inverse statement follows from the definition ̂ ̂ of M. ̂ ̂ Corollary 1. B(T, R) ⊂ M(T, R, 𝜇). Proof. The proof is the same as the proof of Corollary 1 to Lemma 5. Corollary 2. Let 𝜇 be a positive measure on a ring R such that 𝜇× is internally finite. Then, 𝜇̂̂ = (𝜇f× )󸀠 where 𝜇f× ≡ 𝜇× |Lf (T, R, 𝜇).

220 | 3.1 Spaces with semimeasures and measures

Proof. It follows from Corollary 1 to Theorem 6 (3.1.4). ̂ ̂ The measure 𝜇̂̂ : M(T, R, 𝜇) → R+ will be called the large complete strongly saturated extension of the measure 𝜇 : R → R+ . ̂ ̂ For the ensemble M(T, R, 𝜇) and the measure 𝜇,̂̂ consider the following subensembles: ̂ ̂ ̂ ̂̂ = 0}, ̂ 0 (T, R, 𝜇) ≡ M(T, ̂ ̂ – M R, 𝜇)0 (𝜇)̂̂ = {M ∈ M(T, R, 𝜇) | 𝜇M ̂ ̂ ̂ f f ̂ ̂ ̂ ̂ ̂ ̂ < ∞}. – M (T, R, 𝜇) ≡ M(T, R, 𝜇) (𝜇)̂ = {M ∈ M(T, R, 𝜇) | 𝜇M ̂ ̂ ̂ ̂ ̂ 0 (T, R, 𝜇), and co-M ̂ 0 (T, R, 𝜇) will be ̂ ̂ f (T, R, 𝜇), M The elements of M(T, R, 𝜇), M ̂̂ ̂̂ ̂̂ ̂̂ called 𝜇-measurable, 𝜇-squarable, 𝜇-negligible, and 𝜇-conegligible, respectively. The following assertion is very important. Lemma 14. Let 𝜇 be a 𝜎-finite positive measure on a ring R. Then, the large complete ̂ saturated extension 𝜇̂ : M(T, R, 𝜇) → R+ and the large complete strongly saturated ̂ ̂ ̂ extension 𝜇̂ : M(T, R, 𝜇) → R of 𝜇 coincide. +

̂ ̂ and Corollary 1 to Lemma 6. Proof. It follows from the definition of M Now, we shall give some helpful sufficient condition for the measure 𝜇× to be internally finite. Lemma 15. Let R be a 𝜎-ring and 𝜇 be an internally finite positive measure on R. Then, 𝜇× is internally finite as well. ̄ = ∞. Then, Proof. By Corollary 1 to Lemma 12, 𝜇× = 𝜇.̄ Let K ∈ Kl (T, R, 𝜇) and 𝜇K ̄ = 𝜇R. By K = R ∪ N for some R ∈ R and N ∈ Nl (T, R, 𝜇). By Proposition 1 (3.1.4), 𝜇K f the condition, there is S ∈ R (𝜇) such that S ⊂ R and 𝜇S > 0. Since 𝜇̄ extends 𝜇, we get ̄ S ⊂ K and 𝜇S ̄ > 0. S ∈ Kl (T, R, 𝜇)f (𝜇), Consider now the most important cases of the large strongly saturated extension. Lemma 16. 1) If 𝜇 is an internally finite positive measure on a 𝜎-ring R, then 𝜇̂̂ = 𝜇󸀠󸀠̄ . 2) If 𝜇 is a locally complete internally finite positive measure on a 𝜎-ring R, then 𝜇̂̂ = 𝜇󸀠󸀠 . 3) If 𝜇 is a locally complete internally finite strongly saturated positive measure on a 𝜎-ring R, then 𝜇̂̂ = 𝜇. Proof. 1. By Corollary 1 to Lemma 12, 𝜇× = 𝜇.̄ By Lemma 15, 𝜇× is internally finite. As a result, 𝜇̂̂ = (𝜇× )󸀠󸀠 = 𝜇󸀠󸀠̄ . 2. By the condition, 𝜇̄ = 𝜇. Then, by (1), 𝜇̂̂ = 𝜇󸀠󸀠 . 3. By the condition, 𝜇󸀠󸀠 = 𝜇. Hence, by (2), 𝜇̂̂ = 𝜇.

3.1.5 Extension of a positive measure to a wide complete saturated measure | 221

Corollary 1. Let 𝜇 be a positive measure on a ring R such that 𝜇× is internally finite. ̂ ̂ ̂ ̂ ̂ ̂ Then, M(T, M(T, R, 𝜇), 𝜇)̂̂ = M(T, R, 𝜇). Proof. By Theorem 6 (3.1.4) the measure 𝜇̂̂ ≡ (𝜇× )󸀠󸀠 is internally finite. By Proposition 3 and Corollary 1 to Lemma 3 (3.1.4), 𝜇̂̂ is a locally complete strongly saturated positive measure. Therefore, by Lemma 16, we get the necessary equality. Thus, if we repeat the method of the extension of a positive measure 𝜇 with the internally finite 𝜇× starting from the large complete strongly saturated extension 𝜇̂̂ we obtain no new measure. We shall prove now for the extension 𝜇̂̂ the analogues of Lemma 11 and Proposition 2 for the extension 𝜇.̂ Lemma 17. Let 𝜇 be a positive measure on a ring R such that 𝜇× is internally finite. ̂ ̂ f (T, R, 𝜇) iff M = E ∪ N for some E ∈ Rf (T, R, 𝜇) and N ∈ Let M ⊂ T. Then, M ∈ M 𝜎 ̂ ̂ 0 (T, R, 𝜇). M ̂ ̂ ̂ ̂ L, etc. without Proof. We shall denote M(T, R, 𝜇), L(T, R, 𝜇), etc. simply by M, ̂ f ̂ their arguments in round brackets. Let M ∈ M . By virtue of Proposition 3 there ̂̂ = sup{𝜇× L | n ∈ 𝜔}. is some sequence (L ∈ Lf | n ∈ 𝜔) ↑ such that L ⊂ M and 𝜇M n

n

n

Further, the proof is the same as the proof of Lemma 11. ̂ ̂ Corollary 1. In the conditions of Lemma 17 for every M ∈ M(T, R, 𝜇) and every 𝜀 > 0, f ̂ ̂ there is a set F ∈ R𝜎 (T, R, 𝜇) such that F ⊂ M and 𝜇(M\F) < 𝜀. The proof is the same as the proof of Corollary 2 to Lemma 11. Proposition 4. Let 𝜇 be a positive measure on a ring R such that 𝜇× is internally ̂ ̂ finite. If M ∈ M(T, R, 𝜇), then for every 𝜀 > 0 there is a set R ∈ Rf (𝜇) such that ̂̂ 󳵻 R) < 𝜀. 𝜇(M Proof. By Lemma 17, M = E ∪ N for some E ∈ R𝜎f (T, R, 𝜇) ⊂ H𝜎f (T, R, 𝜇) and N ∈ ̂ ̂ 0 (T, R, 𝜇). Further, the proof is the same as the proof of Proposition 2. M Lemma 18. Let 𝜇 be a positive measure on a ring R such that 𝜇× is internally finite. For ̂ ̂ a set M ∈ M(T, R, 𝜇), the following conditions are equivalent: ̂0 ̂ 1) M ∈ M (T, R, 𝜇); 2) L ∈ L0 (T, R, 𝜇) for every L ∈ Lf (T, R, 𝜇) such that L ⊂ M; 3) M ∩ R ∈ L0 (T, R, 𝜇) for every R ∈ Rf (𝜇). Proof. (1) ⊢ (2). This implication follows from assertion 3 of Proposition 3.

222 | 3.1 Spaces with semimeasures and measures

(2) ⊢ (3). If R ∈ Rf (𝜇), then L ≡ M ∩ R ∈ Lf (T, R, 𝜇) and L ⊂ M. Therefore, by (2), L ∈ L0 (T, R, 𝜇). (3) ⊢ (1). Condition 3. By virtue of assertion 4 of Proposition 3 implies M ∈ ̂ ̂ M(T, R, 𝜇). Take any L ∈ Lf (T, R, 𝜇). By the definition of 𝜇∗ and 𝜇× for the set L,

there is a countable collection (R i ∈ R | i ∈ I) such that L ⊂ ⋃⟮R i | i ∈ I⟯ and 𝜇× L ⩽ ∑net (𝜇R i | i ∈ I) < ∞. Hence, R i ∈ Rf (𝜇) for every i ∈ I. By virtue of (3) and Corollary 2 to Theorem 1, M ∩ L = (⋃⟮M ∩ R i | i ∈ I⟯) ∩ L ∈ L0 (T, R, 𝜇). By assertion 3 of ̂̂ = 0. Proposition 3, 𝜇M 3.1.6 Properties of the extended Borel – Lebesgue measure on ℝn The most famous example of an extension of a measure was considered by H. Lebesgue in the beginning of the 20th century. Take the initial Borel – Lebesgue measure 𝜆 : R(Rn , Spar ) → R+ from Example 1 (3.1.4). Consider, according to 3.1.5, for 𝜆, the Lebesgue – Caratheodory extension 𝜆× : L(Rn , R(Rn , Spar ), 𝜆) → R+ and the large saturated extension ̂ n , R(Rn , S ), 𝜆) → R . 𝜆̂ : M(R par + Since 𝜆 is finite, we infer by virtue of Proposition 1 (3.1.5) that 𝜆× is 𝜎-finite, and therefore, internally finite. Hence, we may consider for 𝜆 also the large complete ̂ ̂ ̂ n , R(Rn , S ), 𝜆) → R . strongly saturated extension 𝜆̂ : M(R par

+

Further, we shall denote the ensembles R(Rn , Spar ), L(Rn , R(Rn , Spar ), 𝜆), ̂ ̂ ̂ n , R(Rn , S ), 𝜆) and M(R ̂ n , R(Rn , S ), 𝜆) simply by R , L , M ̂ and M ̂ , M(R par par 𝜆 𝜆 𝜆 𝜆 respectively. ̂ By Lemma 14 (3.1.5), 𝜆̂ coincides with 𝜆.̂ By virtue of Lemma 13 (3.1.5), 𝜆̂ coincides ̂ ̂ ̂ =M ̂ , and therefore, we shall consider the meawith 𝜆× . Thus, 𝜆̂ = 𝜆̂ = 𝜆× , L𝜆 = M 𝜆 𝜆 × sure 𝜆 : L𝜆 → R+ . Besides, we see that L𝜆 is a 𝜎-algebra and 𝜆× is complete and strongly saturated. The measure 𝜆× is called the extended Borel – Lebesgue measure on Rn and elements of L𝜆 is called Borel – Lebesgue measurable subsets of Rn . Let P|s, t| ≡ ∏⟮|a i , b i | | i ∈ n⟯, where s ≡ (a i ∈ R | i ∈ n) and t ≡ (b i ∈ R | i ∈ n) such that a i ⩽ b i , be a parallelepiped of general kind in Rn (see 2.1.1). The set S|s, t| ≡ P[s, t]\P]s, t[∈ H𝜎 (Rn , R𝜆 ) is called the surface (or boundary) of the parallelepiped P|s, t|. It consists of (n − 1)-dimensional faces F|s, t|. For every face F|s, t| ∈ H𝜎 (Rn , R𝜆 ) of P|s, t| and every k ∈ N, there is a half-open parallelepiped P k containing F|s, t| such that v(P k ) < 1/k. Therefore, 𝜆∗ (F|s, t|) = 0. It follows from Lemmas 1 (3.1.5) and 3 (3.1.1) that 𝜆∗ (S|s, t|) = 0. Hence, every face F|s, t| and the whole surface S|s, t| are contained in H𝜎0 (Rn , R𝜆 , 𝜆) (see 3.1.5). By Corollary 2 to Theorem 1 (3.1.5), every F|s, t| and S|s, t| are contained in L0𝜆 . This means that every face F|s, t| and the whole surface S|s, t| are Borel – Lebesgue measurable and 𝜆× -negligible.

3.1.6 Properties of the extended Borel – Lebesgue measure on Rn

|

223

If E is any subset of S|s, t|, then by Corollary 2 to Theorem 1 (3.1.5), E ∈ L0𝜆 , i. e. E is also Borel – Lebesgue measurable and 𝜆× -negligible. If follows from these facts that every parallelepiped P|s, t| is Borel – Lebesgue measurable. Thus, we have proven the first statement of the following Lemma 1. 1) Every parallelepiped P|s, t| in Rn is Borel – Lebesgue measurable. 2) R𝜎 (Rn , Spar ) = R𝜎 (Rn , R𝜆 ) = B(Rn , Spar ) = B(Rn , R𝜆 ) = B(Rn , Ost ) ⊂ L𝜆 , where Ost is the standard topology on Rn . 3) For every L ∈ L𝜆f and every 𝜀 > 0, there is a finite pairwise disjoint collection (P i ∈ Spar | i ∈ I) of half-open parallelepipeds such that 𝜆× (L 󳵻 ⋃⟮P i | i ∈ I⟯) < 𝜀. Proof. According to Statement 1 the countable ensemble of all open parallelepipeds P]s, t[ with the main rational corner points s, t ∈ Qn is a countable base of the topology Ost . Consider the 𝜎-ring R𝜎 ≡ R𝜎 (Rn , Spar ) and the 𝜎-algebra B ≡ B(Rn , Ost ). Every parallelepiped P]s, t[ is a union ⋃⟮P i [s i , t[| i ∈ I⟯ of some countable collection (P i [s i , t[∈ Spar | i ∈ I). Therefore, P]s, t[∈ R𝜎 . Hence, Ost ⊂ R𝜎 and R𝜎 is a 𝜎-algebra. Since L𝜆 is a 𝜎-algebra, we conclude from the proven inclusion Spar ⊂ L𝜆 that Ost ⊂ R𝜎 = B(Rn , Spar ) ⊂ L𝜆 and consequently, B ⊂ R𝜎 . Every parallelepiped P[s, t[ is an intersection ⋂⟮P j ]s j , t[| j ∈ J⟯ of some countable collection (]s j , t[∈ Ost | j ∈ J). Therefore, P[s, t[∈ B, where Spar ⊂ B and (R𝜆 )𝜎 ⊂ B. As a result, we get B = R𝜆𝜎 = B(Rn , Spar ) ⊂ L𝜆 . By Corollary 1 to Lemma 10 (2.1.1), R𝜎 (Rn , R𝜆 ) = R𝜎 . Therefore, R𝜎 (Rn , R𝜆 ) is a 𝜎-algebra and hence R𝜎 (Rn , R𝜆 ) = B(Rn , R𝜆 ). The third statement follows from Proposition 2 (3.1.5) and Theorem 4 (2.1.1). By virtue of Lemma 1, we can consider the restriction 𝜆̃ ≡ 𝜆× |B(Rn , Ost ). The measure 𝜆̃ : B(Rn , Ost ) → R+ is called the Borel measure on Rn . Lemma 2. 1) The local completion 𝜆̄̃ : Kl (Rn , B(Rn , Ost ), 𝜆)̃ → R+ of the Borel measure 𝜆̃ coincides with the extended Borel – Lebesgue measure 𝜆× . ̂ 2) 𝜆̂̃ = 𝜆̂̃ = 𝜆.̄̃ Proof. 1. The assertion follows from Lemma 12 (3.1.5) and Lemma 1. 2. By virtue of Corollary 1 to Lemma 13 (3.1.5), 𝜆̂̃ = (𝜆)̃ × = 𝜆.̄̃ Besides, by ̂ Lemma 15 (3.1.5), 𝜆̂̃ = 𝜆.̂̃ Further on, we shall denote the ensemble co-Ost of all closed sets in the topological space (Rn , Ost ) by F.

224 | 3.1 Spaces with semimeasures and measures

Lemma 3. Let R ∈ R𝜆 . Then, for every 𝜀 > 0 there exist sets R󸀠 , R󸀠󸀠 ∈ R𝜆 , F ∈ F and G ∈ Ost such that R󸀠 ⊂ F ⊂ R ⊂ G ⊂ R󸀠󸀠 and 𝜆(R󸀠󸀠 \R󸀠 ) < 𝜀. Proof. According to Example 1 (3.1.4), R = ⋃⟮P𝛼 | 𝛼 ∈ A⟯ for some finite pairwise disjoint collection (P𝛼 ∈ Spar | 𝛼 ∈ A). Let p = card A. By Lemma 1 (3.1.4) for every parallelepiped P𝛼 , there exist half-open parallelepipeds P󸀠𝛼 and P󸀠󸀠𝛼 , a closed parallelepiped F𝛼 and an open parallelepiped G𝛼 such that P󸀠𝛼 ⊂ F𝛼 ⊂ P𝛼 ⊂ G𝛼 ⊂ P󸀠󸀠𝛼 and vP󸀠󸀠𝛼 − vP󸀠𝛼 < 𝜀/p. Consider the sets F ≡ ⋃⟮F𝛼 | 𝛼 ∈ A⟯ ∈ F, G ≡ ⋃⟮G𝛼 | 𝛼 ∈ A⟯ ∈ Ost , R󸀠 ≡ ⋃⟮P󸀠𝛼 | 𝛼 ∈ A⟯ ∈ R𝜆 , and R󸀠󸀠 = ⋃⟮P󸀠󸀠𝛼 | 𝛼 ∈ A⟯ ∈ R𝜆 . We have R󸀠 ⊂ F ⊂ R ⊂ G ⊂ R󸀠󸀠 . Since the collection (P󸀠𝛼 | 𝛼 ∈ A) is pairwise disjoint, we have 𝜆R󸀠 = ∑(vP󸀠𝛼 | 𝛼 ∈ A). Besides, 𝜆R󸀠󸀠 ⩽ ∑(𝜆P󸀠󸀠𝛼 | 𝛼 ∈ A) = ∑(vP󸀠󸀠𝛼 | 𝛼 ∈ A). As a result, 𝜆(R󸀠󸀠 \R󸀠 ) = 𝜆R󸀠󸀠 − 𝜆R󸀠 ⩽ ∑(vP󸀠󸀠𝛼 − vP󸀠𝛼 | 𝛼 ∈ A) < p(𝜀/p) = 𝜀. The subensemble of F, consisting of all compact sets, will be denoted by C. Theorem 1. Let L ∈ L𝜆 . Then, 𝜆× L = sup{𝜆× K | K ∈ C ∧ K ⊂ L}. Proof. Let M ∈ T\L. Consider the parallelepiped P k ≡] − k, k[n and the sets M k ≡ M ∩ P k and L k ≡ L ∩ P k . According to the definitions of the outer evaluation 𝜆∗ and the Lebesgue – Caratheodory extension 𝜆× for 𝜀 > 0, there is a sequence (R n ∈ R𝜆 | n ∈ 𝜔) such that M k ⊂ ⋃⟮R n ∩ P k | n ∈ 𝜔⟯ ≡ E k and 𝜆× M k ⩽ ∑net (𝜆(R n ∩ P k ) | i ∈ 𝜔) < 𝜆× M k + 𝜀/4. By Lemma 3 for every R n , there is sets G n ∈ Ost and R󸀠󸀠n ∈ R𝜆 such that R n ⊂ G n ⊂ R󸀠n and 𝜆(R󸀠n \R n ) < 𝜀/2n+3 . Consider the sets R kn ≡ P k ∩ R n ∈ R𝜆 , G kn ≡ P k ∩ G n ∈ Ost , R󸀠kn ≡ P k ∩ R󸀠n ∈ R𝜆 . Then, R kn ⊂ G kn ⊂ R󸀠kn and 𝜆(R󸀠kn \R kn ) < 𝜀/2n+3 . Consider also the sets G k ≡ ⋃⟮G kn | n ∈ 𝜔⟯ ∈ Ost and M 󸀠k ≡ ⋃⟮R󸀠kn | n ∈ 𝜔⟯ ∈ L𝜆 . From M 󸀠k \E k ⊂ ⋃⟮R󸀠kn \R kn | n ∈ 𝜔⟯, we infer by virtue of Lemma 3 (3.1.1) that 𝜆× (M 󸀠k \M k ) = 𝜆× (M 󸀠󸀠k \E k ) + 𝜆× (E k \M k ) ⩽ ∑(𝜆(R󸀠kn \R kn ) | n ∈ 𝜔) + 𝜆× (E k \M k ). By Lemma 3 (1.4.8), ∑net (𝜀/2n+3 | n ∈ 𝜔) = ∑(𝜀/2n+3 | n ∈ 𝜔) = 𝜀/4. Thus, 𝜆× (M 󸀠k \M k ) ⩽ 𝜀/4 + 𝜆× E k − 𝜆× M k . By Proposition 1 (3.1.1), ∑net (𝜆R kn | n ∈ 𝜔) = 𝜆× E k . As a result, 𝜆× (M 󸀠k \M k ) < 𝜀/2. Besides, M k ⊂ G k ⊂ M 󸀠k . Take the sets H k ≡ P k \G k and L󸀠k ≡ P k \M 󸀠k . Then, L󸀠k ⊂ H k ⊂ L k and 𝜆× L k − 𝜆× H k ⩽ × 𝜆 L k − 𝜆× L󸀠k = 𝜆× P k − 𝜆× M k − (𝜆× P k − 𝜆× M 󸀠k ) < 𝜀/2. Consider the closed parallelepipeds P ki ≡ [−k + 1/i, k − 1/i] and the closed sets F ki ≡ P ki \G k ⊂ L k . Then, P k = ⋃(P ki | i ∈ N) implies H k = ⋃⟮F ki | i ∈ N⟯. Since the sequence (F ki | i ∈ N) increases, we infer by Lemma 5 (3.1.1) that 𝜆× H k = lim (𝜆× F ki | i ∈ N). Therefore, there is i such that 0 ⩽ 𝜆× H k −𝜆× F ki < 𝜀/2. Finally, 𝜆× L k −𝜆× F ki < 𝜀. By Statement 1 (3.1.4), the bounded closed set F ki is compact. This means that 𝜆× L k = sup{𝜆× K | K ∈ C ∧ K ⊂ L k }. Since the sequence (L k | k ∈ N) increases, we have 𝜆× L = sup{𝜆× L k | k ∈ N}. As a result, 𝜆× L = sup{𝜆× K | K ∈ C ∧ K ⊂ L}.

3.1.6 Properties of the extended Borel – Lebesgue measure on Rn

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225

In conclusion, we shall prove some new delicate property of the measure 𝜆× . Lemma 4. Every subensemble E of L𝜆 consisting of pairwise disjoint sets E with 𝜆× E > 0 is countable. Proof. Consider for k ∈ N the parallelepiped P k ≡ [−k, k]n and the ensembles Ek ≡ {E ∈ E | 𝜆× (E ∩ P k ) > 0} and Lk ≡ {L ∈ L𝜆 | L ⊂ P k }. Since 𝜆P k = (2k)n ≡ m k , we have 𝜆× L ⩽ m k for every L ∈ Lk . By Lemma 9 (3.1.4) the subensemble E󸀠k ≡ {E ∩ P k | E ∈ Ek } of Lk is countable. Since E ∩ F = ⌀ for every E = / F from Ek , we have E ∩ P k = / F ∩ Pk . This means that the mapping u : Ek → E󸀠k such that uE ≡ E ∩ P k is bijective. Hence, Ek is countable as well. Let E ∈ E. Since the sequence (E ∩ P k | k ∈ N) increases to E, we infer by Lemma 5 (3.1.1) that 𝜆× E = lim(𝜆× (E ∩ P k ) | k ∈ N). From 𝜆× E > 0, we deduce that there is k such that 𝜆× (E ∩ P k ) > 0, where E ∈ Ek . As a result, E = ⋃⟮Ek | k ∈ N⟯. By Theorem 1 (1.3.9), the set E is countable. Now, we shall formulate once more some important properties of the extended Borel – Lebesgue measure 𝜆× : 1) the measure 𝜆× is wide, complete, internally finite, and strongly saturated; 2) Ost ⊂ L𝜆 and C ⊂ L𝜆 ; 3) for every point t ∈ Rn there is an open G ∈ Ost such that t ∈ G and 𝜆× G < ∞; 4) 𝜆× L = sup{𝜆× K | K ∈ C ∧ K ⊂ L}. In the sequel, these properties will be put in the base of the definition of Radon measures in Hausdorff topological spaces.

Auxiliary definitions and statements 1∘ Consider the standard topology Ost on Rn consisting of all sets G ⊂ Rn such that for every r ∈ G, there is an open parallelepiped P]s, t[ (see Example 2 (2.1.1)) such that r ∈ P]s, t[⊂ G. It is clear that every open parallelepiped P]s, t[ and every open halfbounded parallelepiped P]s, → [≡ ∏⟮]a i , → [| i ∈ n⟯ or P] ←, t[≡ ∏⟮] ←, b i [| i ∈ n⟯ for any main corner points s ≡ (a i | i ∈ n) and t ≡ (b i | i ∈ n) such that a i ⩽ b i is an open set. Denote by Opar the ensemble of all open parallelepipeds in Rn . Statement 1. 1) The ensemble Opar is a base of the topology Ost . 2) The ensemble of all open parallelepipeds in the set Rn with the rational main corner points is a countable base of Ost .

226 | 3.2 Decompositions of semimeasures and measures

3.2 Decompositions of semimeasures and measures 3.2.1 The Hahn and Jordan decompositions of measures on a 𝛿-ring Let T be a set and R be a ring on T. A finite semimeasure 𝜇 : R → R will be called overfinite if its variation 𝜇̄̄ : P(T) → R+ (see 3.1.3) takes on R only finite real values, ̄̄ ∈ R+ for every R ∈ R. The subfamily of the family SMeasf (T, R) [Measf (T, R)] i. e. 𝜇R consisting of all overfinite semimeasures [measures] will be denoted by SMeasof (T, R) [Measof (T, R)]. Lemma 1. Let R be a ring on a set T. Then, SMeasof (T, R) and Measof (T, R) are ordered linear spaces. Proof. It follows from Corollary 2 to Lemma 1 (3.1.2) and Lemma 10 (3.1.3). By Corollary 1 to Lemma 8 (3.1.3) for a bounded semimeasure 𝜇 : R → [a, b] ⊂ R, we ̄̄ have 𝜇(T) < ∞. By virtue of Lemma 4 (3.1.3), a bounded semimeasure is overfinite. Thus, SMeasb (T, R) ⊂ SMeasof (T, R) and Measb (T, R) ⊂ Measof (T, R). It is evident that SMeasb (T, R) [Measb (T, R)] is an ordered linear subspace of SMeasof (T, R) [Measof (T, R)]. By Corollary 2 to Lemma 7 (3.1.3), every finite positive semimeasure 𝜇 is overfinite ̄̄ = 𝜇. because 𝜇|R Lemma 2. Let 𝜇 ∈ SMeasf (T, R) [Measf (T, R)] and 𝜇̄̄ be its variation. Then, the following conditions are equivalent: 1) 𝜇 ∈ SMeasof (T, R) [Measof (T, R)]; ̄̄ ∈ SMeasof (T, R)0 [Measof (T, R)0 ]; 2) 𝜇|R 3) there is 𝜈 ∈ SMeasf (T, R)0 [Measf (T, R)0 ] such that 𝜇 ⩽ 𝜈; 4) 𝜇 = 𝜇1 − 𝜇2 for some 𝜇1 , 𝜇2 ∈ SMeasf (T, R)0 [Measf (T, R)0 ]. ̄̄ Proof. (1) ⊢ (2). By Corollary 1 to Proposition 2 (3.1.3), v(𝜇) ≡ 𝜇|R is a semimeasure ̄ ̄ ̄̄ implies its ̄ [measure]. By Lemma 9 (3.1.3), var(𝜇|R) = 𝜇.̄ Hence, the finiteness of 𝜇|R overfiniteness. ̄̄ ∈ SMeasof (T, R)0 ⊂ (2) ⊢ (1). By Lemma 4 (3.1.3), it follows from 𝜇|R ̄ ̄ < ∞ for every R ∈ R. Therefore, 𝜇 ∈ SMeasf (T, R). SMeasf (T, R)0 that |𝜇R| ⩽ 𝜇R The same is valid for measures. ̄̄ (2) ⊢ (3). By Lemma 4 (3.1.3), 𝜇 ⩽ 𝜇|R. (3) ⊢ (4). Take 𝜇1 ≡ 𝜈 ∈ SMeasf (T, R)0 and 𝜇2 ≡ 𝜈 − 𝜇 ⩾ 0. By Corollary 1 to Lemma 1 (3.1.2), 𝜇2 is a finite semimeasure [measure]. Thus, we have the necessary decomposition. (4) ⊢ (1). Consider the evaluation 𝜈 ≡ 𝜇1 +𝜇2 . By Corollary 1 to Lemma 1 (3.1.2), 𝜈 is a finite positive semimeasure [measure]. Hence, by Lemma 4 (3.1.1), 𝜈 is increasing. From

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the inequalities −𝜈 ⩽ −𝜇2 ⩽ 𝜇 ⩽ 𝜇1 ⩽ 𝜈, we infer that |𝜇R| ⩽ 𝜈R for every R ∈ R. Let S ∈ ̄̄ ⩽ 2 sup{|𝜇S| | R and S ⊂ R. Then, |𝜇S| ⩽ 𝜈S ⩽ 𝜈R. Therefore, by Lemma 8 (3.1.3), 𝜇R S ∈ R ∧ S ⊂ R} ⩽ 2𝜈R < ∞ for every R ∈ R. The main aim of this subsection is to prove that every finite measure on a 𝛿-ring is overfinite. The key tool for it will be a Hahn decomposition. Let 𝜇 be a fixed measure on a ring R. A set R ∈ R is called essentially positive for 𝜇 if 𝜇S ⩾ 0 for every S ∈ R such that S ⊂ R. If R is essentially positive, then 1. 𝜇R ⩾ 0; 2. 𝜇S1 ⩽ 𝜇S2 for every S1 , S2 ∈ R such that S1 ⊂ S2 ⊂ R, i. e. 𝜇 is increasing on the ensemble {S ∈ R | S ⊂ R} = {P ∩ R | P ∈ R} ≡ RR (see 2.5.1); 3. S is essentially positive for every S ∈ R such that S ⊂ R. A set R ∈ R is called essentially negative for 𝜇 if 𝜇S ⩽ 0 for every S ∈ R such that S ⊂ R. If R is essentially negative, then 1. 𝜇R ⩽ 0; 2. 𝜇S1 ⩾ 𝜇S2 for every S1 , S2 ∈ R such that S1 ⊂ S2 ⊂ R, i. e. 𝜇 is decreasing on the ensemble RR ; 3. S is essentially negative for every S ∈ R such that S ⊂ R. Lemma 3. Let 𝜇 be a measure on a ring R, (R i ∈ R | i ∈ I) be a countable collection of essentially positive sets for 𝜇 and R ≡ ⋃⟮R i | i ∈ I⟯ ∈ R. Then, R is essentially positive. I. Consider the sets S k ≡ R u(k) \ ⋃⟮R u(i) | i ∈ k⟯ for Proof. Take a bijection u : 𝜔 every k ∈ 𝜔. The sets S k are essentially positive, mutually disjoint, belong to R and ⋃⟮S k | k ∈ 𝜔⟯ = R. Let S ∈ R and S ⊂ R. Then, S = R ∩ S = ⋃⟮S k ∩ S | k ∈ 𝜔⟯. Consequently, 𝜇S = ∑(𝜇(S k ∩ S) | k ∈ 𝜔) ⩾ 0 by virtue of Corollary 3 to Proposition 1 (1.4.7), Proposition 1 (1.4.8), and the definition from 1.4.8. Therefore, S is essentially positive. Theorem 1 (Hahn). Let 𝜇 be a measure on a 𝛿-ring R. Then, for every set R ∈ R, there are two disjoint sets R+ , R− ∈ R such that R+ is essentially positive for 𝜇, R− is essentially negative, and R = R+ ∪ R− . Proof. Without loss of generality, we can assume that rng 𝜇 ⊂] − ∞, ∞]. Consider the ensemble S of all S ∈ R such that S ⊂ R and S is essentially positive for 𝜇. Consider also the extended real number x ≡ sup (𝜇S | S ∈ S). If x < ∞, then by Lemma 1 (1.4.5) for every n ∈ N, there is S n ∈ S such that x − 1/n < 𝜇S n ⩽ x. Fix 𝜀 > 0 and take n ∈ N such that 1/n < 𝜀. Then, for every p ⩾ n, we have |x − 𝜇S p | < 1/n < 𝜀. This means that x = lim(𝜇S n | n ∈ N) (see 1.4.4). If x = ∞, then for every n ∈ N, there is S n ∈ S such that 𝜇S n > n. Then, for every p ⩾ n, we have 𝜇S p > n. This means that x = lim(𝜇S n | n ∈ N) (see 1.4.4). Consider the set P ≡ ⋃⟮S n | n ∈ N⟯. By Lemma 9 (2.1.1), P ∈ R, then by Lemma 3 P ∈ S. Since S n ⊂ P, we have 𝜇S n ⩽ 𝜇P. By virtue of Corollary 3 to Proposition 1 (1.4.7), x = lim 𝜇S n ⩽ 𝜇P ⩽ x, where 𝜇P = x.

228 | 3.2 Decompositions of semimeasures and measures

Suppose that there is a non-empty set A ∈ R such that A ⊂ R and A ∩ P = ⌀ and 𝜇A > 0. Consider the ensemble E ≡ {E ∈ R | E ⊂ A ∧ 𝜇E ⩾ 𝜇A > 0}. Introduce an order in E setting E 1 ⩽ E2 if either E2 ⊂ E1 and 𝜇E1 < 𝜇E2 or E1 = E2 . Take any non-empty chain C in E and consider the number y ≡ sup (𝜇E | E ∈ C) ∈ R+ . First, suppose that there is F ∈ C such that y = 𝜇F. Then, F is an upper bound for C in E. In fact, inf E ∈ C, then either E ⩽ F or F ⩽ E. In the second case, if F = / E, then E ⊂ F and y = 𝜇F < 𝜇E ⩽ y, but this is impossible. Hence, F = E. As a result, F is an upper bound. Now, suppose that for every E ∈ C, we have 𝜇E < y. Then, consider the choice mapping p : P(E)\{⌀} → E from the axiom of choice (see 1.1.12). If x < ∞, then for every u ∈ N consider the non-empty subset Cn ≡ {E ∈ C | 𝜇E > y − 1/n} and take the element D n ≡ pCn ∈ Cn . If y = ∞, then for every n ∈ N consider the non-empty subset Cn ≡ {E ∈ C | 𝜇E > n} and take the element D n ≡ pCn ∈ Cn . In both cases, sup (𝜇D n | n ∈ N) = y. By virtue of Corollary 1 to Theorem 3 (1.2.6), for every n ∈ N, the set {D k | k ∈ n + 1} has the greatest element, which will be denoted by E n . Then, E n = sup{D k | k ∈ n + 1} in E and E n ∈ C. Since E n ⩾ D k for every k ∈ n + 1, we conclude that either E n = D k or E n ⊂ D k and 𝜇E n > 𝜇D k . In all the cases, 𝜇E n ⩾ sup{𝜇D k | k ∈ n + 1}. From E n ⩽ E n+1 , we conclude that either E n = E n+1 or E n+1 ⊂ E n and 𝜇E n+1 > 𝜇E n . Thus, (E n | n ∈ N) ↓ and (𝜇E n | n ∈ N) ↑ y. Besides 𝜇E1 < ∞. Consider the set F ≡ ⋂(E n | n ∈ N) ∈ M. By Lemma 6 (3.1.1), 𝜇F = lim(𝜇E n | n ∈ N) = y ⩾ 𝜇E1 ⩾ 𝜇A. Moreover, F ⊂ E n ⊂ A. Hence, F ∈ E. If E ∈ C, then 𝜇E < y = 𝜇F. Therefore, 𝜇E < 𝜇E n for some n. Since C is a choice, we conclude that F ⊂ E n ⊂ E. This means that E ⩽ F. Hence, F is an upper bound for C in E. Thus, the chain C has an upper bound. By virtue of the Kuratowski – Zorn lemma from Theorem 1 (1.2.11), the ordered set E contains some maximal element E. Then, E ⊂ A ⊂ R. Suppose that E ∉ S. Then, there is B ∈ R such that B ⊂ E and 𝜇B < 0. From 0 < 𝜇E = 𝜇B + 𝜇(E\B), we deduce that −∞ < 𝜇B < 0 and −∞ < 𝜇(E\B) < ∞. Therefore, 𝜇(E\B) > 𝜇E ⩾ 𝜇A. Besides, E\B ⊂ E ⊂ A. As a result, E\B ∈ E. With respect to the order in E, we get E < E\B. But this contradicts the maximality of E. Thus, E ∈ S. Therefore, 0 < 𝜇E ⩽ x. Besides E ⊂ A implies E ∩ P = ⌀. Consequently, ∞ > 𝜇(E ∪ P) = 𝜇E + 𝜇P = 𝜇E + x ⩾ x > 0. As a result, 𝜇(E ∪ P) > x. But E ∪ P ∈ S implies 𝜇(E ∪ P) ⩽ x. It follows from this contradiction that the conditions A ∈ R and A ⊂ R\P imply 𝜇A ⩽ 0, i. e. the set N ≡ R\P is essentially negative. Finally, taking R + ≡ P and R− ≡ N, we get the assertion of the theorem. The partition (R+ , R− ) (see 1.1.6) of the set R from Theorem 1 is called a Hahn decomposition of the set R with respect to the measure 𝜇. Note that this decomposition is not unique. Corollary 1. Let 𝜇 be a measure on a 𝛿-ring R and (R+ , R− ) be a Hahn decomposition of R ∈ R. If S ∈ R and S ⊂ R, then (S ∩ R+ , S ∩ R− ) is a Hahn decomposition of S.

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Corollary 2. Let 𝜇 be a measure on a 𝜎-algebra M and (T+ , T− ) be a Hahn decomposition of T. Then, for every S ∈ M the partition (S ∩ T+ , S ∩ T− ) is a Hahn decomposition of S. Measures on a 𝛿-ring have the following important property. Proposition 1. Let 𝜇 be a measure on a 𝛿-ring R. Then, sup{|𝜇S| | S ∈ R ∧ S ⊂ R} ⩽ 𝜇R+ − 𝜇R− < ∞ for every R ∈ Rf (𝜇). Proof. Consider for R the sets R+ and R− from Theorem 1 and the numbers x ≡ 𝜇R+ ∈ R+ and y ≡ 𝜇R− ∈ R− . By the definition of a measure, either rng 𝜇 ∈ R ∪ {∞} or rng 𝜇 ∈ {−∞} ∪ R. Therefore, from 𝜇R = x + y ∈ R, we deduce that in both cases x ∈ R+ and y ∈ R− . For any S ∈ R such that S ⊂ R, we have 0 ⩽ 𝜇(S∩R+ ) ⩽ 𝜇R+ = x and 0 ⩾ 𝜇(S ∩ R− ) ⩾ 𝜇R− = y, where y ⩽ 𝜇(S ∩ R+ ) + 𝜇(S ∩ R− ) ⩽ x. Consequently, y ⩽ 𝜇S ⩽ x implies |𝜇S| ⩽ x − y < ∞. Corollary 1. Let 𝜇 be a measure on a 𝛿-ring R. If R ∈ Rf (𝜇), then v(𝜇)R < ∞. Proof. By Corollary 2 to Lemma 8 (3.1.3), we have v(𝜇)R ⩽ 2 sup{|𝜇S| | S ∈ R ∧ S ⊂ R}. Then, Proposition 1 implies v(𝜇)R < ∞. Corollary 2. Let 𝜇 be a finite measure on a 𝛿-ring R. Then, 𝜇 is overfinite, i. e. v(𝜇) is finite. So, Measf (T, R) = Measof (T, R). Corollary 3. Let 𝜇 be a measure on a 𝜎-algebra M. Then, 𝜇 is finite iff 𝜇 is bounded. So, Measb (T, M) = Measf (T, M) = Measof (T, M). ̄̄ is finite. Then, by Corollary 1 to Proof. By Corollary 1, 𝜇 is overfinite, where 𝜇T Lemma 8 (3.1.3), 𝜇 is bounded. Corollary 4. If 𝜇 and 𝜈 be measures on a 𝛿-ring R, the spaces ⟮T, R, 𝜇⟯ and ⟮T, R, 𝜈⟯ are 𝜎-finite and either 𝜇, 𝜈 ∈ Meas(T, R, ] − ∞, ∞]) or 𝜇, 𝜈 ∈ Meas(T, R, [−∞, ∞[), then the spaces ⟮T, R, 𝜇 + 𝜈⟯ and ⟮T, R, v(𝜇) + v(𝜈)⟯ are 𝜎-finite. Proof. According to Lemma 1 (3.1.2), we can consider the measures 𝜇+𝜈 and 𝜆 ≡ v(𝜇)+ v(𝜈). By the condition, there are countable collections (M i ∈ Rf (𝜇) | i ∈ I) and (N j ∈ Rf (𝜈) | j ∈ J) such that T = ⋃⟮M i | i ∈ I⟯ = ⋃⟮N j | j ∈ J⟯. By Corollary 1, v(𝜇)M i < ∞ and v(𝜈)N j < ∞. Consequently, 𝜆(M i ∩ N j ) ⩽ v(𝜇)M i + v(𝜈)N j < ∞. By Lemma 4 (3.1.3), |(𝜇+𝜈)(M i ∩N j )| ⩽ |𝜇(M i ∩N j )|+|𝜈(M i ∩N j )| ⩽ v(𝜇)(M i ∩N j )+v(𝜈)(M i ∩N j ) = 𝜆(M i ∩N j ) < ∞. Thus, M i ∩ N j ∈ Rf (𝜆) ∩ Rf (𝜇 + 𝜈) and T = ⋃⟮M i ∩ N j | (i, j) ∈ I × J⟯. The Hahn decomposition with respect to 𝜇 allows us to decompose 𝜇 into a sum of a positive measure and a negative measure.

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Lemma 4. Let 𝜇 be a measure on a 𝛿-ring R. If (R+ , R− ) and (R󸀠+ , R󸀠− ) are two Hahn decompositions of a set R ∈ R, then 𝜇(R+ ) = 𝜇(R󸀠+ ) and 𝜇(R− ) = 𝜇(R󸀠− ). Proof. Since A ≡ R+ \R󸀠+ ⊂ R+ ∩ R󸀠− , we deduce that 𝜇A ⩾ 0 and 𝜇A ⩽ 0, where 𝜇A = 0. Similarly, 𝜇B = 0 for B ≡ R󸀠+ \R+ . As a result, 𝜇R+ = 𝜇(R+ ∩ R󸀠+ ) + 𝜇A = 𝜇(R󸀠+ ∩ R+ ) + 𝜇B = 𝜇R󸀠+ . The second equality is checked analogously. It follows from this lemma that we can define correctly some positive and negative evaluations v+ (𝜇) and v− (𝜇) on R, setting v+ (𝜇)R ≡ 𝜇(R+ ) and v− (𝜇)R ≡ 𝜇(R− ) for any Hahn decomposition (R+ , R− ) of R ∈ R. If rng 𝜇 ⊂] − ∞, ∞], then v− (𝜇) is finite, i. e. rng v− (𝜇) ⊂ R− . If rng 𝜇 ⊂ [−∞, ∞[, then v+ (𝜇) is finite, i. e. rng v+ (𝜇) ⊂ R+ . Proposition 2. Let 𝜇 be a measure on a 𝛿-ring R. Then, v+ (𝜇) and v− (𝜇) are positive and, respectively, negative measures on R such that at least one of them is finite, 𝜇 = v+ (𝜇) + v− (𝜇) and v(𝜇) = v+ (𝜇) − v− (𝜇). Proof. Let (R n ∈ R | n ∈ 𝜔) be a countable pairwise disjoint collection such that R ≡ ⋃⟮R n | n ∈ 𝜔⟯ ∈ R. Take some Hahn decompositions (R+n , R−n ) of R n for every n ∈ 𝜔 and consider the sets R󸀠 ≡ ⋃⟮R+n | n ∈ 𝜔⟯ and R󸀠󸀠 ≡ ⋃⟮R−n | n ∈ 𝜔⟯ from R. If A ∈ R and A ⊂ R󸀠 , then for A n ≡ A ∩ R+n ∈ R, we have A = ⋃⟮A n | n ∈ 𝜔⟯ and 𝜇A n ⩾ 0, where 𝜇A = ∑(𝜇A n | n ∈ 𝜔) ⩾ 0. This means that R󸀠 is essentially positive. Similarly, R󸀠󸀠 is essentially negative. Hence, the partition (R󸀠 , R󸀠󸀠 ) of the set R is its Hahn decomposition. Consequently, v+ (𝜇)R ≡ 𝜇R󸀠 = ∑(𝜇(R+n ) | n ∈ 𝜔) = ∑(v+ (𝜇)R n | n ∈ 𝜔) and v− (𝜇)R ≡ 𝜇R󸀠󸀠 = ∑(𝜇(R−n ) | n ∈ 𝜔) = ∑(v− (𝜇)R n | n ∈ 𝜔). Thus, v+ (𝜇) and v− (𝜇) are measures. Now, fix any set R ∈ R. Then, taking the Hahn decomposition (R+ , R− ) of R, we get v+ (𝜇)R + v− (𝜇)R = 𝜇R+ + 𝜇R− = 𝜇R and v+ (𝜇)R − v− (𝜇)R = 𝜇R+ − 𝜇R− = |𝜇R+ | + |𝜇R− | ⩽ v(𝜇)R. Conversely, take any finite partition (R i ∈ R | i ∈ I) of the set R. Then, |𝜇R i | ⩽ |v+ (𝜇)R i | + |v− (𝜇)R i | = v+ (𝜇)R i − v− (𝜇)R i implies ∑(|𝜇R i | | i ∈ I) ⩽ ∑(v+ (𝜇)R i | i ∈ I) − ∑(v− (𝜇)R i | i ∈ I) = v+ (𝜇)R − v− (𝜇)R. Then, by Lemma 2 (3.1.3) v(𝜇)R ⩽ v+ (𝜇)R − v− (𝜇)R. Corollary 1. In the conditions of Proposition 2, if 𝜇 ∈ Meas(T, R, ] − ∞, ∞]), then v− (𝜇) is finite; if 𝜇 ∈ Meas(T, R, [−∞, ∞[), then v+ (𝜇) is finite. Corollary 2. Let 𝜇 be a finite measure on a 𝛿-ring R. Then, the equalities 𝜇 = v+ (𝜇) + v− (𝜇) and v(𝜇) = v+ (𝜇) − v− (𝜇) define the measures v+ (𝜇) and v− (𝜇) in the unique way by the formulas v + (𝜇) = (1/2)(𝜇 + v(𝜇)) and v− (𝜇) = (1/2)(𝜇 − v(𝜇)). Proof. By Corollary 1, v(𝜇) is finite. Therefore, adding our equalities, we get 2v+ (𝜇) = 𝜇 + v(𝜇). Subtracting these equalities, we get 2v− (𝜇) = 𝜇 − v(𝜇).

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The equality 𝜇 = v+ (𝜇) + v− (𝜇) is called the Jordan decomposition of the measure 𝜇 on the 𝛿-ring R. The measures v+ (𝜇) and v− (𝜇) are called the positive variation and the negative variation of the measure 𝜇, respectively. Lemma 5. Let 𝜇 be a measure on a 𝜎-algebra M and (T+ , T− ) be a Hahn decomposition of T with respect to 𝜇. Then, v+ (𝜇)M = 𝜇(M ∩ T+ ) and v− (𝜇)M = 𝜇(M ∩ T− ) for every M ∈ M. Proof. By Corollary 2 to Theorem 1, (M ∩ T+ , M ∩ T− ) is a Hahn decomposition of M. Then, v+ (𝜇)M = 𝜇(M ∩ T+ ) and v− (𝜇)M = 𝜇(M ∩ T− ). Corollary 1. In the conditions of Lemma 5, M ∩ T+ = ⌀ implies v+ (𝜇)M = 0, and M ∩ T− = ⌀ implies v− (𝜇)M = 0. Corollary 2. In the conditions of Lemma 5, v+ (𝜇)T− = 0 and v− (𝜇)T+ = 0.

3.2.2 The Riesz decomposition of overfinite semimeasures and measures on a ring The families of all finite and overfinite semimeasures and measures have good order properties (see 1.1.15 for some notions and notations using below).

The Dedekind completeness Theorem 1. Let R be a ring on a set T, M be a subset of SMeasf (T, R), and R ∈ R. Then, 1) the ordered linear space SMeasf (T, R) is Dedekind complete; 2) if M is bounded above, then (sup M)R = sup{∑(𝜇i R i | i ∈ I) | (R i | i ∈ I) ∈ Parf (R, R) ∧ (𝜇i | i ∈ I) ∈ Map(I, M)}; 3) if M is bounded below, then (inf M)R = inf{∑(𝜇i R i | i ∈ I) | (R i | i ∈ I) ∈ Parf (R, R) ∧ (𝜇i | i ∈ I) ∈ Map(I, M)}. Proof. 2. Let 𝜇 ⩽ 𝜇1 for every 𝜇 ∈ M. Then, ∑(𝜇i R i | i ∈ I) ⩽ ∑(𝜇1 R i | i ∈ I) = 𝜇1 R for every finite partition (R i ∈ R | i ∈ I) of R and every collection (𝜇i ∈ M | i ∈ I). Therefore, we can define correctly a mapping 𝜈 : R → R setting 𝜈R equal to the right part of the equality in (2). Let R, R󸀠 , R󸀠󸀠 ∈ R, R = R󸀠 ∪ R󸀠󸀠 , R󸀠 ∩ R󸀠󸀠 = ⌀, (R󸀠j ∈ R | j ∈ J) be a finite partition 󸀠 of R and (R󸀠󸀠k ∈ R | k ∈ K) be a finite partition of R󸀠󸀠 . Then, ∑(𝜇j󸀠 R󸀠j | j ∈ J) + ∑(𝜇k󸀠󸀠 R󸀠󸀠k | k ∈ K) ⩽ 𝜈R. Therefore, 𝜈R󸀠 ⩽ 𝜈R − ∑(𝜇k󸀠󸀠 R󸀠󸀠k | k ∈ K) and 𝜈R󸀠󸀠 ⩽ 𝜈R − 𝜈R󸀠 . On the other hand, let (R i | i ∈ I) be a finite partition of R, R = R󸀠 ∪ R󸀠󸀠 and 󸀠 R ∩ R󸀠󸀠 = ⌀. Consider the bijection 𝛽 : m + 1 I. Then, I = {i k | k ∈ m + 1}. Consider the collection (S kl ∈ R | (k, l) ∈ (m + 1) × 2) such that S k0 ≡ R i k ∩ R󸀠 and S k1 ≡ R i k ∩ R󸀠󸀠 .

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Then, R i k = S k0 ∪ S k1 and S k0 ∩ S k1 = ⌀. Besides, (S k0 | k ∈ m + 1) is a partition of R󸀠 and (S k1 | k ∈ m + 1) is a partition of R󸀠󸀠 . Therefore, ∑(𝜇i (R i ) | i ∈ I) = ∑(𝜇i k (R i k ) | k ∈ m + 1) = ∑(𝜇i k (S k0 ) | k ∈ m + 1) + ∑(𝜇i k (S k1 ) | k ∈ m + 1) ⩽ 𝜈R󸀠 + 𝜈R󸀠󸀠 . This implies 𝜈R ⩽ 𝜈R󸀠 + 𝜈R󸀠󸀠 . Thus, 𝜈R = 𝜈R󸀠 + 𝜈R󸀠󸀠 , i. e. 𝜈 is binary additive. By Lemma 1 (3.1.1), 𝜈 is finitely additive; hence, 𝜈 is a semimeasure. Besides, 𝜈 ⩽ 𝜇1 . Taking card I = 1, we get 𝜈R ⩾ 𝜇R for every R ∈ R and 𝜇 ∈ M, where 𝜈 ⩾ 𝜇 for every 𝜇 ∈ M. Since 𝜇1 is an arbitrary upper bound, we conclude that 𝜈 = sup M. Thus, assertion 2 is proven. 3. Let 𝜇0 ⩽ 𝜇 for every 𝜇 ∈ M. Then, from −𝜇 ⩽ −𝜇0 for every 𝜇 ∈ M we deduce that defining a mapping 𝜈 : R → R by the equality 𝜈R ≡ sup{∑(−𝜇i R i | i ∈ I) | (R i | i ∈ I) ∈ Parf (R, R) ∧ (𝜇i | i ∈ I) ∈ Map(I, M)}, we get 𝜈 = sup(−M). According to assertion 3󸀠 of Theorem 1 (2.2.2), we get −𝜈 = inf M, and therefore, (inf M)R = (−𝜈)R = −𝜈(R). But −𝜈(R) is equal to the right part of the equality in (3). Thus, assertion 3 is proven. 1. According to the definition of a Dedekind complete ordered class in 1.1.15, assertions 2 and 3 imply assertion 1. Corollary 1. In the conditions of Theorem 1, we have 1) if M is bounded above and upward directed, then (sup M)R = sup(𝜇R | 𝜇 ∈ M); 2) if M is bounded below and downward directed, then (inf M)R = inf(𝜇R | 𝜇 ∈ M). Proof. 1. Let (𝜇i ∈ M | i ∈ I) be a finite collection and (R i ∈ R | i ∈ I) be a partition of R. For the set {𝜇i | i ∈ I}, there is some element 𝜇 ∈ M such that 𝜇 ⩾ 𝜇i for every i. Then, we have ∑ (𝜇i R i | i ∈ I) ⩽ ∑(𝜇R i | i ∈ I) = 𝜇R ⩽ sup(𝜇R | 𝜇 ∈ M). From the second assertion of Theorem 1, we deduce (sup M)R ⩽ sup(𝜇R | 𝜇 ∈ M) ⩽ (sup M)R. For case 2, the arguments are analogous. The Riesz decomposition For overfinite semimeasures the additional (in comparison with Theorem 1) lattice property arises. Namely, the following theorem shows that the ordered set ⟮SMeasof (T, R), ⩽⟯ is lattice-ordered (see 1.1.15). Theorem 2. Let R be a ring on a set T, M be a subset of SMeasof (T, R), R ∈ R, and 𝜆, 𝜈 ∈ SMeasof (T, R). Then, 1) the ordered linear space SMeasof (T, R) is lattice-ordered and Dedekind complete; 2) if M is bounded above, then the supremum of M in SMeasf (T, R) and the supremum of M in SMeasof (T, R) are equal and expressed by the formula from assertion 2) of Theorem 1; 3) if M is bounded below, then the infimum of M in SMeasf (T, R) and the infimum of M in SMeasof (T, R) are equal and expressed by the formula from assertion 3) of Theorem 1;

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4) (𝜆 ∨ 𝜈)(R) = sup{𝜆(E) + 𝜈(F) | (E, F) ∈ Parf (R, R)} = sup{∑(𝜆(R i ) ⊻ 𝜈(R i ) | i ∈ I) | (R i ∈ R | i ∈ I) ∈ Parf (R, R)}; 5) (𝜆 ∧ 𝜈)(R) = inf{𝜆(E) + 𝜈(F) | (E, F) ∈ Parf (R, R)} = inf{∑(𝜆(R i ) ⊼ 𝜈(R i ) | i ∈ I) | (R i ∈ R | i ∈ I) ∈ Parf (R, R)}. Proof. For supremums and infimums in SMeasf (T, R) and SMeasof (T, R) we shall use the corresponding indices “f ” and “of ”. ̄̄ ̄̄ ∈ SMeasof (T, R)0 . Since SMeasof (T, R) is linear by By Lemma 2 (3.2.1), 𝜆|R, 𝜈|R ̄ ̄̄ and ̄ + 𝜈|R ̄̄ ∈ SMeasof (T, R)0 . By Lemma 4 (3.1.3), 𝜆 ⩽ 𝜆|R Lemma 1 (3.2.1), 𝜇0 ≡ 𝜆|R ̄̄ 𝜈 ⩽ 𝜈|R. Therefore, the set {𝜆, 𝜈} is bounded above by 𝜇0 . Similarly, {𝜆, 𝜈} is bounded below by the element −𝜇0 . Consequently, by Theorem 1, there exist 𝜇1 ≡ 𝜆 ∨ 𝜈 ≡ sup{𝜆, 𝜈} and 𝜇2 ≡ 𝜆 ∧ 𝜈 ≡ inf{𝜆, 𝜈} in SMeasf (T, R) such that 𝜇1 R = sup{𝜆(E) + 𝜈(F) | (E, F) ∈ Parf (R, R)} and 𝜇2 R = inf{𝜆(E) + 𝜈(F) | (E, F) ∈ Parf (R, R)}. Since −𝜇0 ⩽ 𝜇2 ⩽ 𝜇1 ⩽ 𝜇0 , we have |𝜇1 S| ⩽ 𝜇0 S and |𝜇2 S| ⩽ 𝜇0 S for every S ∈ R. Consequently, by virtue of Lemma 8 (3.1.3), we get 𝜇̄̄ 1 R ⩽ 2 sup{|𝜇1 S| | S ∈ R ∧ S ⊂ R} ⩽ 2 sup{𝜇0 S | S ∈ R ∧ S ⊂ R} ⩽ 2𝜇0 R < ∞, whereas 𝜇0 is increasing by Lemma 4 (3.1.1). Similarly, 𝜇̄̄ 2 R < ∞. Since R is an arbitrary element of R, this means that 𝜇1 and 𝜇2 belong to SMeasof (T, R). Therefore, 𝜇1 is the supremum and 𝜇2 is the infimum of the set {𝜆, 𝜈} in the ordered set SMeasof (T, R). Thus, the ordered linear subspace SMeasof (T, R) of the ordered linear space SMeasf (T, R) is lattice-ordered and the supremum and the infimum of every set {𝜆, 𝜇} ⊂ SMeasof (T, R) are the same in the ordered set SMeasf (T, R) and in the ordered set SMeasof (T, R). Let M ⊂ SMeasof (T, R)0 and M be bounded above in SMeasof (T, R)0 . By Theorem 1, there is 𝜇1 ≡ supf M in SMeasf (T, R). Since 𝜇1 ⩾ 0, we infer by virtue of Corollary 2 to Lemma 7 (3.1.3) that 𝜇1 ∈ SMeasof (T, R). Hence, 𝜇1 = supof M. Let now L ⊂ SMeasof (T, R) and L is bounded above by 𝜇0 ∈ SMeasof (T, R). Take some 𝜘 ∈ L. Consider the set M ≡ {𝜆 ∨ 𝜘 − 𝜘 | 𝜆 ∈ L} ⊂ SMeasof (T, R)+ . Since M is bounded above by 𝜇0 − 𝜘, by the property proven above, there exists 𝜇1 ≡ supf M and 𝜇1 = supof M. According to assertion 1 of Theorem 1 (2.2.2), there exists 𝜈f ≡ supf {𝜇 + 𝜘 | 𝜇 ∈ M} and 𝜇1 + 𝜘 = 𝜈f = supf {(𝜆 ∨ 𝜘 − 𝜘) + 𝜘 | 𝜆 ∈ L} = supf {𝜆 ∨ 𝜘 | 𝜆 ∈ L}. This implies 𝜇1 + 𝜘 = supf L. By the same reasons, there exists 𝜈of ≡ supof {𝜇 + 𝜘 | 𝜇 ∈ M} in SMeasof (T, R) and 𝜇1 + 𝜘 = 𝜈of = supof {(𝜆 ∨ 𝜘 − 𝜘) + 𝜘 | 𝜆 ∈ L} = supof {𝜆 ∨ 𝜘 | 𝜆 ∈ L}. This implies 𝜇1 + 𝜘 = supof L in SMeasof (T, R). Therefore, supf L = supof L. If K ⊂ SMeasof (T, R) and K is bounded below, then we take the set L ≡ −K, which is bounded above. As was shown above, there exists 𝜆 ≡ supf L = supof L ∈ SMeasof (T, R). Then, −𝜆 1 = inf f K = inf of K ∈ SMeasof (T, R). Thus, the lattice-ordered linear subspace SMeasof (T, R) of the Dedekind complete ordered linear space SMeasf (T, R) is also Dedekind complete and the

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supremums [infimums] of every subset L [K] of SMeasof (T, R), which is bounded above [below] are the same in the ordered set SMeasf (T, R) and in the ordered set SMeasof (T, R). Denote the right parts of the equalities in (4) and (5) by x and y, respectively. It is evident that 𝜇1 R ⩽ x and 𝜇2 R ⩾ y. Let (R i ∈ R | i ∈ I) be a finite partition of R. Consider the subsets J ≡ {j ∈ I | 𝜆R j > 𝜈R j } and K ≡ {k ∈ I | 𝜆R k ⩽ 𝜈R k } and the sets E ≡ ⋃⟮R j | j ∈ J⟯ and F ≡ ⋃⟮R k | k ∈ K⟯. Then, we have E ∪ F = R and E ∩ F = ⌀, where ∑(𝜆(R i ) ⊻ 𝜈(R i ) | i ∈ I) = ∑(𝜆R j | j ∈ J) + ∑(𝜈R k | k ∈ K) = 𝜆E + 𝜈F ⩽ 𝜇1 R and ∑(𝜆(R i ) ⊼ 𝜈(R i ) | i ∈ I) = ∑(𝜈R j | j ∈ J) + ∑(𝜆R k | k ∈ K) = 𝜈E + 𝜆F ⩾ 𝜇2 R. These inequalities imply x ⩽ 𝜇1 R and y ⩾ 𝜇2 R. Thus, 𝜇1 R = x and 𝜇2 R = y. If we consider the family SMeasof (T, R) together with its lattice order (see Theorem 2) and we fix the neutral element 0, then according to 1.1.15, we can single out its positive part SMeasof (T, R)+ ≡ {𝜇+ ≡ 𝜇∨0 | 𝜇 ∈ SMeasof (T, R)} = SMeasof (T, R)0 and its negative part SMeasof (T, R)− ≡ {𝜇− ≡ 𝜇 ∧ 0 | 𝜇 ∈ SMeasof (T, R)} consisting of all positive and negative parts 𝜇+ and 𝜇− of all semimeasures 𝜇 ∈ SMeasof (T, R). Theorem 2 gives also the opportunity to consider for every semimeasure 𝜇 ∈ SMeasof (T, R) its modulus |𝜇| ≡ 𝜇 ∨ (−𝜇) (see 2.2.2). Corollary 1. Let R be a ring on T, 𝜇 ∈ SMeasof (T, R), and R ∈ R. Then, 1) 𝜇+ (R) = sup{𝜇(E) | E ∈ R ∧ E ⊂ R} = sup{∑((𝜇(R i ))+ | i ∈ I) | (R i ∈ R | i ∈ I) ∈ Parf (R, R)}; 2) 𝜇− (R) = inf{𝜇(E) | E ∈ R ∧ E ⊂ R} = inf{∑((𝜇(R i ))− | i ∈ I) | (R i ∈ R | i ∈ I) ∈ Parf (R, R)}; 3) |𝜇|R = sup{𝜇(E) − 𝜇(F) | (E, F) ∈ Parf (R, R)} = sup{∑(𝜇(R i ) ⊻ (−𝜇(R i )) | i ∈ I) | (R i ∈ R | i ∈ I) ∈ Parf (R, R)}; 4) |𝜇| = v(𝜇); 5) sup{|𝜇S| | S ∈ R ∧ S ⊂ R} ⩽ |𝜇|R ⩽ 2 sup{|𝜇S| | S ∈ R ∧ S ⊂ R}. Proof. Formulas 1, 2, and 3 follow from assertions 4 and 5 of Theorem 2. Equality (4) follows from (3) and Lemma 2 (3.1.3). Inequalities (5) follow from (4) and Lemma 8 (3.1.3). Corollary 2. In the conditions of Corollary 1, |𝜇R| ⩽ |𝜇|R. Proof. By Lemma 4 (3.1.3), |𝜇R| ⩽ v(𝜇)R. By Corollary 1, v(𝜇)R = |𝜇|R. Corollary 3. In the conditions of Corollary 1, 𝜇 = 𝜇+ +𝜇− , |𝜇| = 𝜇+ −𝜇− , 𝜇+ = (1/2)(𝜇+|𝜇|) = (1/2)(𝜇 + v(𝜇)), and 𝜇− = (1/2)(𝜇 − |𝜇|) = (1/2)(𝜇 − v(𝜇)). Proof. The first two equalities follow directly from Statement 2 (2.2.4). These equalities and assertion 4 of Corollary 1 imply the remaining equalities.

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The equality 𝜇 = 𝜇+ + 𝜇− is called the Riesz decomposition of 𝜇. The measures 𝜇+ and 𝜇− are called the positive part and the negative part of the overfinite semimeasure 𝜇, respectively (cf. 4∘ (2.2.4) and 2.2.8). Now, we shall connect the Riesz and Jordan decompositions. Proposition 1. Let R be a 𝛿-ring on a set T and 𝜇 be a finite (and, therefore, overfinite) measure on R. Then, 𝜇+ = v+ (𝜇) and 𝜇− = v− (𝜇), i. e. the Jordan decomposition coincides with the Riesz decomposition. Proof. The statement follows from Corollary 2 to Proposition 2 (3.2.1) and Corollary 3 to Theorem 2. Some properties of families of overfinite and bounded semimeasures and measures Recall that SMeasb (T, R) is an ordered linear subspace of SMeasof (T, R). Lemma 1. Let R be a ring on a set T, 𝜇 ∈ SMeasof (T, R), 𝜈 ∈ SMeasb (T, R), and |𝜇| ⩽ |𝜈| in SMeasof (T, R). Then, 𝜇 ∈ SMeasb (T, R). ̄̄ Proof. Since 𝜈 is bounded, by Corollary 1 to Lemma 8 (3.1.3), we get 𝜈(T) ∈ R. By Corollary 2 to Theorem 2, assertion 4 of Corollary 1 to Theorem 2, and assertion 2 ̄̄ of Lemma 4 (3.1.3), we have |𝜇R| ⩽ |𝜇|R ⩽ |𝜈|R ⩽ 𝜈(T) for every R ∈ R. Thus, 𝜇 is bounded. Corollary 1. Let R be a ring on T. Then, the ordered linear subspace SMeasb (T, R) of the lattice-ordered linear space SMeasof (T, R) is an l-ideal in it. Proof. It follows from Lemma 1 and Statement 2 (2.2.8). Corollary 2. Let R be a ring on T. Then, the lattice-ordered linear subspace SMeasb (T, R) of the Dedekind complete lattice-ordered linear space SMeasof (T, R) is Dedekind complete as well; the smallest upper [the greatest lower] bounds of bounded above [below] sets in the ordered sets SMeasof (T, R) and SMeasb (T, R) coincide and they are expressed by the formulas from Theorem 2 (with its Corollaries). Proof. It easily follows from Theorem 2, Corollary 1, and Statements 3 and 1 (2.2.8). Corollary 3. Let R be a ring on T and 𝜇 ∈ SMeasf (T, R). Then, 𝜇 ∈ SMeasb (T, R) iff 𝜇 ∈ SMeasof (T, R) and |𝜇| ∈ SMeasb (T, R). Proof. If 𝜇 ∈ SMeasb (T, R) ⊂ SMeasof (T, R), then assertion 5 of Corollary 1 to Theorem 2 implies that |𝜇| is bounded.

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If 𝜇 ∈ SMeasof (T, R) and |𝜇| ∈ SMeasb (T, R), then, applying Lemma 1 to 𝜇 and 𝜈 ≡ |𝜇|, we see that 𝜇 ∈ SMeasb (T, R). Now, we consider the family Measof (T, R) of all overfinite measures. Proposition 2. Let R be a ring on a set T, 𝜇 ∈ SMeasof (T, R), 𝜈 ∈ Measof (T, R) and |𝜇| ⩽ |𝜈| in SMeasof (T, R). Then, 𝜇 ∈ Measof (T, R). Proof. By Lemma 2 (3.2.1) and Corollary 1 to Theorem 2 |𝜈| = v(𝜈) ∈ Measof (T, R)+ . Take any decreasing sequence (R n ∈ R | n ∈ 𝜔) such that ⋂⟮R n | n ∈ 𝜔⟯ = ⌀. By Lemma 6 (3.1.1), lim(|𝜈|R n | n ∈ 𝜔) = 0. Since 0 ⩽ 𝜇+ R n ⩽ |𝜈|R n and 0 ⩽ −𝜇− R n ⩽ |𝜈|R n for every n, we infer by Lemma 6 (1.4.7) that lim(𝜇+ R n | n ∈ 𝜔) = 0 and lim(−𝜇− R n | n ∈ 𝜔) = 0. By virtue of Corollary 3 to Theorem 2 and Proposition 1 (1.4.7), we conclude that lim(𝜇R n | n ∈ 𝜔) = 0. By Lemma 7 (3.1.1), this means that 𝜇 is a measure. Corollary 1. Let R be a ring on T. Then, the ordered linear subspace Measof (T, R) of the lattice-ordered linear space SMeasof (T, R) is an l-ideal in it. Proof. It follows from Proposition 2 and Statement 2 (2.2.8). Corollary 2. Let R be a ring on a set T. Then, the lattice-ordered linear subspace Measof (T, R) of the Dedekind complete lattice-ordered linear space SMeasof (T, R) is Dedekind complete as well; the smallest upper [the greatest lower] bounds of bounded above [below] sets in the ordered sets SMeasof (T, R) and Measof (T, R) coincide and they are expressed by the formulas from Theorem 1 and Theorem 2 (with its Corollaries). Proof. It easily follows from Theorem 2, Corollary 1, and Statements 3 and 1 (2.2.8). Corollary 3. Let R be a ring on T and 𝜇 ∈ SMeasf (T, R). Then, 𝜇 ∈ Measof (T, R) iff 𝜇 ∈ SMeasof (T, R) and |𝜇| ∈ Measof (T, R). Proof. If 𝜇 ∈ Measof (T, R) ⊂ SMeasof (T, R), then by Corollary 1 to Theorem 2 |𝜇| = ̄̄ ∈ Measof (T, R)+ . v(𝜇). By Lemma 2 (3.2.1), v(𝜇) ≡ 𝜇|R If 𝜇 ∈ SMeasof (T, R) and |𝜇| ∈ Measof (T, R), then applying Proposition 2 to 𝜇 and 𝜈 ≡ |𝜇|, we see that 𝜇 ∈ Measof (T, R). Finally, we consider the family Measb (T, R) of all bounded measures. Lemma 2. Let R be a ring on T, 𝜇 ∈ SMeasof (T, R), 𝜈 ∈ SMeasb (T, R), and |𝜇| ⩽ |𝜈| in SMeasof (T, R). Then, 𝜇 ∈ Measb (T, R). Proof. By Lemma 1 𝜇 ∈ SMeasb (T, R). By Proposition 2 𝜇 ∈ Measof (T, R). Thus, 𝜇 is a bounded measure.

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Corollary 1. Let R be a ring on T. Then, the ordered linear subspace Measb (T, R) of the Dedekind complete lattice-ordered linear spaces SMeasb (T, R) and Measof (T, R) is an l-ideal in both of them. Proof. It follows from Lemma 2 and Statement 2 (2.2.8). Corollary 2. Let R be a ring on T. Then, the lattice-ordered linear subspace Measb (T, R) of the Dedekind complete lattice-ordered linear spaces SMeasb (T, R) and Measof (T, R) is Dedekind complete as well; the smallest upper [the greatest lower] bounds of bounded above [below] sets in the ordered sets SMeasb (T, R), Measof (T, R), and Measb (T, R) coincide and they are expressed by the formulas from Theorem 1 and Theorem 2 (with its Corollaries). Proof. By Corollary 2 to Lemma 1 and Corollary 2 to Proposition 2, the spaces SMeasb (T, R) and Measof (T, R) are Dedekind complete. The assertion now follows from Corollary 1 and Statements 3 and 1 (2.2.8). Corollary 3. Let R be a ring on T and 𝜇 ∈ SMeasf (T, R). Then, 𝜇 ∈ Measb (T, R) iff 𝜇 ∈ SMeasof (T, R) and |𝜇| ∈ Measb (T, R). Proof. If 𝜇 ∈ Measb (T, R) ⊂ SMeasof (T, R) ∩ Measf (T, R), then by Corollary 1 to ̄̄ ∈ Measof (T, R)+ . Therefore, Theorem 2 |𝜇| = v(𝜇), by Lemma 2 (3.2.1) v(𝜇) ≡ 𝜇|R assertion 5 of Corollary 1 to Theorem 2 implies that |𝜇| is bounded. If 𝜇 ∈ SMeasof (T, R) and |𝜇| ∈ Measb (T, R), then applying Lemma 2 to 𝜇 and 𝜈 ≡ |𝜇|, we see that 𝜇 ∈ Measb (T, R). Extension of overfinite measures Now, we shall consider the extension of overfinite measures. Proposition 3. Let 𝜇 be a positive finite measure on a ring R. Then, it has the unique extension to some positive finite measure 𝜇̃ on the 𝛿-ring R𝛿 (T, R). Proof. By Theorem 1 (3.1.5) for 𝜇, there is its Lebesgue – Caratheodory extension 𝜇× : L(T, R, 𝜇) → R+ and the ensemble L(T, R, 𝜇) is a 𝜎-ring containing the ring R. Therefore, by Lemma 9 (3.1.1), Lf (T, R, 𝜇) is a 𝛿-ring containing Rf (𝜇) = R. By Lemma 10 (2.1.1) R𝛿 (T, R) ⊂ Lf (T, R, 𝜇). Consider the finite positive measure 𝜇̃ ≡ 𝜇× |R𝛿 (T, R). By virtue of Theorem 2 (3.1.5), the extension 𝜇̃ is the unique positive 𝜎-additive extension of 𝜇 from R to R𝛿 (T, R). Let R be a ring and 𝜇 be an overfinite measure on R. By Corollary 1 of Proposition 2 for 𝜇, we can take the positive measures 𝜇+ and −𝜇− from Measof (T, R) such that 𝜇 = 𝜇+ + 𝜇− . By Proposition 3, for 𝜇1 ≡ 𝜇+ and 𝜇2 ≡ −𝜇− , there are their unique positive

238 | 3.2 Decompositions of semimeasures and measures

finite extensions 𝜇1̃ and 𝜇2̃ . Then, by Corollary 2 to Proposition 1 (3.2.1), 𝜇1̃ and 𝜇2̃ belong to Measof (T, R𝛿 (T, R)). Therefore, we can consider the new overfinite measure 𝜇̃ ≡ 𝜇1̃ − 𝜇2̃ on the 𝛿-ring R𝛿 (T, R). The mapping 𝜇 󳨃→ 𝜇̃ from Measof (T, R) into Measof (T, R𝛿 (T, R)) will be denoted by Δ. Lemma 3. Let 𝜇 be an overfinite measure on a ring R. Then, 𝜇̃ is the unique extension of 𝜇 to an overfinite measure on R𝛿 (T, R). Proof. Let 𝜈 be an overfinite measure on R𝛿 (T, R) such that 𝜈|R = 𝜇. By Corollary 1 to Proposition 2, 𝜈 = 𝜈+ +𝜈− . Consider the positive measures 𝜆 1 ≡ 𝜈+ |R and 𝜆 2 ≡ (−𝜈− )|R. If R ∈ R, then 𝜇1 R − 𝜇2 R = 𝜇+ R + 𝜇− R = 𝜇R = 𝜈R = 𝜈+ R + 𝜈− R = 𝜆 1 R − 𝜆 2 R means that (𝜇1̃ − 𝜈− )R = 𝜇1 R + 𝜆 2 R = 𝜇2 R + 𝜆 1 R = (𝜇2̃ + 𝜈+ )R for every R ∈ R. By Proposition 3 𝜇1̃ − 𝜈− = 𝜇2̃ + 𝜈+ . As a result, 𝜇̃ = 𝜇1̃ − 𝜇2̃ = 𝜈+ + 𝜈− = 𝜈. Proposition 4. Let R be a ring on a set T. Then, the mapping Δ : Measof (T, R) → Measof (T, R𝛿 (T, R)) such that Δ𝜇 = 𝜇̃ is an isomorphism of these lattice-ordered linear spaces. Proof. Take any 𝜇, 𝜈 ∈ Measof (T, R). For 𝜇, 𝜈 and 𝜆 ≡ 𝜇 + 𝜈, consider 𝜇,̃ 𝜈̃ and 𝜆.̃ ̃ = 𝜆R = 𝜇R + 𝜈R = 𝜇R ̃ + 𝜈R ̃ = (𝜇̃ + 𝜈)R. ̃ For every R ∈ R, we have 𝜆R By virtue of ̃ Lemma 3, 𝜆 = 𝜇̃ + 𝜈,̃ i. e. Δ is additive. Completely in the same way, it is checked that Δ is a homomorphism of the lattice-ordered linear spaces. It is evident that Δ is injective. ̄̄ ⩽ Take any 𝜘 ∈ Measof (T, R𝛿 (T, R)) and 𝜇 ≡ 𝜘|R. Lemma 5 (3.1.3) implies 𝜇R ̄ ̄ < ∞ for every R ∈ R, where 𝜇 is overfinite. By Lemma 3, 𝜘 = 𝜇.̃ This means that 𝜘R Δ is surjective. Finally, according to Statement 5 (2.2.7), Δ is an isomorphism. Corollary 1. Let R be a ring on T, 𝜇, 𝜈 ∈ Measof (T, R) and 𝜇,̃ 𝜈̃ ∈ Measof (T, R𝛿 (T, R)). Then, 1) 𝜈 ⊥ 𝜇 iff 𝜈̃ ⊥ 𝜇;̃ ̃ . 2) 𝜈 ∈ 𝜇⊥⊥ iff 𝜈̃ ∈ 𝜇⊥⊥ Proof. 1. Applying Statement 2 to the isomorphisms Δ and Δ−1 , we get assertion 1. 2. Let 𝜈 ∈ 𝜇⊥⊥ . Take 𝜘 ∈ 𝜇⊥̃ . Then, by (1), Δ−1 𝜘 ⊥ 𝜇 implies Δ−1 𝜘 ⊥ 𝜈. Again, by (1), ̃ . Conversely, let 𝜈̃ ∈ 𝜇⊥⊥ ̃ . Take 𝜆 ∈ 𝜇⊥ . By (1), 𝜆̃ ⊥ 𝜇̃ implies 𝜘 ⊥ 𝜈.̃ This means 𝜈̃ ∈ 𝜇⊥⊥ ⊥⊥ 𝜆̃ ⊥ 𝜈.̃ Again by (1), 𝜆 ⊥ 𝜈. This means 𝜈 ∈ 𝜇 . Auxiliary definitions and statements Let A be a lattice-ordered linear space. Elements a, b ∈ A are called disjoint if |a| ∧ |b| = 0. This relation between a and b is denoted by a ⊥ b. For E ⊂ A, the set E⊥ ≡ {a ∈ A | ∀ e ∈ E (|a| ∧ |e| = 0)} is called the band disjoint to the set E. The set E⊥⊥ ≡ (E⊥ )⊥ is called the band generated by the set E. Obviously, E ⊂ E⊥⊥ . For

3.2.3 Norms on linear spaces of bounded semimeasures and measures |

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a, b ∈ A, the element b is called subordinated to the element a or a-subordinated if b ∈ a⊥⊥ ≡ {a}⊥⊥ . Statement 1 (Riesz). Let |⟮A, R⟯, s lols | be a Dedekind complete lattice-ordered linear space and E ⊂ A. Then, for every element a ∈ A, there are the unique elements a󸀠 ∈ E⊥ and a󸀠󸀠 ∈ E⊥⊥ such that a = a󸀠 + a󸀠󸀠 . Statement 2. Let A and B be lattice-ordered linear spaces and u : A → B be a homomorphism of linear spaces. Then, the following conclusions are equivalent: 1) u is a homomorphism of lattice-ordered linear spaces; 2) ∀ a ∈ A (u(|a|) = |ua|); 3) ∀ a ∈ A (u(a + ) = (ua)+ ); 4) ∀ a󸀠 , a󸀠󸀠 ∈ A (a󸀠 ∧ a󸀠󸀠 = 0 ⇒ ua󸀠 ∧ ua󸀠󸀠 = 0); 5) ∀ a󸀠 , a󸀠󸀠 ∈ A (a󸀠 ∧ a󸀠󸀠 = 0 ⇒ ua󸀠 ∨ ua󸀠󸀠 = u(a󸀠 ∨ a󸀠󸀠 )). 3.2.3 Norms on linear spaces of bounded semimeasures and measures If R is a ring on a set T, then by Corollary 1 to Lemma 10 (3.1.3), the mapping ̄̄ 𝜀 󳨃→ 𝜀(T) is a norm on the linear space Evalb (T, R). Since SMeasb (T, R) is a linear ̄̄ subspace of Evalb (T, R), this norm induces a norm ‖ ⋅ ‖ : 𝜇 󳨃→ 𝜇(T) on the linear space SMeasb (T, R). Similarly, since SMeasb (T, R) is a linear subspace of the normed linear space (F b (R), ‖ ⋅ ‖u ) from 2.2.7. The norm ‖ ⋅ ‖u induces on SMeasb (T, R) the corresponding norm ‖ ⋅ ‖u,SMeasb (T,R) of the uniform convergence (see 2.2.7), which we shall denote simply by ‖ ⋅ ‖u . It follows from Lemma 8 (3.1.3) that ‖𝜇‖u ⩽ ‖𝜇‖ ⩽ 2‖𝜇‖u for every 𝜇 ∈ SMeasb (T, R). Consequently, the normed metrics 𝜌‖⋅‖ and 𝜌‖⋅‖u are equivalent in the sense that G(𝜌‖⋅‖ ) = G(𝜌‖⋅‖u ), i. e. they induce the same topology. Proposition 1. Let R be a ring on T. Then, ⟮SMeasb (T, R), ‖ ⋅ ‖⟯ is a normed latticeordered linear space. Moreover, it is an L-space. Proof. Denote in the proof SMeasb (T, R) by A b . Let 𝜇 ∈ A b . By Lemma 9 (3.1.3) and ̄̄ ̄̄ Corollary 1 to Theorem 2 (3.2.2), ‖|𝜇|‖ = var(|𝜇|)(T) = var(𝜇|R)(T) = 𝜇(T) = ‖𝜇‖. Let 0 ⩽ 𝜇 ⩽ 𝜈 in SMeasb (T, R). Take any finite collection (R i | i ∈ I) ∈ Parf (R, E). Then, ̄̄ ̄̄ ∑(|𝜇R i | | i ∈ I) ⩽ ∑(|𝜈R i | | i ∈ I) ⩽ 𝜈(T) = ‖𝜈‖ implies ‖𝜇‖ = 𝜇(T) ⩽ ‖𝜈‖. This means that if 0 ⩽ |𝜇| ⩽ |𝜈|, then ‖𝜇‖ = ‖|𝜇|‖ ⩽ ‖|𝜈|‖ = ‖𝜈‖, i. e. the norm ‖ ⋅ ‖ is modulusly monotone. Therefore, we have a normed lattice-ordered linear space. Now, let 𝜇, 𝜈 ∈ SMeasb (T, R)+ . Take two arbitrary finite pairwise disjoint collections (R i ∈ R | i ∈ I) and (S j ∈ R | i ∈ I). Then, the sets R ≡ ⋃⟮R i | i ∈ I⟯ and S ≡ ⋃⟮S j | j ∈ J⟯ belong to R, and therefore, R ∪ S ∈ R. Consequently, by virtue of Lemma 4 (3.1.1), we get ∑(|𝜇R i | | i ∈ I) + ∑(|𝜈S j | | j ∈ J) = 𝜇R + 𝜈S ⩽ (𝜇 + 𝜈)(R ∪ S). By Corollary 2 to Lemma 7 (3.1.3), (𝜇 + 𝜈)(R ∪ S) ⩽ sup{(𝜇 + 𝜈)E | E ∈ R} = var(𝜇 + 𝜈)

240 | 3.2 Decompositions of semimeasures and measures

̄̄ ⩽ ‖𝜇 + 𝜈‖ − ∑ |𝜈S j |. (T) = ‖𝜇 + 𝜈‖. Thus, ∑ |𝜇R i | ⩽ ‖𝜇 + 𝜈‖ − ∑ |𝜈S j | implies ‖𝜇‖ = 𝜇T ̄̄ ⩽ ‖𝜇+𝜈‖−‖𝜇‖, where ‖𝜈‖+‖𝜇‖ ⩽ ‖𝜇+𝜈‖. Similarly, ∑ |𝜈S j | ⩽ ‖𝜇+𝜈‖−‖𝜇‖ implies ‖𝜈‖ = 𝜈T Since ‖ ⋅ ‖ is a norm, we conclude that ‖𝜇 + 𝜈‖ = ‖𝜈‖ + ‖𝜇‖. Now, we shall prove that A b is closed in F b (R) with respect to the uniform convergence of sequences. Let (𝜇n ∈ A b | n ∈ 𝜔), 𝜈 ∈ F b (R), and 𝜈 = u-lim(𝜇n | n ∈ 𝜔). Let Q, R ∈ R and Q ∩ R = ⌀. Then, we have 𝜈Q = lim 𝜇n Q, 𝜈R = lim 𝜇n R and 𝜈(Q ∪ R) = lim 𝜇n (Q ∪ R). By Proposition 1 (1.4.7), 𝜈Q + 𝜈R = lim(𝜇n Q + 𝜇n R | n ∈ 𝜔) = lim 𝜇n (Q ∪ R) = 𝜈(Q ∪ R). By virtue of Lemma 1 (3.1.1), 𝜈 is a semimeasure. Take 𝜀 > 0. Then, there is n such that |𝜈R − 𝜇p R| < 𝜀 for every R ∈ R and p ⩾ n. By the condition rng 𝜇n ⊂ [a, b] ⊂ R. Therefore, a − 𝜀 ⩽ 𝜇n R − 𝜀 < 𝜈R < 𝜇n R + 𝜀 ⩽ b + 𝜀 means that 𝜈 is bounded. Thus, 𝜈 ∈ A b . By virtue of Lemma 3 (2.2.7), (A b , ‖ ⋅ ‖u ) is a Banach space. Since the metrics 𝜌‖⋅‖u and 𝜌‖⋅‖ are equivalent, (A b , ‖⋅‖) is also a Banach space. Proposition 2. Let R be a ring on T. Then, ⟮Measb (T, R), ‖ ⋅ ‖⟯ is an L-space. Proof. Let (𝜇n ∈ Measb (T, R) | n ∈ N ⊂ 𝜔) be a sequence for some cofinal subset N of 𝜔, 𝜇 ∈ SMeasb (T, R) and 𝜇 = lim(𝜇n | n ∈ 𝜔) with respect to G(‖ ⋅ ‖). By Corollary 1 to Statement 6 (2.2.7), lim(‖𝜇 − 𝜇n ‖ | n ∈ N) = 0. Therefore, lim(‖𝜇 − 𝜇n ‖u | n ∈ N) = 0. Take any decreasing sequence (R m ∈ R | m ∈ 𝜔) such that ⋂⟮R m | m ∈ 𝜔⟯ = ⌀. Fix 𝜀 > 0. Then, there exists n ∈ N such that ‖𝜇 − 𝜇p ‖u < 𝜀/2 for every p ∈ N\n. Take some p ∈ N\n. By Lemma 6 (3.1.1), lim(𝜇p R m | m ∈ 𝜔) = 0. Hence, there exists m ∈ 𝜔 such that |𝜇p R q | < 𝜀/2 for every q ⩾ m. From ‖𝜇 − 𝜇p ‖ < 𝜀/2, we deduce that |𝜇R q − 𝜇p R q | < 𝜀/2 for every q ∈ 𝜔. Therefore, |𝜇R q | ⩽ |𝜇R q −𝜇p R q |+|𝜇p R q | < 𝜀 for every q ⩾ m. This means that lim(𝜇R m | m ∈ 𝜔) = 0. By Lemma 7 (3.1.1), 𝜇 is a measure. Since by Proposition 1 ⟮SMeasb (T, R), ‖ ⋅ ‖⟯ is an L-space, this implies that ⟮Measb (T, R), ‖ ⋅ ‖⟯ is a Banach space and is therefore an L-space as well.

3.2.4 Absolute continuity and singularity and the Lebesgue decomposition In 3.2.2, we proved that the family SMeasof (T, R) of all overfinite semimeasures on a ring R of subsets of a set T has a structure of Dedekind complete lattice linear space. Now, we shall use this structure. Let 𝜇, 𝜈 ∈ SMeasof (T, R). Proposition 1. Let R be a ring on T and 𝜇 ∈ SMeasof (T, R). Then, for every element 𝜈 ∈ SMeasof (T, R), there are unique elements 𝜈𝜇󸀠 ∈ 𝜇⊥ and 𝜈𝜇󸀠󸀠 ∈ 𝜇⊥⊥ from SMeasof (T, R) such that 𝜈 = 𝜈𝜇󸀠 + 𝜈𝜇󸀠󸀠 . The same assertion is valid for Measof (T, R). Proof. The assertion follows from Theorem 2 (3.2.2) and Statement 1 (3.2.2) applied to the set E = {𝜇}.

3.2.4 Absolute continuity and singularity and the Lebesgue decomposition | 241

The property 𝜈 = 𝜈𝜇󸀠 + 𝜈𝜇󸀠󸀠 is called the Riesz decomposition of 𝜈 with respect to 𝜇. The point of interest now is to give some concrete characterizations of the disjointness and subordinateness and then to find their analogues for non-finite measures.

Singularity of a measure A semimeasure 𝜈 ∈ SMeas(T, R) is called concentrated on a set N ∈ R if the conditions R ∈ R and R ∩ N = ⌀ imply 𝜈R = 0. The semimeasure 𝜈 ∈ SMeas(T, R) will be called singular to a semimeasure 𝜇 or 𝜇-singular if there is a set N ∈ R such that v(𝜇)(N) = 0 and 𝜈 is concentrated on N. Lemma 1. Let R be a ring on T and 𝜇, 𝜈 ∈ Meas(T, R). If there are disjoint sets M, N ∈ R such that 𝜇 is concentrated on M and 𝜈 is concentrated on N, then 𝜇 and 𝜈 are mutually singular. Proof. If S ∈ R and S ⊂ N, then S ∩ M = ⌀ implies 𝜇S = 0. By virtue of Corollary 2 to Lemma 8 (3.1.3), v(𝜇)N ⩽ 2 sup{|𝜇S| | S ∈ R ∧ S ⊂ N} = 0. This means that 𝜈 is singular to 𝜇. In a similar way, it is checked that 𝜇 is singular to 𝜈. Lemma 2. Let R be a ring on T and 𝜇, 𝜈 ∈ Measof (T, R). If 𝜈 is singular to 𝜇, then 𝜈 is disjoint to 𝜇. Proof. Let N ∈ R such that v(𝜇)N = 0 and 𝜈 is concentrated on N. Take any R ∈ R and consider E ≡ R ∩ N and F ≡ R\N. Then, E, F ∈ R. By virtue of Lemma 4 (3.1.3), v(𝜇)E = 0. If S ∈ R and S ⊂ F, then S ∩ N = ⌀ implies 𝜈S = 0. Using Corollary 2 to Lemma 8 (3.1.3) yields v(𝜈)F = 0. Therefore, by assertion 5 of Corollary 1 to Theorem 2 (3.2.2), we have |𝜇|E = 0 and |𝜈|F = 0. Consequently, by Theorem 2 (3.2.2), (|𝜇| ∧ |𝜈|)R = inf{x ∈ R+ | ∃(E, F) ∈ R × R ((x = |𝜇|E + |𝜈|F) ∧ (E ∪ F = R) ∧ (E ∩ F = ⌀))} = 0 for every R ∈ R. Thus, 𝜇 ⊥ 𝜈. Proposition 2. Let M be a 𝜎-algebra on T, 𝜇, 𝜈 ∈ Measb (T, M). If 𝜇 and 𝜈 are disjoint, then there are disjoint sets M, N ∈ M such that M ∪ N = T, 𝜇 is concentrated on M, 𝜈 is concentrated on N, and therefore, 𝜇 and 𝜈 are mutually singular. Proof. By Theorem 2 (3.2.2), 0 = (|𝜇|∧|𝜈|)T = inf{x ∈ R+ | ∃(E, F) ∈ M×M ((x = |𝜇|E+ |𝜈|F) ∧ (E ∪ F = T) ∧ (E ∩ F = 0))}. This means that for every n ∈ 𝜔, there exists E n ∈ M such that |𝜇|E n + |𝜈|(T\E n ) < 2−n . Consider the sets N k ≡ ⋃⟮E n | n ∈ 𝜔\k⟯, N ≡ ⋂⟮N k | k ∈ 𝜔⟯ and M ≡ T\N from M. Then, by Lemma 3 (3.1.1) and Lemma 3 (1.4.8), |𝜇|N k ⩽ ∑net (|𝜇|E n | n ∈ 𝜔\k) ⩽ ∑net (2−n | n ∈ 𝜔\k) = 2−k+1 . Besides, the sequence (N k | k ∈ 𝜔) decreases. Therefore, by Lemma 6 (3.1.1) and Lemma 7 (1.4.4), |𝜇|N = lim(|𝜇|N k | k ∈ 𝜔) ⩽ lim(2−k+1 | k ∈ 𝜔) = 0. Furthermore, |𝜈|M ⩽ |𝜈|(T\N k ) ⩽ |𝜈|(T\E k ) < 2−k for every k implies |𝜈|M = 0.

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If E ∈ M and E ∩ M = ⌀, then |𝜇|E = 0. By Corollary 2 to Theorem 2 (3.2.2), 𝜇E = 0. This means that 𝜇 is concentrated on M. Similarly, 𝜈 is concentrated on N. By Lemma 1, 𝜇 and 𝜈 are mutually singular. Absolute continuity of a measure Let 𝜇 and 𝜈 be semimeasures on a ring R. The semimeasure 𝜈 is called absolutely continuous with respect to the semimeasure 𝜇 or 𝜇-absolutely continuous if for every S ∈ R and every 𝜀 > 0, there is 𝛿 > 0 such that the conditions R ∈ R, R ⊂ S and v(𝜇)R < 𝛿 imply |𝜈R| < 𝜀. It will be symbolized by 𝜈 ≪ 𝜇. Theorem 1. Let R be a 𝛿-ring on a set T, 𝜇, 𝜈 ∈ Meas(T, R). Then, the following assertions are equivalent: 1) 𝜈 is 𝜇-absolutely continuous, i. e. 𝜈 ≪ 𝜇; 2) S ∈ R and v(𝜇)S = 0 imply 𝜈S = 0; 3) v(𝜈) is 𝜇-absolutely continuous, i. e. v(𝜈) ≪ 𝜇; 4) S ∈ R and v(𝜇)S = 0 imply v(𝜈)S = 0. Moreover, if 𝜇 and 𝜈 are overfinite, then these assertions are also equivalent to the following one: 5) 𝜈 is 𝜇-subordinate, i. e. 𝜈 ∈ 𝜇⊥⊥ . Proof. The implication (1) ⊢ (2) is obvious. The equivalence of assertions 1 and 3 and assertions 2 and 4 follows from Corollary 2 to Lemma 8 (3.1.3). (2) ⊢ (1). Suppose that (1) is not fulfilled. Then, there are S ∈ R and 𝜀 > 0 such that for every 𝛿 > 0, there is R ∈ R such that R ⊂ S, v(𝜇)R < 𝛿 and |𝜈R| ⩾ 𝜀. This means that for 𝛿n ≡ 2−n , there is R n ∈ R such that R n ⊂ S, v(𝜇)R n < 2−n and |𝜈R n | ⩾ 𝜀. By Theorem 1 (3.2.1), for every n, there is the Hahn decomposition (R+n , R−n ) of the set R n with respect to 𝜈. Then, the condition |𝜈(R+n ) + 𝜈(R−n )| ⩾ 𝜀 implies either 𝜈(R+n ) ⩾ 𝜀 − 𝜈(R−n ) ⩾ 𝜀 or 𝜈R−n ⩽ −𝜀 − 𝜈(R+n ) ⩽ −𝜀. Consider the subsets M ≡ {m ∈ 𝜔 | 𝜈(R+m ) ⩾ 𝜀} and N ≡ {n ∈ 𝜔 | 𝜈R−n ⩽ −𝜀} of 𝜔. Then, either M is cofinal in 𝜔 or N. In the first case, consider the new sequence (S m ∈ R | m ∈ M) such that S m ≡ R+m . In the second case, consider the new sequence (T n ∈ R | n ∈ N) such that T n ≡ R−n . Suppose first that M is cofinal. For the sequence (S m | m ∈ M), consider the sets A k ≡ ⋃⟮S m | m ∈ M\k⟯ ⊂ S and A ≡ ⋂⟮A k | k ∈ 𝜔⟯ ⊂ S. By Lemma 9 (2.1.1), A k ∈ R, where A ∈ R. Then, by Lemma 3 (3.1.1) and Lemma 3 (1.4.8)v(𝜇)A k ⩽ ∑net (v(𝜇)S m | m ∈ M\k) ⩽ ∑net (2−m | m ∈ M \ k) ⩽ ∑(2−m | m ∈ 𝜔 \ k) = 2−k+1 . Besides, the sequence (A k | k ∈ 𝜔) decreases. Therefore, by Corollary 1 to Proposition 2 (3.1.3), Lemma 6 (3.1.1), and Lemma 7 (1.4.4), we get v(𝜇)A = lim(v(𝜇)A k | k ∈ 𝜔) ⩽ lim(2−k+1 | k ∈ 𝜔) = 0. At the same time, every S m is essentially positive for 𝜈. Hence, by Lemma 3 (3.2.1), the set A0 is essentially positive. Take for every k the number m k ≡ sm(M\k). Then,

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according to 3.2.1, the inclusions S m ⊂ A k ⊂ A0 imply 𝜈A k ⩾ 𝜈S m k ⩾ 𝜀. Hence, 𝜈A = lim(𝜈A k | k ∈ 𝜔) ⩾ 𝜀. If M is not cofinal, then N is cofinal. Consider the sets B k ≡ ⋃⟮T n | n ∈ N\k⟯ ⊂ S and B ≡ ⋂⟮B k | k ∈ 𝜔⟯ ⊂ S from R. Then, v(𝜇)B = 0 and 𝜈B ⩽ −𝜀, because T n k ⊂ B k ⊂ B0 implies 𝜈B k ⩽ 𝜈T n k ⩽ −𝜀, where n k ≡ sm(N\k). In both cases, we get the contradiction. (5) ⊢ (2). Now, suppose that 𝜇 and 𝜈 are overfinite. Suppose that (2) is not fulfilled. Then, there is N ∈ R such that v(𝜇)N = 0 and 𝜈N = / 0. According to Corollary 1 to Theorem 2 (3.2.2) and Lemma 2 (3.2.1), 𝜘 ≡ |𝜈| is an overfinite measure. Consider the new evaluation 𝜆 on R such that 𝜆R ≡ 𝜘(R ∩ N) for every R ∈ R. According to 3.1.1, 𝜆 is a finite measure. Take for R any finite collection (R i | i ∈ I) ∈ Parf (R, R). Then, ̄̄ ̄̄ by virtue of Lemma 2 (3.1.3), ∑(|𝜆R i | | i ∈ I) ⩽ ∑(|𝜘R i | | i ∈ I) ⩽ 𝜘(R) implies 𝜆(R) ⩽ ̄̄ 𝜘(R) < ∞. Hence, 𝜆 is overfinite. By Corollary 2 to Theorem 2 (3.2.2), 𝜆N = |𝜈|N ⩾ |𝜈N| > 0. Besides, 𝜆R ⩽ |𝜈|R for every R ∈ R. Consequently, |𝜈| ⩾ 𝜆 > 0. Since R ∈ R and R ∩ N = ⌀ imply 𝜆R = 0, we infer that 𝜆 is concentrated on N. By Lemma 2 |𝜇| ∧ 𝜆 = 0, where 𝜆 ∈ 𝜇⊥ . By the condition, 𝜈 ∈ 𝜇⊥⊥ . Hence, |𝜈| ∧ 𝜆 = 0. But this contradicts the inequality |𝜈| ⩾ 𝜆 > 0. (3) ⊢ (5). By condition 3, for every 𝜀 > 0, there is 𝛿𝜀 > 0 such that S ∈ R, S ⊂ R, and |𝜇|S < 𝛿𝜀 imply |𝜈|S < 𝜀/2. Take any 𝜆 ∈ Measof (T, R) such that 𝜆 ∈ 𝜇⊥ . By Theorem 2 (3.2.2), we get 0 = (|𝜆|∧|𝜇|)R = inf{x ∈ R+ | ∃(E, F) ∈ Parf (R, R) (x = 𝜆(E)+ 𝜈(F))} for every R ∈ R. This means that for 𝛿 ≡ sm{𝛿𝜀 , 𝜀/2} > 0, there is a partition (E, F) of the set R such that E, F ∈ R and |𝜆|E + |𝜇|F < 𝛿. As a result, F ∈ R, F ⊂ R, and |𝜇|F < 𝛿 ⩽ 𝛿𝜀 imply |𝜈|F < 𝜀/2, where |𝜆|E + |𝜈|F < 𝛿 + 𝜀/2 ⩽ 𝜀. This means that (|𝜆| ∧ |𝜈|)R = {x ∈ R+ | ∃(E, F) ∈ Parf (R, R) (x = 𝜆(E) + 𝜈(F))} = 0. Hence, 𝜈 ⊥ 𝜆, i. e. 𝜈 ∈ 𝜇⊥⊥ . Remark. In the proof of Theorem 1, the deductions (1) ⊢ (2) and (3) ⊢ (5) and the equivalences (1) ⇔ (3) and (2) ⇔ (4) are, in fact, proven for an arbitrary ring R. Corollary 1. Let R be a 𝛿-ring on a set T, 𝜇 and 𝜈 be measures on R. If 𝜈 ≪ 𝜇, then v+ (𝜈) ≪ 𝜇 and v− (𝜈) ≪ 𝜇. Proof. By Theorem 1, v(𝜈) ≪ 𝜇. By virtue of Proposition 2 (3.2.1), v+ (𝜈) ⩽ v(𝜈) and −v− (𝜈) ⩽ v(𝜈). Consequently, v+ (𝜈) ≪ 𝜇 and v− (𝜈) ≪ 𝜇. Consider now the operator of extension Δ : Measof (T, R) from 3.2.2 such that Δ𝜇 = 𝜇.̃

Measof (T, R𝛿 (T, R))

Lemma 3. Let R be a ring on T, 𝜇, 𝜈 ∈ Measof (T, R) and 𝜇,̃ 𝜈̃ ∈ Measof (T, R𝛿 (T, R)). Then, 𝜈 ≪ 𝜇 iff 𝜈̃ ≪ 𝜇.̃

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Proof. Let 𝜈 ≪ 𝜇. By virtue of Theorem 1 and the remark to it, 𝜈 ∈ 𝜇⊥⊥ . By Corollary 1 ̃ . Again, by virtue of Theorem 1, 𝜈̃ ≪ 𝜇.̃ to Proposition 4 (3.2.2), 𝜈̃ ∈ 𝜇⊥⊥ Now, let 𝜈̃ ≪ 𝜇.̃ Take any S ∈ R ⊂ R𝛿 (T, R) and any 𝜀 > 0. By the condition, there ̃ < 𝛿 imply |𝜈E| ̃ < 𝜀. Let R ∈ R, R ⊂ S, is 𝛿 > 0 such that E ∈ R𝛿 (T, R), E ⊂ S, and |𝜇|E and |𝜇|R < 𝛿. By virtue of Proposition 4 (3.2.2), |𝜇|̃ is the extension of |𝜇|. Therefore, ̃ = |𝜇|R < 𝛿. This implies |𝜈R| = |𝜈R| ̃ < 𝜀. As a result, 𝜈 ≪ 𝜇. |𝜇|R After such a preparation, we can prove another version of Theorem 1. Theorem 2. Let R be a ring on a set T, 𝜇, 𝜈 ∈ Measof (T, R). Then, the following assertions are equivalent: 1) 𝜈 is 𝜇-absolutely continuous, i. e. 𝜈 ≪ 𝜇; 2) |𝜈| is 𝜇-absolutely continuous, i. e. |𝜈| ≪ 𝜇; 3) 𝜈 is 𝜇-subordinated, i. e. 𝜈 ∈ 𝜇⊥⊥ . Proof. The equivalence of assertions 1 and 2 and the deduction (1) ⊢ (3) are valid in virtue of Theorem 1 and the remark to it. ̃ . By virtue of (3) ⊢ (1). Let 𝜈 ∈ 𝜇⊥⊥ . By Corollary 1 to Proposition 4 (3.2.2), 𝜈̃ ∈ 𝜇⊥⊥ Theorem 1 𝜈̃ ≪ 𝜇̃ and by Lemma 3, 𝜈 ≪ 𝜇.

The Lebesgue decomposition Let M be a 𝜎-algebra and 𝜈 be a measure on M. Consider the factor-ensemble M ≡ M/M0 (𝜈) of equivalence classes M of elements M ∈ M with respect to the ideal ensemble M0 (𝜈). According to 2.1.4, M 󸀠 ∈ M iff M 󸀠 󳵻 M ∈ M0 (𝜈). Recall that if A ∈ M, then MA ≡ {M ∈ M | M ⊂ A}. Theorem 3. Let M be a 𝜎-algebra on T and 𝜇 be a measure on M. Then, for every measure 𝜈 ∈ Meas(T, M, R ∪ {∞}) [Meas(T, M, {−∞} ∪ R)], there are the unique element a ∈ M ≡ M/M0 (𝜈), some set A ∈ a, and the unique measures 𝜈𝜇s and 𝜈𝜇ac from Meas(T, M, R ∪ {∞}) [Meas(T, M, {−∞} ∪ R)] such that 𝜈𝜇s is 𝜇-singular, 𝜈𝜇ac is 𝜇-absolutely continuous, 𝜈𝜇s M = 𝜈(M ∩ A) and 𝜈𝜇ac M = 𝜈(M\A) for every M ∈ M, v(𝜇)A = 0, and 𝜈 = 𝜈𝜇s + 𝜈𝜇ac . Proof. By Corollary 1 to Proposition 2 (3.1.3), v(𝜇) is a positive measure. First, suppose that 𝜈 ⩾ 0. Consider the ensemble N = {N ∈ M | v(𝜇)N = 0}. Since |𝜇|⌀ = 0, we conclude that N = / ⌀. Then, the subensemble N of M generates the non-empty subensemble N ≡ {N | N ∈ N} in M. The lattice order on M such that M ⩽ N ⇔ M ∨ N = N induces the corresponding order on N. Define an increasing evaluation 𝜆 : M → R+ , setting 𝜆M ≡ 𝜈M 󸀠 for any M 󸀠 ∈ M. Take any non-empty chain C in N and consider the number x ≡ sup{𝜆N | N ∈ C} ∈ R+ . First, suppose that there is M ∈ C such that x = 𝜆M. Then, M is an ̄ In fact, if N ∈ C, then either N ⩽ M or M ⩽ N. In the second upper bound for C in N.

3.2.4 Absolute continuity and singularity and the Lebesgue decomposition | 245

case, x = 𝜆M ⩽ 𝜆N ⩽ x implies 𝜆M = 𝜆N, where 𝜈M = 𝜈N. At the same time, M ∪ N = N means (M ∪ N)\N = (M ∪ N) 󳵻 N ∈ M0 (𝜈). Consequently, 𝜈(M ∪ N) = 𝜈N + 𝜈((M ∪ N)\N) = 𝜈N = 𝜈M gives M ∪ N\M ∈ M0 (𝜈), where M ∨ N = M ∪ N = M, i. e. N ⩽ M. Thus, M is an upper bound. Now, suppose that for every N ∈ C, we have 𝜆N < x. Then, consider the choice mapping p : P(N)\{⌀} → N from the axiom of choice (1.1.12). If x < ∞, then for every n ∈ N, consider the non-empty subset C n ≡ {N ∈ C | 𝜆N > x − 1/n} and take the element M n ≡ pC n ∈ C n . If x = ∞, then for every n ∈ N consider the nonempty subset C n ≡ {N ∈ C | 𝜆N > n} and take the element M n ≡ pC n ∈ C n . In both cases, sup{𝜆M n | n ∈ N} = x. Again, take some choice mapping q : P(M)\{⌀} → M. / ⌀ for every n, we can take the elements M 󸀠n ≡ q(M n ∩ N) ∈ Since M n ∩ N = M n ∩ N. By virtue of Corollary 1 to Theorem 3 (1.2.6), for every n ∈ N the set {M k | k ∈ n + 1} has the greatest element which will be denoted by N n . Then, M n = sup{M k | k ∈ n + 1} and N n ∈ C. Therefore, 𝜆M n ⩽ 𝜆N n ⩽ x. Consider the sets N n󸀠 ≡ ⋃⟮M 󸀠k | k ∈ n + 1⟯ ↑. 󸀠

Then, N n = sup{M k | k ∈ n + 1} = N n ∈ C. Consider also the set M ≡ ⋃⟮N n󸀠 | n ∈ N⟯ = ⋃⟮M 󸀠n | n ∈ N⟯ ∈ M. Using Lemma 5 (3.1.1), we get v(𝜇)M = lim(|𝜇|N n󸀠 |n ∈ N) = 0, i. e. M ∈ N and M ∈ N. Besides, M ⩾ 󸀠

N n = N n ⩾ M n for every n. Check that M is an upper bound for C in N. In fact, if N ∈ C, then either N ⩽ M n for some n ∈ N or N ⩾ M n for every n. In the first case, N ⩽ M. In the second case, x ⩾ 𝜆N ⩾ 𝜆M n for every n implies 𝜆N = x. But this equality contradicts our supposition that 𝜆N < x. This means that the second case is impossible. As a result, M is an upper bound. Thus, the chain C has an upper bound. By virtue of the Kuratowski – Zorn lemma from Theorem 1 (1.2.11), the ordered set N contains some maximal element a. Take some A ∈ a ∩ N. Consider the new positive measures 𝜈1 and 𝜈2 on M such that 𝜈1 R ≡ 𝜈(R ∩ A) and 𝜈2 R ≡ 𝜈(R\A) for every R ∈ M. Then, we have v(𝜇)A = 0 and 𝜈2 A = 0. Let S ∈ M and v(𝜇)S = 0. Then, S ∈ N. Suppose that 𝜈2 S > 0. Consider the set B ≡ A ∪ S ∈ M. Since B\A = S\A and v(𝜇)B = v(𝜇)A = v(𝜇)(S\A) = 0, we get B ∈ N. Furthermore 𝜈(B 󳵻 A) = 𝜈(B\A) = 𝜈(S\A) = 𝜈2 S > 0 means that A < B ∈ N. But this contradicts the maximality of A. Hence, 𝜈2 S = 0. By virtue of Theorem 1, this means that 𝜈2 is 𝜇-absolutely continuous. If R ∈ M and R ∩ A = ⌀, then 𝜈1 R = 0. Hence, 𝜈1 is concentrated on A. Besides, v(𝜇)A = 0. This means that 𝜈1 is 𝜇-singular. Moreover, 𝜈 = 𝜈1 + 𝜈2 . Now, we take an arbitrary measure 𝜈. We may and shall suppose that rng 𝜈 ⊂ ] − ∞, ∞]. By Proposition 2 (3.2.1), 𝜈 = v+ (𝜈) + v− (𝜈) and rng v− (𝜈) ⊂ R− . Denote v+ (𝜈) and −v− (𝜈) by 𝜋 and 𝜘. Then, rng 𝜋 ⊂ R+ and rng 𝜘 ⊂ R+ . It was proven above that 𝜋 = 𝜋1 + 𝜋2 and 𝜘 = 𝜘1 + 𝜘2 , where 𝜋1 , 𝜋2 , 𝜘1 , and 𝜘2 are positive, 𝜋1 and 𝜘1 are 𝜇-singular, and 𝜋2 and 𝜘2 are 𝜇-absolutely continuous. Besides, 𝜘1 and 𝜘2 are finite. Consider the measures 𝜈1 ≡ 𝜋1 − 𝜘1 and 𝜈2 ≡ 𝜋2 − 𝜘2 . It was proven above that there are sets A, B ∈ M such that v(𝜇)A = 0, v(𝜇)B = 0, 𝜋1 is concentrated on A and 𝜘1 is concentrated on B. Consider C ≡ A ∪ B. Then, by Lemma 5 (3.1.1), v(𝜇)C = 0. If

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R ∩ C = ⌀, then 𝜋1 R = 0 = 𝜘1 R implies 𝜈1 R = 0. This means that 𝜈1 is concentrated on C. Thus, 𝜈1 is singular to 𝜇. It is also proven above that S ∈ M and v(𝜇)S = 0 imply 𝜋2 S = 0 and 𝜘2 S = 0, where 𝜈2 S = 0. By virtue of Theorem 1, this means that 𝜈2 ≪ 𝜇. Moreover, 𝜈 = 𝜈1 + 𝜈2 and 𝜈1 , 𝜈2 ∈ Meas(T, M, R ∪ {∞}). Finally, we need to check that this decomposition 𝜈 = 𝜈1 + 𝜈2 is unique. Suppose that there are b ∈ M/M0 (𝜈), B ∈ b, and 𝜆 1 , 𝜆 2 ∈ Meas(T, M, R ∪ {∞}) such that 𝜆 1 is 𝜇-singular, 𝜆 2 is 𝜇-absolutely continuous, 𝜆 1 R = 𝜈(R ∩ B) and 𝜆 2 R = 𝜈(R\B) for every R ∈ M, v(𝜇)B = 0, and 𝜈 = 𝜆 1 + 𝜆 2 . Consider the set C ≡ A\B. We have 𝜈1 C = 𝜈C, 𝜈2 C = 0, 𝜆 1 C = 0 and 𝜆 2 C = 𝜈C. Since v(𝜇) is a positive measure, we have v(𝜇)C = 0. This implies by virtue of Theorem 1 that 𝜆 2 C = 0. Hence, 𝜈1 C = 0. Thus, 𝜈C = 0, i. e. C ∈ M0 (𝜈). Similarly, D ≡ B\A ∈ M0 (𝜈). Therefore, a = A = B = b. Moreover, for any R ∈ M, consider E ≡ (R ∩ B)\(R ∩ A) and F ≡ (R ∩ A)\(R ∩ B). Then, 𝜈1 E = 0, 𝜈2 E = 𝜈E, 𝜆 1 E = 𝜈E, and 𝜆 2 E = 0. From E ⊂ B, we infer that v(𝜇)E = 0, and therefore, 𝜈2 E = 0. Consequently, 𝜆 1 E = 0. Thus, 𝜆E = 0. Similarly, 𝜈F = 0. It follows from these facts that 𝜆 1 R = 𝜈(R ∩ B) = 𝜈((R ∩ B) ∩ (R ∩ A)) + 𝜈E = 𝜈((R ∩ A) ∩ (R ∩ B)) + 𝜈F = 𝜈(R ∩ A) = 𝜈1 R, where 𝜆 1 = 𝜈1 . Consider also the sets G ≡ (R\B)\(R\A) = (R ∩ A)\B and H ≡ (R\A)\(R\B) = (R ∩ B)\A. Then, 𝜈1 G = 𝜈G, 𝜈2 G = 0, 𝜆 1 G = 0, and 𝜆 2 G = 𝜈G. From G ⊂ A, we infer that v(𝜇)G = 0, and therefore, 𝜆 2 G = 0. Consequently, 𝜈1 G = 0. Hence, 𝜈G = 0. Similarly, 𝜈H = 0. This implies 𝜆 2 R = 𝜈(R\B) = 𝜈((R\B) ∩ (R\A)) + 𝜈G = 𝜈((R\A) ∩ (R\B)) + 𝜈H = 𝜈(R\A) = 𝜈2 R, where 𝜆 2 = 𝜈2 . The property 𝜈 = 𝜈𝜇s + 𝜈𝜇ac is called the Lebesgue decomposition of 𝜈 with respect to 𝜇. Corollary 1. Let M be a 𝜎-algebra on T and 𝜇 be a bounded measure on M. Then, for every bounded measure 𝜈 on M, the Riesz decomposition 𝜈 = 𝜈𝜇󸀠 + 𝜈𝜇󸀠󸀠 from Proposition 1 and the Lebesgue decomposition 𝜈 = 𝜈𝜇s + 𝜈𝜇ac from Theorem 3 coincide. Proof. The assertion follows from Proposition 1, Theorem 3, Proposition 2, Lemma 2, and Theorem 1. Let 𝜇 be a measure on a ring R. Consider in the family Measof (T, M) the subfamily Measof (T, M, ≪ 𝜇) of all 𝜇-absolutely continuous measures. Lemma 4. Let 𝜇 be a measure on a ring R. Then, Measof (T, R, ≪ 𝜇) is an l-ideal of the lattice-ordered linear space Measof (T, R). Proof. By Corollary 2 to Proposition 2 (3.2.2), A ≡ Measof (T, R) is a lattice-ordered linear space. Take any x, y ∈ R\{0} and 𝜘, 𝜆 ∈ B ≡ Measof (T, M, ≪ 𝜇). If S ∈ R and 𝜀 > 0, then by assertion 4 of Corollary 1 to Theorem 2 (3.2.2), Theorem 1, and the remark to it, there is 𝛾 > 0 [𝛿 > 0] such that the conditions R ∈ R, R ⊂ S, and v(𝜇)R < 𝛾

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[v(𝜇)R < 𝛿] imply |𝜘|R < 𝜀/(2|x|) [|𝜆|R < 𝜀/(2|y|)]. Consequently, if v(𝜇)R < 𝛾 ⊼ 𝛿, then we get |x𝜘+y𝜆|R ⩽ |x||𝜘|R+|y||𝜆|R < 𝜀. By Theorem 1 and the remark to it, we conclude that x𝜘 + y𝜆 ∈ B. Hence, B is a linear subspace of A. Let 𝜘 ∈ A, 𝜆 ∈ B and |𝜘| ⩽ |𝜆|. If S ∈ R and 𝜀 > 0, then there is 𝛿 > 0 such that the conditions R ∈ R, R ⊂ S and v(𝜇)R < 𝛿 imply |𝜘|R ⩽ |𝜆|R < 𝜀. This means, as above, that 𝜘 ∈ B. Hence, B is an l-ideal. Corollary 1. Let 𝜇 be a measure on a ring R. Then, the lattice-ordered linear subspace Measof (T, R, ≪ 𝜇) of the Dedekind complete lattice-ordered linear space Measof (T, R) is Dedekind complete as well; the smallest upper [the greatest lower] bounds of bounded above [below] sets in the ordered sets Measof (T, R) and Measof (T, R, ≪ 𝜇) coincide and they are expressed by the formulas from Theorem 1 (3.2.2) and Theorem 2 (3.2.2) (with its corollaries). Proof. By Corollary 2 to Proposition 2 (3.2.2), Measof (T, R) is Dedekind complete. The assertion now follows from Lemma 4 and Statements 3 and 1 (2.2.8). Corollary 2. Let 𝜇 be a measure on a ring R. If 𝜈 ∈ Measof (T, R, ≪ 𝜇), then |𝜈|, 𝜈+ , and 𝜈− belong to Measof (T, R, ≪ 𝜇). Proof. By Theorem 1 and the remark to it, v(𝜈) ≪ 𝜇. By Corollary 1 of Theorem 2 (3.2.2), |𝜈| = v(𝜈). Hence, |𝜈| ∈ A ≡ Measof (T, R, ≪ 𝜇). From 𝜈+ ⩽ |𝜈| and |𝜈− | ⩽ |𝜈|, we conclude by Lemma 4 that 𝜈+ ∈ A and 𝜈− ∈ A.

3.3 The Lebesgue integral The idea of construction of an abstract Lebesgue integral belongs to M. Fréchet [1915a], and it is based on the complex of brilliant ideas suggested by J. Radon in his famous paper [Radon, 1913]. 3.3.1 Measurable functions on a space with a positive wide measure Let ⟮T, M, 𝜇⟯ be a measurable space with a 𝜎-algebra M and a positive measure 𝜇 : M → R+ . Instead of M0 (𝜇), Mf (𝜇), and M𝜎f (𝜇) we shall write now simply M0 , Mf , and M𝜎f , respectively. By Lemma 9 (3.1.1), the ensemble M0 is a 𝜎-ring. For us, the most important cases of such a situation will be the following cases. ̂ Let 𝜌 be a positive measure on a ring R and 𝜌̂ : M(T, R, 𝜌) → R+ be its large complete ̂ saturated extension (see 3.1.5). Then, we can take M = M(T, R, 𝜌) and 𝜇 ≡ 𝜌.̂ We also can take any 𝜎-subalgebra M of the 𝜎-algebra M(T, R, 𝜌) containing the ring R ̂ and the measure 𝜇 ≡ 𝜌|M. If 𝜌 is a positive measure on a ring R such that its Lebesgue – Caratheodory extension 𝜌× is internally finite, then we can consider also

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̂ ̂ the large complete strongly saturated extension 𝜌̂̂ : M(T, R, 𝜌) → R+ of 𝜌. In this ̂ ̂ ̂ case, we can take M ≡ M(T, R, 𝜌) and 𝜇 ≡ 𝜌.̂ We also can take any 𝜎-subalgebra M ̂ ̂̂ ̂ of the 𝜎-algebra M(T, R, 𝜌) containing the ring R and the measure 𝜇 ≡ 𝜌|M. According to 2.3.1, functions f from M(T, M) are called M-measurable. Let 𝜑(x) be a formula about classes depending on an argument x (see 1.1.2). If the proposition 𝜑(t) about points t ∈ T is true for every t from some set U ∈ co-M0 , i. e. the formula ∃ U ∈ co-M0 (∀ t ∈ U (𝜑(t))) is true in the sense of 1.1.3, then we shall say that the proposition 𝜑(t) is true M-almost everywhere or 𝜇-almost everywhere. Lemma 1. Let ⟮T, M, 𝜇⟯ be a measurable space with the locally complete wide positive measure 𝜇. Let f ∈ F(T) and g ∈ M(T, M). If f (t) = g(t) 𝜇-almost everywhere, then f ∈ M(T, M). Proof. Let f |U = g|U, U ∈ M and M ≡ T\U ∈ M0 . Then, f −1 (]x, y[) = (g−1 (]x, y[) ∩ U) ∪ (f −1 (]x, y[) ∩ N) ∈ M because M0 is an ideal and a ring by virtue of Lemma 9 (3.1.1). Proposition 1. Let ⟮T, M, 𝜇⟯ be a measurable space with the locally complete wide positive measure 𝜇. Let (f n ∈ M(T, M) | n ∈ 𝜔) be a sequence, f ∈ F(T), and f (t) = lim(f n (t) | n ∈ 𝜔) 𝜇-almost everywhere. Then, f ∈ M(T, M). Proof. Let f |M = p-lim(f n |M | n ∈ 𝜔) (see 2.2.3) for some M ∈ co-M0 . Consider the functions g ≡ f 𝜒(M) ∈ F(T) and g n ≡ f n 𝜒(M). By Corollary 1 to Lemma 2 (2.3.1) and Theorem 1 (2.3.2), g n ∈ M(T, M). By Corollary 3 to Lemma 2 (2.3.4), g = p-lim (g n | n ∈ 𝜔) ∈ M(T, M). By Lemma 1, f ∈ M(T, M). ̂ Corollary 1. Let 𝜌 be a positive measure on a ring R on a set T, M ≡ M(T, R, 𝜌), and 𝜇 ≡ 𝜌.̂ Let f ∈ F(T) and for every R ∈ R there is a sequence (f n ∈ M(T, M) | n ∈ 𝜔) such that f (t) = lim(f n (t) | n ∈ 𝜔) for every t ∈ R\N and for some N ∈ M0 . Then, f ∈ M(T, M). Proof. Since f n 𝜒(R) ∈ M(T, M), we infer from Proposition 1 that f 𝜒(R) ∈ M(T, M). Therefore, M R ≡ (f 𝜒(R))−1 []x, y[] ∈ M, and hence, L R ≡ M R ∩ R ∈ L(T, R, 𝜌). Consequently, for M ≡ f −1 []x, y[] we have M ∩ R = L R ∈ L(T, R, 𝜌). By Lemma 5 (3.1.5), M ∈ M. Corollary 2. In the conditions of the previous corollary for a function f ∈ F(T) the following conditions are equivalent: 1) f ∈ M(T, M); 2) f 𝜒(R) ∈ M(T, M) for every R ∈ R.

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Corollary 3. Let 𝜌 be a positive measure on a ring R on a set T such that the measure 𝜌× ̂ ̂ is internally finite. Let M ≡ M(T, R, 𝜌) and 𝜇 ≡ 𝜌.̂̂ Let f ∈ F(T) and for every R ∈ Rf (𝜌) there is a sequence (f n ∈ M(T, M) | n ∈ 𝜔) such that f (t) = lim(f n (t) | n ∈ 𝜔) for every t ∈ R\N and for some N ∈ M0 . Then, f ∈ M(T, M). Proof. Analogously, to the proof of Corollary 1 for M ≡ f −1 []x, y[], we have M ∩ R ∈ L(T, R, 𝜌) for every R ∈ Rf (𝜌). By virtue of Proposition 3 (3.1.5), M ∈ M. Corollary 4. In the conditions of the previous corollary for a function f ∈ F(T), the following conditions are equivalent: 1) f ∈ M(T, M); 2) f 𝜒(R) ∈ M(T, M) for every R ∈ Rf (𝜌). Theorem 1 (the Egorov theorem). Let 𝜌 be a positive measure on a ring R on a set T, ̂ ̂ R ⊂ M ⊂ M(T, R, 𝜌), 𝜇 ≡ 𝜌|M and 𝜇 be a wide locally complete measure. Let f ∈ F(T), (f n ∈ M(T, M) | n ∈ 𝜔) be a sequence and f (t) = lim(f n (t) | n ∈ 𝜔) 𝜇-almost everywhere. Then, for every set A ∈ Mf and every number 𝜀 > 0 there is a set B ∈ R𝜎f (T, R, 𝜌) such that B ⊂ A, 𝜇(A\B) < 𝜀, and f |B = u-lim(f n |B | n ∈ 𝜔) (see 2.2.3). If, ̂ ̂ R, 𝜌) and moreover, 𝜌× is internally finite then the same is valid for R ⊂ M ⊂ M(T, ̂ ̂ 𝜇 ≡ 𝜌|M. ̂ 0 and f |M = p-lim(f |M | n ∈ 𝜔). By Proposition 1, f is 𝜇-̂ Proof. Let M ∈ co-M n ̂ f . We have measurable. Consider the set A󸀠 ≡ A ∩ M ∈ M A󸀠pqm ≡ {t ∈ A󸀠 | |f p (t) − f q (t)| < 1/m} = ̂ = ⋃ {{t ∈ A󸀠 | |f p (t) − r| < 𝜀 − 𝜀/m} ∩ {t ∈ A󸀠 | |f q (t) − r| < 𝜀/m} | r ∈ Q} ∈ M. { } ̂ It Therefore, for every n ∈ 𝜔 and m ∈ N, we get A nm ≡ ⋂⟮A󸀠pqm | p, q > n⟯ ∈ M. 󸀠 󸀠 󸀠 is clear that (A nm | n ∈ 𝜔) ↑ A , since f |A = p-lim (f n |A | n ∈ 𝜔). It follows from Lemma 6 (3.1.1) that lim(𝜇(A󸀠 \A nm ) | n ∈ 𝜔) = 0. Consequently, we can find a natural ̂ 󸀠 \A m ) < 𝜀/2m+1 , where A m ≡ A n(m)m . The set A0 ≡ number n(m) such that 𝜇(A ̂ ⋂⟮A m | m ∈ N⟯ is 𝜇-measurable. Besides, by Lemma 3 (3.1.1) 𝜇(A\A0 ) = 𝜇(A󸀠 \A0 ) = 󸀠 󸀠 𝜇(⋃⟮A \A m | m ∈ N⟯) ⩽ ∑(𝜇(A \A m ) | m ∈ N) < 𝜀/2. For each m and each t ∈ A0 , we have t ∈ A m , where |f p (t) − f q (t)| < 1/m for p, q > n(m). It means that f |A0 = u-lim (f n |A0 | n ∈ 𝜔). By Corollary 2 to Lemma 11 (3.1.5), there is B ∈ R𝜎 (T, R) such that B ⊂ A0 and 𝜇(A0 \B) < 𝜀/2. If 𝜌 is such that 𝜌× is internally. Finite, then we should apply Corollary 1 to Lemma 17 (3.1.5). As a result, we obtain 𝜇(A\B) < 𝜀 and f |B = u-lim (f n |B | n ∈ 𝜔).

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̂ Corollary 1. In the conditions of Theorem 1, let f (t) = lim(f n (t) | n ∈ 𝜔) 𝜇-almost everŷ 0 and a sequence (A ∈ ̂ 𝜎f , there are a set N ∈ M where. Then, for every set A ∈ M m R𝜎f (T, R, 𝜌) | m ∈ 𝜔) of pairwise disjoint sets such that f |A m = u-lim(f n |A m | n ∈ 𝜔) and A = N ∪ ⋃⟮A m | m ∈ 𝜔⟯. ̂ f . By Theorem 1, for every m ∈ N, there is a set B ∈ Proof. Suppose first that A ∈ M m f ̂ R𝜎 (T, R, 𝜌) such that B m ⊂ A, 𝜇(A\B m ) ⩽ 1/(m + 1), and f |B m = u-lim (f n |B m | n ∈ 𝜔). Take A m ≡ B m \ ⋃⟮B i | i ∈ m⟯. Then, A󸀠 ≡ ⋃⟮A m | m ∈ N⟯ = ⋃⟮B m | m ∈ N⟯, the sets A m are pairwise disjoint and f |A m = u-lim (f n |A m | n ∈ 𝜔). Therefore, N ≡ A\A󸀠 ⊂ A\B m implies 𝜇N ⩽ 1/(m + 1) for every m, where N ∈ M0 . If A is the union of a countable collection (B k ∈ M | k ∈ K), we can take the sets B k ̂ disjoint. Each B k is the union of 𝜇-negligible set N k and a disjoint sequence (B km ∈ R𝜎f (T, R, 𝜌) | m ∈ 𝜔), such that on each B km the sequence (f n | n ∈ 𝜔) converges ̂ and A\N = ⋃⟮B km | (k, m) ∈ uniformly to f . The set N ≡ ⋃⟮N k | k ∈ 𝜔⟯ is 𝜇-negligible 𝜔 × 𝜔⟯. By Lemma 3 (1.3.9) the set 𝜔 × 𝜔 is countable, and therefore, the double sequence (B km | (k, m) ∈ 𝜔 × 𝜔) can be arranged onto a sequence (A l | l ∈ 𝜔). Corollary 2. Let ⟮T, M, 𝜇⟯ be a measurable space with the wide positive locally complete measure 𝜇. Let f ∈ F(T), (f n ∈ M(T, M) | n ∈ 𝜔) be a sequence, and f (t) = lim(f n (t) | n ∈ 𝜔) 𝜇-almost everywhere. Then, for every set A ∈ Mf and every number 𝜀 > 0 there is a set B ∈ Mf such that B ⊂ A, 𝜇(A\B) < 𝜀 and f |B = u-lim(f n |B | n ∈ 𝜔). ̂ Proof. Take in Theorem 1, R ≡ M and 𝜌 ≡ 𝜇. Then, R = M ⊂ M(T, R, 𝜌) and 𝜇 = 𝜌 = f f ̂ = 𝜌|M. ̂ 𝜌|R Since M is 𝜎-algebra, we have R𝜎 (T, R, 𝜌) = M . ̂ Theorem 2. Let 𝜌 be a positive measure on a ring R on a set T, R ⊂ M ⊂ M(T, R, 𝜌), ̂ 𝜇 ≡ 𝜌|M and 𝜇 be locally complete wide measure. Let f ∈ M(T, M). Then, for every set E ∈ M𝜎f there are a set N ⊂ E and a sequence (f n ∈ St(T, R𝜎f (T, R, 𝜌)) | n ∈ 𝜔) (see 2.2.4 and 3.1.5) such that N ∈ M0 and f (t) = lim(f n (t) | n ∈ 𝜔) for every t ∈ E\N. Moreover, we can choose functions f n such that |f n | ⩽ |f | for every n. If, moreover, 𝜌× is ̂ ̂̂ ̂ internally finite then the same is valid for M ⊂ M(T, R, 𝜌) and 𝜇 = 𝜌|M. Proof. By Corollary 2 to Proposition 4 (2.3.4) there is a sequence (f n ∈ M(T, M) | n ∈ 𝜔) such that f = u-lim (f n | n ∈ 𝜔) and rng f n is countable. From the proof of Proposition 4 (2.3.4) we deduce that we can choose the functions f n such that |f n (t) − f (t)| ⩽ 1/(n + 1) for every t ∈ T. From Lemma 4 (2.3.4), it follows that for each n there is a sequence (f nq ∈ St(T, M) | q ∈ 𝜔) such that f n = p-lim(f nq | q ∈ 𝜔). Let A ∈ Mf be a set such that A ⊂ E. Take 𝜀 > 0. Using the Egorov theorem, we can find a set A n ∈ R𝜎f (T, R, 𝜌) such that A n ⊂ A, 𝜇(A\A n ) < 𝜀/2n+1 , and f n |A n = u-lim(f nq |A n | q ∈ 𝜔). Since, by Lemma 9 (3.1.1), the ensemble R𝜎f (T, R, 𝜌) is a 𝛿-ring, the set B ≡ ⋂⟮A n | n ∈ 𝜔⟯ belongs to it. Lemmas 3 (3.1.1) and 3 (1.4.8) imply 𝜇(A\B) ⩽

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∑(𝜇(A\A n ) | n ∈ 𝜔) < 𝜀. Besides, f n |B = u-lim(f nq |B | q ∈ 𝜔). Taking 𝜀 = 1/(j + 1), we can choose a sequence (B j ∈ R𝜎f (R, 𝜇) | j ∈ 𝜔) such that f n |B j = u-lim(f nq |B j | q ∈ 𝜔) and 𝜇(A\B j ) < 1/(j + 1). Hence, A\ ⋃⟮B j | j ∈ 𝜔⟯ ∈ M0 . By the conditions, there is a sequence (A i ∈ Mf | i ∈ 𝜔) of pairwise disjoint sets such that E = ⋃⟮A i | i ∈ 𝜔⟯. By the property proven above for each A i , we can find a sequence (B ij ∈ R𝜎f (T, R, 𝜌) | j ∈ 𝜔) such that N i ≡ A i \ ⋃⟮B ij | j ∈ 𝜔⟯ ∈ M0 and f n |B ij = u-lim(f nq |B ij | q ∈ 𝜔). Consider for each k ∈ 𝜔 the set B k ≡ ⋃⟮B ij | i, j ∈ k + 1⟯ ∈ R𝜎f (T, R, 𝜌). Then, the sequence (B k | k ∈ 𝜔) is increasing and f n |B k = u-lim(f nq |B k | q ∈ 𝜔). The set M ≡ E\ ⋃⟮B k | k ∈ 𝜔⟯ belongs to M and is contained in ⋃⟮N i | i ∈ 𝜔⟯ ∈ M0 . Hence, M belongs to M0 . For each n, we can find an index q n such that |f nq n (t) − f n (t)| < 1/(n + 1) for every t ∈ B n . Then, |f nq n (t) − f (t)| < 2/(n + 1) for every t ∈ B n . Therefore, f (t) = lim(f nq n (t) | n ∈ 𝜔) for every t ∈ ⋃⟮B n | n ∈ 𝜔⟯. Consider the functions g n ≡ f nq n 𝜒(B n ) ∈ St(T, M). Then, g n = ∑(x nu 𝜒(M nu ∩ B n ) | u ∈ U n ) for some finite collections (x nu ∈ R | u ∈ U n ) and (M nu ∈ M | u ∈ U n ) (see 2.2.4). By Lemma 11 (3.1.5), M nu ∩ B n = E nu ∪ N nu for some E nu ∈ R𝜎f (T, R, 𝜌) ̂ 0 . If 𝜌 is such that 𝜌× is internally finite, then we should apply and N nu ∈ M Lemma 17 (3.1.5). Since N nu = (M nu ∩ B n )\E nu ∈ M, we get N nu ∈ M0 . Therefore, N n ≡ ⋃⟮N nu | u ∈ U n ⟯ ∈ M0 , h n ≡ ∑(x nu 𝜒(E nu ) | u ∈ U n ) ∈ St(T, R𝜎f (T, R, 𝜌)), and h n (t) = g n (t) for every t ∉ N n . Take the set N ≡ M ∪ ⋃⟮N n | n ∈ 𝜔⟯ ∈ M0 . Thus, we get f (t) = lim h n (t) for every t ∈ E\N. According to Corollary 2 to Proposition 4 (2.3.4), we can take |f n (t)| ⩽ |f (t)| for every n and t ∈ T. According to Lemma 4 (2.3.4), we can take also the sequence (f nq | q ∈ 𝜔) such that |f nq (t)| ⩽ |f n (t)| for every q, n and t ∈ T. Finally, |h n (t)| ⩽ |f nq n (t)| ⩽ |f (t)| for every n and t ∈ T. Corollary 1. Let ⟮T, M, 𝜇⟯ be a measurable space with the wide positive locally complete measure 𝜇. Let f ∈ M(T, M). Then, for every set E ∈ M𝜎f , there are a set N ⊂ E and a sequence (f n ∈ St(T, Mf ) ⊂ M(T, M) | n ∈ 𝜔) such that N ∈ M0 and f (t) = lim(f n (t) | n ∈ 𝜔) for every t ∈ E\N. Moreover, we can choose the functions f n such that |f n | ⩽ |f | for every n. Proof. It follows from Theorem 2 and Corollary 2 to Lemma 13 (3.1.5). Further, in this subsection, ⟮T, M, 𝜇⟯ will be any measurable space with the wide positive measure 𝜇. Corollary 2. Let ⟮T, M, 𝜇⟯ be any measurable space with the wide positive locally complete measure 𝜇. Let f ∈ M(T, M). Then, for every set A ∈ Mf , there is a sequence (f n ∈ St(T, Mf ) ⊂ M(T, M) | n ∈ 𝜔) such that ⋃⟮coz f n | n ∈ 𝜔⟯ ⊂ A and for every 𝜀 > 0, there is a set B ∈ Mf such that B ⊂ A, 𝜇(A\B) < 𝜀 and f |B = u-lim(f n |B | n ∈ 𝜔). Moreover, we can choose the functions f n such that |f n | ⩽ |f | for every n.

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Proof. By Corollary 1 for g ≡ f 𝜒(A) ∈ M(T, M) and A, there are N ∈ M0 and (g n ∈ St(T, Mf ) ⊂ M(T, M) | n ∈ 𝜔) such that N ⊂ A, g(t) = lim(g n (t) | n ∈ 𝜔) for every t ∈ A\N and |g n | ⩽ |g| for every n. Consider the functions f n ≡ g n 𝜒(A) ∈ St(T, Mf ) ∩ M(T, M). Then, g(t) = lim(f n (t) | n ∈ 𝜔) 𝜇-almost everywhere. Therefore, by Corollary 2 to Theorem 1 for A and 𝜀 > 0, there is a set B ∈ Mf such that B ⊂ A, 𝜇(A\B) < 𝜀 and f |B = g|B = u-lim(f n |B | n ∈ 𝜔). Besides, |f n | ⩽ |g n | ⩽ |g| ⩽ |f | and ⋃⟮coz f n | n ∈ 𝜔⟯ ⊂ A. According to 2.3.2, 2.3.3, and 2.3.4, M(T, M) and M b (T, M) are 𝜔-Dedekind complete lattice-ordered linear algebras with a unit 1. Moreover, the unit-function 1 is a weak order unit in M(T, M) and a strong order unit in M b (T, M). Besides, according to 2.2.8, we can consider on M b (T, M) the norm of uniform convergence ‖ ⋅ ‖u,M b (T,M) = ‖ ⋅ ‖1 , which we shall denote simply by ‖ ⋅ ‖u . By Proposition 2 (2.3.4), M b (T, M) is closed with respect to the uniform convergence of sequences in F b (T). Then, by Lemma 3 (2.2.7), ⟮M b (T, M), ‖ ⋅ ‖u ⟯ is a Banach latticeordered linear algebra with the multiplicative and strong order unit 1; moreover, it is an M-algebra. Owing to Lemma 9 (3.1.1) M0 is a 𝜎-ideal in M. By Lemma 1 (2.1.4), the ensemble I(M0 ) ≡ {P ∈ P(T) | ∃ N ∈ M0 (P ⊂ N)} is a 𝜎-ideal (in P(T)). Consider the equivalence relations 𝜃I(M0 ),M(T,M) on M(T, M) and 𝜃I(M0 ),M b (T,M) on M b (T, M) with respect to the 𝜎-ideal ensemble I(M0 ) on T (see 2.2.6). We shall denote them simply by 𝜃 and 𝜃b . Lemma 3 (2.2.6) implies M 0 (T, M) ≡ {f ∈ M(T, M) | f 𝜃0} = {f ∈ M(T, M) | coz f ∈ M0 } and M 0b (T, M) ≡ {f ∈ M b (T, M) | f 𝜃b 0} = {f ∈ M b (T, M) | coz f ∈ M 0 }. Consider the equivalence relations 𝜃M0 (T,M) and 𝜃M0 (T,M) on M(T, M) and b M b (T, M), respectively. According to 2.2.6, the first one is equal to 𝜃 and the second one is equal to 𝜃b . Therefore, M(T, M)/I(M0 ) = M(T, M)/M 0 (T, M) and M b (T, M)/I(M0 ) = M b (T, M)/M 0b (T, M). The first factor-set will be denoted by L(T, M, 𝜇) and the second one by L b (T, M, 𝜇). It follows from 2.2.6 that L(T, M, 𝜇) and L b (T, M, 𝜇) are lattice-ordered linear algebras with the unit 1.̄ Moreover, 1̄ is a weak order unit in L(T, M, 𝜇) and a strong order unit in L b (T, M, 𝜇). Besides, according to 2.2.7, we can consider on L b (T, M, 𝜇) the norm of the uniform convergence ‖ ⋅ ‖∞ ≡ ‖ ⋅ ‖1̄ = ‖ ⋅ ‖u,L b (T,M,𝜇) . By Corollary 2 to Lemma 5 (2.2.7), (L b (T, M, 𝜇), ‖ ⋅ ‖∞ ) is a Banach lattice-ordered algebra with the unit 1;̄ moreover, it is an M-algebra. Consider now the subset Sl b (T, M, 𝜇) ≡ { f ∈ L b (T, M, 𝜇) | f ∈ St(T, M)} in L b (T, M, 𝜇) consisting of stepwise elements. Proposition 2. Let ⟮T, M, 𝜇⟯ be a measurable space with the wide positive measure 𝜇. Then, the set Sl b (T, M, 𝜇) is a lattice-ordered linear subalgebra with the unit 1̄ of the lattice-ordered linear algebra L b (T, M, 𝜇) with the unit 1,̄ and it is dense in L b (T, M, 𝜇) in the normed topology G(‖ ⋅ ‖∞ ).

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Proof. According to Proposition 1 (2.2.4), St(T, M) is a lattice-ordered linear subalgebra of F b (T) with the unit 1. By Corollary 1 to Lemma 2 (2.3.1), St(T, M) ⊂ M b (T, M). Substituted all mentioned pointwise structures considered everywhere by the corresponding pointwise structures considered almost everywhere on T, we conclude that Sl b (T, M, 𝜇) is a lattice-ordered linear subalgebra of the lattice-ordered linear algebra L b (T, M, 𝜇) with the unit 1.̄ This gives the first assertion of the proposition. According to the definition of density from 2.1.1, we need take an arbitrary non-empty set G ∈ G(‖ ⋅ ‖∞ ). Let f ∈ G. By the definition of the norm topology, there are some ḡ ∈ L b and 0 < x ∈ R such that f ∈ B ≡ {ḡ + x h̄ | ‖h‖∞ < 1}. So y ≡ ‖ f − g‖̄ < x. By Corollary 3 to Proposition 4 (2.3.4), there is a sequence (f n ∈ St | n ∈ N) such that for every 𝜀 > 0 there is n ∈ 𝜔 such that ‖f p − f ‖u < 𝜀 for every p ⩾ n. Take 𝜀 ≡ x − y. Then, ‖ f p − g‖̄ ∞ ⩽ ‖ f p − f ‖∞ + y ⩽ ‖f p − f ‖u + y < 𝜀 + y = x. Thus, f p ∈ B ⊂ G, i. e. G ∩ Sl b ≠ ⌀. Along with the subalgebra M b (T, M) of all bounded M-measurable functions we shall consider with respect to the 𝜎-ideal I(M0 ), the subalgebra M∞ (T, M, 𝜇) ≡ F∞ (T, I(M0 )) ∩ M(T, M) of all essentially bounded M-measurable functions (see 2.2.7). By Lemma 6 (2.2.7), F∞ (T) is a lattice-ordered linear algebra with the unit function 1 and with the modulusly monotone unitary submultiplicative seminorm ‖ ⋅ ‖eu . Moreover, F∞ (T) is an l-ideal of the lattice-ordered linear space F(T). By Theorem 1 (2.3.2), M(T, M) is also a lattice-ordered linear algebra with 1. Therefore, M∞ (T, M, 𝜇) is an l-ideal of the lattice-ordered linear space M(T, M) and ⟮M∞ (T, M, 𝜇), ‖ ⋅ ‖eu ⟯ is a seminormed lattice-ordered linear algebra with 1. Besides, M∞ (T, M, 𝜇) is Dedekind complete. Consider the relation of equivalence 𝜃∞ ≡ 𝜃I(M0 ),M∞ (T,M,𝜇) on M∞ (T, M, 𝜇) with respect to the 𝜎-ideal I(M0 ) on T. By virtue of Lemma 3 (2.2.6), f 𝜃∞ g iff coz(f − g) ∈ M0 . Remind that in 2.2.7 we introduced the notation C x f ≡ {t ∈ T | |f (t)| > x}. It is evident that f ∈ M∞ (T, M, 𝜇) iff C x f ∈ M0 for some x ∈ R+ . Lemma 2. Let 𝜇 be a positive measure on a 𝜎-algebra M on T, f , g ∈ M∞ (T, M, 𝜇) and h ∈ M(T, M). Then, 1) {t ∈ T | |f (t)| > ‖f ‖eu } ∈ M0 ; 2) if coz(f − g) ∈ M0 , then ‖f ‖eu = ‖g‖eu ; 3) coz f ∈ M0 iff ‖f ‖eu = 0; 4) h ∈ M∞ (T, M, 𝜇) iff there is h󸀠 ∈ M b (T, M) such that coz(h − h󸀠 ) ∈ M0 ; 5) if f ∈ M b (T, M) then ‖f ‖eu ⩽ ‖f ‖u ; 6) there is f 󸀠 ∈ M b (T, M) such that coz(f − f 󸀠 ) ∈ M0 and ‖f ‖eu = ‖f 󸀠 ‖u ; 7) if coz(f − h) ∈ M0 , then h ∈ M∞ (T, M, 𝜇). Proof. Assertions 1 – 3 follows from Lemma 7 (2.2.7).

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4. Let h ∈ M∞ , i. e. C x h ∈ M0 for some x ∈ R+ . Consider the function h󸀠 ≡ h𝜒(T\C x f ) ∈ M b . Then, coz(h − h󸀠 ) ∈ M0 . Conversely, if coz(h − h󸀠 ) ∈ M0 , then by Lemma 8 (2.2.7), h ∈ F∞ , where h ∈ M∞ . 5. It follows from Lemma 8 (2.2.7). 6. For f , consider x ≡ ‖f ‖eu . By (1), C x f ∈ M0 . Consider the function f 󸀠 ≡ f 𝜒(T\C x f ) ∈ M b . Then, coz(f − f 󸀠 ) ⊂ C x f implies coz(f − f 󸀠 ) ∈ M0 . By (2) and (5) x = ‖f 󸀠 ‖eu ⩽ ‖f 󸀠 ‖u ⩽ x. 7. By the condition, C x f ∈ M0 for some x > 0. From C x h ⊂ C x f ∪ coz(f − h) ∈ M0 , we conclude that C x h ∈ M0 , where h ∈ M∞ . Proposition 3. Let 𝜇 be a positive measure on a 𝜎-algebra M on a set T. Then, ⟮M∞ (T, M, 𝜇), ‖ ⋅ ‖eu ⟯ is a complete seminormed lattice-ordered linear algebra with the unit 1. Proof. Let (f n ∈ M∞ | n ∈ 𝜔) be inner convergent sequence with respect to the seminormed pseudometric 𝜌‖⋅‖eu , i. e. for every 𝜀 > 0 there is n = n(𝜀) such that ‖f p − f q ‖eu < 𝜀 for all p, q ⩾ u. Then, for every k ∈ N, there is n k such that ‖f p − f q ‖eu < 1/k for all p, q ⩾ n k . Therefore, there is x k ∈ R+ such that x k < 1/k and E p,q = {t ∈ T | |f p (t) − f q (t)| > x k } ∈ I(M0 ). Consider the set E ≡ ⋃⟮⋃⟮E pq | p, q ⩾ n k ⟯ | k ∈ N⟯ ∈ M0 and the functions g n ≡ f n 𝜒(T\E) ∈ M. Since |g n | ⩽ |f n | and M∞ is an l-ideal in M, we conclude that g n ∈ M∞ . Since |f p (t) − f q (t)| ⩽ x k < 1/k for every t ∉ E and every p, q ⩾ n k , the sequence (g n ∈ M∞ | n ∈ 𝜔) is inner uniformly convergent in M (see 2.2.3). Therefore, by Lemma 3 (2.3.4), there is the unique function f ∈ M such that f = u-lim(g n | n ∈ 𝜔). Moreover, for numbers y n ≡ ‖f n ‖eu , we have |y p − y q | ⩽ ‖f p − f q ‖eu < 1/k for all p, q ⩾ n k . Therefore, there is the unique number y ⩾ 0 such that y = lim(y n | n ∈ 𝜔). By Lemma 2, F n ≡ {t ∈ T | |f n (t)| > y n } ∈ M0 . So F ≡ ⋃⟮F n | n ∈ 𝜔⟯ ∈ M0 and |f n (t)| ⩽ y n for every t ∉ F. By Corollary 3 to Proposition 1 (1.4.7), |f (t)| ⩽ y for every t ∉ E ∪ F ∈ M0 . Hence, C y f ⊂ E ∪ F. Since C y f ∈ M, we conclude that C y f ∈ M0 . This means that f ∈ M∞ . By the definition of the uniform convergence in F(T) from 2.2.3 for every 𝜀 > 0 there is m = m(𝜀) such that ‖g n − f ‖u < 𝜀 for all n ⩾ m. Since coz((f n − f ) − (g n − f )) = coz(f n − g n ) ∈ M and coz(f n − g n ) ⊂ E ∈ M0 , we get coz((f n − f ) − (g n − f )) ∈ M0 . By virtue of Lemma 2, ‖f n − f ‖eu = ‖g n − f ‖eu ⩽ ‖g n − f ‖u < 𝜀 for all n ⩾ m. Thus, f = lim⟮f n | n ∈ 𝜔⟯ with respect to 𝜌‖⋅‖eu . 0 Consider the subset M∞ (T, M, 𝜇) ≡ {f ∈ M∞ (T, M, 𝜇) | f 𝜃∞ 0} = {f ∈ M∞ (T, M, 𝜇) | 0 coz f ∈ M } (see Lemma 3 (2.2.6)) and the equivalence relation 𝜃M∞ 0 (T,M,𝜇) on M∞ (T, M, 𝜇). According to 2.2.6, this relation is equal to 𝜃∞ . Hence, 0 (T, M, 𝜇). This factor-set is denoted M∞ (T, M, 𝜇)/I(M0 ) = M∞ (T, M, 𝜇)/M∞ by L∞ (T, M, 𝜇).

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It follows from 2.2.6, that L∞ (T, M, 𝜇) is a lattice-ordered linear algebra with the unit 1,̄ which is a weak order unit in L∞ (T, M, 𝜇). Besides, according to 2.2.7 we can consider on L∞ (T, M, 𝜇), the norm of the essentially uniform convergence ‖ ⋅ ‖eu,L∞ (T,M,𝜇) . By Proposition 2 (2.2.7), the normed latticeordered linear algebras (L∞ (T, M, 𝜇), ‖ ⋅ ‖) and (L b (T, M, 𝜇), ‖ ⋅ ‖) are naturally isomorphic with respect to the (bijective isometrical) isomorphism u : L∞ (T, M, 𝜇) L b (T, M, 𝜇) such that uf ≡ f ∩ M b (T, M). Therefore, we may denote the coinciding norms ‖ ⋅ ‖eu,L∞ (T,M,𝜇) and ‖ ⋅ ‖u,L b (T,M,𝜇) by the same symbol ‖ ⋅ ‖∞ . Hence, we conclude that ⟮L∞ (T, M, 𝜇), ‖ ⋅ ‖∞ ⟯ is a Banach lattice-ordered algebra with the unit 1;̄ moreover, it is an M-algebra. Consider now the subset SL∞ (T, M, 𝜇) ≡ { f ∈ L∞ (T, M, 𝜇) | f ∈ St(T, M)} in L∞ (T, M, 𝜇) consisting of stepwise elements. It follows from Proposition 2 and the above isomorphism u, that SL∞ (T, M, 𝜇) is dense in (L∞ (T, M, 𝜇), G(‖ ⋅ ‖∞ )). 3.3.2 The Lebesgue integral over a space with a positive measure In this subsection, ⟮T, M, 𝜇⟯ is a measurable space with a 𝜎-algebra M and a positive measure 𝜇 : M → R+ (see also the beginning of the previous subsection). Let f be a positive function on T and 𝜋 ≡ (A i ∈ M | i ∈ I) be a finite partition of T (see 1.1.10), i. e. 𝜋 ∈ Parf (M, T) (see 3.1.3). Put 𝜎(f , 𝜋) ≡ ∑(inf f [A i ]𝜇A i | i ∈ I). Note that we use here the equality 0∞ = 0 (see 1.4.3). Then, we can define the extended positive number ∫ f d𝜇 ≡ sup {𝜎(f , 𝜋) | 𝜋 ∈ Parf (M, T)} ∈ R+ = [0, ∞]. A function f ∈ F(T) will be called extendedly (Lebesgue) integrable over the space ⟮T, M, 𝜇⟯ if at least one of the numbers ∫(f+ ) d𝜇 and ∫(−f− ) d𝜇 is finite, where f+ and f− are the positive and the negative parts of f (see 2.2.2). A function f ∈ F(T) is called (Lebesgue) integrable over the space ⟮T, M, 𝜇⟯ if both the numbers ∫(f+ ) d𝜇 and ∫(−f− ) d𝜇 are finite. The family of all extendedly integrable functions over the space ⟮T, M, 𝜇⟯ will be denoted by LI e (T, M, 𝜇). The subset of the set LI e (T, M, 𝜇) consisting of all integrable functions will be denoted by LI(T, M, 𝜇). Obviously, F(T)+ ⊂ LI e (T, M, 𝜇) and if 𝜇 is bounded, then F b (T) ⊂ LI(T, M, 𝜇). We shall also introduce an intermediate family of functions between LI(T, M, 𝜇) and LI e (T, M, 𝜇). A function f ∈ F(T)+ will be called (Lebesgue) 𝜎-integrable over the space ⟮T, M, 𝜇⟯ if there is a countable collection (T i ∈ M | i ∈ I) such that T = ⋃⟮T i | i ∈ I⟯ and f 𝜒(T i ) ∈ LI(T, M, 𝜇)+ for every i ∈ I. A function f ∈ F(T) will be called (Lebesgue) 𝜎-integrable over the space ⟮T, M, 𝜇⟯, if either f + is 𝜎-integrable and −f− is integrable or f+ is integrable and −f− is 𝜎-integrable. The family of all 𝜎-integrable functions over the space ⟮T, M, 𝜇⟯ will be denoted by LI 𝜎 (T, M, 𝜇). It is clear that LI(T, M, 𝜇) ⊂ LI 𝜎 (T, M, 𝜇) ⊂ LI e (T, M, 𝜇).

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If f ∈ LI e (T, M, 𝜇) then we can define the extended number ∫ f d𝜇 ≡ ∫(f+ ) d𝜇 − ∫(−f− ) d𝜇 ∈ R. This number ∫ f d𝜇 is called the (Lebesgue) integral of the function f over the space ⟮T, M, 𝜇⟯. Clearly, ∫(x1) d𝜇 = x𝜇T for every x ∈ R. The mapping Λ : LI e (T, M, 𝜇) → R such that Λf ≡ ∫ f d𝜇 is called the (Lebesgue) integral over the space ⟮T, M, 𝜇⟯. Sometimes we shall use more rigorous notations: Λ(𝜇) : LI e (T, M, 𝜇) → R and i𝜇 f ≡ Λ(𝜇)f ≡ ∫ f d𝜇. The letters L and Λ are employed here in memoriam Henri Lebesgue (1875 – 1941) who was one of the founding fathers of modern measure and integration theory. Along with the phrase “over the space ⟮T, M, 𝜇⟯” we shall use also the phrase “with respect to the measure 𝜇 (on the descriptive space ⟮T, M⟯)”. Lemma 1. Let 𝜇 be a positive measure on a 𝜎-algebra M on T, (M k ∈ M | k ∈ K) be a finite collection of pairwise disjoint sets, (x k ∈ R+ | k ∈ K) and f ≡ ∑(x k 𝜒(M k ) | k ∈ K) ∈ St(T, M)+ . Then, Λf = ∑(x k 𝜇(M k ) | k ∈ K). Proof. We may suppose that the pairwise disjoint collection (M k | k ∈ K) covers T. Since inf f [M k ] = x k , we have Λf ⩾ ∑ x k 𝜇M k . Now, let (A i ∈ M | i ∈ I) be any finite partition of T. Consider for every i the set / ⌀}. Then, 𝜇A i = ∑(𝜇(A i ∩ M k ) | k ∈ K i ) implies ∑(inf f [A i ]𝜇A i | K i ≡ {k ∈ K | A i ∩ M k = i ∈ I) = ∑(∑(inf f [A i ]𝜇(A i ∩ M k ) | k ∈ K i ) | i ∈ I) ⩽ ∑(∑(inf f [A i ∩ M k ]𝜇(A i ∩ M k ) | k ∈ K i ) | i ∈ I) = ∑(∑(x k 𝜇(A i ∩ M k ) | k ∈ K i ) | i ∈ I) = ∑(∑(x k 𝜇(A i ∩ M k ) | k ∈ K) | i ∈ I) ≡ x. Suppose at first that 𝜇(A i ∩ M k ) < ∞ for all i, k. Then, by Corollary 1 to Proposition 1 (1.4.3) the number x is equal to the number y ≡ ∑(x k ∑(𝜇(A i ∩ M k ) | i ∈ I) | k ∈ K) = ∑(x k 𝜇M k | k ∈ K). Suppose now that for some i and k we have 𝜇(A i ∩ M k ) = ∞. If x k = 0 for every k such that 𝜇M k = ∞ then we may also permute the signs of summation. Finally, suppose that there are indices k such that 𝜇M k = ∞ and x k > 0. Then, both the considered numbers x and y are equal to ∞ and consequently x = y. Thus, we obtain Λf ⩽ ∑(x k 𝜇M k | k ∈ K). Proposition 1. Let 𝜇 be a positive measure on a 𝜎-algebra M on T, f , g ∈ St(T, M)+ and x, y ∈ R+ . Then, Λ(xf + yg) = xΛf + yΛg. Proof. By virtue of Lemma 8 (2.2.4) we may suppose that f = ∑(x i 𝜒(A i ) | i ∈ I) and g = ∑(y j 𝜒(B j ) | j ∈ J) for some finite partitions (A i ∈ M | i ∈ I) and (B j ∈ M | j ∈ J) of T. Then, x i , y j ∈ R+ for all i, j. By Lemma 7 (2.2.4), 𝜒(A i ) = ∑(𝜒(A i ∩ B j ) | j ∈ J) and 𝜒(B j ) = ∑(𝜒(A i ∩ B j ) | i ∈ I), where f = ∑(∑(x i 𝜒(A i ∩ B j ) | j ∈ J) | i ∈ I) and g = ∑(∑(y j 𝜒(A i ∩ B j ) | i ∈ I) | j ∈ J). By Corollary 1 to Proposition 1 (1.4.3), f = ∑(x i 𝜒(A i ∩ B j ) | (i, j) ∈ I × J) and g = ∑(y j 𝜒(A i ∩ B j ) | (i, j) ∈ I × J). Therefore, by Corollary 1 to Lemma 4 (1.4.3) and Lemma 6 (1.4.3), xf + yg = ∑((xx i + yy j )𝜒(A i ∩ B j ) | (i, j) ∈ I × J),

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where xx i + yy j ⩾ 0 and (A i ∩ B j | (i, j) ∈ I × J) is a collection of pairwise disjoint sets. By Lemma 1, Λ(xf + yg) = ∑((xx i + yy j )𝜇(A i ∩ B j ) | (i, j) ∈ I × J). On the other hand, by Lemma 1 Λf = ∑(x i 𝜇(A i ∩ B j ) | (i, j) ∈ I × J) and similarly Λg = ∑(y j 𝜇(A i ∩ B j ) | (i, j) ∈ I × J). Therefore, xΛf + yΛg = ∑ ((xx i + yy j )𝜇(A i ∩ B j ) | (i, j) ∈ I × J). Corollary 1. Let f , g ∈ St(T, M)+ and f ⩾ g. Then, Λ(f − g) = Λf − Λg. Proof. Consider h ≡ f − g ∈ St(T, M)+ . Then, f = h + g implies Λf = Λh + Λg. Lemma 2. Let 𝜇 be a positive measure on a 𝜎-algebra M on T, f , g ∈ F(T)+ and f ⩽ g. Then, 0 ⩽ Λf ⩽ Λg. Proof. Let 𝜋 ≡ (A i ∈ M | i ∈ I) be a finite partition of T. Then, 0 ⩽ x i ≡ inf{f (t) | t ∈ A i } ⩽ inf{g(t) | t ∈ A i } ≡ g i for every i. By Proposition 4 (1.4.3), 0 ⩽ 𝜎(f , 𝜋) ≡ ∑(x i 𝜇(A i ) | i ∈ I) ⩽ ∑(y i 𝜇(A i ) | i ∈ I) ≡ 𝜎(g, 𝜋). This implies 0 ⩽ Λf ⩽ Λg. Lemma 3. Let 𝜇 be a positive measure on a 𝜎-algebra M on T and f ∈ M(T, M)+ . Then, 𝜇(coz f ) > 0 implies Λf > 0 and 𝜇(coz f ) = 0 implies Λf = 0. Proof. First, suppose that 𝜇(coz f ) = 0. Let (A i ∈ M | i ∈ I) be a finite partition of T. Then, inf{f (t) | t ∈ A i }𝜇A i ⩽ inf{f (t) | t ∈ A i ∩ coz f }𝜇(A i ∩ coz f ) + inf{f (t) | t ∈ A i ∩ zer f }𝜇(A i ∩ zer f ) = 0 implies Λf = 0. Now, suppose that 𝜇(coz f ) > 0 and consider the sets C n ≡ cozn f (see 2.2.5). Lemma 5 (3.1.1) guarantees that lim 𝜇C n = 𝜇(coz f ) > 0. Hence, 𝜇C n > 0 for some n. Consider the partition (B i ∈ M | i ∈ 2) such that B0 ≡ C n and B1 ≡ T\C n . Then, Λf ⩾ ∑(inf f [B i ]𝜇B i | i ∈ 2) = (1/n)𝜇C n + inf{f (t) | t ∈ T\C n }𝜇C n ⩾ 𝜇C n /n > 0. Lemma 4. Let 𝜇 be a positive measure on a 𝜎-algebra M on T, f ∈ F(T)+ and x ∈ R. Then, Λ(xf ) = xΛf . Proof. If x ∈ R+ , then the statement follows from Lemma 6 (1.4.3) and Lemma 4 (1.4.5). If x ∈ R− , then (xf )+ = 0 and (xf )− = xf . Hence, Λ(xf ) = Λ((xf )+ ) − Λ(−(xf )− ) = 0 − Λ((−x)f ) = xΛf . The following result, which looks harmless enough, is the key to the proof that Λ is countably additive in Theorem 1 (3.3.3). Theorem 1. Let 𝜇 be a positive measure on a 𝜎-algebra M on T, (f n ∈ St(T, M)+ | n ∈ N) ↑ be increasing sequence for some cofinal subset N of 𝜔, g ∈ St(T, M)+ , and g(t) ⩽ sup (f n (t) | n ∈ N) in R+ for every t ∈ T. Then, Λg ⩽ sup (Λf n | n ∈ N) in R+ .

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Proof. Put U ≡ coz g and note that U ∈ M. If 𝜇U = 0, then Λg = 0 due to Lemma 3 and the necessary inequality is trivial. Further, we suppose that 0 < 𝜇U ⩽ ∞. By virtue of Lemma 8 (2.2.4) the function g can be represented in the form g = 0𝜒(T\U) + ∑(x i 𝜒(A i ) | i ∈ m + 1), where 0 < x0 < x1 < . . . < x m and (A i ∈ M | i ∈ m + 1) is a partition of U. First, suppose that 𝜇U < ∞. For any 𝛿 > 0, choose 𝜀 > 0 satisfying the inequality 𝜀 ⩽ (𝛿/(2𝜇U)) ⊼ x0 . Consider the sets V n ≡ {t ∈ U | f n (t) > g(t) − 𝜀} ∈ M and W n ≡ U\V n . From the conditions of the theorem, we deduce that (V n | n ∈ N) ↑ U. By Lemma 5 (3.1.1), (𝜇V n | n ∈ N) ↑ 𝜇U, where (𝜇W n | n ∈ N) ↓ 0. Using Corollary 1 to Proposition 1 and Lemmas 2 and 4, we deduce from g𝜒(U) ⩾ x0 𝜒(U) ⩾ 𝜀𝜒(U) that Λf n ⩾ Λ(f n 𝜒(V n )) ⩾ Λ((g − 𝜀1)𝜒(V n )) = Λ(g𝜒(V n )) − 𝜀𝜇V n . It follows from g = g𝜒(V n ) + g𝜒(W n ) ⩽ g𝜒(V n ) + x m 𝜒(W n ) that Λg ⩽ Λ(g𝜒(V n )) + x m 𝜇W n . Hence, Λf n ⩾ Λg−x m 𝜇W n −𝜀𝜇V n ⩾ Λg−x m 𝜇W n −𝜀𝜇U ⩾ Λg−x m 𝜇W n −𝛿/2. If n is so large that x m 𝜇W n < 𝛿/2, then Λf n > Λg−𝛿. Since 𝛿 is arbitrary, we get Λg ⩽ sup (Λf n | n ∈ N). Now, we suppose that 𝜇U = ∞. Lemma 2 provides Λg ⩾ x0 𝜇U = ∞. Let 𝜀 ∈]0, x0 [. Consider the sets V n and W n as above. Then, (𝜇V n | n ∈ N) ↑ 𝜇U = ∞. For t ∈ V n , we have f n (t) > x0 − 𝜀, where f n 𝜒(V n ) ⩾ (x0 − 𝜀)𝜒(V n ). Therefore, Λf n ⩾ Λ(f n 𝜒(V n )) ⩾ Λ((x0 − 𝜀)𝜒(V n )) = (x0 − 𝜀)𝜇V n implies sup (Λf n | n ∈ N) ⩾ (x0 − 𝜀) sup (𝜇V n | n ∈ N) = ∞ = Λg. Corollary 1. Let 𝜇 be a positive measure on a 𝜎-algebra M on T, (f n ∈ St(T, M)+ | n ∈ N) ↑ be an increasing sequence for some cofinal subset N of 𝜔, f ∈ F(T)+ , and f (t) = sup (f n (t) | n ∈ N) for every t ∈ T. Then, Λf = sup (Λf n | n ∈ N) = lim(Λf n | n ∈ N). Proof. By Lemma 2, sup (Λf n | n ∈ N) ⩽ Λf . Now, let x be any real number such that x < Λf . Then, there is 𝜋 ≡ (A i | i ∈ I) from Parf (M, T) such that x < 𝜎(f , 𝜋) ≡ y. Consider the function g ≡ ∑(inf f [A i ]𝜒(A i ) | i ∈ I) ∈ St(T, M)+ . By Lemma 1, Λg = y. From g ⩽ f , By virtue of Theorem 1 we conclude that y ⩽ sup (Λf n | n ∈ N). Since x is arbitrary, we infer that Λf ⩽ sup (Λf n | n ∈ N). Thus, we get the first necessary equality. The second one follows from Lemma 7 (1.4.7). Corollary 2. Let 𝜇 be a positive measure on a 𝜎-algebra M on T, f , g ∈ M(T, M)+ , and x, y ∈ R+ . Then, Λ(xf + yg) = xΛf + yΛg. Proof. By Theorem 2 (2.3.4) there are sequences (f n | n ∈ N) ↑ and (g n | n ∈ N) ↑ in St(T, M)+ such that f = p-lim (f n | n ∈ N) and g = p-lim (g n | n ∈ N). By virtue of Proposition 1 (1.4.7) and Lemma 7 (1.4.7) (xf + yg)(t) = sup ((xf n + yg n )(t) | n ∈ N) for every t ∈ T. By Corollary 1 to Theorem 1 and Proposition 1 Λ(xf +yg) = lim(Λ(xf n +yg n ) | n ∈ N) = lim (xΛf n + yΛg n | n ∈ N) = x lim (Λf n | n ∈ N) + y lim (Λg n | n ∈ N) = xΛf + yΛg. The very important Corollary 2 shows that the most natural domains for the Lebesgue integral Λ are the family MI e (T, M, 𝜇) ≡ M(T, M) ∩ LI e (T, M, 𝜇) of all 𝜇-measurable

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extendedly integrable functions and its subfamilies MI(T, M, 𝜇) ≡ M(T, M) ∩ LI(T, M, 𝜇) and MI 𝜎 (T, M, 𝜇) ≡ M(T, M) ∩ LI 𝜎 (T, M, 𝜇). Corollary 3. Let 𝜇 be a positive measure on a 𝜎-algebra M on T and f ∈ MI e (T, M, 𝜇). Then, | ∫ f d𝜇| ⩽ ∫ |f | d𝜇. Proof. At first suppose that x ≡ ∫ f+ d𝜇 = ∞ and y ≡ ∫(−f− ) d𝜇 < ∞. Then, by Lemma 2, x ⩽ Λ|f | implies |Λf | = |x − y| = x ⩽ Λ|f |. If x < ∞ and y = ∞, then similarly |Λf | = |x − y| = y ⩽ Λ|f |. Finally, if x < ∞ and y < ∞, then by Corollary 2, |Λf | = |x − y| ⩽ x + y = Λ|f |. Lemma 5. Let 𝜇 be a positive measure on a 𝜎-algebra M on T, f , g ∈ M(T, M), and f (t) = g(t) 𝜇-almost everywhere on T. If f ∈ MI e (T, M, 𝜇), then g ∈ MI e (T, M, 𝜇) and Λg = Λf . Proof. Let E = {t ∈ T | f (t) = / g(t)} and F ≡ T\E. By hypothesis 𝜇E = 0. First, consider the case f ⩾ 0 and g ⩾ 0. By Corollary 2 to Theorem 1 and Lemma 3, Λf = Λ(f 𝜒(E)) + Λ(f 𝜒(F)) = Λ(g𝜒(F)) = Λ(g𝜒(F)) + Λ(g𝜒(E)) = Λg. Now, consider the general case. For t ∈ F, we have f + (t) = g+ (t) and f− (t) = g− (t). Therefore, f+ = g+ and f− = g− 𝜇-almost everywhere. By the first case, Λg+ = Λf+ and Λ(−g− ) = Λ(−f− ). Therefore, g is extendedly integrable and Λg = Λf . Corollary 1. Let 𝜇 be a positive measure on a 𝜎-algebra M on T, f , g ∈ M(T, M), and f (t) = g(t) 𝜇-almost everywhere on T. If f ∈ MI(T, M, 𝜇), then g ∈ MI(T, M, 𝜇) and Λg = Λf . Lemma 6. Let 𝜇 be a positive measure on a 𝜎-algebra M on T and f , g ∈ MI(T, M, 𝜇)+ . If f − g ∈ MI(T, M, 𝜇), then Λ(f − g) = Λf − Λg. Proof. Denote f − g by h. By the definition, Λh ≡ Λh+ − Λ(−h− ). Since f − g = h+ + h− , we have f − h− = g + h+ . From Corollary 2 to Theorem 1, we infer that Λf + Λ(−h− ) = Λg + Λ(h+ ), where Λh = Λf − Λg. Theorem 2. Let 𝜇 be a positive measure on a 𝜎-algebra M on T, f , g ∈ MI(T, M, 𝜇), and x ∈ R. Then, f + g, xf , f ∨ g, and f ∧ g belong to MI(T, M, 𝜇). Besides, Λ(f + g) = Λf + Λg and Λ(xf ) = x Λ f . Thus, MI(T, M, 𝜇) is a lattice-ordered linear subspace of the lattice-ordered linear spaces F(T) and M(T, M) and the Lebesgue integral Λ is a positive linear functional on it. Proof. By the definition, Λ(f+ ), Λ(−f− ), Λ(g+ ), and Λ(−g− ) are real numbers. By virtue of Corollary 1 to Lemma 1 (2.2.2) we have (f + g)+ ⩽ f+ + g+ and −(f + g)− ⩽ −f− − g− . By Lemma 2 and Corollary 2 to Theorem 1, Λ((f + g)+ ) ⩽ Λ(f+ + g+ ) = Λ(f+ ) + Λ(g+ ) and

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Λ(−(f + g)− ) ⩽ Λ(−f− − g− ) = Λ(−f− )+Λ(−g− ). These inequalities mean that f + g, f+ + g+ , and −f− − g− belong to LI. By virtue of Theorem 1 (2.3.2) and Proposition 1 (2.3.3) these functions belong to MI. From the equality f + g = (f+ + g+ ) − (−f− − g− ) and Lemma 6, we infer that Λ(f + g) = Λ(f+ + g+ ) − Λ(−f− − g− ) = Λf+ + Λg+ − Λ(−f− ) − Λ(−g− ) = Λf + Λg. Let x ⩾ 0. By virtue of Theorem 1 (2.2.2), (xf )+ = x(f+ ) and −(xf )− = x(−f− ). By Lemma 4, Λ((xf )+ ) = xΛ(f+ ) and Λ(−(xf )− ) = xΛ(−f− ) are real numbers, so that xf ∈ LI. Let x < 0. By virtue of Theorem 1 (2.2.2), (xf )+ = xf− and (xf )− = xf+ . By Lemma 4, Λ((xf )+ ) = −xΛ(−f− ) and Λ(−(xf )− ) = −xΛf+ are real numbers, so that xf ∈ LI. In both cases, Λ(xf ) = xΛf . We proved that MI is a linear space. Since g−f ∈ MI, by the definition, (g−f )+ ∈ LI and −(g − f )− ∈ LI. Theorem 1 (2.3.2) and Proposition 1 (2.3.3) guarantee that these functions belong to MI. By Lemma 5 (1.4.5), we have f ∨ g = f + (g − f )+ and f ∧ g = f + (g − f )− . This imply that f ∨ g and f ∧ g belong to MI. Thus, MI is a lattice-ordered linear space. Corollary 1. Let 𝜇 be a positive measure on a 𝜎-algebra M on T. Then, the latticeordered linear subspace MI(T, M, 𝜇) of the lattice-ordered linear space M(T, M) is an l-ideal. Proof. Let f ∈ MI, g ∈ M and |g| ⩽ |f | in M. Then, |f | ∈ MI. From g+ ⩽ |g| ⩽ |f | and −g− ⩽ |f | we infer by Lemma 2 that Λg+ ⩽ Λ|f | < ∞ and similarly Λ(−g− ) < ∞. Hence, g ∈ LI, where g ∈ MI. According to Statement 2 (2.2.8), this property means that MI is an l-ideal. Corollary 2. The lattice-ordered linear subspace MI(T, M, 𝜇) of the 𝜔-Dedekind complete lattice-ordered linear space M(T, M) is also 𝜔-Dedekind complete and the smallest upper [greatest lower] bounds of bounded above [below] countable sets in the ordered sets M(T, M) and MI(T, M, 𝜇) coincide. Proof. By Proposition 2 (2.3.4), M(T, M) is 𝜔-Dedekind complete. The assertion now follows from Corollary 1 and Statements 3 and 1 (2.2.8). Corollary 3. Let 𝜇 be a positive measure on a 𝜎-algebra M on T. Then, a function f ∈ M(T, M) belongs to MI(T, M, 𝜇) iff |f | belongs to MI(T, M, 𝜇). Proof. If f ∈ MI then |f | = f + + f− = f ∨ 0 − f ∧ 0 ∈ MI. If |f | ∈ MI, then |f | ⩽ |f | implies f ∈ MI by virtue of Corollary 1. Lemma 7 (the Chebyshev inequality). Let 𝜇 be a positive measure on a 𝜎-algebra M on T, f ∈ MI(T, M), U ∈ M, and f (t) ⩾ 0 for every t ∈ U. Then, c𝜇{t ∈ U | f (t) ⩾ c} ⩽ Λ(f 𝜒(U)) for every c ∈ R+ .

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Proof. Take the set A = {t ∈ U | f (t) ⩾ c} ∈ M. Then, by Lemmas 2 and 1 Λ(f 𝜒(U)) ⩾ Λ(f 𝜒(A)) ⩾ Λ(c𝜒(A)) ⩾ c𝜇A. Compare now the integrals over the measurable space (T, M, 𝜇) and over any measurable “subspace” (U, MU , 𝜇U ), where U ∈ M, MU = {M ∩ U | M ∈ M} = {M ∈ M | M ⊂ U}, and 𝜇U = 𝜇|MU . The Lebesgue integral on the space (U, MU , 𝜇U ) will be denoted by Λ U . Proposition 2. Let 𝜇 be a positive measure on a 𝜎-algebra M on T, f ∈ M(T, M), and U ∈ M. Then, 1) f |U ∈ M(U, MU ) and f 𝜒(U) ∈ M(T, M); 2) f |U ∈ MI e (U, MU , 𝜇U ) iff f 𝜒(U) ∈ MI e (T, M, 𝜇); 3) if f 𝜒(U) ∈ MI e (T, M, 𝜇), then ∫(f |U) d𝜇U = ∫ f 𝜒(U) d𝜇. Proof. 1. Lemma 3 (2.5.1) guarantees that 𝜑 ≡ f |U ∈ M(U, MU ). Since 𝜒(U) ∈ M(T, M), we get g ≡ f 𝜒(U) ∈ M(T, M). 2. First, suppose that f ⩾ 0. Let x < Λ U 𝜑 and (B j ∈ MU | j ∈ J) be a finite partition of U such that x < ∑(inf 𝜑[B j ]𝜇U B j | j ∈ J). Then, B j ∈ M and x < ∑(inf g[B j ]𝜇B j | j ∈ J) + inf g[T\U]𝜇(T\U) ⩽ Λg. This means that Λ U 𝜑 ⩽ Λg ⩽ Λf . Let x < Λg and (A i ∈ M | i ∈ I) be a finite partition of T such that x < ∑(inf g[A i ]𝜇A i | i ∈ I). Then, x < ∑(inf g[A i ∩ U]𝜇(A i ∩ U) | i ∈ I) + ∑(inf g(A i \U) 𝜇(A i \U) | i ∈ I) = ∑(inf 𝜑[A i ∩ U]𝜇U (A i ∩ U) | i ∈ I) ⩽ Λ U 𝜑. Hence, Λg ⩽ Λ U 𝜑. Finally Λ U 𝜑 = Λg. Now, let f be an arbitrary function. Then, from 𝜑+ = f+ |U and −𝜑− = (−f− )|U, using the facts proven above, we conclude that Λ U (𝜑+ ) = Λ(f+ 𝜒(U)) and Λ U (−𝜑− ) = Λ((−f− )𝜒(U)). But f+ 𝜒(U) = g+ and (−f− )𝜒(U) = −g− . As a result, Λ U (𝜑+ ) = Λ(g+ ) and Λ U (−𝜑− ) = Λ(−g− ). These equalities imply the assertion. 3. If 𝜑 ∈ MI e (U, MU , 𝜇U ), then by (2), g ∈ MI e (T, M, 𝜇). Consequently, by the definition of the Lebesgue integral Λ U 𝜑 ≡ Λ U 𝜑+ − Λ U (−𝜑− ) = Λg+ − Λ(−g− ) ≡ Λg. Besides, |𝜑| = |f ||U implies Λ U |𝜑| = Λ(|f |𝜒(U)) ⩽ Λ|f |. Corollary 1. Let 𝜇 be a positive measure on a 𝜎-algebra M on T, f ∈ M(T, M), and U ∈ M. Then, f |U ∈ MI(U, MU , 𝜇U ) iff f 𝜒(U) ∈ MI(T, M, 𝜇). Corollary 2. Let 𝜇 be a positive measure on a 𝜎-algebra M on T, f ∈ MI e (T, M, 𝜇), and U ∈ M. Then, f 𝜒(U) ∈ MI e (T, M, 𝜇), f |U ∈ MI e (U, MU , 𝜇U ), and ∫ f 𝜒(U) d𝜇 = ∫(f |U) d𝜇U . Proof. By the condition, either Λf+ or Λ(−f− ) is finite. Since (f 𝜒(U))+ ⩽ f+ and −(f 𝜒(U))− ⩽ −f− , we infer by virtue of Lemma 2 that either Λ((f 𝜒(U))+ ) or Λ(−(f 𝜒(U))− ) is finite. Therefore, f 𝜒(U) ∈ MI e (T, M, 𝜇). The other assertions follow from this fact and Proposition 2.

262 | 3.3 The Lebesgue integral

Corollary 3. Let 𝜇 be a positive measure on a 𝜎-algebra M on T, f ∈ MI(T, M, 𝜇), and U ∈ M. Then, f 𝜒(U) ∈ MI(T, M, 𝜇), f |U ∈ MI(U, MU , 𝜇U ) and ∫ f 𝜒(U) d𝜇 = ∫(f |U) d𝜇U . Proof. By the condition, Λf+ and Λ(−f− ) are finite. As in the proof of Corollary 2 this implies that Λ((f 𝜒(U))+ ) and Λ(−(f 𝜒(U))− ) are finite. Therefore, f 𝜒(U) ∈ MI(T, M, 𝜇). The other assertions follow from Corollaries 1 and 2. Lemma 8. Let f ∈ MI(T, M, 𝜇). Then, C x ≡ {t ∈ T | |f (t)| ⩾ x} ∈ Mf (𝜇) for every x > 0. Proof. By Corollary 3 to Theorem 2 |f | ∈ MI(T, M, 𝜇). Then, Lemma 7 yields 𝜇C x ⩽ (1/x)Λ|f | < ∞. Corollary 1. Let f ∈ MI(T, M, 𝜇). Then, coz f ∈ M𝜎f (𝜇). Proposition 3. Let 𝜇 be a positive measure on a 𝜎-algebra M on T, f ∈ MI(T, M, 𝜇), (f n ∈ MI(T, M, 𝜇) | n ∈ N) be a sequence for some cofinal subset N in 𝜔, and lim(Λ|f − f n | | n ∈ N) = 0. Then, there is a subsequence (f n k | k ∈ 𝜔) such that n k+1 > n k ⩾ k, Λ|f − f n k | < 2−2k , and f (t) = lim(f n k (t) | k ∈ 𝜔) 𝜇-almost everywhere. Proof. For every pair (m, k) ∈ N × 𝜔, consider the non-empty set I mk ≡ {(n ∈ N | n > m⊻k)∧(Λ(|f −f n |) < 2−2(k+1) )}. Take n0 ≡ sm N. Define a mapping V : N ×𝜔 → N setting V(m, k) ≡ sm I mk . By Theorem 1 (1.2.8), there is the unique mapping u : 𝜔 → N such that u(0) = n0 and u(k + 1) = V(u(k), k) for every k ∈ 𝜔. Denote u(k) by n k . Then, we have n k+1 = V(n k , k) = sm{n ∈ N | (n > n k ⊻ k) ∧ (Λ(|f − f n |) < 2−2(k+1) )}. This provides n k+1 > n k , n k+1 ⩾ k + 1, and Λ(|f − f n k+1 |) < 2−2(k+1) . Consider the function g k ≡ f n k and the set E k ≡ {t ∈ T | |f (t) − g k (t)| ⩾ 2−k }. By Lemma 8, E k ∈ Mf . By virtue of Lemmas 1 and 2, 2−k 𝜇E k = Λ(2−k 𝜒(E k )) ⩽ Λ|f − g k | < 2−2k , where 𝜇E k < 2−k . Consider the sets F k ≡ ⋃⟮E i | i ∈ 𝜔\k⟯ and G k ≡ T\F k . By Lemma 3 (3.1.1), 𝜇F k ⩽ ∑net (𝜇E i | i ∈ 𝜔\k) ⩽ ∑net (2−i | i ∈ 𝜔\k). Using Lemma 3 (1.4.8), we infer that 𝜇F k ⩽ 2−k ∑(2−n | n ∈ 𝜔) ⩽ 2−k+1 . If t ∈ G k , then t ∉ E i for every i ⩾ k, where |f (t) − g i (t)| < 2−i for every i ⩾ k. Consider the set F ≡ ⋂⟮F k | k ∈ 𝜔⟯. Since 𝜇F ⩽ 𝜇F k ⩽ 21−k for every k, we infer that 𝜇F = 0. If t ∈ T\F = ⋃(G k | k ∈ 𝜔) then t ∈ G k for some k. Fix 𝜀 > 0 and take n ⩾ k such that 2−n ⩽ 𝜀. Then, for every i ⩾ n, we get |f (t) = g i (t)| < 2−i ⩽ 2−n ⩽ 𝜀. This means that f (t) − lim(g i (t) | i ∈ 𝜔). In conclusion, we shall make some remarks about notations of the integral ∫ f d𝜇. If we want to show the set T, then we write ∫T f d𝜇. If we want to show the argument of the function f , then we write ∫ f (t) d𝜇 and ∫T f (t) d𝜇.

3.3.3 Sequential properties of the Lebesgue integral | 263

If T is an interval |a, b| (see 1.1.15) of one of the following four kinds: (1) [a, b] for a, b ∈ R; (2) [a, b[ for a ∈ R and b ∈ R; (3) ]a, b] for a ∈ R and b ∈ R; (4) ]a, b[ for a, b ∈ R, and 𝜆 is the Lebesgue measure on R (see Example 1 (3.1.4) and subsection 3.1.6), then the integral Λ |a,b| f of the function f over the subspace b

⟮|a, b|, (M𝜆 )|a,b| , 𝜆 |a,b| ⟯ (see 3.1.2) is denoted by ∫a f (x) dx.

3.3.3 Sequential properties of the Lebesgue integral Here we use the notations from 3.3.2. In particular, (T, M, 𝜇) is a measurable space with the wide positive measure 𝜇. Theorem 1 (the Lebesgue theorem on the countable additivity of integral). Let 𝜇 be a positive measure on a 𝜎-algebra M on T and (f n ∈ MI(T, M, 𝜇)+ | n ∈ N) be a sequence for some cofinal subset N of 𝜔 such that ∑(Λf n | n ∈ N) < ∞ (see 1.4.8). Then, there is a function f ∈ MI(T, M, 𝜇)+ such that f (t) = ∑(f n (t) | n ∈ N) 𝜇-almost everywhere on T and Λf = ∑(Λf n | n ∈ N). Proof. Define the mapping 𝜑 : T → R+ and the functions 𝜑n ∈ MI+ setting 𝜑(t) ≡ ∑(f n (t) | n ∈ N) and 𝜑n ≡ ∑(f k | k ∈ N ∩ (n + 1)). Consider the sets A ≡ {t ∈ T | 𝜑(t) = ∞} and A nm ≡ {t ∈ T | 𝜑n (t) > m} ∈ M. Then, A ≡ ⋂⟮⋃⟮A nm | n ∈ 𝜔⟯ | m ∈ 𝜔⟯ ∈ M. By Lemma 7 (3.3.2), m𝜇A nm ⩽ Λ𝜑n . By Lemma 5 (3.1.1), 𝜇 ⋃⟮A nm | n ∈ 𝜔⟯ = lim(𝜇A nm | n ∈ 𝜔). Therefore, by Proposition 1 (1.4.8) and Theorem 2 (3.3.2), m𝜇 ⋃⟮A nm | n ∈ 𝜔⟯ ⩽ lim(Λ𝜑n | n ∈ 𝜔) = ∑(Λf n | n ∈ N) ≡ y. From A ⊂ ⋃⟮A nm | n ∈ 𝜔⟯, we conclude that m𝜇A ⩽ y for every m ∈ 𝜔, where, by Lemma 13 (1.4.3), 𝜇A = 0. Thus, B ≡ T\A ∈ co-M0 . Define the function f ∈ F(T) setting f (t) ≡ 𝜑(t) for t ∈ B and f (t) ≡ 0 for t ∈ A. Since f = p-lim(𝜑n 𝜒(B) | n ∈ 𝜔), Corollary 3 to Lemma 2 (2.3.4) provides that f ∈ M(T, M). By Lemma 5 (3.3.2), Λ𝜑n = Λ(𝜑n 𝜒(B)). So f ⩾ 𝜑n 𝜒(B) implies Λf ⩾ Λ𝜑n for every n ∈ 𝜔, by virtue of Lemma 2 (3.3.2), where Λf ⩾ y. By Theorem 2 (2.3.4), there are sequences (s nk ∈ St(T, M)+ | k ∈ N) such that s nk ⩽ s nk+1 ⩽ f n and f n = p-lim(s nk | k ∈ N). Consider the functions 𝜓k ≡ 𝜒(B) ∑(s ik | i ∈ N ∩ (k + 1)) ∈ St(T, M)+ . The sequence (𝜓k | k ∈ 𝜔) increases and for every m ⩽ k, we have 𝜒(B) ∑(s ik | i ∈ N ∩ (m + 1)) ⩽ 𝜓k ⩽ 𝜑k 𝜒(B) ⩽ f . Then, by Lemma 1 (2.2.3), this sequence has the pointwise limit p-lim(𝜓k | k ∈ N) ≡ g. Taking the limit with respect to k, we find that 𝜑m 𝜒(B) ∑ (f i | i ∈ N ∩ (m + 1)) = ∑⟮p-lim⟮S ik | k ∈ 𝜔⟯ | i ∈ N ∩ (m + 1)⟯ = p-lim (∑ (S ik | i ∈ N ∩ (m + 1)) | k ∈ 𝜔) ⩽ g ⩽ f for every t ∈ B. Taking the limit with respect to m, we obtain f ⩽ g ⩽ f for t ∈ B. This means that f = g.

264 | 3.3 The Lebesgue integral

Now, by virtue of Corollary 1 to Theorem 1 (3.3.2), Λf = lim(Λ𝜓k | k ∈ N) ⩽ lim(Λ(𝜑k 𝜒(B)) | k ∈ N) = lim(Λ𝜑k | k ∈ N) = y. Thus, Λf = y. Hence, f ∈ MI(T, M, 𝜇). Theorem 2 (the Beppo Levi theorem). Let 𝜇 be a positive measure on a 𝜎-algebra M on T and (f n ∈ MI(T, M, 𝜇) | n ∈ N ⊂ 𝜔) be an infinite increasing sequence such that sup (Λf n | n ∈ N) < ∞. Then, there is a function f ∈ MI(T, M, 𝜇) such that (f n (t) | n ∈ N) ↑ f (t) 𝜇-almost everywhere on T and (Λf n | n ∈ N) ↑ Λf . N. Proof. By Theorem 1 (1.2.7) for N, there is an isotone (see 1.1.15) bijection u : 𝜔 Therefore, we can consider the functions g i ≡ f u(i+1) − f u(i) ∈ MI+ (T, M, 𝜇), for which we have f u(i) = f u(0) + ∑(g k | k ∈ i) for i ∈ N, where ∑(Λg k | k ∈ i) = Λf u(i) − Λf u(0) . Hence, ∑ (Λg k | k ∈ 𝜔) = sup (∑(Λg k | k ∈ i) | i ∈ N) = sup (Λf u(i) | i ∈ N) − Λf u(0) < ∞. By Theorem 1, there is a function g ∈ MI(T, M, 𝜇)+ such that g(t) = ∑(g k (t) | k ∈ 𝜔) 𝜇-almost everywhere and Λg = ∑(Λg k | k ∈ 𝜔) = sup (∑(Λg k | k ∈ i) | i ∈ N) = sup (Λf u(i) − Λf u(0) | i ∈ N) = sup (Λf u(i) | i ∈ N) − Λf u(0) . Take the function f ≡ g + f u(0) . Then, f (t) = sup (∑(g k (t) | k ∈ i) + f u(0) (t) | i ∈ N) = = sup (f u(i) (t) | i ∈ N) = sup (f n (t) | n ∈ N) 𝜇-almost everywhere, and therefore, Λf = Λg + Λf u(0) = sup (Λf n | n ∈ N). Corollary 1. Let 𝜇 be a positive measure on a 𝜎-algebra M on T, f ∈ M(T, M), (f n ∈ MI(T, M, 𝜇) | n ∈ N ⊂ 𝜔) be an infinite increasing sequence such that sup (Λf n | n ∈ N) < ∞, and (f n (t) | n ∈ N) ↑ f (t) 𝜇-almost everywhere on T. Then, f ∈ MI(T, M, 𝜇) and (Λf n | n ∈ N) ↑ Λf . Proof. By Theorem 2, there is g ∈ MI(T, M, 𝜇) such that (f n (t) | n ∈ N) ↑ g(t) 𝜇-almost everywhere. Therefore, f (t) = g(t) 𝜇-almost everywhere. Consequently, Corollary 1 to Lemma 5 (3.3.2) implies f ∈ MI(T, M, 𝜇) and Λf = Λg. Theorem 3 (the Fatou lemma). Let 𝜇 be a positive measure on a 𝜎-algebra M on T and (f n ∈ MI(T, M, 𝜇)+ | n ∈ N ⊂ 𝜔) be an infinite sequence such that lim (Λf n | n ∈ N) < ∞. Then, there is a function f ∈ MI(T, M, 𝜇)+ such that f (t) = lim(f n (t) | n ∈ N) (see 1.1.15) 𝜇-almost everywhere on T and Λf ⩽ lim(Λf n | n ∈ N). Proof. Consider the functions g n ≡ inf (f k | k ∈ N ∧ k ⩾ n) in F(T)+ for n ∈ N. It is clear that (g n | n ∈ N) ↑. By Proposition 1 (2.3.3), g n ∈ M(T, M). Since g n ⩽ f k for all k ⩾ n, we infer by Lemma 2 (3.3.2) that that Λg n ⩽ Λf k , where Λg n ⩽ inf{Λf k | k ∈ N ∧ k ⩾ n} and g n ∈ MI(T, M, 𝜇).

3.3.3 Sequential properties of the Lebesgue integral | 265

By virtue of Theorem 2, there is a function f ∈ MI(T, M, 𝜇) such that f (t) = sup (g n (t) | n ∈ N) 𝜇-almost everywhere and Λf = sup (Λg n | n ∈ N). Therefore, Λf ⩽ lim(Λf n | n ∈ N). Besides, f (t) = lim(f n (t) | n ∈ N) 𝜇-almost everywhere. Corollary 1. Let (f n ∈ MI(T, M, 𝜇) | n ∈ N) be a sequence for some cofinal subset N of 𝜔, f ∈ F(T) and f (t) = lim(f n (t) | n ∈ N) 𝜇-almost everywhere on T. If lim(Λ(|f n |) | n ∈ N) < ∞, then f ∈ MI(T, M, 𝜇). Proof. Since f = p-lim (f n | n ∈ N) and |f | = p-lim (|f n | | n ∈ N) 𝜇-almost everywhere, Proposition 1 (3.3.1) guarantees that the functions f and |f | belong to M(T, M). By Theorem 3, |f | ∈ MI(T, M, 𝜇), where f ∈ MI(T, M, 𝜇) by virtue of Corollary 3 to Theorem 2 (3.3.2). Theorem 4 (the Lebesgue dominated convergence theorem). Let 𝜇 be a positive measure on a 𝜎-algebra M on T, (f n ∈ M(T, M) | n ∈ N ⊂ 𝜔) be an infinite sequence, g ∈ MI(T, M, 𝜇)+ , f ∈ F(T), |f n (t)| ⩽ g(t) 𝜇-almost everywhere on T for every n ∈ N, and f (t) = lim(f n (t) | n ∈ N) 𝜇-almost everywhere on T. Then, f ∈ MI(T, M, 𝜇) and Λf = lim(Λf n | n ∈ N). Proof. By Corollary 1 to Theorem 2 (3.3.2) and Statement 2 (2.2.8) all functions f n are integrable. By Proposition 1 (3.3.1), f ∈ M(T, M). According to Lemma 9 (3.1.1) M0 is a 𝜎-ideal in M. Then, using Corollaries 2 and 3 to Proposition 1 (1.4.7), we get the inequality |f (t)| ⩽ g(t) 𝜇-almost everywhere on T. Consequently, f (t) is integrable. Consider the functions g n ≡ f n + g ∈ MI(T, M, 𝜇)+ . By Lemmas 2 and 5 (3.3.2) Λf n ⩽ Λg for every n ∈ N. Since sup (Λg n | n ∈ N) ⩽ sup (Λf n | n ∈ N) + Λg ⩽ 2Λg < ∞, by the Fatou lemma (Theorem 3) there is a function g 󸀠 ∈ MI(T, M, 𝜇)+ such that g󸀠 (t) = lim (g n (t) | n ∈ N). Then, according to the definition of the limit inferior in 1.1.15 and Lemma 4 (1.4.5), g󸀠 (t) = lim (f n (t) + g(t) | n ∈ 𝜔) = lim (f n (t) | n ∈ 𝜔) + g(t) = lim (f n (t) | n ∈ N) + g(t) 𝜇-almost everywhere and Λg 󸀠 ⩽ lim(Λg n | n ∈ N) = lim(Λf n | n ∈ N) + Λg. Consider the function 𝜑 ≡ g󸀠 − g ∈ MI(T, M, 𝜇). Then, 𝜑(t) = lim (f n (t) | n ∈ N) and Λ𝜑 ⩽ lim (Λf n | n ∈ N). On the other hand, consider the functions h n ≡ g − f n ∈ MI(T, M, 𝜇)+ . Then, Λ(−f n ) ⩽ Λg implies sup Λh n ⩽ 2Λg < ∞. Hence, there is a function h󸀠 such that h󸀠 (t) = lim (h n (t) | n ∈ N) = g(t) − lim (f n (t) | n ∈ N) 𝜇-almost everywhere and Λh󸀠 ⩽ lim (Λh n | n ∈ N) = Λg − lim (Λf n | n ∈ N). Consider the function 𝜓 ≡ g − h󸀠 . Then, 𝜓(t) = lim (f n (t) | n ∈ N) and Λ𝜓 ⩾ lim (Λf n | n ∈ N). By virtue of Lemma 4 (1.1.15) and Lemma 3 (1.4.7), f (t) = lim (f n (t) | n ∈ N) = lim (f n (t) | n ∈ N) 𝜇-almost everywhere. Therefore, f (t) = 𝜑(t) = 𝜓(t) 𝜇-almost everywhere and Λf = Λ𝜑 = Λ𝜓. In conclusion we get Λf ⩽ lim (Λf n | n ∈ N) ⩽ lim (Λf n | n ∈ N) ⩽ Λf . Using Lemmas 3 (1.4.7) and 4 (1.1.15) again, we get Λf = lim (Λf n | n ∈ N).

266 | 3.3 The Lebesgue integral

Note that the presence of the dominating function g in the above theorem is of the utmost importance. If no such function exists, the conclusion may fail. For example, let T = R, 𝜇 = 𝜆× (the usual Lebesgue measure on R, see 3.1.6), and f n = n𝜒(]0, 1/n]) for every n ∈ N. Then, ∫ f n d𝜇 = n(1/n) = 1, by virtue of Lemma 1 (3.3.2), while lim (f n (t) | n ∈ N) = 0 for all t ∈ R. Therefore, lim (Λf n | n ∈ N) = 1 = / 0. Theorem 5. Let 𝜇 be a positive measure on a 𝜎-algebra M, (f n ∈ M(T, M)+ | n ∈ N ⊂ 𝜔) be an increasing infinite sequence, f ∈ M(T, M)+ , and f (t) = lim(f n (t) | n ∈ N) 𝜇-almost everywhere on T. Then, lim(Λf n | n ∈ N) = Λf . Proof. By the condition, there is E ∈ M0 (𝜇) such that f (t) = lim(f n (t) | n ∈ N) for every t ∈ U ≡ T\E. First, suppose that Λf < ∞. Then, by Lemma 2 (3.3.2), Λ(f n 𝜒(U)) ⩽ Λf . By Lemma 5 (3.3.2), Λ(f n ) = Λ(f n 𝜒(U)) < Λf < ∞ for every n. Hence, f n ∈ MI(T, M, 𝜇). Theorem 2 implies that there are a function g ∈ MI(T, M, 𝜇) and F ∈ M0 (𝜇) such that g(t) = lim (f n (t) | n ∈ N) for every t ∈ V ≡ T\F and Λg = lim(Λf n | n ∈ N). Consequently, f |U ∩ V = g|U ∩ V and, in view of Lemma 5 (3.3.2), Λf = Λg. Now, suppose that Λf = ∞. Assume that there is x ∈ R+ such that Λf n ⩽ x for every n ∈ N. Then, as proven above ∞ = Λf = lim(Λf n | n ∈ N) ⩽ x, but this is impossible. It follows from this contradiction that for every x there is n such that Λf p ⩾ Λf n > x for every p ⩾ n. According to 1.4.7 this means that lim(Λf n | n ∈ N) = ∞ = Λf .

3.3.4 Properties of density and completeness for the family and the factor-family of integrable functions Now, we shall consider the family MI(T, M, 𝜇) for a positive wide measure 𝜇 : M → R+ with the point of view of a seminormed space. Here we use the notations from 3.3.1 and 3.3.2. In particular, (T, M, 𝜇) is a measurable space with the wide positive measure 𝜇. By Corollary 2 to Theorem 2 (3.3.2), MI(T, M, 𝜇) is a Dedekind 𝜔-complete latticeordered linear space. Therefore, we can consider on MI(T, M, 𝜇) the functional r : MI(T, M, 𝜇) → R+ such that rf ≡ Λ|f |. Lemma 1. The functional r is a monotone seminorm on the lattice-ordered linear space MI(T, M, 𝜇) such that r(f + g) = rf + rg for every f , g ∈ MI(T, M, 𝜇)+ . Proof. By Theorem 2 (3.3.2), r(xf ) = Λ(|xf |) = Λ(|x||f |) = |x|rf , i. e. r is absolutely homogeneous. Further, by Lemma 2 (3.3.2) and Theorem 2 (3.3.2) the inequality |f + g| ⩽ |f | + |g| implies r(f + g) = Λ|f + g| ⩽ Λ(|f | + |g|) = rf + rg, i. e. r is subadditive. Thus, r is a seminorm.

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Again by Lemma 2 (3.3.2), |f | ⩽ |g| implies rf = Λ|f | ⩽ Λ|g| = rg, i. e. r is modulusly monotone. By virtue of the additivity of integral Λ we have r(f + g) = rf + rg for all f , g ∈ MI(T, M, 𝜇)+ . This seminorm r is denoted by ‖⋅‖i . It is called the seminorm of the integral convergence on MI(T, M, 𝜇). Corollary 1. ⟮MI(T, M, 𝜇), ‖⋅‖i ⟯ is a seminormed lattice-ordered linear space. Moreover, it is a pre-L-space. Consider on MI(T, M, 𝜇), the seminormed topology G(‖⋅‖i ). Corollary 2. The functional Λ : ⟮MI(T, M, 𝜇), ‖⋅‖i ⟯ → ⟮R, | ⋅ |⟯ is (seminorm) bounded and ‖Λ‖󸀠 ≡ sup{|Λf | | ‖f ‖i ⩽ 1} ⩽ 1. Proof. By Corollary 3 to Theorem 1 (3.3.2) |Λf | ⩽ ‖f ‖i . This means that Λ is (seminorm) bounded in sense of 7∘ (2.2.7). Corollary 3. The functional Λ : ⟮MI(T, M, 𝜇), G(‖⋅‖i )⟯ → ⟮R, Gu ⟯ is continuous. Proof. It follows from Corollaries 1 and 2 and Statement 12 (2.2.7). Corollary 4. The seminorm ‖⋅‖i : ⟮MI(T, M, 𝜇), G(‖⋅‖i )⟯ → ⟮R, Gu ⟯ is continuous. Proof. It follows from Corollary 1, the equality |‖f ‖i | = ‖f ‖i , and Statement 12 (2.2.7).

Proposition 1. Let ⟮T, M, 𝜇⟯ be a measurable space with the wide positive measure 𝜇. Then, the family of all integrable step functions St(T, M) ∩ MI(T, M, 𝜇) is a latticeordered linear subspace of the lattice-ordered linear space MI(T, M, 𝜇) and it is dense in MI(T, M, 𝜇) in the seminormed topology G(‖⋅‖i ). Moreover, for every f ∈ MI(T, M, 𝜇), there is a sequence (f n ∈ St(T, M) ∩ MI(T, M, 𝜇) | n ∈ N) such that |f n−1 | ⩽ |f n | ⩽ |f |, f = p-lim (f n | n ∈ N), and lim (‖f n − f ‖i | n ∈ N) = 0. If f ⩾ 0, then the functions f n can be chosen such that 0 ⩽ f n−1 ⩽ f n ⩽ f . Proof. The first statement follows from the fact that St(T, M) is a lattice-ordered linear subspace of F(T) (see Proposition 1 (2.2.4)). By Theorem 2 (2.3.4) for every f ∈ MI(T, M, 𝜇)+ , there is a sequence (f n ∈ St(T, M)+ | n ∈ N) ↑ such that f (t) = lim (f n (t) | n ∈ N) = sup (f n (t) | n ∈ N) for every t ∈ T. By Corollary 1 to Theorem 1 (3.3.2) Λf = sup (Λf n | n ∈ N) = lim(Λf n | n ∈ N). Therefore, all f n ∈ MI(T, M, 𝜇) and lim(Λ(f − f n ) | n ∈ N) = 0. Since f − f n = |f − f n |, we infer that lim (‖f − f n ‖i | n ∈ N) = 0. By Corollary 1 to Statement 6 (2.2.7), f = lim (f n | n ∈ N) in G(‖⋅‖i ).

268 | 3.3 The Lebesgue integral

Now, let f ∈ MI(T, M, 𝜇). As was shown above, for the functions f+ and (−f− ), there are sequences (g n | n ∈ N) ↑ f+ and (h n | n ∈ N) ↑ (−f− ) from St(T, M)+ ∩ MI(T, M, 𝜇) such that lim (‖f+ − g n ‖i | n ∈ N) = 0 and lim (‖ − f− − h n ‖i | n ∈ N) = 0. Consider the functions f n ≡ g n − h n . Since g n ∧ h n = 0, coz g n ∩ coz h n = ⌀ and T = coz g n ∪ coz h n ∪ (zer g n ∩ zer h n ). This implies |f n | = g n + h n . Then, we have |f n | ⩽ g n + h n ⩽ f+ − f− = |f |, |f n−1 | = g n−1 + h n−1 ⩽ g n + h n = |f n |, and ‖f − f n ‖i ⩽ ‖f+ − g n ‖i + ‖f− + h n ‖i . By virtue of Proposition 1 (1.4.7) f (t) = lim (f n (t) | n ∈ N) for every t ∈ T and lim (‖f+ − g n ‖i + ‖f− + h n ‖i | n ∈ N) = 0. Finally, Lemma 6 (1.4.7) implies lim (‖f − f n ‖i | n ∈ N) = 0. Proposition 2. Let 𝜌 be a positive measure on a ring R on a set T, R ⊂ M ⊂ ̂ ̂ M(T, R, 𝜌), 𝜇 ≡ 𝜌|M be a wide measure. The family of all integrable step functions St(T, R) ∩ MI(T, M, 𝜇) is a lattice-ordered linear subspace of the lattice-ordered linear space MI(T, M, 𝜇), and it is dense in MI(T, M, 𝜇) in the normed topology G(‖⋅‖i ). Moreover, for every f ∈ MI(T, M, 𝜇), there is a sequence (f n ∈ St(T, R) ∩ MI(T, M, 𝜇) | n ∈ N) such that lim (‖f − f n ‖i | n ∈ N) = 0. If, besides, 𝜌× is internally finite, then ̂ ̂̂ ̂ the same is valid for R ⊂ M ⊂ M(T, R, 𝜌) and 𝜇 ≡ 𝜌|M. Proof. By Proposition 1 for f ∈ MI(T, M, 𝜇)+ there is (g n ∈ St(T, M)+ ∩ MI(T, M, 𝜇) | n ∈ N), |g n | ⩽ f and lim (‖f − g n ‖i | n ∈ N) = 0, that is for 𝜀 > 0 there is n such that ‖f − g n ‖i < 𝜀/2 for every n ⩾ m. Fix a number n. We have g n = ∑(x k 𝜒(M k ) | k ∈ n ∈ N) for some finite collections (x k ∈ R+ | k ∈ n) and (M k ∈ M | k ∈ n). Since g n is integrable, we conclude by virtue of Lemma 1 (3.3.2) that ∑ (x k 𝜇M k | k ∈ n) = Λg n < ∞. Then, we get M k ∈ Mf for every k, whereas 𝜇 is positive. By Proposition 2 (3.1.5) for every M k there is R k ∈ Rf such that 𝜇(M k 󳵻 R k ) < 𝜀/(2nx k ). If 𝜌× is internally finite, then we should apply Proposition 4 (3.1.5). Consider the function h n ≡ ∑(x k 𝜒(R k ) | k ∈ n) ∈ St(T, R)+ . Since |g n − h n | ⩽ ∑ (x k |𝜒(M k ) − 𝜒(R k )| | k ∈ n) = ∑ (x k 𝜒(M k 󳵻 R k ) | k ∈ n), we infer by Lemma 1 (3.3.2) that Λ|g n − h n | ⩽ ∑ (x k 𝜇(M k 󳵻 R k ) | k ∈ n) < 𝜀/2. Hence, ‖f − h n ‖i ⩽ ‖f − g n ‖i + ‖g n − h n ‖i < 𝜀. Now, let f ∈ MI(T, M, 𝜇). As was shown above, there are sequences (h󸀠n | n ∈ N) and (h󸀠󸀠n | n ∈ N) from St(T, M)+ ∩ MI(T, M, 𝜇) such that lim (‖f+ − h󸀠n ‖i | n ∈ N) = 0

and lim (‖ − f− − h󸀠󸀠n ‖i | n ∈ N) = 0. Consider the functions f n ≡ h󸀠n − h󸀠󸀠n . Then, we have ‖f − f n ‖i ⩽ ‖f+ − h󸀠n ‖i + ‖f− + h󸀠󸀠n ‖i → 0. By virtue of Corollary 1 to Statement 6 (2.2.7) this gives the necessary density. The following result gives a good criterion for the integrability of functions. ̂ Theorem 1. Let 𝜌 be a positive measure on a ring R on a set T, R ⊂ M ⊂ M(T, R, 𝜇), ̂ 𝜇 ≡ 𝜌|M be a wide measure, and f ∈ M(T, M). Then, the following assertions are equivalent:

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1) f ∈ MI(T, M, 𝜇); 2) there is a sequence (f n ∈ St(T, R) ∩ MI(T, M, 𝜇) | n ∈ N) such that f (t) = lim(f n (t) | n ∈ N) 𝜇-almost everywhere and for every 𝜀 > 0 there is n such that ‖f p − f q ‖i < 𝜀 for every p, q ⩾ n. ̂ × ̂̂ ̂ R, 𝜌) and 𝜇 ≡ 𝜌|M. If 𝜌 is internally finite, then the same is valid for R ⊂ M ⊂ M(T, Proof. (1) ⊢ (2). By Proposition 2 there is a sequence (g n ∈ St(T, R) ∩ MI(T, M, 𝜇) | n ∈ N) such that lim (‖f − g n ‖i | n ∈ N) = 0. By Proposition 3 (3.3.2), there is a subsequence (g n k | k ∈ 𝜔) such that n k ⩾ n, ‖f − g n k ‖i < 2−2k and f (t) = lim(g n k (t) | k ∈ 𝜔) 𝜇-almost everywhere. Consider the functions f k ≡ g n k . Take 𝜀 > 0 and the number k such that 21−2k < 𝜀. Then, for every p, q ⩾ k we have ‖f p − f q ‖i ⩽ ‖f − f p ‖i + ‖f − f q ‖i < 𝜀. (2) ⊢ (1). The condition implies that the set I nk ≡ {q ∈ N | (q > n ⊻ k) ∧ (∀ p ∈ N (p ⩾ q ⇒ (‖f p − f q ‖i < 2−k−1 )))} is non-empty for every k ∈ 𝜔 and n ∈ N. Therefore, we can correctly define the mapping v : N × 𝜔 → N, setting v(n, k) ≡ sm I nk . By Theorem 1 (1.2.8), there is a unique mapping u : 𝜔 → N such that u(0) = 1 and u(k+1) = v(u(k), k) = sm{q ∈ N | (q > u(k) ⊻ k) ∧ (∀ p ∈ N (p ⩾ q ⇒ (‖f p − f q ‖i < 2−k−1 )))}. It is clear that u(k + 1) > u(k) and u(k + 1) > k for every k ∈ 𝜔. Besides, p ⩾ u(k + 1) implies ‖f p − f u(k+1) ‖i < 2−k−1 for every k ∈ 𝜔. This yields ‖f u(k+2) − f u(k+1) ‖i < 2−k−1 for every k ∈ 𝜔. Consider the new step function g n ≡ ∑ (|f u(k+2) − f u(k+1) | | k ∈ n + 1) ∈ St(T, R) ∩ MI(T, M, 𝜇) for every n ∈ N. We have g n ↑ and Λg n ⩽ ∑ (2−k−1 | k ∈ 𝜔) = 1 (see Lemma 3 (1.4.8)). Consider the mapping g : T → R+ such that g(t) = sup (g n (t) | n ∈ 𝜔). Consider also the sets A ≡ {t ∈ T | g(t) = ∞} and A nm ≡ {t ∈ T | g n (t) > m} ∈ M. Then, A = ⋂⟮⋃⟮A nm | n ∈ 𝜔⟯ | m ∈ 𝜔⟯ ∈ M. By Lemma 7 (3.3.2), m𝜇A nm ⩽ Λg n ⩽ 1. By Lemma 5 (3.1.1), 𝜇 ⋃⟮A nm | n ∈ 𝜔⟯ = lim(𝜇A nm | n ∈ 𝜔). Therefore, m𝜇 ⋃⟮A nm | n ∈ 𝜔⟯ ⩽ 1. From A ⊂ ⋃⟮A nm | n ∈ 𝜔⟯, we conclude that m𝜇A ⩽ 1 for every m ∈ 𝜔, where, by Lemma 13 (1.4.3), 𝜇A = 0. Thus, B ≡ T\A ∈ co-M0 . By the condition, there is a set C ≡ co-M0 such that f (t) = lim(f n (t) | n ∈ N) for every t ∈ C. Let t ∈ D ≡ B ∩ C ∈ co-M0 and 𝜀 > 0. Then, there is n ∈ N such that p ⩾ n implies |f p (t) − f (t)| < 𝜀. If k ⩾ n then u(k) ⩾ k ⩾ n implies |f u(k) (t) − f (t)| < 𝜀. This means that f (t) = lim(f u(k) (f ) | k ∈ 𝜔). Thus, f (t) = lim(f u(k) | k ∈ 𝜔) for every t ∈ D. Define a function h ∈ F(T)+ setting h(t) ≡ g(t) for t ∈ D and h(t) ≡ 0 for t ∈ T\D. Consider the increasing functions h n ≡ g n 𝜒(D) ∈ St(T, M)+ . Since h(t) = lim(h n (t) | n ∈ 𝜔) = sup (h n (t) | n ∈ 𝜔) for every t ∈ T, we conclude by virtue of Corollary 3 to Lemma 2 (2.3.4) that h ∈ M(T, M). By Corollary 1 to Theorem 1 (3.3.2), Λh = sup (Λh n | n ∈ 𝜔). By Lemma 5 (3.3.2), Λh n = Λg n . Thus, Λh ⩽ 1, where h ∈ MI(T, M, 𝜇). Now, we have f (t) − f u(1) (t) = lim(f u(k) (t) − f u(1) (t) | k ∈ 𝜔). Therefore, by Corollary 2 to Proposition 1 (1.4.7), |f (t) − f u(1) (t)| = lim(|f u(k) (t) − f u(1) (t)| | k ∈ 𝜔). Since |f u(k) (t) − f u(1) (t)| ⩽ g k−2 (t) for every k ⩾ 2, by Corollary 3 to Proposition 1 (1.4.7) we obtain

270 | 3.3 The Lebesgue integral

|f (t) − f u(1) (t)| ⩽ lim(g k (t) | k ∈ 𝜔) = lim(h k (t) | k ∈ 𝜔) = h(t). Thus, |f − f u(1) |𝜒(D) ⩽ h. Lemma 2 (3.3.2) guarantees (f − f u(1) )𝜒(D) ∈ MI(T, M, 𝜇). By Lemma 5 (3.3.2), f − f u(1) ∈ MI(T, M, 𝜇), where f ∈ MI(T, M, 𝜇). This result is very important. It gives another method of a construction of the Lebesgue integral. In such a method, the integral is defined at first on a subset of the set St(T, R), and then it is extended on a subset of the set M(T, M) by means of Theorem 1 (or in some other similar way). Corollary 1. Let ⟮T, M, 𝜇⟯ be a measurable space with the wide positive measure 𝜇. Then, for a function f ∈ M(T, M) the following assertions are equivalent. 1) f ∈ MI(T, M, 𝜇); 2) there is a sequence (f n ∈ St(T, M) ∩ MI(T, M, 𝜇) | n ∈ N) such that f (t) = lim(f n (t) | n ∈ N) 𝜇-almost everywhere and for every 𝜀 > 0 there is n such that ‖f p − f q ‖i < 𝜀 for every p, q ⩾ n. ̂ Proof. Take in Theorem 1, R ≡ M and 𝜌 ≡ 𝜇. Then, R = M ⊂ M(T, R, 𝜌) and 𝜇 = 𝜌 = ̂ = 𝜌|M. ̂ 𝜌|R In this case, St(T, R) = St(T, M). Further, ⟮T, M, 𝜇⟯ will be again any measurable space with the wide positive measure 𝜇. Theorem 2. The seminormed lattice-ordered linear space ⟮MI(T, M, 𝜇), ‖⋅‖i ⟯ is a complete pre-L-space. Proof. Let (f n ∈ MI(T, M, 𝜇) | n ∈ 𝜔) be an inner convergent sequence with respect to the seminorm ‖⋅‖i . Then, the set I nk ≡ {q ∈ 𝜔 | (q > n ⊻ k) ∧ (∀ p ∈ 𝜔 (p ⩾ q ⇒ (‖f p − f q ‖i < 2−k−3 )))} is non-empty for every k, n ∈ 𝜔. So that we can define correctly the mapping v : 𝜔 × 𝜔 → 𝜔, setting v(n, k) ≡ sm I nk . By Theorem 1 (1.2.8) there is a unique mapping u : 𝜔 → 𝜔 such that u(0) = 0 and u(k + 1) = v(u(k), k) = sm{q ∈ 𝜔 | (q > u(k) ⊻ k) ∧ (∀ p ∈ 𝜔 (p ⩾ q ⇒ (‖f p − f q ‖i < 2−k−3 )))}. It is clear, that u(k + 1) > u(k) and u(k + 1) > k for every k ∈ 𝜔. Besides, p ⩾ u(k + 1) implies ‖f p − f u(k+1) ‖i < 2−k−3 for every k ∈ 𝜔. This gives ‖f u(k+2) − f u(k+1) ‖i < 2−k−3 for every k ∈ 𝜔. By Proposition 1 for every function f u(k) there is an integrable step function 𝜑u(k) ∈ St(T, M) such that ‖f u(k) − 𝜑u(k) ‖i < 2−k−1 . Thus, we have ‖𝜑u(k+2) − 𝜑u(k+1) ‖i < 2−k−1 for every k ∈ 𝜔. Consider the new integrable step function g n ≡ ∑(|𝜑u(k+2) − 𝜑u(k+1) | | k ∈ n + 1) for every n ∈ 𝜔. We have (g n | n ∈ 𝜔) ↑ and Λg n ⩽ ∑(2−k−1 | k ∈ 𝜔) = 1 (see Lemma 3 (1.4.8)). Consider the mapping g : T → R+ such that g(t) ≡ sup (g n (t) | n ∈ 𝜔). Consider also the set A ≡ {t ∈ T | g(t) = ∞}. As in the proof of Theorem 1, it is checked that B ≡ T\A ∈ co-M0 . Let t ∈ B and 𝜀 > 0. By Proposition 1 (1.4.4), there is n such that |g p (t) − g q (t)| < 𝜀 for every p, q ⩾ n. Let q > p. Then, |𝜑u(q+2) (t) − 𝜑u(p+2) (t)| ⩽ ∑(|𝜑u(k+2) (t) − 𝜑u(k+1) (t)| |

3.3.4 Density and completeness for the families of integrable functions

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p + 1 ⩽ k ⩽ q) = g q (t) − g p (t) < 𝜀. If p > q, then we get the similar inequality. This means that the sequence (𝜑u(k) (t) | k ∈ 𝜔) is inner convergent in R. Therefore, there is a unique number x ∈ R+ such that x = lim(𝜑u(k) (t) | k ∈ 𝜔) = lim((𝜑u(k) 𝜒(B))(t) | k ∈ 𝜔). Thus, we can define correctly the function f ∈ F(T) setting f (t) ≡ lim((𝜑u(k) 𝜒(B))(t) | t ∈ 𝜔) for every t ∈ T. Define a function h ∈ F(T)+ setting h(t) ≡ g(t) for t ∈ B and h(t) ≡ 0 for t ∈ T\B. Consider the increasing sequence of functions h n ≡ g n 𝜒(B) ∈ St(T, M)+ . By Lemma 5 (3.3.2) all h n are integrable. Since h(t) = lim(h n (t) | n ∈ 𝜔) = sup (h n (t) | n ∈ 𝜔) for every t ∈ T, we conclude by virtue of Corollary 3 to Lemma 2 (2.3.4) that h ∈ M(T, M). By Corollary 1 to Theorem 1 (3.3.2), Λh = sup (Λh n | n ∈ 𝜔) = lim(Λh n | n ∈ 𝜔). By Lemma 5 (3.3.2), Λh n = Λg n . Thus, Λh ⩽ 1, where h ∈ MI(T, M, 𝜇). Now, for t ∈ B, we have f (t) − 𝜑u(m) (t) = lim(𝜑u(k) (t) − 𝜑u(m) (t) | k ∈ 𝜔). Therefore, Corollary 2 to Proposition 1 (1.4.7) guarantees that |f (t) − 𝜑u(m) (t)| = lim(|𝜑u(k) (t) − 𝜑u(m) (t)| | k ∈ 𝜔). Since |𝜑u(k) − 𝜑u(m) | ⩽ ∑(|𝜑u(i+1) − 𝜑u(i) | | m ⩽ i ⩽ k − 1) = g k−2 − g m−2 ⩽ g − g m−2 for every k ⩾ 2, we conclude using Corollary 3 to Proposition 1 (1.4.7) that |f (t) − 𝜑u(m) (t)| ⩽ g(t) − g m−2 (t). Thus, |f − 𝜑u(m) |𝜒(B) ⩽ h − h m−2 . Lemmas 2 and 5 (3.3.2) and Theorem 2 (3.3.2) provide ‖f − 𝜑u(m) ‖i ⩽ Λh − Λh m−2 . Let 𝜀 > 0. Then, there is m ∈ 𝜔 such that 2−m−2 < 𝜀/3 and Λh − Λh m−1 < 𝜀/3. ‖f − f u(k) ‖i ⩽ ‖f − 𝜑u(k) ‖i + ‖𝜑u(k) − f u(k) ‖i ⩽ 𝜀/2 + 2−k−1 < 𝜀 for every k ⩾ m. If p ⩾ u(m + 1) then we have ‖f − f p ‖i ⩽ ‖f − 𝜑u(m+1) ‖i + ‖𝜑u(m+1) − f u(m+1) ‖i + ‖f u(m+1) − f p ‖i ⩽ Λh − Λh m−1 + 2−m−2 + 2−m−3 < 𝜀. This means that f ∈ lim (f n | n ∈ 𝜔) in G(‖⋅‖i ). Use of Corollary 1 to Lemma 1 concludes the proof. Consider the equivalence relation 𝜃I(M0 ),MI(T,M,𝜇) on MI(T, M, 𝜇) with respect to the 𝜎-ideal ensemble I(M0 ) on T (cf. 3.3.1). We shall denote it simply by 𝜃. Lemma 3 (2.2.6) implies MI 0 (T, M, 𝜇) ≡ {f ∈ MI(T, M, 𝜇) | f 𝜃0} = {f ∈ MI(T, M, 𝜇) | ‖f ‖i = 0}. Consider the relation of equivalence 𝜃MI 0 (T,M,𝜇) on MI(T, M, 𝜇). According to 2.2.6, it is equal to 𝜃. Therefore, MI(T, M, 𝜇)/I(M0 ) = MI(T, M, 𝜇)/MI 0 (T, M, 𝜇). This factor-set is denoted by L1 (T, M, 𝜇). It follows from 2.2.6 that L1 (T, M, 𝜇) is a lattice-ordered linear space. Besides, we can consider on L1 (T, M, 𝜇) the seminorm ‖⋅‖1 ≡ ‖⋅‖i,L1 (T,M,𝜇) . We have ‖ f ‖1 = ‖f ‖i for all f ∈ f ∈ L1 (T, M, 𝜇). Lemma 2. The subset MI 0 (T, M, 𝜇) is closed in ⟮MI(T, M, 𝜇), G(‖⋅‖i )⟯. Proof. Let f ∈ MI(T, M, 𝜇) and f ∈ lim(f n ∈ MI 0 (T, M, 𝜇) | n ∈ N) for some cofinal subset N of 𝜔. From the inequality |‖f ‖i − ‖f n ‖i | ⩽ ‖f − f n ‖i and Corollary 1 to Statement 6 (2.2.7), it follows that ‖f ‖i = lim (‖f n ‖i | n ∈ N) = 0, where f ∈ MI 0 (T, M, 𝜇). By Statement 7 (2.2.7), this gives the necessary assertion. Corollary 1. The seminorm ‖⋅‖1 on L1 (T, M, 𝜇) is a norm such that ‖ f + g‖̄ 1 = ‖ f ‖1 + ‖g‖̄ 1 for every f , ḡ ∈ L1 (T, M, 𝜇)+ .

272 | 3.3 The Lebesgue integral

Proof. It follows from Lemma 2 and Statement 9 (2.2.7) that ‖⋅‖1 is a norm. If f , ḡ ∈ L1 (T, M, 𝜇)+ , then 0 ⩽ f 󸀠 ≡ f ∨ 0 ∈ f and 0 ⩽ g󸀠 ≡ g ∨ 0 ∈ g.̄ Thus, we have ‖f + g‖̄ 1 = ‖f 󸀠 + g󸀠 ‖i = ‖f 󸀠 ‖i + ‖g󸀠 ‖i = ‖f ‖1 + ‖g‖̄ 1 . Corollary 2. The normed lattice-ordered linear space (L1 (T, M, 𝜇), ‖⋅‖1 ) is a Banach lattice-ordered space. Moreover, it is an L-space. Proof. It follows from Corollary 1 to Lemma 2, Statements 10 (2.2.7), and 11 (2.2.7). The norm ‖⋅‖1 on L1 (T, M, 𝜇) is called the norm of the convergence in mean. Consider now the subset SL1 (T, M, 𝜇) ≡ { f ∈ L1 (T, M, 𝜇) | f ∈ St(T, M) ∩ MI(T, M, 𝜇)} in L1 (T, M, 𝜇) consisting of stepwise elements. Proposition 3. The set SL 1 (T, M, 𝜇) is a lattice-ordered linear subspace of the latticeordered linear space L1 (T, M, 𝜇) and it is dense in L 1 (T, M, 𝜇) in the normed topology G(‖⋅‖1 ). Proof. The first statement follows from Proposition 1 and the fact that the factorL1 (T, M, 𝜇) such that pf = f is a homomorphism of mapping p : MI(T, M, 𝜇) lattice-ordered linear spaces. Now, let f ∈ L1 (T, M, 𝜇). By Proposition 1 there is a sequence (f n ∈ St(T, M) ∩ MI(T, M, 𝜇) | n ∈ N) such that lim (‖f − f n ‖i | n ∈ N) = 0. Then, lim (‖f ̄ − pf n ‖1 | n ∈ N) = lim (‖pf − pf n ‖1 | n ∈ N) = lim (‖f − f n ‖i | n ∈ N) = 0. Besides, pf n ∈ SL1 (T, M, 𝜇). Lemma 3. Let 𝜇 be a positive measure on 𝜎-algebra M on T. If h ∈ M∞ (T, M, 𝜇) and f ∈ MI(T, M, 𝜇), then hf ∈ MI(T, M, 𝜇). Proof. Let h ∈ M∞ (T, M) and f ∈ MI(T, M, 𝜇). Then, by Lemma 2 (3.3.1) there is h󸀠 ∈ M b (T, M) such that coz(h − h󸀠 ) ∈ M0 and ‖h‖eu = ‖h󸀠 ‖u . Since |h󸀠 f | ⩽ |‖h󸀠 ‖u f | and ‖h󸀠 ‖u f ∈ MI(T, M, 𝜇), by Corollary 1 to Theorem 2 (3.3.2) and Statement 2 (2.2.8) we conclude that h󸀠 f ∈ MI(T, M, 𝜇). By Corollary 1 to Lemma 5 (3.3.2), hf ∈ MI(T, M, 𝜇). Lemma 4. Let 𝜇 be a positive measure on a 𝜎-algebra M on T, f 󸀠 ∈ f ∈ L1 (T, M, 𝜇) and h󸀠 ∈ h̄ ∈ L∞ (T, M, 𝜇). Then, coz(hf − h󸀠 f 󸀠 ) ∈ M0 . Proof. Since |h󸀠 f 󸀠 − hf | ⩽ |h󸀠 − h||f 󸀠 | + |h||f 󸀠 − f |, we get coz |h󸀠 f 󸀠 − hf | ⊂ coz |h󸀠 − h| ∪ coz |f 󸀠 − f | ∈ M0 . By Lemma 3, hf and h󸀠 f 󸀠 belong to MI(T, M, 𝜇).

3.3.5 Comparison of some Lebesgue integrals over spaces with positive wide measures | 273

Lemma 5. Let 𝜇 be a positive measure on a 𝜎-algebra M on T, f ∈ L1 (T, M, 𝜇) and h̄ ∈ L∞ (T, M, 𝜇). Then, ‖hf ‖i = ‖hf̄ ‖1 ⩽ ‖h‖eu ‖f ‖i = ‖h‖̄ ∞ ‖ f ‖1 . Proof. By Lemma 2 (3.3.1) there is h󸀠 ∈ M b (T, M) such that h󸀠 ∈ h̄ and ‖h‖eu = ‖h󸀠 ‖u . By virtue of Lemma 4 and Corollary 1 to Lemma 5 (3.3.2), we have ‖hf̄ ‖i = ‖hf̄ ‖1 = ‖hf ‖1 = ∫ |hf | d𝜇 = ∫ |h 󸀠 f | d𝜇 ⩽ ∫ ‖h󸀠 ‖u |f | d𝜇 = ‖h‖eu ‖f ‖i = ‖h‖̄ ∞ ‖ f ‖1 . Lemma 6. Let 𝜇 be a positive measure on a 𝜎-algebra M on T, a ∈ L1 (T, M, 𝜇) and c ∈ L∞ (T, M, 𝜇). If a ∩ c = / ⌀, then a = c. Proof. Let f ∈ a ∩ c ⊂ MI ∩ M∞ , g ∈ a and h ∈ c. Then, g ∈ MI, h ∈ M∞ , g ∼ f mod I(M0 ) and h ∼ f mod I(M0 ). Therefore, g ∼ h mod I(M0 ). By Lemma 3 (2.2.7), coz(g − h) ∈ M0 . By Lemma 2 (3.3.1), g ∈ M∞ . By Corollary 1 to Lemma 5 (3.3.2), h ∈ MI. As a result, g ∈ c and h ∈ a. This means that a = c.

3.3.5 Comparison of some Lebesgue integrals over spaces with positive wide measures Proposition 1. Let 𝜇1 and 𝜇2 be positive measures on a 𝜎-algebra M on a set T. Let Λ 1 be the integral over the space ⟮T, M, 𝜇1 ⟯ and Λ 2 be the integral over the space ⟮T, M, 𝜇2 ⟯. If 𝜇1 ⩽ 𝜇2 , then MI e (T, M, 𝜇2 ) ⊂ MI e (T, M, 𝜇1 ), MI(T, M, 𝜇2 ) ⊂ MI(T, M, 𝜇1 ) and Λ 1 f ⩽ Λ 2 f for every f ∈ F(T)+ . Proof. Let f ∈ MI(T, M, 𝜇2 ). By Corollary 1 to Theorem 1 (3.3.4), there is a sequence (f n ∈ St(T, M) ∩ MI(T, M, 𝜇2 ) | n ∈ N) such that f (t) = lim(f n (t) | n ∈ N) for every t ∈ U ∈ co-M0 (𝜇2 ) and for every 𝜀 > 0 there is n such that Λ 2 (|f p − f q |) < 𝜀 for every p, q ⩾ n. By Lemma 8 (2.2.4), |f p − f q | = ∑(x i 𝜒(R i ) | i ∈ I) for some finite collection (R i ∈ M | i ∈ I) of pairwise disjoint sets and for some collection (x i ∈ R | i ∈ I). By Lemma 1 (3.3.2), 𝜀 > Λ 2 (|f p −f q |) = ∑(x i 𝜇2 (R i ) | i ∈ I) ⩾ ∑(x i 𝜇1 (R i ) | i ∈ I) = Λ 1 (|f p −f q |) for every p, q ⩾ n. Similarly, |f n | = ∑(y j 𝜒(S j ) | j ∈ J) for some finite collection (S j ∈ M | j ∈ J) of pairwise disjoint sets and for some collection (y j ∈ R | j ∈ J). From |f n | ∈ MI(T, M, 𝜇2 ) we conclude by Lemma 1 (3.3.2) that ∞ > Λ 2 |f n | = ∑(y j 𝜇2 (S j ) | j ∈ J) ⩾ ∑(y j 𝜇1 (S j ) | j ∈ J) = Λ 1 |f n |. Thus, |f n | ∈ MI(T, M, 𝜇1 ). By Corollary 3 to Theorem 2 (3.3.2), f n ∈ MI(T, M, 𝜇1 ) ∩ St(T, M). Further, U ∈ M and 𝜇2 (T\U) = 0 imply 𝜇1 (T\U) = 0, i. e. U ∈ co-M0 (𝜇1 ). Since f ∈ M(T, M), again by Corollary 1 to Theorem 1 (3.3.4), f ∈ MI(T, M, 𝜇1 ). Thus, MI(T, M, 𝜇2 ) ⊂ MI(T, M, 𝜇1 ).

274 | 3.3 The Lebesgue integral

Now, let f ∈ MI e (T, M, 𝜇2 ). Then, at least one of the numbers Λ 2 f+ and Λ 2 (−f− ) is finite. It follows from the inclusion proven above that at least one of the numbers Λ 1 f+ and Λ 1 (−f− ) is finite. Hence, f ∈ MI e (T, M, 𝜇1 ). Finally, the inequality Λ 1 f ⩽ Λ 2 f for every f ∈ F(T)+ follows directly from the definition of the Lebesgue integral in 3.3.2. Proposition 2. Let M1 and M2 be 𝜎-algebras, 𝜇1 and 𝜇2 be positive measures on M1 and M2 , respectively. Let Λ 1 be the integral over the space ⟮T, M1 , 𝜇1 ⟯ and Λ 2 be the integral over the space ⟮T, M2 , 𝜇2 ⟯. If M1 ⊂ M2 and 𝜇2 is an extension of 𝜇1 , then M01 (𝜇1 ) ⊂ M02 (𝜇2 ), M(T, M1 ) ⊂ M(T, M2 ), MI(T, M1 , 𝜇1 ) ⊂ MI(T, M2 , 𝜇2 ) and Λ 1 f = Λ 2 f for every f ∈ MI(T, M1 , 𝜇1 ). Proof. The inclusions M01 (𝜇1 ) ⊂ M02 (𝜇2 ) and M(T, M1 ) ⊂ M(T, M2 ) are evident. Let f ∈ MI(T, M1 , 𝜇1 )+ . By Theorem 2 (2.3.4) there is a sequence (f n ∈ St(T, M1 )+ | n ∈ N) ↑ such that f (t) = lim (f n (t) | n ∈ N) = sup (f n (t) | n ∈ N) for every t ∈ T. By Corollary 1 to Theorem 1 (3.3.2), Λ 1 f = sup (Λ 1 f n | n ∈ N) = lim(Λ 1 f n | n ∈ N). Therefore, all f n ∈ MI(T, M1 , 𝜇1 ). By Lemma 8 (2.2.4), f n = ∑(x i 𝜒(R i ) | i ∈ I) for some finite collection (R i ∈ M1 ⊂ M2 | i ∈ I) of pairwise disjoint sets and for some collection (x i ∈ R | i ∈ I). Then, by Lemma 1 (3.3.2), Λ 2 f n = ∑(x i 𝜇2 (R i ) | i ∈ I) = ∑(x i 𝜇1 (R i ) | i ∈ I) = Λ 1 f n for every n ∈ N. Applying Theorem 5 (3.3.3), we get Λ 2 f = lim (Λ 2 f n | n ∈ N) = lim (Λ 1 f n | n ∈ N) = Λ 1 f , and therefore, f ∈ M(T, M2 , 𝜇2 ). Now, let f ∈ MI(T, M1 , 𝜇1 ). As was shown above, for the functions f+ and f− , we have f+ ∈ M(T, M2 , 𝜇2 ), −f− ∈ M(T, M2 , 𝜇2 ), Λ 2 f+ = Λ 1 f+ , and Λ 2 (−f− ) = Λ 1 (−f− ). Theorem 2 (3.3.2) implies that f = f+ −(−f− ) ∈ M(T, M2 , 𝜇2 ) and Λ 2 f = Λ 2 f+ −Λ 2 (−f− ) = Λ 1 f+ − Λ 1 (−f− ) = Λ 1 f . This proposition is often used in the following situation. Let 𝜇1 be a positive measure on a 𝜎-algebra M1 . Consider the local completion 𝜇1̄ : Kl (T, M1 , 𝜇1 ) → R+ of the measure 𝜇1 . By Lemma 4 (3.1.4), Kl (T, M1 , 𝜇1 ) is a 𝜎-algebra. By Proposition 1 (3.1.4), 𝜇1̄ is a locally complete measure. Therefore, we can take M2 ≡ Kl (T, M1 , 𝜇1 ) and 𝜇2 ≡ 𝜇1̄ . For this particular situation, we can reinforce Proposition 2. Theorem 1. Let M1 and M2 be 𝜎-algebras on T, 𝜇1 and 𝜇2 be positive measures on M1 and M2 , respectively, and 𝜇2 be the local completion of 𝜇1 . Let Λ 1 and Λ 2 be the integrals over the spaces ⟮T, M1 , 𝜇1 ⟯ and ⟮T, M2 , 𝜇2 ⟯, respectively. Then, 1) M01 (𝜇1 ) ⊂ M02 (𝜇2 ) = Nl (T, M1 , 𝜇1 ) = {N ⊂ T | ∃ N0 ∈ M01 (𝜇1 )(N ⊂ N0 )} and M02 (𝜇2 ) is an ideal and a 𝜎-ring; 2) M(T, M1 ) ⊂ M(T, M2 ); moreover, for every function f ∈ M(T, M2 ) there is a function f 󸀠 ∈ M(T, M1 ) such that f (t) = f 󸀠 (t) for every point t from some set U ∈ co-M02 (𝜇2 ), i. e. 𝜇2 -almost everywhere; 3) MI(T, M1 , 𝜇1 ) ⊂ MI(T, M2 , 𝜇2 ) and Λ 1 f = Λ 2 f for every f ∈ MI(T, M1 , 𝜇1 ); moreover, for every function f ∈ MI(T, M2 , 𝜇2 ) there is a function g ∈ MI(T, M1 , 𝜇1 ) such that f (t) = g(t) 𝜇2 -almost everywhere and Λ 2 f = Λ 1 g.

3.3.5 Comparison of some Lebesgue integrals over spaces with positive wide measures | 275

Proof. 1. By Proposition 2, M01 (𝜇1 ) ⊂ M02 (𝜇2 ). By virtue of Proposition 1 (3.1.4), M02 (𝜇2 ) = N(T, M1 , 𝜇1 ) = {N ⊂ T | ∃N0 ∈ M01 (𝜇1 )(N ⊂ N0 )}. According to Lemma 9 (3.1.1) it is an ideal and a 𝜎-ring. 2. By Theorem 2 (2.3.4) for f ∈ M(T, M2 ) there is a sequence (f n ∈ St(T, M2 ) | n ∈ N) such that (|f n | | n ∈ N) ↑ and f = p-lim (f n | n ∈ N). By Lemma 8 (2.2.4), f n = ∑(x ni 𝜒(M ni ) | i ∈ I n ) for some finite partition (M ni ∈ M2 | i ∈ I n ) of T and some collection (x ni ∈ R | i ∈ I n ). According to Proposition 1 (3.1.4), M ni = M 󸀠ni ∪ N ni for some M 󸀠ni ∈ M1 and N ni ∈ Nl (T, M1 , 𝜇1 ) = M02 (𝜇2 ). Consider the sets A ni ≡ N ni \M 󸀠ni ⊂ N ni . Since M02 (𝜇2 ) is a 𝜎-ideal, we have A ni ∈ 0 M2 (𝜇2 ), and, therefore, A ≡ ⋃⟮⋃⟮A ni | i ∈ I n ⟯ | n ∈ N⟯ ∈ M02 (𝜇2 ). Take the set U ≡ T\A ∈ co-M02 (𝜇2 ). Consider the step functions f n󸀠 ≡ ∑(x ni 𝜒(M 󸀠ni ) | i ∈ I n ) ∈ St(T, M1 ). If t ∈ U, then t ∈ M ni \A for every n and some i ∈ I n . Hence, t ∈ M 󸀠ni . Therefore, f n󸀠 (t) = x ni 𝜒(M 󸀠ni )(t) = x ni 𝜒(M ni ) = f n (t). Thus, f (t) = lim (f n (t) | n ∈ N) = lim (f n󸀠 (t) | n ∈ N).

Applying Lemma 4 (1.1.15) and Lemma 3 (1.4.7), we get f (t) = lim (f n󸀠 (t) | n ∈ N) for

every t ∈ U. Consider the function f 󸀠 ≡ lim (f n󸀠 | n ∈ N) in F(T). By Proposition 1 (2.3.3),

f 󸀠 ∈ M(T, M1 ). Finally, f 󸀠 (t) = lim (f n󸀠 (t) | n ∈ N) = f (t) for every t ∈ U. 3. Let f ∈ MI(T, M2 , 𝜇2 ). Then, by Theorem 2 (2.3.4) the functions f n from the proof of assertion 2 can be chosen such that 0 ⩽ f n ⩽ f n+1 ⩽ f . Therefore, f n ∈ MI(T, M2 , 𝜇2 ) due to Lemma 5 (3.3.2). By Lemma 1 (3.3.2), Λ 2 f n = ∑(x ni 𝜇2 (M ni ) | i ∈ I n ). Since M ni = M 󸀠ni ∪ A ni , we have 𝜇2 (M ni ) = 𝜇2 (M 󸀠ni ) = 𝜇1 (M 󸀠ni ), where Λ 2 f n = ∑(x ni 𝜇1 (M 󸀠ni ) | i ∈ I n ) = Λ 1 f n󸀠 . By Lemma 7 (1.4.7), f (t) = sup (f n (t) | n ∈ N) for every t ∈ T. Then, by Corollary 1 to Theorem 1 (3.3.2) Λ 2 f = sup (Λ 2 f n | n ∈ N). Thus, Λ 2 f = sup (Λ 1 f n󸀠 | n ∈ N), where f n󸀠 ∈ MI(T, M1 , 𝜇1 ). Since 𝜒(M 󸀠ni ) ⩽ 𝜒(M ni ) and x ni ⩾ 0, we get f n󸀠 ⩽ f n ⩽ f . Consider the new step functions g n ≡ sup (f m󸀠 | 1 ⩽ m ⩽ n) ∈ St(T, M1 ) ∩ MI(T, M1 , 𝜇1 ). From g n ⩽ f , we conclude that there is a function g ∈ F(T) such that g = sup (g n | n ∈ N). By Proposition 1 (2.3.3), g ∈ M(T, M1 ) and by Corollary 1 to Theorem 1 (3.3.2), Λ 1 g = sup (Λ 1 g n | n ∈ N). Proposition 2 implies that g n ∈ MI(T, M2 , 𝜇2 ) and Λ 1 g n = Λ 2 g n . If t ∈ U, then g n (t) = sup (f m󸀠 (t) | 1 ⩽ m ⩽ n) = sup (f m (t) | 1 ⩽ m ⩽ n) = f n (t). Therefore, Lemma 5 (3.3.2) provides the equality Λ 2 g n = Λ 2 f n . Thus, Λ 1 g = sup (Λ 2 f n | n ∈ N) = Λ 2 f , where g ∈ MI(T, M1 , 𝜇1 ). Besides, g(t) = sup (g n (t) | n ∈ N) = sup (f n (t) | n ∈ N) = f (t) for every t ∈ U. Now, we shall consider the Lebesgue integral with respect to cone combinations of positive measures. Proposition 3. Let 𝜇 and 𝜈 be positive measures on a 𝜎-algebra M on T and x, y ∈ R+ . Let Λ(𝜇) be the integral over the space ⟮T, M, 𝜇⟯ and Λ(𝜈) be the integral over the space ⟮T, M, 𝜈⟯. Then, 1) MI e (T, M, x𝜇) = MI e (T, M, 𝜇) and MI(T, M, x𝜇) = MI(T, M, 𝜇) for x > 0 and Λ(x𝜇)f = xΛ(𝜇)f for every f ∈ MI e (T, M, 𝜇);

276 | 3.3 The Lebesgue integral

2)

MI e (T, M, x𝜇 + y𝜈) ⊂ MI e (T, M, 𝜇) ∩ MI e (T, M, 𝜈), MI(T, M, x𝜇 + y𝜈) = MI(T, M, 𝜇) ∩ MI(T, M, 𝜈), and Λ(x𝜇 + y𝜈)f = xΛ(𝜇)f + yΛ(𝜈)f for every f ∈ MI e (T, M, x𝜇 + y𝜈).

Proof. 1. It follows directly from the definition that Λ(x𝜇)f = xΛ(𝜇)f for every positive function f on T. Therefore, MI(T, M, x𝜇) = MI(T, M, 𝜇) and MI e (T, M, x𝜇) = MI e (T, M, 𝜇) for every x > 0 and Λ(x𝜇)f = xΛ(𝜇)f for every f ∈ MI e (T, M, 𝜇). 2. The inclusions MI e (T, M, x𝜇 + y𝜈) ⊂ MI e (T, M, 𝜇) ∩ MI e (T, M, 𝜈) and MI(T, M, x𝜇 + y𝜈) ⊂ MI(T, M, 𝜇) ∩ MI(T, M, 𝜈) follow from assertion 1 and Proposition 1. Take any function f ∈ M(T, M)+ and any finite partition (A i ∈ M | i ∈ I) of T. Then, by the definition of the integral ∑(inf f [A i ](x𝜇 + y𝜈)A i | i ∈ I) = x ∑(inf f [A i ]𝜇A i | i ∈ I) + y ∑(inf f [A i ]𝜈A i | i ∈ I) ⩽ xΛ(𝜇)f + yΛ(𝜈)f , where Λ(x𝜇 + y𝜈)f ⩽ (xΛ(𝜇) + yΛ(𝜈))f . If Λ(x𝜇 + y𝜈)f = ∞, then Λ(x𝜇 + y𝜈)f = (xΛ(𝜇)+ yΛ(𝜈))f . Further on, suppose that f ∈ MI(T, M, x𝜇 + y𝜈)+ . By Proposition 1, f ∈ MI(T, M, 𝜇)+ ∩MI(T, M, 𝜈)+ . Take any 𝜀 > 0. Then, there are finite partitions (B j ∈ M | j ∈ J) and (C k ∈ M | k ∈ K) of T such that a ≡ ∑(inf f [B j ]𝜇B j | j ∈ J) > Λ(𝜇)f − 𝜀/(2x) and b ≡ ∑(inf f [C k ]𝜈(C k ) | k ∈ K) > Λ(𝜈)f − 𝜀/(2y). Using Proposition 4 (1.4.3), we get xΛ(𝜇)f + yΛ(𝜈)f − 𝜀 < xa + yb = ∑(∑(inf f [B j ]x𝜇(B j ∩ C k ) | k ∈ K) | j ∈ J)+ ∑(∑(inf f [C k ]x𝜈(B j ∩ C k ) | j ∈ J) | k ∈ K) ⩽ ∑(∑(inf f [B j ∩ C k ]x𝜇(B j ∩ C k ) | k ∈ K) | j ∈ J) + ∑(∑(inf f [B j ∩ C k ]y𝜈(B j ∩ C k ) | j ∈ J) | k ∈ K) = ∑(inf f [B j ∩ C k ](x𝜇 + y𝜈) (B j ∩ C k ) | (j, k) ∈ J × K) ⩽ Λ(x𝜇 + y𝜈)f . Since 𝜀 is arbitrary, we infer that xΛ(𝜇)f + yΛ(𝜈)f ⩽ Λ(x𝜇 + y𝜈)f . Now, let f be an arbitrary function from MI e (T, M, x𝜇 + y𝜈). Suppose that f− ∈ MI(T, M, x𝜇 + y𝜈). Then, f− ∈ MI(T, M, 𝜇) ∩ MI(T, M, 𝜈). Therefore, Λ(x𝜇 + y𝜈)f ≡ Λ(x𝜇 + y𝜈)f + − Λ(x𝜇 + y𝜈)(−f− ) = xΛ(𝜇)f+ + yΛ(𝜈)f+ − xΛ(𝜇)(−f− ) − yΛ(𝜈)(−f− ) = xΛ(𝜇)f + yΛ(𝜈)f . If f+ ∈ MI(T, M, x𝜇 + y𝜈), then the arguments are the same. Finally, let f ∈ MI(T, M, 𝜇) ∩ MI(T, M, 𝜈). Then, by the definition f+ and (−f− ) are integrable with respect to 𝜇 and 𝜈 simultaneously. Therefore, the inequalities 0 ⩽ Λ(x𝜇 + y𝜈)f+ ⩽ xΛ(𝜇)f+ + yΛ(𝜈)f+ < ∞ and 0 ⩽ Λ(x𝜇 + y𝜈)(−f− ) ⩽ xΛ(𝜇)(−f− ) + yΛ(𝜈)(−f− ) < ∞ means that f ∈ MI(T, M, x𝜇 + y𝜈). This gives the necessary inverse inclusion.

3.3.6 The Lebesgue integral over a space with an arbitrary wide measure. The problem of characterization of Lebesgue integrals as linear functionals In 3.3.2 the Lebesgue integral was constructed for a positive wide measure. Here we shall define the Lebesgue integral for an arbitrary wide measure.

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Definition and basic properties Let ⟮T, M, 𝜇⟯ be a measurable space with a 𝜎-algebra M and a measure 𝜇 : M → R. By Proposition 2 (3.2.1), 𝜇 = v+ (𝜇) + v− (𝜇), where v+ (𝜇) is the positive measure, v− (𝜇) is the negative measure and at least one of them is finite. We shall denote them sometimes simply by v+ and v− . The variation v(𝜇) = v+ (𝜇) − v− (𝜇) of 𝜇 also will be denoted sometimes simply by v. By Corollary 1 to Proposition 2 (3.1.3), v(𝜇) is a measure as well. A function f ∈ F(T) will be called extendedly (Lebesgue) integrable over the space ⟮T, M, 𝜇⟯ if either f ∈ LI e (T, M, v+ (𝜇)) ∩ LI(T, M, −v− (𝜇)) or f ∈ LI(T, M, v+ (𝜇)) ∩ LI e (T, M, −v− (𝜇)) (see 3.3.2). A function f ∈ F(T) is called (Lebesgue) integrable over the space ⟮T, M, 𝜇⟯ if f ∈ LI(T, M, v + (𝜇)) ∩ LI(T, M, −v− (𝜇)). The family of all extendedly integrable functions over the space ⟮T, M, 𝜇⟯ will be denoted by LI e (T, M, 𝜇). And the subfamily of the family LI e (T, M, 𝜇) consisting of all integrable functions will be denoted by LI(T, M, 𝜇). Now, we shall introduce an intermediate family of functions situated between LI(T, M, 𝜇) and LI e (T, M, 𝜇). A function f ∈ F(T) will be called (Lebesgue) 𝜎-integrable over the space ⟮T, M, 𝜇⟯ if either f ∈ LI 𝜎 (T, M, v+ (𝜇)) ∩ LI(T, M, −v− (𝜇)) or f ∈ LI(T, M, v+ (𝜇)) ∩ LI 𝜎 (T, M, −v− (𝜇)) (see 3.3.2). The family of all 𝜎-integrable functions over the space ⟮T, M, 𝜇⟯ will be denoted by LI 𝜎 (T, M, 𝜇). It is clear that LI(T, M, 𝜇) ⊂ LI 𝜎 (T, M, 𝜇) ⊂ LI e (T, M, 𝜇). If f ∈ LI e (T, M, 𝜇), then either ∫ f dv+ (𝜇) ∈ R and ∫ f d(−v− (𝜇)) ∈ R or ∫ f dv+ (𝜇) ∈ R and ∫ f d(−v− (𝜇)) ∈ R. Therefore, we can define the extended number ∫ f d𝜇 ≡ ∫ f dv+ (𝜇) − ∫ f d(−v− (𝜇)) ∈ R. This number ∫ f d𝜇 is called the (Lebesgue) integral of the function f over the space ⟮T, M, 𝜇⟯. If f ∈ LI(T, M, 𝜇) then ∫ f d𝜇 ∈ R. The mapping Λ : LI e (T, M, 𝜇) → R such that Λf ≡ ∫ f d𝜇 is called the (Lebesgue) integral over the space ⟮T, M, 𝜇⟯. Sometimes we shall use more rigorous notations: Λ(𝜇) : LI e (T, M, 𝜇) → R and Λ(𝜇)f ≡ ∫ f d𝜇. Along with the phrase “over the space ⟮T, M, 𝜇⟯” we shall use also the phrase “with respect to the measure 𝜇 (on the descriptive space ⟮T, M⟯)”. According to 3.3.2, the most natural domains for the Lebesgue integral Λ are the family MI e (T, M, 𝜇) ≡ M(T, M) ∩ LI e (T, M, 𝜇) of all M-measurable extendedly integrable functions and its subfamilies MI(T, M, 𝜇) ≡ M(T, M) ∩ LI(T, M, 𝜇) = MI(T, M, v+ (𝜇)) ∩ MI(T, M, −v− (𝜇)) and MI 𝜎 (T, M, 𝜇) ≡ M(T, M) ∩ LI 𝜎 (T, M, 𝜇). Now, let 𝜇 be a finite measure. By virtue of Corollary 3 to Proposition 1 (3.2.1), every measure on a 𝜎-algebra is finite iff it is bounded. According to Proposition 1 (3.2.2), for such a measure, v+ (𝜇) = 𝜇+ and v− (𝜇) = 𝜇− . Therefore, for a finite measure 𝜇 the definition of the Lebesgue integral uses only the positive part 𝜇+ and the negative part 𝜇− of 𝜇.

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Proposition 1. Let 𝜇 be a measure on a 𝜎-algebra M on T. Then, MI(T, M, 𝜇) = MI(T, M, v(𝜇)) and Λ(v(𝜇)) = Λ(v + (𝜇)) + Λ(−v− (𝜇)). Proof. Let f ∈ MI(T, M, 𝜇)+ . For any finite partition (A i ∈ M | i ∈ I) of T, we have ∑(inf f [A i ]vA i | i ∈ I) = ∑(inf f [A i ]v+ (A i ) | i ∈ I) + ∑(inf f [A i ](−v− )A i | i ∈ I) ⩽ ∫ f dv+ + ∫ f d(−v− ). From this inequality, we conclude that ∫ f dv ⩽ ∫ f dv+ + ∫ f d(−v− ) < ∞. Hence, f ∈ MI(T, M, 𝜇)+ . This implies MI(T, M, 𝜇) ⊂ MI(T, M, v). Let f ∈ MI(T, M, v)+ . Take two finite partitions (A i ∈ M | i ∈ I) and (B j ∈ M | j ∈ J) of T. Then, v+ A i = ∑(v+ (A i ∩ B j ) | j ∈ J) and (−v− )B j = ∑((−v− )(A i ∩ B j ) | i ∈ I imply x ≡ ∑(inf f [A i ]v+ A i | i ∈ I) + ∑(inf f [B j ](−v− )B j | j ∈ J) ⩽ ∑(∑(inf f [A i ∩ B j ] v+ (A i ∩ B j ) | j ∈ J) | i ∈ I) + ∑(∑(inf f [A i ∩ B j ](−v− )(A i ∩ B j ) | i ∈ I) | j ∈ J). Using Corollary 1 to Proposition 1 (1.4.3), we get x ⩽ ∑(inf f [A i ∩B j ]v(A i ∩B j ) | (i, j) ∈ I×J), whereas v(A i ∩ B j ) = v+ (A i ∩ B j ) + (−v− )(A i ∩ B j ). Therefore, x ⩽ ∫ f dv, where ∑(inf f [A i ]v+ A i | i ∈ I) ⩽ ∫ f dv−∑(inf f [B j ](−v− )B j | j ∈ J) implies ∫ f dv+ ⩽ ∫ f dv−∑(inf f [B j ](−v− )B j | j ∈ J) < ∞. Thus, f is v+ -integrable. In turn, in the same manner, this implies ∫ f d(−v− ) ⩽ ∫ f dv − ∫ f dv+ < ∞. This means that f is (−v− )-integrable. As a result, we obtain f ∈ MI(T, M, 𝜇)+ . Thus, MI(T, M, v) ⊂ MI(T, M, 𝜇). We proved for f ⩾ 0 that ∫ f dv = ∫ f dv+ + ∫ f d(−v− ). Since all of these integrals are linear and MI(T, M, v) is a lattice-ordered linear space, we infer that this equality is valid for every f ∈ MI(T, M, v). Corollary 1. Let 𝜇 be a measure on a 𝜎-algebra M on T. Then, MI(T, M, 𝜇) is a latticeordered linear space. Consider in the lattice-ordered linear space MI(T, M, 𝜇) the l-ideal MI 0 (T, M, 𝜇) ≡ {f ∈ MI(T, M, 𝜇) | coz f ∈ M0 (v(𝜇))}. Denote by L1 (T, M, 𝜇) the lattice-ordered linear space MI(T, M, 𝜇)/MI 0 (T, M, 𝜇). Corollary 2. Let 𝜇 be a measure on a 𝜎-algebra M on T. Then, MI 0 (T, M, 𝜇) = MI 0 (T, M, v(𝜇)) and L1 (T, M, 𝜇) = L1 (T, M, v(𝜇)). Lemma 1. Let 𝜇 be a measure on a 𝜎-algebra M on T and f ∈ MI e (T, M, 𝜇). Then, |f | ∈ MI e (T, M, 𝜇) and | ∫ f d𝜇| ⩽ ∫ |f | dv(𝜇). Proof. Let f ∈ MI e (T, M, v+ ) ∩ MI(T, M, −v− ). Then, by Corollary 3 to Theorem 2 (3.3.2), |f | belongs to the same family. Consider the extended real numbers x ≡ ∫ f dv + and y ≡ ∫ f d(−v− ). If |x| = ∞ then by Corollary 3 to Theorem 1 (3.3.2), | ∫ f d𝜇| = |x − y| = |x| ⩽ ∫ |f | dv + . Since v+ ⩽ v it follows directly from the definition of the Lebesgue integral that ∫ |f | dv + ⩽ ∫ |f | dv. As a result, | ∫ f d𝜇| ⩽ ∫ |f | dv. If |x| < ∞ then by Corollary 3 to Theorem 1 (3.3.2) and Proposition 1, | ∫ f d𝜇| = |x − y| ⩽ |x| + |y| ⩽ ∫ |f | dv + + ∫ |f | d(−v− ) = ∫ |f | dv. If f ∈ MI(T, M, v + ) ∩ MI e (T, M, −v− ), then the arguments are the same.

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Lemma 2. Let 𝜇 be a measure on a 𝜎-algebra M on T. Then, the lattice-ordered linear space MI(T, M, 𝜇) is truncatable (in the sense of 2.2.9). Proof. Let f ∈ MI(T, M, 𝜇). By the definition, f ∈ MI(T, M, v + ). Since |f ∧ 1| ⩽ |f |, we infer by virtue of Corollary 1 to Theorem 2 (3.3.2) that f ∧ 1 ∈ MI(T, M, v + ). The same is valid for the measure −v− . Then, by the definition f ∧ 1 ∈ MI (T, M, 𝜇). Proposition 2. Let 𝜇 be a measure on a 𝜎-algebra M on T. Then, Λ(𝜇) is a linear functional on MI(T, M, 𝜇), Λ(v+ (𝜇)) and Λ(v− (𝜇)) are positive and, respectively, negative linear functionals on MI(T, M, 𝜇), Λ(v(𝜇)) : MI(T, M, 𝜇) → R is a positive linear functional, Λ(𝜇) = Λ(v+ (𝜇)) + Λ(v− (𝜇)), and Λ(𝜇) ⩽ Λ(v(𝜇)) = Λ(v+ (𝜇)) − Λ(v− (𝜇)). Proof. If f , g ∈ MI(T, M, 𝜇) and x, y ∈ R then by Theorem 2 (3.3.2), Λ(𝜇)(xf + yg) = Λ(v+ )(xf + yg) − Λ(−v− )(xf + yg) = xΛ(v+ )f + yΛ(v+ )g − xΛ(−v− )f − yΛ(−v− )g = xΛ(𝜇)f + yΛ(𝜇)g. Thus, Λ(𝜇) is linear. Besides, By virtue of Theorem 2 (3.3.2) the linear functionals Λ(v+ ) and Λ(−v− ) are positive. Using also Proposition 1, we see that Λ(v) is a positive linear functional on the lattice linear space MI(T, M, 𝜇) = MI(T, M, v). If f ∈ MI(T, M, 𝜇)+ , then by Lemma 1 Λ(𝜇)f ⩽ Λ(v)f . This means that Λ(𝜇) ⩽ Λ(v). Integration on measurable subspaces Now, we shall consider the integration on a measurable subspace of the measurable space ⟮T, M, 𝜇⟯. Let U ∈ M. Consider the 𝜎-algebra MU ≡ {M ∈ M | M ⊂ U} and the measures 𝜇U ≡ 𝜇|MU , (v(𝜇))U ≡ v(𝜇)|MU , (v+ (𝜇))U ≡ v+ (𝜇)|MU , and (−v− (𝜇))U ≡ (−v− (𝜇))|MU on MU . Lemma 3. Let 𝜇 be a measure on a 𝜎-algebra M on T and U ∈ M. Then, v(𝜇)U = v(𝜇U ), (v+ (𝜇))U = v+ (𝜇U ) and (−v− (𝜇))U = −v− (𝜇U ). ̄̄ Proof. According to 3.1.3, var(𝜇U ) ⩽ 𝜇.̄̄ Therefore, v(𝜇U ) ≡ var(𝜇U )|MU ⩽ 𝜇|M U = v(𝜇)|MU = v(𝜇)U . On the other hand, for any M ∈ MU take any finite partition (M i ∈ MU | i ∈ I) of the set M. Then, ∑(|𝜇M i | | i ∈ I) = ∑(|𝜇U M i | | i ∈ I) ⩽ v(𝜇U )(M) implies v(𝜇)U (M) = v(𝜇)(M) ⩽ v(𝜇U )(M). Consequently, v(𝜇)U ⩽ v(𝜇U ) and, as a result, v(𝜇)U = v(𝜇U ). Let M ∈ MU and (M+ , M− ) be a Hahn decomposition of M with respect 𝜇 (see 3.2.1). Then, M+ , M− ∈ MU . If S ∈ MU , then 𝜇U S = 𝜇S ⩾ 0 for S ⊂ M+ and 𝜇U S = 𝜇S ⩽ 0 for S ⊂ M− . Consequently, (M+ , M− ) is a Hahn decomposition of M with respect to 𝜇U .

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Therefore, by the definition v+ (𝜇U )M = 𝜇U (M+ ) = 𝜇(M+ ) = v+ (𝜇)M = (v+ (𝜇))U M. Similarly, v− (𝜇U )M = (v− (𝜇))U M. Proposition 3. Let 𝜇 be a measure on a 𝜎-algebra M on T, U ∈ M, and f ∈ M(T, M). Then, 1) f |U ∈ M(U, MU ) and f 𝜒(U) ∈ M(T, M); 2) f |U ∈ MI e (U, MU , 𝜇U ) iff f 𝜒(U) ∈ MI e (T, M, 𝜇); 3) if f |U ∈ MI e (U, MU , 𝜇U ), then ∫(f |U) d𝜇U = ∫ f 𝜒(U) d𝜇. Proof. 1. Lemma 3 (2.5.1) implies that 𝜑 ≡ f |U ∈ M(U, MU ). According to Theorem 1 (2.3.2) and Corollary 1 to Lemma 2 (2.3.1), g ≡ f 𝜒(U) ∈ M(T, M). 2. Let 𝜑 ∈ MI e (U, MU , 𝜇U ) and suppose that 𝜑 ∈ MI e (U, MU , v+ (𝜇U )) ∩ MI(U, MU , −v− (𝜇U )). Then, Lemma 3 implies that 𝜑 ∈ MI e (U, MU , (v+ (𝜇))U ) ∩ MI(U, MU , (−v− (𝜇))U ). Therefore, by Proposition 2 (3.3.2) and Corollary 1 to it g ∈ MI e (T, M, v+ (𝜇)) ∩ MI(T, M, −v− (𝜇)). Hence, g ∈ MI e (T, M, 𝜇). In the case of the second opportunity for 𝜑, the arguments are the same. Conversely, let g ∈ MI e (T, M, 𝜇) and suppose that g ∈ MI(T, M, v+ (𝜇)) ∩ e MI (T, M, −v− (𝜇)). Then, by Proposition 2 (3.3.2) and Corollary 1 to it, 𝜑 ∈ MI(U, MU , (v+ (𝜇))U ) ∩ MI e (U, MU , (−v− (𝜇))U ). Therefore, by Lemma 3, 𝜑 ∈ MI(U, MU , v+ (𝜇U )) ∩ MI(U, MU , −v− (𝜇)). Hence, 𝜑 ∈ MI e (U, MU , 𝜇U ). In the case of the second opportunity for g, the arguments are the same. 3. If 𝜑 ∈ MI e (U, MU , 𝜇U ) then by (2), g ∈ MI e (T, M, 𝜇). Using the definition of the Lebesgue integral, Proposition 2 (3.3.2), and Lemma 3, we get ∫ 𝜑 d𝜇U ≡ ∫ 𝜑 dv+ (𝜇U )−∫ 𝜑 d(−v− (𝜇U )) = ∫ 𝜑 d(v+ (𝜇))U −∫ 𝜑 d(−v− (𝜇))U = ∫ g dv+ (𝜇)− ∫ g d(−v− (𝜇)) ≡ ∫ g d𝜇. Corollary 1. Let 𝜇 be a measure on a 𝜎-algebra M on T, U ∈ M, and f ∈ M(T, M). Then, f |U ∈ MI(U, MU , 𝜇U ) iff f 𝜒(U) ∈ MI(T, M, 𝜇). Corollary 2. Let 𝜇 be a measure on a 𝜎-algebra M on T, f ∈ MI e (T, M, 𝜇), and U ∈ M. Then, f 𝜒(U) ∈ MI e (T, M, 𝜇), f |U ∈ MI e (U, MU , 𝜇U ), and ∫ f 𝜒(U) d𝜇 = ∫(f |U) d𝜇U . Proof. Suppose that f ∈ MI e (T, M, v+ (𝜇)) ∩ MI(T, M, −v− (𝜇)). Then, by Corollaries 2 and 3 to Proposition 2 (3.3.2), f 𝜒(U) belongs to the same family. Hence, f 𝜒(U) ∈ MI e (T, M, 𝜇). In the case of the second opportunity for f , the arguments are the same. The other assertions follow from this fact and Proposition 3. Corollary 3. Let 𝜇 be a measure on a 𝜎-algebra M on T, U ∈ M, and f ∈ MI(T, M, 𝜇). Then, f 𝜒(U) ∈ MI(T, M, 𝜇), f |U ∈ MI(U, MU , 𝜇U ) and ∫ f 𝜒(U) d𝜇 = ∫(f |U) d𝜇U .

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Proof. By the condition, f ∈ MI(T, M, v+ (𝜇)) ∩ MI(T, M, −v− (𝜇)). Then, by Corollary 3 to Proposition 2 (3.3.2), f 𝜒(U) belongs to the same family. Hence, f 𝜒(U) ∈ MI(T, M, 𝜇). The other assertions follow from Corollaries 1 and 2. Corollary 4. Let 𝜇 be a measure on a 𝜎-algebra M on T, U ∈ M and f ∈ MI(T, M, 𝜇). Then, f 𝜒(U) ∈ MI(T, M, v(𝜇)), f |U ∈ MI(U, MU , v(𝜇)U ) and ∫ f 𝜒(U) dv(𝜇) = ∫(f |U) dv(𝜇U ). Proof. It follows from Proposition 1 and Corollary 3. Now, we shall compare the integration with respect to the measure 𝜇 and the measure v(𝜇). Proposition 4. Let 𝜇 be a measure on a 𝜎-algebra M on T. Then, 1) MI e (T, M, 𝜇) ⊂ MI e (T, M, v(𝜇)); 2) ∫ f dv(𝜇) = ∫ f dv + (𝜇) + ∫ fd(−v− (𝜇)) for every f ∈ MI e (T, M, 𝜇). Proof. By Corollary 1 to Lemma 5 (3.2.1), v+ is concentrated on T+ and v− is concentrated on T− . If A ∈ M and A ⊂ T+ , then (−v− )T+ A = −v− A = 0 implies by virtue of Proposition 2 (3.2.1) that (v+ )T+ A = v+ A = v+ A − v− A = vA = v T+ A. Hence, (−v− )T+ = 0 and (v+ )T+ = v T+ . Similarly, if B ∈ M and B ⊂ T− , then (v+ )T− B = v+ B = 0 implies (−v− )T− B = −v− B = v+ B − v− B = vB = v T− B. Thus, (v+ )T− = 0 and (−v− )T− = v T− . Take any function g ∈ M(T, M)+ . Corollary 2 to Theorem 1 (3.3.2) and Corollary 2 to Proposition 3 guarantees that ∫ g dv = ∫ g𝜒(T+ ) dv + ∫ g𝜒(T− ) dv = ∫(g|T+ ) dv T+ + ∫(g|T− ) dv T− . Using the equalities proven above, we get ∫ g dv = ∫(g|T+ )(v+ )T+ + ∫(g|T− )(−v− )T− . Again by Corollary 2 to Proposition 3, ∫ g dv = ∫ g𝜒(T+ ) dv+ + ∫ g𝜒(T− ) d(−v− ). Since C ≡ coz(g𝜒(T− )) ⊂ T− , we have v+ C = 0. Therefore, by Lemma 3 (3.3.2), ∫ g𝜒(T− ) dv+ = 0. Similarly, ∫ g𝜒(T+ ) d(−v− ) = 0. From these facts, again by Corollary 2 to Theorem 1 (3.3.2), we conclude that ∫ g dv = ∫ g𝜒(T+ ) dv+ + ∫ g𝜒(T− ) dv+ + ∫ g𝜒(T− ) d(−v− ) + ∫ g𝜒(T+ ) d(−v− ) = ∫ g dv+ + ∫ gd(−v− ). Let f ∈ MI e (T, M, 𝜇). First, suppose that f ∈ MI e (T, M, v+ ) ∩ MI(T, M, −v− ). Then, ∫ f+ dv− ∈ R− and ∫ f− dv− ∈ R+ . Suppose also that ∫ f+ dv+ ∈ R+ and ∫(−f− ) dv+ ∈ R+ . Taking g = f+ , we get ∫ f+ dv = ∫ f+ dv+ + ∫ f+ d(−v− ) ∈ R+ . Taking g = −f− , we get ∫(−f− ) dv = ∫(−f− ) dv+ + ∫(−f− ) d(−v− ) ∈ R+ . According to 3.3.2, this means that f ∈ MI e (T, M, v) and ∫ f dv ≡ ∫ f+ dv − ∫(−f− ) dv = ∫ f+ dv+ + ∫ f+ d(−v− ) − ∫(−f− ) dv+ − ∫(−f− ) d(−v− ). Besides, from the supposition about the function f , we infer that ∫ f dv+ = ∫ f+ dv+ − ∫(−f− ) dv+ and ∫ f d(−v− ) = ∫ f+ d(−v− ) − ∫(−f− ) d(−v− ) (see 3.3.2). Thus, ∫ f dv = ∫ f dv+ + ∫ f d(−v− ). If ∫ f+ dv+ ∈ R+ and ∫(−f− ) dv+ ∈ R+ , then the arguments are the same. If f ∈ MI(T, M, v+ ) ∩ MI e (T, M, −v− ), then all the arguments are analogous.

282 | 3.3 The Lebesgue integral

The problem of characterization of Lebesgue integrals as linear functionals Since there are many Lebesgue integrals Λ(𝜇) with respect to different measures 𝜇 and they are defined on different functional families MI(T, M, 𝜇), it is natural to consider them on some common linear space A(T). For this purpose, we consider some subset M(T) of the set Mw (T) of all wide measures on T from 3.1.2. This subset determines the corresponding family I(T, M(T)) of Lebesgue integrals Λ(𝜇) : MI(T, M, 𝜇) → R for all measures 𝜇 ∈ M(T) and the corresponding family UI(T, M(T)) ≡ ⋂⟮MI(T, M, 𝜇) | 𝜇 ∈ M(T)⟯ of M(T)universally integrable functions. Every family A(T) ⊂ UI(T, M(T)) determines the family I(T, A(T), M(T)) ≡ {Λ(𝜇)|A(T) | 𝜇 ∈ M(T)} of all Lebesgue integrals on the family A(T) with respect to the set M(T). According to Proposition 2, all these functionals are linear, i.e. I(T, A(T), M(T)) ⊂ A(T)× (see 2.2.7). In this connection, the following problem of characterization of Lebesgue integrals as linear functionals is most interesting: for various families M(T) and for appropriate linear spaces A(T) to describe the corresponding families I(T, A(T), M(T)) in the general family A(T)× of all linear functionals on A(T).

Some necessary condition crucial for solving the problem has been found by H. Lebesgue himself. Theorem 1. Let 𝜇 be a measure on a 𝜎-algebra M on a set T. Then, the functionals Λ(v+ (𝜇)), Λ(−v− (𝜇)), Λ(v(𝜇)), and Λ(𝜇) are pointwise 𝜎-continuous (in the sense of 2.2.8) on the corresponding lattice-ordered linear spaces of measurable integrable functions. If 𝜇 is bounded, then all the functionals are uniformly bounded. Proof. Let f ∈ MI(T, M, v+ ) and f = p-lim (f n | n ∈ N) for some infinite increasing sequence (f n ∈ MI(T, M, v+ ) | n ∈ N ⊂ 𝜔). Since Λf n ⩽ Λf < ∞, we infer that sup (Λf n | n ∈ N) < ∞. It follows now from Corollary 1 to Theorem 2 (3.3.3) that (Λf n | n ∈ N) ↑ Λf . Thus, Λ(v+ ) is pointwise 𝜎-continuous. The same is valid for Λ(−v− ). Consequently, by Proposition 2 and Theorem 1 (2.2.8), the functionals Λ(𝜇) = Λ(v+ ) + Λ(v− ) and Λ(v(𝜇)) = Λ(v+ (𝜇)) + Λ(−v− (𝜇)) are pointwise 𝜎-continuous. Suppose now that 𝜇 is bounded. Take any subset F ⊂ MI(T, M, v+ ) such that |f | ⩽ x1 for some x ∈ R+ and f ∈ F. Then, by Corollary 3 to Theorem 1 (3.3.2) and Lemmas 1 and 2 (3.3.2) |Λ(v+ )f | ⩽ Λ(v+ )|f | ⩽ ∫ x1 dv+ = xv+ T ⩽ xvT < ∞ for every f ∈ F. So Λ(v+ ) is uniformly bounded. The same is valid for Λ(−v− ). Since Λ(𝜇) = Λ(v+ ) − Λ(−v− ) and Λ(v) = Λ(v+ ) + Λ(−v− ), we deduce that these functionals also have the same property. It follows from this theorem that Λ(𝜇) is pointwise 𝜎-continuous on every latticeordered linear subspace A(T) of MI(T, M, 𝜇). A complete solution of this problem is presented in 3.4.2.

3.3.7 Wide measures defined by densities | 283

3.3.7 Wide measures defined by densities The previous subsections show that the properties of wideness and 𝜎-additivity of a measure played the crucial role in creation of such a reach and fruitful integration theory. Wide positive measures are obtained by some types of extensions studied in 3.1.4 and 3.1.5. But the method of extensions is not suitable for the construction of arbitrary (not necessarily positive) wide measures. Along with the method of extensions there is an other method of construction of wide measures. This is the method of the composition of functions and wide measures. Let ⟮T, M, 𝜇⟯ be a measurable space with the 𝜎-algebra M and the measure 𝜇 : M → R. If f ∈ MI e (T, M, 𝜇), then by Corollary 2 to Proposition 3 (3.3.6), f 𝜒(M) ∈ MI e (T, M, 𝜇), f |M ∈ MI e (M, MM , 𝜇M ) and ∫ f 𝜒(M) d𝜇 = ∫(f |M) d𝜇M for every M ∈ M. Therefore, we may define an evaluation f ⋅ 𝜇 : M → R setting (f ⋅ 𝜇)M ≡ ∫ f 𝜒(M) d𝜇 = ∫(f |M) d𝜇M for every M ∈ M. According to Proposition 2 (3.2.1), v+ (𝜇) and v− (𝜇) are positive and, respectively, negative measures on M. According to the definition, either f ∈ MI e (T, M, v+ (𝜇))∩ MI(T, M, −v− (𝜇)) or f ∈ MI(T, M, v+ (𝜇)) ∩ MI e (T, M, −v− (𝜇)). Consequently, in both cases, we can consider the evaluations f ⋅ v+ (𝜇) and f ⋅ v− (𝜇). By Corollary 1 to Proposition 2 (3.1.3) v(𝜇) is a measure on M as well. By virtue of Proposition 4 (3.3.6), f ∈ MI e (T, M, v(𝜇)). Hence, we can consider also the evaluation f ⋅ v(𝜇). Proposition 1. Let 𝜇 be a measure on a 𝜎-algebra M on T and f ∈ MI e (T, M, 𝜇). Then, 1) the evaluation f ⋅ 𝜇 is presented in the form of the sum f ⋅ 𝜇 = f ⋅ v+ (𝜇) + f ⋅ v− (𝜇) of the two natural evaluations at least one of which is finite; 2) the evaluation f ⋅ 𝜇 is presented in the form of the sum f ⋅ 𝜇 = f+ ⋅ v+ (𝜇) + f− ⋅ v+ (𝜇) + f+ ⋅ v− (𝜇) + f− ⋅ v− (𝜇) of the two positive f+ ⋅ v+ (𝜇) and f− ⋅ v− (𝜇) and the two negative f− ⋅ v+ (𝜇) and f+ ⋅ v− (𝜇) evaluations at least three of which are finite; 3) the evaluation f ⋅ 𝜇 is natural. Proof. Take any M ∈ M. First, suppose that f ∈ MI e (T, M, v+ ) ∩ MI(T, M, −v− ). Then, by Corollary 3 to Proposition 3 (3.3.6), f 𝜒(M) ∈ MI(T, M, −v− ). According to the definition from 3.3.2, we get (f+ ⋅ (−v− ))M = ∫(f 𝜒(M))+ d(−v− ) ∈ R+ and ((−f− ) ⋅ (−v− ))M = ∫(−(f 𝜒(M))− ) d(−v− ) ∈ R+ . Therefore, (f+ ⋅ v− )M ∈ R− , (f− ⋅ v− )M ∈ R+ , and (f ⋅ v− )M = (f+ ⋅ v− )M + (f− ⋅ v− )M ∈ R. This means that f+ ⋅ v− is finite and negative, f− ⋅ v− is finite and positive and f ⋅ v− = f+ ⋅ v− + f− ⋅ v− is finite. If f+ ∈ MI e (T, M, v+ ) and f− ∈ MI(T, M, v+ ), then by Corollaries 2 and 3 to Proposition 2 (3.3.2) we have f+ 𝜒(M) ∈ MI e (T, M, v+ ) and f− 𝜒(M) ∈ MI(T, M, v+ ). Therefore, (f+ ⋅v+ )M ∈ R+ , (f− ⋅v+ )M ∈ R− , and (f ⋅v+ )M = (f+ ⋅v+ )M+(f− ⋅v+ )M ∈ R∪{∞}. This means that f+ ⋅ v+ is positive, f− ⋅ v+ is finite and negative, and f ⋅ v+ = f+ ⋅ v+ + f− ⋅ v+ is ] − ∞, ∞]-valued, and therefore, natural.

284 | 3.3 The Lebesgue integral

Further, (f ⋅ 𝜇)M ≡ ∫ f 𝜒(M) d𝜇 ≡ ∫ f 𝜒(M) dv+ − ∫ f 𝜒(M) d(−v− ) = (f ⋅ v+ )M+ (f ⋅ v− )M ∈ R ∪ {∞}. This means that f ⋅ 𝜇 = f ⋅ v+ + f ⋅ v− = f+ ⋅ v+ + f− ⋅ v+ + f+ ⋅ v− + f− ⋅ v− and f ⋅ 𝜇 is natural. If f+ ∈ MI(T, M, v+ ) and f− ∈ MI e (T, M, v+ ), then by the same reason f+ 𝜒(M) ∈ MI(T, M, v+ ) and f− 𝜒(M) ∈ MI e (T, M, v+ ). Therefore, (f+ ⋅ v+ )M ∈ R+ , (f− ⋅ v+ )M ∈ R− and (f ⋅ v+ )M = (f+ ⋅ v+ )M + (f− ⋅ v+ )M ∈ {−∞} ∪ R. This means that f+ ⋅ v+ is finite and positive, f− ⋅ v+ is negative and f ⋅ v+ = f+ ⋅ v+ + f− ⋅ v+ is [−∞, ∞[-valued, and therefore, natural. Then, (f ⋅𝜇)M = (f ⋅ v + )M +(f ⋅ v− )M ∈ {−∞}∪R. This means that f ⋅𝜇 = f ⋅ v+ + f ⋅ v− = f+ ⋅ v+ + f− ⋅ v+ + f+ ⋅ v− + f− ⋅ v− and f ⋅ 𝜇 is natural. In the case of the second opportunity for f , the arguments are the same. Corollary 1. Let 𝜇 be a measure on a 𝜎-algebra M on T and f ∈ MI e (T, M, 𝜇). Then, the evaluation f ⋅ v(𝜇) is natural. Proof. It follows from assertion 1 of Proposition 4 (3.3.6) and assertion 3 of Proposition 1. Proposition 2. Let 𝜇 be a measure on a 𝜎-algebra M on T and f ∈ MI e (T, M, 𝜇). Then, all the natural evaluations f + ⋅ v+ (𝜇), f− ⋅ v+ (𝜇), f+ ⋅ v− (𝜇), f− ⋅ v− (𝜇), f ⋅ v+ (𝜇), f ⋅ v− (𝜇), and f ⋅ 𝜇 are measures. Proof. First, suppose that f ⩾ 0 and 𝜇 ⩾ 0. Let (M k ∈ M | k ∈ K) be a finite partition of a set M ∈ M. Then, by Corollary 2 to Theorem 1 (3.3.2), (f ⋅ 𝜇)M = ∫ f 𝜒(M) d𝜇 = ∑(∫ f 𝜒(M k ) d𝜇 | k ∈ K) = ∑((f ⋅ 𝜇)M k | k ∈ K). This means that f ⋅ 𝜇 is a semimeasure. Take any increasing sequence (R n ∈ M | n ∈ N ⊂ 𝜔) and the set R ≡ ⋃⟮R n | n ∈ N⟯ ∈ M. Then, (f 𝜒(R n ) | n ∈ N) ↑ and f 𝜒(R) = p-lim(f 𝜒(R n ) | n ∈ N). By Theorem 5 (3.3.3), (f ⋅ 𝜇)R = ∫ f 𝜒(R) d𝜇 = lim(∫ f 𝜒(R n ) d𝜇 | n ∈ N) = lim((f ⋅ 𝜇)R n | n ∈ N). Consequently, by Lemma 5 (3.1.1), f ⋅ 𝜇 is a measure. It follows from this fact that f ⋅ 𝜇 is a measure in the other three cases, i. e. when f ⩾ 0 and 𝜇 ⩽ 0, f ⩽ 0 and 𝜇 ⩾ 0, f ⩽ 0 and 𝜇 ⩽ 0. Let now f and 𝜇 be arbitrary. Then, by Lemma 1, 𝜈 ≡ f ⋅ 𝜇 = f+ ⋅ v+ + f− ⋅ v+ + f+ ⋅ v− + f− ⋅ v− . As was shown above, all the four summands are measures and, according to Lemma 1, at least three of them are finite. If all the four summands are finite, then by Corollary 1 to Lemma 1 (3.1.2), 𝜈 is a measure. Suppose now that, for instance, the measure 𝜘 ≡ f− ⋅v+ ⩽ 0 is not finite. Then, 𝜆 ≡ f+ ⋅v+ +f+ ⋅v− +f− ⋅v− is a finite measure. Let a sequence (M k ∈ M | k ∈ K ⊂ 𝜔) be a countable partition of a set M ∈ M and k0 ≡ / −∞ for every k. Therefore, using Proposition 1 (1.4.7), sm K. If 𝜘M = / −∞, then 𝜘M k = we get 𝜈M = 𝜘M + 𝜆M = lim(∑(𝜘M k | k ∈ K ∩ (n + 1)) | n ∈ 𝜔\k0 ) + lim(∑(𝜆M k | k ∈ K ∩ (n + 1)) | n ∈ 𝜔\k0 ) = lim(∑(𝜈M k | k ∈ K ∩ (n + 1)) | n ∈ 𝜔\k0 ) = ∑(𝜈M k | k ∈ K) (see 1.4.8).

3.3.7 Wide measures defined by densities | 285

If 𝜘M = −∞, then according to 1.4.7, for every number 𝛿 < 0 there is l ∈ 𝜔\k0 such that x n ≡ ∑(𝜘M k | k ∈ K ∩ (n + 1)) < 𝛿 for every n ⩾ l. Take some number 𝜀 > 0. Then, there is m ∈ 𝜔\k0 such that y n ≡ ∑(𝜆M k | k ∈ K ∩ (n + 1)) ∈]𝜆M − 𝜀, 𝜆M + 𝜀[ for every n ⩾ m. Consequently, z n ≡ ∑(𝜈M k | k ∈ K ∩ (n + 1)) = x n + y n < 𝛿 + 𝜆M + 𝜀 for every n ⩾ l ⊻ m. Since 𝛿 is arbitrary, we conclude that lim(z n | n ∈ 𝜔\k0 ) = −∞. Besides, 𝜈M = 𝜘M + 𝜆M = −∞. Finally, 𝜈M = lim(z n | n ∈ 𝜔\k0 ) = ∑(𝜈M k | k ∈ K). Thus, 𝜈 is a measure. Analogously, f ⋅v− = f+ ⋅v− +f− ⋅v− is a finite measure and f ⋅v+ = f+ ⋅v+ +f− ⋅v+ is a measure. In the case when one of the measures f+ ⋅ v+ , f+ ⋅ v− and f− ⋅ v− is not finite the arguments are analogous. Corollary 1. Let 𝜇 be a measure on a 𝜎-algebra M on T and f ∈ MI e (T, M, 𝜇). Then, f ⋅ v(𝜇) and |f | ⋅ v(𝜇) are measures. Proof. By Lemma 1 (3.3.6), |f | ∈ MI e (T, M, 𝜇). By Proposition 4 (3.3.6), f , |f | ∈ MI e (T, M, v). Consequently, Proposition 2 and Corollary 1 to Proposition 1 imply that the natural evaluations f ⋅ v and |f | ⋅ v are measures. The measure f ⋅𝜇 is called the composition (or product) of the function f ∈ MI e (T, M, 𝜇) and the measure 𝜇. The function f is called the density of the measure f ⋅𝜇 with respect to the measure 𝜇 or 𝜇-density of f ⋅ 𝜇. Lemma 1. Let 𝜇 be a measure on a 𝜎-algebra M on T and f ∈ MI e (T, M, 𝜇). Then, 1) |(f ⋅ 𝜇)M| ⩽ (|f | ⋅ v(𝜇))M for every M ∈ M; 2) the measures f ⋅ 𝜇, f ⋅ v+ (𝜇), f ⋅ v− (𝜇), f ⋅ v(𝜇) and |f | ⋅ v(𝜇) are absolutely continuous with respect to the measure 𝜇. Proof. 1. The assertion follows from Lemma 1 (3.3.6). 2. Let S ∈ M and v(𝜇)S = 0. Then, C ≡ coz(|f |𝜒(S)) ⊂ S implies vC = 0. By virtue of Lemma 3 (3.3.2), 0 ⩽ |(f ⋅ 𝜇)S| ⩽ (|f | ⋅ v)S = ∫ |f |𝜒(S) dv = 0. By Theorem 1 (3.2.4), f ⋅ 𝜇 ≪ 𝜇 and |f | ⋅ v ≪ 𝜇. According to the definition, f ∈ MI e (T, M, v+ ) ∩ MI e (T, M, v− ). By Proposition 4 (3.3.6), f ∈ MI e (T, M, v) and by Corollary 2 to Proposition 3 (3.3.6), f 𝜒(S) ∈ MI e (T, M, v+ ). Therefore, by Corollary 3 to Theorem 1 (3.3.2), 0 ⩽ |(f ⋅ v+ )S| ⩽ (|f |⋅ v+ )S. Since v+ ⩽ v, it follows from the definition of the Lebesgue integral in 3.3.2 that (|f |⋅v+ )S ⩽ (|f |v)S = 0. Hence, |(f ⋅v+ )S| = 0, where f ⋅v+ ≪ 𝜇. Analogously, f ⋅(−v− ) ≪ 𝜇 and f ⋅ v ≪ 𝜇. Consequently, f ⋅ v− ≪ 𝜇. Proposition 3. Let 𝜇 be a measure on a 𝜎-algebra M on T and f ∈ MI e (T, M, 𝜇). Then, 1) the measure f ⋅ v(𝜇) is presented in the form of the difference f ⋅ v(𝜇) = f ⋅ v+ (𝜇) − f ⋅ v− (𝜇) of the two measures at least one of which is finite;

286 | 3.3 The Lebesgue integral

2) the measure f ⋅ v(𝜇) is presented in the form of the sum and difference f ⋅ v(𝜇) = f+ ⋅ v+ (𝜇) + f− ⋅ v+ (𝜇) − f+ ⋅ v− (𝜇) − f− ⋅ v− (𝜇) of the two positive f+ ⋅ v+ (𝜇) and f− ⋅ v− (𝜇) and the two negative f− ⋅ v+ (𝜇) and f+ ⋅ v− (𝜇) measures at least three of which are finite. Proof. By Corollary 1 to Lemma 5 (3.2.1), v+ is concentrated on T+ and v− is concentrated on T− . If A, K ∈ M and A ⊂ K ⊂ T+ , then (−v− )K A = −v− A = 0 implies (v+ )K A = v+ A = v+ A − v− A = vA = v K A. Thus, (−v− )K = 0 and (v+ )K = v K . Similarly, if B, L ∈ M and B ⊂ L ⊂ T− , then (v+ )L B = v+ B = 0 implies (−v− )L B = −v− B = v+ B − v− B = vB = v L B. Hence, (v+ )L = 0 and (−v− )L = v L . For a set M ∈ M, consider the sets K ≡ M ∩ T+ and L ≡ M ∩ T− . According to Proposition 2 and Corollary 1 to it f ⋅ v+ , f ⋅ v− , and f ⋅ v are measures. By virtue of Proposition 1 at least one of the measures f ⋅ v+ and f ⋅ v− is finite. Therefore, using the equalities proven above we get (f ⋅ v+ )M + (f ⋅ (−v− ))M = (f ⋅ v+ )K + (f ⋅ v+ )L + (f ⋅ (−v− ))K + (f ⋅ (−v− ))L = ∫(f |K) d(v+ )K + ∫(f |L) d(v+ )L + ∫(f |K) d(−v− )K + ∫(f |L) d(−v− )L = ∫(f |K) dv K + ∫(f |L) dv L = (f ⋅ v)K + (f ⋅ v)L = (f ⋅ v)M. Thus, f ⋅ v + − f ⋅ v− = f ⋅ v. But f ⋅ v+ = f+ ⋅ v+ + f− ⋅ v+ and f ⋅ v− = f+ ⋅ v− + f− ⋅ v− . Therefore, f ⋅ v = f+ ⋅ v+ + f− ⋅ v+ − f+ ⋅ v− − f− ⋅ v− . All the other assertions follow from Proposition 1. Theorem 1. Let 𝜇 be a measure on a 𝜎-algebra M on T and f ∈ MI e (T, M, 𝜇). Then, v(f ⋅ 𝜇) = |f | ⋅ v(𝜇), v+ (f ⋅ 𝜇) = f+ ⋅ v+ (𝜇) + f− ⋅ v− (𝜇) and v− (f ⋅ 𝜇) = f+ ⋅ v− (𝜇) + f− ⋅ v+ (𝜇). Proof. By Corollary 1 to Lemma 5 (3.2.1), v+ is concentrated on M ≡ T+ and v− is concentrated on N ≡ T− . If A, K ∈ M and A ⊂ K ⊂ M, then v− A = 0 implies 𝜇K A = 𝜇A = v+ A + v− A = v+ A − v− A = vA = v K A. Thus, 𝜇K = v K . Similarly, if B, L ∈ M and B ⊂ L ⊂ N, then v+ B = 0 implies 𝜇L B = 𝜇B = v+ B + v− B = −v+ B + v− B = −vB = −v L B. Hence, 𝜇L = −v L . Consider the sets P ≡ {t ∈ T | f (t) ⩾ 0} and Q ≡ {t ∈ T | f (t) < 0} from M. Take any R ∈ M. Then, for R0 ≡ R ∩ M ∩ P ⊂ M we have (f ⋅ 𝜇)R0 = ∫ f |R0 d𝜇R = ∫ |f ||R0 dv R0 = (|f | ⋅ v)R0 . For R1 ≡ R ∩ M ∩ Q ⊂ M, we have (f ⋅ 𝜇)R1 = − ∫ |f ||R1 dv R1 = −(|f | ⋅ v)R1 . For R2 ≡ R ∩ N ∩ P ⊂ N, we have (f ⋅ 𝜇)R2 = − ∫ |f ||R2 dv R2 = −(|f | ⋅ v)R2 . Finally, for R3 ≡ R ∩ N ∩ Q ⊂ N we have (f ⋅ 𝜇)R3 = ∫ |f ||R3 dv R3 = (|f | ⋅ v)R3 . In all cases, we obtain the same equality |(f ⋅ 𝜇)R k | = (|f | ⋅ v)R k for every k ∈ 4. Fix any set E ∈ M and take any finite partition (R i ∈ M | i ∈ I) of E. For every R i consider the corresponding disjoint sets R0i , R1i , R2i and R3i . Since R i = ⋃⟮R ki | k ∈ 4⟯ for every i ∈ I, we obtain the new partition (R ki ∈ M | (i, k) ∈ I × 4) of E. Since f ⋅ 𝜇 is a measure, we have |(f ⋅ 𝜇)R i | = | ∑((f ⋅ 𝜇)R ki | k ∈ 4) ⩽ ∑(|(f ⋅ 𝜇)R ki | | k ∈ 4). Consequently, for our two partitions of E, we have x ≡ ∑(|(f ⋅ 𝜇)R i | | i ∈ I) ⩽ ∑(∑(|(f ⋅ 𝜇)R ki | | k ∈ 4) | i ∈ I) ≡ y. If all the summands in y are finite, then Corollary 1 to Proposition 1 (1.4.3) guarantees that y = z, where z ≡ ∑(|(f ⋅ 𝜇)R ki | | (i, k) ∈ I × 4). By virtue of Lemma 2 (3.1.3), we get x ⩽ y = z ⩽ v(f ⋅𝜇)E. If |(f ⋅𝜇)R ki | = ∞ for some i and k, then y = ∞ and z = ∞, where y = z again.

3.3.7 Wide measures defined by densities | 287

By Lemma 1 (3.3.6), |f | ∈ MI e (T, M, 𝜇) and by Corollary 1 to Proposition 2, |f | ⋅ v is a measure. Therefore, using the property |(f ⋅ 𝜇)R ki | = (|f | ⋅ v)R ki proven above, we deduce that y = ∑(∑((|f | ⋅ v)R ki | k ∈ L) | i ∈ I) = ∑((|f | ⋅ v)R i | i ∈ I) = (|f | ⋅ v)E. As a result, (|f | ⋅ v)E = y = z ⩽ v(f ⋅ 𝜇)E. Again by Lemma 2 (3.1.3) from x ⩽ y = (|f | ⋅ v)E we infer that v(f ⋅ 𝜇)E ⩽ (|f | ⋅ v)E. Thus, we get the first necessary equality. Now, consider the sets X ≡ (M ∩ P) ∪ (N ∩ Q) and Y ≡ (M ∩ Q) ∪ (N ∩ P). For the set R ∈ M, we have that (R0 , R3 ) is a partition of R ∩ X and (R1 , R2 ) is a partition of R ∩ Y. Since E ≡ coz(f− 𝜒(R3 )) ⊂ N and F ≡ coz(f+ 𝜒(R0 )) ⊂ M, we have v+ E = 0 and v− F = 0. Therefore, by Lemma 3 (3.3.2), ∫ f− 𝜒(R3 ) dv+ = 0 and ∫ f+ 𝜒(R0 ) dv− = 0. Besides, ∫ f− 𝜒(R0 ) dv+ = 0 and ∫ f+ 𝜒(R3 ) dv− = 0. Consequently, (f− ⋅ v+ )(R ∩ X) = (f− ⋅ v+ )R0 + (f− ⋅ v+ )R3 = 0. Similarly, (f+ ⋅ v− )(R ∩ X) = 0. Thus, Proposition 1 gives (f ⋅ 𝜇)(R ∩ X) = (f+ ⋅ v+ )(R ∩ X) + (f− ⋅ v− )(R ∩ X) ⩾ 0. This means that X is essentially positive for f ⋅ 𝜇. It is checked in the same way that (f+ ⋅ v+ )(R ∩ Y) = 0, (f− ⋅ v− )(R ∩ Y) = 0 and (f ⋅ 𝜇)(R ∩ Y) = (f− ⋅ v+ )(R ∩ Y) + (f+ ⋅ v− )(R ∩ Y) ⩽ 0. This means that Y is essentially negative for f ⋅ 𝜇. Since (X, Y) is a partition of T, we conclude that (X, Y) is a Hahn decomposition of T with respect to f ⋅ 𝜇 (see 3.2.1). Therefore, by Lemma 5 (3.2.1), v+ (f ⋅ 𝜇)R = (f ⋅ 𝜇)(R ∩ X) and v− (f ⋅ 𝜇)R = (f ⋅ 𝜇)(R ∩ Y). Finally, using the equalities proven above, we get v+ (f ⋅ 𝜇)R = (f+ ⋅ v+ )(R ∩ X) + (f− ⋅ v− )(R ∩ X) = (f+ ⋅ v+ )R + (f− ⋅ v− )R and v− (f ⋅ 𝜇)R = (f− ⋅ v+ )(R ∩ Y) + (f+ ⋅ v− )(R ∩ Y) = (f− ⋅ v+ )R + (f+ ⋅ v− )R. Corollary 1. Let 𝜇 be a measure on a 𝜎-algebra M on T and f ∈ MI(T, M, 𝜇). Then, |f ⋅ 𝜇| = |f | ⋅ |𝜇|, (f ⋅ 𝜇)+ = f+ ⋅ 𝜇+ + f− ⋅ 𝜇− and (f ⋅ 𝜇)− = f+ ⋅ 𝜇− + f− ⋅ 𝜇+ . Proof. By Proposition 1 (3.2.1), 𝜇 and f ⋅ 𝜇 are overfinite. Therefore, by Corollary 1 to Theorem 2 (3.2.2), v(𝜇) = |𝜇| and v(f ⋅ 𝜇) = |f ⋅ 𝜇|. The other two equalities follow from Proposition 1 (3.2.2). Proposition 4. Let 𝜇 be a measure on a 𝜎-algebra M on T and f ∈ MI 𝜎 (T, M, 𝜇). Then, the measurable space (T, M, f ⋅ 𝜇) is 𝜎-finite. Proof. Suppose that f ∈ MI 𝜎 (T, M, v+ ) ∩ MI(T, M, −v− ), f+ ∈ MI 𝜎 (T, M, v+ ) and −f− ∈ MI(T, M, v+ ). By the definition, there is a countable collection (T i ∈ M | i ∈ I) such that T = ⋃⟮T i | i ∈ I⟯ and f+ 𝜒(T i ) ∈ MI(T, M, v+ ), i. e. (f+ ⋅ v+ )T i ∈ R+ . According to Proposition 1, (f ⋅ 𝜇)T i = (f+ ⋅ v+ )T i + (f− ⋅ v+ )T i + (f+ ⋅ v− )T i + (f− ⋅ v− )T i ∈ R, i. e. T i ∈ Mf (f ⋅ 𝜇) for every i. For the other three opportunities for f the arguments are the same. Corollary 1. Let 𝜇 be a positive measure on a 𝜎-algebra M on T and f ∈ MI 𝜎 (T, M, 𝜇)+ . Then, the measure f ⋅ 𝜇 is 𝜎-finite. Proof. It follows from Proposition 4 and Lemma 11 (3.1.1).

288 | 3.3 The Lebesgue integral

Now, we shall consider the integration with respect to the measure f ⋅ 𝜇. Proposition 5. Let 𝜇 be a positive measure on a 𝜎-algebra M on T, f ∈ M(T, M)+ , and g ∈ M(T, M). Then, for the measures 𝜇 and f ⋅ 𝜇 the following conditions are equivalent: 1) g ∈ MI e (T, M, f ⋅ 𝜇) [g ∈ MI(T, M, f ⋅ 𝜇)]; 2) gf ∈ MI e (T, M, 𝜇) [gf ∈ MI(T, M, 𝜇)]. In both cases, ∫ g d(f ⋅ 𝜇) = ∫ gf d𝜇. Proof. At first, suppose that g ⩾ 0. By virtue of Theorem 2 (2.3.4), there is a sequence (g n ∈ St(T, M)+ | n ∈ N) ↑ such that g = p-lim (g n | n ∈ N). Let (M k ∈ M | k ∈ K) be a finite collection of pairwise disjoint sets, (x k ∈ R+ | k ∈ K) and 𝜑 ≡ ∑(x k 𝜒(M k ) | k ∈ K) ∈ St(T, M)+ . Then, 𝜑f = ∑(x k 𝜒(M k )f | k ∈ K) ∈ M(T, M)+ . By Corollary 2 to Theorem 1 (3.3.2) and Lemma 1 (3.3.2) we get ∫ 𝜑f d𝜇 = ∑(x k ∫ 𝜒(M k )f d𝜇 | k ∈ K) = ∑(x k (f ⋅ 𝜇)M k | k ∈ K) = ∫ 𝜑 d(f ⋅ 𝜇). Consequently, 𝜑 ∈ MI(T, M, f ⋅ 𝜇). Taking 𝜑 = g n , we get ∫ g n f d𝜇 = ∫ g n d(f ⋅ 𝜇). Since (g n f ∈ M(T, M)+ | n ∈ N) ↑ and p-lim(g n f | n ∈ N) = gf ∈ M(T, M)+ , we infer by virtue of Theorem 5 (3.3.3) that ∫ gf d𝜇 = lim ∫ g n f d𝜇 = lim(∫ g n d(f ⋅𝜇) | n ∈ N) = ∫ g d(f ⋅𝜇). Thus, gf ∈ MI(T, M, 𝜇) iff g ∈ MI(T, M, f ⋅ 𝜇). Let now g = g+ + g− . Then, ∫(gf )+ d𝜇 = ∫ g+ f d𝜇 = ∫ g+ d(f ⋅ 𝜇) and ∫(−(gf )− ) d𝜇 = ∫(−g− )f d𝜇 = ∫(−g− ) d(f ⋅ 𝜇). It follows from these equalities that conditions 1 and 2 are equivalent. In both cases, ∫ g d(f ⋅ 𝜇) ≡ ∫ g+ d(f ⋅ 𝜇) − ∫(−g− ) d(f ⋅ 𝜇) = ∫(gf )+ d𝜇 − ∫(−(gf )− ) d𝜇 ≡ ∫ gf d𝜇. Corollary 1. Let 𝜆 be a positive measure on a 𝜎-algebra M on T and f , g ∈ M(T, M)+ . Then, for the positive measures 𝜇 ≡ f ⋅ 𝜆 ≪ 𝜆 and 𝜈 ≡ g ⋅ 𝜇 ≪ 𝜇 we have 𝜈 = gf ⋅ 𝜆 ≪ 𝜆. Proof. If M ∈ M, then by Proposition 5, 𝜈M = (g⋅𝜇)M = (g⋅(f ⋅𝜆))M = ∫ g𝜒(M) d(f ⋅𝜆) = ∫ g𝜒(M)f d𝜆 = (gf ⋅ 𝜆)M. By Lemma 1, 𝜇 ≪ 𝜆, 𝜈 ≪ 𝜇, and 𝜈 ≪ 𝜆. Theorem 2. Let 𝜇 be a measure on a 𝜎-algebra M on T, f ∈ MI e (T, M, 𝜇) and g ∈ M(T, M). Then, for the measures 𝜇 and f ⋅ 𝜇 the following conditions are equivalent: 1) g ∈ MI(T, M, f ⋅ 𝜇); 2) gf ∈ MI(T, M, 𝜇). In both cases, ∫ g d(f ⋅ 𝜇) = ∫ gf d𝜇. Proof. Denote the measures v+ (𝜇) and v− (𝜇) simply by v+ and v− . By Theorem 1, v+ (f ⋅ 𝜇) = f+ ⋅ v+ + f− ⋅ v− and v− (f ⋅ 𝜇) = f+ ⋅ v− + f− ⋅ v+ . Let g ∈ MI(T, M, f ⋅ 𝜇). By the definition of the integral from 3.3.6, ∫ g d(f ⋅ 𝜇) ≡ ∫ g dv+ (f ⋅ 𝜇) − ∫ g d(−v− )(f ⋅ 𝜇). According to Proposition 3 (3.3.5), g is integrable with respect to f+ ⋅ v+ , f− ⋅ v− , f+ ⋅ (−v− ) and (−f− ) ⋅ v+ and ∫ g d(f ⋅ 𝜇) = ∫ g d(f+ ⋅ v+ ) + ∫ g d(f− ⋅ v− ) − ∫ g d(f+ ⋅ (−v− )) −

3.3.8 The Lebesgue – Radon – Nikodym theorem | 289

∫ g d((−f− ) ⋅ v+ ). By Proposition 5, gf+ ∈ MI(T, M, v+ ), g(−f− ) ∈ MI(T, M, −v− ), gf+ ∈ MI(T, M, −v− ), g(−f− ) ∈ MI(T, M, v+ ), and ∫ g d(f ⋅ 𝜇) = ∫ gf+ dv+ + ∫ g(−f− ) d(−v− ) − ∫ gf+ d(−v− ) − ∫ g(−f− ) dv+ . Again by the definition from 3.3.6, gf+ and g(−f− ) are integrable with respect to 𝜇 and ∫ g d(f ⋅ 𝜇) = ∫ gf+ d𝜇 − ∫ g(−f− ) d𝜇 = ∫ gf d𝜇. Thus, gf ∈ MI(T, M, 𝜇). Conversely, let gf ∈ MI(T, M, 𝜇). Since |gf+ | ⩽ |gf | and |gf− | ⩽ |gf |, we infer by virtue of Corollary 1 to Theorem 2 (3.3.2) that gf+ and gf− are integrable with respect to 𝜇. Then, by Theorem 2 (3.3.2), ∫ gf d𝜇 = ∫ gf+ d𝜇 − ∫ g(−f− ) d𝜇. Besides, gf+ and gf− are integrable with respect to v+ and (−v− ) simultaneously and ∫ gf d𝜇 = ∫ gf+ dv+ − ∫ gf+ d(−v− ) − ∫ g(−f− ) dv+ + ∫ g(−f− ) d(−v− ). Again, by Proposition 5, g is integrable with respect to f+ ⋅ v+ , f+ ⋅ (−v− ), (−f− ) ⋅ v+ and (−f− ) ⋅ (−v− ) and ∫ gf d𝜇 = ∫ g d(f+ ⋅ v+ ) + ∫ g d(f+ ⋅ v− ) + ∫ g d(f− ⋅ v+ ) + ∫ g d(f− ⋅ v− ). Then, by Proposition 3 (3.3.5), g is integrable with respect to the sums v+ (f ⋅ 𝜇) = f+ ⋅ v+ + f− ⋅ v− and −v− (f ⋅ 𝜇) = f+ ⋅ (−v− ) + (−f− ) ⋅ v+ , and ∫ gf d𝜇 = ∫ g dv+ (f ⋅ 𝜇) − ∫ g d(−v− (f ⋅ 𝜇)). This means that g ∈ MI(T, M, f ⋅ 𝜇) and ∫ gf d𝜇 = ∫ g d(f ⋅ 𝜇). 3.3.8 The Lebesgue – Radon – Nikodym theorem Let ⟮T, M, 𝜇⟯ be a measurable space with the measure 𝜇 on the 𝜎-algebra M on the set T. By Proposition 2 (3.3.7) and Lemma 1 (3.3.7), every function f ∈ MI e (T, M, 𝜇) generates the measure f ⋅ 𝜇 on M, which is absolutely continuous with respect to 𝜇. If f ∈ MI(T, M, 𝜇), then the measure f ⋅ 𝜇 is finite. Note that by Corollary 3 to Proposition 1 (3.2.1), a measure is finite iff it is bounded. Now, we shall consider some converse versions of these statement. Let Λ be the Lebesgue integrals over the measurable space ⟮T, M, 𝜇⟯. For every M ∈ M, consider the 𝜎-algebra MM ≡ {R ∈ M | R ⊂ M}, the measure 𝜇M ≡ 𝜇|MM and the measurable “subspace” ⟮M, MM , 𝜇M ⟯. Let Λ M be the Lebesgue integral over the space ⟮M, MM , 𝜇M ⟯. Along with v(𝜇), v+ (𝜇) and v− (𝜇) we shall write sometimes simply v, v+ and v− . By Proposition 2 (3.3.7) and its Corollary 1 every function f ∈ MI e (T, M, 𝜇) generates also the measures f ⋅ v+ , f ⋅ v− , and f ⋅ v. Lemma 1. Let 𝜇 be a measure on a 𝜎-algebra M on T. If f 󸀠 ∈ f ∈ L1 (T, M, 𝜇) (see 3.3.6) then f ⋅ 𝜇 = f 󸀠 ⋅ 𝜇. Proof. By the definition, from 3.3.6, v(coz |f − f 󸀠 |) = 0. Take any M ∈ M. Then, v(coz |(f − f 󸀠 )𝜒(M)|) = 0. Therefore, Lemma 3 (3.3.2) implies ∫ |(f − f 󸀠 )𝜒(M)| dv = 0. By virtue of Lemma 1 (3.3.6) we get | ∫(f − f 󸀠 )𝜒(M) d𝜇| ⩽ ∫ |(f − f 󸀠 )𝜒(M)| dv = 0. Consequently, (f ⋅ 𝜇)M = ∫ f 𝜒(M) d𝜇 = ∫ f 󸀠 𝜒(M) d𝜇 + ∫(f − f 󸀠 )𝜒(M) d𝜇 = (f 󸀠 ⋅ 𝜇)M. It follows from Lemma 1 that we can define correctly the bounded measure f ⋅ 𝜇 on M, setting f ⋅ 𝜇 ≡ f 󸀠 ⋅ 𝜇 for any f 󸀠 ∈ f ∈ L1 (T, M, 𝜇). It is clear that f ⋅ 𝜇 ≪ 𝜇. Similarly, we can define the bounded measures f ⋅ v+ , f ⋅ v− , and f ⋅ v on M. It follows from

290 | 3.3 The Lebesgue integral

Propositions 1 and 3 (3.3.7) that f ⋅ 𝜇 = f ⋅ v+ + f ⋅ v− = f + ⋅ v+ + f − ⋅ v+ + f + ⋅ v− + f − ⋅ v− and f ⋅ v = f ⋅ v+ − f ⋅ v− = f + ⋅ v+ + f − ⋅ v+ − f + ⋅ v− − f − ⋅ v− . Consider the subset Measb (T, M, ≪ 𝜇) of the set Measb (T, M) consisting of all bounded 𝜇-absolutely continuous measures on M. By Corollary 1 to Lemma 4 (3.2.4), Measb (T, M, ≪ 𝜇) is a Dedekind complete lattice-ordered linear space. Consequently, U : f 󳨃→ f ⋅ 𝜇 is a mapping from the lattice-ordered linear space L1 (T, M, 𝜇) into the lattice-ordered linear space Measb (T, M, ≪ 𝜇). Similarly, consider the mappings U1 : f 󳨃→ f ⋅ v+ , U2 : f 󳨃→ f ⋅ (−v− ) and V : f 󳨃→ f ⋅ v. Proposition 1. Let 𝜇 be a measure on a 𝜎-algebra M on T. Then, Measb (T, M, ≪ 𝜇) is an injective linear operator between 1) U : L 1 (T, M, 𝜇) the given lattice-ordered linear spaces; 2) U1 , U2 , and V are positive linear operators such that U = U1 − U2 ⩽ V = U1 + U2 . Proof. 1. Let a, b ∈ A ≡ L1 (T, M, 𝜇), x, y ∈ R, f ∈ a, g ∈ b, and c ≡ xa + yb. Then, h ≡ xf + yg ∈ c. If M ∈ M, then (Uc)M = (c ⋅ 𝜇)M = (h ⋅ 𝜇)M = ∫ h𝜒(M) d𝜇 = x ∫ f 𝜒(M) d𝜇 + y ∫ g𝜒(M) d𝜇 = x(f ⋅ 𝜇)M + y(g ⋅ 𝜇)M = x(a ⋅ 𝜇)M + y(b ⋅ 𝜇) M= (xUa + yUb)M. Hence, U is linear. Suppose that Ua = Ub, i. e. (f ⋅ 𝜇)M = (g ⋅ 𝜇)M for every M ∈ M. Consider the sets P ≡ {t ∈ T | f (t) − g(t) ⩾ 0} and Q ≡ T\P from M. By Corollary 1 to Lemma 5 (3.2.1), v+ is concentrated on R ≡ T+ and v− is concentrated on S ≡ T− . If E, K ∈ M and E ⊂ K ⊂ R, then v− E = 0 implies 𝜇K E = 𝜇E = v+ E + v− E = v+ E − v− E = vE = v K E. Thus, 𝜇K = v K . Similarly, if E, K ∈ M and E ⊂ K ⊂ S, then v+ E = 0 implies 𝜇K E = 𝜇E = v+ E + v− E = −v+ E + v− E = −vE = −v K E. As a result, 𝜇K = −v K . For K ≡ P ∩ R, we have 0 = ∫(f − g)|K d𝜇K = ∫ |f − g||K dv K . Therefore, by Lemma 3 (3.3.2), v K coz(|f − g||K) = 0. For L ≡ P ∩ S, we have 0 = ∫(f − g)| L d𝜇L = − ∫ |f − g||L dv L , where v L coz(|f − g||L) = 0. For M ≡ Q ∩ R, we have 0 = ∫(g − f )|M d𝜇M = ∫ |f − g||M dv M , where v M coz(|f − g||M) = 0. Finally, for N ≡ Q ∩ S, we have 0 = ∫ |g − f ||N d𝜇N = − ∫ |f − g||N dv N , where v N coz(|f − g||N) = 0. Since K, L, M, and N form a partition of T and v is a measure, we deduce that v coz |f −g| = 0. Hence, a = f = ḡ = b. 2. In the same way, it is checked that U1 , U2 , and V are linear operators. Let a ∈ A+ and M ∈ M. Then, there is f ∈ a such that f ⩾ 0. Therefore, by Lemma 2 (3.3.2), (U1 a)M = ∫ f 𝜒(M) dv+ ⩾ 0. This means that U1 a ⩾ 0. Hence, U1 is positive. Similarly, U2 and V are positive. Furthermore, for every a ∈ A, we have Ua = a ⋅ 𝜇 = a ⋅ v+ − a ⋅ (−v− ) = (U1 − U2 )a, where U = U1 − U2 . Besides, U ⩽ U1 + U2 = V. Theorem 1. Let ⟮T, M, 𝜇⟯ be a measurable space with the bounded positive measure 𝜇 on the 𝜎-algebra M. Then, for every bounded positive, 𝜇-absolutely continuous measure 𝜈 on M, there is the unique element f ∈ L1 (T, M, 𝜇)+ such that 𝜈 = f ⋅ 𝜇 and ‖𝜈‖ = ‖f ‖1 .

3.3.8 The Lebesgue – Radon – Nikodym theorem

| 291

Proof. Consider the non-empty subfamily H of the family MI(T, M, 𝜇)+ consisting of functions h such that h ⋅ 𝜇 ⩽ 𝜈. Consider the lattice-ordered linear space L1 ≡ L1 (T, M, 𝜇). Then, the subfamily H of MI(T, M, 𝜇) generates the nonempty subset H̄ ≡ {h̄ | h ∈ H} in L1 . The order on L1 induces the corresponding order on H.̄ Take any non-empty chain C in H̄ and consider the number x ≡ sup{(h̄ ⋅ 𝜇)(T) | h̄ ∈ C} ⩽ 𝜈T < ∞. First, suppose that there is ḡ ∈ C such that x = (ḡ ⋅ 𝜇)(T). Then, ḡ is an upper bound for C in H.̄ In fact, if h̄ ∈ C, then either h̄ ⩽ ḡ or ḡ ⩽ h.̄ In the second ̄ ̄ ̄ ̄ case, x = (g⋅𝜇)(T) ⩽ (h⋅𝜇)(T) ⩽ x implies (g⋅𝜇)(T) = (h⋅𝜇)(T), where ∫ g d𝜇 = (g⋅𝜇)T = (h ⋅ 𝜇)T = ∫ h d𝜇. By virtue of Lemma 3 (3.3.2), 𝜇(coz(g − h)) = 0, where ḡ = h,̄ i. e. ḡ is an upper bound. Now, suppose that for every h̄ ∈ C, we have (h̄ ⋅ 𝜇)(T) < x. Then, consider ̄ the choice mapping p : P(H)\{⌀} → H̄ from the axiom of choice. For every n ∈ N, consider the non-empty subset C n ≡ {h̄ ∈ C | (h̄ ⋅ 𝜇)(T) > x − 1/n} and take the element ḡ n ≡ pC n ∈ C n . Then, sup ((ḡ n ⋅ 𝜇)(T) | n ∈ N) = x. Again take some choice mapping q : P(MI(T, M, 𝜇))\{⌀} → MI(T, M, 𝜇). Since ḡ n ∩ H = / ⌀ for every n, we can take the elements g 󸀠n ≡ q(ḡ n ∩ H) ∈ ḡ n ∩ H. By virtue of Corollary 1 to Theorem 3 (1.2.6) for every n ∈ N the set {ḡ k | k ∈ n + 1} has the greatest element, which will be denoted by h̄ n . Then, h̄ n = sup (ḡ k | k ∈ n + 1) and h̄ n ∈ C. Therefore, (ḡ n ⋅ 𝜇)(T) ⩽ (h̄ n ⋅ 𝜇)(T) ⩽ x. Consider the functions h󸀠n ≡ sup (g 󸀠k | k ∈ n + 1) ↑. Then, h̄ 󸀠n = sup (ḡ k | k ∈ n + 1) = h̄ n ∈ C. Take any M ∈ M. We have Λ M (h󸀠n )(M) = (h󸀠n ⋅ 𝜇)(M) = (h̄ n ⋅ 𝜇)(M) ⩽ 𝜈M ⩽ 𝜈T < ∞ for every n. By virtue of Theorem 2 (3.3.3) there is a function f M ∈ MI(M, MM , 𝜇M ) such that f M (t) = sup ((h󸀠n |M)(t) | n ∈ N) for some P M ∈ M0M (𝜇M ) and every t ∈ M\P M , and ∫ f M d𝜇M = sup (∫(h󸀠n |M) d𝜇M | n ∈ N).

In particular, g ≡ f T ∈ MI(T, M, 𝜇), g(t) = sup (h󸀠n (t) | n ∈ N) for Q ≡ P T ∈ M0 (𝜇) and every t ∈ T\Q, and ∫ g d𝜇 = sup (∫ h󸀠n d𝜇 | n ∈ N). By Corollary 3 to Proposition 2 (3.3.2), g|M ∈ MI(M, MM , 𝜇M ). Consider the set Q M ≡ Q ∩ M ∈ MM . From the inequalities 0 ⩽ 𝜇M Q M = 𝜇Q M ⩽ 𝜇Q = 0, we conclude that Q M ∈ M0M (𝜇M ). Then, R M = P M ∪ Q M ∈ M0M (𝜇M ) and for every t ∈ M\R M ⊂ M\Q, we have (g|M)(t) = sup ((h󸀠n |M)(t) | n ∈ N) = f M (t). Consequently, by Lemma 5 (3.3.2), (g ⋅ 𝜇)(M) = Λ M (g|M) = ∫ f M d𝜇M = sup (Λ M (h󸀠n |M) | n ∈ N) ⩽ 𝜈M for every M ∈ M. This means that g ∈ H and ḡ ∈ H.̄ Besides, ḡ ⩾ h̄ 󸀠n = h̄ n ⩾ ḡ n for every n. Check that ḡ is an upper bound for C in H.̄ In fact, if h̄ ∈ C, then either h̄ ⩽ ḡ n for some n ∈ N or h̄ ⩾ ḡ n for every n. In the first case, h̄ ⩽ g.̄ In the second case, x ⩾ (h̄ ⋅𝜇)(T) ⩾ (ḡ n ⋅𝜇)(T) for every n implies (h̄ ⋅𝜇)(T) = x. But this equality contradicts our supposition that (h̄ ⋅ 𝜇)(T) < x. This means that the second case is impossible, and therefore, ḡ is an upper bound. Thus, the chain C has an upper bound. By virtue of the Kuratowski – Zorn lemma from Theorem 1 (1.2.11), the ordered set H̄ contains some minimal element a. Take some f ∈ a ∩ H. Consider the new bounded positive measure 𝜆 on M such that

292 | 3.3 The Lebesgue integral

𝜆U ≡ 𝜈U − (f ⋅ 𝜇)(U) for every U ∈ M. Since 𝜈 and (f ⋅ 𝜇) are 𝜇-absolutely continuous, we infer by virtue of Lemma 4 (3.2.4) that 𝜆 ≪ 𝜇. Finally, we shall prove that 𝜆 = 0. Suppose that 𝜆 = / 0. Then, 𝜆T > 0. By the Archimedes principle (see Lemma 13 (1.4.3)) there is n ∈ N such that 𝜇T < n𝜆T. Consider the new finite measure 𝜘 ≡ 𝜇 − n𝜆. By Theorem 1 (3.2.1) for the set T and measure 𝜘, there is a Hahn decomposition T = T + ∪ T− . Take any U ∈ M. Then, 𝜇(U ∩ T− ) − n𝜆U ⩽ 𝜘(U ∩ T− ) ⩽ 0 and 𝜘(U ∩ T+ ) ⩾ 0. Suppose that 𝜇T− = 0. Then, by Theorem 1 (3.2.4), 𝜆T− = 0. Consequently, 𝜇T = 𝜇T+ and 𝜆T = 𝜆T+ imply 0 ⩽ 𝜘T+ = 𝜘T = 𝜇T − n𝜆T < 0. This contradiction shows that 𝜇T− > 0. Consider the step function 𝜑 ≡ (1/n)𝜒(T− ) and the function g ≡ f + 𝜑 from MI(T, M, 𝜇)+ . Using Lemma 1 (3.3.2), we get Λ M (𝜑|M) = (1/n)𝜇M (M∩T− ) = (1/n)𝜇(M∩ T− ) ⩽ 𝜆M = 𝜈M − (f ⋅ 𝜇)M = 𝜈M − Λ M (f |M), where (g ⋅ 𝜇)(M) = Λ M ((𝜑 + f )|M) ⩽ 𝜈M for every M ∈ M. Therefore, g ∈ H and ḡ ∈ H.̄ From 𝜇(coz 𝜑) = 𝜇T− > 0, we conclude that 𝜑̄ > 0 and ḡ > f . But this contradicts the maximality of f . Thus, 𝜆 = 0 and 𝜈 = f ⋅ 𝜇 = f ⋅ 𝜇. The uniqueness of f follows from Proposition 1. ̄̄ According to 3.2.3 and 3.3.4, ‖𝜈‖ ≡ 𝜈(T) = (f ⋅ 𝜇)(T) = Λf = ‖f ‖i = ‖ f ‖1 . Theorem 2. Let ⟮T, M, 𝜇⟯ be a 𝜎-finite measurable space (see 3.1.1) with the positive measure 𝜇 on the 𝜎-algebra M. Then, for every bounded 𝜇-absolutely continuous measure 𝜈 on M, there is the unique element f ∈ L1 (T, M, 𝜇) such that 𝜈 = f ⋅ 𝜇. Proof. At first suppose that 𝜈 ⩾ 0. By the definition, there is a countable collection (M i ∈ Mf (𝜇) | i ∈ I) such that T = ⋃⟮M i | i ∈ I⟯. This collection may be rearranged into a sequence (M n | n ∈ 𝜔). Moreover, we may regard that it consists of pairwise disjoint sets. For each n, consider the finite positive measures 𝜇n and 𝜈n on M such that 𝜇n A ≡ 𝜇(A ∩ M n ) and 𝜈n A ≡ 𝜈(A ∩ M n ) for every A ∈ M. Then, 𝜈n ≪ 𝜇n for each n. By Theorem 1, for every n, there is a function f n ∈ MI(T, M, 𝜇n )+ such that 𝜈n = f n ⋅ 𝜇n . Since (M n | n ∈ 𝜔) is a partition of T, we may define a function f ∈ F(T)+ setting f |M n ≡ f n |M n for each n. It is easy to see that f is M-measurable. Take any U ∈ M and consider the subsets U n ≡ U ∩ M n . Then, by Proposition 2 (3.3.2), ∫ f 𝜒(U n ) d𝜇 = ∫ f |U n d𝜇U n = ∫ f n |U n d𝜇U n . If A ∈ MU n = {M ∈ M | M ⊂ U n }, then (𝜇n )U n = 𝜇n A = 𝜇A = 𝜇U n A. Hence, 𝜇U n = (𝜇n )U n . Thus, ∫ f 𝜒(U n ) d𝜇 = ∫ f n | U n d(𝜇n )U n = (f n ⋅ 𝜇n )U n = 𝜈n U n = 𝜈U n ∈ R, and therefore, f 𝜒(U n ) ∈ MI(T, M, 𝜇)+ . Since 𝜈 is a finite positive measure, we get ∑(∫ f 𝜒(U n ) d𝜇 | n ∈ 𝜔) = ∑(𝜈U n | n ∈ 𝜔) = 𝜈U < 𝜈T < ∞. By Theorem 1 (3.3.3), there is a function g U ∈ MI(T, M, 𝜇) such that g U (t) = ∑(f 𝜒(U n )(t) | n ∈ 𝜔) for some N U ∈ M0 (𝜇) and every t ∈ T\N U and ∫ g U d𝜇 = ∑(∫ f 𝜒(U n ) d𝜇 | n ∈ 𝜔) = 𝜈U. If t ∈ U\N U , then t ∈ U m \N U for a single m ∈ 𝜔. Since U n ∩ U m = ⌀, we conclude that f 𝜒(U n )(t) = 0 for every n = / m and f 𝜒(U m )(t) = f (t). As a result, we obtain g U (t) = ∑(f 𝜒(U n )(t) | n ∈ 𝜔) = f (t) for every t ∈ U\(N U ∩ U). In particular for U = T we have g T (t) = f (t) for every t ∈ T\N T . Then, by Lemma 5 (3.3.2), f ∈ MI(T, M, 𝜇).

3.3.8 The Lebesgue – Radon – Nikodym theorem | 293

If t ∈ T\U, then f 𝜒(U n )(t) = 0 for every n. Consequently, g U (t) = 0. Thus, g U = g U 𝜒(U). Then, by Proposition 2 (3.3.2), ∫ g U d𝜇 = ∫ g U 𝜒(U) d𝜇 = ∫ g U |U d𝜇U . Since 𝜇U (N U ∩ U) = 𝜇(N U ∩ U) = 0 and (g U |U)(t) = (f |U)(t) for every t ∈ U\(N U ∩ U), by Lemma 5 (3.3.2) that ∫ g U |U d𝜇U = ∫ f |U d𝜇U = (f ⋅ 𝜇)U. Finally, we obtain 𝜈U = ∫ g U d𝜇 = (f ⋅ 𝜇)U, and therefore, 𝜈 = f ⋅ 𝜇 = f ⋅ 𝜇. Now, we take an arbitrary measure 𝜈. By Corollary 3 to Theorem 2 (3.2.2), 𝜈 = 𝜈+ + 𝜈− . By Corollary 2 to Lemma 4 (3.2.4), 𝜈+ ≪ 𝜇 and 𝜈− ≪ 𝜇. As was shown above, 𝜈+ = f 1 ⋅ 𝜇 and −𝜈− = f 2 ⋅ 𝜇 for some f 1 , f 2 ∈ L1 (T, M, 𝜇)+ . Consider the function f ≡ f1 − f2 and the element f = f 1 − f 2 ∈ L1 (T, M, 𝜇). Then, for every U ∈ M, we have 𝜈U = ∫ f1 𝜒(U) d𝜇 − ∫ f2 𝜒(U) d𝜇 = ∫ f 𝜒(U) d𝜇 = (f ⋅ 𝜇)U, i. e. 𝜈 = f ⋅ 𝜇. The uniqueness of f follows from Proposition 1. Proposition 2. Let ⟮T, M, 𝜇⟯ be a 𝜎-finite measurable space with the positive measure 𝜇 on the 𝜎-algebra M. Then, the positive linear operator U : L1 (T, M, 𝜇) Measb (T, M, ≪ 𝜇) is an isomorphism of the given lattice-ordered linear spaces. Proof. It follows from Proposition 1 and Theorem 2 that U is bijective and increasing because U = V. Let Uf ⩽ U ḡ and suppose that f ⩽̸ g,̄ i. e. f ∨ ḡ − ḡ > 0.̄ Then, there is a function 𝜑 ∈ MI 0 (T, M, 𝜇)+ such that h ≡ f ∨ g − g + 𝜑 ⩾ 0. According to 3.3.4, we have Q ≡ coz 𝜑 ∈ M0 (𝜇). Consider the set R ≡ coz h ∈ M. Suppose that P ≡ R\Q ∈ M0 (𝜇). Then, R ∈ M0 (𝜇) implies f ∨ ḡ − ḡ = 0.̄ But this is not true. Hence, 𝜇P P = 𝜇P > 0. If t ∈ P, then h(t) = f (t) ∨ g(t) − g(t) > 0 means f (t) > g(t), where P = coz((f − g)|P). Applying Lemma 3 (3.3.2) to the function (f − g)|P ∈ M(P, MP ), we conclude that ∫(f − g)|P d𝜇P > 0. On the other hand, Uf ⩽ U ḡ gives ∫(f − g)|P d𝜇P ⩽ 0. This contradiction proves that f ⩽ g.̄ Thus, U is isotone (see 1.1.15). By Lemma 2 (1.1.15) the mapping U preserves any exact bounds. In particular, U is a homomorphism of lattice-ordered linear spaces (see 5∘ (2.2.7)). Since U is bijective, U is a necessary isomorphism by virtue of Statement 5 (2.2.7). Now, we shall consider on the family Measb (T, M, ≪ 𝜇) the norm ‖⋅‖ induced from ⟮Measb (T, M), ‖⋅‖⟯ (see 3.2.3). Corollary 1. The mapping U : ⟮L1 (T, M, 𝜇), ‖⋅‖1 ⟯ ⟮Measb (T, M, ≪ 𝜇), ‖⋅‖⟯ from Proposition 2 is isometric, i.e. U is an isomorphism of the given normed lattice-ordered linear spaces. Now, we shall extend Theorem 2 on not necessarily positive measures 𝜇. Theorem 3 (the Lebesgue – Radon – Nikodym theorem for bounded measures). Let ⟮T, M, 𝜇⟯ be a 𝜎-finite measurable space with the measure 𝜇 on the 𝜎-algebra M. Then, for every bounded 𝜇-absolutely continuous measure 𝜈 on M there is the unique element f ∈ L1 (T, M, 𝜇) such that 𝜈 = f ⋅ 𝜇.

294 | 3.3 The Lebesgue integral

Proof. By Proposition 2 (3.2.1), 𝜇 = v+ + v− and v = v+ − v− . Take some Hahn decomposition (P, N) of T from Theorem 1 (3.2.1) with respect to 𝜇. By Lemma 5 (3.2.1), v+ A = 𝜇(A ∩ P) and v− A = 𝜇(A ∩ N) for every A ∈ M. Consider the bounded measures 𝜈󸀠 and 𝜈󸀠󸀠 on M such that 𝜈󸀠 A ≡ 𝜈(A ∩ P) and 󸀠󸀠 𝜈 A ≡ 𝜈(A ∩ N) for every A ∈ M. If v+ A = 0, then by Corollary 1 to Lemma 5 (3.2.1), 0 ⩽ v(A ∩ P) = v + (A ∩ P)− − v− (A ∩ P) = v+ (A ∩ P) ⩽ v+ A = 0. By virtue of Theorem 1 (3.2.4), 𝜈󸀠 A ≡ 𝜈(A ∩ P) = 0, and therefore, 𝜈󸀠 ≪ v+ . Analogously, 𝜈󸀠󸀠 ≪ −v− . By the definition, there is a countable collection (M i ∈ Mf (𝜇) | i ∈ I) such that T = ⋃⟮M i | i ∈ I⟯. By virtue of Proposition 1 (3.2.1), v+ M i = 𝜇(M i ∩ P) ∈ R. This means that the measurable space (T, M, v+ ) is 𝜎-finite. The same is valid for (T, M, −v− ). Therefore, by virtue of Theorem 2 there are functions f 󸀠 ∈ X ≡ MI(T, M, v+ ) and f 󸀠󸀠 ∈ Y ≡ MI(T, M, −v− ) such that 𝜈󸀠 = f 󸀠 ⋅ v+ and 𝜈󸀠󸀠 = f 󸀠󸀠 ⋅ (−v− ). Define a function f on T setting f ≡ f 󸀠 𝜒(P) − f 󸀠󸀠 𝜒(N). Evidently, f ∈ M(T, M). Corollary 1 to Proposition 1 (3.3.6) guarantees that f+󸀠 , f−󸀠 ∈ X and f+󸀠󸀠 , f−󸀠󸀠 ∈ Y. Then, by Corollary 3 to Proposition 3 (3.3.6), f+󸀠 𝜒(P) ∈ X and −f−󸀠󸀠 𝜒(N) ∈ Y. By Corollary 2 to Lemma 5 (3.2.1), v+ N = −v− P = 0. Lemma 3 (3.3.2) entails (f+ ⋅ v+ )N ≡ ∫ f+ 𝜒(N) dv+ = 0 and (f+ ⋅ (−v− ))P ≡ ∫ f+ 𝜒(P) d(−v− ) = 0. Therefore, by Corollary 2 to Theorem 1 (3.3.2), we get ∫ f+ dv+ = ∫ f+ 𝜒(P) dv+ + ∫ f+ 𝜒(N) dv+ = ∫ f+󸀠 𝜒(P) dv+ ∈ R and ∫ f+ d(−v− ) = ∫ f+ 𝜒(P) d(−v− ) + ∫ f+ 𝜒(N) d(−v− ) = ∫(−f−󸀠󸀠 )𝜒(N) d(−v− ) ∈ R. Analogously, ∫(−f− ) dv+ and ∫(−f− ) d(−v− ) are finite. This means that f ∈ MI(T, M, 𝜇). Hence, we can consider the bounded measures f ⋅ v+ , f ⋅ (−v− ), and f ⋅ 𝜇. Let A ∈ M. Consider B ≡ A ∩ P and C ≡ A ∩ N. Then, by virtue of Proposition 1 (3.3.7), (f ⋅ 𝜇)A = (f ⋅ 𝜇)B + (f ⋅ 𝜇)C = (f ⋅ v + )B − (f ⋅ (−v− ))B + (f ⋅ v+ ) C − (f ⋅ (−v− ))C. By Corollary 1 to Lemma 5 (3.2.1), v+ C = (−v− )B = 0. This implies (f ⋅ v+ )C = 0 and (f ⋅ (−v− ))B = 0. As a result, (f ⋅ 𝜇)A = ∫ f 𝜒(B) dv+ − ∫ f 𝜒(C) d(−v− ) = ∫ f 󸀠 𝜒(B) dv+ + ∫ f 󸀠󸀠 𝜒(C) d(−v− ) = 𝜈󸀠 B + 𝜈󸀠󸀠 C = 𝜈(A ∩ P) + 𝜈(A ∩ N) = 𝜈A. Thus, f ⋅ 𝜇 = f ⋅ 𝜇 = 𝜈. The uniqueness of f follows from Proposition 1. Corollary 1. Let ⟮T, M, 𝜇⟯ be a 𝜎-finite measurable space with the measure 𝜇 on the 𝜎-algebra M. Then, U : L 1 (T, M, 𝜇) Measb (T, M, ≪ 𝜇) is a bijective linear operator between the given lattice-ordered linear spaces. Proof. This follows from Proposition 1 and Theorem 3. Now, we shall extend Theorem 2 on not necessarily bounded measures 𝜈. According to Proposition 4 (3.3.7), if f ∈ MI 𝜎 (T, M, 𝜇), then the measurable space (T, M, f ⋅ 𝜇) is 𝜎-finite and, according to Lemma 1 (3.3.7), f ⋅ 𝜇 ≪ 𝜇. We shall prove the converse statement. Theorem 4 (the Lebesgue – Radon – Nikodym theorem for general measures). Let ⟮T, M, 𝜇⟯ be a 𝜎-finite measurable space with the positive measure 𝜇 on the

3.3.8 The Lebesgue – Radon – Nikodym theorem | 295

𝜎-algebra M. Then, for every 𝜎-finite measurable space ⟮T, M, 𝜈⟯ with the 𝜇-absolutely continuous measure 𝜈 on M there is a function f ∈ MI 𝜎 (T, M, 𝜇) such that 𝜈 = f ⋅ 𝜇. Proof. At first suppose that 𝜈 ⩾ 0. By the definition, there are countable collections (M i ∈ Mf (𝜇) | i ∈ I) and (N j ∈ Mf (𝜈) | j ∈ J) such that T = ⋃⟮M i | i ∈ I⟯ = ⋃⟮N j | j ∈ J⟯. The collection (N j | j ∈ J) may be rearranged into a sequence (N n | n ∈ 𝜔). Moreover, we may regard that it consists of pairwise disjoint sets. For each n, consider the measures 𝜇n and 𝜈n on M such that 𝜇n A ≡ 𝜇(A ∩ N n ) and 𝜈n A ≡ 𝜈(A ∩ N n ) for every A ∈ M. Then, 𝜈n ≪ 𝜇n for each n. Proposition 1 (3.2.1) and its corollary 2 guarantee that all the measures 𝜈n are finite, and therefore, bounded. By the same reason 𝜇n M i = 𝜇(N n ∩ M i ) ∈ R for all i and n. Therefore, all the measurable spaces ⟮T, M, 𝜇n ⟯ are 𝜎-finite. By Theorem 2 for every n, there is a function f n ∈ MI(T, M, 𝜇n )+ such that 𝜈n = f n ⋅ 𝜇n . Since (N n | n ∈ 𝜔) is a partition of T we may define a function f ∈ M(T, M)+ setting f |N n ≡ f n |N n for each n. Take any U ∈ M and consider the subsets U n ≡ U ∩N n . Then, by Proposition 2 (3.3.2), ∫ f 𝜒(U n ) d𝜇 = ∫ f |U n d𝜇U n = ∫ f n |U n d𝜇U n . If A ∈ MU n then (𝜇n )U n A = 𝜇n A = 𝜇A = 𝜇U n A. Therefore, 𝜇U n = (𝜇n )U n . As a result, ∫ f 𝜒(U n ) d𝜇 = ∫ f n |U n d(𝜇n )U n = (f n ⋅ 𝜇n )U n = 𝜈n U n ∈ R whereas 𝜈n is a finite measure. In particular, for U = T, we have ∫ f 𝜒(N n ) d𝜇 ∈ R for every n, where f ∈ MI 𝜎 (T, M, 𝜇)+ . Furthermore, 𝜈U = ∑(𝜈U n | n ∈ 𝜔) = ∑(𝜈n U n | n ∈ 𝜔) = ∑(∫ f 𝜒(U n ) d𝜇 | n ∈ 𝜔) = ∑((f ⋅ 𝜇)U n | n ∈ 𝜔) = (f ⋅ 𝜇)U whereas f ⋅ 𝜇 is a measure by virtue of Proposition 2 (3.3.7). Now, we take an arbitrary measure 𝜈. Take some Hahn decomposition (P, N) of T from Theorem 1 (3.2.1) with respect to 𝜈. By Proposition 2 (3.2.1), 𝜈 = v + (𝜈) + v− (𝜈) and by Lemma 5 (3.2.1), v+ (𝜈)A = 𝜈(A ∩ P) and v− (𝜈)A = 𝜈(A ∩ N) for every A ∈ M. Suppose that rng 𝜈 ⊂ R ∪ {∞}. Then, v − (𝜈) is finite, and, therefore, bounded. Since 𝜈N j ∈ R, we conclude that v+ (𝜈)N j ∈ R as well. Consequently, the measurable space (T, M, v+ (𝜈)) is 𝜎-finite. By Corollary 1 to Theorem 1 (3.2.4), v+ (𝜈) ≪ 𝜇 and −v− (𝜈) ≪ 𝜇. It was proven above that v+ (𝜈) = f1 ⋅ 𝜇 for some f1 ∈ MI 𝜎 (T, M, 𝜇)+ . By Theorem 2, −v− (𝜈) = f2 ⋅ 𝜇 for some f2 ∈ MI(T, M, 𝜇)+ . Consider the function f ≡ f1 𝜒(P) − f2 𝜒(N). From the inequalities f+ ⩽ f1 and −f− = (−f ) ∨ 0 ⩽ f2 we infer that f+ ∈ MI 𝜎 (T, M, 𝜇) and −f− ∈ MI(T, M, 𝜇). By the definition, from 3.3.2 this means that f ∈ MI 𝜎 (T, M, 𝜇). If rng 𝜈 ⊂ {−∞} ∪ R, then the arguments are analogous. Therefore, we may consider the measure f ⋅ 𝜇. Then, for every U ∈ M, we get by Corollary 2 to Theorem 1 (3.3.2) that 𝜈U = v+ (𝜈)U + v− (𝜈)U = ∫ f1 𝜒(U) d𝜇 − ∫ f2 𝜒(U) d𝜇 = ∫ f1 𝜒(U)𝜒(P) d𝜇 + ∫ f1 𝜒(U)𝜒(N) d𝜇 − ∫ f2 𝜒(U)𝜒(P) d𝜇 − ∫ f2 𝜒(U)𝜒(N) d𝜇. Since ∫ f1 𝜒(U ∩ N) d𝜇 = (f1 ⋅ 𝜇)(U ∩ N) = v+ (𝜈)(U ∩ N) = 0 and ∫ f2 𝜒(U ∩ P) d𝜇 = −v− (𝜈)(U ∩ P) = 0, we get 𝜈U = ∫ f1 𝜒(P)𝜒(U) d𝜇 − ∫ f2 𝜒(N)𝜒(U) d𝜇. Since f1 ⩾ 0, f2 ⩾ 0, and (P, N) is a partition of T, we conclude that f+ = f1 𝜒(P) and f− = f2 𝜒(N). Finally, according to the definition of the integral from 3.3.2, we obtain 𝜈U = ∫ f+ 𝜒(U) d𝜇 − ∫(−f− )𝜒(U) d𝜇 = ∫(f 𝜒(U))+ d𝜇 − ∫(−(f 𝜒(U))− ) d𝜇 ≡ ∫ f 𝜒(U) d𝜇 = (f ⋅ 𝜇)U, where 𝜈 = f ⋅ 𝜇.

296 | 3.3 The Lebesgue integral

Now, with the help of the Lebesgue – Radon – Nikodym theorem, we shall prove some analogue of Proposition 3 (3.3.5) for non-positive measures. Theorem 5 (the Zakharov theorem on the Lebesgue integral with respect to linear combinations of finite measures). Let 𝜇 and 𝜈 be finite, (and therefore, bounded) measures on a 𝜎-algebra M on a set T and x, y ∈ R. If f ∈ MI(T, M, 𝜇) ∩ MI(T, M, 𝜈), then f ∈ MI(T, M, x𝜇 + y𝜈) and ∫ f d(x𝜇 + y𝜈) = x ∫ f d𝜇 + y ∫ f d𝜈. Proof. Consider the bounded measure 𝜆 ≡ |x||𝜇| + |y||𝜈|. If M ∈ M and 𝜆M = 0, then |𝜇|M = 0 = |𝜈|M. Therefore, by Corollary 2 to Theorem 2 (3.2.2), 𝜇M = 0 = 𝜈M = (x𝜇 + y𝜈)M = 0. Then, Theorem 1 (3.2.4) implies that 𝜇 and 𝜈 are 𝜆-absolutely continuous. Consequently, by virtue of Theorem 2, there are functions 𝜑, 𝜓 ∈ MI(T, M, 𝜆) such that 𝜇 = 𝜑 ⋅ 𝜆 and 𝜈 = 𝜓 ⋅ 𝜆. From the linearity of the Lebesgue integral, we conclude that x𝜇 + y𝜈 = (x𝜑 + y𝜓) ⋅ 𝜆. By Proposition 1 (3.3.6), f is integrable with respect to |𝜇| and |𝜈|. By Proposition 3 (3.3.5), f ∈ MI(T, M, 𝜆). From |x𝜇 + y𝜈| ⩽ 𝜆, we infer by Proposition 1 (3.3.5) that f ∈ MI(T, M, |x𝜇 + y𝜈|). Again by Proposition 1 (3.3.6), f ∈ MI(T, M, x𝜇 + y𝜈). Thus, f is integrable with respect to 𝜑 ⋅ 𝜆, 𝜓 ⋅ 𝜆 and (x𝜑 + y𝜓) ⋅ 𝜆. Then, by Theorem 2 (3.3.7), the functions f 𝜑, f 𝜓 and f (x𝜑 + y𝜓) are integrable with respect to 𝜆 and ∫ f d(𝜑 ⋅ 𝜆) = ∫ f 𝜑 d𝜆, ∫ f d(𝜑 ⋅ 𝜆) = ∫ f 𝜑 d𝜆 and ∫ f d((x𝜑 + y𝜓) ⋅ 𝜆) = ∫ f (x𝜑 + y𝜓) d𝜆. As a result, we obtain ∫ f d(x𝜇 + y𝜈) = ∫ f (x𝜑 + y𝜓) d𝜆 = x ∫ f 𝜑 d𝜆 + y ∫ f 𝜓 d𝜆 = x ∫ f d𝜇 + y ∫ f d𝜈.

3.3.9 Dual to the factor-space of integrable functions Let ⟮T, M, 𝜇⟯ be a measurable space with the positive measure 𝜇 on the 𝜎-algebra M on the set T. In 3.3.1 and 3.3.4 we introduced the Banach lattice-ordered linear spaces ⟮L ∞ (T, M, 𝜇), ‖⋅‖∞ ⟯ and ⟮L1 (T, M, 𝜇), ‖⋅‖1 ⟯. Now, using the Lebesgue – Radon – Nikodym theorem we shall connect them. Lemma 1. Let 𝜇 be a positive measure on a 𝜎-algebra T on T, f 󸀠 ∈ f ∈ L1 (T, M, 𝜇) and h󸀠 ∈ h̄ ∈ L∞ (T, M). Then, 1) hf and h󸀠 f 󸀠 belong to MI(T, M, 𝜇); 2) ∫ hf d𝜇 = ∫ h󸀠 f 󸀠 d𝜇; 3) ∫ hf d𝜇 ⩽ ‖h‖eu ‖f ‖i = ‖h‖̄ ∞ ‖ f ‖1 . Proof. 1. The first assertion follows from Lemma 3 (3.3.4). 2. The second assertion follows from Lemma 4 (3.3.4) and Lemma 5 (3.3.2). 3. By Corollary 3 to Theorem 1 (3.3.2) and Lemma 5 (3.3.4), | ∫ hf d𝜇| ⩽ ∫ |fh| d𝜇 ⩽ ‖h‖eu ‖f ‖i = ‖h‖̄ ∞ ‖ f ‖1 .

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It follows from Lemma 1 that for every element c ∈ L∞ (T, M, 𝜇), we may define correctly a functional 𝜑c : L1 (T, M, 𝜇) → R setting 𝜑c (a) ≡ ∫ hf d𝜇 for any f ∈ a ∈ L1 (T, M, 𝜇) and h ∈ c. Proposition 1. Let 𝜇 be an internally finite (see 3.1.1) positive measure on a 𝜎-algebra M on T and c ∈ L∞ (T, M, 𝜇). Then, 𝜑c is a continuous linear functional on L 1 (T, M, 𝜇) and ‖𝜑c ‖󸀠 ≡ sup{|𝜑c f ̄| | ‖f ̄‖1 ⩽ 1} = ‖c‖∞ (see 7∘ (2.2.7)). Proof. It is evident that 𝜑c is linear. By virtue of Lemma 1, |𝜑c a| ⩽ ‖c‖∞ ‖a‖1 for every a ∈ L1 (T, M, 𝜇). This means that 𝜑c is (norm) bounded. By Statement 12 (2.2.7) 𝜑c is continuous. Besides, ‖𝜑c ‖󸀠 ⩽ x ≡ ‖c‖∞ . Suppose that x > 0 and take any 𝜀 such that 0 < 𝜀 < x. Consider the set E ≡ {t ∈ T | |h(t)| > x − 𝜀}. If 𝜇E = 0, then h󸀠 ≡ h𝜒(T\E) ∈ c and by Lemma 2 (3.3.1), x = ‖h‖eu = ‖h󸀠 ‖eu ⩽ ‖h󸀠 ‖u ⩽ x − 𝜀, but this is impossible. Hence, 𝜇E > 0. By the condition, there is F ⊂ E such that 0 < 𝜇F < ∞. Consider the function f ≡ (1/𝜇F)𝜒(F) sign(h) (see 2.2.4). By Lemmas 1 and 2 (3.3.2), a ≡ f ∈ L1 (T, M, 𝜇), ‖a‖1 ⩽ 1, and 𝜑c a = ∫ fh d𝜇 = (1/𝜇F) ∫ |h|𝜒(F) d𝜇 ⩾ (1/𝜇F) ∫(x − 𝜀)𝜒(F) d𝜇 = x − 𝜀. This implies ‖𝜑c ‖󸀠 ⩾ x − 𝜀, where ‖𝜑c ‖󸀠 ⩾ x. Further, Consider the norm dual ⟮⟮L1 (T, M, 𝜇), ‖⋅‖1 ⟯, ‖⋅‖󸀠 ⟯ of the normed linear space ⟮L1 (T, M, 𝜇), ‖⋅‖1 ⟯ (see 7∘ (2.2.7)). By virtue of Proposition 1 we may consider the mapping W : L∞ (T, M, 𝜇) → L1 (T, M, 𝜇)󸀠 such that W(c) ≡ 𝜑c for every c ∈ L∞ (T, M, 𝜇). Theorem 1. Let ⟮T, M, 𝜇⟯ be a 𝜎-finite measurable space with the positive measure 𝜇 on the 𝜎-algebra M on T. Then, W : ⟮L ∞ (T, M, 𝜇), ‖⋅‖∞ ⟯ → ⟮⟮L1 (T, M, 𝜇), ‖⋅‖1 ⟯, ‖⋅‖󸀠 ⟯ is an isomorphism of the given normed lattice-ordered linear spaces. Proof. Denote L1 (T, M, 𝜇) by A and L∞ (T, M, 𝜇) by C. Let p : MI(T, M, 𝜇) A be the factor-mapping such that pf ≡ f . Take any 𝜑 ∈ A. At first suppose that 𝜇T < ∞. Then, by Lemma 1 (3.3.2), 𝜒(M) ∈ MI(T, M, 𝜇) ∩ M∞ (T, M, 𝜇) for every M ∈ M. According to Lemma 6 (3.3.4), the classes of equivalence of the function 𝜒(M) in MI(T, M, 𝜇) and M∞ (T, M, 𝜇) coincide. Denote this class of equivalence by c(M). Then, c(M) ∈ A ∩ C. Define an evaluation 𝜈 : M → R setting 𝜈M ≡ 𝜑(c(M)). It is a positive finite measure. In fact, let (E n ∈ M | n ∈ 𝜔) be a partition of a set E ∈ M, and let F n ≡ ⋃⟮E i | i ∈ 𝜔\(n + 1)⟯. It is clear that (F n | n ∈ 𝜔) ↓ ⌀. Therefore, by Lemma 6 (3.1.1), lim(𝜇F n | n ∈ 𝜔) = 0. However, by Lemma 1 (3.3.2), |𝜈E − ∑(𝜈E i | i ∈ n + 1)| = |𝜑(c(E)) − 𝜑(∑(c(E i ) | i ∈ n + 1))| = |𝜑(c(F n ))| ⩽ ‖𝜑‖󸀠 ‖c (F n )‖1 = ‖𝜑‖󸀠 ‖𝜒(F n )‖i = ‖𝜑‖󸀠 𝜇F n . This guarantees that 𝜈E = lim(∑(𝜈E i | i ∈ n + 1) | n ∈ 𝜔) = ∑(𝜈E n | n ∈ 𝜔). If 𝜇M = 0, then c(M) = 0̄ and 𝜈M = 0. Therefore, by Theorem 1 (3.2.4), 𝜈 ≪ 𝜇. According to Theorem 2 (3.3.8), there is the unique element c ∈ A such that 𝜈 = c ⋅ 𝜇. Take any h󸀠 ∈ c. By Corollary 1 to Theorem 1 (3.3.7), |𝜈| = |h󸀠 | ⋅ 𝜇.

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We assert that |h󸀠 (t)| ⩽ ‖𝜑‖󸀠 𝜇-almost everywhere. To check this consider the set M ≡ {t ∈ T | |h󸀠 (t)| > ‖𝜑‖󸀠 } and suppose that 𝜇M > 0. By Lemmas 1 and 3 (3.3.2), |𝜈|M = ∫ |h󸀠 |𝜒(M) d𝜇 > ‖𝜑‖󸀠 𝜇M ≡ x. By Lemma 2 (3.1.3), there is a finite partition (M i ∈ M | i ∈ I) of M such that ∑(|𝜈M i | | i ∈ I) > x. Using the definition of 𝜈, we get x < ∑(|𝜑(c(M i ))| | i ∈ I) ⩽ ∑(‖𝜑‖󸀠 ‖𝜒(M i )‖i | i ∈ I) = ‖𝜑‖󸀠 ∑(𝜇M i | i ∈ I) = x. This contradiction shows that our assertion is true. Therefore, h ≡ h󸀠 𝜒(T\M) ∈ c and |h(t)| ⩽ ‖𝜑‖󸀠 for every t ∈ T. Then, by Lemma 6 (3.3.4), c ∈ C. Let s ≡ ∑(x k 𝜒(R k ) | k ∈ K) ∈ St(T, M). Then, 𝜑s̄ = ∑(x k 𝜑(c(R k )) | k ∈ K) = ∑(x k 𝜈 R k | k ∈ K) = ∑(x k ∫ h𝜒(R k ) d𝜇 | k ∈ K) = ∫ hs d𝜇. Now, let f ∈ a ∈ A. By Theorem 2 (2.3.4), there is a sequence (s n ∈ St(T, M) | n ∈ N) such that |s n | ⩽ |s n+1 | ⩽ |f | and f = p-lim (s n | n ∈ N). Then, fh = p-lim(s n h | n ∈ N), 0 = p-lim(|f − s n | | n ∈ N), |s n h| ⩽ ‖𝜑‖󸀠 |f |, and |f − s n | ⩽ 2|f |. By Theorem 4 (3.3.3), lim(‖a − s̄ n ‖1 | n ∈ N) = lim(∫ |f − s n | d𝜇 | n ∈ N) = 0 and lim(∫ s n h d𝜇 | n ∈ N) = ∫ fh d𝜇. Thus, a = lim(s̄ n | n ∈ N) in (A, G(‖⋅‖1 )). Since 𝜑 is continuous, we infer that 𝜑a = lim(𝜑s̄ n | n ∈ N) = lim(∫ hs n d𝜇 | n ∈ N) = ∫ hf d𝜇 = 𝜑c (a), where 𝜑 = 𝜑c ≡ W(c). Now, we consider the general case. By the condition, there is a countable collection (M i ∈ Mf (𝜇) | i ∈ I) such that T = ⋃⟮M i | i ∈ I⟯. This collection may be rearranged into a sequence (M n | n ∈ 𝜔). Moreover, we may regard that it consists of pairwise disjoint sets. For each n, consider the finite positive measure 𝜇n on M such that 𝜇n R ≡ 𝜇(R ∩ M n ) for every R ∈ M. Also consider the normed linear spaces A n ≡ L1 (T, M, 𝜇n ) and C n ≡ L∞ (T, M, 𝜇n ). Denote the norm in A n by ‖⋅‖1n and n . Let f ∈ a ∈ A n . Then, 𝜇n coz(f − f 𝜒(M n )) = 𝜇⌀ = 0 implies by the norm in C n by ‖⋅‖∞ Lemma 5 (3.3.2) that ∫ f d𝜇n = ∫ f 𝜒(M n ) d𝜇n . By Corollary 3 to Proposition 3 (3.3.6), f |M n ∈ MI(M n , MM n , (𝜇n )M n ) and ∫ f 𝜒(M n ) d𝜇n = ∫(f |M n ) d(𝜇n )M n , where MM n ≡ {R ∈ M | R ⊂ M n } and (𝜇n )M n ≡ 𝜇n |MM n . If R ∈ MM n , then (𝜇n )M n R = 𝜇n R = 𝜇R = 𝜇M n R, where 𝜇M n ≡ 𝜇|MM n . This means that (𝜇n )M n = 𝜇M n . Consequently, ∫ f d𝜇n = ∫(f |M n ) d𝜇M n . Now, by Corollary 1 to Proposition 1 (3.3.6), f 𝜒(M n ) ∈ MI(T, M, 𝜇) and by Proposition 1 (3.3.6), ∫(f |M n ) d𝜇M n = ∫ f 𝜒(M n ) d𝜇. Thus, f 𝜒(M n ) ∈ MI(T, M, 𝜇) and ∫ f d𝜇n = ∫ f 𝜒(M n ) d𝜇. Applying this equality to the function |f |, we get ∫ |f | d𝜇n = ∫ |f 𝜒(M n )| d𝜇. If f 󸀠 ∈ a, then 𝜇 coz(f 𝜒(M n ) − f 󸀠 𝜒(M n )) = 𝜇n coz(f − f 󸀠 ) = 0. Therefore, we can define correctly a mapping v n : A n → A setting v n a ≡ (f 𝜒(M n )) for any f ∈ a. Consequently, we can consider the functional 𝜑n : A n → R such that 𝜑n a ≡ 𝜑(v n a). Using the equality of the integrals proven above, we get |𝜑n a| ⩽ ‖𝜑‖󸀠 ‖v n a‖1 = ‖𝜑‖󸀠 ∫ |f 𝜒(M n )| d𝜇 = ‖𝜑‖󸀠 ∫ |f | d𝜇n = ‖𝜑‖󸀠 ‖a‖1n . Therefore, according to the definitions in 7∘ (2.2.7), 𝜑n is norm bounded, 𝜑n ∈ A󸀠n , and ‖𝜑n ‖󸀠 ⩽ ‖𝜑‖󸀠 . As was shown above, for the functional 𝜑n , there is an element c n ∈ C n such that 𝜑n a = ∫ h󸀠n f 󸀠 d𝜇n for every f 󸀠 ∈ a ∈ A n and h󸀠n ∈ c n . Besides, there is h n ∈ c n such that h n (t) ⩽ ‖𝜑n ‖󸀠 for every t ∈ T. Moreover, it was proven above that ∫ h󸀠n f 󸀠 d𝜇n = ∫ h󸀠n f 󸀠 𝜒(M n ) d𝜇 whereas by Lemma 1, h󸀠n f 󸀠 ∈ MI(T, M, 𝜇n ). Since 𝜇n ⩽ 𝜇, by Proposition 1 (3.3.5) we see that MI(T, M, 𝜇) ⊂ MI(T, M, 𝜇n ). Let p n : MI(T, M, 𝜇n ) A n be the factor-mapping. If g󸀠 ∈ ḡ ∈ A, then 𝜇n coz(g󸀠 − g) ⩽

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𝜇 coz(g󸀠 − g) = 0. Hence, p n g󸀠 = p n g. Therefore, we can define correctly a mapping w n : A → A n setting w n ḡ ≡ p n g󸀠 for any g󸀠 ∈ g.̄ Then, 𝜑n w n ḡ = ∫ h󸀠n g󸀠 d𝜇n = ∫ h󸀠n g󸀠 𝜒(M n ) d𝜇 for any g󸀠 ∈ g.̄ Consider the function h on T such that h|M n ≡ h n |M n . Evidently, h ∈ M(T, M). If t ∈ T, then t ∈ M n for some n. Therefore, |h(t)| = |h n (t)| ⩽ ‖𝜑n ‖󸀠 ⩽ ‖𝜑‖󸀠 . Thus, h ∈ M b (T, M) and ‖h‖u ⩽ ‖𝜑‖󸀠 . Let c ≡ h̄ ∈ C. Take any g ∈ b ∈ A. By Corollary 3 to Proposition 3 (3.3.6), g n ≡ hg𝜒(M n ) ∈ MI(T, M, 𝜇). Since gh = p-lim(∑(g n | n ∈ k + 1) | k ∈ 𝜔) and | ∑(g n | n ∈ k + 1)| ⩽ |gh| ∈ MI(T, M, 𝜇)+ , we conclude by virtue of Theorem 4 (3.3.3) that ∫ hg d𝜇 = lim(∫ ∑(g n | n ∈ k + 1) d𝜇 | k ∈ 𝜔) = lim (∑ (∫ h n g𝜒(M n ) d𝜇 | n ∈ k + 1) | k ∈ 𝜔). As was shown above, hg, g ∈ MI(T, M, 𝜇n ) and ∫ h n g𝜒(M n ) d𝜇 = ∫ h n g d𝜇n = 𝜑n w n b = 𝜑(v n w n b) = 𝜑(p(g𝜒(M n ))). Consequently, ∫ hg d𝜇 = lim(∑(𝜑(p(g𝜒(M n )) | n ∈ k + 1) | k ∈ 𝜔)) = lim(𝜑(p(∑(g𝜒(M n ) | n ∈ k + 1) | k ∈ 𝜔))). Again by Theorem 4 (3.3.3), lim(‖b − p(∑(g𝜒(M n ) | n ∈ k + 1))‖1 | k ∈ 𝜔) = lim(∫ |g − g ∑(𝜒(M n | n ∈ k + 1))|d𝜇 | k ∈ 𝜔) = lim(∫ |g|𝜒(T\ ⋃⟮M n | n ∈ k + 1⟯) d𝜇 | k ∈ 𝜔) = 0, where we get lim(p(∑(g𝜒M n ) | n ∈ k + 1) | k ∈ 𝜔) = b in ⟮A, G(‖⋅‖1 )⟯ by virtue of Corollary 1 to Statement 6 (2.2.7). Since by Statement 12 (2.2.7), 𝜑 is continuous, we get ∫ hg d𝜇 = 𝜑b. As a result, 𝜑b = ∫ hg d𝜇 = 𝜑c b for every b ∈ A, where 𝜑 = 𝜑c . Thus, W is surjective. By Proposition 1, W is isometric. Let f ∈ a ∈ A, (h i ∈ c i ∈ C | i ∈ n ∈ N), (x i ∈ R | i ∈ n), c ≡ ∑(x i c i | i ∈ n) ∈ C and h ≡ ∑(x i h i | i ∈ n) ∈ c. By the definition, 𝜑c i (a) = ∫ h i f d𝜇 and 𝜑c (a) = ∫ hf d𝜇. From the linearity of the integral we get 𝜑c (a) = ∑(x i ∫ h i f d𝜇 | i ∈ n) = ∑(x i 𝜑c i (a) | i ∈ n). This means that Wc = ∑(x i Wc i | i ∈ n), i. e. W is a linear operator. It follows from the established properties that W is injective. Thus, according to Statements 3 and 5 (2.2.7), W is an isomorphism of linear spaces. Let c ∈ C+ and a ∈ A + . Then, there are positive functions h ∈ c and f ∈ a. From 𝜑c (a) = ∫ hf d𝜇 ⩾ 0 we conclude that Wc ⩾ 0. Hence, W is positive and therefore increasing. Let b, c ∈ C and Wb ⩽ Wc. This means that 𝜑b (a) ⩽ 𝜑c (a) for every a ∈ A+ . Suppose that b ⩽̸ c, i. e. b ∨ c = / c. Take any functions g ∈ b and h ∈ c. Then, g ∨ h ∉ c. Consider the function u ≡ g ∨ h − h ∈ M∞ (T, M, 𝜇)+ . From 𝜇 coz u > 0, we conclude by virtue of Lemma 11 (3.1.1) that there is E ∈ Mf (𝜇) such that E ⊂ coz u and 𝜇E > 0. Consider the function f ≡ 𝜒(E). By Lemma 1 (3.3.2), f ∈ MI(T, M, 𝜇)+ . Therefore, 0 < a ≡ f ∈ A. From coz(uf ) = E we conclude by Lemma 3 (3.3.2) that ∫ uf d𝜇 > 0. Then, uf = (g − h)f implies 𝜑b a = ∫ gf d𝜇 > ∫ hf d𝜇 = 𝜑c a ⩾ 𝜑b a, but this is impossible. It follows from this contradiction that b ⩽ c. Thus, W is isotone (see 1.1.15). By Lemma 2 (1.1.15) the mapping W preserves any exact bounds. In particular, W is a homomorphism of lattice-ordered linear spaces (see 5∘ (2.2.7)). Since W is bijective and isometric, W is a necessary isomorphism by virtue of Statement 5 (2.2.7).

300 | 3.4 Representation of a functional by the Lebesgue integral

3.4 Representation of a functional by the Lebesgue integral In this chapter, we shall consider the connection between functionals and measures having its roots in the famous Riesz representation theorem (see Appendix D) for ⟮C([0, 1]), ‖ ⋅ ‖u ⟯󸀠 = ⟮C([0, 1]), ⩽⟯∼ = ⟮C([0, 1]), ⩽⟯† (see Corollary 1 to Lemma 3 (2.2.8)).

3.4.1 Regularity and continuity of evaluations. The key theorem for integral representations Let T be a set, S be an ensemble on T, and K and L be subensembles of S. An increasing evaluation 𝜀 : S → R is called inner K-regular if 𝜀S = sup{𝜀K | (K ∈ K ∩ Sf (𝜀)) ∧ (K ⊂ S)} for every S ∈ S. It is called outer L-regular if 𝜀S = inf{𝜀L | L ∈ L ∧ S ⊂ L} for every S ∈ S. An increasing evaluation 𝜀 : S → R is called upper K-𝜎-continuous if we have (𝜀K n | n ∈ N) ↓ 𝜀K for every set K ∈ K and every infinite sequence (K n ∈ K | n ∈ N ⊂ 𝜔) ↓ K in P(T). It is called upper K-continuous if(𝜀K𝛾 | 𝛾 ∈ Γ) ↓ 𝜀K for every set K ∈ K and every net (K𝛾 ∈ K | 𝛾 ∈ Γ) ↓ K in P(T). An increasing evaluation 𝜀 : S → R is called lower L-𝜎-continuous if (𝜀L n | n ∈ N) ↑ 𝜀L for every set L ∈ L and every infinite sequence (L n ∈ L | n ∈ N ⊂ 𝜔) ↑ L in P(T). It is called lower L-continuous if (𝜀L𝛾 | 𝛾 ∈ Γ) ↑ 𝜀L for every set L ∈ L and every net (L𝛾 ∈ L | 𝛾 ∈ Γ) ↑ L in P(T). Let E be a subensemble of the ensemble S. An arbitrary evaluation 𝜀 : S → R will be called (inner) E-regular if it has the following properties: 1) for every set S ∈ S such that 𝜀S ∈ R and every real number 𝛿 > 0, there is a set E ∈ Ef (𝜀) ≡ E ∩ Sf (𝜀) such that E ⊂ S and |𝜀S − 𝜀E| < 𝛿; 2) for every set S ∈ S such that 𝜀S = ∞ [𝜀S = −∞] and every real number 𝛿 > 0, there is a set E ∈ Ef (𝜀) such that E ⊂ S and 𝜀E > 𝛿 [𝜀E < 𝛿]. For an increasing evaluation 𝜀, this definition of inner regularity is equivalent to the previous one. Lemma 1. Let K and G be some ensembles on T such that K\G ∈ K for every K ∈ K and G ∈ G. Let 𝜇 : A → R+ be a positive inner K-regular and upper K-continuous semimeasure on an algebra A, containing K and G. Then, 𝜇 is lower G-continuous. Proof. Let G ∈ G ⊂ A and (G𝛾 ∈ G | 𝛾 ∈ Γ) ↑ G in P(T). Take any K ∈ K ∩ Af (𝜇) such that K ⊂ G. Then, K𝛾 ≡ K\G𝛾 ∈ K and (K𝛾 | 𝛾 ∈ Γ) ↓ ⌀ in P(T). Therefore, (𝜇K𝛾 | 𝛾 ∈ Γ) ↓ 0. Consequently, 𝜇K = 𝜇K − 0 = 𝜇K − inf (𝜇K𝛾 | 𝛾 ∈ Γ) = sup (𝜇K − 𝜇K𝛾 | 𝛾 ∈ Γ) = sup (𝜇(K ∩ G𝛾 ) | 𝛾 ∈ Γ) ⩽ sup (𝜇G𝛾 | 𝛾 ∈ Γ) ≡ x.Finally,𝜇G = sup (𝜇K | K ∈ K ∩ Af (𝜇) ∧ K ⊂ G) ⩽ x ⩽ 𝜇G.

3.4.1 Regularity and continuity of evaluations | 301

The following theorem is the key result for all the further studies in integral representations in this chapter and in 3.6. It is taken from the book [Fremlin, 1974] (see also [Topsøe, 1970]). Theorem 1. Let K be an additive and 𝛿-multiplicative ensemble on a set T and 𝜆 : K → R+ be an increasing overadditive natural evaluation on K (see 3.1.1) such that 1) 𝜆L ⩾ 𝜆H + sup{𝜆K | K ∈ K ∧ K ⊂ L\H} for every H, L ∈ K such that H ⊂ L; 2) (𝜆K n | n ∈ 𝜔) ↓ 0 for every sequence (K n ∈ K | n ∈ 𝜔) ↓ ⌀ in P(T). Let 𝜈 : P(T) → R+ be an evaluation such that 𝜈E ≡ sup{𝜆K | K ∈ K ∧ K ⊂ E}, M ≡ {M ⊂ T | ∀ L ∈ K (𝜆L ⩽ 𝜈(L ∩ M) + 𝜈(L\M))}, and 𝜇 ≡ 𝜈|M. Then, M is a 𝜎-algebra and 𝜇 : M → R+ is a complete, strongly saturated (see 3.1.4), inner K-regular, and K-saturated (see 2.1.1) measure extending the evaluation 𝜆. Proof. Since 𝜆 is increasing, we conclude that 𝜈 is an extension of 𝜆 and 𝜈 is increasing. If H ∈ K, then by the definition 𝜆L ⩽ 𝜆(L ∩ H) + sup{𝜆K | K ∈ K ∧ K ⊂ L\H} = 𝜈(L ∩ H) + 𝜈(L\H) for every L ∈ K. This means that H ∈ M. Thus, K ⊂ M and 𝜇 is an extension of 𝜆. If A and B are disjoint subsets of T, then by Lemma 4 (1.4.5) and Corollary 1 to Proposition 2 (1.1.15), we have 𝜈A + 𝜈B = sup{sup{𝜆H + 𝜆K | H ∈ K ∧ H ⊂ A} | K ∈ K ∧ K ⊂ B} ⩽ sup{𝜆(H ∪ K) | (H, K) ∈ K × K ∧ H ⊂ A ∧ K ⊂ B} ⩽ 𝜈(A ∪ B). If M ∈ M and A ⊂ T, then 𝜈A = sup{𝜆K | K ∈ K ∧ K ⊂ A} ⩽ sup{𝜈(K ∩ M) + 𝜈(K\M) | K ∈ K ∧ K ⊂ A} ⩽ 𝜈(A ∩ M) + 𝜈(A\M) because 𝜈 is increasing. On the other hand, as was shown above, 𝜈(A ∩ M) + 𝜈(A\M) ⩽ 𝜈A. Thus, 𝜈A = 𝜈(A ∩ M) + 𝜈(A\M). It follows immediately from the definition that T\M ∈ M for every M ∈ M and T ∈ M. Let M, N ∈ M. If L ∈ K, then 𝜆L ⩽ 𝜈(L ∩ N) + 𝜈(L\N). Using the property proven above, for M and A = L\N, we get 𝜈(L\N) = 𝜈((L\N) ∩ M) + 𝜈((L\N)\M). Similarly, for N and A = L ∩ (M ∪ N), we get 𝜈(L ∩ (M ∪ N)) = 𝜈(L ∩ N) + 𝜈((L\N) ∩ M) because L ∩ (M ∪ N)\N = (L\N) ∩ M. Hence, 𝜆L ⩽ 𝜈(L ∩ (M ∪ N)) + 𝜈(L\(N ∪ M)). This means that M ∪ N ∈ M, and therefore, M is an algebra. Moreover, if M and N are disjoint then for M and A = M∪N, then we get 𝜇(M∪N) = 𝜈((M ∪ N) ∩ M) + 𝜈((M ∪ N)\M) = 𝜇M + 𝜇N. This means that 𝜇 is (binary) additive. By Lemma 1 (3.1.1), 𝜇 is a positive semimeasure. Then, by Lemma 9 (3.1.1), Mf (𝜇) is a ring containing K, and 𝜇 is a finite positive semimeasure on Mf (𝜇). Now, let (M n ∈ M | n ∈ 𝜔) ↑ and M ≡ ⋃⟮M n | n ∈ 𝜔⟯. Take any L ∈ K and 𝜀 > 0. Then, L\M n ∈ Mf (𝜇). From 𝜇(L\M n ) = sup{𝜆K | K ∈ K ∧ K ⊂ L\M n }, we infer that Kn ≡ {K ∈ K | K ⊂ L\M n ∧ 𝜆K > 𝜇(L\M n ) − 2−n 𝜀} = / ⌀. Using the choice mapping p : P(K)\{⌀} → K from the axiom of choice, we choose the sets K n ≡ pKn ∈ Kn . Since K n ∪ K n+1 ⊂ L\M n , we conclude that 𝜆(K n ∪ K n+1 ) ⩽ 𝜇(L\M n ) < 𝜆K n + 2−n 𝜀. Therefore, 𝜇(K n+1 \K n ) = 𝜆(K n ∪ K n+1 ) − 𝜆K n ⩽ 2−n 𝜀.

302 | 3.4 Representation of a functional by the Lebesgue integral

Consider the sets H n ≡ ⋂⟮K i | i ∈ n + 1⟯ ∈ K. Now, we shall prove by the natural induction that K n+1 \H n ⊂ ⋃⟮K i+1 \K i | i ∈ n + 1⟯ for every n ∈ 𝜔. It is clear that for n = 0, it is valid. Suppose that we have this inclusion for n. Then, K n+2 \H n+1 = K n+2 \(K n+1 ∩ H n ) = (K n+2 \K n+1 ) ∪ (K n+2 \H n ) and K n+2 \H n = ((K n+2 \K n+1 ) ∪ (K n+2 ∩ K n+1 ))\H n ⊂ (K n+2 \K n+1 ) ∪ (K n+1 \H n ) imply K n+2 \H n+1 ⊂ (K n+2 \K n+1 ) ∪ ⋃⟮K i+1 \K i | i ∈ n + 1⟯ = ⋃⟮K i+1 \K i | i ∈ n + 2⟯. Thus, we have the same inclusion for n + 1. By Theorem 1 (1.2.6), our assertion is valid for every n ∈ 𝜔. The proven inclusion and Lemma 3 (1.4.8) guarantee that 𝜇(K n+1 \H n ) ⩽ ∑(𝜇(K i+1 \K i ) | i ∈ n + 1) ⩽ 𝜀 ∑(2−i | i ∈ 𝜔) = 2𝜀. Consequently, from K n+1 = H n+1 ∪ (K n+1 \H n ), we get 𝜆H n+1 = 𝜆K n+1 − 𝜇(K n+1 \H n ) ⩾ 𝜆K n+1 − 2𝜀 ⩾ 𝜇(L\M n+1 ) − 2−n−1 𝜀 − 2𝜀 ⩾ 𝜇(L\M n+1 ) − 3𝜀. Consider now the set H ≡ ⋂⟮H n | n ∈ 𝜔⟯ ⊂ L\M. Then, we have x ≡ 𝜇(L ∩ M) + 𝜇(L\M) ⩾ 𝜇(L ∩ M n ) + 𝜆H for every n. Hence, x ⩾ sup (𝜇(L ∩ M n ) | n ∈ 𝜔) + 𝜆H. If K ∈ K and K ⊂ L\H, then (K ∩ H n | n ∈ 𝜔) ↓ ⌀. Therefore, by property 2 from the conditions of the theorem (𝜆(K ∩ H n ) | n ∈ 𝜔) ↓ 0. Then from the equality 𝜆K = 𝜆(K ∩ H n ) + 𝜇(K\H n ) we conclude that 𝜆K = sup{𝜇(K\H n ) | n ∈ 𝜔} ⩽ sup{𝜇(L\H n ) | n ∈ 𝜔} ≡ y. By the definition of 𝜈, we get 𝜆L − 𝜆H = 𝜇(L\H) = sup{𝜆K | K ∈ K ∧ K ⊂ L\H} ⩽ y = sup{𝜆L − 𝜆H n | n ∈ 𝜔} = 𝜆L − inf{𝜆H n | n ∈ 𝜔}, where 𝜆H ⩾ inf{𝜆H n | n ∈ 𝜔}. Using the estimation proven at the end of the previous indentation, we get 𝜆H ⩾ inf (𝜇(L\M n ) | n ∈ 𝜔) − 3𝜀. Thus, by Lemma 7 (1.4.7) and Proposition 1 (1.4.7), x ⩾ sup (𝜇(L ∩ M n ) | n ∈ 𝜔) + inf (𝜇(L\M n ) | n ∈ 𝜔) − 3𝜀 = lim(𝜇(L ∩ M n ) + 𝜇(L\M n ) | n ∈ 𝜔) − 3𝜀 = 𝜆L − 3𝜀. Since 𝜀 and L are arbitrary, we get 𝜆L ⩽ x, where M ∈ M. This means that M is a 𝜎-algebra. Take now in the previous arguments L ∈ K such that L ⊂ M. Then, H = ⌀ and 0 = 𝜆H ⩾ inf (𝜇(L\M n ) | n ∈ 𝜔) − 3𝜀, where inf (𝜇(L\M n ) | n ∈ 𝜔) ⩽ 3𝜀. Since 𝜀 is arbitrary, we get inf (𝜇(L\M n ) | n ∈ 𝜔) = 0. Then, from 𝜆L = 𝜇(L ∩ M n ) + 𝜇(L\M n ), we deduce 𝜆L − sup (𝜇(L ∩ M n ) | n ∈ 𝜔) = inf (𝜆L − 𝜇(L ∩ M n ) | n ∈ 𝜔) ⩽ 0, where 𝜆L ⩽ sup (𝜇(L ∩ M n ) | n ∈ 𝜔) ⩽ sup (𝜇M n | n ∈ 𝜔). Since L is arbitrary, we get 𝜇M ⩽ sup (𝜇M n | n ∈ 𝜔) ⩽ 𝜇M. By virtue of Lemma 7 (1.4.7) and Lemma 5 (3.1.1), this means that 𝜇 is a measure. If M ∈ M and 𝜇M = ∞, then there is K ∈ K such that K ⊂ M and 0 < 𝜆K < ∞. Thus, 𝜇 is internally finite. If N ⊂ M ∈ M0 (𝜇) then for any L ∈ K, we have 𝜈(L ∩ N) + 𝜈(L\N) ⩾ 𝜈(L\M) = 𝜆L − 𝜇(L ∩ M) ⩾ 𝜆L − 𝜇M = 𝜆L, where N ∈ M. Thus, 𝜇 is locally complete. By Corollary 2 to Lemma 4 (3.1.4), it is complete. Finally, if E ⊂ T and E∩K ∈ M for every K ∈ K, then K\E = K\(E∩K) ∈ M implies 𝜈(K ∩ E) + 𝜈(K\E) = 𝜇(K ∩ E) + 𝜇(K\E) = 𝜆K, where E ∈ M. Thus, we have the property of K-saturation. Since K ⊂ Mf (𝜇), this also means that 𝜇 is strongly saturated. Denote the family of all evaluations 𝜆 : K → R+ on T, satisfying the conditions of Theorem 1 by L(T, K)0 . According to Theorem 1, we can correctly define a mapping e(K) : L(T, K)0 → Mw (T)0 (see 3.1.2), setting e(K)𝜆 ≡ 𝜇. The mapping e(K) will be called extending.

3.4.2 Representation of pointwise 𝜎-continuous functionals by Lebesgue integrals |

303

3.4.2 Representation of pointwise 𝜎-continuous functionals by Lebesgue integrals. The solution of the problem of characterization of Lebesgue integrals as linear functionals Let A(T) be a lattice-ordered linear space of functions on T and 𝜑 be a linear functional on A(T). Let ⟮T, M, 𝜇⟯ be a measurable space with the measure 𝜇 on the 𝜎-algebra M on T. The functional 𝜑 will be called representable by the Lebesgue integral over the measurable space ⟮T, M, 𝜇⟯ if A(T) ⊂ MI(T, M, 𝜇) and 𝜑 = Λ(𝜇)|A(T). It follows from Theorem 1 (3.3.6) that such a functional is necessarily pointwise 𝜎-continuous. Our aim is to prove the inverse result for truncatable lattice-ordered linear spaces A(T). The Alexandrov – Stone – Fremlin integral representation theorem The following result is taken from the book [Fremlin, 1974]. Proposition 1. Let 𝜇 be a positive measure on a 𝜎-algebra M on T and A(T) be a truncatable lattice-ordered linear space of functions on T such that Sf ≡ {t ∈ T | f (t) ⩾ 1} ∈ Mf (𝜇) for every f ∈ A(T). Let 𝜑 be a pointwise 𝜎-continuous positive linear functional on A(T) such that 𝜇(Sf ) = inf{𝜑g | g ∈ A(T)+ ∧ g ⩾ 𝜒(Sf )}. Then, 𝜑 is representable by the Lebesgue integral over the measurable space ⟮T, M, 𝜇⟯. Proof. We shall denote the integral Λ(𝜇) simply by Λ and Mf (𝜇) by Mf . Let f ∈ A(T)+ . For i, n, r ∈ N, consider the sets M ni ≡ {t ∈ T | f (t) ⩾ i2−n } = {t ∈ T | 2n f (t)/i ⩾ 1} ∈ Mf and the functions f nr ≡ ∑(2−n 𝜒(M ni ) | i ⩽ r) ∈ M(T, M)+ . By Lemma 1 (3.3.2), Λf nr = ∑(2−n 𝜇M ni | i ⩽ r) < ∞, and therefore, f nr ∈ MI(T, M, 𝜇)+ . We have f nr (t) = 0 for t ∉ M n1 , f nr (t) = i2−n for t ∈ M ni \M n(i+1) and i < r, and f nr (t) = r2−n for t ∈ M nr . It is clear that f nr ⩽ f and (f nr | r ∈ N) ↑. Since 2−n 𝜒(M ni ) ⩽ f ∧ i2−n 1 − f ∧ (i − 1)2−n 1, we conclude from the condition that −n 2 𝜇M ni ⩽ 𝜑(f ∧ i2−n 1) − 𝜑(f ∧ (i − 1)2−n 1). Therefore, Λf nr ⩽ ∑(𝜑(f ∧ i2−n 1) − 𝜑(f ∧ (i − 1)2−n 1 | i ⩽ r)) = 𝜑(f ∧ r2−n 1) ⩽ 𝜑f < ∞. According to Proposition 1 (2.3.3), we may consider the functions f n ≡ sup (f nr | r ∈ N) from M(T, M)+ . By virtue of Theorem 2 (3.3.3) and Corollary 1 to Lemma 5 (3.3.2), f n ∈ MI(T, M, 𝜇) and Λf n = sup (Λf nr | r ∈ N) ⩽ 𝜑f . Let t ∈ T be a point such that f (t) ⩾ 2−n . Then, t ∈ M ni \M n i+1 for some i. Therefore, f n (t) = f nr (t) = i2−n for every r and f (t) − f n (t) < 2−n . For t, we have either t ∈ M(n+1)2i \M(n+1)(2i+1) or t ∈ M(n+1)(2i+1) \M(n+1)(2i+2) . In the first case, f n+1 (t) = 2i ⋅ 2−n−1 = f n (t). In the second case, f n+1 (t) = (2i + 1)2−n−1 > f n (t). If t is a point such that f (t) < 2−n , then f n (t) = f n1 (t) = 0 ⩽ f n+1 (t) and f (t) − f n (t) < 2−n . Consequently, (f n (t) | n ∈ N) ↑ f (t) for every t ∈ T. Proposition 1 (2.3.3) implies that f ∈ M(T, M). Again by Theorem 2 (3.3.3) and Corollary 1 to Lemma 5 (3.3.2), f ∈ MI(T, M, 𝜇)+ and Λf = sup (Λf n | n ∈ N) ⩽ 𝜑f . Thus, A(T) ⊂ MI(T, M, 𝜇) and Λf ⩽ 𝜑f for every f ∈ A(T)+ .

304 | 3.4 Representation of a functional by the Lebesgue integral

Further, consider the functions g n ≡ f ∧ 2n 1 − f ∧ 2−n 1 ∈ A(T) ∩ MI(T, M, 𝜇). We have (g n | n ∈ N) ↑ f (pointwise). By Lemma 1 (2.2.8), (𝜑g n | n ∈ N) ↑ 𝜑f . Therefore, for any 𝜀 > 0, there is p ∈ N such that 0 ⩽ 𝜑f − 𝜑g n < 𝜀 for every n ⩾ p. Consider the set S ≡ {t ∈ T | f (t) ⩾ 2−n } ∈ Mf (𝜇) and the function g ≡ 2n 𝜒(S) ∈ MI(T, M, 𝜇)+ . By the condition, 𝜇S = inf{𝜑h | h ∈ A(T)+ ∧ h ⩾ 𝜒(S)}. Therefore, there is h ∈ A(T)+ such that h ⩾ 𝜒(S) and 𝜇S + 𝜀2−n ⩾ 𝜑h. From h n ≡ 2n h − g n ∈ A(T)+ , we conclude that h n ∈ MI(T, M, 𝜇) and Λh n ⩽ 𝜑h n . Consequently, by Lemma 1 (3.3.2), we get 𝜑f ⩽ 𝜑g n + 𝜀 = 𝜀 + 𝜑(2n h) − 𝜑h n ⩽ 𝜀 + 2n 𝜑h − Λh n ⩽ 𝜀 + 2n 𝜇S + 𝜀 − Λh n ⩽ 2𝜀 + 2n 𝜇S − Λg + Λg n = 2𝜀 + 2n 𝜇S − 2n 𝜇S + Λg n ⩽ 2𝜀 + Λf . Since 𝜀 is arbitrary, we infer that 𝜑f ⩽ Λf . As a result, 𝜑f = Λf . Clearly now that 𝜑f = Λf for every f ∈ A(T). Now, we can prove the first result about representations of functionals by Lebesgue integrals. This result may be called the Alexandrov – Stone – Fremlin integral representation theorem, because it has the origin in the papers [Alexandrov, 1940; 1941; 1943] and [Stone, 1948a; 1948b; 1948c; 1949]. M. H. Stone showed the importance of the pointwise 𝜎-continuity for the existence of a measure 𝜇. A. D. Alexandrov showed the importance of the inner regularity for the uniqueness of 𝜇. D. H. Fremlin invented some powerful and general techniques for the measure construction. Remind that the definitions of the ensembles I(A(T)) and I𝜎 (A(T)) are given in 2.2.9. Theorem 1. Let A(T) be a truncatable lattice-ordered linear space of functions on a set T and 𝜑 ∈ (A(T)∧ )+ . Define on the ensemble K ≡ I𝜎 (A(T)), an evaluation 𝜆 : K → R+ setting 𝜆K ≡ inf{𝜑f | f ∈ A(T)+ ∧ f ⩾ 𝜒(K)}. Then, 1) 𝜆 ∈ L(T, K)0 (see 3.4.1); 2) the measure 𝜇 ≡ e(K)𝜆 : M → R+ from 3.4.1 is positive, wide, complete, strongly saturated, K-saturated, inner K-regular, and upper K-𝜎-continuous; 3) 𝜑 is representable by the Lebesgue integral over the measurable space ⟮T, M, 𝜇⟯; 4) the measure 𝜇 is unique in the following sense: if there is another positive, wide, complete, strongly saturated and inner K-regular measure 𝜈 : N → R+ such that 𝜑 is representable by the Lebesgue integral over the space ⟮T, N, 𝜈⟯, then N = M and 𝜈 = 𝜇; 5) if 𝜑 is uniformly bounded (see 2.2.9), then 𝜇 is bounded. Proof. According to the definition in 2.2.9, for every K ∈ K, there exists a decreasing sequence (f n ∈ A(T)+ | n ∈ 𝜔) such that (f n | n ∈ 𝜔) ↓ 𝜒(K) in F(T). If A(T) = / {0}, then by Lemma 1 (2.2.9) K = / {⌀}. By Proposition 1 (2.2.9), K is additive and 𝛿multiplicative. Let (f n | n ∈ 𝜔) ↓ 𝜒(K). For every n ∈ 𝜔, there is g n ∈ A(T) such that g n ⩾ 𝜒(K) and 𝜆K + 1/(n + 1) > 𝜑g n . Consider the functions h n ≡ f n ∧ inf (g i | i ∈ n + 1) ∈ A(T). Then, h n ⩽ g n , (h n | n ∈ 𝜔) ↓ 𝜒(K), and 𝜆K ⩽ 𝜑h n ⩽ 𝜑g n < 𝜆K + 1/(n + 1). By Lemma 6 (1.4.7),

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we get 𝜆K = lim (𝜑h n | n ∈ 𝜔). Consider the functions w n ≡ f n − h n . By virtue of Proposition 1 (1.4.7), p-lim(w n | n ∈ 𝜔) = 0. Since 𝜑 is pointwise 𝜎-continuous, we have lim(𝜑w n | n ∈ 𝜔) = 0. Then, lim (𝜑f n | n ∈ 𝜔) = lim (𝜑h n | n ∈ 𝜔) = 𝜆K. Thus, using Lemma 7 (1.4.7), we obtain that (f n | n ∈ 𝜔) ↓ 𝜒(K) implies (𝜑f n | n ∈ 𝜔) ↓ 𝜆K. It is evident that 𝜆 is natural and increasing. Let K, L ∈ K, (f n | n ∈ 𝜔) ↓ 𝜒(K), (g n | n ∈ 𝜔) ↓ 𝜒(L), and K ∩ L = ⌀. Take any function f ∈ A(T)+ such that f ⩾ 𝜒(K ∪ L). Consider the functions h n ≡ f ∧ f n ∧ g n . Since (h n | n ∈ 𝜔) ↓ 0 in F(T), then by Lemma 1 (2.2.8), (𝜑h n | n ∈ 𝜔) ↓ 0. Lemma 5 (1.4.5) implies that p ∨ q + p ∧ q = p + q for any p, q ∈ F(T). Take p ≡ f ∧ f n and q ≡ f ∧ g n . Then, f ⩾ p ∨ q = p + q − p ∧ q = p + q − h n guarantees that 𝜑f ⩾ 𝜑(f ∧ f n ) + 𝜑(f ∧ g n ) − 𝜑h n ⩾ 𝜆K + 𝜆L − 𝜑h n . As a result, we get 𝜑f ⩾ 𝜆H+𝜆K. Since f is arbitrary, we conclude that 𝜆(K∪L) ⩾ 𝜆K+𝜆L, i. e., 𝜆 is (binary) overadditive. Let now H, L ∈ K, (f n | n ∈ 𝜔) ↓ 𝜒(L), and H ⊂ L. For 𝜀 > 0, there is g ∈ A(T)+ such that g ⩾ 𝜒(H) and 𝜆H+𝜀 > 𝜑g. Consider the set K ≡ {t ∈ L | g(t) ⩽ 1/(1+𝜀)} ⊂ L\H. If t ∈ K, then f n (t) − g(t) ⩾ 𝜀/(1 + 𝜀) for every n. Conversely, if t ∈ L and f n (t) − g(t) ⩾ 𝜀/(1 + 𝜀) for every n, then g(t) ⩽ inf{f n (t) − 𝜀/(1 + 𝜀) | n ∈ 𝜔} = 1/(1 + 𝜀) means t ∈ K. Consequently, K = {t ∈ L | ∀n ∈ 𝜔 (f n (t) − g(t) ⩾ 𝜀/(1 + 𝜀))} = L ∩ ⋂⟮{t ∈ T | (1 + 1/𝜀)(f n − g)(t) ⩾ 1} | n ∈ 𝜔⟯. By Lemma 1 (2.2.9), K n ≡ {t ∈ T | (1 + 1/𝜀)(f n − g)(t) ⩾ 1} ∈ K for every n. Hence, K = L ∩ ⋂⟮K n | n ∈ 𝜔⟯ ∈ K. Let h ∈ A(T)+ and h ⩾ 𝜒(K). Then, h + (1 + 𝜀)g ⩾ 𝜒(L) implies 𝜆L ⩽ 𝜑h + (1 + 𝜀)𝜑g. Since h is arbitrary, we infer that 𝜆L − (1 + 𝜀)𝜑g ⩽ 𝜆K, where 𝜆L − 𝜆K ⩽ (1 + 𝜀)(𝜆H + 𝜀) = 𝜆H + 𝜀(1 + 𝜆H) + 𝜀. Consequently, 𝜆L − 𝜆H − 𝜀(1 + 𝜆H) + 𝜀2 ⩽ sup{𝜆K | K ∈ K ∧ K ⊂ L\H} ≡ x. Since 𝜀 is arbitrary, we conclude that 𝜆L ⩽ 𝜆H + x. Thus, 𝜆 has property 1 from the condition of Theorem 1 (3.4.1). Finally, let K ∈ K and (K m ∈ K | m ∈ 𝜔) ↓ K in P(T) and (f mn | n ∈ 𝜔) ↓ 𝜒(K m ). Consider the functions f n ≡ inf (f ij | (i, j) ∈ (n + 1) × (n + 1)) ∈ A(T). Then, by Corollary 1 to Proposition 2 (1.1.15), 𝜒(K) = inf (𝜒(K m ) | m ∈ 𝜔) = inf (f mn | (m, n) ∈ 𝜔 × 𝜔) = inf (f n | n ∈ 𝜔). As was shown before, (𝜑f n | n ∈ 𝜔) ↓ 𝜆K. But f n ⩾ inf (f ij | (i, j) ∈ (n + 1) × 𝜔) = inf (𝜒(K i ) | i ∈ n + 1) = 𝜒(K n ) implies 𝜆K ⩽ 𝜆K n ⩽ 𝜑f n . Hence, (𝜆K n | n ∈ 𝜔) ↓ 𝜆K. Thus, 𝜆 is upper K-𝜎-continuous. Therefore, for 𝜆, we may consider the corresponding measure 𝜇 : M → R+ from Theorem 1 (3.4.1). By Lemma 1 (2.2.9), Sf ∈ K ⊂ Mf (𝜇) for every f ∈ A(T). Furthermore, 𝜇Sf = 𝜆Sf ≡ inf{𝜑g | g ∈ A(T)+ ∧ g ⩾ 𝜒(Sf )}. By Proposition 1, 𝜑 is representable by the Lebesgue integral over the space ⟮T, M, 𝜇⟯. Now, we need to check the uniqueness of 𝜇. Let K ∈ K. Then, (f n ∈ A(T) | n ∈ 𝜔) ↓ 𝜒(K) in F(T). By the condition, A(T) ⊂ MI(T, N, 𝜈). Consequently, by Theorem 4 (3.3.3), 𝜒(K) ∈ MI(T, N, 𝜈) and ∫ 𝜒(K) d𝜈 = lim(∫ f n d𝜈 | n ∈ 𝜔). By Lemma 1 (3.3.2), 𝜈K = ∫ 𝜒(K) d𝜈 < ∞, where K ∈ N f (𝜈). As a result, we obtain 𝜈K = lim(∫ f n d𝜈 | n ∈ 𝜔). By the same reason, K ∈ Mf (𝜇) and 𝜇K = lim(∫ f n d𝜇 | n ∈ 𝜔). By the condition, ∫ f n d𝜈 = 𝜑f n = ∫ f n d𝜇. Thus, 𝜈K = 𝜇K for every K ∈ K and K ∈ Mf (𝜇) ∩ N f (𝜈).

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Let N ∈ N and H ∈ K. By the property of the inner K-regularity of 𝜈, we can find some sequences (K n ⊂ H ∩ N | n ∈ N) ↑ and (L n ⊂ H\N | n ∈ N) ↑ of elements of K such that 𝜈(H ∩ N) = sup (𝜈K n | n ∈ N) and 𝜈(H\N) = sup (𝜈L n | n ∈ N). Consider the disjoint sets K ≡ ⋃⟮K n | n ∈ N⟯ ⊂ K ∩ N and L ≡ ⋃⟮L n | n ∈ N⟯ ⊂ F\N belonging to M. We have 𝜈((H\K m )\L n ) = 𝜈(H\K m ) − 𝜈((H\K m ) ∩ L n ) = 𝜈H − 𝜈(H ∩ K m ) − 𝜈L n = 𝜈H − 𝜈K m − 𝜈L n . Consequently, by virtue of Lemma 5 (3.1.1), 𝜇K = sup (𝜇K n | n ∈ N) = sup (𝜈K n | n ∈ N) = 𝜈(H ∩ N) and 𝜇L = sup (𝜇L n | n ∈ N) = sup (𝜈L n | n ∈ N) = 𝜈 (H\N). Then, 𝜇(K ∪ L) = 𝜇K + 𝜇L = 𝜈(H ∩ N) + 𝜈(H\N) = 𝜈H, and therefore, 𝜈H = 𝜇H = 𝜇(H\(K ∪ L)) + 𝜇(K ∪ L) = 𝜇(H\(K ∪ L)) + 𝜈H, where 𝜇(H\(K ∪ L)) = 0. From (H ∩ N)\K ⊂ H\(K ∪ L) due to the local completeness of 𝜇, we infer that (H ∩ N)\K ∈ M. Hence, N ∩ H = ((H ∩ N)\K) ∪ K ∈ M. Since by virtue of Theorem 1 (3.4.1), the ensemble M is K-saturated, we get N ∈ M. Thus, N ⊂ M. If M ∈ M and H ∈ K, then completely by the same reason M ∩ H ∈ N. Take any L ∈ N f (𝜈). Since 𝜈 is inner K-regular, we have 𝜈L = sup{𝜈H | H ∈ K ∧ H ⊂ L}. Therefore, there exists a sequence (H n ∈ K | n ∈ N) ↑ such that ⋃⟮H n | n ∈ N⟯ ⊂ L and 𝜈L = sup (𝜈H n | n ∈ N). Consider the sets E ≡ ⋃⟮H n | n ∈ N⟯ and F ≡ L\E from N. Then, 𝜈H n ⩽ 𝜈E = 𝜈L − 𝜈F for every n implies 𝜈L ⩽ 𝜈L − 𝜈F. Hence, 𝜈F = 0. From (M ∩ L)\E ⊂ F due to the local completeness of 𝜈, we infer that (M ∩ L)\E ∈ N. On the other hand, by the condition, (M ∩ L)∩ E = ⋃⟮(M ∩ H n )∩ L | n ∈ N⟯ ∈ N. Therefore, M ∩ L ∈ N. It follows now from the strong saturation of 𝜈, that M ∈ N. Thus, M ⊂ N. Now, from the property of the inner K-regularity, it follows that 𝜇 = 𝜈. Finally, suppose that 𝜑 is uniformly bounded. Consider the set B ≡ {f ∈ A(T)+ | f ⩽ 1}. Then, there is a number x such that 0 ⩽ 𝜑f ⩽ x for every f ∈ B. If K ∈ K, then 𝜆K ≡ inf{𝜑f | f ∈ A(T) ∧ (𝜒(K) ⩽ f ⩽ 1)} ⩽ x. Therefore, 𝜇T = sup{𝜇K | K ∈ K} ⩽ x. Corollary 1. In the conditions of Theorem 1 for the space ⟮T, M, 𝜇⟯, the pointwise 𝜎-continuous functional Λ(𝜇) on the truncatable lattice-ordered linear space MI(T, M, 𝜇) ⊃ A(T) is an extension of the functional 𝜑 and has the Beppo Levi property from Theorem 2 (3.3.3). The solution of the problem of characterization of Lebesgue integrals as linear functionals Having the previous results, we can prove the theorem on characterization of Lebesgue integrals. The family of all positive inner I𝜎 (A(T))-regular and upper I𝜎 (A(T))-𝜎-continuous measures 𝜇 : M → R+ on T, defined on all possible 𝜎-algebras M ⊃ I𝜎 (A(T)), will be denoted by M(T, I𝜎 (A(T)))0 . Its subfamily consisting of all complete and strongly saturated measures will be denoted by Me (T, I𝜎 (A(T)))0 . To denote the corresponding subfamilies of bounded measures, we shall use the lower index “b”. Theorem 1 implies that we can correctly define a mapping r : (A(T)∧ )+ → L(T, I𝜎 (A(T)))0 , setting r𝜑 ≡ 𝜆. Moreover, according to Theorem 1, we can correctly

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define a mapping V ∧ : (A(T)∧ )+ → Mw (T)0 setting V ∧ (𝜑) ≡ 𝜇 ≡ e(I𝜎 (A(T)))(r𝜑) (see 3.4.1) and a mapping V b∧ : (A(T) ∧ )+ → Mwb (T)0 setting V b∧ ≡ V ∧ |(A(T) ∧ )+ . It is evident that V ∧ and V b∧ are injective. Moreover, rng V ∧ ⊂ Me (T, I𝜎 (A(T)))0 and rng V b∧ ⊂ Meb (T, I𝜎 (A(T)))0 . If we shall take another pointwise 𝜎-continuous positive linear functional 𝜓 on A(T), then for 𝜓 by Theorem 1, there exists its own measurable space ⟮T, N, 𝜈⟯ such that 𝜓 is representable by the Lebesgue integral over the space ⟮T, N, 𝜈⟯. Therefore, it is important to consider for all such functionals an integral representation over one and the same measurable space. For this reason, we shall consider for the ensemble K ≡ I𝜎 (A(T)) its Borel envelope B(T, K) in P(T) (see 2.1.1). Lemma 1. Let A(T) be a truncatable lattice-ordered linear space of functions on a set T. Then, Coz A(T) ⊂ B(T, I𝜎 (A(T))). Proof. Let f ∈ A(T)+ and D ≡ coz f . By Lemma 1 (2.2.9), S n ≡ {t ∈ T | f (t) ⩾ 1/n} = S(nf ) ∈ I𝜎 (A(T)). Consequently, D = ⋃⟮S n | n ∈ N⟯ ∈ B(T, I𝜎 (A(T))). If 𝜓 ∈ (A(T)∧ )+ , 𝜇 ≡ V ∧ 𝜑 ∈ Mw (T)0 and M ≡ dom 𝜇, then K ⊂ M implies B(T, K) ⊂ M because M is a 𝜎-algebra. For the future utilization in 3.6, we shall consider an arbitrary lattice-ordered linear subspace A(T) of the lattice-ordered linear space A(T)∧ (see Corollary 2 to Lemma 4 (2.2.8)) and the corresponding lattice-ordered linear subspace A(T) ≡ A(T) ∩ A(T) ∧ of the lattice-ordered linear space A(T) . Take any 𝜎-algebra M0 such that B(T, K) ⊂ M0 ⊂ ⋂⟮dom V ∧ 𝜑 | 𝜑 ∈ (A(T) )+ ⟯. Then, for every functional 𝜑 ∈ (A(T) )+ and the corresponding measure 𝜇 ≡ V ∧ 𝜑, we can consider the measure 𝜇0 ≡ 𝜇|M0 . This means that we can correctly define a mapping V : (A(T) )+ → Meas(T, M0 )0 setting V𝜑 ≡ (V ∧ 𝜑)|M0 , and a mapping V b : (A(T) )+ → Measb (T, M0 )+ setting V b ≡ V|(A(T) )+ . They are injective. Theorem 2. Let A(T) be a truncatable lattice-ordered linear space of functions on a set T, K ≡ I𝜎 (A(T)), A(T) be a lattice-ordered linear subspace of A(T)∧ , 𝜑 ∈ (A(T) )+ , and 𝜇0 ≡ V𝜑. Then, 1) 𝜇0 is a positive inner K-regular and upper K-𝜎-continuous measure; 2) 𝜑 is representable by the Lebesgue integral over the measurable space ⟮T, M0 , 𝜇0 ⟯; 3) the measure 𝜇0 is unique in the following sense: if there is another positive inner Kregular measure 𝜘 : M0 → R+ such that 𝜑 is representable by the Lebesgue integral over the space ⟮T, M0 , 𝜘⟯, then 𝜘 = 𝜇0 ; 4) if 𝜑 is uniformly bounded, then the measure 𝜇0 is bounded. Proof. Consider 𝜆 and 𝜇 from the proof of Theorem 1. Let f ∈ A(T). According to the proof of Theorem 1, Sf ∈ K ⊂ Mf (𝜇). Therefore, Sf ∈ M0 and 𝜇0 Sf = 𝜇Sf < ∞.

308 | 3.4 Representation of a functional by the Lebesgue integral

Furthermore, 𝜇0 Sf = 𝜇Sf = inf{𝜀g | g ∈ A(T)+ ∧ g ⩾ 𝜒(Sf )}. Thus, by Proposition 1, 𝜑 is representable by the Lebesgue integral over the space ⟮T, M0 , 𝜇0 ⟯. Now, we need to check the uniqueness of 𝜇0 . If K ∈ K, then 𝜒(K) = p-lim(f n | n ∈ 𝜔) for some decreasing sequence (f n ∈ A(T)+ | n ∈ 𝜔). By the condition, A(T) ⊂ MI(T, M0 , 𝜘). Consequently, by Theorem 4 (3.3.3) 𝜒(K) ∈ MI(T, M0 , 𝜘) and ∫ 𝜒(K) d𝜘 = lim(∫ f n d𝜘 | n ∈ 𝜔). By Lemma 1 (3.3.2), 𝜘K = ∫ 𝜒(K) d𝜘 < ∞, where K ∈ M0f (𝜘). As a result, we obtain 𝜘K = lim(∫ f n d𝜘 | n ∈ 𝜔). By the same reason, K ∈ M0f (𝜇0 ) and 𝜇0 K = lim(∫ f n d𝜇0 | n ∈ 𝜔). But by the condition, ∫ f n d𝜘 = 𝜑f n = ∫ f n d𝜇0 . Thus, 𝜘K = 𝜇0 K for every K ∈ K. Now, from the property of the inner K-regularity it follows that 𝜘 = 𝜇0 . If 𝜑 is uniformly bounded, then by virtue of Theorem 2 𝜇0 is bounded. The family of all positive, inner I𝜎 (A(T))-regular, and upper I𝜎 (A(T))-𝜎-continuous measures 𝜇 : M0 → R+ on T will be denoted by Meas(T, M0 , I𝜎 (A(T)))0 . Its subfamily consisting of all bounded measures will be denoted by Measb (T, M0 , I𝜎 (A(T)))0 . Theorem 2 implies that rng V ⊂ Meas(T, M0 , I𝜎 (A(T)))0 and rng V b ⊂ Measb (T, M0 , I𝜎 (A(T)))0 . Lemma 2. The family Measb (T, M0 , I𝜎 (A(T)))0 is a conic space. Proof. Take 𝜇, 𝜈 ∈ Measb (T, M0 , I𝜎 (A(T)))0 and x, y ∈ R+ . By virtue of Corollary 2 to Lemma 2 (3.2.2), 𝜆 ≡ x𝜇 + y𝜈 is a positive bounded measure on M0 . Let M ∈ M0 and x > 0, y > 0. By virtue of the inner regularity for any 𝜀 > 0, there are K and L from K ≡ I𝜎 (A(T)) such that 𝜇M − 𝜇K < 𝜀/(2x), 𝜈M − 𝜈L < 𝜀/(2y) and K ∪ L ⊂ M. Therefore, 𝜆M − 𝜆(K ∪ L) = x(𝜇M − 𝜇(K ∪ L)) + y(𝜈M − 𝜈(K ∪ L)) ⩽ x(𝜇M − 𝜇K) + y(𝜈M − 𝜈L) < 𝜀. By Proposition 1 (2.2.9), K ∪ L ∈ K. Consequently, 𝜆 is inner K-regular. Let K ∈ K and (K 𝛾 ∈ K | 𝛾 ∈ Γ ⊂ 𝜔) ↓ K in P(T). Then, by virtue of the upper Kcontinuity and Lemma 7 (1.4.7) for any 𝜀 > 0, there are 𝜉, 𝜂 ∈ Γ such that 𝜇K + 𝜀/(2x) > 𝜇K𝜉 and 𝜈K + 𝜀/(2y) > 𝜈K𝜂 . Take any 𝛾 such that 𝛾 ⩾ 𝜉 and 𝛾 ⩾ 𝜂. Then, 𝜆K𝛾 ⩽ x𝜇K𝜉 + y𝜈K𝜂 < 𝜆K + 𝜀. Now, again by Lemma 7 (1.4.7), we conclude that 𝜆K = inf{𝜆K𝛾 | 𝛾 ∈ Γ}. Thus, 𝜆 is upper K-𝜎-continuous. It follows from Lemma 2 that we may consider the family Measb (T, M0 , I𝜎 (A(T))) ≡ {𝜇1 − 𝜇2 | 𝜇1 , 𝜇2 ∈ Measb (T, M0 , I𝜎 (A(T)))0 }. Corollary 1. The family Measb (T, M0 , I𝜎 (A(T))) is a linear subspace of the linear space Measb (T, M0 ). Proposition 2. Let A(T) be a truncatable lattice-ordered linear space of functions on a set T. Then, Measb (T, M0 , I𝜎 (A(T))) is an l-ideal in the Dedekind complete latticeordered linear space Measb (T, M0 ).

3.4.2 Representation of pointwise 𝜎-continuous functionals by Lebesgue integrals

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Proof. Denote the first family by X and the second one by Y. By Corollary 2 to Lemma 2 (3.2.2), Y is a Dedekind complete lattice-ordered linear space. By Corollary 1 to Lemma 2, X is a linear subspace of Y. Let 𝜇 ∈ X, 𝜈 ∈ Y, and 0 ⩽ 𝜈 ⩽ 𝜇. Then, for every M ∈ M0 and every 𝜀 > 0, there is K ∈ K ≡ I𝜎 (A(T)) such that K ⊂ M and 𝜇M − 𝜀 < 𝜇K. Therefore, 𝜈M −𝜈K = 𝜈(M\K) ⩽ 𝜇(M\K) = 𝜇M −𝜇K < 𝜀 means that 𝜈M = sup{𝜈K | K ∈ K ∧ K ⊂ M}. Thus, 𝜈 is inner K-regular. Let K ∈ K and (K𝛾 ∈ K | 𝛾 ∈ Γ ⊂ 𝜔) ↓ K in P(T). Then, for every 𝜀 > 0, there is 𝛾 such that 𝜇K+𝜀 > 𝜇K𝛾 . Therefore, 𝜈K𝛾 −𝜈K = 𝜈(K𝛾 \K) ⩽ 𝜇(K𝛾 \K) = 𝜇K𝛾 −𝜇K < 𝜀 means by Lemma 7 (1.4.7) that 𝜈K = inf{𝜈K𝛾 | 𝛾 ∈ Γ}. Thus, 𝜈 is upper K-𝜎-continuous. As a result, 𝜈 ∈ X. Now, let 𝜇 ∈ X, 𝜈 ∈ Y, and |𝜈| ⩽ |𝜇| in Measb (T, M0 ). By the definition 𝜇 = 𝜇1 − 𝜇2 , for some 𝜇1 , 𝜇2 ∈ X+ . Therefore, |𝜇| ⩽ 𝜇1 +𝜇2 . By Lemma 2, 𝜆 ≡ 𝜇1 +𝜇2 ∈ X+ . By the property proven above, the inequality 0 ⩽ 𝜈+ ⩽ |𝜈| ⩽ 𝜆 implies 𝜈+ ∈ X+ . Analogously, −𝜈− ∈ X+ . As a result, by Corollary 3 to Theorem 2 (3.2.2), 𝜈 = 𝜈+ − (−𝜈− ) ∈ X. According to Statement 2 (2.2.8), X is an l-ideal of Y. Corollary 1. Let A(T) be a truncatable lattice-ordered linear space of functions on T. Then, Measb (T, M0 , I𝜎 (A(T))) is a Dedekind complete lattice-ordered linear space, the smallest upper [the greatest lower] bounds of bounded above [below] sets in the ordered sets Measb (T, M0 ) and Measb (T, M0 , I𝜎 (A(T))) coincide and they are expressed by the formulas from Theorem 2 (3.2.2) and its Corollaries. Proof. The assertion follows from Proposition 2 and Statements 3 and 1 (2.2.8). Lemma 3. Let A(T) be a truncatable lattice-ordered linear space of functions on T. If 𝜑, 𝜓 ∈ (A(T) )+ and x, y ∈ R+ , then V b (x𝜑 + y𝜓) = xV b 𝜑 + yV b 𝜓, i. e. V b is a conic operator. Moreover, V b is injective. Proof. Let 𝜇 ≡ V b 𝜑, 𝜈 ≡ V b 𝜓 and 𝜆 ≡ x𝜇 + y𝜈. By Theorem 2, A(T) ⊂ M(T, M0 ) and any function f ∈ A(T) is integrable with respect to 𝜇 and 𝜈. herefore, by virtue of Proposition 3 (3.3.5), we have (x𝜑 + y𝜓)f = x ∫ f d𝜇 + y ∫ f d𝜈 = ∫ f d(x𝜇 + y𝜈) = ∫ f d𝜆. This means that 𝜆 = V b (x𝜑 + y𝜓). Suppose that V b 𝜑 = V b 𝜓. Then, 𝜑f = ∫ f d𝜇 = ∫ f d𝜈 = 𝜓f for every f ∈ A(T), where 𝜑 = 𝜓. Hence, V b is injective. Proposition 3. Let A(T) be a truncatable lattice-ordered linear space of functions on a set T and A(T) be a lattice-ordered linear subspace of A(T)∧ . Then, 1) the mapping V b has the unique extension to the injective linear operator I b : A(T) Measb (T, M0 ) between the given lattice-ordered linear spaces such that I b 𝜑 = V b (𝜑+ ) − V b (−𝜑− ) for every 𝜑 ∈ A(T) ;

310 | 3.4 Representation of a functional by the Lebesgue integral

2) every functional, 𝜑 ∈ A(T) is representable by the Lebesgue integral over the measurable space ⟮T, M0 , I b 𝜑⟯; 3) rng I b ⊂ Measb (T, M0 , I𝜎 (A(T))). Proof. 1. By the condition, A(T) is a lattice-ordered linear space. By Corollary 2 to Lemma 2 (3.2.2), Measb (T, M0 ) is a lattice-ordered linear space as well. Therefore, by virtue of Lemma 3 and Statement 1, V b has the unique extension to the injective linear Measb (T, M0 ) such that I b 𝜑 = V b (𝜑+ ) − V b (−𝜑− ). operator I b : A(T) 2. If f ∈ A(T), then f is integrable with respect to V b (𝜑+ ) and V b (−𝜑− ). Therefore, by Theorem 5 (3.3.8), f ∈ MI(T, M0 , I b 𝜑) and 𝜑f = (𝜑+ )f − (−𝜑− )f = ∫ f dV b (𝜑+ ) − ∫ f dV b (−𝜑− ) = ∫ f dV b 𝜑. 3. The assertion follows from Corollary 1 to Proposition 2, and the inclusion rng V b ⊂ Measb (T, M0 , I𝜎 (A(T)))0 . The following theorem solves the problem of characterization of Lebesgue integrals as linear functionals from 3.3.6. Theorem 3 (the Zakharov theorem on characterization of Lebesgue integrals as linear functionals). Let A(T) be a truncatable lattice-ordered linear subspace of F b (T) and M0 be any 𝜎-algebra such that B(T, I𝜎 (A(T))) ⊂ M0 ⊂ ⋂⟮dom V ∧ 𝜑 | 𝜑 ∈ A(T)∧ ⟯. Then, the mapping I b : A(T) ∧ Measb (T, M0 , I𝜎 (A(T))) from Proposition 3 is an isomorphism of the given lattice-ordered linear spaces. Proof. Let 𝜇 ∈ Measb (T, M0 , I𝜎 (A(T))). Then, 𝜇 = 𝜇+ + 𝜇− . By Theorem 1 (3.3.6), the functionals 𝜓󸀠 ≡ Λ(𝜇+ ) : MI(T, M0 , 𝜇+ ) → R+ , 𝜓󸀠󸀠 ≡ Λ(−𝜇− ) : MI(T, 𝜇0 , −𝜇− ) → R+ , and 𝜓 ≡ Λ(𝜇) : MI(T, M0 , 𝜇) → R are pointwise 𝜎-continuous and uniformly bounded. Besides, 𝜓 = 𝜓󸀠 − 𝜓󸀠󸀠 . By Lemma 1 (3.3.2), ∫ 1 d𝜇+ = 𝜇+ T < ∞. By the condition for every f ∈ A(T), there is a number x such that |f | ⩽ x1. Therefore, ∫ |f | d𝜇+ ⩽ x ∫ 1 d𝜇+ < ∞ implies |f | ∈ MI(T, M0 , 𝜇+ ). By Corollary 3 to Theorem 2 (3.3.2), f ∈ MI(T, M0 , 𝜇+ ). The same is valid for −𝜇− . According to the definition from 3.3.6, the same is valid for 𝜇. Thus, A(T) ⊂ MI(T, M0 , 𝜇+ ) ∩ MI(T, M0 , −𝜇− ) = MI(T, M0 , 𝜇). That is why we may consider the restrictions 𝜑󸀠 , 𝜑󸀠󸀠 , and 𝜑 on A(T) of the functionals 𝜓󸀠 , 𝜓󸀠󸀠 , and 𝜓 correspondingly. All of them belong to A(T)∧ and 𝜑 = 𝜑󸀠 − 𝜑󸀠󸀠 . By this definition, 𝜑󸀠 is representable by the Lebesgue integral over the space ⟮T, M0 , 𝜇+ ⟯. According to assertion 3 of Theorem 2, 𝜇+ is unique and 𝜇+ = V b 𝜑󸀠 . This means that V b is surjective. Similarly, −𝜇− = V b 𝜑󸀠󸀠 . By virtue of Proposition 3, I b is linear, and therefore, I b 𝜑 = V b 𝜑󸀠 − V b 𝜑󸀠󸀠 = 𝜇+ + 𝜇− = 𝜇. This means that I b is surjective. By virtue of Proposition 3, I b is bijective. It is clear that I b is positive and increasing. Let 𝜑, 𝜓 ∈ (A(T) ∧ )+ and 𝜇 ≡ V b 𝜑 ⩽ 𝜈 ≡ V b 𝜓. Suppose that 𝜑 ⩽̸ 𝜓, i. e., 𝜑 ∨ 𝜓 = / 𝜓. Consider the functional 𝜒 ≡ 𝜑 ∨ 𝜓 − 𝜓 > 0 from A(T) ∧ . Then, there exists f ∈ A(T)+ such that 𝜒f > 0. By virtue of Corollary 2 to Statement 5 (2.2.8) for 𝜀 = 𝜒f /2, there

3.4.3 Representation of pointwise continuous functionals by Lebesgue integrals |

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exists a function g ∈ A(T)+ such that 𝜑g + 𝜓(f − g) + 𝜀 > (𝜑 ∨ 𝜓)f . This implies 𝜒f = (𝜑 ∨ 𝜓)f − 𝜓f < 𝜑g + 𝜓f − 𝜓g + 𝜀 − 𝜓f , where 0 < 𝜑g − 𝜓g. But according to Proposition 1 (3.3.5) 𝜑g = ∫ g d𝜇 ⩽ ∫ g d𝜈 = 𝜓g. It follows from this contradiction that 𝜑 ⩽ 𝜓. This means that V b is isotone. By Lemma 2 (1.1.15), V b preserves any exact bounds. In particular, 𝜑1 ∧ 𝜑2 = 0 implies I b 𝜑1 ∧ I b 𝜑2 = V b 𝜑1 ∧ V b 𝜑2 = V b (𝜑1 ∧ 𝜑2 ) = 0. Then, by Statement 2 (3.2.2) I b is an homomorphism of lattice-ordered linear spaces. Finally, according to Statement 5 (2.2.7), I b is an isomorphism. Auxiliary definitions and statements 1∘ In 3.4 – 3.6, we use also the following generalization of the notion of a linear space from 2∘ (2.2.4). Let A be a set. The system U ≡ |⟮A, R+ ⟯, 0A , +A , ⋅⟮A,R+ ⟯ | with the structures 0A ∈ A, +A : A ∗ A → A, and ⋅⟮A,R+ ⟯ : R+ ∗ A → A will be called a conic space over the system ⟮R+ , w⟯ of positive real numbers with the structure w ≡ ⟮0R+ , 1R+ , +R+ , ⋅R+ ⟯ or a (real) (R+ -)conic space if the following properties hold: 1) ∀ a ∈ A (0A + a = a); 2) ∀ a, b, c ∈ A (a + (b + c) = (a + b) + c); 3) ∀ a, b ∈ A (a + b = b + a); 4) ∀ a ∈ A (0R+ a = 0A ); 5) ∀ a ∈ A (1R+ a = a); 6) ∀ x ∈ R+ ∀ a, b ∈ A (x(a + b) = xa + xb); 7) ∀ x, y ∈ R+ ∀ a ∈ A ((x +R+ y)a = xa + ya); 8) ∀ x, y ∈ R+ ∀ a ∈ A ((xy)a = x(ya)). The element 0A is called the neutral element of the conic space U with respect to addition or the zero element. Let U ≡ |⟮A, R+ ⟯, s cs | and V ≡ |⟮B, R+ ⟯, t cs | be conic spaces. A mapping u : A → B will be called conic (or a conic operator) if u(a1 + a2 ) = ua1 + ua2 for every a1 , a2 ∈ A and u(xa) = xua for every a ∈ A and x ∈ R+ . Statement 1. Let A be a lattice-ordered linear space, B be a linear space, and w : A+ → B be a conic operator. Then w has the unique extension to the linear operator v : A → B such that va = w(a+ ) − w(−a− ) for every a ∈ A. Moreover, if w is injective, then v is injective as well.

3.4.3 Representation of pointwise continuous functionals by Lebesgue integrals Here, we shall consider the analogs of the results from 3.4.2 for pointwise continuous functionals (see 2.2.9). Remind that the definition of the ensemble I(A(T)) is given in 2.2.9.

312 | 3.4 Representation of a functional by the Lebesgue integral

Theorem 1. Let A(T) be a truncatable lattice-ordered linear space of functions on a set T and 𝜑 ∈ (A(T)∨ )+ . Define on the ensemble K ≡ I(A(T)) an evaluation 𝜆 : K → R+ setting 𝜆K ≡ inf{𝜑f | f ∈ A(T)+ ∧ f ⩾ 𝜒(K)}. Then, 1) 𝜆 ∈ L(T, K) (see 3.4.1); 2) the measure 𝜇 ≡ e(K)(𝜆) : M → R+ from 3.4.1 is a positive, wide, complete, strongly saturated, K-saturated, inner-K-regular, and upper K-continuous; 3) 𝜑 is representable by the Lebesgue integral over the measurable space ⟮T, M, 𝜇⟯; 4) the measure 𝜇 is unique in the following sense: if there is another positive, complete, strongly saturated, inner K-regular, and upper K-continuous measure 𝜈 : N → R+ such that 𝜑 is representable by the Lebesgue integral over the space ⟮T, N, 𝜈⟯, then N = M and 𝜈 = 𝜇; 5) if 𝜑 is uniformly bounded, then 𝜇 is bounded. Proof. By Proposition 1 (2.2.9), K is additive and completely multiplicative. If A(T) = / {0}, then by Lemma 1 (2.2.9), K = / {⌀}. Consider the families E(K) ≡ {f ∈ A(T)+ | f ⩾ 𝜒(K)} and Γ(K) ≡ −E(K). By the definition of K, we have inf{f (t) | f ∈ E(K)} = 𝜒(K)(t). Since Γ(K)is upwarddirected, we may consider the net (u 𝛾 | 𝛾 ∈ Γ(K)) ↓ such that u𝛾 ≡ f for 𝛾 ≡ −f . From the equality 𝜒(K)(t) = inf (u𝛾 (t) | 𝛾 ∈ Γ(K)), we conclude that (u𝛾 | 𝛾 ∈ Γ(K)) ↓ 𝜒(K) in F(T). Take any n ∈ N. Then, there is f ∈ E(K) such that 𝜆K + 1/n > 𝜑f = 𝜑u 𝛾 ⩾ 𝜆K. By Lemma 1 (1.4.5), this means that 𝜆K = inf (𝜑u𝛾 | 𝛾 ∈ Γ(K)). Suppose now that there is another net (v𝛿 ∈ A(T)+ | 𝛿 ∈ Δ) ↓ 𝜒(K). Consider as above the upward directed set Z ≡ Γ(K) × Δ and the net (w𝜁 | 𝜁 ∈ Z) such that w𝜁 ≡ u𝛾 − v𝛿 . Take t ∈ T and 𝜀 > 0. Then, there is 𝜁0 ≡ (𝛾0 , 𝛿0 ) such that u𝛾0 (t) − 𝜒(K)(t) < 𝜀/2 and v𝛿0 (t) − 𝜒(K)(t) < 𝜀/2. Therefore, for every 𝜁 ≡ (𝛾, 𝛿) ⩾ 𝜁0 , we have |w𝜁 (t)| ⩽ u𝛾 (t) − 𝜒(K)(t) + v𝛿 (t) − 𝜒(K)(t) ⩽ u𝛾0 (t) − 𝜒(K)(t) + v𝛿0 (t) − 𝜒(K)(t) < 𝜀. This means that lim(w𝜁 (t) | 𝜁 ∈ Z) = 0 for every t ∈ T, i. e., p-lim(w𝜁 | 𝜁 ∈ Z) = 0 in F(T). Since 𝜑 is pointwise continuous, we have lim(𝜑w𝜁 | 𝜁 ∈ Z) = 0. Suppose that x ≡ inf (𝜑v𝛿 | 𝛿 ∈ Δ) > 𝜆K and take 𝜀1 ≡ x − 𝜆K. Then, there is 𝜁1 ≡ (𝛾1 , 𝛿1 ), such that |𝜑u𝛾 − 𝜑v𝛿 | = |𝜑w𝜁 | < 𝜀1 /2 for every 𝜁 ≡ (𝛾, 𝛿) < 𝜁1 . Besides, there is 𝛾2 such that 𝜆K + 𝜀1 /2 > 𝜑u𝛾2 . Take 𝛾 ∈ Γ(K) such that 𝛾 ⩾ 𝛾1 and 𝛾 ⩾ 𝛾2 . Then, for 𝜁 ≡ (𝛾, 𝛿1 ) ⩾ 𝜁1 , we have x ⩽ 𝜑v𝛿1 < 𝜑u𝛾 + 𝜀1 /2 < 𝜑u𝛾2 + 𝜀1 /2 < 𝜆K + 𝜀1 = x. It follows from this contradiction that x = 𝜆K. Thus, (v𝛿 | 𝛿 ∈ Δ) ↓ 𝜒(K) implies (𝜑v𝛿 | 𝛿 ∈ Δ) ↓ 𝜆K. It is evident that 𝜆 is natural and increasing. Let K, L ∈ K, (f 𝛾 | 𝛾 ∈ Γ(K)) ↓ 𝜒(K), (g𝛿 | 𝛿 ∈ Δ) ↓ 𝜒(L), and K ∩ L = ⌀. Consider the upward directed set Z ≡ Γ(K) × Δ and the net (h𝜁 | 𝜁 ∈ Z) ↓ such that h𝜁 ≡ f𝛾 ∧ g𝛿 for every 𝜁 ≡ (𝛾, 𝛿). Proposition 1 (1.4.7) guarantees that (h𝜁 | 𝜁 ∈ Z) ↓ 0 in F(T). Then, by Lemma 1 (2.2.9), (𝜑h𝜁 | 𝜁 ∈ Z) ↓ 0. Therefore, by Lemma 7 (1.4.7), for every 𝜀 > 0, there is 𝜁 = (𝛾, 𝛿) such that 𝜑h𝜁 < 𝜀. Take some h ∈ A(T)+ such that 𝜒(K ∪ L) ⩽ h and 𝜆(K ∪ L) + 𝜀 > 𝜑h. Then, h ⩾ h ∧ (f𝛾 ∨ g𝛿 ) = (h ∧ f𝛾 ) ∨ (h ∧ g𝛿 ) = h ∧ f𝛾 + h ∧ g𝛿 − h ∧ h𝜁 guarantees that 𝜆(K ∪ L) > 𝜑h − 𝜀 ⩾ 𝜑(h∧(f𝛾 ∨g𝛿 ))−𝜀 = 𝜑((h∧f𝛾 )∨(h∧g𝛿 ))−𝜀 = 𝜑(h∧f𝛾 )+𝜑(h∧g𝛿 )−𝜑(h∧h𝜁 )−𝜀 ⩾ 𝜆K+𝜆L−2𝜀. Since 𝜀 is arbitrary, we conclude that 𝜆(K ∪L) ⩾ 𝜆K +𝜆L, i. e., 𝜆 is (binary) overadditive.

3.4.3 Representation of pointwise continuous functionals by Lebesgue integrals

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Let now H, L ∈ K, (f𝛾 | 𝛾 ∈ Γ) ↓ 𝜒(L), and K ⊂ L. For 𝜀 > 0, there is g ∈ A(T+ ) such that g ⩾ 𝜒(H) and 𝜆H + 𝜀 > 𝜑g. Consider the set K ≡ {t ∈ L | g(t) ⩽ 1/(1 + 𝜀)} ⊂ L\H. If t ∈ K, then f 𝛾 (t) − g(t) ⩾ 𝜀/(1 + 𝜀) for every 𝛾. Conversely, if t ∈ L and f𝛾 (t) − g(t) ⩾ 𝜀/(1 + 𝜀) for every 𝛾, then g(t) ⩽ inf{f 𝛾 (t) − 𝜀/(1 + 𝜑) | 𝛾 ∈ Γ} = 1/(1 + 𝜀) means that t ∈ K. Consequently, K = {t ∈ L | ∀ 𝛾 ∈ Γ (f 𝛾 (t) − g(t) ⩾ 𝜀/(1 + 𝜀))} = L ∩ ⋂⟮{t ∈ T | (1 + 1/𝜀)(f𝛾 − g)(t) ⩾ 1} | 𝛾 ∈ Γ⟯. By Lemma 1 (2.2.9), K𝛾 ≡ {t ∈ T | (1 + 1/𝜀)(f𝛾 − g)(t) ⩾ 1} ∈ K for every 𝛾. Thus, K = L ∩ ⋂⟮K𝛾 | 𝛾 ∈ Γ⟯ ∈ K. Let h ∈ A(T)+ and h ⩾ 𝜒(K). Then, h + (1 + 𝜀)g ⩾ 𝜒(L) implies 𝜆L ⩽ 𝜑h + (1 + 𝜀)𝜑g. Since h is arbitrary, we infer that 𝜆L−(1+𝜀)𝜑g ⩽ 𝜆K, where 𝜆L−𝜆K ⩽ 𝜆H +𝜀(1+𝜆H)+𝜀2 . Since 𝜀 is arbitrary, we conclude that 𝜆L ⩽ 𝜆H + sup{𝜆K | K ∈ K ∧ K ⊂ L\H}. Thus, 𝜆 has property 1 from the condition of Theorem 1 (3.4.1). Finally, let K ∈ K and (K𝜁 ∈ K | 𝜁 ∈ Z) ↓ K in P(T). Consider the set E ≡ {f ∈ A(T)+ | ∃ 𝜁 ∈ Z (f ⩾ 𝜒(K𝜁 ))}. For every 𝜁, the corresponding sets E(K𝜁 ) and Γ(K𝜁 ) and the corresponding net (u𝛾 | 𝛾 ∈ Γ(K𝜁 )) defined above. Then, (u𝛾 | 𝛾 ∈ Γ(K𝜁 )) ↓ 𝜒(K𝜁 ). We have E = ⋃⟮E(K𝜁 ) | 𝜁 ∈ Z⟯. Let f , g ∈ E. Then, f ⩾ 𝜒(K𝜉 ) and g ⩾ 𝜒(K𝜂 ) for some 𝜉, 𝜂 ∈ Z. Take 𝜁 such that 𝜁 ⩾ 𝜉 and 𝜁 ⩾ 𝜂. Then, f ∧g ⩾ 𝜒(K𝜉 ∩K𝜂 ) ⩾ 𝜒(K𝜁 ) means that f ∧ g ∈ E. Therefore, the set Γ ≡ −E is upward directed, and we may consider the net (u 𝛾 | 𝛾 ∈ Γ) ↓ such that u𝛾 ≡ f for 𝛾 ≡ −f . Since Γ = ⋃⟮Γ(K𝜁 ) | 𝜁 ∈ Z⟯, we conclude by virtue of Corollary 1 to Proposition 2 (1.1.15) that inf{u𝛾 (t) | 𝛾 ∈ Γ} = inf{inf{u𝛾 (t) | 𝛾 ∈ Γ(K𝜁 )} | 𝜁 ∈ Z} = inf{inf{f (t) | f ∈ E(K𝜁 )} | 𝜁 ∈ Z} = inf{𝜒(K𝜁 )(t) | 𝜁 ∈ Z} = 𝜒(K)(t) for every t ∈ T. Thus, (u𝛾 | 𝛾 ∈ Γ) ↓ 𝜒(K) in F(T). As was shown before, this implies (𝜑(u𝛾 ) | 𝛾 ∈ Γ) ↓ 𝜆K. Also, it was proven above that 𝜆K𝜁 = inf{𝜑u𝛾 | 𝛾 ∈ Γ(K𝜁 )} for every 𝜁. Therefore, again by this corollary, we get 𝜆K = inf{inf{𝜑(u𝛾 ) | 𝛾 ∈ Γ(K𝜁 )} | 𝜁 ∈ Z} = inf{𝜆K𝜁 | 𝜁 ∈ Z}. Hence, (𝜆K𝜁 ) ↓ 𝜆K. Thus, 𝜆 is upper K-continuous. Therefore, for 𝜆, we may consider the corresponding measure 𝜇 : M → R+ from Theorem 1 (3.4.1). By Lemma 1 (2.2.9), Sf ∈ K ⊂ Mf (𝜇) for every f ∈ A(T). Furthermore, 𝜇Sf = 𝜆Sf = inf{𝜑g | g ∈ A(T)+ ∧ g ⩾ 𝜒(Sf )}. By Proposition 1 (3.4.2), 𝜑 is representable by the Lebesgue integral over the space ⟮T, M, 𝜇⟯. Now, we need to check the uniqueness of 𝜇. Let f ∈ A(T). Then, by Lemma 1 (2.2.9) for the set Sf ≡ {t ∈ T | f (t) ⩾ 1}, there is a sequence (f n ∈ A(T)+ | n ∈ 𝜔) ↓ 𝜒(Sf ) in F(T). By the condition, A(T) ⊂ MI(T, N, 𝜈). Consequently, by Theorem 4 (3.3.3), 𝜒(Sf ) ∈ MI(T, N, 𝜈) and ∫ 𝜒(Sf ) d𝜈 = lim(∫ f n d𝜈 | n ∈ 𝜔). By Lemma 1 (3.3.2), 𝜈Sf = ∫ 𝜒(Sf ) d𝜈 < ∞, where Sf ∈ N f (𝜈). As a result, we get 𝜈Sf = lim(∫ f n d𝜈 | n ∈ 𝜔). By the same reason, Sf ∈ Mf (𝜇) and 𝜇Sf = lim(∫ f n d𝜇 | n ∈ 𝜔). By the condition, ∫ f n d𝜈 = 𝜑f n = ∫ f n d𝜇. Thus, 𝜈Sf = 𝜇Sf for every f ∈ A(T) and Sf ∈ N f (𝜈)∩ Mf (𝜇). Let K ∈ K. Then, (f𝜁 ∈ A(T) | 𝜁 ∈ Z) ↓ 𝜒(K) in F(T). Consider the sets K𝜁 ≡ Sf [𝜁] ∈ K. Then, (K𝜁 | 𝜁 ∈ Z) ↓ K. Since 𝜈 is upper K-continuous, we have (𝜈K𝜁 | 𝜁 ∈ Z) ↓ 𝜈K. By the same reason, (𝜇K𝜁 | 𝜁 ∈ Z) ↓ 𝜇K. As a result, 𝜈K = 𝜇K for every K ∈ K and K ⊂ N f (𝜈) ∩ Mf (𝜇). Further, the proof is completely the same as the proof of Theorem 1 (3.4.2).

314 | 3.4 Representation of a functional by the Lebesgue integral

The family of all positive inner I(A(T))-regular and upper I(A(T))-continuous measures 𝜇 : M → R+ on T defined on all possible 𝜎-algebras M ⊃ I(A(T)) will be denoted by M(T, I(A(T)))0 . Its subfamily consisting of all complete and strongly saturated measures will be denoted by Me (T, I(A(T)))0 . To denote the corresponding subfamilies of bounded measures, we shall use the lower index “b”. Theorem 1 implies that we can correctly define a mapping s : (A(T)∨ )+ → L(T, I(A(T)))+ setting s𝜑 ≡ 𝜆. Moreover, according to Theorem 1, we can correctly define mappings W ∨ : (A(T)∨ )+ → M(T)+ setting W𝜑∨ ≡ 𝜇 ≡ e(I(A(T)))(s𝜑)

(see 3.4.1) and W b∨ : (A(T) ∨ )+ → Mwb (T)+ setting W b∨ ≡ W ∨ |(A(T) ∨ )+ . It is evident that W ∨ and W b∨ are injective. Moreover, rng W ∨ ⊂ Me (T, I(A(T)))0 and rng W b∨ ⊂ Meb (T, I(A(T)))0 . To find as in 3.4.2 for all such functionals an integral representation over one and the same measurable space, we shall consider for the ensemble K ≡ I(A(T)) its Borel envelope B(T, K) in P(T) (see 2.1.1). Lemma 1 (3.4.2) implies Coz A(T) ⊂ B(T, I(A(T))). If 𝜑 ∈ (A(T)∨ )+ , 𝜇 ≡ W ∨ 𝜑 ∈ M(T)+ and M ≡ dom 𝜇, then K ⊂ M guarantees that B(T, K) ⊂ M because M is a 𝜎-algebra. For the future utilization in 3.6, we shall consider an arbitrary lattice-ordered linear subspace A(T) of the lattice-ordered linear space A(T)∨ (see Corollary 2 to Lemma 4 (2.2.8)) and the corresponding lattice-ordered linear subspace A(T) ≡ A(T) ∩ A(T) ∨ of the lattice-ordered linear space A(T) . Take any 𝜎-algebra M0 such that B(T, K) ⊂ M0 ⊂ ⋂⟮dom W ∨ 𝜑 | 𝜑 ∈ (A(T) )+ ⟯. Then, for every functional 𝜑 ∈ (A(T) )+ and for the corresponding measure 𝜇 ≡ W ∨ 𝜑, we can consider the measure 𝜇0 ≡ 𝜇|M0 . This means that we can correctly define a mapping W : (A(T) )+ → Meas(T, M0 )+ setting W𝜑 ≡ (W ∨ 𝜑)|M0 and a mapping W b : (A(T) )+ → Measb (T, M0 )+ setting W b ≡ W|(A(T) )+ . They are injective. Theorem 2. Let A(T) be a truncatable lattice-ordered linear space of functions on a set T, A(T) be a lattice-ordered linear subspace of A(T)∨ , K ≡ I(A(T)) ∈ (A(T) )+ , 𝜇 ≡ W𝜑 and 𝜇0 ≡ W𝜑. Then, 1) 𝜇0 is a positive, inner K-regular,and upper K-continuous measure and 𝜇0 = 𝜇|M0 ; 2) 𝜑 is representable by the Lebesgue integral over the measurable space ⟮T, M0 , 𝜇0 ⟯; 3) the measure 𝜇0 is unique in the following sense: if there is another positive, inner K-regular, and upper K-continuous measure 𝜘 : M0 → R+ such that 𝜑 is representable by the Lebesgue integral over the space ⟮T, M0 , 𝜘⟯, then 𝜘 = 𝜇0 ; 4) if 𝜑 is uniformly bounded, then the measure 𝜇0 is bounded.

Proof. Consider 𝜆 and 𝜇 from the proof of Theorem 1. Let f ∈ A(T). According to the proof of Theorem 1 Sf ∈ K ⊂ Mf (𝜇). Therefore, Sf ∈ M0 and 𝜇0 Sf = 𝜇Sf < ∞. Furthermore, 𝜇0 Sf = 𝜇Sf = inf{𝜑g | g ∈ A(T)+ ∧ g ⩾ 𝜒(Sf )}. Thus, by Proposition 1 (3.4.2), 𝜑 is representable by the Lebesgue integral over the space ⟮T, M0 , 𝜇0 ⟯.

3.4.3 Representation of pointwise continuous functionals by Lebesgue integrals

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Now, we need to check the uniqueness of 𝜇0 . As in the proof of Theorem 1, we deduce that 𝜘K = 𝜇0 K for every K ∈ K. Now, from the property of the inner K-regularity, it follows that 𝜘 = 𝜇0 . If 𝜑 is uniformly bounded, then by virtue of Theorem 1, 𝜇0 is bounded. The family of all positive, inner I(A(T))-regular, and upper I(A(T))-continuous measures 𝜇 : M0 → R+ on T will be denoted by Meas(T, M0 , I(A(T)))0 . Its subfamily consisting of all bounded measures will be denoted by Measb (T, M0 , I(A(T)))0 . It follows from Theorem 2 that rng W ⊂ Meas(T, M0 , I(A(T)))0 and rng W b0 ⊂ Measb (T, M0 , I(A(T)))0 . Lemma 1. The family Meas0b (T, M0 , I(A(T)))0 is a conic space. Proof. All the arguments are similar to the arguments in the proof of Lemma 2 (3.4.2). It follows from Lemma 1 that we may consider the family Measb (T, M0 , I(A(T))) ≡ {𝜇1 − 𝜇2 | 𝜇1 , 𝜇2 ∈ Measb (T, I(A(T)))0 }. Corollary 1. The family Measb (T, M0 , I(A(T))) is a linear subspace of the linear space Measb (T, M0 ). Proposition 1. Let A(T) be a truncatable lattice-ordered linear space of functions on a set T. Then, 1) the family Measb (T, M0 , I(A(T))) is an l-ideal in the Dedekind complete latticeordered linear space Measb (T, M0 ); 2) Measb (T, M0 , I(A(T))) is a Dedekind complete lattice-ordered linear space; the smallest upper [the greatest lower] bounds of bounded above [below] sets in the ordered sets Measb (T, M0 ) and Measb (T, M0 , I(A(T))) coincide and they are expressed by the formulas from Theorem 2 (3.2.2) and its corollaries. Proof. All the arguments are similar to the arguments in the proofs of Proposition 2 (3.4.2) and its Corollary 1. Proposition 2. Let A(T) be a truncatable lattice-ordered linear space of functions on a set T and A(T) be a lattice-ordered linear subspace of A(T)∨ . Then, 1) the mapping W b has the unique extension to an injective linear operator J b : A(T) Measb (T, M0 ) between the given lattice-ordered linear spaces such that J b 𝜑 = W b (𝜑+ ) − W b (−𝜑− ) for every 𝜑 ∈ A(T) ; 2) every functional 𝜑 ∈ A(T) is representable by the Lebesgue integral over the measurable space ⟮T, M0 , J b 𝜑⟯; 3) rng J b ⊂ Measb (T, M0 , I(A(T))).

316 | 3.5 Topological spaces with measures. The Radon integral

Proof. 1. Using the same arguments as in the proof of Lemma 3 (3.4.2), we check that the mapping W b is injective and it is a conic operator, i. e. W b (x𝜑+y𝜓) = xW b 𝜑+yW b 𝜓 for every 𝜑, 𝜓 ∈ (A(T) ∨ )+ and every x, y ∈ R+ . By the condition, A(T) is a lattice-ordered linear space. By Corollary 2 to Lemma 2 (3.2.2), Meas(T, M0 ) is a lattice-ordered linear space as well. Therefore, by virtue of the mentioned properties and Statement 1 (3.4.2), W b has the unique extension to an injective linear operator J b : A(T) Measb (T, M0 ) such that J b 𝜑 = W b (𝜑+ ) − W b (−𝜑− ). 2. The assertion is checked in just the same way as in the proof of Proposition 3 (3.4.2). 3. The assertion follows from Proposition 1 and the inclusion rng W b ⊂ Measb (T, M0 , I(A(T))).

3.5 Topological spaces with measures. The Radon integral 3.5.1 Topological spaces with evaluations, semimeasures, and measures Let ⟮T, G⟯ be a topological space (see 2.1.1) and G, F, and C be the ensembles of all open, closed, and compact (≡ G-compact) subsets of T, respectively. Let S be an ensemble on T and 𝜀 : S → R be an evaluation (see 3.1.1). If there is a base G󸀠 of the topology G such that G󸀠 ⊂ S, then the evaluation 𝜀 is called topological or defined on the topological space ⟮T, G⟯. In this case, the topological space ⟮T, G⟯ with the evaluation 𝜀 will be called the topological space with evaluation or evaluable topological space and will be denoted by ⟮T, G, S, 𝜀⟯. If G ⊂ S, then 𝜀 is called quite topological. A topological space with evaluation ⟮T, G, R, 𝜇⟯ is called a topological space with a semimeasure [measure] or semimeasurable [measurable] topological space if the evaluation 𝜇 : R → R is a semimeasure [measure] on the ring R. Every (inner) C-regular (see 3.4.1) evaluation is called also compactly regular; G-regular and F-regular evaluations are called also openly regular and closedly regular, respectively. An evaluation 𝜀 is called locally bounded if for every point t ∈ T there is an open neighborhood G of t such that sup{|𝜀S| | S ∈ S ∧ S ⊂ G} < ∞. An evaluation 𝜀 will be called topologically internally finite, if for every S ∈ S such that 𝜀S > 0 [𝜀S < 0] there is G ∈ G ∩ S such that 0 < 𝜀G < ∞ [−∞ < 𝜀G < 0] and 0 < 𝜀(S ∩ G) < ∞ [−∞ < 𝜀(S ∩ G) < 0]. It is clear that topologically internally finite evaluation is internally finite. For the justification of this term, see Lemma 1. Every bounded evaluation such that T ∈ S is topologically internally finite. The fact that the extended Borel – Lebesgue measure 𝜆× on Rn has all enumerated properties (see the end of 3.1.6) justifies the naturalness of these notions.

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317

Let ⟮T, G, R, 𝜇⟯ be a topological space with an overfinite semimeasure [arbitrary measure] 𝜇 on a ring R [𝛿-ring R]. A subset S of T, consisting of all points t ∈ T such that |𝜇|G > 0 [v(𝜇)G > 0] for every set G ∈ G ∩ R containing the point t, is called the support (or carrier) of 𝜇 and is denoted by supp 𝜇 (or car 𝜇). Lemma 1. Let ⟮T, G, R, 𝜇⟯ be a topological space with a positive and topologically internally finite measure on a 𝜎-ring R. Then, 𝜇M = sup{𝜇(M ∩ G) | G ∈ G ∩ Rf (𝜇)} for every M ∈ R. Proof. Consider the number x ≡ sup{𝜇(M ∩ G) | G ∈ G ∩ Rf (𝜇)} ⩽ 𝜇M. If x = ∞, then x = 𝜇M. Suppose that x < ∞. Then, using any choice mapping p : P(G ∩ Rf )\{⌀} → G ∩ Rf from 1.1.12, we can find some sequence (G n ∈ G ∩ Rf | n ∈ N) such that 𝜇(M ∩ G n ) ⩾ x − 1/n. Consider the sets G ≡ ⋃ ⟮G n | n ∈ N⟯ and L ≡ M\G from R. Suppose that 𝜇L > 0. Then, by the condition, there is H ∈ G∩Rf such that 𝜇(L∩H) > 0. Therefore, 𝜇(L ∩ H) > 1/n for some n. Since H ∪ G n ∈ G ∩ Rf , we conclude by virtue of the additivity of 𝜇 that x ⩾ 𝜇(M ∩ (H ∪ G n )) = 𝜇((M ∩ H) ∪ (M ∩ G n )) ⩾ 𝜇((L ∩ H) ∪ (M ∩ G n )) = 𝜇(L ∩ H) + 𝜇(M ∩ G n ) > 1/n + x − 1/n = x, but this is impossible. It follows from this contradiction that 𝜇L = 0. Consider the sets Q n ≡ ⋃ ⟮G i | i ∈ n + 1⟯ ∈ G ∩ Rf . Since (M ∩ Q n | n ∈ N) ↑ M ∩ G in P(T), we infer by Lemma 5 (3.1.1) that 𝜇M = 𝜇(M ∩ G) + 𝜇L = lim(𝜇(M ∩ Q n ) | n ∈ N) ⩽ x. Hence, x = 𝜇M. Lemma 2. Let ⟮T, G, M, 𝜇⟯ be a topological space with a wide, positive, locally complete, strongly saturated, and topologically internally finite measure. If M ⊂ T and M ∩ G ∈ M for every G ∈ G ∩ Mf (𝜇), then M ∈ M. Proof. Take any L ∈ Mf . Using Lemma 1 and the arguments from its proof, we can find a sequence (G n ∈ G∩Mf | n ∈ N) such that 𝜇L = sup{𝜇(L ∩ G n ) | n ∈ N}. Consider the sets G ≡ ⋃ ⟮G n | n ∈ N⟯ and K ≡ L\G from M. Then, 𝜇(L ∩ G n ) ⩽ 𝜇(L ∩ G) = 𝜇L − 𝜇K for every n implies 𝜇L ⩽ 𝜇L − 𝜇K, and so 𝜇K = 0. Since (M ∩ L)\G ⊂ K, by virtue of the local completeness of 𝜇, we infer that (M ∩ L)\G ∈ M. On the other hand, by the condition, (M ∩L)∩G = ⋃ ⟮(M ∩G n )∩L | n ∈ N⟯ ∈ M. As a result, M ∩L ∈ M. It follows now from the strong saturation of 𝜇 that M ∈ M. Lemma 3. Let ⟮T, G, M, 𝜇⟯ and ⟮T, G, N, 𝜈⟯ be topological spaces with wide, positive, locally complete, strongly saturated, quite topological, topologically internally finite and closedly regular measures. Let Gf ≡ G ∩ Mf (𝜇) = G ∩ N f (𝜈). If 𝜇|Gf = 𝜈|Gf , then M = N and 𝜇 = 𝜈. Proof. Let N ∈ N and G ∈ Gf . By the property of the closed regularity of 𝜈 using the arguments from the proof of Lemma 1, we can find some sequences (K n ⊂ G ∩ N | n ∈ N) and (L n ⊂ G\N | n ∈ N) of closed sets such that 𝜈(G ∩ N) = sup{𝜈K n | n ∈ N}

318 | 3.5 Topological spaces with measures. The Radon integral

and 𝜈(G\N) = sup{𝜈L n | n ∈ N}. Consider the disjoint sets K ≡ ⋃ ⟮K n | n ∈ N⟯ ⊂ G ∩ N and L ≡ ⋃ ⟮L n | n ∈ N⟯ ⊂ G\N from M. We have 𝜈((G\K m )\L n ) = 𝜈(G\K m ) − 𝜈((G\K m ) ∩ L n ) = 𝜈G − 𝜈(G ∩ K m ) − 𝜈L n = 𝜈G − 𝜈K m − 𝜈L n . Consequently, Lemma 4 (1.4.5) and Corollary 1 to Proposition 2 (1.1.15) imply that 𝜇(G\(K ∪L)) ⩽ inf{𝜇(G\(K m ∪L n )) | (m, n) ∈ N×N} = inf{𝜈(G\(K m ∪ L n )) | (m, n) ∈ N × N} = inf{𝜈G − 𝜈K m − 𝜈L n | (m, n) ∈ N × N} = 𝜈G − sup{sup{𝜈K m | m ∈ N} + L n | n ∈ N} = 𝜈G − 𝜈(G ∩ N) − 𝜈(G\N) = 0. From (G ∩ N)\K ⊂ G\(K ∪ L), by virtue of the local completeness of 𝜇, we infer that (G ∩ N)\K ∈ M. Hence, N ∩ G = ((G ∩ N)\K) ∪ K ∈ M. Now, by Lemma 2, we get N ∈ M. In a similar way, we get M ∈ N. Let F ∈ F. Then, Lemma 1 implies that 𝜇F = sup{𝜇(F ∩ G) | G ∈ Gf } = sup{𝜇G − 𝜇(G\F) | G ∈ Gf } = sup{𝜈G − 𝜈(G\F) | G ∈ Gf } = 𝜈F. Now, it follows from the property of the closed regularity that 𝜇 = 𝜈. Lemma 4. Let ⟮T, G, M, 𝜇⟯ and ⟮T, G, N, 𝜈⟯ be topological spaces with wide, positive, quite topological, topologically internally finite, and lower G-continuous measures. If 𝜇 and 𝜈 coincide on some additive base of G, then 𝜇 and 𝜈 coincide on the ensembles G and K ≡ {G ∩ F | G ∈ G ∧ F ∈ F} (see 2.1.1), and therefore, on F. Proof. Let H be an additive base in G such that 𝜇H = 𝜈H for every H ∈ H. For the set G ∈ G, consider the set H󸀠 ≡ {H ∈ H | H ⊂ G} and the identical collection 𝜘 ≡ idH󸀠 ≡ (H H | H ∈ H󸀠 ). By the condition, 𝜘 ↑ G. Since 𝜇 and 𝜈 are lower G-continuous, we get 𝜇G = sup{𝜇H | H ∈ H󸀠 } = sup{𝜈H | H ∈ H󸀠 } = 𝜈G. Thus, 𝜇 and 𝜈 coincide on open sets. Hence, G ∩ Mf (𝜇) = G ∩ Mf (𝜈). Denote this set by Gf . Let F ∈ F and H ∈ G. Then, by Lemma 1, we obtain 𝜇(F ∩ H) = sup{𝜇(F ∩ H ∩ G) | G ∈ Gf } = sup{𝜇(H ∩ G) − 𝜇((H ∩ G)\F) | G ∈ Gf } = sup{𝜈(H ∩ G) − 𝜈((H ∩ G)\F) | G ∈ Gf } = sup{𝜈(F ∩ H ∩ G) | G ∈ Gf } = 𝜈(F ∩ H). This means that 𝜇 and 𝜈 coincide on the ensemble K. Obviously, F ⊂ K. Lemma 5. Let ⟮T, G, R, 𝜇⟯ be a topological space with a positive and closedly regular measure on a 𝛿-ring R. If R ∈ R and x ≡ inf{𝜇F | G ∈ G∩R ∧ R ⊂ G}, then either x = ∞ or x = 𝜇R. Moreover, if 𝜇 is bounded, then it is openly regular. Proof. Let x < ∞. Then, there exists a sequence (G n ∈ G ∩ Rf (𝜇) | n ∈ N) such that R ⊂ G n for every n and x = inf{𝜇G n | n ∈ N}. Consider the set S ≡ ⋂⟮G n | n ∈ N⟯ ∈ R. Suppose that 𝜇R < 𝜇S ⩽ x and consider the set Q ≡ S\R ∈ R. Since 𝜇Q = 𝜇S − 𝜇R > 0, we conclude by virtue of the closed regularity that there is F ∈ F ∩ R such that F ⊂ Q and 𝜇F > 0. Since H n ≡ G n \F ∈ G ∩ R and R ⊂ H n for every n ∈ N, we get x ⩽ 𝜇H n = 𝜇G n − 𝜇F. Hence, x ⩽ inf{𝜇G n | n ∈ N} − 𝜇F < x. From this contradiction, we infer that 𝜇R = 𝜇S. Now, consider the sets S n = ⋂⟮G i | i ∈ n + 1⟯ ∈ G ∩ Rf (𝜇). Then, (S n | n ∈ N) ↓ S in P(T), and therefore, Lemma 6 (3.1.1) and Lemma 7 (1.4.7) guarantee that

3.5.1 Topological spaces with evaluations, semimeasures, and measures |

𝜇S = lim(𝜇S n | n ∈ N) = inf{𝜇S n | n ∈ N} ⩽ inf{𝜇G n | n ∈ N} = x. implies x ⩽ inf{𝜇S n | n ∈ N} = 𝜇S. Thus, x = 𝜇S = 𝜇R.

Besides,

319

R ⊂ Sn

Lemma 6. Let ⟮T, G, R, 𝜇⟯ be a topological space with a positive, topologically internally finite, and lower G-continuous measure on a 𝜎-ring R. If M ∈ R, G ∈ G ∩ R and (G𝛼 ∈ G ∩ R | 𝛼 ∈ A) is a net such that (G𝛼 | 𝛼 ∈ A) ↑ G in P(T), then (𝜇(M ∩ G𝛼 ) | 𝛼 ∈ A) ↑ 𝜇(M ∩ G). Proof. First, consider any set H ∈ H ≡ G ∩ Rf (𝜇) and any net (H𝛼 ∈ H | 𝛼 ∈ A) such that (H𝛼 | 𝛼 ∈ A) ↑ H in P(T). By the lower G-continuity, (𝜇H𝛼 | 𝛼 ∈ A) ↑ 𝜇H. Consider the number x ≡ sup{𝜇(M ∩ H𝛼 ) | 𝛼 ∈ A} ⩽ 𝜇(M ∩ H). Then, for any 𝜀 > 0, there is 𝛽 ∈ A such that 𝜇H < 𝜇H𝛽 + 𝜀. Therefore, 𝜇(M ∩ H) = 𝜇H − 𝜇(H\M) < 𝜇H𝛽 + 𝜀 − 𝜇(H𝛽 \M) = 𝜇(M ∩ H𝛽 ) + 𝜀 ⩽ x + 𝜀. Since 𝜀 is arbitrary, we get 𝜇(M ∩ H) ⩽ x. Thus, x = 𝜇(M ∩ H). Now, let G and (G𝛼 | 𝛼 ∈ A) satisfy the condition. By Lemma 1, we obtain 𝜇(M ∩ G) = sup{𝜇(M ∩ G ∩ E) | E ∈ H} and 𝜇(M ∩ G𝛼 ) = sup{𝜇(M ∩ G𝛼 ∩ E) | E ∈ H}. Fix a set E. Consider the sets H ≡ G ∩ E and H𝛼 ≡ G𝛼 ∩ E from H. Since (H𝛼 | 𝛼 ∈ A) ↑ H in P(T), we conclude by the property proven above that sup{𝜇(M ∩ G𝛼 ∩ E) | 𝛼 ∈ A} = 𝜇(M ∩ G ∩ E). Consider the sets I ≡ A × H, I𝛼 ≡ {𝛼} × H, and I E ≡ A × {E} for every 𝛼 ∈ A and E ∈ H. We have I = ⋃ ⟮I𝛼 | 𝛼 ∈ A⟯ = ⋃ ⟮I E | E ∈ H⟯. Denote 𝜇(M ∩ G𝛼 ∩ E) by x i for i ≡ (𝛼, E). In these notations, we have 𝜇(M ∩ G𝛼 ) = sup{x i | i ∈ I𝛼 } and sup{x i | i ∈ I E } = 𝜇(M ∩ G ∩ E). Now, Proposition 2 (1.1.15) implies 𝜇(M ∩ G) = sup{𝜇(M ∩ G ∩ E) | E ∈ H} = sup{sup{x i | i ∈ I E } | E ∈ H} = sup{x i | i ∈ I} = sup{sup{x i | i ∈ I𝛼 } | 𝛼 ∈ A} = sup{𝜇(M ∩ G𝛼 ) | 𝛼 ∈ A}. Lemma 7. Let ⟮T, G, M, 𝜇⟯ be a topological space with a wide, positive, locally complete, strongly saturated, and closedly regular [compactly regular] measure. If M ⊂ T and M ∩ F ∈ M for every F ∈ F ∩ Mf (𝜇) [F ∈ C ∩ Mf (𝜇)], then M ∈ M. Proof. Take any L ∈ Mf . Since 𝜇 is closedly regular, we have 𝜇L = sup{𝜇F | F ∈ F ∩ Mf ∧ F ⊂ L}. Therefore, there exists a sequence (F n ∈ F ∩ Mf | n ∈ N) ↑ such that ⋃ ⟮F n | n ∈ N⟯ ⊂ L and 𝜇L = sup{𝜇F n | n ∈ N}. Consider the sets E ≡ ⋃ ⟮F n | n ∈ N⟯ and K ≡ L\E from M. Since 𝜇F n ⩽ 𝜇E = 𝜇L − 𝜇K for every n, we get 𝜇L ⩽ 𝜇L − 𝜇K. Hence, 𝜇K = 0. From (M ∩ L)\E ⊂ K by virtue of the local completeness of 𝜇, we infer that (M ∩ L)\E ∈ M. On the other hand, by the condition, (M ∩ L)∩ E = ⋃ ⟮(M ∩ F n )∩ L | n ∈ N⟯ ∈ M, and so M ∩ L ∈ M. It follows now from the strong saturation of 𝜇 that M ∈ M. For the compactly regular 𝜇, the arguments are the same. Lemma 8. Let ⟮T, G, M, 𝜇⟯ and ⟮T, G, N, 𝜈⟯ be topological spaces [Hausdorff spaces] with wide, positive, locally complete, strongly saturated, quite topological, and closedly regular [compactly regular] measures. If 𝜇|F = 𝜈|F [𝜇|C = 𝜈|C], then M = N and 𝜇 = 𝜈.

320 | 3.5 Topological spaces with measures. The Radon integral

Proof. By the condition, F ∩ Mf (𝜇) = F ∩ N f (𝜈). Denote this ensemble by F f . Let N ∈ N and F ∈ F f . By the property of the closed regularity of 𝜈, we can find some sequences (K n ⊂ F ∩ N | n ∈ N) ↑ and (L n ⊂ F\N | n ∈ N) ↑ of closed sets from F f such that 𝜈(F ∩ N) = sup{𝜈K n | n ∈ N} and 𝜈(F\N) = sup{𝜈L n | n ∈ N}. Consider the disjoint sets K ≡ ⋃ ⟮K n | n ∈ N⟯ ⊂ F ∩ N and L ≡ ⋃ ⟮L n | n ∈ N⟯ ⊂ F\N from M. We have 𝜈((F\K m )\L n ) = 𝜈(F\K m ) − 𝜈((F\K m ) ∩ L n ) = 𝜈F − 𝜈(F ∩ K m ) − 𝜈L n = 𝜈F− 𝜈K m − 𝜈L n . Consequently, Lemma 5 (3.1.1) implies 𝜇K = sup (𝜇K n | n ∈ N) = sup (𝜈K n | n ∈ N) = 𝜈(F ∩ N) and 𝜇L = sup (𝜇L n | n ∈ N) = sup (𝜈L n | n ∈ N) = 𝜈(F\N). Therefore, 𝜇(K ∪ L) = 𝜇K + 𝜇L = 𝜈(F ∩ N) + 𝜈(F\N) = 𝜈F. As a result, 𝜈F = 𝜇F = 𝜇(F\(K ∪ L)) + 𝜇(K ∪ L) = 𝜇(F\(K ∪ L)) + 𝜈F, where 𝜇(F\(K ∪ L)) = 0. From (F ∩ N)\K ⊂ F\(K ∪ L), by virtue of the local completeness of 𝜇, we infer that (F ∩ N)\K ∈ M. Hence, N ∩ F = ((F ∩ N)\K) ∪ K ∈ M. Now, by Lemma 7, we get N ∈ M. Thus, N ⊂ M. In a similar way, we get M ⊂ N. Now, it follows from the property of the closed regularity that 𝜇 = 𝜈. For the compact sets, the arguments are the same. Lemma 9. Let ⟮T, G, M, 𝜇⟯ be a Hausdorff space with a wide, positive, quite topological, locally bounded measure 𝜇. Then, C ⊂ Mf (𝜇), and for every C ∈ C, there is G ∈ Gf (𝜇) such that C ⊂ G. In addition, if 𝜇 is lower G-continuous, then it is upper C-continuous. Proof. Let C be a compact subset of T. By Statement 5, the set C is closed, and there/ ⌀} and the identical fore, C ∈ M. Consider the ensemble Q ≡ {Q ∈ Gf (𝜇) | Q ∩ K = collection 𝜘 ≡ idQ ≡ (Q Q | Q ∈ Q) of elements of Q (see 1.1.8 and 1.1.9). Since 𝜇 is locally bounded, idQ is a cover of C. By the property of compactness, there is a finite subcover (Q Q | Q ∈ 𝜃 ⊂ 𝜘) of C. Consider the open set G ≡ ⋃ ⟮Q Q | Q ∈ 𝜃⟯. By Lemmas 3 and 4 (3.1.1), we get 𝜇C ⩽ 𝜇G ⩽ ∑(𝜇Q | Q ∈ 𝜃) < ∞. Now, suppose that 𝜇 is lower G-continuous. Let C ∈ C and (C𝛾 ∈ C | 𝛾 ∈ Γ) ↓ C in P(T). Fix any 𝜁 ∈ Γ and consider the set Δ ≡ {𝛾 ∈ Γ | 𝛾 ⩾ 𝜁}. Take G ∈ Gf (𝜇) such that C𝜁 ⊂ G. Consider the open sets G𝛿 ≡ G\C𝛿 . Since (C𝛿 | 𝛿 ∈ Δ) ↓ C, we conclude that (G𝛿 | 𝛿 ∈ Δ) ↑ G\C. By the condition, (𝜇G𝛿 | 𝛿 ∈ Δ) ↑ 𝜇(G\C). As a result, inf{𝜇C𝛿 | 𝛿 ∈ Δ} = inf{𝜇G − 𝜇G𝛿 | 𝛿 ∈ Δ} = 𝜇G − sup{𝜇G𝛿 | 𝛿 ∈ Δ} = 𝜇G − 𝜇(G\C) = 𝜇C. This implies (𝜇C𝛾 | 𝛾 ∈ Γ) ↓ 𝜇C. Corollary 1. In the conditions of Lemma 9, 𝜇C = inf{𝜇G | G ∈ G ∩ Mf (𝜇) ∧ C ⊂ G} for every C ∈ C. Proof. The assertion follows from Lemma 9 and Lemma 5. Corollary 2. Let ⟮T, G, M, 𝜇⟯ be a Hausdorff space with a wide, positive, quite topological, locally bounded, and compactly regular measure. Then, for every M ∈ Mf (𝜇) and every 𝜀 > 0, there is G ∈ G ∩ Mf (𝜇) such that 𝜇(M\G) = 0 and 𝜇(G\M) < 𝜀.

3.5.1 Topological spaces with evaluations, semimeasures, and measures |

321

Proof. By the condition, there is a sequence (K n ∈ C | n ∈ 𝜔) such that E ≡ ⋃{K n | n ∈ 𝜔} ⊂ M and 𝜇M = sup{𝜇K n | n ∈ 𝜔}. By Corollary 1, there is a sequence (G n ∈ G ∩ Mf (𝜇) | n ∈ 𝜔) such that K n ⊂ G n and 𝜇K n + 𝜀/2n+1 > 𝜇G n . Consider the open set G ≡ ⋃ ⟮G n | n ∈ 𝜔⟯. From G\M ⊂ ⋃ ⟮G n \K n | n ∈ 𝜔⟯ we deduce by Lemma 3 (3.1.1) and Lemma 3 (1.4.8) that 𝜇(G\M) ⩽ ∑net (𝜇(G n \K n ) | n ∈ 𝜔) = ∑(𝜇G n − 𝜇K n | n ∈ 𝜔) < ∑(𝜀/2n+1 | n ∈ 𝜔) = 𝜀. On the other hand, 𝜇(M\G) ⩽ 𝜇(M\K n ) = 𝜇M − 𝜇K n for every n implies 𝜇(M\G) ⩽ inf{𝜇M − 𝜇K n | n ∈ 𝜔} = 𝜇M − sup{𝜇K n | n ∈ 𝜔} = 0. Lemma 10. Let ⟮T, G, M, 𝜇⟯ be a topological space with a quite topological and lower G-continuous measure. Then, the set supp 𝜇 is closed and 𝜇(T\ supp 𝜇) = 0. Proof. Consider the ensemble Q ≡ {Q ∈ G | v(𝜇)Q = 0}, the identical collection 𝜘 ≡ idQ ≡ (Q Q | Q ∈ Q), and the set R ≡ ⋃ ⟮Q | Q ∈ Q⟯ ∈ G. By virtue of Lemma 4 (3.1.1), the collection 𝜘 is increasing, and so 𝜘 ↑ R. Since 𝜇 is lower G-continuous, we conclude that 𝜇R = sup{v(𝜇)Q | Q ∈ 𝜘} = 0. It is evident that R = T\ supp 𝜇. Therefore, supp 𝜇 = T\R is closed. Now, we shall consider for a positive measure 𝜇 on a ring R its extensions from 3.1.4. For a positive measure 𝜇, consider its completion μ̌ : K(T, R, 𝜇) → R+ . Lemma 11. Let ⟮T, G, R, 𝜇⟯ be a topological space [Hausdorff space] with a positive measure 𝜇 on a ring R. If 𝜇 is topologically internally finite, then 𝜇̌ is topologically internally finite as well. If 𝜇 is closedly regular [compactly regular], then 𝜇̌ is the same. ̌ > 0. Proof. Denote K(T, R, 𝜇) simply by K and N(T, R, 𝜇) by N. Let K ∈ K and 𝜇K Then, according to 3.1.4, K = R ∪ N for some R ∈ R and N ∈ N(T, R, 𝜇). Besides, ̌ = 𝜇R and 𝜇N ̌ = 0. If 𝜇 N ⊂ N0 for some N0 ∈ N0 (T, R, 𝜇). By Theorem 4 (3.1.4), 𝜇K is topologically internally finite, then there is G ∈ G ∩ R such that 0 < 𝜇G < ∞ and 0 < 𝜇(R ∩ G) < ∞. Consider the sets S ≡ R ∩ G ∈ R and P ≡ K ∩ G\S ⊂ N0 . Since P ∈ N(T, R, 𝜇), we conclude that K ∩ G = S ∪ P ∈ K. Therefore, by Theorem 4 (3.1.4), ̌ ∩ G) = 𝜇S. Then, 𝜇̌ is topologically internally finite. 𝜇(K If 𝜇 is closedly regular, then 𝜇R = sup{𝜇F | F ∈ F ∩ Rf (𝜇) ∧ F ⊂ R}. Therefore, ̌ = 𝜇R ⩽ sup{𝜇F ̌ | F ∈ F ∩ Kf (𝜇)̌ ∧ F ⊂ K} ⩽ 𝜇K ̌ that 𝜇̌ is closedly it follows from 𝜇K regular as well. If ⟮T, G⟯ is Hausdorff and 𝜇 is compactly regular, then the arguments are the same. If a positive measure 𝜇 on a 𝛿-ring R is internally finite, then according to 3.1.4, we may consider the complete strongly saturated extension 𝜇#: M#(T, R, 𝜇) → R+ of 𝜇. Lemma 12. Let ⟮T, G, R, 𝜇⟯ be a topological space [Hausdorff space] with a positive internally finite measure 𝜇 on a 𝛿-ring R. If 𝜇 is topologically internally finite, then 𝜇# is topologically internally finite as well. If 𝜇 is closedly regular [compactly regular], then 𝜇# is the same.

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Proof. Denote K(T, R, 𝜇) simply by K and M#(T, R, 𝜇) by M#. Let M ∈ M#. Accorď | L ∈ Kf (𝜇)̌ ∧ L ⊂ M}. If 𝜇#M > 0, then 0 < 𝜇L ̌ for some ing to 3.1.4, 𝜇#M = sup{𝜇L ̌ < ∞ and L ∈ K such that L ⊂ M. By Lemma 11, there is G ∈ G ∩ K such that 𝜇G ̌ ∩ G). Hence, 0 < 𝜇(L ̌ ∩ G) = 𝜇#(L ∩ G) ⩽ 𝜇#(M ∩ G) ⩽ 𝜇#G = 𝜇G ̌ < ∞. Thus, 0 < 𝜇(L # 𝜇 is topologically internally finite. If 𝜇 is closedly regular, then by Lemma 11, 𝜇̌ is the same. Take any real number, ̌ > x. By virtue of closed x < 𝜇#M. Then, there is L ∈ Kf (𝜇) such that L ⊂ M and 𝜇L f ̌ > x. Therefore, regularity, of 𝜇,̌ there is F ∈ F ∩ K (𝜇) such that F ⊂ L and 𝜇#F = 𝜇F x < sup{𝜇#F | F ∈ F ∩ (M#)f (𝜇#) ∧ F ⊂ M} = y. Since x is arbitrary, we deduce that 𝜇#M ⩽ y ⩽ 𝜇#M. This means that 𝜇# is closedly regular. If ⟮T, G⟯ is Hausdorff and 𝜇 is compactly regular, then the arguments are the same. Lemma 13. Let ⟮T, G, R, 𝜇⟯ be a topological space with a positive, quite topological, and internally finite measure 𝜇 on a 𝛿-ring R. If 𝜇 is lower G-continuous, then 𝜇̌ and 𝜇# are the same. Proof. Let G ∈ G, (G𝛼 ∈ G | 𝛼 ∈ A) be a net, and (G𝛼 | 𝛼 ∈ A) ↑ G in P(T). Since G ⊂ R ⊂ K(T, R, 𝜇) ⊂ M(T, R, 𝜇), all three measures 𝜇, 𝜇,̌ and 𝜇# coincide on G. By the condition, (𝜇G𝛼 | 𝛼 ∈ A) ↑ 𝜇G. Hence, the same is valid for 𝜇̌ and 𝜇#. Auxiliary definitions and statements 1∘ Let ⟮T, G⟯ be a topological space (see 2.1.1) and G, F, and C be the ensembles of all open, closed, and compact (≡ G-compact) subsets of T, respectively. If t ∈ T, V ∈ G, and t ∈ V, then V is called an (open) neighborhood of the point t. It is easy to see that a set G ⊂ T is open (i. e. G ∈ G) iff for every t ∈ G there is its (open) neighborhood V such that V ⊂ G. An ensemble G󸀠 on T is called a base of the topological space ⟮T, G⟯ or a base of the topology G if for every non-empty set G ∈ G there is a collection (G i ∈ G󸀠 | i ∈ I) such that G = ⋃ ⟮G i | i ∈ I⟯. It is easily proven that an ensemble S is a base iff for every t ∈ T and every neighborhood V ∋ t there is S ∈ S such that t ∈ S ⊂ V. Statement 1. Let ⟮T, G⟯ be a topological space, C ∈ C, F ∈ F, and F ⊂ C. Then, F ∈ C. Let S ⊂ T. Consider the ensemble F S ≡ {F ∈ F | S ⊂ F} of all closed sets containing S. The set cl S ≡ ⋂⟮F | F ∈ F S ⟯ is called the closure of the set S. Similarly, consider the ensemble GS ≡ {G ∈ G | S ⊃ G}. The set int S ≡ ⋃ ⟮G | G ∈ GS ⟯ is called the interior of the set S. The set fr S ≡ cl S\ int S is called the boundary or the frontier of the set S in the topological space ⟮T, G⟯. Clearly, cl S ∈ F, int S ∈ G, and fr S ∈ F. Statement 2. Let ⟮T, G⟯ be a topological space and S ⊂ T. Then: 1) S ∈ F iff S = cl S; 2) S ∈ G iff S = int S.

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Statement 3. Let ⟮T, G⟯ be a topological space, G ∈ G, P ⊂ T, and G ∩ P = ⌀. Then, G ∩ cl P = ⌀. Statement 4. Let ⟮T, G⟯ be a topological space and S, S󸀠 , S󸀠󸀠 ⊂ T. Then, 1) fr S = cl S ∩ cl(T\S); 2) fr S = fr(T\S); 3) fr S󸀠 ∪ fr S󸀠󸀠 = fr(S󸀠 ∪ S󸀠󸀠 ) ∪ fr(S󸀠 ∩ S󸀠󸀠 ) ∪ (fr S󸀠 ∩ fr S󸀠󸀠 ). 2∘ A topological space ⟮T, G⟯ is called a Hausdorff space if for every t󸀠 , t󸀠󸀠 ∈ T such that t󸀠 ≠ t󸀠󸀠 there are neighborhoods G󸀠 and G󸀠󸀠 such that G󸀠 ∩ G󸀠󸀠 = ⌀. Statement 5. Let ⟮T, G⟯ be a Hausdorff space. Then, C ⊂ F. Statement 6. Let ⟮T, G⟯ be a compact Hausdorff space. Then, it is normal (see 2.1.1). Statement 7. Let ⟮T, G⟯ be a Hausdorff space and (K i | i ∈ I) be a finite collection of pairwise disjoint compact subsets of T. Then, there exists a finite collection (U i | i ∈ I) of pairwise disjoint open sets such that K i ⊂ U i . A Hausdorff space ⟮T, G⟯ is called a locally compact space if for every t ∈ T there is a neighborhood U of the point t such that the space ⟮cl U, Gcl U ⟯ is compact in the sense of 2.1.1. A Hausdorff space ⟮T, G⟯ is called a Tychonoff space or a completely regular space if for every t ∈ T and every F ∈ F such that t ∈ ̸ F there is f ∈ C b (T, G) such that f (t) = 0 and f [F] = {1}. Statement 8. Let ⟮T, G⟯ be a Tychonoff space, K ∈ C, F ∈ F, and F ⊂ T\K. Then, there is a continuous function f : T → [0, 1] such that f [K] = {0} and f [F] = {1}. Statement 9. Let ⟮T, G⟯ be a locally compact space. Then, it is a Tychonoff space. A subset P of T is called precompact if cl P is compact. Statement 10. Let ⟮T, G⟯ be a locally compact space, C ∈ C, G ∈ G, and C ⊂ G. Then, there exists U ∈ G such that C ⊂ U ⊂ cl U ⊂ G and cl U ∈ C.

3.5.2 Measurable and integrable functions on topological spaces with measures Let ⟮T, G⟯ be a topological space. A function f ∈ F(T) is called a function with a compact support if the set supp f ≡ cl coz f (the support of the function f ) is compact. The family of all such

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functions on T is denoted by F c (T, G). The family F b (T) ∩ F c (T, G) of all bounded functions with compact supports is denoted by F bc (T, G). If A(T) is any subfamily in F(T), then the family A(T)∩F c (T, G) will be denoted by A c (T, G). It is called the part of A(T) with compact supports. Analogously the family A(T) ∩ F bc (T, G) will be denoted by A bc (T, G). It is called the bounded part of A(T) with compact supports. Recall that a G-measurable (see 2.3.1) function f ∈ F(T) is called continuous on the topological space ⟮T, G⟯. The family M(T, G) of all continuous functions on ⟮T, G⟯ is denoted usually by C(T, G). It follows from Lemma 3 (2.5.1) and Corollary 1 to Lemma 2 (2.4.1) that C c (T, G) = C bc (T, G). Similarly, a lower [upper] G-semimeasurable function f on the set T is called lower [upper] semicontinuous on the topological space ⟮T, G⟯ (see 2.3.8). The families SM l (T, G) and SM u (T, G) of all lower semicontinuous and upper semicontinuous functions on ⟮T, G⟯ is denoted by SC l (T, G) and SC u (T, G), respectively. Consider also the family S(T, G) of symmetrizable function on the space ⟮T, G⟯ (see 2.4.5). Recall that a function f : T → R is called symmetrizable if for every 𝜀 > 0 there is a finite cover (K i ∈ K(T, G) | i ∈ I) of the set T such that 𝜔(f , K i ) ≡ sup{|f (s) − f (t)| | s, t ∈ K i } < 𝜀 for every i ∈ I. According to 2.1.1 a set K in T is symmetrizable iff K = F ∩ G for some sets F ∈ F and G ∈ G and K ≡ K(T, G) denotes the ensemble of all symmetrizable sets. Consider on T the algebra A ≡ A(T, G) of all Aleksandrov sets in T, i. e. the algebra generated by the ensemble G. According to Proposition 3 (2.1.1), A(T, G) = K𝜑 . By Theorem 1 (2.4.5), the family St(T, A(T, G)) of all A-step functions on T is uniformly dense in the family S(T, G). Lemma 1 (2.4.5) guarantees that SC lb (T, G) ∪ SC ub (T, G) ⊂ S(T, G). A function f on T is called separating a point s ∈ T and a closed set F ⊂ T if f (s) = 1 and f (t) = 0 for every t ∈ F. If A(T) is a functional family on T, then it is called separating points and closed sets in the topological space ⟮T, G⟯ if for every s ∈ T and every F ∈ F such that s ∉ F there is a function f ∈ A(T) separating s and F. Proposition 1. Let ⟮T, G⟯ be a topological space. Then, 1) if it is Hausdorff, then the families SC lb (T, G), SC uc (T, G), and S(T, G) separate points and closed sets; 2) if it is Tychonoff, then the family C b (T, G) separates points and closed sets; 3) if it is locally compact, then the family C c (T, G) separates points and closed sets. Proof. Let F ∈ F, G ≡ T\F, and s ∉ F. 1. The function 𝜒(G) belongs to SC lb and separates s and F. On the other hand, the function 𝜒({s}) belongs to SC uc and separates s and F. By Lemma 1 (2.4.5)m SC lb (T, G) ∪ SC ub (T, G) ⊂ S(T, G). 2. The assertion follows immediately from the definition of a Tychonoff space.

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3. Since the set {s} is compact, by Statement 10 (3.5.1), there is a precompact open set H such that s ∈ H ⊂ cl H ⊂ G. By Statement 9 (3.5.1), the space ⟮T, G⟯ is Tychonoff, and therefore, there is a continuous function f : T → [0, 1] such that f (s) = 1 and f (t) = 0 for every t ∈ T\H ⊃ T\G = F. From coz f ⊂ H, it follows that supp f ⊂ cl H. Thus, f has the compact support. Now, we shall consider a measurable topological space ⟮T, G, M, 𝜇⟯. The measurable functions on a measurable topological space ⟮T, G, M, 𝜇⟯ have some additional (topological) properties in comparison with properties of measurable functions on an abstract measurable space ⟮T, M, 𝜇⟯ considered in 3.3.1. Theorem 1 (Lusin). Let ⟮T, G, M, 𝜇⟯ be a topological space [Hausdorff space] with a wide, positive, complete, strongly saturated, quite topological, and closedly regular [compactly regular] measure. Then, for a function f ∈ F(T), the following conclusions are equivalent: 1) f ∈ M(T, M); 2) for every set M ∈ Mf (𝜇) and every number 𝜀 > 0, there is a closed [compact] set F ⊂ M such that 𝜇(M\F) < 𝜀 and f |F ∈ C(F, GF ). Proof. (1) ⊢ (2). By Corollary 2 to Theorem 2 (3.3.1), for M, there is a sequence (f n ∈ St(T, Mf (𝜇)) ⊂ M(T, M) | n ∈ 𝜔) such that for 𝜀 > 0 there is a set N ∈ Mf such that B ⊂ A, 𝜇(M\B) < 𝜀/2, and f |B = u-lim(f n |B | n ∈ 𝜔). Fix for a moment a number n. By Lemma 4 (2.2.5), f n 𝜒(B) = ∑(x i 𝜒(M i ) | i ∈ I) for some inite collection (M i ∈ M | i ∈ I) of pairwise disjoint sets and some collection (x i ∈ R | i ∈ I) such that p ≡ P(I) ∈ N. We may regard that ⋃ ⟮M i | i ∈ I⟯ = B. Since 𝜇 is closedly regular, there are closed sets H i ⊂ M i such that 𝜇(M i \H i ) < 𝜀/(p2n+2 ). Consider the closed set F ≡ ⋃ ⟮H i | i ∈ I⟯. Then, we have 𝜇(B\F n ) = 𝜇(⋃ ⟮M i \H i | i ∈ I⟯) = ∑(𝜇(M i \H i ) | i ∈ I) ⩽ 𝜀/2n+2 . Besides, f n (t) = x i for every t ∈ H i . Consider the function g ≡ f n |F n . If x < y in R and P ≡ g−1 []x, y[], then either P ∩ H i = H i or P ∩ H i = ⌀ for every i ∈ I. Consider the open sets G i ≡ T\ ⋃ ⟮H j | j ∈ I\{i}⟯. Define a collection (Q i ∈ G | i ∈ I) setting Q i ≡ G i if P ∩ H i = H i and Q i ≡ ⌀ if P ∩ H i = ⌀. Consider the open set Q ≡ ⋃ ⟮Q i | i ∈ I⟯. Then, we have P = ⋃ ⟮P ∩ H i | i ∈ I⟯. If P ∩ H i = H i , then P ∩ H i = H i ∩ Q i = H i ∩ Q. If P ∩ H i = ⌀, then P ∩ H i = H i ∩ Q i = H i ∩ Q as well. Therefore, P = ⋃ ⟮P ∩ H i | i ∈ I⟯ = ⋃ ⟮H i ∩ Q | i ∈ I⟯ = F n ∩ Q ∈ GF n . This means that g ∈ C(F n , GF n ). Consider now the closed set F ≡ ⋂⟮F n | n ∈ 𝜔⟯. Then, by Lemma 3 (3.1.1) and Lemma 3 (1.4.8), 𝜇(B\F) = 𝜇(⋃ ⟮B\F n | n ∈ 𝜔⟯) ⩽ ∑net (𝜇(B\F n ) | n ∈ 𝜔) ⩽ ∑(𝜀/2n+2 | n ∈ 𝜔) = 𝜀/2. Besides, f n |F ∈ C(F, GF ). Since f |F = u-lim (f n |F | n ∈ 𝜔), we conclude by Proposition 2 (2.3.4) that f |F ∈ C(F, GF ). As a result, we obtain 𝜇(M\F) = 𝜇(M\B) + 𝜇(B\F) < 𝜀. If ⟮T, G⟯ is Hausdorff and 𝜇 is compactly regular, then by Statement 5 (3.5.1), C ⊂ F and the arguments are quite similar. (2) ⊢ (1). Let M ∈ Mf (𝜇). Then, by condition for M, there is a sequence (F n ∈ F | n ∈ N) such that F n ⊂ M, 𝜇(M\F n ) < 1/n, and f |F n ∈ C(F n , GF n ). Consider

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the sets L ≡ ⋃ ⟮F n | n ∈ N⟯ and N = M\L from M. From 𝜇N ⩽ 𝜇(M\F n ) < 1/n, we conclude that 𝜇N = 0. Take any x < y and consider the set P ≡ f −1 []x, t[]. Then, Q n ≡ F n ∩ P = (f |F n )−1 []x, y[] is an open set in the topological space (F n , GF n ), and so Q n = F n ∩ G for some G ∈ G, where Q n ∈ M. Thus, L ∩ P = ⋃ ⟮Q n | n ∈ N⟯ ∈ M. Since 𝜇 is locally complete and N ∩ P ⊂ N ∈ M0 (𝜇), we conclude that N ∩ P ∈ M. As a result, we obtain M ∩ P = (L ∩ P) ∪ (N ∩ P) ∈ M. Since 𝜇 is strongly saturated, we get P ∈ M. This means that f ∈ M(T, M). For a measurable topological space, ⟮T, G, M, 𝜇⟯ with a quite topological measure 𝜇, we shall consider the ensembles Gf (𝜇) ≡ G ∩ Mf (𝜇), F f (𝜇) ≡ F ∩ Mf (𝜇), and Cf (𝜇) ≡ C ∩ Mf (𝜇). Integrable functions on some measurable topological spaces also have additional topological properties. Lemma 1. Let ⟮T, G⟯ be a topological space and M be a 𝜎-algebra on T containing G. Then, 1) the functional families C(T, G), SC l (T, G), SC u (T, G), S(T, G), and BM(T, G) ≡ M(T, B(T, G)) are contained in M(T, M); 2) for every bounded measure 𝜇 on M the functional families C b (T, G), SC lb (T, G), SC ub (T, G), S(T, G), and BM b (T, G) are contained in M b (T, M) ⊂ MI(T, M, 𝜇); 3) for every measure, 𝜇 on M such that C = Cf (𝜇) the functional families C c (T, G), SC lbc (T, G), SC ubc (T, G), S c (T, G), and BM bc (T, G) are contained in M bc (T, M) ⊂ MI(T, M, 𝜇). Proof. 1. Since M is a 𝜎-algebra containing G and B is the smallest 𝜎-algebra containing G, we have B ⊂ M, and therefore, BM ≡ M(T, B) ⊂ M(T, M). Lemma 1 (2.3.8) implies that C(T, G) ⊂ SC l (T, G) ∩ SC u (T, G). Theorem 2 (2.3.8) guarantees that SC l (T, G) ∪ SC u (T, G) ⊂ M(T, K𝜎 ), where K ≡ K(T, G) (see 2.1.1). By Lemma 4 (2.4.1), S(T, G) ⊂ M(T, K𝜎 ). Since K𝜎 ⊂ B, we infer that SC l ∪ SC u ⊂ BM and S(T, G) ⊂ BM. 2. Let f ∈ M b (T, M) and |f | ⩽ x1. Then, by Lemmas 1 and 2 (3.3.2), ∫ f+ d𝜇+ ⩽ x ∫ 1 d𝜇+ = x𝜇+ T < ∞. Similarly, ∫(−f− ) d𝜇+ ⩽ x𝜇+ T < ∞. By the definition, from 3.3.2, f ∈ MI(T, M, 𝜇+ ). Analogously, f ∈ MI(T, M, −𝜇− ). By the definition, from 3.3.6, f ∈ MI(T, M, 𝜇). Thus, M b (T, M) ⊂ MI(T, M, 𝜇). Now, the necessary inclusions follow from (1). 3. Let f ∈ M bc (T, M) and |f | ⩽ x𝜒(K) for some K ∈ C. Then, ∫ f+ d𝜇+ ⩽ x ∫ 𝜒(K) d𝜇+ = x𝜇+ K < ∞. Similarly, ∫(−f− ) d𝜇+ ⩽ x𝜇+ K < ∞. By definition, f ∈ MI (T, M, 𝜇+ ). Analogously, f ∈ MI(T, M, −𝜇− ). Then, by the definition from 3.3.6, f ∈ MI(T, M, 𝜇). Thus, M bc (T, M) ⊂ MI(T, M, 𝜇). Now, the necessary inclusions follow from (1). Theorem 2. Let ⟮T, G, M, 𝜇⟯ be a topological [Hausdorff ] space with a wide, positive, quite topological, and closedly regular [compactly regular] measure and f ∈

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MI(T, M, 𝜇). Then, there is a sequence (f n | n ∈ N) such that f n = ∑(x ni 𝜒(K ni ) | i ∈ I n ) for some finite collection (K ni | i ∈ I n ) of pairwise disjoint sets from F f (𝜇) [Cf (𝜇)] and some collection (x ni ∈ R | i ∈ I n ) of real numbers, |f n | ⩽ |f |, and ‖f n − f ‖i ⩽ z n for some sequence (z n ∈ R | n ∈ N) ↓ 0. Moreover, 1) if ⟮T, G⟯ is Hausdorff and 𝜇 is compactly regular and locally bounded, then there are collections (G ni ∈ Gf (𝜇) | i ∈ I n ) of pairwise disjoint sets such that K ni ⊂ G ni and the functions h n ≡ ∑(x ni 𝜒(G ni ) | i ∈ I n ) satisfy the inequalities ‖f n − h n ‖i ⩽ z n ; 2) if ⟮T, G⟯ is Tychonoff, then there are collections (𝜑ni ∈ C b (T, G) | i ∈ I) of func−1 [{1}] ⊂ coz 𝜑ni ⊂ G ni , 0 ⩽ 𝜑ni ⩽ 1, and the functions g n ≡ tions such that K ni ⊂ 𝜑ni ∑(x ni 𝜑ni | i ∈ I n ) satisfy the inequalities ‖f n − g n ‖i ⩽ ‖f n − h n ‖i ⩽ z n ; 3) if ⟮T, G⟯ is locally compact, then the functions 𝜑ni can be chosen from the family C c (T, G); 4) in all the cases for f ⩾ 0, all the functions f n , g n , and h n can be chosen such that 0 ⩽ fn ⩽ gn ⩽ hn ⩽ f . Proof. By Proposition 1 (3.3.4), there is a sequence (s n ∈ St(T, M) ∩ MI(T, M, 𝜇) | n ∈ N) such that s n = ∑(x ni 𝜒(M ni ) | i ∈ I n ) for some finite collection (M ni ∈ M | i ∈ I n ) of pairwise disjoint sets, |s n | ⩽ |f | and lim (‖s n − f ‖i | n ∈ N) = 0. If f ⩾ 0, then the functions s n can be chosen such that 0 ⩽ s n ⩽ f . From |s n | ∈ MI(T, M, 𝜇) we conclude by Lemma 1 (3.3.2) that ∑(|x ni |𝜇M ni | i ∈ I n ) = ∫ |s n | d𝜇 < ∞. According to 3.3.2, we use the equality 0∞ = 0 (see 1.4.3). Therefore, we can regard that 𝜇M ni < ∞ and p n ≡ P(I n ) ⩾ 1 for all indices. By Lemma 3 (1.4.7), o-lim (‖s n − f ‖i | n ∈ N) = 0. By the definition, from 1.1.15, there is a sequence (z n ∈ R | n ∈ N) ↓ 0 such that x n ≡ ‖s n − f ‖i ⩽ z n /2. Since 𝜇 is closedly regular, we can take closed sets K ni ⊂ M ni such that 𝜇(M ni \K ni ) = 𝜇M ni − 𝜇K ni < z n /(2p n |x ni |). Then, consider the functions f n ≡ ∑(x ni 𝜒(K ni ) | i ∈ I n ). We have |f n | ⩽ |s n | ⩽ |f | and |s n − f n | = ∑(|x ni |𝜒(M ni \K ni ) | i ∈ I n ). Therefore, by Lemma 1 (3.3.2), ‖s n − f n ‖i = ∑(|x ni |𝜇(M ni \F ni ) | i ∈ I n ) ⩽ z n /2. This implies ‖f n − f ‖i ⩽ ‖f n − s n ‖i + x n ⩽ z n ↓ 0, where o-lim (‖f n − f ‖i | n ∈ N) = 0. Again, by Lemma 3 (1.4.7), (lim ‖f n − f ‖i | n ∈ N) = 0. For compact sets, the arguments are the same. 1. Now, let ⟮T, G⟯ be Hausdorff and 𝜇 be a compactly regular and locally bounded. By Statement 7 (3.5.1), there exist collections (U ni | i ∈ I n ) of pairwise disjoint open sets such that K ni ⊂ U ni . Using Corollary 1 to Lemma 9 (3.5.1) and applying the axiom of choice from 1.1.12 as, for example, in the proof of Theorem 1 (3.4.1), we can take some collections (V ni ∈ Gf (𝜇) | i ∈ I n ) such that K ni ⊂ V ni and 𝜇V ni − 𝜇K ni < z n /(p n |x ni |). Consider the sets G ni ≡ U ni ∩ V ni ∈ Gf (𝜇) and the functions h n ≡ ∑(x ni 𝜒(G ni ) | i ∈ I n ). If s n ⩾ 0, then f n ⩾ 0 and h n ⩾ 0 as well. Since |f n − h n | = ∑(|x ni |𝜒(G ni \K ni ) | i ∈ I n ) and 𝜇(G ni \K ni ) < z n /(p n |x ni |), we deduce by virtue of Lemma 1 (3.3.2) that ‖f n − h n ‖i = ∑(|x ni |𝜇(G ni \K ni ) | i ∈ In ) ⩽ zn .

328 | 3.5 Topological spaces with measures. The Radon integral

2. Now, suppose that ⟮T, G⟯ is Tychonoff. Then, Statement 8 (3.5.1) implies that there are continuous functions 𝜑ni : T → [0, 1] such that 𝜑ni [K ni ] = {1} and coz f ⊂ G ni . Therefore, 𝜑ni − 𝜒(K ni ) ⩽ 𝜒(G ni \K ni ) implies |f n − g n | = ∑(|x ni |(𝜑ni − 𝜒(K ni )) | i ∈ I n ) ⩽ ∑(|x ni |𝜒(G ni \K ni ) | i ∈ I n ) = |f n − h n |. As a result, ‖f n − g n ‖i ⩽ ‖f n − h n ‖ ⩽ z n . 3. Finally, suppose that ⟮T, G⟯ is locally compact. Then, using Statement 10 (3.5.1) and again applying the axiom of choice, we can take some collections (H ni ∈ G | i ∈ I n ) such that K ni ⊂ H ni ⊂ cl H ni ⊂ G ni and L ni ≡ cl H ni is compact. By the same procedure and Statement 8 (3.5.1), we can take some collections (𝜑ni ∈ C(T, G) | i ∈ I n ) such that 𝜑ni [K ni ] = {1} and coz 𝜑ni ⊂ H ni ⊂ L ni ∈ C. Therefore, 𝜑ni ∈ C c (T, G) for all indices. 4. If s n ⩾ 0, then g n ⩾ 0 as well. Corollary 1. Let ⟮T, G, M, 𝜇⟯ be a topological [Hausdorff ] space with a wide, positive, quite topological, and closedly regular [compactly regular] measure. Then, the family St(T, F f (𝜇)) is [the families St(T, Cf (𝜇)) and St(T, Gf (𝜇)) are] dense in MI(T, M, 𝜇) in the normed topology G(‖ ⋅ ‖i ). Moreover, if ⟮T, G⟯ is Tychonoff, then the family CI b (T, G, M, 𝜇) ≡ C b (T, G) ∩ MI(T, M, 𝜇) is dense in MI(T, M, 𝜇). Lemma 2. Let ⟮T, G⟯ be a locally compact space. Then, Sm (T, G, C c (T, G)) = S c (T, G). Proof. If f ∈ E ≡ Sm (T, G, C c (T, G)), u ∈ A ≡ C c (T, G), and |f | ⩽ u, then obviously f ∈ S c (T, G). Conversely, let g ∈ S c (T, G). Consider the compact set K ≡ supp g. By Proposition 1, the family A separates points and closed sets. Hence, for every t ∈ K, there is a function u ∈ A + such that u(t) > 1. Therefore, the set B t ≡ {u ∈ A+ | u(t) > 1} is non-empty. Take some choice mapping c : P(A+ )\{⌀} → A+ and consider the mapping d : K → A+ such that d(t) ≡ c(B t ) ∈ B t . Since d(t) is continuous, the sets G t ≡ {s ∈ T | d(t)(s) > 1} are open. The collection ⟮G t | t ∈ K⟯ is a cover of the set K. Since K is compact, there is a finite set L ⊂ K such that ⟮G t | t ∈ L⟯ is a cover of K. Consider the function u ≡ sup (d(t) | t ∈ L) ∈ A+ . Then, 𝜒(K) ⩽ u. Since |g| ⩽ x1 for some real number x, we conclude that |g| ⩽ x𝜒(K) ⩽ xu ∈ A + . Hence, g ∈ E.

3.5.3 Wide Radon measures on Hausdorff spaces. The problem of characterization of Radon integrals as linear functionals Here we shall study the best topological measures with properties which are similar to the properties of the extended Borel – Lebesgue measure 𝜆× on Rn , enumerated in 3.1.6.

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Definitions of a Radon measure A measure 𝜇 : M → R on a Hausdorff space ⟮T, G⟯ will be called a (wide) Radon measure if it is wide, quite topological, finite on compact sets (i. e. C ⊂ Mf (𝜇)), and compactly regular (see 3.5.1). If 𝜇 is complete and strongly saturated, then it will be called an (wide) extended Radon measure. If M = B(T, G), then 𝜇 will be called a (wide) Borel – Radon measure. The family of all (wide) Radon measures 𝜇 : M → R on ⟮T, G⟯ defined on all possible 𝜎-algebras M will be denoted by RMw (T, G). The subfamily of RMw (T, G) consisting of all extended measures will be denoted by RMwe (T, G). The subfamily consisting of all measures defined on a fixed 𝜎-algebra M will be denoted by RMw (T, G, M). The family RMw (T, G, B(T, G)) will be denoted by RMw⋆ (T, G). To denote the corresponding subfamilies of positive or bounded measures we shall use the lower indices “0” or “b”, respectively. It follows from 3.1.6 that 𝜆× is a positive (wide) Radon measure. Finite Radon measures on compact topological spaces and positive Radon measures on locally compact spaces were considered initially in [Saks, 1938; Kakutani, 1941] and [Halmos, 1950; Hewitt, 1952; Edwards, 1953], respectively. The notion of an arbitrary, i. e. not necessary finite and not necessary positive Radon measure was introduced in papers [Zakharov, 2002b; 2005a]. Lemma 1. Let ⟮T, G⟯ be a Hausdorff space and 𝜇 : M → R be a wide measure on it. Then, the following conclusions are equivalent: 1) 𝜇 is a Radon measure; 2) the variations v+ (𝜇) and v− (𝜇) of the measure 𝜇 (see 3.2.1) are positive and negative Radon measures, respectively; 3) 𝜇 = 𝜇1 − 𝜇2 for some positive Radon measures 𝜇1 , 𝜇2 ∈ RMw (T, G, M) at least one of which is finite; 4) 𝜇 is quite topological and finite on compact sets and has the following properties: a) for every set M ∈ M such that 𝜇M = ∞ [𝜇M = −∞] and every real number 𝛿 > 0 there is a compact set C ⊂ M such that 𝜇C󸀠 > 𝛿 [𝜇C󸀠 < 𝛿] for every compact set C󸀠 satisfying the condition C ⊂ C󸀠 ⊂ M; b) for every set M ∈ M such that 𝜇M ∈ R and every real number 𝜀 > 0 there is a compact set C ⊂ M such that |𝜇M − 𝜇C󸀠 | < 𝜀 for every compact set C󸀠 satisfying the condition C ⊂ C󸀠 ⊂ M. Proof. (1) ⊢ (2). Take any M ∈ M. By virtue of a Hahn decomposition of T (Theorem 1 (3.2.1)) and the Jordan decomposition of 𝜇 (Lemma 5 (3.2.1)), there are sets T+ and T− and positive wide measures v+ (𝜇) and −v− (𝜇) such that v+ (𝜇)M = 𝜇(M ∩ T+ ) and v− (𝜇)M = 𝜇(M ∩ T− ). Take any compact set C. Then, 𝜇C = v+ (𝜇)C + v− (𝜇)C and at least one of these variations is finite. Since 𝜇C ∈ R, we infer that v+ (𝜇)C and v− (𝜇)C are finite. Thus, these variations are finite on compact sets.

330 | 3.5 Topological spaces with measures. The Radon integral

If v+ (𝜇)M = 𝜇(M ∩ T+ ) = ∞, then, by definition, for every a > 0, there is a compact set C ⊂ M ∩ T+ such that v+ (𝜇)C = 𝜇(C ∩ T+ ) = 𝜇C > a. If v+ (𝜇)M < ∞, then, by definition, for every 𝜀 > 0, there is a compact set C ⊂ M ∩ T+ such that 0 ⩽ v+ (𝜇)M − v+ (𝜇)C = |𝜇(M ∩ T+ ) − 𝜇C| < 𝜀. This means that v+ (𝜇) is a positive Radon measure. For v− (𝜇), the arguments are the same. (2) ⊢ (3). We can take 𝜇1 ≡ v+ (𝜇) and 𝜇2 ≡ −v− (𝜇). (3) ⊢ (4). It is clear that 𝜇 is finite on compact sets. Take any M ∈ M. Let x ≡ 𝜇M = ∞. Then, y ≡ 𝜇1 M = ∞ and z ≡ 𝜇2 M < ∞. Take any 𝛿 > 0. Then, there is a compact set C ⊂ M such that 𝜇1 C > 𝛿 + z. If C󸀠 ∈ C and C ⊂ C󸀠 ⊂ M, then 𝜇C󸀠 = 𝜇1 C󸀠 − 𝜇2 C󸀠 > 𝜇1 C − z > 𝛿 + z − z = 𝛿. In the case x = −∞, the arguments are the same. Let x ∈ R. Then, y, z ∈ R+ . Take any 𝜀 > 0. By the condition, there are compact sets C1 , C2 ⊂ M such that 0 ⩽ y − 𝜇1 C1 < 𝜀/2 and 0 ⩽ z − 𝜇2 C2 < 𝜀/2. Take C ≡ C1 ∪ C2 and any C󸀠 ∈ C such that C ⊂ C󸀠 ⊂ M. Then, |x − 𝜇C󸀠 | ⩽ (y − 𝜇1 C󸀠 ) + (z − 𝜇2 C󸀠 ) ⩽ y − 𝜇1 C1 + z− 𝜇2 C2 < 𝜀. (4) ⊢ (1). This deduction is evident. There is an equivalent definition of a Radon measure in the case of finite measures. According to Corollary 3 to Proposition 1 (3.2.1), every finite wide measure is bounded. As was noted in 3.2.1, by virtue of Corollary 1 to Lemma 8 (3.1.3) and Lemma 4 (3.1.3), we have Measb (T, R) ⊂ Measof (T, R). Therefore, Corollary 1 to Theorem 2 (3.2.2) implies that |𝜇| = v(𝜇) for every finite wide measure 𝜇. Proposition 1. Let ⟮T, G⟯ be a Hausdorff space and 𝜇 be a finite quite topological wide measure on it. Then, the following conclusions are equivalent: 1) 𝜇 ∈ RMw (T, G, M); 2) for every M ∈ M and every 𝜀 > 0 there are C ∈ C and G ∈ G such that C ⊂ M ⊂ G and |𝜇|(G\C) < 𝜀. Proof. Put 𝜇1 ≡ v+ (𝜇) and 𝜇2 ≡ −v− (𝜇). By Corollary 1 to Proposition 2 (3.2.1), these measures are finite. Then, 𝜇M, 𝜇1 M, 𝜇2 M ∈ R for all M ∈ M. (1) ⊢ (2). By Lemma 1 𝜇1 , 𝜇2 ∈ RM(T, G, M)0 . Let M ∈ M and 𝜀 > 0. By definition of a Radon measure, there are C1 , C2 ∈ C such that C1 ⊂ M, C2 ⊂ M, 0 ⩽ 𝜇1 (M) < 𝜇1 (C1 )+𝜀/4, and 0 ⩽ 𝜇2 (M) < 𝜇2 (C2 )+𝜀/4. Since T\M ∈ M, we have 𝜇1 (T\M), 𝜇2 (T\M) ∈ R. Therefore, for every 𝜀 > 0, there exist D1 , D2 ∈ C such that D1 ⊂ T\M, D2 ⊂ T\M, 0 ⩽ 𝜇1 (T\M) < 𝜇1 (D1 ) + 𝜀/4, and 0 ⩽ 𝜇2 (T\M) < 𝜇2 (D2 ) + 𝜀/4. Put C ≡ C1 ∪ C2 ∈ C, G ≡ T\(D1 ∪ D2 ) ∈ G. Then, C ⊂ M ⊂ G and 𝜇1 (G\C) ⩽ 𝜇1 ((T\D1 )\C1 ) = 𝜇1 (((T\D1 )\C1 ) ∩ M) + 𝜇1 (((T\D1 )\C1 ) ∩ (T\M)) = 𝜇1 (M\C1 ) + 𝜇1 ((T\M)\(T\D1 )) < 𝜀/4 + 𝜇1 ((T\M)\D1 ) < 𝜀/4 + 𝜀/4 = 𝜀/2. Similarly, 𝜇2 (G\C) < 𝜀/2. Consequently, using Proposition 2 (3.2.1), we get |𝜇|(G\C) = 𝜇1 (G\C) + 𝜇2 (G\C) < 𝜀. (2) ⊢ (1). Let M ∈ M and 𝜀 > 0. Consider C ∈ C and G ∈ G such that C ⊂ M ⊂ G and |𝜇|(G\C) < 𝜀. Then, 0 ⩽ 𝜇1 (M) − 𝜇1 (C) = 𝜇1 (M\C) < 𝜇1 (G\C) ⩽ |𝜇|(G\C) < 𝜀 and 0 ⩽

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𝜇2 (M) − 𝜇2 (C) = 𝜇2 (M\C) < 𝜇2 (G\C) ⩽ |𝜇|(G\C) < 𝜀. Thus, 𝜇1 and 𝜇2 are positive Radon measures. By virtue of Lemma 1, this implies that 𝜇 = 𝜇1 − 𝜇2 is also Radon. Corollary 1. Let ⟮T, G⟯ be a compact topological space and 𝜇 be a finite quite topological wide measure on it. Then, the following conclusions are equivalent: 1) 𝜇 ∈ RMw (T, G, M); 2) for every M ∈ M and every 𝜀 > 0 there are F ∈ F and G ∈ G such that F ⊂ M ⊂ G and |𝜇|(G\F) < 𝜀. Proof. It is sufficiently to note that, according to Statement 1 (3.5.1), every closed set in a compact space is compact. Namely, property 2 was laid by J. Radon, S. Saks, and S. Kakutani in the base of the definition of a finite Radon measure on a compact space (see [Semadeni, 1971, 18.2.1]). Generalizing their definition for the case of finite measure on a non-compact space, it is naturally to replace the approximation by closed subsets by the approximation by compact subsets. This idea is realized in Proposition 1. A positive Radon measure has some good additional properties. Lemma 2. Let ⟮T, G⟯ be a Hausdorff space and 𝜇 be a positive wide Radon measure on ⟮T, G⟯. Then, 𝜇 is internally finite and lower G-continuous. If, besides, ⟮T, G⟯ is locally compact, then 𝜇 is locally bounded, topologically internally finite, and upper C-continuous. Proof. Let S ∈ S and 𝜇S = ∞. Since 𝜇 is compactly regular, there is C ∈ C ⊂ Sf such that C ⊂ S and 0 < 𝜇C < ∞. Thus, 𝜇 is internally finite. Let G ∈ G and (G𝜁 ∈ G | 𝜁 ∈ Z) ↑ G in P(T). Take a compact K ⊂ G. Then, there is a finite subcover (G𝜁 | 𝜁 ∈ Y ⊂ Z) of K for some finite set Y. By Theorem 3 (1.2.6), there is 𝜉 ∈ Z such that 𝜉 is an upper bound of Y. Therefore, K ⊂ G𝜉 . Consequently, 𝜇G = sup{𝜇K | K ∈ C ∧ K ⊂ G} ⩽ sup (𝜇G𝜁 | 𝜁 ∈ Z) ⩽ 𝜇G. This means that 𝜇 is lower G-continuous. Further, suppose that ⟮T, G⟯ is locally compact. Let t ∈ T. Take a neighborhood U of t such that K ≡ cl U is compact. Then, 𝜇K ∈ R. By Lemma 4 (3.1.1), the positive measure 𝜇 is increasing. Therefore, sup{|𝜇M| | M ∈ M ∧ M ⊂ U} ⩽ sup{𝜇M | M ∈ M ∧ M ⊂ K} ⩽ 𝜇K < ∞, i. e. 𝜇 is locally bounded. Let M ∈ M and 𝜇M > 0. Since 𝜇 is compactly regular, there is a compact set C ⊂ M such that 𝜇C > 0. By Lemma 9 (3.5.1), there is G ∈ Gf (𝜇) such that C ⊂ G. Then, we have 0 < 𝜇C ⩽ 𝜇(M ∩ G) ⩽ 𝜇G < ∞. Thus, 𝜇 is topologically internally finite. As was shown above, 𝜇 is lower G-continuous and locally bounded. Then, by Lemma 9 (3.5.1), it is upper C-continuous. In a locally compact Hausdorff space, we have something close to a converse result to Lemma 2.

332 | 3.5 Topological spaces with measures. The Radon integral

Lemma 3. Let ⟮T, G, M, 𝜇⟯ be a locally compact Hausdorff space with a positive, wide, quite topological, closedly regular, locally bounded, topologically internally finite, and lower G-continuous measure 𝜇 : M → R+ . Then, 𝜇 is a Radon measure. Proof. Let M ∈ M and 0 < x < 𝜇M. Then, by the closed regularity of 𝜇 there is a closed set F ⊂ M such that x < 𝜇F. By Lemma 1 (3.5.1), there is G ∈ G ∩ Mf (𝜇) such that x < 𝜇(F ∩ G). Consider the ensemble Q ≡ {Q ∈ G | Q ⊂ G ∧ cl Q ∈ C} and the identical collection idQ ≡ (Q Q | Q ∈ Q) (see 1.1.8 and 1.1.9). Using the local compactness of ⟮T, G⟯, we get G = ⋃ ⟮Q Q | Q ∈ Q⟯, and therefore, idQ ↑ G in P(T). Since 𝜇 is lower G-continuous, for a number 𝜀 ≡ 𝜇(F ∩ G) − x, there is Q ∈ Q such that 𝜇G − 𝜀 < 𝜇Q. Consequently, 𝜇(G\Q) = 𝜇G−𝜇Q < 𝜀. Consider the set C ≡ F∩cl Q ⊂ M. It is compact by virtue of Statement 1 (3.5.1). We have 𝜇C ⩾ 𝜇(F∩Q) ⩾ 𝜇(F∩G)−𝜇(G\Q) > 𝜇(F∩G)−𝜀 = x. Thus, x ⩽ sup{𝜇C | C ∈ C ∧ C ⊂ M} ⩽ 𝜇M. Since x is arbitrary, we get the necessary equality.

Extensions of a positive Radon measure According to Lemma 2, every positive (wide) Radon measure 𝜇 : M → R+ is internally finite. Therefore, we can consider the complete strongly saturated extension 𝜇# from 3.5.1. Consider also the Borel measure 𝜇0 ≡ 𝜇|B(T, G) and its extension 𝜇0#. Lemma 4. Let ⟮T, G⟯ be a Hausdorff space and 𝜇 : M → R+ be a wide positive Radon measure on it. Then, 𝜇0 : B(T, G) → R+ , (𝜇0 )#: M#(T, B(T, G), 𝜇0 ) → R+ , and 𝜇#: M#(T, M, 𝜇) → R+ are wide Radon measures on ⟮T, G⟯ and (𝜇0 )# = 𝜇# and all three measures are bounded simultaneously. Moreover, 𝜇# is a unique wide positive, complete, and strongly saturated extension of 𝜇 and 𝜇0 among all wide Radon measures 𝜈 : N → R+ on ⟮T, G⟯ extending 𝜇0 . Besides, if 𝜇 is 𝜎-finite, then 𝜇# = 𝜇̄ = 𝜇0# = 𝜇0̄ and 𝜇# is 𝜎-finite as well. Proof. It is evident that 𝜇0 is a Radon measure. By virtue of Lemma 2 and Lemma 12 (3.5.1), 𝜇0# and 𝜇# are positive, complete, and strongly saturated Radon measures on ⟮T, G⟯ and 𝜇0# = 𝜇#. Evidently, if 𝜇 is bounded, then they are bounded. The necessary uniqueness now follows from Corollary 1 to Lemma 3 (3.1.4) and Lemma 8 (3.5.1). The assertion about the 𝜎-finiteness follows from Lemma 12 (3.1.4). L. Schwartz [1973] considered only Radon measures from RMw⋆ (T, G)0 ; D. H. Fremlin [1974] considered only Radon measures from RMwe (T, G)0 . But there is a bijective correspondence between these families of Radon measures. Lemma 4 implies that we can define correctly a mapping 𝜎w : RMw⋆ (T, G)0 → RMwe (T, G)0 setting 𝜎w 𝜇 ≡ 𝜇#.

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Corollary 1. Let ⟮T, G⟯ be a Hausdorff space. Then, the mapping 𝜎w : RMw⋆ (T, G)0 → RMwe (T, G)0 is bijective. Proof. Let 𝜇, 𝜈 ∈ RMw⋆ (T, G)0 and 𝜎w 𝜇 = 𝜎w 𝜈. Then, for every B ∈ B(T, G) we have 𝜇B = (𝜎w 𝜇)B = (𝜎w 𝜈)B = 𝜈B. Hence, 𝜇 = 𝜈, i. e. 𝜎w is injective. If 𝜆 ∈ RMwe (T, G)0 , then by Lemma 4, 𝜆 0 ∈ RMw⋆ (T, G)0 and 𝜆 = 𝜆#0 = 𝜎w 𝜆 0 . Corollary 2. Let ⟮T, G⟯ be a Hausdorff space and 𝜇 : M → R+ be a 𝜎-finite positive Radon measure on it. Then, for the pair of measures 𝜇1 ≡ 𝜇 and 𝜇2 ≡ 𝜇#, all the assertions of Theorem 1 (3.3.5) are valid. Properties of some families of Radon measures Lemma 5. Let ⟮T, G⟯ be a Hausdorff space and M be a 𝜎-algebra on T containing the ensemble G. Then, RMwb (T, G, M)0 is a conic space. Proof. All the necessary arguments are similar to the arguments in the proof of Lemma 2 (3.4.2). Corollary 1. Let ⟮T, G⟯ be a Hausdorff space and M be a 𝜎-algebra on T containing G. Then, the family RMwb (T, G, M) is a linear subspace of the linear space Measb (T, M). Proposition 2. Let ⟮T, G⟯ be a Hausdorff space and M be a 𝜎-algebra on T containing the ensemble G. Then, 1) the family RMwb (T, G, M) is an l-ideal in the Dedekind complete lattice-ordered linear space Measb (T, M); 2) RMwb (T, G, M) is a Dedekind complete lattice-ordered linear space; the smallest upper [greatest lower] bounds of bounded above [below] sets in the ordered sets Measb (T, M) and RMwb (T, G, M) coincide and they are expressed by the formulas from Theorem 2 (3.2.2) and its Corollaries; 3) RMwb (T, G, M)+ = RMwb (T, G, M)0 . Proof. All the necessary arguments are similar to the arguments in the proofs of Proposition 2 (3.4.2) and its corollary 1. Corollary 1. Let ⟮T, G⟯ be a Hausdorff space and ‖ ⋅ ‖ be the norm induced on RMw⋆ b (T, G) by the norm on Measb (T, B(T, G)) from Proposition 2 (3.2.3). Then, ⟮RMw⋆ b (T, G), ‖ ⋅ ‖⟯ is a normed lattice-ordered linear space. Proposition 3. Let ⟮T, G⟯ be a Hausdorff space and M be a 𝜎-algebra on T containing the ensemble G. Then, the family RMw (T, G, M)0 is a conic space.

334 | 3.5 Topological spaces with measures. The Radon integral

Proof. Take 𝜇, 𝜈 ∈ RMw (T, G, M)0 and x, y ∈ R+ . By Lemma 1 (3.1.2), 𝜆 ≡ x𝜇 + y𝜈 is a positive measure. Hence, 𝜆 is finite on compact sets. Suppose that x, y > 0. Let M ∈ M. Consider the extended number z ≡ sup{𝜆E | E ∈ C ∧ E ⊂ M}. Take any positive real numbers a < x𝜇M and b < y𝜈M. Then, there are compact sets C and D such that C∪D ⊂ M, a/x < 𝜇C, and b/y < 𝜈D. Therefore, 𝜆(C∪D) < ∞. Besides, 𝜆M ⩾ 𝜆(C∪D) ⩾ x𝜇C + y𝜈D > a + b. Since a and b are arbitrary, we deduce that x𝜇M + b ⩽ z ⩽ 𝜆M. If 𝜆M < ∞, then z < ∞ and x𝜇M < ∞ imply 𝜆M = x𝜇M + y𝜈M ⩽ z ⩽ 𝜆M, where 𝜆M = z. If 𝜆M = ∞ and z = ∞, then 𝜆M = z. Finally, if 𝜆M = ∞ and z < ∞, then b ⩽ z − x𝜇M implies y𝜈M ⩽ z − x𝜇M, where 𝜆M = x𝜇M + y𝜈M ⩽ z ⩽ 𝜆M, i. e. 𝜆M = z. This means that 𝜆 is compactly regular. Corollary 1. Let ⟮T, G⟯ be a Hausdorff space. Then, the family RMw⋆ (T, G)0 is a conic space. The Riesz – Radon – Fréchet problem of characterization of Radon integrals as linear functionals The Lebesgue integral Λ(𝜇) over the topological space ⟮T, G, M, 𝜇⟯ with the Radon measure 𝜇 will be called the Radon integral. Then, the general problem from 3.3.6 generates the following Riesz – Radon – Fréchet problem of characterization of Radon integrals as linear functionals: for various classes T of Hausdorff topological spaces ⟮T, G⟯, for various families M(T) ⊂ RMw⋆ (T, G) of wide Borel – Radon measures 𝜇 on the space ⟮T, G⟯, and for appropriate linear spaces A(T) ⊂ ⋂⟮MI(T, B(T, G), 𝜇) | 𝜇 ∈ M(T)⟯ of Borel (measurable) and M(T)-universally integrable functions on the space ⟮T, G⟯ to describe the corresponding families I(T, A(T), M(T)) of Radon integrals in the general family A(T)× of all linear functionals on A(T).

To solve the problem in the capacity of the first step (necessity), it is necessary to select and investigate some special properties of Radon integrals Λ(𝜇)|A(T) for some important triplets ⟮T, M(T), A(T)⟯. It is done in 3.6.1. In the capacity of the second step (sufficiency), it is necessary to construct for an arbitrary linear functional 𝜑 : A(T) → R with these special properties some unique Radon measure 𝜇𝜑 ∈ M(T) such that 𝜑 = Λ(𝜇𝜑 )|A(T). This difficult process occupies subsections 3.6.2 – 3.6.4. Note that the solution of the Riesz – Radon – Fréchet problem has a long and rich history, which is presented in Appendix D.

3.5.4 Narrow Radon measures on Hausdorff spaces Consider some lattice-ordered linear spaces of narrow Radon measures connected with the families RMwe (T, G)0 and RMw⋆ (T, G)0 of positive (wide) Radon measures on a Hausdorff space ⟮T, G⟯.

3.5.4 Narrow Radon measures on Hausdorff spaces

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The subensemble of the ensemble B(T, G) consisting of all precompact (see 2∘ (3.5.1)) Borel sets will be denoted by Bc (T, G). It is easily proven that Bc (T, G) is a 𝛿-ring. According to Statement 5 (3.5.1), C ⊂ F. Therefore, every compact set is precompact and C ⊂ Bc (T, G). A finite measure 𝜇 : M → R on a Hausdorff space ⟮T, G⟯ will be called a narrow Radon measure if 𝜇 is compactly regular and M is a 𝛿-ring containing the 𝛿-ring Bc (T, G) and such that M ∩ G ∈ M for every M ∈ M and G ∈ G. If M = Bc (T, G), then 𝜇 will be called a narrow Borel – Radon measure. The family of all narrow Radon measures 𝜇 : M → R on a Hausdorff space ⟮T, G⟯ defined on all possible 𝛿-rings M will be denoted by RMn (T, G). Its subfamily consisting of all measures defined on a fixed 𝛿-ring M will be denoted by RMn (T, G, M). The family RMn (T, G, Bc (T, G)) will be denoted by RMn⋆ (T, G). To denote the corresponding subfamilies of bounded or positive narrow Radon measures, we shall use the lower indices “b” or “0”, respectively. Lemma 1. Let ⟮T, G⟯ be a Hausdorff space and 𝜇 : M → R be a finite narrow measure on it. Then, the following conclusions are equivalent: 1) 𝜇 is a narrow Radon measure; 2) the variations v + (𝜇) and v− (𝜇) of the measure 𝜇 are positive and negative narrow Radon measures, respectively; 3) 𝜇 = 𝜇1 − 𝜇2 for some positive narrow Radon measures 𝜇1 , 𝜇2 ∈ RMn (T, G, M); 4) Bc (T, G) ⊂ M and for every set M ∈ M every real number 𝜀 > 0, there is a compact set C ⊂ M such that |𝜇M − 𝜇C󸀠 | < 𝜀 for every compact set C󸀠 satisfying the condition C ⊂ C󸀠 ⊂ M. Proof. (1) ⊢ (2). Take any M ∈ M. By virtue of a Hahn decomposition of T (Theorem 1 (3.2.1)) and the Jordan decomposition of 𝜇 (see Proposition 2 (3.2.1) and its Corollaries) there are sets M+ ⊂ M and M− ⊂ M and positive narrow measures v+ (𝜇) and −v− (𝜇) such that v+ (𝜇)M = 𝜇(M+ ) and v− (𝜇)M = 𝜇(M− ). For every 𝜀 > 0, there is a compact set C ⊂ M+ such that 0 ⩽ v+ (𝜇)M − v+ (𝜇)C = |𝜇(M+ ) − 𝜇C| < 𝜀. This means that v+ (𝜇) is a positive narrow Radon measure. For v− (𝜇), the arguments are the same. (2) ⊢ (3). We can take 𝜇1 ≡ v+ (𝜇) and 𝜇2 ≡ −v− (𝜇). (3) ⊢ (4). Take any M ∈ M. Let x ≡ 𝜇M, y ≡ 𝜇1 M, and z ≡ 𝜇2 M. Take any 𝜀 > 0. By the condition, there are compact sets C1 , C2 ⊂ M such that 0 ⩽ y − 𝜇1 C1 < 𝜀/2 and 0 ⩽ z − 𝜇2 C2 < 𝜀/2. Take C ≡ C1 ∪ C2 and any C󸀠 ∈ C such that C ⊂ C󸀠 ⊂ M. Then, |x − 𝜇C󸀠 | ⩽ (y − 𝜇1 C󸀠 ) + (z − 𝜇2 C󸀠 ) ⩽ y − 𝜇1 C1 + z − 𝜇2 C2 < 𝜀. (4) ⊢ (1). This deduction is evident. Proposition 1. Let ⟮T, G⟯ be a Hausdorff space. Then, 1) the family RMn (T, G, M) [RMnb (T, G, M)] is an l-ideal in the Dedekind complete lattice-ordered linear space Measf (T, M) = Measof (T, M) [Measb (T, M)];

336 | 3.5 Topological spaces with measures. The Radon integral

2) RMn (T, G, M)+ = RMn (T, G, M)0 and RMnb (T, G, M)+ = RMnb (T, G, M)0 ; 3) RMn (T, G, M) [RMnb (T, G, M)] is a Dedekind complete lattice-ordered linear space; the smallest upper (greatest lower) bounds of bounded above (below, respectively) sets in the ordered sets Measf (T, M) [Measb (T, M)] and RMn (T, G, M) [RMnb (T, G, M)] coincide and they are expressed by the formulas from Theorem 2 (3.2.2) and its Corollaries; 4) RMnb (T, G, M) is an l-ideal in RMn (T, G, M). Proof. 1. Since M is a 𝛿-ring, Corollary 2 to Proposition 1 (3.2.1) guarantees that B ≡ Measf (T, M) = Measof (T, M). By Corollary 2 to Proposition 2 (3.2.2), B is a Dedekind complete lattice-ordered linear space. Denote A ≡ RMn (T, G, M). Let 𝜇, 𝜈 ∈ A and x, y ∈ R. Then, for every M ∈ M and every 𝜀 > 0 there are C, D ∈ C such that C ∪ D ⊂ M, |𝜇(M\C)| = 𝜇M − 𝜇C| < 𝜀/|2x|, and |𝜈(M\D)| < 𝜀/|2y|. Then, for the finite measure 𝜆 ≡ x𝜇 + y𝜈 on the 𝛿-ring M, we have |𝜆M − 𝜆(C ∪ D)| = |𝜆(M\(C ∪ D))| ⩽ |x𝜇(M\C)| + |y𝜈(M\D)| < 𝜀. Thus, 𝜆 is compactly regular, and therefore, 𝜆 ∈ A. It follows from the proven property that A is a linear subspace of the linear space B. Let 𝜇 ∈ A 0 , 𝜈 ∈ B, and 0 ⩽ 𝜈 ⩽ 𝜇. Then, for every M ∈ M and every 𝜀 > 0 there is C ∈ C such that C ⊂ M and 𝜇M − 𝜀 < 𝜇C. Hence, 𝜈M − 𝜈C = 𝜈(M\C) ⩽ 𝜇(M\C) = 𝜇M − 𝜇C < 𝜀. Thus, 𝜈 is compactly regular, and therefore, 𝜈 ∈ A0 . Now, let 𝜇 ∈ A, 𝜈 ∈ B, and |𝜈| ⩽ |𝜇| in B. By Lemma 1, 𝜇 = 𝜇1 − 𝜇2 for some 𝜇1 , 𝜇2 ∈ A0 . Therefore, |𝜇| = 𝜇1 + 𝜇2 ≡ 𝜆 ∈ A0 . By the property proven above, the inequality 0 ⩽ 𝜈+ ⩽ |𝜈| ⩽ 𝜆 implies 𝜈+ ∈ A0 . Similarly, −𝜈− ∈ A0 . Hence, by Corollary 3 to Theorem 2 (3.2.2), 𝜈 = 𝜈+ − (−𝜈− ) ∈ A. Then, according to Statement 2 (2.2.8), A is an l-ideal of B. For bounded measures, the arguments are the same. 2. Now, let 𝜇 ∈ A+ . By the definition, 𝜇 = 𝜇1 −𝜇2 for some 𝜇1 , 𝜇2 ∈ A0 . Then, by virtue of the property proven above, the inequality 0 ⩽ 𝜇 ⩽ 𝜇1 implies 𝜇 ∈ A0 . This means that A+ = A0 . The same is valid for bounded measures. Assertion 3 follow from (1) and Statements 3 and 1 (2.2.8). 4. We need to check that A b is an l-ideal of A. Let a1 ∈ A b , a2 ∈ A and |a2 | ⩽ |a1 |. By Corollary 1 of Lemma 2 (3.2.2), the measure a2 is bounded. Then, a2 ∈ A b . Thus, narrow Radon measures form a linear space. But unfortunately, we have the integration technique only for wide measures. By this reason, we are constrained to extend narrow Radon measures to 𝜎-algebras. For every narrow positive Radon measure 𝜇 : M → R+ on a 𝛿-ring M, consider its complete strongly saturated extension 𝜇#: M# → R+ from 3.1.4. Proposition 2. Let ⟮T, G⟯ be a Hausdorff space and 𝜇 : M → R+ be a positive narrow Radon measure. Then, 𝜇#: M# → R+ is a positive, wide, complete, and strongly

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saturated Radon measure and (𝜇#)0 : B(T, G) → R+ is a positive wide Borel – Radon measure. Besides, if 𝜇 is bounded, then 𝜇# and (𝜇#)0 are bounded. Proof. Denote M# by N and 𝜇# by 𝜈. By Lemma 12 (3.5.1), 𝜈 is compactly regular. If 𝜇 is bounded, then 𝜈 is bounded as well. Now, we need to check that 𝜈 is quite topological. Denote K(T, M, 𝜇) by K and 𝜇̌ by 𝜘. According to Theorem 4 (3.1.4), 𝜘 is finite on K. Let G ∈ G and K ∈ K. Then, K = R ∪ N for some R ∈ M and N ∈ N(T, M, 𝜇). Therefore, G ∩ K = (G ∩ R) ∪ (G ∩ N). By the definition, of narrow Radon measures, S ≡ G ∩ R ∈ M. By virtue of Corollary 3 to Lemma 2 (3.1.4), P ≡ G ∩ N ∈ N(T, M, 𝜇). As a result, G ∩ K = S ∪ P ∈ K. This means that G ∈ N. By virtue of Lemma 11 (3.1.4), 𝜈 is complete. According to Theorem 6 (3.1.4) and Corollary 2 to Theorem 4 (3.1.4), 𝜈 is a strongly saturated wide measure. By Lemma 4 (3.5.3), 𝜈0 is also a positive wide Radon measure. If 𝜇 is bounded, then 𝜈0 is bounded as well. Corollary 1. Let 𝜇 : M → R+ be a positive wide Radon measure. Then, 𝜇00 ≡ 𝜇|Bc (T, G) is a positive narrow Borel – Radon measure and 𝜇# = (𝜇00 )#. Proof. It is evident that 𝜇00 is a narrow positive Borel – Radon measure. If C is a compact set, then 𝜇#C = 𝜇C = 𝜇00 C = (𝜇00 )#C. Therefore, by Lemma 8 (3.5.1), 𝜇# = (𝜇00 )#. Proposition 2 implies that we can define correctly the mapping R e : RMn⋆ (T, G)+ → RMwe (T, G)0 setting R e 𝜇 ≡ 𝜇# and the mapping R eb ≡ R e |RMn⋆ b (T, G)+ . The composition mappings 𝜇 󳨃→ (𝜇#)0 will be denoted by R : RMn⋆ (T, G)+ → RMw⋆ (T, G)0 w⋆ and R b : RMn⋆ b (T, G)+ → RM (T, G)0 correspondingly. According to Proposition 1, w⋆ RMw⋆ b (T, G)0 = RMb (T, G)+ . The measure R𝜇 will be called the wide Borel – Radon extension of the narrow positive Borel – Radon measure 𝜇. Lemma 2. Let ⟮T, G⟯ be a Hausdorff space. Then, the mappings R e , R, R eb , and R b are bijective. Proof. Let 𝜇, 𝜈 ∈ RMn⋆ (T, G)+ and 𝜇# = 𝜈#. Then, for every B ∈ Bc (T, G) we have 𝜇B = 𝜇#B = 𝜈#B = 𝜈B. Hence, 𝜇 = 𝜈, i. e. R e is injective. The same is valid for all the other mappings. If 𝜆 ∈ RMwe (T, G)+ , then by Lemma 4 (3.5.3), 𝜆 00 ∈ RMn⋆ (T, G)+ . Therefore, by Corollary 1 to Proposition 2, 𝜆 = 𝜆# = (𝜆 00 )# = R e 𝜆 00 . Thus, R e is surjective. If 𝜆 is a wide Borel – Radon measure, then analogously 𝜆 = (𝜆#)0 = ((𝜆 00 )#)0 = R𝜆 00 . Thus, R is surjective. If 𝜆 is bounded, then 𝜆 00 is bounded as well. So R eb and R b are surjective.

338 | 3.5 Topological spaces with measures. The Radon integral

Proposition 3. Let ⟮T, G⟯ be a Hausdorff space. Then, R and R b are conic operators. Proof. Take 𝜇, 𝜈 ∈ RMn⋆ (T, G)+ and x, y ∈ R+ . Consider the positive narrow Borel – Radon measure 𝜆 ≡ x𝜇 + y𝜈. By Proposition 3 (3.5.3), 𝜋 ≡ xR𝜇 + yR𝜈 is a positive wide Borel – Radon measure. By Proposition 2, 𝜘 ≡ R𝜆 is also the a positive wide Borel – Radon measure. If C is a compact set, then 𝜋C = x𝜇C + y𝜈C = 𝜆C = 𝜘C. From the compact regularity of 𝜋 and 𝜘, we deduce that 𝜋 = 𝜘. For bounded measures, there is a stronger statement than Proposition 3. Theorem 1. Let ⟮T, G⟯ be a Hausdorff space. Then, the mapping R b has a unique extension up to an isomorphism S b : RMn⋆ RMw⋆ b (T, G) b (T, G) between the given lattice-ordered linear spaces such that S b 𝜇 = R b (𝜇+ ) − R b (−𝜇− ). Proof. By Proposition 1 A ≡ RMn⋆ b (T, G) is a lattice-ordered linear space. By Proposition 2 (3.5.3), B ≡ RMw⋆ b (T, G) is a lattice-ordered linear space as well. Therefore, Proposition 3, Lemma 2, and Statement 1 (3.4.2) imply that the mapping R b has a unique extension to the injective linear operator S b : A B such that S b 𝜇 = R b (𝜇+ ) − R b (−𝜇− ). Let 𝜇, 𝜈 ∈ A+ . If 𝜇 ⩽ 𝜈, then (R b 𝜇)C ⩽ (R b 𝜈)C for every compact set C. Using the compact regularity of these measures, we deduce that (R b 𝜇)M ⩽ (R b )M for every M ∈ B(T, G). This means that R b is increasing. If R b 𝜇 ⩽ R b 𝜈, then obviously 𝜇 ⩽ 𝜈. Thus, R b is isotone. Now, by Lemma 2 (1.1.15) R b preserves any exact bounds. Therefore, 𝜇 ∧ 𝜈 = 0 in A+ implies S b 𝜇 ∧ S b 𝜈 = R b 𝜇 ∧ R b 𝜈 = R b (𝜇 ∧ 𝜈) = 0. Then, by Statement 2 (3.2.2), S b is an homomorphism of lattice-ordered linear spaces. Let 𝜋 ∈ B. By Proposition 2 (3.5.3) and Statement 2 (2.2.4), we get 𝜋 = 𝜋+ + 𝜋− . By Lemma 2, 𝜋+ = R b 𝜇󸀠 and −𝜋− = R b 𝜇󸀠󸀠 for some 𝜇󸀠 , 𝜇󸀠󸀠 ∈ A+ . Consider the measure 𝜇 = 𝜇󸀠 − 𝜇󸀠󸀠 ∈ A. Then, 𝜋 = R b 𝜇󸀠 − R b 𝜇󸀠󸀠 = S b 𝜇. Thus, S b is bijective. Therefore, according to Statement 5 (2.2.7), S b is a necessary isomorphism. The measure S b 𝜇 will be called the wide Borel – Radon extension of the narrow bounded Borel – Radon measure 𝜇. Now, we shall prove that S b preserves norm. Consider on RMn⋆ b (T, G) and w⋆ RMb (T, G) the norms induced from the normed lattice-ordered linear spaces ⟮Measb (T, Bc (T, G)), ‖ ⋅ ‖⟯ and ⟮Measb (T, B(T, G)), ‖ ⋅ ‖⟯, respectively. Proposition 4. Let ⟮T, G⟯ be a Hausdorff space. Then, w⋆ 1) ⟮RMn⋆ b (T, G), ‖ ⋅ ‖⟯ and ⟮RMb (T, G), ‖ ⋅ ‖⟯ are normed lattice-ordered linear spaces; moreover, they are pre-L-spaces; 2) S b : ⟮RMn⋆ ⟮RMw⋆ b (T, G), ‖ ⋅ ‖⟯ b (T, G), ‖ ⋅ ‖⟯ is an (isometric) isomorphism of the given normed lattice-ordered linear spaces (see 6∘ (2.2.7)).

3.5.5 Radon bimeasures on Hausdorff spaces

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Proof. 1. Denote the first family by Y, the second by Z and Measb (T, Bc (T, G)) by X. ̄̄ By Proposition 2 (3.2.3), ⟮X, ‖ ⋅ ‖⟯ is an L-space with respect to the norm ‖𝜇‖ = 𝜇T. By Proposition 1, the smallest upper [greatest lower] bounds coincide in X and Y. Therefore, |𝜇|X = |𝜇|Y for every 𝜇 ∈ Y. As a result, |𝜇|Y ⩽ |𝜈|Y in Y implies ‖𝜇‖ ⩽ ‖𝜈‖, whereas the norm ‖ ⋅ ‖ is monotone in X. Then, (Y , ‖ ⋅ ‖) is a normed lattice-ordered linear spaces. Besides, if 𝜇, 𝜈 ∈ Y+ ⊂ X+ , then ‖𝜇+𝜈‖ = ‖𝜇‖+‖𝜈‖. Hence, (Y , ‖ ⋅ ‖) is a pre-L-space. Analogously, (Z, ‖ ⋅ ‖) is a pre-L-space. 2. Let 𝜇 ∈ Y. Then, by (1), |𝜇| = 𝜇+ + (−𝜇− ) implies ‖𝜇‖ = ‖ |𝜇| ‖ = ‖𝜇+ ‖ + ‖−𝜇− ‖ = var(𝜇+ )(T) + var(−𝜇− )(T). By Corollary 2 to Lemma 7 (3.1.3), var(𝜇+ )T = sup{𝜇+ B | B ∈ Bc (T, G)} = sup{𝜇+ C | C ∈ C}. Consider the measure 𝜋 ≡ S b 𝜇. By virtue of Theorem 1, 𝜋+ = S b (𝜇+ ) = R b (𝜇+ ) and −𝜋− = S b (−𝜇− ) = R b (−𝜇− ). By Corollary 2 to Lemma 7 (3.1.3), ‖𝜋+ ‖ = var(𝜋+ )T = 𝜋+ T. Since 𝜋+ is compactly regular, we infer that ‖𝜋+ ‖ = sup{𝜋+ C | C ∈ C} = sup{𝜇+ C | C ∈ C} = var(𝜇+ )T. In the same way, we check that ‖ − 𝜋− ‖ = var(−𝜇− )T. Since Z is a pre-L-space, we conclude that ‖𝜋‖ = ‖𝜋+ ‖+‖−𝜋− ‖ = ‖𝜇‖. Thus, S b is isometrical. By virtue of Theorem 1, S b is a necessary isomorphism. Thus, the mappings R : RMn⋆ (T, G)+ RMw⋆ (T, G)0 and S b : RMn⋆ b (T, G) w⋆ w⋆ RMb (T, G) give the family RM (T, G)0 ∪ RMw⋆ (T, G). But unfortunately, this b family as well as the family of all wide Borel – Radon measures RMw⋆ (T, G) are not linear spaces. In the following subsection, we shall construct some lattice-ordered linear space “including” all these families and isomorphic to the lattice-ordered linear space RMn⋆ (T, G) of all narrow Borel – Radon measures.

3.5.5 Radon bimeasures on Hausdorff spaces Radon triplets and bimeasures Let ⟮T, G⟯ be a Hausdorff space. Any triplet 𝛽 ≡ (𝜇, 𝜇1 , 𝜇2 ) consisting of measures 𝜇, 𝜇1 , and 𝜇2 on ⟮T, G⟯ will be called a Radon triplet on the Hausdorff space ⟮T, G⟯ if: 1) 𝜇1 : M1 → R+ and 𝜇2 : M2 → R+ are positive wide Radon measures; 2) 𝜇 : M → R is a narrow Radon measure on a 𝛿-ring M ≡ M1f (𝜇1 ) ∩ M2f (𝜇2 ); 3) 𝜇M = 𝜇1 M − 𝜇2 M for every M ∈ M. If 𝜇1 and 𝜇2 belong to RMwe (T, G)0 , then 𝛽 will be called an extended Radon triplet. If 𝜇1 and 𝜇2 belong to RMw⋆ (T, G)0 , then 𝛽 will be called a Borel – Radon triplet. If 𝜇1 and 𝜇2 are bounded, then 𝛽 will be called bounded. The set of all Radon triplets on ⟮T, G⟯ will be denoted by R(T, G). The subsets of all extended Radon triplets and all Borel – Radon triplets will be denoted

340 | 3.5 Topological spaces with measures. The Radon integral

correspondingly by Re (T, G) and R⋆ (T, G). To denote the corresponding subfamilies of bounded Radon triplets, we shall use the lower index “b”. Radon triplet 𝛽 ≡ (𝜇, 𝜇1 , 𝜇2 ) and 𝛾 ≡ (𝜈, 𝜈1 , 𝜈2 ) will be called equivalent (𝛽 ∼ 𝛾 or 𝛽𝜃𝛾), if 𝜇|Bc (T, G) = 𝜈|Bc (T, G). The equivalence class in R(T, G) of a Radon triplet 𝛽 will be denoted by 𝛽 ̄ ≡ 𝜃𝛽. The set of all classes 𝛽 ̄ for all Radon triplets 𝛽 ∈ R(T, G) will be denoted by RB(T, G). An element m ≡ 𝛽 ̄ of RB(T, G) will be called a Radon bimeasure. If 𝛽 ∈ Rb (T, G), then m ≡ 𝛽 ̄ will be called a bounded Radon bimeasure. The family of all bounded Radon bimeasures will be denoted by RBb (T, G). For a zero element in RB(T, G), take the element 0 ≡ 𝜃(𝜁, 𝜁1 , 𝜁2 ), where 𝜁, 𝜁1 , and 𝜁2 are zero measures on the 𝜎-algebra B(T, G). Lemma 1. Let 𝛽 ≡ (𝜇, 𝜇1 , 𝜇2 ) ∼ 𝛾 ≡ (𝜈, 𝜈1 , 𝜈2 ) in R(T, G). Then, 1) (𝜇1 + 𝜈2 )M = (𝜇2 + 𝜈1 )M for every M ∈ M1 ∩ M2 ∩ N1 ∩ N2 ; 2) 𝜇M = 𝜈M for every M ∈ M ∩ N. Proof. 1. Consider the 𝜎-algebra L ≡ M1 ∩ M2 ∩ N1 ∩ N2 and the positive measures 𝜘 ≡ 𝜇1 |L + 𝜈2 |L and 𝜋 ≡ 𝜇2 |L + 𝜈1 |L on L. By Proposition 3 (3.5.3), 𝜘 and 𝜋 are Radon measures. If C ∈ C ⊂ Bc (T, G) ⊂ M ∩ N ⊂ L, then by the definition 𝜇C = 𝜇1 C − 𝜇2 C, 𝜈C = 𝜈1 C − 𝜈2 C, and 𝜇C = 𝜈C. Therefore, 𝜘C = 𝜋C for every C. From the compact regularity of 𝜘 and 𝜋, we conclude that 𝜘 = 𝜋. 2. If M ∈ M ∩ N ⊂ L, then 𝜇1 M + 𝜈2 M = 𝜘M = 𝜋M = 𝜇2 M + 𝜈1 M implies 𝜇M = 𝜈M. Linear space structures on RB(T , G) Let 𝛽 ≡ (𝜇, 𝜇1 , 𝜇2 ) ∼ 𝛽󸀠 ≡ (𝜇󸀠 , 𝜇󸀠 1 , 𝜇󸀠 2 ) and 𝛾 ≡ (𝜈, 𝜈1 , 𝜈2 ) ∼ 𝛾󸀠 ≡ (𝜈󸀠 , 𝜈󸀠 1 , 𝜈󸀠 2 ) in R(T, G). Consider the 𝜎-algebras L1 ≡ M1 ∩ N1 , L2 ≡ M2 ∩ N2 , L󸀠 1 ≡ M󸀠 1 ∩ N󸀠 1 and L󸀠 2 ≡ M󸀠 2 ∩ N󸀠 2 and the 𝛿-rings L ≡ M ∩ N and L󸀠 ≡ M󸀠 ∩ N󸀠 . Consider 𝜆 1 ≡ 𝜇1 |L1 + 𝜈1 |L1 , 𝜆 2 ≡ 𝜇2 |L2 + 𝜈2 |L2 , 𝜆󸀠 1 ≡ 𝜇󸀠 1 |L󸀠 1 + 𝜈󸀠 1 |L󸀠 1 , 𝜆󸀠 2 ≡ 𝜇󸀠 2 |L󸀠 2 + 𝜈󸀠 2 |L󸀠 2 , 𝜆 = 𝜇|L + 𝜈|L, and 𝜆󸀠 ≡ 𝜇󸀠 |L󸀠 + 𝜈󸀠 |L󸀠 and the triplets 𝛼 ≡ (𝜆, 𝜆 1 , 𝜆 2 ) and 𝛼󸀠 ≡ (𝜆󸀠 , 𝜆󸀠 1 , 𝜆󸀠 2 ). Lemma 2. The triplets 𝛼 and 𝛼󸀠 are Radon triplets and 𝛼 ∼ 𝛼󸀠 . Proof. According to Proposition 3 (3.5.3), 𝜆 1 and 𝜆 2 are positive wide Radon measures. Consequently, by Corollary 2 to Lemma 1 (3.1.2), 𝜆 is a measure on L ⊂ L1 ∩ L2 . Further, L1f (𝜆 1 ) = L1f (𝜇1 ) ∩ L1f (𝜈1 ) = M1f (𝜇1 ) ∩ N1 ∩ M1 ∩ N1f (𝜈1 ) = M1f (𝜇1 )∩ N1f (𝜈1 ). Analogously L2f (𝜆 2 ) = M2f (𝜇2 ) ∩ N2f (𝜈2 ). Therefore, L1f (𝜆 1 ) ∩ L2f (𝜆 2 ) = M1f (𝜇1 ) ∩ N1f (𝜈1 ) ∩ M2f (𝜇2 ) ∩ N f (𝜈2 ) = M ∩ N = L ⊃ Bc (T, G). If L ∈ L and G ∈ G, then L ∩ G ∈ M∩N ≡ L. Hence, 𝜆̃ 1 ≡ 𝜆 1 |L and 𝜆̃ 2 ≡ 𝜆 2 |L are positive finite Radon measures. Then, Proposition 1 (3.5.4) implies that 𝜆 = 𝜆󸀠1 − 𝜆󸀠2 is a narrow Radon measure.

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Finally, if L ∈ L, then 𝜆 1 L and 𝜆 2 L are finite and 𝜆L = 𝜇L +𝜈L = 𝜇1 L −𝜇2 L +𝜈1 −𝜈2 L. All the proven properties mean that 𝛼 is a Radon triplet. Similarly, 𝛼󸀠 is a Radon triplet. By the condition, 𝜇B = 𝜇󸀠 B and 𝜈N = 𝜈󸀠 B for every B ∈ Bc (T, G). Therefore, 𝜆B = 𝜆 1 B−𝜆 2 B = 𝜇1 B+𝜈1 B−𝜇2 B−𝜈2 B = 𝜇B+𝜈B = 𝜇󸀠 B+𝜈󸀠 B = 𝜇󸀠 1 B−𝜇󸀠 2 B+𝜈󸀠 1 B−𝜈󸀠 2 B = 𝜆󸀠 1 B − 𝜆󸀠 2 B = 𝜆󸀠 B. Thus, 𝛼 ∼ 𝛼󸀠 . It follows from Lemma 2 that we can define correctly an operation of addition “+” in RB(T, G) setting 𝛽 ̄ + 𝛾̄ ≡ 𝛼.̄ Let 𝛽 ≡ (𝜇, 𝜇1 , 𝜇2 ) ∼ 𝛽󸀠 ≡ (𝜇󸀠 , 𝜇󸀠 1 , 𝜇󸀠 2 ) and x ∈ R. If x ⩾ 0, then consider the evaluations 𝜋1 ≡ x𝜇1 , 𝜋2 ≡ x𝜇2 , 𝜋󸀠 1 ≡ x𝜇󸀠 1 , 𝜋󸀠 2 ≡ x𝜇󸀠 2 , 𝜋 ≡ x𝜇 and 𝜋󸀠 ≡ x𝜇󸀠 . If x < 0, then consider the evaluations 𝜘1 ≡ −x𝜇2 , 𝜘2 ≡ −x𝜇1 , 𝜘󸀠 1 ≡ −x𝜇󸀠 2 , 𝜘󸀠 2 ≡ −x𝜇󸀠 1 , 𝜘 ≡ x𝜇 and 𝜘󸀠 ≡ x𝜇󸀠 . Consider the triplets 𝛿 ≡ (𝜋, 𝜋1 , 𝜋2 ), 𝛿󸀠 ≡ (𝜋󸀠 , 𝜋󸀠 1 , 𝜋󸀠 2 ), 𝜀 ≡ (𝜘, 𝜘1 , 𝜘2 ) and 𝜀󸀠 ≡ (𝜘󸀠 , 𝜘󸀠 1 , 𝜘󸀠 2 ). Lemma 3. The triplets 𝛿, 𝛿󸀠 , 𝜀, and 𝜀󸀠 are Radon triplets, 𝛿 ∼ 𝛿󸀠 and 𝜀 ∼ 𝜀󸀠 . Proof. By Proposition 3 (3.5.3), 𝜘1 and 𝜘2 are positive wide Radon measures. By Proposition 1 (3.5.4), 𝜘 is a narrow Radon measure. If M ∈ M, then 𝜇M = 𝜇1 M − 𝜇2 M. Consequently, 𝜘M = 𝜘1 M − 𝜘2 M. This means that 𝜀 is a Radon triplet. Similarly, 𝜀󸀠 is a Radon triplet. By the condition, 𝜇B = 𝜇󸀠 B for every B ∈ Bc (T, G). Therefore, 𝜘B = 𝜘1 B − 𝜘2 B = −x𝜇2 B + x𝜇1 B = x𝜇B = x𝜇󸀠 B = −x𝜇󸀠 2 B + x𝜇󸀠 1 B = 𝜘󸀠 1 B − 𝜘󸀠 2 B = 𝜘󸀠 B. Thus, 𝜀 ∼ 𝜀󸀠 . For 𝛿 and 𝛿󸀠 , the arguments are the same. It follows from Lemma 3 that we can define correctly a composition of multiplication “∗” of elements of RB(T, G) by real numbers setting x ∗ 𝛽 ̄ ≡ 𝛿̄ for x ⩾ 0 and x ∗ 𝛽 ̄ ≡ 𝜀 ̄ for x < 0. Our aim now is to prove that with respect to the operation 𝛽 ̄ + 𝛾̄ and the composition x ∗ 𝛽 ̄ the family RB(T, G) is a linear space. For that, we shall use the family RMn⋆ (T, G) of all narrow Borel – Radon measures. Let 𝜋 ∈ RMn⋆ (T, G). Then, 𝜋 = 𝜋1 − 𝜋2 for some 𝜋1 , 𝜋2 ∈ RMn⋆ (T, G)+ . Therefore, according to Proposition 2 (3.5.4), the measures 𝜇1 ≡ 𝜋1# and 𝜇2 ≡ 𝜋2# belong to RMwe (T, G)0 . Consider on the 𝛿-ring M ≡ domf 𝜇1 ∩ domf 𝜇2 , the finite measure 𝜇 ≡ 𝜇1 |M − 𝜇2 |M. Obviously, it is a narrow Radon measure. Thus, we can consider for 𝜋 the triplet 𝛽 ≡ (𝜇, 𝜇1 , 𝜇2 ). # Similarly, if 𝜋 = 𝜋󸀠 1 − 𝜋󸀠 2 for some 𝜋󸀠 1 , 𝜋󸀠 2 ∈ RMn⋆ (T, G)+ , then 𝜇󸀠 1 ≡ 𝜋󸀠 1 and # 𝜇󸀠 2 ≡ 𝜋󸀠 2 belong to RMwe (T, G)+ . Consider on the 𝛿-ring M󸀠 ≡ domf 𝜇󸀠 1 ∩ domf 𝜇󸀠 2 the narrow Radon measure 𝜇󸀠 ≡ 𝜇󸀠 1 |M󸀠 − 𝜇󸀠 2 |M󸀠 . As a result, we obtain for 𝜋 the other triplet 𝛽󸀠 ≡ (𝜇󸀠 , 𝜇󸀠 1 , 𝜇󸀠 2 ). Lemma 4. 𝛽 and 𝛽󸀠 are Radon triplets and 𝛽 ∼ 𝛽󸀠 .

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Proof. By the definition, of a Radon measure, C ⊂ domf 𝜇1 . Therefore, B(T, G) ⊂ dom 𝜇1 implies Bc (T, G) ⊂ domf 𝜇1 , where Bc (T, G) ⊂ M. Consequently, 𝛽 is a Radon triplet. The same is valid for 𝛽󸀠 . If B ∈ Bc (T, G), then according to the definition of narrow Radon measures from 3.5.4, 𝜇B = 𝜇1 B − 𝜇2 B = 𝜋1 B − 𝜋2 B = 𝜋B = 𝜋󸀠 1 B − 𝜋󸀠 2 B = 𝜇󸀠 1 B − 𝜇󸀠 2 B = 𝜇󸀠 B. This means that 𝛽 ∼ 𝛽󸀠 . It follows from Lemma 4 that we can define correctly a mapping H : RMn⋆ (T, G) → RB(T, G) setting H𝜋 ≡ 𝛽 ̄ = 𝛽󸀠̄ . # # If 𝜋 ∈ RMn⋆ b (T, G) and 𝜋 = 𝜋1 − 𝜋2 , then by Proposition 2 (3.5.4), 𝜋1 and 𝜋2 are bounded. Therefore, 𝛽 ≡ (𝜇, 𝜋1#, 𝜋2#) ∈ Reb (T, G) and 𝛽 ̄ ∈ RBb (T, G). Consequently, we ̄ can define correctly a mapping H b : RMn⋆ b (T, G) → RBb (T, G) setting H b 𝜋 ≡ 𝛽. It is n⋆ clear, that H b = H|RMb (T, G). Proposition 1. Let ⟮T, G⟯ be a Hausdorff space. Then, the mappings H : RMn (T, G) → RB(T, G) and H b : RMn⋆ b (T, G) → RBb (T, G) are bijective. Proof. Let 𝜋, 𝜘 ∈ RMn (T, G) and suppose that H𝜋 = H𝜘. By the definition, H𝜋 = 𝜃𝛽 for 𝛽 ≡ (𝜇, 𝜇1 , 𝜇2 ), where 𝜇1 ≡ (𝜋+ )#, 𝜇2 ≡ (−𝜋− )#, and 𝜇M ≡ 𝜇1 M − 𝜇2 M for every M ∈ M ≡ domf 𝜇1 ∩ domf 𝜇2 . Similarly, H𝜘 = 𝜃𝛾 for 𝛾 ≡ (𝜈, 𝜈1 , 𝜈2 ), where 𝜈1 ≡ (𝜘+ )#, 𝜈2 ≡ (−𝜘− )#, and 𝜈N ≡ 𝜈1 N − 𝜈2 N for every N ∈ N ≡ domf 𝜈1 ∩ domf 𝜈2 . From 𝛽 ∼ 𝛾, we conclude that 𝜇B = 𝜈B for every B ∈ Bc (T, G). As a result, 𝜋B = 𝜋+ B − (−𝜋− )B = 𝜇1 B − 𝜇2 B = 𝜇B = 𝜈B = 𝜈1 B − 𝜈2 B = 𝜘+ B − (−𝜘− )B = 𝜘B. Thus, 𝜋 = 𝜘, i. e. H is injective. Let 𝛼 ≡ (𝜆, 𝜆 1 , 𝜆 2 ) ∈ R(T, G), where 𝜆L = 𝜆 1 L − 𝜆 2 L for every L ∈ L ≡ domf 𝜆 1 ∩ domf 𝜆 2 . Consider the finite measures 𝜂 ≡ 𝜆, 𝜂1 ≡ 𝜆 1 |L and 𝜂2 ≡ 𝜆 2 |L. By the definition, 𝜂1 and 𝜂2 are compactly regular. It is obvious that they are locally bounded. Consequently, 𝜂1 and 𝜂2 are narrow positive Radon measures. Consider the measures 𝜇1 ≡ 𝜂1# and 𝜇2 = 𝜂2# from RMwe (T, G)0 . Let M ≡ domf 𝜇1 ∩ domf 𝜇2 . Consider the measure 𝜇 ≡ 𝜇1 |M − 𝜇2 |M and the triplet 𝛽 ≡ (𝜇, 𝜇1 , 𝜇2 ). By Lemma 4, H𝜂 = 𝛽.̄ If B ∈ Bc (T, G), then 𝜆 1 B = 𝜂1 B = 𝜇1 B and 𝜆 2 B = 𝜂2 B = 𝜇2 B. By Lemma 4, we have 𝜇B = 𝜇1 B − 𝜇2 B = 𝜆 1 B − 𝜆 2 B = 𝜆B. This means that 𝛽 ∼ 𝛼, where H𝜂 = 𝛼.̄ Thus, H is surjective. If 𝛼 ∈ Rb (T, G), then 𝜆 1 and 𝜆 2 are bounded and 𝜂 is bounded as well. So H b is surjective. Proposition 2. Let ⟮T, G⟯ be a Hausdorff space, 𝜋, 𝜘 ∈ RMn⋆ (T, G) and x, y ∈ R. Then, H(x𝜋 + y𝜘) = x ∗ H𝜋 + y ∗ H𝜘. The similar property is valid for H b . Proof. By Proposition 1, 𝜌 ≡ x𝜋 + y𝜘 ∈ A ≡ RMn⋆ (T, G). By the definition, 𝜋 = 𝜋1 − 𝜋2 and 𝜘 = 𝜘1 −𝜘2 for some 𝜋i , 𝜘i ∈ A+ . Consider the measures 𝜇1 ≡ 𝜋1#, 𝜇2 ≡ 𝜋2#, 𝜈1 ≡ 𝜘1# and 𝜈2 ≡ 𝜘2# from B ≡ RMwe (T, G)0 . Let M ≡ domf 𝜇1 ∩domf 𝜇2 and N ≡ domf 𝜈1 ∩domf 𝜈2 .

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Consider the measures 𝜇 ≡ 𝜇1 |M − 𝜇2 |M and 𝜈 ≡ 𝜈1 |N − 𝜈2 |N. Then, for the triplets 𝛽 ≡ (𝜇, 𝜇1 , 𝜇2 ) and 𝛾 ≡ (𝜈, 𝜈1 , 𝜈2 ), we have H𝜋 = 𝛽 ̄ and H𝜘 = 𝛾.̄ Suppose that x ⩾ 0 and y < 0. Then, by the definition x ∗ 𝛽 ̄ = 𝜃(x𝜇, x𝜇1 , x𝜇2 ) and y∗ 𝛾̄ = 𝜃(y𝜈, −y𝜈2 , −y𝜈1 ). Consider the 𝜎-algebras L1 ≡ dom 𝜇1 ∩dom 𝜈2 , L2 ≡ dom 𝜇2 ∩ dom 𝜈1 and the 𝛿-ring L ≡ M ∩ N. Consider the measures 𝜆 1 ≡ x𝜇1 |L1 + (−y𝜈2 )|L1 , 𝜆 2 ≡ x𝜇2 |L2 +(−y𝜈1 )|L2 and 𝜆 ≡ x𝜇|L+y𝜈|L. Then, for the Radon triplet 𝛼 ≡ (𝜆, 𝜆 1 , 𝜆 2 ) we have x ∗ 𝛽 ̄ + y ∗ 𝛾̄ = 𝛼.̄ Furthermore, consider the measures 𝜌1 ≡ x𝜋1 − y𝜘2 and 𝜌2 ≡ x𝜋2 − y𝜘1 . Then, 𝜌 = 𝜌1 − 𝜌2 and by Proposition 1 (3.5.4), 𝜌1 , 𝜌2 ∈ A+ . Consider the measures 𝜆󸀠 1 ≡ 𝜌1# and 𝜆󸀠 2 ≡ 𝜌2# from B. Let L󸀠 ≡ domf 𝜆󸀠 1 ∩ domf 𝜆󸀠 2 and 𝜆󸀠 ≡ 𝜆󸀠 1 |L󸀠 − 𝜆󸀠 2 |L󸀠 . Then, by the definition for the triplet 𝛼󸀠 ≡ (𝜆󸀠 , 𝜆󸀠 1 , 𝜆󸀠 2 ) we have H𝜌 = 𝜃𝛼󸀠 . If B ∈ Bc (T, G), then 𝜆󸀠 B = 𝜆󸀠 1 B − 𝜆󸀠 2 B = 𝜌1 B − 𝜌2 B = x𝜋1 B − y𝜘2 B − x𝜋2 B + y𝜘1 B = x𝜇1 B − y𝜈2 B − x𝜇2 B + y𝜈1 B = x𝜇B + y𝜈B = 𝜆B. This means that 𝛼󸀠 ∼ 𝛼. As a result, x ∗ H𝜋 + y ∗ H𝜘 = 𝜃𝛼 = 𝜃𝛼󸀠 = H(x𝜋 + y𝜘). For the other three opportunities for x and y, the arguments are the same. Corollary 1. Let ⟮T, G⟯ be a Hausdorff space. Then, with respect to the operation 𝛽 ̄ + 𝛾 ̄ and the composition x ∗ 𝛽 ̄ the families RB(T, G) and RBb (T, G) are linear spaces. Proof. By Proposition 1 (3.5.4), the family A ≡ RMn⋆ (T, G) is a linear space. Using the properties of the mapping H, from Propositions 1 and 2, we can deduce that the family B ≡ RB(T, G) is also a linear space. Check for example that x ∗ (y ∗ b) = (xy) ∗ b. By Proposition 1, b = Ha for some a ∈ A. From x(ya) = (xy)a we conclude by Proposition 2 that x ∗ (y ∗ b) = x ∗ H(ya) = x ∗ (y ∗ Ha) = H(x(ya)) = H((xy)a) = (xy) ∗ Ha = (xy) ∗ b. For the bounded case, the arguments are the same. Corollary 2. Let ⟮T, G⟯ be a Hausdorff space. Then, H : RMn⋆ (T, G) RB(T, G) (T, G) RB (T, G) are isomorphisms of the given linear spaces. and H b : RMn⋆ b b Proof. By virtue of Proposition 2 and Corollary 1 to it, H is a homomorphism of the given linear spaces. Then, Proposition 1 and Statement 5 (2.2.7) imply that H is an isomorphism. For H b , the arguments are the same. Further, instead of x ∗ 𝛽,̄ we shall write simply x𝛽.̄ The canonical order on RB(T , G) Now, we shall define an order on the linear space RB(T, G). Lemma 5. Let 𝛽 ≡ (𝜇, 𝜇1 , 𝜇2 ), 𝛽󸀠 ≡ (𝜇󸀠 , 𝜇󸀠 1 , 𝜇󸀠 2 ), 𝛾 ≡ (𝜈, 𝜈1 , 𝜈2 ) and 𝛾 ≡ (𝜈󸀠 , 𝜈󸀠 1 , 𝜈󸀠 2 ) be Radon triplets such that 𝛽 ∼ 𝛽󸀠 and 𝛾 ∼ 𝛾󸀠 . Then, the following conditions are equivalent:

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1) 𝜇B ⩽ 𝜈B for every B ∈ Bc (T, G); 2) 𝜇󸀠 B ⩽ 𝜈󸀠 B for every B ∈ Bc (T, G). Proof. The assertion follows from the equalities 𝜇B = 𝜇󸀠 B and 𝜈B = 𝜈󸀠 B for every B ∈ Bc (T, G). It follows from Lemma 5, that we can define correctly a binary relation 𝜔 on RB(T, G) ̄ 𝛾 ̄ if 𝜇󸀠 B ⩽ 𝜈󸀠 B for every B ∈ B (T, G) and some 𝛽󸀠 ≡ (𝜇󸀠 , 𝜇󸀠 , 𝜇󸀠 ) ∈ 𝛽 ̄ and setting 𝛽𝜔 c 1 2 󸀠 󸀠 𝛾 ≡ (𝜈 , 𝜈󸀠 1 , 𝜈󸀠 2 ) ∈ 𝛾.̄ Lemma 6. The relation 𝜔 is an order relation. ̄ 𝛽.̄ Let 𝛽𝜔 ̄ 𝛾̄ and 𝛾𝜔 ̄ 𝛽.̄ Then, 𝜇B ⩽ 𝜈B and 𝜈󸀠 B ⩽ 𝜇󸀠 B for Proof. It is obvious that 𝛽𝜔 every B ∈ Bc (T, G) and some 𝛽 ≡ (𝜇, 𝜇1 , 𝜇2 ) ∈ 𝛽,̄ 𝛽󸀠 ≡ (𝜇󸀠 , 𝜇󸀠 1 , 𝜇󸀠 2 ) ∈ 𝛽,̄ 𝛾 ≡ (𝜈, 𝜈1 , 𝜈2 ) ∈ 𝛾̄ and 𝛾󸀠 ≡ (𝜈󸀠 , 𝜈󸀠 1 , 𝜈󸀠 2 ) ∈ 𝛾.̄ From 𝜇B = 𝜇󸀠 B and 𝜈B = 𝜈󸀠 B we conclude that 𝜈B = 𝜈󸀠 B ⩽ 𝜇󸀠 B = 𝜇B, where 𝜈B = 𝜇B for every B ∈ Bc (T, G). Hence, 𝛾 ∼ 𝛽 and 𝛾 ̄ = 𝛽.̄ ̄ 𝛾̄ and 𝛾𝜔 ̄ 𝛿.̄ Then, 𝜇B ⩽ 𝜈B and 𝜈󸀠 B ⩽ 𝜋󸀠 B for every B ∈ Finally, let 𝛽𝜔 Bc (T, G) and some 𝛽 ≡ (𝜇, 𝜇1 , 𝜇2 ) ∈ 𝛽,̄ 𝛾 ≡ (𝜈, 𝜈1 , 𝜈2 ) ∈ 𝛾,̄ 𝛾󸀠 ≡ (𝜈󸀠 , 𝜈󸀠 1 , 𝜈󸀠 2 ) ∈ 𝛾̄ and 𝛿󸀠 ≡ (𝜋󸀠 , 𝜋󸀠 1 , 𝜋󸀠 2 ) ∈ 𝛿.̄ From 𝜈B = 𝜈󸀠 B, we infer that 𝜇B ⩽ 𝜈B = 𝜈󸀠 B ⩽ 𝜋󸀠 B. So by ̄ 𝛿.̄ the definition of 𝜔, we get 𝛽𝜔 Theorem 1. Let ⟮T, G⟯ be a Hausdorff space. Then, 1) the mappings H and H b are isotone; 2) the families RB(T, G) and RBb (T, G) are Dedekind complete lattice-ordered linear spaces and the second one is an l-ideal in the first one; 3) the mapping H : RMn⋆ (T, G) RB(T, G) is an isomorphism of the given lattice-ordered linear spaces. 4) the mapping H b : RMn⋆ RBb (T, G) is an isomorphism of the given b (T, G) lattice-ordered linear spaces. Proof. 1. Let 𝜋, 𝜘 ∈ RMn⋆ (T, G) ≡ A. By the definition, 𝜋 = 𝜋1 − 𝜋2 and 𝜘 = 𝜘1 − 𝜘2 for some positive Borel – Radon measures. Consider 𝜇i ≡ 𝜋i#, 𝜈i ≡ 𝜘i#, Mi ≡ dom 𝜇i , Ni ≡ dom 𝜈i , M ≡ M1f (𝜇1 ) ∩ M2f (𝜇2 ), N ≡ N1f (𝜈1 ) ∩ N2f (𝜈2 ), 𝜇 ≡ 𝜇1 |M − 𝜇2 |M and 𝜈 ≡ 𝜈1 |N − 𝜈2 |N. Then, by the definition of H for the triplets 𝛽 ≡ (𝜇, 𝜇1 , 𝜇2 ) and 𝛾 ≡ (𝜈, 𝜈1 , 𝜈2 ), we have H𝜋 = 𝛽 ̄ and H𝜘 = 𝛾.̄ If 𝜋 ⩽ 𝜘, then B ∈ Bc (T, G) implies 𝜇B = 𝜇1 B − 𝜇2 B = 𝜋1 B − 𝜋2 B ⩽ 𝜘1 B − 𝜘2 B = 𝜈1 B − 𝜈2 B = 𝜈B. This gives H𝜋 = 𝛽 ̄ ⩽ 𝛾̄ = H𝜘. Now, let 𝛽 ̄ = H𝜋 ⩽ H𝜘 = 𝛾.̄ Then, by Lemma 6, 𝜇B ⩽ 𝜈B for every B ∈ Bc (T, G). Therefore, 𝜋B = 𝜋1 B − 𝜋2 B = 𝜇1 B − 𝜇2 B = 𝜇B ⩽ 𝜈B = 𝜈1 B − 𝜈2 B = 𝜘1 B − 𝜘2 B = 𝜘B, i. e. 𝜋 ⩽ 𝜘. 2. By Corollary 2 to Proposition 2, H is bijective. Then, by Lemma 1 (1.1.15), the mapping H −1 is isotone, and according to Lemma 2 (1.1.15), H and H −1 preserve any

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exact bounds. Again, by Corollary 2 to Proposition 2, H is an isomorphism of the given linear spaces. Let E ⊂ RB(T, G) ≡ B and b is an upper bound of E. Consider the set D ≡ H −1 [E] and the element a ≡ H −1 b. If d ∈ D, then Hd ⩽ b = Ha implies d ⩽ a. Therefore, by Proposition 1 (3.5.4), there is d0 ≡ sup D. As a result, Hd0 = sup H[D] = sup E. If E is lower bounded, then the arguments are the same. Thus, B is Dedekind complete. Let b1 , b2 ∈ B. Consider the elements a i ≡ H −1 b i , a󸀠 = a1 ∨ a2 and a󸀠󸀠 = a1 ∧ a2 . Then, Ha󸀠 = Ha1 ∨ Ha2 = b1 ∨ b2 and Ha󸀠󸀠 = Ha1 ∧ Ha2 = b1 ∧ b2 . Consequently, the ordered set B is lattice-ordered. In a similar way, it is checked that B is a lattice-ordered linear space. For example, let b, b1 , b2 ∈ B and b1 ⩽ b2 . Take a i ≡ H −1 b i and a ≡ H −1 b. Then, a1 ⩽ a2 implies a1 + a ⩽ a2 + a. Applying H, we get b1 + b = H(a1 + a) ⩽ H(a2 + a) = b2 + b. 3. From the equalities H(a1 ∨ a2 ) = Ha1 ∨ Ha2 and H(a1 ∧ a2 ) = Ha1 ∧ Ha2 and Corollary 2 to Proposition 2 it follows that H : A → B is a homomorphism of the given lattice-ordered linear spaces. Since H is bijective by Proposition 1, Statement 5 (2.2.7) implies that H is the necessary isomorphism. 4. For the families A b ≡ RMn⋆ b (T, G) and B b ≡ RBb (T, G) and for mapping H b , the arguments are the same. We need only to check that B b is an l-ideal in B. Let b ∈ B b , d ∈ B and |d| ⩽ |b|. Take a ≡ H −1 b ∈ A and c ≡ H −1 d ∈ A. Since H b is the restriction of H, we have a ∈ A b . Since H is the isomorphism, we deduce that |c| ⩽ |a| in A. Now, by Proposition 1 (3.5.4), c ∈ A b . As a result, d = H b c ∈ B b . The canonical embedding of RMw⋆ (T , G)0 into RB(T , G) Now, we shall connect wide positive Radon measures and Radon bimeasures. The cones of positive elements of the lattice-ordered linear spaces RB(T, G) and RBb (T, G) will be denoted as usual by RB(T, G)+ and RBb (T, G)+ , respectively. For every measure 𝜇 ∈ RMw⋆ (T, G)0 , take the narrow positive Radon measure f 𝜇 ≡ 𝜇| domf 𝜇, the Radon triplet 𝛽 ≡ (𝜇f , 𝜇, 0) and the positive Radon bimeasure m = 𝛽.̄ This procedure defines a mapping E0 : RMw⋆ (T, G)0 → RB(T, G) such that E𝜇 = m. If 𝜇 ∈ RMw⋆ b (T, G)+ , then E𝜇 ∈ RBb (T, G)+ . Therefore, we can define a mapping E0b : RMw⋆ (T, G)+ → RBb (T, G) setting E0b ≡ E0 |RMw⋆ b b (T, G)+ . Theorem 2. Let ⟮T, G⟯ be a Hausdorff space. Then, 1) the mappings E0 : RMw⋆ (T, G)0 → RB(T, G)+ and E0b : RMw⋆ b (T, G)+ → RBb (T, G)+ are bijective; 2) the mappings E0 and E0b are conic operators; 3) the mappings E0 and E0b are isotone and preserve any exact bounds; 4) for every bimeasure m ∈ RB(T, G) there are the unique measures 𝜇󸀠 , 𝜇󸀠󸀠 ∈ RMw⋆ (T, G)0 such that m = E0 𝜇󸀠 − E0 𝜇󸀠󸀠 , E0 𝜇󸀠 = m+ and E0 𝜇󸀠󸀠 = −m− ; if m is bounded, then 𝜇󸀠 and 𝜇󸀠󸀠 are bounded as well.

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Proof. Let E0 𝜇 = E0 𝜈. This means, that 𝜇f B = 𝜈f B for every B ∈ Bc (T, G). This implies 𝜇C = 𝜈C for every C ∈ C. By virtue of the compact regularity of these measures 𝜇 = 𝜈. Thus, E0 is injective. Take 𝜇, 𝜈 ∈ RMw⋆ (T, G)0 and x, y ∈ R+ . By Corollary 1 of Proposition 3 (3.5.3), 𝜋 = x𝜇+y𝜈 ∈ RMw⋆ (T, G)0 . Suppose that x, y > 0. By the definition, E0 𝜇 = 𝛽,̄ E0 𝜈 = 𝛾̄ and E0 𝜋 = 𝛿̄ for 𝛽 ≡ (𝜇f , 𝜇, 0), 𝛾 ≡ (𝜈f , 𝜈, 0) and 𝛿 ≡ (𝜋f , 𝜋, 0). It is clear that 𝜋f = x𝜇f + y𝜈f . Consider the 𝜎-algebra L1 ≡ L2 ≡ B(T, G) and the 𝛿-ring L ≡ domf (x𝜇) ∩ domf (y𝜈) = domf 𝜇 ∩ domf 𝜈 = domf 𝜋. Consider the measures 𝜆 1 ≡ 𝜋 on L1 , 𝜆 2 ≡ 0 on L2 and 𝜆 = 𝜋f on L and the corresponding Radon triplet 𝛼 ≡ (𝜆, 𝜆 1 , 𝜆 2 ). By the definition, x𝛽 ̄ = 𝜃(x𝜇f , x𝜇, 0), y𝛾̄ = 𝜃(y𝜈f , y𝜈, 0) and xE0 𝜇 + yE0 𝜈 = x𝛽 ̄ + y𝛾 ̄ = 𝛼̄ = 𝜃(𝜋f , 𝜋, 0) = E0 𝜋. Let m ∈ RB(T, G) and m = 𝛽 ̄ for some 𝛽 ≡ (𝜇, 𝜇1 , 𝜇2 ) ∈ R(T, G). Consider the wide Borel measures 𝜈󸀠 ≡ 𝜇1 |B(T, G) and 𝜈󸀠󸀠 ≡ 𝜇2 |B(T, G). By Lemma 4 (3.5.3), 𝜈󸀠 , 𝜈󸀠󸀠 ∈ RMw⋆ (T, G)0 . Consider the 𝛿-ring N ≡ domf 𝜈󸀠 ∩ domf 𝜈󸀠󸀠 ⊂ Mf (𝜇) and the finite measure 𝜈 ≡ 𝜈󸀠 |N − 𝜈󸀠󸀠 |N. Then, we get Radon triplet 𝛾 ≡ (𝜈, 𝜈󸀠 , 𝜈󸀠󸀠 ). From 𝜈B = 𝜈󸀠 B − 𝜈󸀠󸀠 B = 𝜇1 B − 𝜇2 B = 𝜇B for every B ∈ Bc (T, G) we infer that 𝛽 ∼ 𝛾. Thus, m = 𝛾.̄ f

By the definition, of the mapping E0 we have E0 𝜈󸀠 = 𝜃(𝜈󸀠 , 𝜈󸀠 , 0) and E0 𝜈󸀠󸀠 = 𝜃(𝜈 , 𝜈󸀠󸀠 , 0). Respectively, by the definition of the composition, we have −E0 𝜈󸀠󸀠 = f 𝜃(−𝜈󸀠󸀠 , 0, 𝜈󸀠󸀠 ). Consider the 𝜎-algebras L1 ≡ L2 ≡ B(T, G) and the measures 𝜆 1 ≡ 𝜈󸀠 f f on L1 and 𝜆 2 ≡ 𝜈󸀠󸀠 on L2 . Consider also the 𝛿-ring L ≡ domf (𝜈󸀠 ) ∩ domf (−𝜈󸀠󸀠 ) = f f N, the measure 𝜆 ≡ 𝜈󸀠 |L + (−𝜈󸀠󸀠 )|L = 𝜈󸀠 |N − 𝜈󸀠󸀠 |N = 𝜈 and the Radon triplet 𝛼 ≡ (𝜆, 𝜆 1 , 𝜆 2 ). By the definition, of the addition, we have E0 𝜈󸀠 − E0 𝜈󸀠󸀠 = 𝛼̄ = 𝜃(𝜈, 𝜈󸀠 , 𝜈󸀠󸀠 ) = m. If m is bounded, then 𝛽 ∈ Rb (T, G) and consequently 𝜈󸀠 and 𝜈󸀠󸀠 are bounded. Now, suppose, that m = 𝛾̄ ∈ RB(T, G)+ , i. e. 𝛾 ̄ ⩾ 𝜃(𝜁, 𝜁1 , 𝜁2 ). By the definition, of the order in RB(T, G), we have 𝜈B ⩾ 𝜁B = 0 for every B ∈ Bc (T, G). By Corol󸀠 󸀠󸀠 , 𝜈00 ∈ RMn⋆ (T, G)+ . By the definition, the equality lary 1 of Proposition 2 (3.5.4), 𝜈00 󸀠 󸀠󸀠 𝜈00 = 𝜈00 − 𝜈00 means that 𝜈00 ∈ RMn⋆ (T, G)+ . Consider the measure 𝜋 ≡ (𝜈00 )# ∈ RMwe (T, G)0 from Proposition 2 (3.5.4). If m is bounded, then 𝜋 ∈ RMwe b (T, G)+ . w⋆ Finally consider the measure 𝜆 ≡ 𝜋|B(T, G) ∈ RM (T, G)0 (see Lemma 4 (3.5.3)). For 𝜆, take the corresponding Radon triplet 𝛼 ≡ (𝜆f , 𝜆, 0). If B ∈ Bc (T, G), then f 𝜆 B = 𝜆B = 𝜋B = 𝜈00 B = 𝜈B. This means that 𝛼 ∼ 𝛾. As a result, m = 𝛾̄ = 𝛼̄ = E0 𝜆. Thus, E0 is surjective and so bijective. If m is bounded, then 𝜆 and 𝛼 are bounded as well. Consequently, E0b is bijective. Let 𝜇 ⩽ 𝜈 in RMw⋆ (T, G)0 . Then, 𝜇f B ⩽ 𝜈f B for every B ∈ Bc (T, G) means that 0 E 𝜇 = 𝜃(𝜇f , 𝜇, 0) ⩽ 𝜃(𝜈f , 𝜈, 0) = E0 𝜈. Conversely, let E0 𝜇 ⩽ E0 𝜈 in RB(T, G)+ . By the definition, 𝜇󸀠 B ⩽ 𝜈󸀠 B for every B ∈ Bc (T, G) and some 𝛽󸀠 ≡ (𝜇󸀠 , 𝜇󸀠 1 , 𝜇󸀠 2 ) ∈ E0 𝜇 and 𝛾󸀠 ≡ (𝜈󸀠 , 𝜈󸀠 1 , 𝜈󸀠 2 ) ∈ E0 𝜈. According to Lemma 5, 𝜇f B ⩽ 𝜈f B for every B ∈ Bc (T, G). Therefore, for every compact set C, we have 𝜇C ⩽ 𝜈C. If M ∈ B(T, G), then by virtue of the compact regularity of these measures, we get 𝜇M = sup{𝜇C | C ∈ C ∧ C ⊂ M} ⩽ sup{𝜈C | C ∈ C ∧ C ⊂ M} = 𝜈M. Therefore, 𝜇 ⩽ 𝜈. Thus, E 0 is isotone. According to Lemma 2 (1.1.15), E0 preserves any exact bounds. The same is valid for E0b . 󸀠󸀠 f

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Let m ∈ RB(T, G). Then, by virtue of Theorem 1, m = m+ +m− . Since E0 is bijective, 𝜇+ = E0 𝜇󸀠 and −m− = E0 𝜇󸀠󸀠 for unique 𝜇󸀠 , 𝜇󸀠󸀠 ∈ RMw⋆ (T, G)0 . If m is bounded, then by (1) 𝜇󸀠 and 𝜇󸀠󸀠 are bounded as well. The mapping E0 : RMw⋆ (T, G)0 RB(T, G) will be called the canonical embedding of the conic space of all positive wide Borel – Radon measures into the latticeordered linear space of all Radon bimeasures. The measures 𝜇󸀠 and 𝜇󸀠󸀠 will be called the parent positive wide Borel – Radon measures of the Radon bimeasure m and will be denoted by p󸀠 m and p󸀠󸀠 m correspondingly. Between the bimeasure m and its parent measures p󸀠 m and p󸀠󸀠 m there is also the following connection. Proposition 3. Let ⟮T, G⟯ be a Hausdorff space and m ∈ RB(T, G). Then, (𝜇, p 󸀠 m, p󸀠󸀠 m) ∈ R(T, G) and m = 𝜃(𝜇, p 󸀠 m, p󸀠󸀠 m), where 𝜇 ≡ p󸀠 m|M − p󸀠󸀠 m|M is a narrow Radon measure on the 𝛿-ring M ≡ domf (p󸀠 m) ∩ domf (p󸀠󸀠 m). Proof. Denote p󸀠 m and p󸀠󸀠 m by 𝜇󸀠 and 𝜇󸀠󸀠 , respectively. It is obvious that 𝜇1 ≡ 𝜇󸀠 |M and 𝜇2 ≡ 𝜇󸀠󸀠 |M are positive narrow Radon measures. Therefore, by the definition, 𝜇 = 𝜇1 − 𝜇2 is a narrow Radon measure. It is clear also that (𝜇, 𝜇󸀠 , 𝜇󸀠󸀠 ) is a Radon triplet. f By the definition, of the mapping E0 , we have E𝜇󸀠 = 𝜃(𝜇󸀠 , 𝜇󸀠 , 0) and E𝜇󸀠󸀠 = f f 𝜃(𝜇󸀠󸀠 , 𝜇󸀠󸀠 , 0). By the definition, of the composition, we have −E𝜇󸀠󸀠 = 𝜃(−𝜇󸀠󸀠 , 0, 𝜇󸀠󸀠 ). From the definition of the addition, we deduce as in the proof of Theorem 2 that m = E𝜇󸀠 − E𝜇󸀠󸀠 = 𝜃(𝜇, 𝜇󸀠 , 𝜇󸀠󸀠 ). The canonical embedding of RMw⋆ (T, G) into RB(T , G) Now, we shall consider some extension of the embedding E 0 . Let 𝜇 be a Borel – Radon measure. By Lemma 1 (3.5.3), 𝜇 = 𝜇1 − 𝜇2 for some positive Borel – Radon measures 𝜇1 and 𝜇2 at least one of which is finite. Consider the Borel – Radon triplet 𝛽 ≡ (𝜇|Bc , 𝜇1 , 𝜇2 ), where Bc ≡ Bc (T, G) is the 𝛿-ring of precompact Borel sets. If 𝜇 = 𝜈1 − 𝜈2 for some other positive Borel – Radon measures 𝜈1 and 𝜈2 , then (𝜇|Bc , 𝜈1 , 𝜈2 ) ∼ 𝛽. This means that we can define correctly a mapping E : RMw⋆ (T, G) → RB(T, G) by setting E𝜇 ≡ 𝜃𝛽. If 𝜇 ∈ RMw⋆ b (T, G), then E𝜇 ∈ w⋆ RBb (T, G). Therefore, a mapping E b ≡ E|RMw⋆ (T, G) : RM (T, G) → RBb (T, G) b b is well defined. Theorem 3. Let ⟮T, G⟯ be a Hausdorff space. Then, 1) if 𝜇 ∈ RMw⋆ (T, G) and 𝜇 = 𝜇1 − 𝜇2 for some positive Borel – Radon measures 𝜇1 and 𝜇2 at least one of which is finite, then E𝜇 = E0 𝜇1 − E0 𝜇2 ; 2) if 𝜆, 𝜇, 𝜈 ∈ RMw⋆ (T, G), x, y ∈ R, measures x𝜇 and y𝜈 take the infinite values of one and the same sign, and 𝜆 = x𝜇 + y𝜈, then E𝜆 = xE𝜇 + yE𝜈; 3) the mapping E is injective, isotone, and preserves any exact bounds.

348 | 3.5 Topological spaces with measures. The Radon integral

Proof. 1. Consider 𝜈 ≡ 𝜇|Bc , 𝜈1 ≡ 𝜇1 |Bc , 𝜈2 ≡ 𝜇2 |Bc , m ≡ 𝜃(𝜈, 𝜇1 , 𝜇2 ), m1 ≡ 𝜃(𝜈1 , 𝜇1 , 0), and m2 ≡ 𝜃(𝜈2 , 0, 𝜇2 ). Then, −m2 ≡ 𝜃(−𝜈2 , 𝜇2 , 0). Since 𝜈 = 𝜈1 + (−𝜈2 ), 𝜇1 = 𝜇1 + 0, and 𝜇2 = 0 + 𝜇2 , we infer that m = m1 − m2 . Consequently, E𝜇 = E0 𝜇1 − E0 𝜇2 . 2. By Lemma 1 (3.5.3) 𝜇 = 𝜇1 − 𝜇2 and 𝜈 = 𝜈1 − 𝜈2 for some measures satisfying the condition of assertion 1. Then, according to (1), we obtain E𝜇 = E0 𝜇1 − E0 𝜇2 and E𝜈 = E0 𝜈1 − E0 𝜈2 . Assume that x ⩾ 0, y ⩽ 0, and x𝜇 and y𝜈 take the value ∞. Then, x𝜇2 and (−y)𝜈1 are finite. Consequently, 𝜆 = x𝜇1 − x𝜇2 + y𝜈1 − y𝜈2 = (x𝜇1 + (−y)𝜈2 ) − (x𝜇2 + (−y)𝜈1 ). Applying (1) and Theorem 2, we get E𝜆 = E0 (x𝜇1 + (−y)𝜈2 ) − E0 (x𝜇2 + (−y)𝜈1 ) = xE0 𝜇1 + (−y)E0 𝜈2 − (xE0 𝜇2 + (−y)E0 𝜈1 ) = x(E0 𝜇1 − E0 𝜇2 ) + y(E0 𝜈1 − E0 𝜈2 ) = xE𝜇 + yE𝜈. All the other cases are considered in the same way. 3. Let 𝜇 and 𝜈 are Borel – Radon measures such that E𝜇 = E𝜈. As above 𝜇 = 𝜇1 − 𝜇2 and 𝜈 = 𝜈1 − 𝜈2 for some measures satisfying the condition of assertion 1. Then, by (1) we obtain E𝜇 = E0 𝜇1 − E0 𝜇2 and E𝜈 = E0 𝜈1 − E0 𝜈2 , where, by Theorem 2, we get E0 (𝜇1 + 𝜈2 ) = E0 (𝜇2 + 𝜈1 ). Since E0 is injective, we infer that 𝜇1 + 𝜈2 = 𝜇2 + 𝜈1 . Using the definition of E and the equality E𝜇 = E𝜈, we infer that (𝜇|Bc , 𝜇1 , 𝜇2 ) ∼ (𝜈|Bc , 𝜈1 , 𝜈2 ). This means that 𝜇B = 𝜈B for every B ∈ Bc . Take any B ∈ B ≡ B(T, G). First, assume that 𝜇2 is finite and 𝜇B = ∞. Then, 𝜇1 = ∞. Take any a > 0. Then, 𝜇2 B < ka for some k ∈ N. By the definition, of a Radon measure, there is a compact set C ⊂ B such that 𝜇1 C > (k + 1)a. Therefore, 𝜈1 C − 𝜈2 C = 𝜈C = 𝜇C = 𝜇1 C − 𝜇2 C > (k + 1)a − 𝜇2 B ⩾ (k + 1)a − ka = a implies 𝜈1 B ⩾ 𝜈1 C > x. This means that 𝜈1 B = ∞. Therefore, 𝜈2 is finite and 𝜈B = ∞ = 𝜇B. If 𝜇B = −∞, then the arguments are similar. If 𝜇B ∈ R, then 𝜇1 B ∈ R and the equality 𝜇1 B + 𝜈2 B = 𝜇2 B + 𝜈1 B means that 𝜈2 B and 𝜈1 B can be only finite. Consequently, 𝜇B = 𝜇1 B − 𝜇2 B = 𝜈1 B − 𝜈2 B = 𝜈B. If 𝜇1 is finite, then all the arguments are analogous. So, E is injective. Let 𝜇 and 𝜈 are as above and 𝜇 ⩽ 𝜈. Then, by definition, E𝜇 = 𝜃(𝜇|Bc , 𝜇1 , 𝜇2 ) ⩽ 𝜃(𝜈|Bc , 𝜈1 , 𝜈2 ) = E𝜈. Conversely, if E𝜇 ⩽ E𝜈, then 𝜇󸀠 ⩽ 𝜈󸀠 for some Radon triplets (𝜇󸀠 , 𝜇1󸀠 , 𝜇2󸀠 ) ∈ E𝜇 and (𝜈󸀠 , 𝜈1󸀠 , 𝜈2󸀠 ) ∈ E𝜈. By definition, 𝜇B = 𝜇󸀠 B and 𝜈B = 𝜈󸀠 B for every B ∈ Bc . Therefore, 𝜇C ⩽ 𝜈C for every C ∈ C. Assume that 𝜇B > 𝜈B for some B ∈ Bc . Therefore, 𝜈B < ∞ and 𝜈1 B < ∞ in all the cases below. First, consider the case 𝜇B = ∞. Then, 𝜇1 B = ∞ and 𝜇2 B < ∞. Suppose 𝜈2 B = ∞ and take any x > 0. Then, kx > 𝜇2 B and kx > 𝜈1 B for some k ∈ N. By the condition, 𝜇1 C > (k + 1)x and 𝜈2 C > (k + 1)x for some compact set C ⊂ B. As a result, 𝜇C = 𝜇1 C − 𝜇2 C > (k + 1)x − 𝜇2 B > (k + 1)x − kx = x and 𝜈C = 𝜈1 C − 𝜈2 C > (k + 1)x − 𝜇2 B < 𝜈1 B−(k+1)x < kx−(k+1)x = −x imply 𝜇C > x > −x > 𝜈C, but this inequality contradicts to the inequality proven above. Suppose 𝜈2 B < ∞ and take some y > 𝜈1 B. Then, ky > 𝜇2 B for some k ∈ N. By the condition, 𝜇1 D > (k + 1)y for some compact set D ⊂ B. As a result, 𝜇D = 𝜇1 D − 𝜇2 D > (k + 1)y − 𝜇2 B > (k + 1)y − ky = y > 𝜈1 B ⩾ 𝜈1 D ⩾ 𝜈1 D − 𝜈2 D = 𝜈D, but this inequality is also impossible.

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Now, consider the case 𝜇B < ∞. Then, 𝜇1 B < ∞. Therefore, 𝜇B > 𝜈B implies 𝜇2 B < ∞. Suppose 𝜈2 B = ∞ and take some y > 𝜇2 B. Then, ky > 𝜈1 B for some k ∈ N. By the condition, 𝜈2 D > (k + 1)y for some compact set D ⊂ B. As a result, 𝜈D = 𝜈1 D − 𝜈2 D < 𝜈1 B − (k + 1)y < ky − (k + 1)y = −y < −𝜇2 B ⩽ −𝜇2 D ⩽ 𝜇1 D − 𝜇2 D = 𝜈D, but this inequality is impossible. Suppose 𝜈2 B < ∞ and take some 𝜀 > 0 such that 𝜇B − 𝜀 > 𝜈B. It follows from the condition that there is a compact set C ⊂ B such that 0 ⩽ 𝜇1 B − 𝜇1 C < 𝜀/2 and 0 ⩽ 𝜈2 B − 𝜈2 C < 𝜀/2. As a result, 𝜇C = 𝜇1 C − 𝜇2 C ⩾ (𝜇1 C − 𝜇1 B) + 𝜇1 B − 𝜇2 B > −𝜀/2 + 𝜇B > 𝜈B + 𝜀/2 = 𝜈1 B − 𝜈2 B + 𝜀/2 > 𝜈1 C − 𝜈2 C = 𝜈C, but this is also impossible. Thus, in all the cases, we come to the contradiction. Consequently, our assumption 𝜇B > 𝜈B is not true. So 𝜇 ⩽ 𝜈. Thus, E is isotone. According to Lemma 2 (1.1.15), E preserves any exact bounds. Corollary 1. Let ⟮T, G⟯ be a Hausdorff space. Then, E𝜇 = E0 (v+ (𝜇)) − E0 (−v− (𝜇)) for every 𝜇 ∈ RMw⋆ (T, G). Proof. By Lemma 1 (3.5.3) 𝜇 = v+ (𝜇)+ v− (𝜇) and at least one of these variations is finite. Now, assertion 1 of Theorem 3 gives the necessary equality. Corollary 2. Let ⟮T, G⟯ be a Hausdorff space. Then, 1) the mapping E b : RMw⋆ b (T, G) → RBb (T, G) is bijective, isotone, and preserves any exact bounds; 2) E b is an isomorphism between the given lattice-ordered linear spaces; 3) E b 𝜇 = E0b (𝜇+ ) − E0b (−𝜇− ) for every 𝜇 ∈ RMw⋆ b (T, G). Proof. 1. Let m be a bounded Radon bimeasure. By Theorem 2 m = E0 𝜇󸀠 −E0 𝜇󸀠󸀠 for some positive bounded Borel – Radon measures 𝜇󸀠 and 𝜇󸀠󸀠 . Consider the bounded Borel – Radon measure 𝜇 ≡ 𝜇󸀠 − 𝜇󸀠󸀠 . Then, by Theorem 3 E b 𝜇 = E𝜇 = E0 𝜇󸀠 − E0 𝜇󸀠󸀠 = m. So, E b is surjective. By virtue of Theorem 3 E b is injective, isotone, and preserves any exact bounds. 2. This follows from (1) and assertion 2 of Theorem 3. 3. Take the variations v+ (𝜇) and v− (𝜇) from Lemma 1 (3.5.3). For bounded measure 𝜇, we have v+ (𝜇) = 𝜇+ and v− (𝜇) = 𝜇− due to Proposition 1 (3.2.2). Therefore, the necessary equality follows from Corollary 1. The mapping E : RMw⋆ (T, G) RB(T, G) will be called the canonical embedding of the family of all positive wide Borel – Radon measures into the lattice-ordered linear space of all Radon bimeasures. Proposition 4. Let ⟮T, G⟯ be a Hausdorff space. Then, 1) if m ∈ RB(T, G), 𝛾 ≡ (𝜈, 𝜈1 , 𝜈2 ) ∈ m, and 𝜇 ≡ 𝜈|Bc (T, G), then p󸀠 m = (𝜇+ )# 0 and p󸀠󸀠 m = (−𝜇− )# ; 0

350 | 3.5 Topological spaces with measures. The Radon integral

2) if 𝜈 ∈ RMw⋆ (T, G), 𝜇 ≡ 𝜈|Bc (T, G), and m ≡ 𝜃(𝜈c , v+ (𝜈), −v− (𝜈)), then p󸀠 m = v+ (𝜈) and p󸀠󸀠 m = −v− (𝜈). Proof. 1. Denote p󸀠 m and p󸀠󸀠 m by 𝜇󸀠 and 𝜇󸀠󸀠 , respectively. By virtue of Proposition 1 (3.5.4), 𝜇 = 𝜇+ + 𝜇− and by virtue of Theorems 1 and 2, E𝜇󸀠 = H(𝜇+ ) and E𝜇󸀠󸀠 = H(−𝜇− ). For 𝜇+ , take 𝜘1 ≡ 𝜇+ and 𝜘2 ≡ 0. Then, 𝜘1# = (𝜇+ )# and 𝜘2# = 0. We can suppose that dom 𝜘2# = dom 𝜘1# = dom(𝜇+ )#. Consider on the 𝛿-ring K = domf (𝜇+ )# the positive narrow Radon measure 𝜘 ≡ 𝜘1#|K = (𝜇+ )#f . Then, by the definition of H, we have H(𝜇+ ) = 𝜃((𝜇+ )#f , (𝜇+ )#, 0). Analogously, H(−𝜇− ) = 𝜃((−𝜇− )#f , (−𝜇− )#, 0). f

f

Also, by definition of E, we have E𝜇󸀠 = 𝜃(𝜇󸀠 , 𝜇󸀠 , 0) and E𝜇󸀠󸀠 = 𝜃(𝜇󸀠󸀠 , 𝜇󸀠󸀠 , 0). From E𝜇󸀠 = H(𝜇+ ) and E𝜇󸀠󸀠 = H(−𝜇− ), we deduce now that 𝜇󸀠 B = (𝜇+ )#B and 𝜇󸀠󸀠 B = (−𝜇− )#B for every B ∈ R ≡ Bc (T, G). From the compact regularity of these measures we infer 󸀠󸀠 # that 𝜇󸀠 = (𝜇+ )# 0 and 𝜇 = (−𝜇− )0 . 2. By Lemma 1 (3.5.3) 𝜈1 ≡ v+ (𝜈) and 𝜈2 ≡ −v− (𝜈) are positive Radon measures. By Theorem 2 𝜇󸀠 ≡ p󸀠 m and 𝜇󸀠󸀠 ≡ p󸀠󸀠 m are also positive Radon measures. It follows from (1) that 𝜇󸀠 |R = 𝜇+ and 𝜇󸀠󸀠 |R = −𝜇− . By Proposition 1 (3.2.2) 𝜇+ = v+ (𝜇) and 𝜇− = v− (𝜇). Take any Hahn decomposition (T+ , T− ) of T with respect to 𝜈. Then, by Lemma 5 (3.2.1) 𝜈1 (B) = 𝜈(B ∩ T+ ) and 𝜈2 (B) = −𝜈(B ∩ T− ) for every B ∈ B(T, G). Take any R ∈ R and consider R+ ≡ R ∩ T+ and R− ≡ R ∩ T− . If S ∈ R and S ⊂ R+ , then 𝜇S = 𝜈S ⩾ 0. If S ∈ R and S ⊂ R− , then 𝜇S = 𝜈S ⩽ 0. Besides, R = R+ ∪ R− , hence (R+ , R− ) is a Hahn decomposition of R for 𝜈. Therefore, by definition of the variations, v+ (𝜇)R ≡ 𝜇R+ and v− (𝜇)R ≡ 𝜇R− . This implies 𝜇󸀠 R = 𝜇+ R = 𝜇(R ∩ T+ ) = 𝜈(R ∩ T+ ) = 𝜈1 R and 𝜇󸀠󸀠 R = −𝜇− R = −𝜇(R ∩ T− ) = −𝜈(R∩T− ) = 𝜈2 R. From the compact regularity of these positive Radon measures, we infer that 𝜇󸀠 = 𝜈1 and 𝜇󸀠󸀠 = 𝜈2 . Corollary 1. Let ⟮T, G⟯ be Hausdorff space and 𝜇 ∈ RMw⋆ (T, G). Then, p󸀠 (E𝜇) = v+ (𝜇) and p󸀠󸀠 (E𝜇) = −v− (𝜇). Proof. By Lemma 1 (3.5.3) 𝜇 = v+ (𝜇)+ v− (𝜇) and at least one of these variations is finite. Therefore, by definition, we have E𝜇 = 𝜃(𝜇|Bc , v+ (𝜇), −v− (𝜇)). Now, Proposition 4 gives the necessary equalities. 󸀠 Corollary 2. Let ⟮T, G⟯ be Hausdorff space and 𝜇 ∈ RMw⋆ b (T, G). Then, p (E b 𝜇) = 𝜇+ 󸀠󸀠 and p (E b 𝜇) = −𝜇− .

Proof. For a bounded Radon measure 𝜇, we have v+ (𝜇) = 𝜇+ and v− (𝜇) = 𝜇− due to Proposition 1 (3.2.2). Therefore, the necessary equalities follow from Corollary 1.

3.5.5 Radon bimeasures on Hausdorff spaces

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Theorems 1, 2, and 3 show that RB(T, G) forms a natural lattice-ordered linear “envelope” for the family of all wide Borel – Radon measures. Moreover, in 3.5.6 we shall establish that the mappings E0 , E, and E b preserve the Lebesgue integral.

Norm on RBb (T , G) Now, we shall introduce a norm in RBb (T, G) and prove that E b preserves norm. Theorem 2 implies that we can define correctly a mapping 𝜌 : RBb (T, G) → R+ setting 𝜌m ≡ (p󸀠 m)T + (p󸀠󸀠 m)T. In 3.5.4 we considered on RMn⋆ b (T, G) the norm ‖ ⋅ ‖ induced from ⟮Measb (T, Bc (T, G)), ‖ ⋅ ‖⟯. Proposition 5. Let ⟮T, G⟯ be a Hausdorff space. Then, 1) ‖𝜇‖ = 𝜌(H b 𝜇) for every 𝜇 ∈ RMn⋆ b (T, G); 2) 𝜌 is a monotone norm on the lattice-ordered linear space RBb (T, G). Proof. 1. Denote RMn⋆ b (T, G) by Y. By Proposition 4 (3.5.4) ⟮Y , ‖ ⋅ ‖⟯ is a pre-L-space. Let 𝜇 ∈ Y. Then, |𝜇| = 𝜇+ + (−𝜇− ) implies ‖𝜇‖ = ‖ |𝜇| ‖ = ‖𝜇+ ‖ + ‖ − 𝜇− ‖ = var(𝜇+ )T + var(−𝜇− )T. By Corollary 2 to Lemma 7 (3.1.3) var(𝜇+ )T = sup{𝜇+ B | B ∈ Bc (T, G)} = sup{𝜇+ C | C ∈ C}. Consider the positive wide Radon measures 𝜈1 ≡ (𝜇+ )# and 𝜈2 = (−𝜇− )#, the 𝛿-ring N ≡ domf 𝜈1 ∩ domf 𝜈2 , and the narrow Radon measure 𝜈 ≡ 𝜈1 |N − 𝜈2 |N. By the definition of H b , we have H b 𝜇 = 𝜃(𝜈, 𝜈1 , 𝜈2 ). For every B ∈ Bc (T, G), we have 𝜈00 B = 𝜈B = 𝜈1 B − 𝜈2 B = 𝜇+ B − (−𝜇− )B = 𝜇B. Therefore, by Proposition 4, p󸀠 (H b 𝜇) = ((𝜇+ )#)0 and p󸀠󸀠 (H b 𝜇) = ((−𝜇− )#)0 . As a result, x ≡ (p󸀠 H b 𝜇)T = (𝜇+ )#T and y ≡ (p󸀠󸀠 H b 𝜇)T = (−𝜇− )#T. By Proposition 2 (3.5.4), (𝜇+ )# is compactly regular. Therefore, x = sup{(𝜇+ )#C | C ∈ C} = sup{𝜇+ C | C ∈ C} = var(𝜇+ )T. In the same way, y = var(−𝜇− )T. As a conclusion we get ‖𝜇‖ = x + y = 𝜌(H b 𝜇). 2. Denote RBb (T, G) by Z. By Theorem 1 H b : Y Z is an isomorphism of the given lattice-ordered linear spaces. Therefore, 𝜌(z1 + z2 ) = ‖H b−1 z1 + H b−1 z2 ‖ ⩽ ‖H b−1 z1 ‖ + ‖H b−1 z2 ‖ = 𝜌z1 + 𝜌z2 for every z1 , z2 ∈ Z. If a ∈ R and z ∈ Z, then 𝜌(az) = ‖H b−1 (az)‖ = |a|‖H b−1 z‖ = |a|𝜌z. If 𝜌z = 0, then ‖H b−1 z‖ = 0 implies H b−1 z = 0; where z = 0. Thus, 𝜌 is a norm on Z. If |z1 | ⩽ |z2 | in Z, then by the assertion 1 |H b−1 z1 | ⩽ |H b−1 z2 | implies 𝜌z1 = ‖H b−1 z1 ‖ ⩽ ‖H b−1 z2 ‖ = 𝜌z2 . So the norm 𝜌 is monotone. Further, we shall denote 𝜌 by ‖ ⋅ ‖. Corollary 1. Let ⟮T, G⟯ be a Hausdorff space. Then, 1) ⟮RBb (T, G), ‖ ⋅ ‖⟯ is a normed lattice-ordered linear space and, moreover, a pre-L-space; 2) H b : ⟮RMn⋆ ⟮RBb (T, G), ‖ ⋅ ‖⟯ is an (isometric) isomorphism of b (T, G), ‖ ⋅ ‖⟯ the given normed lattice-ordered linear spaces.

352 | 3.5 Topological spaces with measures. The Radon integral

Proof. We shall use the notations from the proof of Proposition 5. According to Proposition 5, ⟮Z, ‖ ⋅ ‖⟯ is a normed lattice-ordered linear space and H b : ⟮Y , ‖ ⋅ ‖⟯ ⟮Z, ‖ ⋅ ‖⟯ is an (isometric) isomorphism of the given lattice-ordered linear spaces. So H b is a necessary isomorphism. By Proposition 4 (3.5.4), ⟮Y , ‖ ⋅ ‖⟯ is a pre-L-space. Using the mentioned isomorphism H b , we check that ⟮Z, ‖ ⋅ ‖⟯ is also a pre-L-space. In conclusion, we shall connect all the considered bijections and isomorphisms. Proposition 6. Let ⟮T, G⟯ be a Hausdorff space. Then, 1) H𝜇 = (E ∘ R)𝜇 for every 𝜇 ∈ RMn⋆ (T, G)+ ; 2) H b = E b ∘ S b . Proof. 1. For 𝜇, consider the measures 𝜘1 ≡ 𝜇 and 𝜘2 ≡ 0. Then, 𝜘1# = 𝜇# and 𝜘2# = 0. We can suppose that dom 𝜘2# = dom 𝜘1# = dom 𝜇#. Consider on the 𝛿-ring domf 𝜇#, the positive narrow Radon measure 𝜘 ≡ 𝜘1#| domf 𝜇# = (𝜇#)f . Then, by the definition of H, we have H𝜇 = 𝜃((𝜇#)f , 𝜇#, 0). On the other hand, R𝜇 = (𝜇#)0 and ER𝜇 = 𝜃(((𝜇#)0 )f , (𝜇#)0 , 0). Since (𝜇#)f B = 𝜇B = # ((𝜇 )0 )f B for every B ∈ Bc (T, G), we conclude that H𝜇 = ER𝜇. If 𝜇 is bounded and positive, then H b 𝜇 = E b R b 𝜇. 2. If 𝜇 ∈ RMn⋆ b (T, G), then by Theorem 1 and Theorem 1 (3.5.4), we get H b 𝜇 = H b (𝜇+ ) − H b (−𝜇− ) = E b R b (𝜇+ ) − E b R b (−𝜇− ) = E b S b 𝜇. Corollary 1. Let ⟮T, G⟯ be a Hausdorff space. Then, E b : ⟮RMw⋆ b (T, G), ‖ ⋅ ‖⟯ ⟮RBb (T, G), ‖ ⋅ ‖⟯ is an (isometric) isomorphism of the given normed lattice-ordered linear spaces. Proof. If 𝜇 is a wide bounded Borel – Radon measure, then by Corollary 1 of Proposi−1 −1 tion 5, ‖S−1 b 𝜇‖ = ‖H b S b 𝜇‖ = ‖E b 𝜇‖. Similarly, by Proposition 4 (3.5.4), ‖S b 𝜇‖ = ‖𝜇‖. As a result, ‖𝜇‖ = ‖E b 𝜇‖. 3.5.6 The Radon integral over a Hausdorff space with a Radon bimeasure Here we shall introduce an integral with respect to a Radon bimeasure. Let ⟮T, G⟯ be a Hausdorff topological space. For every Radon triplet 𝛽 ≡ (𝜇, 𝜇1 , 𝜇2 ), consider the corresponding 𝜎-algebra dom 𝛽 ≡ dom 𝜇1 ∩dom 𝜇2 and the corresponding 𝛿-ring domf 𝛽 ≡ domf 𝜇1 ∩ domf 𝜇2 . Consider also the wide Borel – Radon measures 𝜆 1 ≡ (𝜇1 )0 and 𝜆 2 ≡ (𝜇2 )0 and the complete strongly saturated Radon measures 𝜈1 ≡ 𝜇1# and 𝜈2 ≡ 𝜇2# from Lemma 4 (3.5.3). Consider the 𝛿-rings L ≡ domf 𝜆 1 ∩ domf 𝜆 2 and N ≡ domf 𝜈1 ∩ domf 𝜈2 and the corresponding finite measures 𝜆 ≡ 𝜆 1 |L − 𝜆 2 |L and 𝜈 ≡ 𝜈1 |N − 𝜈2 |N. Then, we get the Borel – Radon triplet 𝛽0 ≡ (𝜆, 𝜆 1 , 𝜆 2 ) and the extended Radon triplet 𝛽e ≡ (𝜈, 𝜈1 , 𝜈2 ).

3.5.6 The Radon integral over a Hausdorff space with a Radon bimeasure |

353

Lemma 1. Let ⟮T, G⟯ be a Hausdorff space and 𝛽 ∈ R(T, G). Then, 𝛽0 ∈ R⋆ (T, G), 𝛽e ∈ Re (T, G), and 𝛽 ∼ 𝛽0 ∼ 𝛽e . Proof. If B ∈ Bc (T, G), then 𝜇B = 𝜇1 B − 𝜇2 B = 𝜆 1 B − 𝜆 2 B = 𝜆B and 𝜇B = 𝜇1 B − 𝜇2 B = 𝜈1 B − 𝜈2 B = 𝜈B. This means that 𝛽 ∼ 𝛽0 and 𝛽 ∼ 𝛽e . Corollary 1. For every Radon bimeasure m, there are Radon triplets 𝛼 ∈ R⋆ (T, G) and 𝛾 ∈ Re (T, G) such that m = 𝛼̄ = 𝛾.̄ Lemma 2. Let 𝛽 ≡ (𝜇, 𝜇1 , 𝜇2 ) and 𝛾 ≡ (𝜈, 𝜈1 , 𝜈2 ) be Radon triplets and 𝛽 ∼ 𝛾. Then, for every 𝜎-algebra L ⊂ dom 𝛽 ∩ dom 𝛾 such that Bc (T, G) ⊂ L and every function f ∈ MI(T, L, 𝜇1 |L) ∩ MI(T, L, 𝜇2 |L) ∩ MI(T, L, 𝜈1 |L) ∩ MI(T, L, 𝜈2 |L), the equality ∫ f d𝜇1 − ∫ f d𝜇2 = ∫ f d𝜈1 − ∫ f d𝜈2 is valid. Proof. Denote the restrictions of the measures 𝜇1 , 𝜇2 , 𝜈1 , and 𝜈2 on the 𝜎-algebra L by 𝜇1󸀠 , 𝜇2󸀠 , 𝜈1󸀠 , and 𝜈2󸀠 , correspondingly. By the definition, 𝜇1󸀠 B − 𝜇2󸀠 B = 𝜇B = 𝜈B = 𝜈1󸀠 B − 𝜈2󸀠 B for every B ∈ Bc (T, G). According to Proposition 3 (3.5.3) we can consider the positive wide Radon measures 𝜋 ≡ 𝜇1󸀠 + 𝜈2󸀠 and 𝜘 ≡ 𝜇2󸀠 + 𝜈1󸀠 on L. Then, 𝜋C = 𝜘C for every compact set C. From the compact regularity of these measures we deduce that 𝜋 = 𝜘. Therefore, by Proposition 3 (3.3.5), X ≡ MI(T, L, 𝜇1󸀠 ) ∩ MI(T, L, 𝜈2󸀠 ) = MI(T, L, 𝜋) = MI(T, L, 𝜘) = MI(T, L, 𝜇2󸀠 )∩MI(T, L, 𝜈1󸀠 ) ≡ Y and ∫ f d𝜇1󸀠 +∫ f d𝜈2󸀠 = ∫ f d𝜋 = ∫ f d𝜘 = ∫ f d𝜇2󸀠 + ∫ f d𝜈1󸀠 for every function f from these family. If f ∈ X ∩ Y, then by Proposition 2 (3.3.5), f is integrable with respect to all the measures 𝜇1 , 𝜇2 , 𝜈1 and 𝜈2 and ∫ f d𝜇1󸀠 = ∫ f d𝜇1 , ∫ f d𝜇2󸀠 = ∫ f d𝜇2 , ∫ f d𝜈1󸀠 = ∫ f d𝜈1 and ∫ f d𝜈2󸀠 = ∫ f d𝜈2 . As a result, for f ∈ X ∩ Y we get ∫ f d𝜇1 − ∫ f d𝜇2 = ∫ f d𝜈1 − ∫ f d𝜈2 . We associate with every measure 𝜇 ∈ RMw⋆ (T, G) the family BI(T, G, 𝜇) ≡ MI(T, B(T, G), 𝜇) of all Borel (measurable) 𝜇-integrable functions (see 3.3.6). Corollary 1. Let 𝛽 ≡ (𝜇, 𝜇1 , 𝜇2 ) and 𝛾 ≡ (𝜈, 𝜈1 , 𝜈2 ) be Radon triplets and 𝛽 ∼ 𝛾. Then, ∫ f d𝜇1 − ∫ f d𝜇2 = ∫ f d𝜈1 − ∫ f d𝜈2 for every f ∈ BI(T, G, (𝜇1 )0 ) ∩ BI(T, G, (𝜇2 )0 ) ∩ BI(T, G, (𝜈1 )0 ) ∩ BI(T, G, (𝜈2 )0 ). Corollary 2. Let 𝛽 ≡ (𝜇, 𝜇1 , 𝜇2 ) and 𝛾 ≡ (𝜈, 𝜈1 , 𝜈2 ) be Borel – Radon triplets and 𝛽 ∼ 𝛾. Then, ∫ f d𝜇1 − ∫ f d𝜇2 = ∫ f d𝜈1 − ∫ f d𝜈2 for every f ∈ BI(T, G, 𝜇1 ) ∩ BI(T, G, 𝜇2 ) ∩ BI(T, G, 𝜈1 ) ∩ BI(T, G, 𝜈2 ). Now, let m be a Radon bimeasure. According to Corollary 1 to Lemma 1, we get m ∩ R⋆ (T, G) = / ⌀. Therefore, we can consider for the bimeasure m the family WI(T, G, m) ≡ ⋂{BI(T, G, 𝜇1 ) ∩ BI(T, G, 𝜇2 ) | (𝜇, 𝜇1 , 𝜇2 ) ∈ m ∩ R⋆ (T, G)} of all Borel measurable functions that is integrable with respect to the wide positive Borel – Radon measures 𝜇1 and 𝜇2 of every Borel – Radon triplet (𝜇, 𝜇1 , 𝜇2 ) from the bimeasure m.

354 | 3.5 Topological spaces with measures. The Radon integral

Consider for m also the wider family BI(T, G, m) ≡ BI(T, G, p󸀠 m)∩ BI(T, G, p󸀠󸀠 m). Every function from the family BI(T, G, m) will be called Borel (measurable) integrable with respect to the bimeasure m. And every function from the family WI(T, G, m) will be called Borel (measurable) well-integrable with respect to the bimeasure m. The number ∫ f dp󸀠 m − ∫ f dp󸀠󸀠 m will be called the (Radon) integral of the function f ∈ BI(T, G, m) over the topological bimeasurable space ⟮T, G, m⟯ and will be denoted by ∫ f dm. Lemma 3. Let ⟮T, G⟯ be a Hausdorff space, m ∈ RB(T, G) and (𝜈, 𝜈1 , 𝜈2 ) ∈ m ∩ R⋆ (T, G). Then, ∫ f dm = ∫ f d𝜈1 − ∫ f d𝜈2 for every f ∈ WI(T, G, m). Proof. Consider the 𝛿-ring M ≡ domf (p󸀠 m) ∩ domf (p󸀠󸀠 m) and the narrow Borel – Radon measure 𝜇 ≡ p󸀠 m|M − p󸀠󸀠 m|M. By Proposition 3 (3.5.5), (𝜇, p󸀠 m, p󸀠󸀠 m) ∈ m ∩ R⋆ (T, G). Consequently, by Corollary 2 to Lemma 2, ∫ f dm = ∫ f d𝜈1 − ∫ f d𝜈2 . Proposition 1. Let ⟮T, G⟯ be a Hausdorff space. Then, for every measure 𝜇 ∈ RMw⋆ (T, G) and every function f ∈ BI(T, G, 𝜇) = BI(T, G, E𝜇) the equality ∫ f d𝜇 = ∫ f dE𝜇 is fulfilled. Proof. By Corollary 1 to Proposition 4 (3.5.5), we get p 󸀠 (E𝜇) = v+ (𝜇) and p󸀠󸀠 (E𝜇) = −v− (𝜇). Therefore, BI(T, G, 𝜇) ≡ BI(T, G, v+ (𝜇)) ∩ BI(T, G, −v− (𝜇)) = BI(T, G, p󸀠 (E𝜇)) ∩ BI(T, G, p󸀠󸀠 (E𝜇)) ≡ BI(T, G, E𝜇) and ∫ f d𝜇 ≡ ∫ f dv+ (𝜇) − ∫ f d(−v− (𝜇)) = ∫ f dp󸀠 (E𝜇) − ∫ f dp󸀠󸀠 (E𝜇) ≡ ∫ f dE𝜇. Thus, we see that the bijection E0 : RMw⋆ (T, G)0 RB(T, G)+ , the embedding E : w⋆ RM (T, G) 󴀚󴀠 RB(T, G), and the isomorphism E b : RMw⋆ RBb (T, G) b (T, G) preserve the Lebesgue integral. This shows that the definitions of the integrals for wide measures (see 3.3.6) and for bimeasures are concordant with each other. For bimeasures, there is the exhausting analogue of Proposition 3 (3.3.5) and Theorem 5 (3.3.8). Proposition 2. Let ⟮T, G⟯ be a Hausdorff space, m, n ∈ RB(T, G), and x, y ∈ R. Then, ∫ f d(xm + yn) = x ∫ f dm + y ∫ f dn for every f ∈ WI(T, G, m) ∩ WI(T, G, n) ∩ WI(T, G, xm + yn). Proof. Take any Borel – Radon triplets 𝛽 ≡ (𝜇, 𝜇1 , 𝜇2 ) ∈ m and 𝛾 ≡ (𝜈, 𝜈1 , 𝜈2 ) ∈ n. Suppose that x ⩾ 0 and y < 0. By the definition, xm = 𝜃(x𝜇, x𝜇1 , x𝜇2 ) and yn = 𝜃(y𝜈, −y𝜈2 , −y𝜈1 ). Consider the 𝛿-ring L ≡ domf 𝛽∩domf 𝛾. Consider the wide positive Borel – Radon measures 𝜆 1 ≡ x𝜇1 + (−y𝜈2 ) and 𝜆 2 ≡ x𝜇2 + (−y𝜈1 ) and the narrow Radon measure 𝜆 ≡ x𝜇|L + y𝜈|L. Then, for Borel – Radon triplet 𝛼 ≡ (𝜆, 𝜆 1 , 𝜆 2 ) we have xm + yn = 𝛼.̄

3.6.1 𝜎-Exact linear functionals on spaces of symmetrizable functions

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Take any f ∈ WI(T, G, m) ∩ WI(T, G, n) ∩ WI(T, G, xm + yn). By Lemma 3, a ≡ ∫ f dm = ∫ f d𝜇1 − ∫ f d𝜇2 , b ≡ ∫ f dn = ∫ f d𝜈1 − ∫ f d𝜈2 and c ≡ ∫ f d(xm + yn) = ∫ f d𝜆 1 −∫ f d𝜆 2 . By Proposition 3 (3.3.5), c = x ∫ f d𝜇1 − y ∫ f d𝜈2 − x ∫ f d𝜇2 + y ∫ f d𝜈1 = xa + yb. The following assertion is very important for general representation theorems in 3.6.3 and 3.6.4. Remind that BM bc (T, G) denotes the family of all bounded Borel measurable functions with compact supports. Proposition 3. Let ⟮T, G⟯ be a Hausdorff space. Then, BM bc (T, G) ⊂ WI(T, G, m) for every m ∈ RB(T, G). Proof. If f ∈ BM bc (T, G), then the set S ≡ supp f is compact. Let (𝜇, 𝜇1 , 𝜇2 ) ∈ m ∩ R⋆ (T, G). Then, by definition of a Radon measure, 𝜇1 S < ∞ and 𝜇2 S < ∞. Since f is bounded, there is x ∈ R such that |f | ⩽ x1. Therefore, Lemmas 1 and 2 (3.3.2) imply ∫ f+ d𝜇1 ⩽ ∫ x𝜒(S) d𝜇1 = x𝜇1 S < ∞. This means that f+ ∈ BI(T, G, 𝜇1 ). The same is valid for −f− . By the definition, from 3.3.2, f ∈ BI(T, G, 𝜇1 ). Analogously, f ∈ BI(T, G, 𝜇2 ). As a result, we obtain that f ∈ WI(T, G, m).

3.6 Representation of a functional by the Radon integral 3.6.1 𝜎-Exact linear functionals on spaces of symmetrizable functions It was established in 3.3.6 (Proposition 2 and Theorem 1) that the Lebesgue integral Λ(𝜇) ≡ i𝜇 for an arbitrary (wide) measure 𝜇 is a linear and pointwise 𝜎-continuous functional on the lattice-ordered linear space MI(T, M, 𝜇) and every its latticeordered linear subspace A(T). It is remarkable that the Radon integral, i. e. the Lebesgue integral over the Hausdorff space ⟮T, G, M, 𝜇⟯ with the Radon measure 𝜇, considered on appropriate lattice-ordered linear spaces of symmetrizable integrable functions acquires some additional properties. These properties and spaces are considered in this subsection. They will be used in 3.6.3 and 3.6.4 to give a solution of the Riesz – Radon – Fréchet problem of characterization of Radon integrals as linear functionals from 3.5.3 (see also Appendix D).

Exact functionals on lattice-ordered linear spaces Let ⟮T, G⟯ be a Hausdorff space and A(T) be a lattice-ordered linear space of functions on T. A functional 𝜑 on A(T) is called tight or a functional with the Prokhorov property

356 | 3.6 Representation of a functional by the Radon integral

if for every 𝜀 > 0 there is a compact set C ⊂ T such that the conditions f ∈ A(T) and |f | ⩽ 𝜒(T\C) imply |𝜑f | < 𝜀. The set of all tight bounded linear functionals on A(T) will be denoted by A(T)𝜋 , its subset of uniformly (order) bounded functionals will be denoted by A(T) 𝜋 (see 2.2.8). A functional 𝜑 on A(T) will be called locally tight or a functional with the local Prokhorov property [Zakharov, 2002b] if for every G ∈ G, u ∈ A(T)+ , and 𝜀 > 0 there is a compact subset C ⊂ G such that the conditions f ∈ A(T) and |f | ⩽ 𝜒(G\C) ∧ u imply |𝜑f | < 𝜀. A functional 𝜑 will be called quite locally tight [Zakharov, 2005a] if for every G ∈ G, u ∈ A(T)+ , and 𝜀 > 0 there are a compact subset C ⊂ G and a positive number 𝛿 such that the conditions f ∈ A(T), |f | ⩽ 𝜒(G) ∧ u, and sup(|f (t)| | t ∈ C) ⩽ 𝛿 imply |𝜑f | < 𝜀. A functional 𝜑 on A(T) is said to be exact [𝜎-exact] if it is pointwise continuous [𝜎-continuous] and quite locally tight. The set of all exact [𝜎-exact] linear functionals on A(T) will be denoted by A(T)󳶋 [A(T)󳵻 ], its subset of all uniformly (order) bounded functionals will be denoted by A(T) 󳶋 [A(T) 󳵻 ]. It is clear that A(T)󳶋 ⊂ A(T)󳵻 . It will be proven at the end of this subsection that Radon integral is 𝜎-exact on the lattice-ordered linear space of integrable symmetrizable functions. According to the definitions from 2.2.7 and 2.2.8, Proposition 1 (2.2.8), and Statement 4 (2.2.8), A(T)󳶋 ⊂ A(T)∨ ⊂ A(T)∼ = A(T)† and A(T)󳵻 ⊂ A(T)∧ ⊂ A(T)∼ = A(T)† . The structures of an ordered linear space on A(T)∼ are introduced in 2.2.8. These structures are inherited by A(T)󳶋 and A(T)󳵻 . Proposition 1. Let ⟮T, G⟯ be a Hausdorff space, A(T) be a lattice-ordered linear space of functions on T, and 𝜑 be an (order) bounded linear functional on A(T). Then, the following conclusions are equivalent: 1) the functional 𝜑 is quite locally tight [tight, locally tight, exact, 𝜎-exact]; 2) the functionals 𝜑+ and −𝜑− are positive quite locally tight [tight, locally tight, exact, 𝜎-exact]; 3) 𝜑 = 𝜑1 − 𝜑2 for some positive quite locally tight [tight, locally tight, exact, 𝜎-exact] linear functionals 𝜑1 and 𝜑2 on A(T). Proof. Consider the case of quite locally tight functionals. In the cases of tight, locally tight, exact, and 𝜎-exact functionals, the arguments are quite similar (see also Proposition 1 (2.2.8)). (3) ⊢ (1). By assumption, 𝜑 = 𝜑1 − 𝜑2 for some positive quite locally tight linear functionals 𝜑1 and 𝜑2 . Then, for every open set G, every function u ∈ A(T)+ , and every 𝜀 > 0 there are compact sets C1 ⊂ G and C2 ⊂ G and numbers 𝛿1 > 0 and 𝛿2 > 0 such that the conditions f ∈ A(T), |f | ⩽ 𝜒(G) ∧ u, and sup(|f (t)| | t ∈ C i ) ⩽ 𝛿i imply the estimations |𝜑i f | < 𝜀. Consider C ≡ C1 ∪ C2 and 𝛿 ≡ 𝛿1 ∧ 𝛿2 . If |f | ⩽ 𝜒(G) ∧ u and sup(|f (t)| | t ∈ C) ⩽ 𝛿, then |𝜑f | ⩽ |𝜑1 f | + |𝜑2 f | < 𝜀. (1) ⊢ (2). Consider some G ∈ G, u ∈ A(T)+ , and 𝜀 > 0. By assumption there exist a compact subset C ⊂ G and 𝛿 > 0 such that the condition f ∈ A(T), |f | ⩽ 𝜒(G) ∧ u,

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and sup(|f (t)| | t ∈ C) ⩽ 𝛿 imply |𝜑f | < 𝜀. For any function g ∈ A(T)+ , it follows from 0 ⩽ g ⩽ |f | that g ⩽ 𝜒(G) ∧ u and sup(|g(t)| | t ∈ C) ⩽ 𝛿. Hence, 𝜑g < 𝜀. According to Statement 5 (2.2.8), A(T)∼ is a lattice-ordered linear space. Therefore, the functionals 𝜑+ ≡ 𝜑 ∨ 0 ∈ A(T)∼ and 𝜑− ≡ 𝜑 ∧ 0 ∈ A(T)∼ are defined. Corollary 3 to Statement 5 (2.2.8) implies that |𝜑+ f | ⩽ 𝜑+ |f | ≡ sup{𝜑g | g ∈ A(T)+ ∧ g ⩽ |f |} ⩽ 𝜀. Similarly, |𝜑− f | ⩽ 𝜀. Thus, 𝜑+ and −𝜑− are quite locally tight. (2) ⊢ (3). By Corollary 3 to Statement 5 (2.2.8) 𝜑 = 𝜑+ +𝜑− . Thus, we can take 𝜑1 ≡ 𝜑+ and 𝜑2 ≡ −𝜑− . Proposition 2. Let ⟮T, G⟯ be a Hausdorff space and A(T) be a lattice-ordered linear space of functions on T. Then, A(T)󳶋 and A(T)󳵻 are l-ideals in the Dedekind complete lattice-ordered linear space A(T)∼ . Proof. Denote the set (A(T)󳶋 )0 of all positive exact linear functionals on A(T) by P. Let 𝜑, 𝜓 ∈ P and x, y ∈ R+ . According to Theorem 1 (2.2.8), x𝜑 + b𝜓 ∈ (A(T)∨ )0 . Consider some G ∈ G, u ∈ A(T)+ , and 𝜀 > 0. By definition, there exist compact subsets C, D ⊂ G and numbers 𝛿, 𝛾 > 0 such that the conditions f ∈ A(T), |f | ⩽ 𝜒(G) ∧ u, and sup(|f (t)| | t ∈ C) ⩽ 𝛿 and sup(|f (t)| | t ∈ D) ⩽ 𝛾 imply |𝜑f | < 𝜀/(2x) and |𝜓f | < 𝜀/(2y), respectively. Therefore, the conditions f ∈ A(T), |f | ⩽ 𝜒(G) ∧ u, and sup(|f (t)| | t ∈ C ∪ D) ⩽ 𝛿 ⊼ 𝛾 imply |(x𝜑 + y𝜓)f | < 𝜀/2 + 𝜀/2 = 𝜀. Thus, x𝜑 + b𝜓 ∈ P. Consequently, A(T)󳶋 = P − P is a linear subspace of the linear space A(T)∼ . Let 𝜑 ∈ A(T)∼ , 𝜓 ∈ P, and 0 ⩽ 𝜑 ⩽ 𝜓. Theorem 1 (2.2.8) implies that 𝜑 ∈ (A(T)∨ )0 . Then, it follows directly from the definition of a quite locally tight functional that 𝜑 ∈ P. Now, let 𝜑 ∈ A(T)∼ , 𝜓 ∈ A(T)󳶋 , and |𝜑| ⩽ |𝜓|. By Proposition 1 𝜓 = 𝜓1 − 𝜓2 for some 𝜓1 , 𝜓2 ∈ P. Therefore, by the property proven above the conditions 0 ⩽ 𝜑+ ⩽ |𝜑| ⩽ 𝜓1 + 𝜓2 ∈ P imply 𝜑+ ∈ P. Analogously, −𝜑− ∈ P. Hence, 𝜑 = 𝜑+ − (−𝜑− ) ∈ A(T)󳶋 . Then, Statement 2 (2.2.8) implies that A(T)󳶋 is an l-ideal in A(T)∼ . For A(T)󳵻 , all the arguments are quite similar. Corollary 1. Let ⟮T, G⟯ be a Hausdorff space and A(T) be a lattice-ordered linear space of functions on T. Then, the sets A(T)󳶋 and A(T)󳵻 are Dedekind complete lattice-ordered linear spaces, the smallest upper [the greatest lower] bounds of bounded above [below] sets in the ordered sets A(T)∼ and A(T)󳶋 (and in the ordered sets A(T)∼ and A(T)󳵻 , respectively) coincide, and they are expressed by the formulas from Statement 5 (2.2.8) and its Corollaries. Proof. The assertion follows from Proposition 2 and Statements 3 and 1 (2.2.8). Lemma 1. Let ⟮T, G⟯ be a Hausdorff space and A(T) be a lattice-ordered linear space of functions on T. Then, A(T) 󳶋 and A(T) 󳵻 are l-ideals in the Dedekind complete lattice-ordered linear spaces A(T)󳶋 and A(T)󳵻 , respectively.

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Proof. All the arguments are similar to the arguments in the proof of Corollary 1 to Lemma 4 (2.2.8). Corollary 1. Let ⟮T, G⟯ be a Hausdorff space and A(T) be a lattice-ordered linear space of functions on T. Then, the sets A(T)󳶋 and A(T)󳵻 are Dedekind complete lattice-ordered linear spaces, the smallest upper [the greatest lower] bounds of bounded above [below] sets in the ordered sets A(T)󳶋 and A(T) 󳶋 (and A(T)󳵻 and A(T) 󳵻 , respectively) coincide and they are expressed by the formulas from Statement 5 (2.2.8) and its Corollaries. Proof. The assertion follows from Lemma 1 and Statements 3 and 1 (2.2.8). A family A(T) ⊂ F(T) is said to have the Dini property or to have property (D) if for every decreasing net (f m ∈ A(T) | m ∈ M) the condition p-lim (f m | m ∈ M) = 0 implies u-lim (f m |C | m ∈ M) = 0|C for every compact subset C ⊂ T. It follows from the Dini theorem (Theorem 1 (2.3.4)) that if A(T) is contained in the lattice-ordered linear space C(T, G) of all continuous functions on a Hausdorff space ⟮T, G⟯, then A(T) has property (D). Lemma 2. Let ⟮T, G⟯ be a Hausdorff space, A(T) be a lattice-ordered linear subspace of F b (T) with property (D), and 𝜑 be a quite locally tight bounded linear functional on A(T). Then 𝜑 is pointwise continuous. Proof. Suppose 𝜑 is positive. Let (f m ∈ A(T)+ | m ∈ M) ↓ 0 in F(T). Take any 𝜀 > 0 and m0 ∈ M. Then, for u ≡ f m0 /‖f m0 ‖u , G ≡ T, and 𝛽 ≡ 𝜀/‖f m0 ‖u there are a compact set C ⊂ G and 𝛿 > 0 such that f ∈ A(T), |f | ⩽ 𝜒(G) ∧ u, and |f (t)| ⩽ 𝛿 for every t ∈ C imply 𝜑f < 𝛽. Take 𝛾 ≡ 𝛿‖f m0 ‖u . By the Dini property there is n0 ∈ M such that f m (t) < 𝛾 for every t ∈ C and every m ⩾ n0 . Take some l ∈ M such that l ⩾ m0 and l ⩾ n0 . If m ⩾ l, then f m ⩽ f m0 implies g m ≡ f m /‖f m0 ‖u ⩽ u ⩽ 𝜒(G). Therefore, g m ∈ A(T), g m ⩽ 𝜒(G)∧ u, and g m (t) ⩽ f m (t)/‖f m0 ‖u < 𝛾/‖f m0 ‖u = 𝛿 for every t ∈ C imply 𝜑g m < 𝛽. Consequently, 𝜑f m < 𝛽‖f m0 ‖u = 𝜀. Using Lemma 1 (2.2.8), we obtain that 𝜑 is pointwise continuous. If 𝜑 is quite locally tight bounded functional, then by Proposition 1 𝜑 = 𝜑1 − 𝜑2 , where 𝜑1 and 𝜑2 are positive quite locally tight bounded functionals. As was shown above, they are pointwise continuous. Therefore, 𝜑 has this property. Corollary 1. Let A(T) be a lattice-ordered linear subspace in F b (T) with property (D). Then, A(T)󳶋 = A(T)󳵻 . Characterizations of 𝜎-exact functionals on spaces of symmetrizable functions We are going to obtain some useful characterizations of locally tight and quite locally tight functionals on some lattice-ordered linear subspaces of the space S(T, G) of

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symmetrizable functions (see 2.4.5). Here the notions of majorized functions and majorized sets introduced in 2.2.4 will be used at last. Lemma 3. Let ⟮T, G⟯ be a Hausdorff space, A(T) be a lattice-ordered linear subspace in S(T, G) such that A(T) = Sm (T, G, A(T)) and C = Cm (A(T)), 𝜑 be a bounded linear functional on A(T). Then, the following conclusions are equivalent: 1) 𝜑 is locally tight; 2) 𝜑 is quite locally tight. Proof. (1) ⊢ (2). Take some G ∈ G, u ∈ A(T)+ , and 𝜀 > 0. By the condition, there is a compact set C ⊂ G such that the conditions f ∈ A(T) and |f | ⩽ 𝜒(G\C) ∧ u imply the estimation |𝜑f | < 𝜀/2. Put v ≡ 𝜒(G\C) and w ≡ 𝜒(C). Since C ∈ C ⊂ F ⊂ K, we get w ∈ S(T, G). Combining this with the equality C = Cm (A(T)), we have w ∈ Sm (T, G, A(T)) = A(T). Therefore, the number 𝛿 ≡ 𝜀/(2(𝜑(w) + 1)) is defined. Let f ∈ A(T), |f | ⩽ 𝜒(G) ∧ u, and |f (t)| ⩽ 𝛿 for every t ∈ C. Consider the functions h ≡ |f | ∧ w ∈ A(T) and g ≡ |f | ∧ v. Since |f | = |f | ∧ 𝜒(G) = |f | ∧ (v + w) = g + h, we have g ∈ A(T). Then, the inequalities g ⩽ v ∧ u and h ⩽ 𝛿w imply |𝜑f | ⩽ 𝜑|f | = 𝜑g + 𝜑h < 𝜀/2 + 𝛿𝜑w < 𝜀/2 + 𝜀/2 = 𝜀. Thus, 𝜑 is quite locally tight. (2) ⊢ (1). It is clear from the definitions. Lemma 4. Let ⟮T, G⟯ be a Hausdorff space, A(T) be a lattice-ordered linear subspace in S(T, G) such that A(T) = Sm (T, G, A(T)), let 𝜑 be a positive pointwise 𝜎-continuous linear functional on A(T). Then, the following conclusions are equivalent: 1) 𝜑 is locally tight; 2) for every K ∈ Km (A(T)) and every 𝜀 > 0 there is C ∈ C such that C ⊂ K and 𝜑𝜒(K\C) < 𝜀; 3) for every A ∈ Am (A(T)) and every 𝜀 > 0 there is C ∈ C such that C ⊂ A and 𝜑𝜒(A\C) < 𝜀. Proof. (1) ⊢ (2). Take an arbitrary K ≡ F ∩ G ∈ Km ≡ Km (A(T)) and 𝜀 > 0. Then, 𝜒(K) ⩽ u ∈ A(T). By definition, there is a compact set B ⊂ G such that f ∈ A(T) and |f | ⩽ 𝜒(G\B) ∧ u imply |𝜑f | < 𝜀. Consider the compact set C ≡ F ∩ B ∈ Km . Then, 𝜒(K\C) = |𝜒(F) ∧ 𝜒(G) − 𝜒(F) ∧ 𝜒(B)| = |𝜒(F) ∧ 𝜒(G) − 𝜒(F) ∧ 𝜒(B)| ∧ u ⩽ |𝜒(G) − 𝜒(B)| ∧ u = 𝜒(G\B) ∧ u. By the condition, it follows from 𝜒(G\B) ∈ S(T, G) and u ∈ A(T) ⊂ S(T, G) that f ≡ 𝜒(G\B) ∧ u ∈ S m ≡ Sm (T, G, A(T)) ⊂ A(T). Besides, 𝜒(K\C) ∈ Sm ⊂ A(T), therefore 𝜑𝜒(K\C) ⩽ 𝜑f < 𝜀. (2) ⊢ (3). Take a set A ∈ Am ≡ Am (A(T)) and 𝜀 > 0. Then, A = ⋃ ⟮K i | i ∈ I⟯ for some finite collection ⟮K i ∈ Km | i ∈ I⟯. The condition implies that for every K i there is a compact set C i ⊂ K i such that 𝜑𝜒(K i \C i ) < 𝜀/n, where n ≡ card I. Consider C ≡ ⋃ ⟮C i | i ∈ I⟯. Then, it follows from A\C ⊂ ⋃ ⟮K i \C i | i ∈ I⟯ that 𝜑𝜒(A\C) ⩽ ∑ (𝜑𝜒(K i \C i ) | i ∈ I) < 𝜀.

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(3) ⊢ (1). Take some G ∈ G, u ∈ A(T)+ , and 𝜀 > 0. By Lemma 2 (2.4.5) there exists a sequence (u n ∈ St(T, A)+ | n ∈ N) ↑ such that u = u-lim (u n | n ∈ N). By the definition, u n = ∑ (x i 𝜒(A i ) | i ∈ I) for some finite collections (x i ∈ R | i ∈ I) and ⟮A i ∈ A | i ∈ I⟯. Since A is an algebra, we may say that x i > 0 for all i ∈ I and the sets A i are mutually disjoint. Taking into account the inequality u n ⩽ u we conclude that u n ∈ Sm = A(T). Besides, we see that 𝜒(A i ) ⩽ (1/x i )u ∈ A(T) and therefore A i ∈ Am . Since 𝜑u = lim (𝜑u n | n ∈ N) ↑, there is a number n ∈ N such that 𝜑u −𝜑u n < 𝜀/2. By the condition, for every A i there exists a compact subset C i ⊂ A i such that 𝜑𝜒(A i \C i ) < 𝜀/(2n), where n ≡ card I. Consider C ≡ ⋃ ⟮C i | i ∈ I⟯ and A ≡ coz u n = ⋃ ⟮A i | i ∈ I⟯ ∈ Am . Then, it follows from 𝜒(A\C) ⩽ 𝜒(⋃ ⟮A i \C i | i ∈ I⟯) ⩽ ∑ (𝜒(A i \C i ) | i ∈ I) that 𝜑𝜒(A\C) ⩽ ∑ (𝜑𝜒(A i \C i ) | i ∈ I) < 𝜀/2. From G\C ∈ G ⊂ K we get 𝜒(G\C) ∈ S(T, G) and 𝜒(G\C) ∧ v ∈ Sm ⊂ A(T) for every v ∈ A(T)+ ; this imply 𝜑(𝜒(G\C) ∧ u) = 𝜑(𝜒(G\C) ∧ (u − u n + u n )) ⩽ 𝜑(𝜒(G\C) ∧ (u − u n ))+𝜑(𝜒(G\C)∧ u) ⩽ 𝜑(u − u n )+𝜑(𝜒(A\C)∧ u n ) ⩽ 𝜀/2+𝜑𝜒(A\C) < 𝜀/2+𝜀/2 = 𝜀. Proposition 3. Let ⟮T, G⟯ be a Hausdorff space, A(T) be a lattice-ordered linear subspace in S(T, G) such that A(T) = Sm (T, G, A(T)), and 𝜑 be a pointwise 𝜎-continuous linear functional on A(T). Then, the following conclusions are equivalent: 1) 𝜑 is locally tight; 2) for every set K ∈ Km (A(T)) and every 𝜀 > 0 there is a compact set C ⊂ K such that the conditions f ∈ A(T) and |f | ⩽ 𝜒(K\C) imply |𝜑f | < 𝜀; 3) for every set A ∈ Am (A(T)) and every 𝜀 > 0 there is a compact set C ⊂ A such that the conditions f ∈ A(T) and |f | ⩽ 𝜒(A\C) imply |𝜑f | < 𝜀. Proof. (1) ⊢ (2). Consider positive locally tight 𝜎-continuous linear functionals 𝜑1 and 𝜑2 such that 𝜑 = 𝜑1 −𝜑2 (see Proposition 1). Take a set K ∈ Km and a number 𝜀 > 0. By Lemma 4 there are compact sets C1 and C2 such that C i ⊂ K and 𝜑i 𝜒(K\C i ) < 𝜀/2. Consider C = C1 ∪ C2 , then 𝜑i 𝜒(K\C) < 𝜀/2. Therefore, the conditions f ∈ A(T) and |f | ⩽ 𝜒(K\C) imply |𝜑f | ⩽ |𝜑1 f | + |𝜑2 f | ⩽ 𝜑1 |f | + 𝜑2 |f | ⩽ 𝜑1 𝜒(K\C) + 𝜑2 𝜒(K\C) < 𝜀. (2) ⊢ (3). For any set A ∈ Am , there is a finite collection ⟮K i ∈ Km | i ∈ I⟯ such that A = ⋃ ⟮K i | i ∈ I⟯. Denote card I by n. Consider positive locally tight 𝜎-continuous linear functionals 𝜑1 and 𝜑2 such that 𝜑 = 𝜑1 − 𝜑2 (see Proposition 1). By the condition, for every 𝜀 > 0 there are compact sets C1i and C2i such that C ki ⊂ K i and it follows from f ∈ A(T) and f ⩽ 𝜒(K i \C ki ) that 𝜑k f < 𝜀/2n. Since C ki ∈ C ⊂ F ⊂ K, we have K i \C ki ∈ Km ⊂ Am . Using Lemma 11 (2.2.4), we make sure of 𝜒(K i \C ki ) ∈ St(T, Am (A(T))) = Stm (T, A, A(T)) ⊂ Sm ≡ Sm (T, G, A(T)) = A(T), what implies 𝜑k 𝜒(K i \C ki ) < 𝜀/2n. Take C ≡ ⋃ ⟮C1i ∪ C2i | i ∈ I⟯. From A\C ⊂ ⋃ ⟮K i \C ki | i ∈ I⟯ we get 𝜑k 𝜒(A\C) ⩽ 𝜑k (∑ (𝜒(K i \C ki ) | i ∈ I)) = ∑ (𝜑k 𝜒(K i \C ki ) | i ∈ I) < 𝜀/2. Then, the conditions f ∈ A(T) and |f | ⩽ 𝜒(A\C) imply |𝜑f | ⩽ |𝜑1 f | + |𝜑2 f | ⩽ 𝜑1 |f | + 𝜑2 |f | ⩽ 𝜑1 𝜒(A\C) + 𝜑2 𝜒(A\C) < 𝜀. (3) ⊢ (1). By the condition, there is a compact set C ⊂ A such that |𝜑f | < 𝜀 as soon as f ∈ A(T) and |f | ⩽ 𝜒(A\C). From A\C ∈ Am by Lemma 11 (2.2.4) we conclude

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that 𝜒(A\C) ∈ St(T, Am ) = Stm (T, A, A(T)) ⊂ Sm = A(T). According to Corollary 3 to Statement 5 (2.2.8) 𝜑 = 𝜑+ + 𝜑− , where the functionals 𝜑+ and −𝜑− are positive, 𝜑+ 𝜒(A\C) = sup{𝜑f | f ∈ A(T) ∧ 0 ⩽ f ⩽ 𝜒(A\C)} < 𝜀 and 𝜑− 𝜒(A\C) = inf{𝜑f | f ∈ A(T) ∧ 0 ⩽ f ⩽ 𝜒(A\C)} ⩾ −𝜀. This imply (−𝜑− )𝜒(A\C) ⩽ 𝜀. Then, by Lemma 4 we see that 𝜑+ and 𝜑− are locally tight; therefore, the functional 𝜑 is also locally tight. Proposition 4. Let (T, G) be a Hausdorff space, A(T) be a lattice-ordered linear subspace in S(T, G) such that A(T) = Sm (T, G, A(T)), and 𝜑 be a pointwise 𝜎-continuous linear functional on A(T). Then, the following conclusions are equivalent: 1) 𝜑 is locally tight; 2) for every 𝜀 > 0 and every decreasing sequence (A n ∈ Am (A(T)) | n ∈ 𝜔) there are a number n0 ∈ 𝜔 and a compact set C ⊂ ⋂ ⟮A n | n ∈ 𝜔⟯ such that the conditions f ∈ A(T) and |f | ⩽ 𝜒(A n0 \C) imply |𝜑f | < 𝜀. Proof. (1) ⊢ (2). Consider positive locally tight 𝜎-continuous linear functionals 𝜑1 and 𝜑2 such that 𝜑 = 𝜑1 − 𝜑2 (see Proposition 1). By Lemma 11 (2.2.4) Stm (T, A, A(T)) = St(T, Am (A(T))) ⊂ Sm (T, G, A(T)) = A(T). By virtue of Proposition 3 for any 𝜀 > 0 and n ∈ 𝜔, there are compact sets C1n and C2n such that 𝜑i 𝜒(A n \C in ) < 𝜀/2n+3 . Consider the compact sets C n ≡ C1n ∪ C2n , D n ≡ ⋂ ⟮C k | k ∈ n + 1⟯, and C ≡ ⋂ ⟮C k | k ∈ 𝜔⟯. Since (𝜒(D n ) | n ∈ 𝜔) ↓ 𝜒(C) and the functionals 𝜑i are pointwise 𝜎-continuous, we obtain (𝜑i 𝜒(D n ) | n ∈ 𝜔) ↓ 𝜑i 𝜒(C). Therefore, there are numbers m1 and m2 such that 𝜑1 𝜒(C) + 𝜀/4 > 𝜑1 𝜒(D m1 ) and 𝜑2 𝜒(C) + 𝜀/4 > 𝜑2 𝜒(D m2 ). For m ≡ m1 ∨ m2 , we get 𝜑i 𝜒(C) + 𝜀/4 > 𝜑i 𝜒(D m ). Since we have A m \C = ((A m \D m ) ∪ D m )\C = (A m \D m ) ∪ (D m \C) = (⋃ ⟮A m \C k | k ∈ m + 1⟯) ∪ (D m \C) ⊂ (⋃ ⟮A k \C k | k ∈ m + 1⟯) ∪ (D m \C), we conclude that 𝜒(A m \C) ⩽ ∑ (A k \C k | k ∈ m + 1) + 𝜒(D m ) − 𝜒(C), where 𝜑i 𝜒(A m \C) ⩽ ∑ (𝜒(A k \C k ) | k ∈ m + 1) + 𝜀/4 < ∑ (𝜀/2k+3 | k ∈ m + 1) + 𝜀/4 < 𝜀/4 + 𝜀/4 = 𝜀/2. Therefore, the conditions f ∈ Sm (T, G, A(T)) and |f | ⩽ 𝜒(A m \C) imply the inequality |𝜑f | ⩽ |𝜑1 f | + |𝜑2 f | ⩽ 𝜑1 |f | + 𝜑2 |f | ⩽ 𝜑1 𝜒(A m \C) + 𝜑2 𝜒(A m \C) < 𝜀. (2) ⊢ (1). By Proposition 3 the functional 𝜑 is locally tight. 𝜎-Exactness of the Radon integral on the space of symmetrizable integrable functions Let ⟮T, G⟯ be a Hausdorff space, M be a 𝜎-algebra on T containing G, and 𝜇 be a Radon measure on M. Consider the lattice-ordered linear space SI(T, M, 𝜇) ≡ S(T, G) ∩ MI(T, M, 𝜇) of all symmetrizable integrable functions. Lemma 1 (3.5.2) guarantees that S(T, G) ⊂ M(T, G), S c (T, G) ⊂ SI(T, M, 𝜇), and, for a bounded measure 𝜇, S(T, G) = SI(T, M, 𝜇). Proposition 5. Let ⟮T, G⟯ be a Hausdorff space with a wide Radon measure 𝜇 : M → R. Then, the Radon integral Λ(𝜇) is a 𝜎-exact functional on SI(T, M, 𝜇).

362 | 3.6 Representation of a functional by the Radon integral

Proof. Denote SI(T, M, 𝜇) by A(T). First, consider the case of positive measure 𝜇. It is clear that A(T) = Sm (T, G, A(T)). By virtue of Theorem 1 (3.3.6), the functional Λ(𝜇) is pointwise 𝜎-continuous. Take any K ∈ Km (A(T)) and 𝜀 > 0. Then, v ≡ 𝜒(K) ⩽ u ∈ A(T) implies v ∈ A(T). Therefore, 𝜇K = ∫ v d𝜇 < ∞. Consequently, there is a compact set C ⊂ K such that 𝜇K−𝜀 < 𝜇C. Since w ≡ 𝜒(K\C) ∈ A(T), we conclude that Λ(𝜇)w = ∫ w d𝜇 = 𝜇(K\C) < 𝜀, where, by Lemma 4, we have that Λ(𝜇) is locally tight. Take any G ∈ G, u ∈ A(T)+ , and 𝜀 > 0. Then, by the proof above, there is C ⊂ G such that the conditions f ∈ A(T) and |f | ⩽ 𝜒(G\C) ∧ u imply Λ(𝜇)f < 𝜀/2. Consider 𝜀 𝛿 ≡ 2𝜇C . Let f ∈ A(T) and |f | ⩽ 𝜒(G) ∧ u, and |f (t)| ⩽ 𝛿 for every t ∈ C. Then, |f | ∧ 𝜒(C) ⩽ 𝛿𝜒(C). Therefore, |Λ(𝜇)f | = | ∫ f d𝜇| ⩽ ∫ |f | d𝜇 = ∫ |f | ∧ 𝜒(G) d𝜇 = ∫ |f | ∧ (𝜒(G\C) + 𝜒(C)) d𝜇 ⩽ ∫ |f |∧𝜒(G\C) d𝜇+∫ |f |∧𝜒(C) d𝜇 ⩽ ∫ 𝜒(G\C)∧u d𝜇+∫ 𝛿𝜒(C) d𝜇 < 𝜀/2+𝛿𝜇C = 𝜀/2 + 𝜀/2 = 𝜀. So, Λ(𝜇) is quite locally tight. Thus, Λ(𝜇) is 𝜎-exact. Now, suppose that 𝜇 is an arbitrary Radon measure. Then, we have the Jordan decomposition 𝜇 = 𝜇1 − 𝜇2 , where 𝜇1 and 𝜇2 are positive (Lemma 1 (3.5.3)). As proven above, the functionals Λ(𝜇1 ) and Λ(𝜇2 ) are 𝜎-exact. Therefore, the functional Λ(𝜇) = Λ(𝜇1 ) − Λ(𝜇2 ) is also 𝜎-exact. Thus, the Radon integral Λ(𝜇) is 𝜎-exact on lattice-ordered linear subspaces of SI(T, M, 𝜇). By this reason the property of 𝜎-exactness is necessary for the integral representation of a functional. Further, (in 3.6.3 and 3.6.4), we show that this property is also sufficient.

3.6.2 Extensions of 𝜎-exact functionals on spaces of symmetrizable functions by the Young – Daniell method The Young extensions of the initial functional domain Fix a truncatable lattice-ordered linear space A(T) of functions on a set T. Consider the family S𝜏 (T, A(T)) of all functions g ∈ F(T) such that g = sup(f m | m ∈ M) in F(T) for some increasing net (f m ∈ A(T) | m ∈ M) and g ⩽ f for some function f ∈ A(T). Such functions will be called supremal for A(T). In a similar way, consider the family I𝜏 (T, A(T)) of all functions h ∈ F(T) such that h = inf(f m | m ∈ M) in F(T) for some decreasing net (f m ∈ A(T) | m ∈ M) and h ⩾ f for some function f ∈ A(T). Such functions will be called infimal for A(T). Note that, according to Lemma 7 (1.4.7), for increasing nets (f m | m ∈ M) ↑ the property g = sup (f m | m ∈ M) is equivalent to the property g = p-lim (f m | m ∈ M). The same is valid for decreasing nets (f m | m ∈ M) ↓ and the property h = inf (f m | m ∈ M). If only sequences (f m ∈ A(T) | m ∈ M ⊂ 𝜔) are used instead of arbitrary nets, then consider the family S𝜎 (T, A(T)) of all 𝜎-supremal functions for A(T) and the family I𝜎 (T, A(T)) of all 𝜎-infimal functions for A(T). It is easy to see that

3.6.2 Extensions of 𝜎-exact functionals by the Young – Daniell method | 363

A(T) ⊂ S𝜎 (T, A(T)) ⊂ S𝜏 (T, A(T)) ⊂ Fm (T, A(T)) (see 2.2.4) and A(T) ⊂ I𝜎 (T, A(T)) ⊂ I𝜏 (T, A(T)) ⊂ Fm (T, A(T)). Lemma 1. Let g, h ∈ F(T). Then, a) the following conclusions are equivalent: 1) g ∈ S𝜏 (T, A(T)) [g ∈ S𝜎 (T, A(T))]; 2) g = sup (f i | i ∈ I) for some [some countable] collection (f i ∈ A(T) | i ∈ I) and g ⩽ f for some function f ∈ A(T); b) the following conclusions are equivalent: 1) h ∈ I𝜏 (T, A(T)) [g ∈ I𝜎 (T, A(T))]; 2) h = inf (f i | i ∈ I) for some [some countable] collection (f i ∈ A(T) | i ∈ I) and h ⩾ f for some function f ∈ A(T). Proof. We prove only assertion a). The proof of assertion b) is quite similar. (1) ⊢ (2). This deduction is evident. (2) ⊢ (1). First, consider the 𝜏-case. Consider the upward directed set Pf (I) of all non-empty subsets of I. Define the increasing net (𝜑p ∈ A(T) | p ∈ Pf (I)) setting 𝜑p ≡ sup (f i | i ∈ p). Fix a point t ∈ T. By Lemma 1 (1.4.5) for every k ∈ N there is i ∈ I such that g(t) ⩾ f i (t) > g(t) − 1/k. Then, g(t) ⩾ 𝜑p (t) = f i (t) > g(t) − 1/k for p ≡ {i}. By the same lemma g(t) = sup (𝜑p | p ∈ Pf (I)). Now, consider the 𝜎-case. We can reckon that I = 𝜔. Consider the increasing sequence (𝜑n ∈ A(T) | n ∈ 𝜔) setting 𝜑n ≡ sup (f m | m ∈ n + 1). As above g(t) ⩾ f n (t) > g(t) − 1/k for some n ∈ 𝜔. Then, g(t) ⩾ 𝜑n (t) ⩾ f n (t) > g(t) − 1/k implies g(t) = sup (𝜑n | n ∈ 𝜔). Lemma 2. Let A(T) be a truncatable lattice-ordered linear space of functions on T. Then, the families Y ≡ S𝜏 (T, A(T)) [Y ≡ S𝜎 (T, A(T))] and Z ≡ I𝜏 (T, A(T)) [Z ≡ I𝜎 (T, A(T))] have the following properties with respect to the operations in F(T): 1) −g ∈ Z and −h ∈ Y for all g ∈ Y and h ∈ Z; 2) ∑(a i g i | i ∈ I) ∈ Y and ∑(a i h i | i ∈ I) ∈ Z for any finite collections (a i ∈ R+ | i ∈ I), (g i ∈ Y | i ∈ I), and (h i ∈ Z | i ∈ I), i. e. Y and Z are conic spaces; 3) inf(g i | i ∈ I) ∈ Y and sup(h i | i ∈ I) ∈ Z for any finite collections (g i ∈ Y | i ∈ I) and (h i ∈ Z | i ∈ I); 4) sup(g i | i ∈ I) ∈ Y and inf(h i | i ∈ I) ∈ Z for any [any countable] collections (g i ∈ Y | i ∈ I) and (h i ∈ Z | i ∈ I) such that g i ⩽ u for some u ∈ A(T) and h i ⩾ v for some v ∈ A(T); 5) g ∧ 1 ∈ Y, g ∨ (−1) ∈ Y, h ∧ 1 ∈ Z, and h ∨ (−1) ∈ Z for all g ∈ Y and h ∈ Z; 6) for every g ∈ Y there are f ∈ A(T) and g 󸀠 ∈ Y0 such that g = f + g󸀠 ; 7) for every h ∈ Z there are f ∈ A(T) and h󸀠 ∈ Y0 such that h = f − h󸀠 . Proof. 1. This assertion follows from Theorem 1 (2.2.2). By virtue of this assertion we need to check the first parts of assertions 2 – 4 only.

364 | 3.6 Representation of a functional by the Radon integral

̃ ∈ A(T) | m ∈ M ) and g ⩽ f ∈ A(T). 2. First, consider the 𝜏-case. Let g i = sup (f im i i i i i Consider the upward directed set N ≡ ∏⟮M i | i ∈ I⟯. Define the functions f in ∈ A(T) ̃ for every n ∈ N such that n(i) = m . setting f in ≡ f im i i It follows from properties 1 and 2 of Proposition 1 (2.2.3) that g ≡ ∑ (a i g i | i ∈ I) = ∑ (a i p-lim (f in | n ∈ N) | i ∈ I) = = p-lim (∑ (a i f in | i ∈ I) | n ∈ N) = sup (∑ (a i f in | i ∈ I) | n ∈ N). Since 𝜑n ≡ ∑ (a i f in | i ∈ I) ∈ A(T), the net (𝜑n | n ∈ N) increases, and g ⩽ ∑(a i f i | i ∈ I) ∈ A(T), we infer that g ∈ Y. In the 𝜎-case, we suppose that M i ⊂ 𝜔. But by virtue of Theorem 1 (1.3.7) we can ̃ . reckon that M i = 𝜔 for every i ∈ I. Therefore, we can take simply N ≡ 𝜔 and f in ≡ f in 3. First, consider the 𝜏-case. We start as above. Consider the function g ≡ inf (g i | i ∈ I) in F(T). It follows from Corollary 1 to Theorem 3 (1.4.5) that g(t) = ̃ (t) | m ∈ M ) | i ∈ I) = sup (inf (f (t) | i ∈ I) | n ∈ N). Since 𝜑 ≡ inf inf (sup (f im i i in n i (f in (t) | i ∈ I) ∈ A(T), (𝜑n | n ∈ N) increases, and g ⩽ inf (f i | i ∈ I) ∈ A(T), we infer that g ∈ Y. Now, consider the 𝜎-case. As above we can reckon that M i = 𝜔 for i ∈ I, and so we ̃ . can take N ≡ 𝜔 and f in ≡ f in Applying property 6 of Proposition 1 (2.2.3), we get g ≡ inf (g i | i ∈ I) = inf (p-lim (f in | n ∈ 𝜔) | i ∈ I) = = p-lim (inf (f in | i ∈ I) | n ∈ 𝜔) = sup (inf (f in | i ∈ I) | n ∈ 𝜔). Since 𝜑n ≡ inf (f in | i ∈ I) ∈ A(T), the sequence (𝜑n | n ∈ 𝜔) increases, and g ⩽ inf (f i | i ∈ I) ∈ A(T), we infer that g ∈ Y. 4. First, consider the 𝜏-case. Let g i = sup (f im i ∈ A(T) | m i ∈ M i ) and g i ⩽ u ∈ A(T). Consider the function g ≡ sup (g i | i ∈ I) ⩽ u in F(T) and the set K ≡ ⋃ ⟮{i} × M i | i ∈ I⟯. It follows from Proposition 2 (1.1.15) that g(t) = sup (g i (t) | i ∈ I) = sup (sup (f im i (t) | m i ∈ M i ) | i ∈ I) = sup (f k (t) | k ∈ K). By Lemma 1 we conclude that g ∈ Y. Note that in the 𝜎-case the conditions I 𝜔 and M i ⊂ 𝜔 imply the property card K ⩽ 𝜔 by virtue of Theorem 1 (1.3.9). 5. The belongings g ∧ 1 ∈ Y and h ∧ 1 ∈ Z follow from Proposition 1 (1.4.5) because A(T) is truncatable. Applying Theorem 1 (2.2.2) and assertion 1, we infer that −(g∨(−1)) = (−g)∧1 ∈ Z, and therefore, g∨(−1) ∈ Y. In the similar manner, we deduce that h ∨ (−1) ∈ Z. 6. Let g = sup (f m | m ∈ M) for an increasing net (f m ∈ A(T) | m ∈ M) ↑ and g ⩽ f ̃ ∈ A(T). Take some m0 ∈ M and f ≡ f m0 . Consider g m = (f m ∨ f ) − f ∈ A(T)+ and g 󸀠 ≡ g − f . Since (g m | m ∈ M) ↑ g󸀠 and 0 ⩽ g 󸀠 ⩽ f ̃ − f ∈ A(T), we infer that g󸀠 ∈ Y. 7. This assertion follows from assertions 6 and 1.

3.6.2 Extensions of 𝜎-exact functionals by the Young – Daniell method | 365

The Young – Daniell extensions of functionals: the first step For a fixed functional 𝜑 : A(T) → R, we shall construct some remarkable extensions onto the Young domains S𝜏 (T, A(T)), I𝜏 (T, A(T)), S𝜎 (T, A(T)), and I𝜎 (T, A(T)) using the Daniell method [Daniell, 1918] taking its origin in papers [Young, 1905; 1911; 1914]. Define the mappings 𝜑 : Y → R and 𝜑 : Z → R for Y ≡ S𝜏 (T, A(T)) and Z ≡ I𝜏 (T, A(T)) setting 𝜑g ≡ sup{𝜑f | f ∈ A(T) ∧ f ⩽ g} for every g ∈ Y and 𝜑h ≡ inf{𝜑f | f ∈ A(T) ∧ f ⩾ h} for every h ∈ Z. The definition of a supremal [an infimal] function implies that for every g ∈ Y [h ∈ Z] there is a function f 󸀠 ∈ A(T) such that g ⩽ f 󸀠 [h ⩾ f 󸀠 ]. Hence, 𝜑g ⩽ 𝜑f 󸀠 < ∞ [𝜑h ⩾ 𝜑f 󸀠 > −∞]. This means that we have the functionals 𝜑 : Y → R and 𝜑 : Z → R. Clearly, 𝜑 and 𝜑 are extensions of 𝜑. Lemma 3. Let A(T) be a truncatable lattice-ordered linear space of functions on T and 𝜑 be a pointwise continuous [𝜎-continuous] positive linear functional on A(T). If g ∈ Y ≡ S𝜏 (T, A(T)) [g ∈ Y ≡ S𝜎 (T, A(T))] and g = p-lim (f m | m ∈ M) for some increasing net (f m ∈ A(T)+ | m ∈ M) [sequence (f m ∈ A(T)+ | m ∈ M ⊂ 𝜔)], then (𝜑f m | m ∈ M) ↑ 𝜑g. Proof. Since f m ⩽ g, we have 𝜑f m ⩽ 𝜑g ≡ b for every m ∈ M. Consider the number a ≡ sup (𝜑f m | m ∈ M) ⩽ b. Take a function f ∈ A(T)+ such that f ⩽ g. Then, ((f − f m )+ | m ∈ M) ↓ (f − g)+ = 0 in F(T). By Lemma 1 (2.2.8), we get (𝜑((f − f m )+ ) | m ∈ M) ↓ 0. Since 𝜑 is increasing, we have (𝜑f m | m ∈ M) ↑ and 𝜑((f − f m )+ ) ⩾ 𝜑(f − f m ). Therefore, 𝜑f = 𝜑f − 0 = 𝜑f − inf (𝜑((f − f m )+ ) | m ∈ M) = = sup (𝜑f − 𝜑((f − f m )+ ) | m ∈ M) ⩽ sup (𝜑f − 𝜑(f − f m ) | m ∈ M) = = sup (𝜑f m | m ∈ M) ≡ a. This implies that b ⩽ a. Lemma 4. Let A(T) be a truncatable lattice-ordered linear space of functions on T, Y ≡ S𝜏 (T, A(T)) [Y ≡ S𝜎 (T, A(T))], and 𝜑 be a pointwise continuous [𝜎-continuous] positive linear functional on A(T). Then, the functional 𝜑 on Y0 has the following properties: 1) 𝜑 is increasing; 2) 𝜑(ag) = a𝜑g for every a ∈ R+ and g ∈ Y0 ; 3) 𝜑(g󸀠 + g󸀠󸀠 ) = 𝜑g󸀠 + 𝜑g󸀠󸀠 for every g󸀠 , g󸀠󸀠 ∈ Y0 .

366 | 3.6 Representation of a functional by the Radon integral

Proof. 1. The assertion follows directly from the definition. 2. The assertion follows from the definition and Lemma 4 (1.4.5). 3. First, consider the 𝜏-case. Consider the non-empty upward directed sets M 󸀠 ≡ {f ∈ A(T)+ | f ⩽ g󸀠 } and M 󸀠󸀠 ≡ {f ∈ A(T)+ | f ⩽ g 󸀠󸀠 }. By definition, 𝜑g󸀠 = sup{𝜑f 󸀠 | f 󸀠 ∈ M 󸀠 } and 𝜑g󸀠󸀠 = sup{𝜑f 󸀠󸀠 | f 󸀠󸀠 ∈ M 󸀠󸀠 }. Consider the upward directed set M ≡ M 󸀠 × M 󸀠󸀠 . Using Corollary 1 to Proposition 2 (1.1.15) and Lemma 4 (1.4.5), we get sup (f 󸀠 + f 󸀠󸀠 | (f 󸀠 , f 󸀠󸀠 ) ∈ M) = sup (sup (f 󸀠 + f 󸀠󸀠 | f 󸀠 ∈ M 󸀠 ) | f 󸀠󸀠 ∈ M 󸀠󸀠 ) = sup (f 󸀠󸀠 + sup (f 󸀠 | f 󸀠 ∈ M 󸀠 ) | f 󸀠󸀠 ∈ M 󸀠󸀠 ) = g󸀠 + sup (f 󸀠󸀠 | f 󸀠󸀠 ∈ M 󸀠󸀠 ) = g󸀠 + g󸀠󸀠 , where (f 󸀠 + f 󸀠󸀠 | (f 󸀠 , f 󸀠󸀠 ) ∈ M) ↑ (g 󸀠 + g󸀠󸀠 ). Then, Lemma 3 implies that (𝜑(f 󸀠 + f 󸀠󸀠 ) | (f 󸀠 , f 󸀠󸀠 ) ∈ M) ↑ 𝜑(g󸀠 + g󸀠󸀠 ). In a similar way, we get (𝜑f 󸀠 + 𝜑f 󸀠󸀠 | (f 󸀠 , f 󸀠󸀠 ) ∈ M) ↑ (𝜑g󸀠 + 𝜑g󸀠󸀠 ). Therefore, 𝜑(g󸀠 + g󸀠󸀠 ) = 𝜑g󸀠 + 𝜑g󸀠󸀠 . Now, consider the 𝜎-case. By definition, g󸀠 = p-lim (f m󸀠 | m ∈ 𝜔) and g󸀠󸀠 = p-lim (f m󸀠󸀠 | m ∈ 𝜔). Then, by Lemma 3, (𝜑f m󸀠 | m ∈ 𝜔) ↑ 𝜑g󸀠 and (𝜑f m󸀠󸀠 | m ∈ 𝜔) ↑

𝜑g󸀠󸀠 . Applying Proposition 1 (1.4.7), we get (𝜑f m󸀠 + 𝜑f m󸀠󸀠 | m ∈ 𝜔) ↑ 𝜑g󸀠 + 𝜑g󸀠󸀠 . By

Proposition 1 (2.2.3), we get g󸀠 + g󸀠󸀠 = p-lim (f m󸀠 + f m󸀠󸀠 | m ∈ 𝜔). Then, Lemma 3 guarantees that (𝜑(f m󸀠 + f m󸀠󸀠 ) | m ∈ 𝜔) ↑ 𝜑(g󸀠 + g󸀠󸀠 ). Since 𝜑 is linear, we infer that 𝜑(g 󸀠 + g󸀠󸀠 ) = 𝜑g󸀠 + 𝜑g󸀠󸀠 . Lemma 5. Let A(T) be a truncatable lattice-ordered linear space of functions on T, Y ≡ S𝜏 (T, A(T)) [Y ≡ S𝜎 (T, A(T))], Z ≡ I𝜏 (T, A(T)) [Z ≡ I𝜎 (T, A(T))], and 𝜑 be a pointwise continuous [𝜎-continuous] positive linear functional on A(T). Then, the functional 𝜑 on Y and the functional 𝜑 on Z have the following properties: 1) 𝜑 and 𝜑 are increasing; 2) 𝜑(−g) = −𝜑(g) and 𝜑(−h) = −𝜑(h) for any functions g ∈ Y and h ∈ Z; 3) 𝜑(f + g) = 𝜑f + 𝜑g and 𝜑(f − g) = 𝜑f − 𝜑g for every f ∈ A(T) and g ∈ Y0 ; 4) 𝜑(∑(a i g i | i ∈ I)) = ∑(a i 𝜑g i | i ∈ I) and 𝜑(∑(a i h i | i ∈ I)) = ∑(a i 𝜑h i | i ∈ I) for any finite collections (a i ∈ R+ | i ∈ I), (g i ∈ Y | i ∈ I), and (h i ∈ Z | i ∈ I); 5) (𝜑g m | m ∈ M) ↑ 𝜑g for every net (g m ∈ Y | m ∈ M) [respectively, every sequence (g m ∈ Y | m ∈ M ⊂ 𝜔)] and every function g ∈ Y such that (g m | m ∈ M) ↑ g in F(T); 6) 𝜑v ⩽ 𝜑u ⩽ 𝜑w for all functions u ∈ Y and v, w ∈ Z such that v ⩽ u ⩽ w. Proof. 1. The assertion follows directly from the definitions. 2. The assertion follows from the definitions and assertion 1 of Lemma 2. 3. By definition, 𝜑(f + g) = sup{𝜑u | u ∈ A(T) ∧ u ⩽ f + g} and 𝜑g = sup{𝜑v | v ∈ A(T) ∧ v ⩽ f + g}. Since u ⩽ f + g and 𝜑 is an increasing extension of 𝜑, u − f ⩽ g implies that 𝜑u − 𝜑f ⩽ 𝜑g, i. e. 𝜑u ⩽ 𝜑f + 𝜑g. Hence, 𝜑(f + g) ⩽ 𝜑f + 𝜑g. Since v ⩽ g, f + v ⩽ f + g implies that 𝜑f + 𝜑v ⩽ 𝜑(f + g), i. e. 𝜑v ⩽ 𝜑(f + g) − 𝜑f . Therefore, 𝜑g ⩽ 𝜑(f + g) − 𝜑f , i. e. 𝜑g + 𝜑f ⩽ 𝜑(f + g). The second equality follows from the first one and assertion 2.

3.6.2 Extensions of 𝜎-exact functionals by the Young – Daniell method | 367

4. The first equality follows from assertions 6 and 2 of Lemma 2, the previous assertion, and assertions 2 and 3 of Lemma 4. The second equality follows from the first one and assertion 2. 5. First, consider the 𝜏-case. Consider the upward directed sets L l ≡ {u ∈ A(T) | u ⩽ g l } for all l ∈ M. Consider the set P ≡ {{m} × L m | m ∈ M}. Define an order on P setting (l, u) ⩽ (m, v) if l ⩽ m and u ⩽ v. If p ≡ (l, u) ∈ P and q ≡ (m, v) ∈ P, then there is n ∈ M such that l ⩽ n and m ⩽ n. Since u ⩽ g l ⩽ g n and v ⩽ g m ⩽ g n , we have w ≡ u ∨ v ∈ L n . Therefore, P is upward directed. Consider the net (f p ∈ A(T)+ | p ∈ P) such that f p ≡ u for every p ≡ (l, u). Let p ≡ (l, u), q ≡ (m, v), and p ⩽ q. Then, u ⩽ v implies f p ≡ u ⩽ v ≡ f q . Hence, (f p | p ∈ P) ↑. Let t ∈ T and 𝜀 > 0. Then, g(t) − 𝜀/2 < g l (t) for some l ∈ M. According to the definition of a supremal function, g l (t) − 𝜀/2 < u(t) = f p (t) for some u ∈ L l and p ≡ (l, u). Therefore, g(t) − 𝜀 < f p (t). Since 𝜀 is arbitrary, we have g(t) = sup (f p (t) | p ∈ P). Thus, (f p | p ∈ P) ↑ g. Then, Lemma 3 implies (𝜑f p | p ∈ P) ↑ 𝜑g. Let 𝜀 > 0. Then, 𝜑g − 𝜀 < 𝜑f p for some p ≡ (l, u). Since f p ≡ u ∈ L l , 𝜑f p = 𝜑u ⩽ 𝜑g l implies 𝜑g − 𝜀 < 𝜑g l . Thus, 𝜑g = sup (𝜑g m | m ∈ M). Now, consider the 𝜎-case. By definition, g m = sup (f mk ∈ A(T) | k ∈ 𝜔). Then, by Lemma 3 (𝜑f mk | k ∈ 𝜔) ↑ 𝜑g m . Consider the functions f k ≡ sup (f mk | m ∈ k + 1) ∈ A(T) for k ∈ 𝜔. Clearly, (f k | k ∈ 𝜔) ↑. Fix t ∈ T. By the condition, (g m | m ∈ 𝜔) ↑ g in F(T). Consequently, for every n ∈ N, there is p such that g(t) ⩾ g p (t) > g(t) − 1/n. Then, there is k ∈ 𝜔 such that g(t) ⩾ g p (t) ⩾ f pk (t) > g(t) − 1/n. Hence, g(t) − 1/n < f pk (t) ⩽ f k (t) ⩽ sup (f mk (t) | m ∈ 𝜔) ⩽ sup (g m (t) | m ∈ 𝜔) = g(t). This means that (f k | k ∈ 𝜔) ↑ g. Now, by Lemma 3, we infer that (𝜑f k | k ∈ 𝜔) ↑ 𝜑g. But f k ⩽ sup (g m | m ∈ k + 1) = g k by assertion 1 implies that 𝜑f k ⩽ 𝜑g k ⩽ 𝜑g. Using Lemma 6 (1.4.7), we conclude that (𝜑g k | k ∈ 𝜔) ↑ 𝜑g. 6. Assertions 6 and 7 of Lemma 2 guarantees that u = f 󸀠 + g󸀠 and v = f 󸀠󸀠 − g󸀠󸀠 for some f 󸀠 , f 󸀠󸀠 ∈ A(T) and g󸀠 , g󸀠󸀠 ∈ Y0 . It follows from v − u ⩾ 0 that f 󸀠󸀠 − f 󸀠 ⩽ g󸀠 + g󸀠󸀠 ∈ Y. Then, by assertions 1 and 3, we get 𝜑f 󸀠󸀠 −𝜑f 󸀠 ⩽ 𝜑(g󸀠 + g󸀠󸀠 ) = 𝜑g󸀠 +𝜑g󸀠󸀠 . Finally, assertion 2 implies that 𝜑v = 𝜑f 󸀠󸀠 − 𝜑g󸀠󸀠 ⩽ 𝜑f 󸀠 + 𝜑g󸀠 = 𝜑u. Further, consider the non-empty sets M 󸀠 ≡ {f ∈ A(T) | f ⩽ u} and M 󸀠󸀠 ≡ {f ∈ A(T) | w ⩾ f }. Then, f 󸀠 ⩽ u ⩽ w ⩽ f 󸀠󸀠 for every f 󸀠 ∈ M 󸀠 and f 󸀠󸀠 ∈ M 󸀠󸀠 . Hence, 𝜑f 󸀠 ⩽ 𝜑f 󸀠󸀠 . By definition, 𝜑u = sup{𝜑f 󸀠 | f 󸀠 ∈ M 󸀠 } and 𝜑w = sup{𝜑f 󸀠󸀠 | f 󸀠󸀠 ∈ M 󸀠󸀠 }. Therefore, 𝜑u ⩽ 𝜑f 󸀠󸀠 and 𝜑u ⩽ 𝜑w. The Young – Daniell extensions of functionals: the second step For a fixed pointwise continuous positive linear functional 𝜑 : A(T) → R define the mappings 𝜑∨ : Fm (T, A(T)) → R and 𝜑∨ : Fm (T, A(T)) → R setting 𝜑∨ f ≡ inf{𝜑g | g ∈ S𝜏 (T, A(T)) ∧ g ⩾ f } and 𝜑∨ f ≡ sup{𝜑h | h ∈ I𝜏 (T, A(T)) ∧ h ⩽ f }

368 | 3.6 Representation of a functional by the Radon integral

for every function f ∈ Fm (T, A(T)). By definition, for every function f ∈ Fm (T, A(T)) there is a function u ∈ A(T) such that −u ⩽ f ⩽ u. Since 𝜑 and 𝜑 are increasing, this implies that for every g ∈ Y [h ∈ Z] such that g ⩾ f [h ⩽ f ] we have 𝜑g ⩾ 𝜑(−u) = −𝜑u > ∞ [𝜑h ⩽ 𝜑u = 𝜑u < ∞]. This means that we have the functionals 𝜑∨ : Fm (T, A(T)) → R and 𝜑∨ : Fm (T, A(T)) → R. It is clear that 𝜑∨ and 𝜑∨ are extensions of the functionals 𝜑 and 𝜑, respectively. Proposition 1. Let A(T) be a truncatable lattice-ordered linear space of functions on T and 𝜑 be a pointwise continuous positive linear functional on A(T). Then, the functionals 𝜑∨ and 𝜑∨ on Fm (T, A(T)) have the following properties: 1) 𝜑∨ and 𝜑∨ are increasing; 2) 𝜑∨ f ⩽ 𝜑∨ f for every f ∈ Fm (T, A(T)); 3) 𝜑∨ (−f ) = −𝜑∨ f and 𝜑∨ (−f ) = −𝜑∨ f for every f ∈ Fm (T, A(T)); 4) 𝜑∨ (af ) = a𝜑∨ f and 𝜑∨ (af ) = a𝜑∨ f for all a ∈ R+ and f ∈ Fm (T, A(T)); 5) 𝜑∨ (f 󸀠 + f 󸀠󸀠 ) ⩽ 𝜑∨ f 󸀠 + 𝜑∨ f 󸀠󸀠 and 𝜑∨ (f 󸀠 + f 󸀠󸀠 ) ⩾ 𝜑∨ f 󸀠 + 𝜑∨ f 󸀠󸀠 for all f 󸀠 , f 󸀠󸀠 ∈ Fm (T, A(T)); 6) 𝜑∨ (f 󸀠 ∨ f 󸀠󸀠 ) + 𝜑∨ (f 󸀠 ∧ f 󸀠󸀠 ) ⩽ 𝜑∨ f 󸀠 + 𝜑∨ f 󸀠󸀠 for all f 󸀠 , f 󸀠󸀠 ∈ Fm (T, A(T)); 7) (𝜑∨ f n | n ∈ 𝜔) ↑ 𝜑∨ f for every sequence (f n ∈ Fm (T, A(T)) | n ∈ 𝜔) and every function f ∈ Fm (T, A(T)) such that (f n | n ∈ 𝜔) ↑ f in F(T). Proof. 1. The assertion follows directly from the definitions. 2. By definition, 𝜑∨ f ≡ inf{𝜑g | g ∈ Y ∧ g ⩾ f } and 𝜑∨ f ≡ sup{𝜑h | h ∈ Z ∧ h ⩽ f }. Since h ⩽ f ⩽ g, assertion 6 of Lemma 5 implies that 𝜑h ⩽ 𝜑g. Therefore, 𝜑∨ f ⩽ 𝜑g and 𝜑∨ f ⩽ 𝜑∨ f . 3. Since 𝜑∨ (−f ) ≡ inf{𝜑g | g ∈ Y ∧ g ⩾ −f }, it follows from f ⩾ −g ∈ Z that 𝜑∨ f ⩾ 𝜑(−g). By assertion 2 of Lemma 5, we get 𝜑(−g) = −𝜑g. Therefore, −𝜑∨ f ⩽ 𝜑g implies −𝜑∨ f ⩽ 𝜑∨ (−f ). On the other hand, since 𝜑∨ f ≡ sup{𝜑h | h ∈ Z ∧ h ⩽ f }, it follows from −f ⩽ −h ∈ Y that 𝜑∨ (−f ) ⩽ 𝜑(−h) = −𝜑h. Therefore, −𝜑∨ (−f ) ⩾ 𝜑h implies −𝜑∨ (−f ) ⩾ 𝜑∨ f , i. e. 𝜑∨ (−f ) ⩽ −𝜑∨ f . 4. The assertion follows from assertion 4 of Lemma 5 and Lemma 4 (1.4.5). 5. By definition, 𝜑∨ (f 󸀠 + f 󸀠󸀠 ) ≡ inf{𝜑g | g ∈ Y ∧ g ⩾ f 󸀠 + f 󸀠󸀠 }, 𝜑∨ f 󸀠 ≡ inf{𝜑g󸀠 | g󸀠 ∈ Y ∧ g󸀠 ⩾ f 󸀠 }, and 𝜑∨ f 󸀠󸀠 ≡ inf{𝜑g󸀠󸀠 | g󸀠󸀠 ∈ Y ∧ g󸀠󸀠 ⩾ f 󸀠󸀠 }, Since g󸀠 + g󸀠󸀠 ⩾ f 󸀠 + f 󸀠󸀠 and g 󸀠 + g󸀠󸀠 ∈ Y, by assertion 4 of Lemma 5, we have 𝜑∨ (f 󸀠 + f 󸀠󸀠 ) ⩽ 𝜑(g󸀠 + g󸀠󸀠 ) = 𝜑g󸀠 + 𝜑g󸀠󸀠 . This implies 𝜑∨ (f 󸀠 + f 󸀠󸀠 ) − 𝜑g󸀠 ⩽ 𝜑g󸀠󸀠 , and therefore, 𝜑∨ (f 󸀠 + f 󸀠󸀠 ) − 𝜑g󸀠 ⩽ 𝜑∨ f 󸀠󸀠 . Further, 𝜑∨ (f 󸀠 + f 󸀠󸀠 ) − 𝜑∨ f 󸀠󸀠 ⩽ 𝜑g󸀠 implies 𝜑∨ (f 󸀠 + f 󸀠󸀠 ) − 𝜑∨ f 󸀠󸀠 ⩽ 𝜑∨ f 󸀠 . The second equality follows from the first one and assertion 3. 6. Take some g󸀠 , g󸀠󸀠 ∈ Y such that g󸀠 ⩾ f 󸀠 and g󸀠󸀠 ⩾ f 󸀠󸀠 . Then, g󸀠 ∨ g󸀠󸀠 ⩾ f 󸀠 ∨ f 󸀠󸀠 and g󸀠 ∧ g󸀠󸀠 ⩾ f 󸀠 ∧ f 󸀠󸀠 . Assertion 4 of Lemma 5 (1.4.5) implies the equality g󸀠 ∨ g󸀠󸀠 + g󸀠 ∧ g󸀠󸀠 = g󸀠 + g󸀠󸀠 . Therefore, using assertion 4 of Lemma 5, we get x ≡ 𝜑∨ (f 󸀠 ∨ f 󸀠󸀠 ) + 𝜑∨ (f 󸀠 ∧ f 󸀠󸀠 ) ⩽ 𝜑∨ (g󸀠 ∨ g󸀠󸀠 ) + 𝜑∨ (g󸀠 ∧ g󸀠󸀠 ) = 𝜑(g󸀠 ∨ g󸀠󸀠 ) + 𝜑(g󸀠 ∧ g󸀠󸀠 ) = 𝜑(g 󸀠 ∨ g󸀠󸀠 + g󸀠 ∧ g󸀠󸀠 ) = 𝜑(g󸀠 + g󸀠󸀠 ) = 𝜑g󸀠 + 𝜑g󸀠󸀠 . Since g󸀠 and g󸀠󸀠 are arbitrary, we obtain x ⩽ inf (𝜑g󸀠 + 𝜑g󸀠󸀠 | g󸀠 ∈ Y ∧ g󸀠 ⩾ f 󸀠 ) = inf (𝜑g󸀠 | g󸀠 ∈ Y ∧ g󸀠 ⩾ f 󸀠 ) + 𝜑g󸀠󸀠 =

3.6.2 Extensions of 𝜎-exact functionals by the Young – Daniell method | 369

𝜑∨ f 󸀠 + 𝜑g󸀠󸀠 . Similarly, x ⩽ inf (𝜑∨ f 󸀠 + 𝜑g󸀠󸀠 | g󸀠󸀠 ∈ Y ∧ g󸀠󸀠 ⩾ f 󸀠󸀠 ) = 𝜑∨ f 󸀠 + inf(𝜑g󸀠󸀠 | g󸀠󸀠 ∈ Y ∧ g󸀠󸀠 ⩾ f 󸀠󸀠 ) = 𝜑∨ f 󸀠 + 𝜑∨ f 󸀠󸀠 . 7. Since 𝜑∨ is increasing, we have (𝜑∨ f n | n ∈ 𝜔) ↑ and 𝜑∨ f n ⩽ 𝜑∨ f ≡ b. Hence, there exists a ≡ sup (𝜑∨ f n | n ∈ 𝜔) ⩽ b. Fix 𝜀 > 0. Then, for every n ∈ 𝜔 there is g󸀠n ∈ Y such that g 󸀠n ⩾ f n and 𝜑∨ f n + 𝜀/2n+1 > 𝜑g󸀠n ⩾ 𝜑∨ f n . Since f ∈ Fm (T, A(T)), there is a function u ∈ A(T) such that f ⩽ u. Consider the functions g󸀠󸀠n ≡ g󸀠n ∧ u. Assertion 3 of Lemma 2 guarantees that g󸀠󸀠n ∈ Y, and therefore, assertion 4 of this Lemma implies that g ≡ sup (g 󸀠󸀠n | n ∈ 𝜔) ∈ Y. It follows from the inequality g󸀠󸀠n ⩽ g󸀠n and assertion 1 of Lemma 5 that 𝜑g󸀠󸀠n ⩽ 𝜑g󸀠n < 𝜑∨ f n + 𝜀/2n+1 . Besides, g󸀠󸀠n ⩾ f n ∧ f = f n . Consider the functions g n ≡ sup (g 󸀠󸀠i | i ∈ n + 1) ∈ Y. Then, (g n | n ∈ 𝜔) ↑ g, and therefore, assertion 5 of Lemma 5 implies that (𝜑g n | n ∈ 𝜔) ↑ 𝜑g. Consider the set N ⊂ 𝜔 consisting of all numbers n such that 𝜑g n ⩽ 𝜑∨ f n +𝜀−𝜀/2n+1 . If n = 0, then 𝜑g0 = 𝜑g0󸀠󸀠 ⩽ 𝜑∨ f0 + 𝜀/21 = 𝜑∨ f0 + 𝜀 − 𝜀/21 , and so 0 ∈ N. Suppose that n ∈ N. By assertion 4 of Lemma 5 (1.4.5) we get g n + g󸀠󸀠n+1 = g n ∧ g󸀠󸀠n+1 + g n ∨ g󸀠󸀠n+1 . Since g n ∧ g󸀠󸀠n+1 ⩾ g󸀠󸀠n ∧ g󸀠󸀠n+1 ⩾ f n ∧ f n+1 = f n , we obtain 𝜑(g n ∧ g󸀠󸀠n+1 ) ⩾ 𝜑∨ f n . Besides, g n ∨ g󸀠󸀠n+1 = g n+1 . Then, 𝜑g n + 𝜑g󸀠󸀠n+1 = 𝜑(g n ∧ g󸀠󸀠n+1 ) + 𝜑g n+1 ⩾ 𝜑∨ f n + 𝜑g n+1 . Hence, 𝜑g n+1 ⩽ (𝜑g n −𝜑∨ f n )+𝜑g󸀠󸀠n+1 ⩽ 𝜀−𝜀/2n+1 +𝜑g󸀠󸀠n+1 < 𝜀−2𝜀/2n+2 +𝜑∨ f n+1 +𝜀/2n+2 = 𝜑∨ f n+1 +𝜀−𝜀/2n+2 . This means that n + 1 ∈ N. By the principle of natural induction (Theorem 1 (1.2.6)) we conclude that N = 𝜔. Thus, 𝜑g n ⩽ 𝜑∨ f n + 𝜀 − 𝜀/2n+1 for every n ∈ 𝜔. Applying Corollary 3 to Proposition 1 (1.4.7) and Lemma 7 (1.4.4), we get 𝜑g = lim (𝜑g n | n ∈ 𝜔) ⩽ lim (𝜑∨ f n + 𝜀 − 𝜀/2n+1 | n ∈ 𝜔) = lim (𝜑∨ f n | n ∈ 𝜔) + 𝜀 − lim (𝜀/2n+1 | n ∈ 𝜔) = a + 𝜀. Besides, f = p-lim (f n | n ∈ 𝜔) ⩽ p-lim (g󸀠󸀠n | n ∈ 𝜔) ⩽ p-lim (g n | n ∈ 𝜔) = g ∈ Y. Then, b ≡ 𝜑∨ f ⩽ 𝜑g, and therefore, b ⩽ a + 𝜀. Since 𝜀 is arbitrary, we have b ⩽ a ⩽ b. In the similar manner for a fixed pointwise 𝜎-continuous positive linear functional 𝜑 : A(T) → R, we define the functionals 𝜑∧ : Fm (T, A(T)) → R and 𝜑∧ : Fm (T, A(T)) → R setting 𝜑∧ f ≡ inf{𝜑g | g ∈ S𝜎 (T, A(T)) ∧ g ⩾ f } and 𝜑∧ f ≡ sup{𝜑h | h ∈ I𝜎 (T, A(T)) ∧ h ⩽ f } for f ∈ Fm (T, A(T)). These functionals are also extensions of functionals 𝜑 and 𝜑. Naturally, in the indicated condition, all the assertions of Proposition 1 are also valid for 𝜑∧ and 𝜑∧ . The Young – Daniell domains Let T be a set, A(T) be a truncatable lattice-ordered linear space of functions on T, 𝜑 be a positive pointwise continuous [𝜎-continuous] linear functional on A(T). A function f ∈ Fm (T, A(T)) will be called Daniell [𝜎-Daniell] with respect to the family A(T) and the functional 𝜑 if 𝜑∨ f = 𝜑∨ f [𝜑∧ f = 𝜑∧ f ]. The family of all Daniell [𝜎-Daniell] functions

370 | 3.6 Representation of a functional by the Radon integral

will be denoted by D𝜏 (T, A(T), 𝜑) [D𝜎 (T, A(T), 𝜑)]. These families may be called the Young – Daniell domains for the system ⟮A(T), 𝜑⟯. If 𝜑 is pointwise continuous, then we can define the functional 𝜑̌ on D𝜏 (T, A(T), 𝜑) setting 𝜑f̌ ≡ 𝜑∨ f = 𝜑∨ f . If 𝜑 is pointwise 𝜎-continuous, then we can define the funĉ ≡ 𝜑∧ f = 𝜑∧ f . tional 𝜑̂ on D𝜎 (T, A(T), 𝜑) setting 𝜑f Theorem 1 (Young – Daniell). Let A(T) be a truncatable lattice-ordered linear space of functions on a set T, 𝜑 be a positive pointwise continuous [𝜎-continuous] linear functional on A(T), Y ≡ S𝜏 (T, A(T)), Z ≡ I𝜏 (T, A(T)), and D ≡ D𝜏 (T, A(T), 𝜑) [Y ≡ S𝜎 (T, A(T)), Z ≡ I𝜎 (T, A(T)), and D ≡ D𝜎 (T, A(T), 𝜑)]. Then, 1) D = {f ∈ Fm (T, A(T)) | ∀ 𝜀 > 0 ∃ g ∈ Y ∃ h ∈ Z (h ⩽ f ⩽ g ∧ 𝜑g − 𝜑h < 𝜀)}; 2) the family D is a lattice-ordered linear space; 3) Y ∪ Z ⊂ D; ̂ is a positive linear extension of the functional 𝜑; 4) the functional 𝜑̌ [the functional 𝜑] ̂ is pointwise 𝜎-continuous on D. 5) the functional 𝜑̌ [the functional 𝜑] Proof. 1. Denote the right part of the desired equality by X. By definition, 𝜑∨ f ≡ inf{𝜑g | g ∈ Y ∧ g ⩾ f } and 𝜑∨ f ≡ sup{𝜑h | h ∈ Z ∧ h ⩽ f }. For every f ∈ Fm (T, A(T)) and 𝜀 > 0, there are g ∈ Y and h ∈ Z such that h ⩽ f ⩽ g and 𝜑∨ f + 𝜀/2 > 𝜑g and 𝜑∨ f − 𝜀/2 < 𝜑h. If f ∈ D, then 𝜑g − 𝜑h < (𝜑∨ f + 𝜀/2) − (𝜑∨ f − 𝜀/2) = 𝜀, and so f ∈ X. Conversely, if f ∈ X and 𝜀 > 0, then it follows from 𝜑∨ f ⩽ 𝜑g and 𝜑∨ f ⩾ 𝜑h that ∨ 𝜑 f ⩽ 𝜑g < 𝜑h + 𝜀 ⩽ 𝜑∨ f + 𝜀. Hence, 𝜑∨ f ⩽ 𝜑∨ f . Using assertion 2 of Proposition 1, we get 𝜑∨ f = 𝜑∨ f , i. e. f ∈ D. 2. Let f ∈ D and a ∈ R. By assertion 3 of Proposition 1, we get 𝜑∨ (−f ) = −𝜑∨ f = ∨ −𝜑 f = 𝜑∨ (−f ), and so −f ∈ D. If a > 0, then by assertion 4 of Proposition 1 we get 𝜑∨ (af ) = a𝜑∨ f = a𝜑∨ f = 𝜑∨ (af ). If a < 0, then 𝜑∨ (af ) = −a𝜑∨ (−f ) = −a𝜑∨ (−f ) = 𝜑∨ ((−a)(−f ))𝜑∨ (af ). Thus, af ∈ D. Let f 󸀠 , f 󸀠󸀠 ∈ D. Then, assertions 2 and 5 of Proposition 1 imply that 𝜑∨ (f 󸀠 + f 󸀠󸀠 ) ⩽ ∨ 󸀠 𝜑 (f + f 󸀠󸀠 ) ⩽ 𝜑∨ f 󸀠 + 𝜑∨ f 󸀠󸀠 = 𝜑∨ f 󸀠 + 𝜑∨ f 󸀠󸀠 ⩽ 𝜑∨ (f 󸀠 + f 󸀠󸀠 ). Thus, f 󸀠 + f 󸀠󸀠 ∈ D. By Lemma 1, we get 𝜑∨ (f 󸀠 ∨ f 󸀠󸀠 ) + 𝜑∨ (f 󸀠 ∧ f 󸀠󸀠 ) ⩽ 𝜑∨ f 󸀠 + 𝜑∨ f 󸀠󸀠 = 𝜑∨ f 󸀠 + 𝜑∨ f 󸀠󸀠 . Similarly, ∨ 𝜑 (−(f 󸀠 ∧ f 󸀠󸀠 )) + 𝜑∨ (−(f 󸀠 ∨ f 󸀠󸀠 )) = 𝜑∨ ((−f 󸀠 ) ∨ (−f 󸀠󸀠 )) + 𝜑∨ ((−f 󸀠 ) ∧ (−f 󸀠󸀠 )) ⩽ 𝜑∨ (−f 󸀠 ) + 𝜑∨ (−f 󸀠󸀠 ), i. e. −𝜑∨ (f 󸀠 ∧ f 󸀠󸀠 ) − 𝜑∨ (f 󸀠 ∨ f 󸀠󸀠 ) ⩽ −𝜑∨ f 󸀠 − 𝜑∨ f 󸀠󸀠 . Adding these inequalities we get (𝜑∨ (f 󸀠 ∨ f 󸀠󸀠 ) − 𝜑∨ (f 󸀠 ∨ f 󸀠󸀠 )) + (𝜑∨ (f 󸀠 ∧ f 󸀠󸀠 ) − 𝜑∨ (f 󸀠 ∧ f 󸀠󸀠 )) ⩽ 0. But each of these differences is non-negative, where they are both zero. This means that f 󸀠 ∨f 󸀠󸀠 and f 󸀠 ∧f 󸀠󸀠 belong to D. 3. Let g ∈ Y. It follows from 𝜑g ≡ sup{𝜑f | f ∈ A(T) ∧ f ⩽ g} that for every 𝜀 > 0 there is f ∈ A(T) ⊂ Z such that f ⩽ g ⩽ g and 𝜑g − 𝜀 < 𝜑f = 𝜑f . Hence, assertion 1 implies that g ∈ D. Similarly, for h ∈ Z, it follows from 𝜑h ≡ sup{𝜑f | f ∈ A(T) ∧ f ⩾ h} that for every 𝜀 > 0 there is f ∈ A(T) ⊂ Y such that h ⩽ h ⩽ f and 𝜑h + 𝜀 > 𝜑f = 𝜑f . Hence, h ∈ D. 4. Let f ∈ A(T) ⊂ Y ⊂ D. Then, 𝜑f̌ ≡ 𝜑∨ f = 𝜑f = 𝜑f . If f ∈ D and f ⩾ 0, then 𝜑f̌ ≡ 𝜑∨ f ⩾ 𝜑∨ 0 = 0.

3.6.2 Extensions of 𝜎-exact functionals by the Young – Daniell method | 371

̌ )= Let f ∈ D and a ∈ R. If a > 0, then by assertion 4 of Proposition 1, we get 𝜑(af ̌ ) = 𝜑∨ ((−a)(−f )) = −a𝜑∨ (−f ) = 𝜑∨ (af ) = a𝜑∨ f = a𝜑f̌ . Similarly, if a < 0, then 𝜑(af a𝜑∨ f = a𝜑f̌ . Let f 󸀠 , f 󸀠󸀠 ∈ D. As was shown in the proof of assertion 2, 𝜑∨ (f 󸀠 + f 󸀠󸀠 ) = 𝜑∨ f 󸀠 + 𝜑∨ f 󸀠󸀠 , ̌ 󸀠 + f 󸀠󸀠 ) = 𝜑f̌ 󸀠 + 𝜑f̌ 󸀠󸀠 . i. e. 𝜑(f 5. Since 𝜑̌ = 𝜑∨ |D, it follows from assertion 7 of Proposition 1 that the condition (f n ∈ D | n ∈ 𝜔) ↑ f ∈ D implies that (𝜑f̌ n | n ∈ 𝜔) ↑ 𝜑f̌ . Therefore, by Lemma 1 (2.2.8), the functional 𝜑̌ is pointwise 𝜎-continuous. For the 𝜎-case, all the arguments are completely analogous. Envelopment properties (E) and (E𝜎 ) We need in two new properties of function families to extend continuous and 𝜎continuous functionals from some subspaces A(T) ⊂ S(T, G) to wider subspaces (see Theorem 5 below). A family A(T) is said to have property (E) [respectively, (E𝜎 )] if the following three conditions are fulfilled: (i) for every G ∈ G and u ∈ A(T)+ the family A(T) envelopes [𝜎-envelopes] (see 2.2.9) from below the function 𝜒(G) ∧ u, (ii) for every F ∈ F, C ∈ C, and u ∈ A(T)+ the family A(T) envelopes [𝜎-envelopes] from above the functions 𝜒(F) ∧ u and 𝜒(C), (iii) for every G ∈ G and every compact subset C ⊂ G there is a function v ∈ A(T) such that 𝜒(C) ⩽ v ⩽ 𝜒(G). It is clear that property (E 𝜎 ) is stronger than property (E). The following proposition shows that the lattice-ordered linear spaces S(T, G), S c (T, G), C b (T, G), and C c (T, G) we are interested in have these properties on appropriate topological spaces ⟮T, G⟯. Proposition 2. The following conclusions are valid: 1) if ⟮T, G⟯ is a Hausdorff space, then S(T, G) and S c (T, G) have property (E𝜎 ); 2) if ⟮T, G⟯ is a Tychonoff space, then C b (T, G) has property (E); 3) if ⟮T, G⟯ is a locally compact space, then C c (T, G) has property (E). Proof. 1. Since all open and closed (including all compact) sets are symmetrizable, we have 𝜒(G), 𝜒(F) ∈ S(T, G) and 𝜒(C) ∈ S c (T, G) for all G ∈ G, F ∈ F, and C ∈ C. Then, for any u ∈ S(T, G)+ [u ∈ S c (T, G)+ ] all functions 𝜒(G) ∧ u, 𝜒(F) ∧ u, and 𝜒(C) belong to S(T, G) [S c (T, G)]. 2. Let G ∈ G, u ∈ C b (T, G)+ . Consider the sets Φ ≡ {f ∈ C b (T, G) | 0 ⩽ f ⩽ 1 ∧ ∃ t ∈ G (f (t) = 1) ∧ f [T\G] = {0} } and Ψ ≡ {f ∧ u | f ∈ Φ}.

372 | 3.6 Representation of a functional by the Radon integral

It is clear that fΨ ≡ sup Ψ ⩽ 𝜒(G) ∧ u. Take any t0 ∈ G. By the definition, of a Tychonoff space there is a function f ∈ C b (T, G) such that 0 ⩽ f ⩽ 1, f (t0 ) = 1, and f [T\G] = {0}. Then, (𝜒(G) ∧ u)(t0 ) = u(t0 ) ∧ 1 = (f ∧ u)(t0 ) ⩽ fΨ (t0 ) ⩽ (𝜒(G) ∧ u)(t0 ). Therefore, 𝜒(G) ∧ u = sup Ψ. Put Ξ ≡ {h ∈ C b (T, G) | ∃ n ∈ N ∃ g0 , g1 , . . . , g n−1 ∈ Ψ (h = g0 ∨ g1 ∨ . . . ∨ g n−1 )}. Since h ⩾ h1 and h ⩾ h2 for any h1 , h2 ∈ Ξ and h ≡ h1 ∨ h2 ∈ Ξ, we have that (h | h ∈ Ξ) is an increasing net. Thus, the family C b (T, G) envelopes from below the function 𝜒(G) ∧ u. Suppose now F ∈ F, u ∈ C b (T, G)+ . Put G ≡ T\F and consider the families Φ1 ≡ {f ∈ C b (T, G) | ∃ g ∈ Φ (f = 1 − g)} = {f ∈ C b (T, G) | 0 ⩽ f ⩽ 1 ∧ ∃ t ∈ G (f (t) = 0) ∧ f [F] = {1} } and Ψ1 ≡ {f ∧ u | f ∈ Φ1 }. It is clear that fΨ1 ≡ inf Ψ1 ⩾ 𝜒(F) ∧ u and for every t ∈ F we get (𝜒(F) ∧ u)(t) = u(t) ∧ 1 = fΨ1 (t). Take any t0 ∈ G. By the definition, of a Tychonoff space there is a function f ∈ C b (T, G) such that 0 ⩽ f ⩽ 1, f (t0 ) = 0, and f [F] = {1}. Then, 0 = (𝜒(F) ∧ u)(t0 ) = (f ∧ u)(t0 ) ⩾ fΨ1 (t0 ) ⩾ (𝜒(F) ∧ u)(t0 ) = 0. Therefore, 𝜒(F) ∧ u = inf Ψ1 . If we put Ξ1 ≡ {h ∈ C b (T, G) | ∃ n ∈ N ∃ g0 , g1 , . . . , g n−1 ∈ Ψ1 (h = g0 ∧ g1 ∧ . . . ∧ g n−1 )}, we obtain a decreasing net (h | h ∈ Ξ1 ) ↓ 𝜒(F) ∧ u, i. e. the family C b (T, G) envelopes from above the function 𝜒(F) ∧ u. In particular, we have seen that for every C ∈ C ⊂ F the family C b (T, G) envelopes from above the function 𝜒(C) ∧ 1 = 𝜒(C). Finally, let G ∈ G, C ∈ C, C ⊂ G. Since ⟮T, G⟯ is Tychonoff, for any point s ∈ C there is a function f ∈ C b (T, G) such that 0 ⩽ f ⩽ (3/2)1, f (s) = 3/2, and f [T\G] = {0}. Consider the open neighborhood U ≡ f −1 []1, 2[] of the point s and the cover of compact set C by all these neighborhoods. Take a finite subcover U1 , . . . , U n and denote by f1 , . . . , f n the functions corresponding to the sets U1 , . . . , U n . The function v ≡ (f1 ∨ . . . ∨ f n ) ∧ 1 ∈ C b (T, G) satisfies the inequality 𝜒(C) ⩽ v ⩽ 𝜒(G). 3. Note that for any u ∈ C c (T, G)+ the families Ψ, Ξ, Ψ1 , and Ξ1 constructed in the previous part of the proof are subfamilies of C c (T, G). Therefore, we see that the family C c (T, G) envelopes from below functions 𝜒(G) ∧ u for all G ∈ G and envelopes from above functions 𝜒(F) ∧ u for all F ∈ F. Take any C ∈ C and G ∈ G such that C ⊂ G and find a function v ∈ C c (T, G) such that 𝜒(C) ⩽ v ⩽ 𝜒(G). According to Statement 10 (3.5.1) there exists an open set U such that C ⊂ U ⊂ cl U ⊂ G and K ≡ cl U ∈ C. Take the sets H ≡ T\K ∈ G and F ≡ K\U ∈ F. By virtue of Statement 6 (3.5.1), the subspace ⟮K, GK ⟯ of the space ⟮T, G⟯ is normal, i. e. the ensemble GK is separable (see 2.1.1). Then, Corollary 1 to Theorem 1 (2.3.5) guarantees that there is a function f ∈ C(K, GK ) ≡ M(K, GK ) such that 0 ⩽ f ⩽ 1, f [F] = {0}, and f [C] = {1}. Extent f to H by zero and denote the resulting function on T by v. Show that this function is continuous on T. For any interval I ≡]a, b[, we see that f −1 [I] ∈ GK , i. e. f −1 [I] = O ∩ K for some O ∈ G. If a < b ⩽ 0, then v −1 [I] = ⌀ ∈ G. If 0 ⩽ a < b, then v −1 [I] = (T\F) ∩ f −1 [I] = (H ∪ U) ∩ (O ∩ K) = ((H ∩ K) ∪ (U ∩ K)) ∩ O =

3.6.2 Extensions of 𝜎-exact functionals by the Young – Daniell method | 373

U ∩ O ∈ G. If a < 0 < b, then v −1 [I] = f −1 [I] ∪ H = (O ∩ K) ∪ H = (O ∪ H) ∩ (K ∪ H) = O∪H ∈ G. Thus, the function v ∈ C c (T, G) because its support contains in the compact set K. Besides, 0 ⩽ v ⩽ 1, v[C] = {1}, and v[H] = {0}. Hence, 𝜒(C) ⩽ v ⩽ 𝜒(G). Since v ∈ C c (T, G) and, according to Statement 5 (3.5.1), C ⊂ F, this implies that the family C c (T, G) envelopes from above the function 𝜒(C) ∧ v = 𝜒(C). The families C c (T, G) and C b (T, G) may not have property (E𝜎 ) but in the same time they have important property (D) (see Corollary 1 to Lemma 2 (3.6.1) and Theorem 1 (2.3.4)). The fact that a family A(T) possesses property (E𝜎 ) or property (E)&(D) is essential for representations of 𝜎-exact functionals on A(T) by Radon integrals in 3.6.3.

The Young – Daniell extensions for some initial domains of symmetrizable functions The abstract method of Young – Daniell is getting fruitful for initial domains A(T) lying in the remarkable lattice-ordered linear space S(T, G) of symmetrizable functions (see 2.4.5) and having the envelopment properties considered above. Proposition 3. Let ⟮T, G⟯ be a Hausdorff space, A(T) be a truncatable lattice-ordered linear space of functions on it, and 𝜑 be a positive linear functional on A(T). Then, 1) if A(T) has property (E) and 𝜑 is pointwise continuous, then the inclusion St(T, Am (T, G, A(T))) ⊂ D𝜏 (T, A(T), 𝜑) holds; 2) if A(T) has property (E𝜎 ) and 𝜑 is pointwise 𝜎-continuous, then the inclusion St(T, Am (T, G, A(T))) ⊂ D𝜎 (T, A(T), 𝜑) holds. Proof. 1) Let K ∈ Km (T, G, A(T)). Then, K = F ∩ G for some F ∈ F and G ∈ G. By definition, 𝜒(K) ⩽ u ∈ A(T). By the condition, there is a net (h n ∈ A(T) | n ∈ N) ↓ such that h ≡ 𝜒(F)∧u = inf(h n | n ∈ N). Therefore, h ∈ I𝜏 (T, A(T)) ⊂ D𝜏 (T, A(T), 𝜑). Also by condition there is a net (g m ∈ A(T) | m ∈ M) ↑ such that g ≡ 𝜒(G) ∧ u = sup(g m | m ∈ M) ∈ S𝜏 (T, A(T)) ⊂ D𝜏 (T, A(T), 𝜑). Then, 𝜒(K) = h ∧ g ∈ D𝜏 (T, A(T), 𝜑). Now, suppose A ∈ Am (T, G, A(T)). Then, A = ⋃ ⟮K i ∈ Km | i ∈ I⟯ implies 𝜒(A) = sup (𝜒(K i ) | i ∈ I) ∈ D𝜏 (T, A(T), 𝜑). Finally, if f ∈ St(T, Am ), then f = ∑ (x j 𝜒(A j ) | j ∈ J) for some finite collections ⟮x j ∈ R | j ∈ J⟯ and ⟮A j ∈ Am | j ∈ J⟯. From the property proven above and the linearity of D𝜏 (T, A(T), 𝜑), we conclude that f ∈ D𝜏 (T, A(T), 𝜑). 2. For the 𝜎-case, all the arguments are completely analogous. Proposition 4. Let ⟮T, G⟯ be a Hausdorff space, A(T) be a truncatable lattice-ordered linear space on T, and 𝜑 be a positive linear functional on it. Then, 1) if A(T) has property (E) and 𝜑 is pointwise continuous, then Sm (T, G, A(T))) ⊂ D𝜏 (T, A(T), 𝜑); 2) if A(T) has property (E𝜎 ) and 𝜑 is pointwise 𝜎-continuous, then Sm (T, G, A(T)) ⊂ D𝜎 (T, A(T), 𝜑).

374 | 3.6 Representation of a functional by the Radon integral

Proof. 1. Denote Sm (T, G, A(T)) by B(T) and D𝜏 (T, A(T), 𝜑) by D(T). Take any function f ∈ B(T)+ and a number 𝜀 > 0. By the definition, there is a function u ∈ A(T)+ such that f ⩽ u ∈ A(T)+ . According to Lemma 1 (2.2.9), u ∧ ((1/n)1) ∈ A(T) for every n ∈ N. It is clear that (u ∧ ((1/n)1) | n ∈ N) ↓ 0 in F(T). Since the functional 𝜑 is pointwise continuous, by Lemma 1 (2.2.8) we get lim (𝜑(u ∧ ((1/n)1)) | n ∈ N) = 0. Hence, there exists n ∈ N such that 𝜑(u ∧ ((1/n)1)) < 𝜀/3. Since f ∈ S(T, G), there is a finite cover (A i ∈ A | i ∈ I) of the set T such that 𝜔(f , A i ) < 1/n. Consider the numbers y i ≡ inf{f (t) | t ∈ A i } and z i ≡ sup{f (t) | t ∈ A i } and the sets J ≡ {i ∈ I | y i > 0} and A ≡ ⋃ ⟮A j | j ∈ J⟯ ∈ A. Consider the functions v j ≡ y j 𝜒(A j ) and the function v ≡ sup(h j | j ∈ J) ⩽ f ⩽ u. Then, v ∈ Stm (T, A, A(T)) = St(T, Am (A(T))) ⊂ D(T) by Lemma 11 (2.2.4) and Proposition 3. If t ∈ A, then t ∈ A j for some j ∈ J. Then, f (t)−v(t) ⩽ f (t)−v j (t) = f (t)−y j ⩽ 1/n. If t ∈ ̸ A, then t ∈ A i for some i such that y i = 0. Then, f (t) − v(t) = f (t) ⩽ z i − y i ⩽ 1/n. Hence, 0 ⩽ f − v ⩽ (1/n)1. Consider the function w ≡ v + u ∧ ((1/n)1). Since u ∧ ((1/n)1) ∈ A(T) and D(T) is linear, we get w ∈ D(T). Using Corollary 1 to Lemma 4 (1.4.5), we obtain w − f = v + u ∧ ((1/n)1)− f = v +(u − f )∧((1/n)1− f ) ⩾ v +0∧((1/n)1− f ) = v ∧(v − f +(1/n)1) ⩾ v ∧0 = 0 and w − f = v + u ∧ ((1/n)1) − f ⩽ u ∧ ((1/n)1) ⩽ (1/n)1. Since v, w ∈ D(T), there are g󸀠 , g󸀠󸀠 ∈ S𝜏 (T, A(T)) and h󸀠 , h󸀠󸀠 ∈ I𝜏 (T, A(T)) such that h󸀠 ⩽ v ⩽ g󸀠 , h󸀠󸀠 ⩽ w ⩽ g 󸀠󸀠 , 𝜑g󸀠 − 𝜑h󸀠 < 𝜀/3, and 𝜑g󸀠󸀠 − 𝜑h󸀠󸀠 < 𝜀/3. It is clear that ̌ 󸀠󸀠 − h󸀠 ⩽ f ⩽ g󸀠󸀠 . Besides, 𝜑g󸀠󸀠 − 𝜑h󸀠 = (𝜑g󸀠󸀠 − 𝜑h󸀠󸀠 ) + (𝜑h󸀠󸀠 − 𝜑g󸀠 ) + (𝜑g󸀠 − 𝜑h󸀠 ) < 2𝜀/3 + 𝜑h ̌ 󸀠 ⩽ 2𝜀/3 + 𝜑w ̌ − 𝜑v̌ = 2𝜀/3 + 𝜑(u ∧ ((1/n)1)) < 𝜀. Consequently, Theorem 1 guarantees 𝜑g that f ∈ D(T). If f is any function from B(T), then f = f1 − f2 for some f1 , f2 ∈ B(T)+ . Since by Theorem 1 D(T) is a linear space, this implies the required inclusion B(T) ⊂ D(T). 2. For the 𝜎-case, all the arguments are completely analogous. Thus, we can consider on the lattice-ordered linear space Sm (T, G, A(T)) ̌ m (T, G, A(T)) and 𝜑̂S ≡ (see Corollary 1 to Lemma 9 (2.2.4)) the functionals 𝜑Š ≡ 𝜑|S ̂ m (T, G, A(T)). In the important case A(T) ⊂ S(T, G), this space Sm (T, G, A(T)) is 𝜑|S an l-ideal in the lattice-ordered linear space S(T, G) generated by the family A(T) and these functionals are good extensions of the functional 𝜑. Proposition 5. Let ⟮T, G⟯ be a Hausdorff space, A(T) be a truncatable lattice-ordered linear subspace in S(T, G) with property (E) [(E𝜎 )], and 𝜑 be a positive linear pointwise continuous [𝜎-continuous] functional on A(T). Then, the functional 𝜑Š [𝜑̂S ] is a positive pointwise 𝜎-continuous extension of a functional 𝜑 on the family Sm (T, G, A(T)). Besides, if 𝜑 is uniformly (order) bounded, then 𝜑Š [𝜑̂S ] is uniformly (order) bounded. Proof. The main conclusion follows from Theorem 1 and Proposition 4.

3.6.2 Extensions of 𝜎-exact functionals by the Young – Daniell method | 375

Denote Sm (T, G, A(T)) by B(T). Suppose 𝜑 is uniformly (order) bounded, i. e. x ≡ sup{|𝜑f | | f ∈ A(T) ∧ |f | ⩽ 1} < ∞. Take any function g ∈ B(T) satisfying the condition |g| ⩽ 1. By the definition, |g| ⩽ u ∈ A(T)+ . Since the family A(T) is truncatable, we ̌ ⩽ 𝜑|g| ̌ ⩽ 𝜑f̌ = 𝜑f ⩽ x. This implies that sup{|𝜑g| ̌ |g∈ get f ≡ u ∧ 1 ∈ A(T)+ . Then, |𝜑g| B(T) ∧ |g| ⩽ 1} ⩽ x < ∞, i. e. 𝜑Š is uniformly (order) bounded on B(T). For the 𝜎-case, all the arguments are completely analogous. The Young – Daniell extensions of 𝜎-exact functionals for some initial domains of symmetrizable functions Consider the Young – Daniell extensions of exact and 𝜎-exact linear functionals from some subspaces A(T) of S(T, G) on the space Sm (T, G, A(T)). Theorem 2. Let ⟮T, G⟯ be a Hausdorff space, A(T) be a lattice-ordered linear subspace in S(T, G) with property (E)[(E𝜎 )], and 𝜑 positive pointwise continuous [𝜎-continuous] linear functional on A(T). The following conclusions are equivalent: 1) 𝜑 is locally tight on A(T); 2) for every set K ∈ Km (A(T)) and every 𝜀 > 0 there is C ∈ C such that C ⊂ K and 𝜑Š 𝜒(K\C) < 𝜀 [𝜑̂S 𝜒(K\C) < 𝜀]; 3) 𝜑Š [respectively, 𝜑̂S ] is quite locally tight on Sm (T, G, A(T)); 4) 𝜑 is quite locally tight on A(T). Proof. (1) ⊢ (2). Take any K ≡ F ∩ G ∈ Km ≡ Km (A(T)) and 𝜀 > 0. Then, there is u ∈ A(T) such that 𝜒(K) ⩽ u. The condition guarantees that there exisits a set B ∈ C such that B ⊂ G and |𝜑f | < 𝜀/2 for all f ∈ A(T) satisfying the inequality |f | ⩽ 𝜒(G\B) ∧ u. Note that 𝜒(G\B) ∈ S𝜏 (T, A(T)) as A(T) has property (E) and G\B ∈ G. Then, 𝜑(𝜒(G\B) ∧ u) ≡ sup{𝜑f | f ∈ A(T) ∧ f ⩽ 𝜒(G\B) ∧ u} ⩽ 𝜀/2 < 𝜀. For the compact set C ≡ F ∩ B ⊂ K, we get 𝜒(K\C) = |𝜒(F) ∧ 𝜒(G) − 𝜒(F) ∧ 𝜒(B)| = |𝜒(F) ∧ 𝜒(G) − 𝜒(F) ∧ 𝜒(B)| ∧ u ⩽ |𝜒(G)−𝜒(B)| ∧ u = 𝜒(G\B) ∧ u, where 𝜑̌S 𝜒(K\C) ⩽ 𝜑(𝜒(G\B) ∧ u) < 𝜀. (2) ⊢ (3). Lemma 10 (2.2.4) implies that B(T) ≡ Sm (T, G, A(T)) = Sm (T, G, B(T)) and Km (A(T)) = Km (B(T)). By virtue of Lemma 4 (3.6.1), the functional 𝜑Š is locally tight on B(T). Property (E) of the family A(T) implies that this family envelopes from above every function 𝜒(C) for C ∈ C. Thus, 𝜒(C) ∈ B(T), and therefore, C = Cm (B(T)). Finally, by Lemma 3 (3.6.1), we conclude that 𝜑Š is quite locally tight on B(T). (3) ⊢ (4). It is sufficient to note that 𝜑 = 𝜑Š |A(T). (4) ⊢ (1). This directly follows from the definitions. For the 𝜎-case, all the arguments are completely analogous. Lemma 6. Let T be a set, P ⊂ Q ⊂ T, and u ∈ F(T)+ . Then, 𝜒(Q\P) ∧ u = 𝜒(Q) ∧ u − 𝜒(P) ∧ u.

376 | 3.6 Representation of a functional by the Radon integral

Proof. Denote the left part of the desired equality by f and the right part by g. Denote 𝜒(P), 𝜒(Q), and 𝜒(Q\P) by p, q and r, respectively. Let t ∈ P. Then, p(t) = q(t) and r(t) = 0. If p(t) ⩽ u(t), then f (t) = r(t) ∧ u(t) = 0 = 1−1 = q(t)−p(t) = q(t) ∧ u(t)−p(t) ∧ u(t) = g(t). If p(t) > u(t) ⩾ 0, then f (t) = 0 = u(t) − u(t) = q(t) ∧ u(t) − p(t) ∧ u(t) = g(t). If t ∈ Q\P, then f (t) = 1 ∧ u(t) = 1 ∧ u(t)−0 ∧ u(t) = q(t) ∧ u(t)− p(t) ∧ u(t) = g(t). Finally, if t ∈ ̸ Q, then f (t) = 0 = 0 − 0 = q(t) ∧ u(t) − p(t) ∧ u(t) = g(t). Proposition 6. Let ⟮T, G⟯ be a Hausdorff space and A(T) be a truncated lattice-ordered linear subspace in S(T, G). Suppose A(T) has property (E𝜎 ) or property (E)&(D). Then, 1) for every positive 𝜎-exact linear functional 𝜑 on A(T) there exists a unique positive 𝜎-exact linear functional 𝜑S on Sm (T, G, A(T)) extending the functional 𝜑; 2) if 𝜑 is uniformly (order) bounded, then 𝜑S is uniformly (order) bounded; 3) the mapping P : 𝜑 󳨃→ 𝜑S is a monotone conic bijection from (A(T)󳵻 )+ onto (Sm (T, G, A(T))󳵻 )+ . Proof. 1. Put B(T) ≡ Sm (T, G, A(T)). If A(T) has property (E𝜎 ), then for every 𝜑 ∈ (A(T)󳵻 )+ put 𝜑S ≡ 𝜑̂S . If A(T) has property (E)&(D), then Corollary 1 to Lemma 2 (3.6.1) implies that A(T)󳵻 = A(T)󳶋 , and therefore, we can put 𝜑S ≡ 𝜑Š for every 𝜑 ∈ (A(T)󳵻 )+ . By virtue of Proposition 5 and Theorem 2, we get 𝜑S ∈ (B(T)󳵻 )+ . Let 𝜓 be is a positive 𝜎-exact linear functional on B(T) and 𝜓 be an extension of 𝜑. Prove that 𝜓 = 𝜑S . First, show that 𝜓(𝜒(G) ∧ u) = 𝜑S (𝜒(G) ∧ u) for all open sets G and functions u ∈ A(T)+ . Let G ∈ G, u ∈ A(T)+ , g ≡ 𝜒(G) ∧ u. For every 𝜀 > 0, there is C ∈ C such that 𝜓(𝜒(G\C) ∧ u) < 𝜀. Put h ≡ 𝜒(C) ∧ u. According to Lemma 6, we have 𝜓g − 𝜓h < 𝜀. By property (E) [(E𝜎 )] of the family A(T), there is a function v ∈ A(T) such that 𝜒(C) ⩽ v ⩽ 𝜒(G), hence h ⩽ f ⩽ g, where f ≡ v ∧ u ∈ A(T). Therefore, 𝜑f ⩽ 𝜓g < 𝜓h + 𝜀 ⩽ 𝜑f + 𝜀; this implies that 𝜓g = sup{𝜑f | f ∈ A(T)+ ∧ f ⩽ g} ≡ 𝜑g = 𝜑S g. Let F ∈ F, u ∈ A(T)+ , f ≡ 𝜒(F) ∧ u. Consider the open set G ≡ T\F. From the facts proven above and Lemma 6 we conclude that 𝜓f = 𝜓((1 − 𝜒(G)) ∧ u) = 𝜓u − 𝜓(𝜒(G) ∧ u) = 𝜑u−𝜑S (𝜒(G) ∧ u) = 𝜑S f . In particular, this implies that 𝜓 and 𝜑S coincide on functions of the form 𝜒(C) ∧ u, where C ∈ C and u ∈ A(T)+ . Take now A ∈ Am (A(T)) and u ∈ A(T)+ such that 𝜒(A) ⩽ u. By Lemma 4 (3.6.1) for every 𝜀 > 0 there are compact sets K ⊂ A and L ⊂ A such that 𝜑S 𝜒(A\K) < 𝜀/2 and 𝜓𝜒(A\L) < 𝜀/2. Consider the compact set C ≡ K ∪ L ⊂ A. Then, 𝜑S 𝜒(A\C) < 𝜀/2 and 𝜓𝜒(A\C) < 𝜀/2, because 𝜓 and 𝜑 are positive. Therefore, |𝜓𝜒(A) − 𝜑S 𝜒(A)| = |𝜓𝜒(A\C) + 𝜓𝜒(C) − 𝜑S 𝜒(A\C) − 𝜑S 𝜒(C)| ⩽ 𝜓𝜒(A\C) + 𝜑S 𝜒(A\C) + |𝜓(𝜒(C) ∧ u) − 𝜑S (𝜒(C) ∧ u)| < 𝜀. Since 𝜀 is arbitrary, we obtain 𝜓𝜒(A) = 𝜑S 𝜒(A). Hence 𝜓 and 𝜑S coincide on the family St(T, Am (A(T)). Finally, take f ∈ B(T) and a function u ∈ A(T)+ such that |f | ⩽ u. Proposition 2 (2.2.4) and Proposition 2 (2.4.3) guarantee that there exists increasing sequence (f n ∈ St(T, A) | n ∈ N) such that f = u-lim (f n | n ∈ N). It follows from the inequality

3.6.2 Extensions of 𝜎-exact functionals by the Young – Daniell method | 377

f n ⩽ f ⩽ u and Lemma 11 (2.2.4) that f n ∈ Stm (T, A, A(T)) = St(T, Am (A(T)) for every n ∈ N. The facts proven above and the 𝜎-continuity of functionals 𝜓 and 𝜑S imply that 𝜓f = lim (𝜓f n | n ∈ N) = lim (𝜑S f n | n ∈ N) = 𝜑S f . ̂ The functional 𝜑̌ [𝜑] ̂ is 2. If A(T) has property (E)&(D) [(E𝜎 )], then 𝜑S ≡ 𝜑̌ [𝜑S ≡ 𝜑]. uniformly bounded by virtue of Proposition 5. 3. If 𝜑1 ≠ 𝜑2 , then, obviously, (𝜑1 )S ≠ (𝜑2 )S . Thus, the mapping P is injective. Let 𝜓 ∈ (B(T)󳵻 )+ . Consider 𝜑 ≡ 𝜓|A(T) ∈ (A(T)󳵻 )+ . According to (1), we have 𝜓 = 𝜑S = P𝜑. Thus, P is surjective. Prove now that P is a conic mapping. Let a > 0. Denote a𝜑 by 𝜓. Take any g ∈ Y ≡ S𝜎 (T, A(T)) and 𝜀 > 0. Consider the set L g ≡ {f ∈ A(T) | f ⩽ g}. Then, 𝜑g = sup(𝜑f | f ∈ L g ) and 𝜓g = sup(𝜓f | f ∈ L g ) imply that there are u, v ∈ L g such that 𝜑g − 𝜀/(2a) < 𝜑u and 𝜓g − 𝜀/2 < 𝜓v. Consider the function w ≡ u ∨ v ∈ L g . Then, |(a𝜑)g − 𝜓g| ⩽ |a𝜑g − a𝜑w| + |𝜓w − 𝜓g| < a|𝜑g − 𝜑w| + 𝜀/2 < 𝜀. Since 𝜀 is arbitrary, we obtain that (a𝜑)g = 𝜓g. In the analogous way, we prove that (a𝜑)h = 𝜓h for every h ∈ Z ≡ I𝜎 (T, A(T)). Now, let f ∈ B(T) and 𝜀 > 0. Consider the set Z f ≡ {h ∈ Z | h ⩽ f }. Then, as above, ̂ = sup(𝜓h | h ∈ Z f ) imply that (a𝜑)f ̂ . ̂ = 𝜓f ̂ = 𝜑∧ f = sup(𝜑h | h ∈ Z f ) and 𝜓f 𝜑f Denote 𝜑1 + 𝜑2 by 𝜓. Let g ∈ Y and 𝜀 > 0. Then, 𝜑1 g = sup(𝜑1 f | f ∈ L g ), 𝜑2 g = sup(𝜑2 f | f ∈ L g ), and 𝜓g = sup(𝜓f | f ∈ L g ) imply that there are u, v, w ∈ L g such that 𝜑1 g − 𝜀/3 < 𝜑1 u, 𝜑2 g − 𝜀/3 < 𝜑2 v, and 𝜓g − 𝜀/3 < 𝜓w. Consider the function x ≡ u ∨ v ∨ w ∈ L g . Then, |(𝜑1 +𝜑2 )g−𝜓g| ⩽ |𝜑1 g−𝜑1 x|+|𝜑2 g −𝜑2 x|+|𝜓x−𝜓g| < 𝜀. Therefore, (𝜑1 + 𝜑2 )g = 𝜓g. In the same way, we prove that (𝜑1 + 𝜑2 )h = 𝜓h for all h ∈ Z. ̂1 f = (𝜑1 )∧ f = sup(𝜑1 h | h ∈ Z f ), 𝜑 ̂2 f = Suppose now f ∈ B(T) and 𝜀 > 0. Then, 𝜑 ̂ = sup(𝜓h | h ∈ Z f ). This implies, as above, that (̂ ̂2 )f = sup(𝜑2 h | h ∈ Z f ), and 𝜓f 𝜑1 + 𝜑 ̂ . 𝜓f Finally, prove that P is monotone. Let 𝜑, 𝜓 ∈ A(T) and 𝜑 ⩽ 𝜓. Take any g ∈ Y and consider the set L g ≡ {f ∈ A(T) | f ⩽ g}. Then, 𝜑g = sup(𝜑f | f ∈ L g ) and 𝜓g = sup(𝜓f | f ∈ L g ) imply that 𝜑g ⩽ 𝜓g. Take now a function f ∈ (Sm (T, G, A(T))󳵻 )+ and consider the set Y f ≡ {g ∈ Y | ̂ = inf(𝜓g | g ∈ Y f ) imply that 𝜑f ̂ = 𝜑∧ f = inf(𝜑g | g ∈ Y f ) and 𝜓f ̂ ⩽ g ⩾ f }. Then, 𝜑f ̂ 𝜓f . Prove now the main theorem about extension of 𝜎-exact linear functionals. Theorem 3. Let ⟮T, G⟯ be a Hausdorff space and A(T) be a truncated lattice-ordered linear subspace in S(T, G). Suppose A(T) has property (E𝜎 ) or property (E)&(D). Then, 1) for every 𝜎-exact linear functional 𝜑 on A(T) there exists a unique 𝜎-exact linear functional 𝜑S on Sm (T, G, A(T)) extending the functional 𝜑; 2) if 𝜑 uniformly (order) bounded, then 𝜑S is uniformly (order) bounded; 3) the mapping Q : 𝜑 󳨃→ 𝜑S is an isomorphism between the lattice-ordered linear spaces A(T)󳵻 and Sm (T, G, A(T))󳵻 ;

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4) the mapping Q b ≡ Q|A(T) 󳵻 is an isomorphism between the lattice-ordered linear spaces A(T) 󳵻 and Sm (T, G, A(T)) 󳵻 . Proof. 1 and 3. By virtue of Proposition 6, there is a conic bijective mapping P from (A(T)󳵻 )+ onto (Sm (T, G, A(T))󳵻 )+ . By Statement 1 (3.4.2), it has the unique linear injective extension Q : A(T)󳵻 Sm (T, G, A(T))󳵻 . Put 𝜑S ≡ Q𝜑 for every 𝜑 ∈ A(T)󳵻 . 󳵻 Let 𝜓 ∈ Sm (T, G, A(T)) . Consider 𝜃 ≡ 𝜓|A(T) ∈ A(T)󳵻 . Since the extension 𝜃S is unique, we get 𝜓 = 𝜃S ≡ Q𝜃. Hence, Q is surjective. Thus, the mapping Q is an isomorphism of the given linear spaces. Note that 𝜑S ≡ Q𝜑 = Q(𝜑+ + 𝜑− ) = P𝜑+ − P(−𝜑− ) = (𝜑+ )S − (−𝜑− )S , where 𝜑+ and −𝜑− are positive functionals from Corollary 3 to Statement 5 (2.2.8). Check now that 𝜑 ∧ 𝜓 = 0 implies Q𝜑 ∧ Q𝜓 = 0 for all 𝜑, 𝜓 ∈ A(T)󳵻 . According to Statement 2 (3.2.2), it will be sufficient to prove that Q is an isomorphism of the given lattice-ordered linear spaces. Suppose 𝜑, 𝜓 ∈ A(T)󳵻 and 𝜑 ∧ 𝜓 = 0. Then, 𝜑, 𝜓 ∈ (A(T)󳵻 )+ and Q𝜑 ∧ Q𝜓 = P𝜑 ∧ P𝜓. Since P is surjective, there is 𝜉 ∈ (A(T)󳵻 )+ such that P𝜉 = P𝜑 ∧ P𝜓 ∈ (Sm (T, G, A(T))󳵻 )+ .Then, P𝜉 ⩽ P𝜑and P𝜉 ⩽ P𝜓,whereweget𝜉 ⩽ 𝜑and𝜉 ⩽ 𝜓because P is injective and monotone. Therefore, 𝜉 ⩽ 𝜑 ∧ 𝜓 and 0 ⩽ P𝜑 ∧ P𝜓 = P𝜉 ⩽ P(𝜑 ∧ 𝜓) = 0. 2 and 4. If 𝜑 is uniformly bounded, then functionals 𝜑+ and 𝜑− are also By virtue of Corollary 3 to Statement 5 (2.2.8). Then, assertion 2 of Proposition 6 implies that their extensions (𝜑+ )S and (−𝜑− )S are uniformly bounded. Therefore, the functional 𝜑S = Q(𝜑+ + 𝜑− ) = (𝜑+ )S − (−𝜑− )S has the same property.

Exactness of bounded functionals on the classical space C c (T , G) on a locally compact space The families of supremal and infimal functions and the functionals 𝜑 and 𝜑 introduced in the beginning of this subsection help us to prove that bounded linear functionals on C c (T, G) for a locally compact topological space ⟮T, G⟯ are exact. Lemma 7. Let ⟮T, G⟯ be a locally compact space, 𝜑 be an (order) bounded linear functional on C c (T, G). Then 𝜑 is pointwise continuous. Proof. First, suppose that 𝜑 is positive. Let (f m ∈ C c (T, G)+ | m ∈ M) ↓ 0 in F(T). Fix m0 ∈ M and consider the function g ≡ f m0 and the compact set C ≡ supp g ≡ cl coz g. Proposition 2 implies that there exists a function h ∈ C c (T)+ such that h ⩾ 𝜒(C). Take 𝜀 > 0. Then, there is a > 0 such that a𝜑(h) < 𝜀. Consider open sets G m ≡ {t ∈ T | f m (t) < a}. Since (f m | m ∈ M) ↓ 0, then (G m | m ∈ M) ↑ T. Whence there is a finite cover G n | n ∈ N ⊂ M) of the set C. The directness of M imply that there is m ∈ M such that n ⩽ m for all n ∈ N. Therefore, G m ⊃ G n for any n. So, C ⊂ G m . Take l ∈ M such that l ⩾ m0 and l ⩾ m. Then, f l ⩽ g and f l ⩽ f m . The first inequality implies H ≡ coz f l ⊂ C ⊂ G m , the second inequality implies f l (t) ⩽ f m (t) < a = a𝜒(C)(t) ⩽ ah(t) for

3.6.2 Extensions of 𝜎-exact functionals by the Young – Daniell method | 379

every point t ∈ H. Hence, it follows from f l ⩽ ah that 𝜑f l ⩽ a𝜑(h) < 𝜀. If k < l, then 𝜑f k ⩽ 𝜑f l < 𝜀. Thus, (𝜑f m | m ∈ M) ↓ 0, and Lemma 1 (2.2.8) implies that 𝜑 is pointwise continuous. Finally, if 𝜑 ∈ C∼c , then Corollary 3 to Statement 5 (2.2.8) guarantees that 𝜑 = 𝜑+ + 𝜑− , where 𝜑+ and −𝜑− are bounded and positive. Therefore, the functionals 𝜑+ and −𝜑− are pointwise continuous. Then, by Proposition 1 (2.2.8) the functional 𝜑 is also pointwise continuous. Proposition 7. Let ⟮T, G⟯ be a locally compact space and 𝜑 be an (order) bounded linear functional on C c (T, G). Then, 𝜑 is locally tight. Proof. Denote C c (T, G) by A. First, suppose that 𝜑 is positive. Let G ∈ G, u ∈ A + , and 𝜀 > 0. Consider the set M ≡ {f ∈ A | 0 ⩽ f ⩽ 𝜒(G)}. Then, (f | f ∈ M) ↑ 𝜒(G) implies that (f ∧ u | f ∈ M) ↑ 𝜒(G) ∧ u, hence 𝜒(G) ∧ u ∈ S𝜏 (T, A). Therefore, (𝜑(f ∧ u) | f ∈ M) ↑ 𝜑(𝜒(G) ∧ u) and there exists f ∈ M such that 𝜑(𝜒(G) ∧ u) − 𝜀/2 < 𝜑(f ∧ u). Define functions f k ≡ (f − k1 1) ∨ 0. Since (f k ∧ u | k ∈ N) ↑ f ∧ u and, according to Lemma 7, the functional 𝜑 is pointwise continuous, we conclude that (𝜑(f k ∧ u) | k ∈ N) ↑ 𝜑(f ∧ u). Whence there is k ∈ N such that 𝜑(f ∧ u)−𝜀/2 < 𝜑(f k ∧ u). Consider the closed set C ≡ supp f k ≡ cl coz f k . It follows from f k ⩽ f ⩽ 𝜒(G) ⩽ 1 and C ⊂ {t ∈ T | f (t) ⩾ 1/k} ⊂ coz f that f k ⩽ 𝜒(C), C ⊂ G, and C is compact. Since by Proposition 2 the family A envelopes from above the function 𝜒(C), we infer that 𝜒(C) ∈ I𝜏 (T, A). This implies 𝜒(C) ∧ u ∈ I𝜏 (T, A). Therefore, 𝜑(f k ∧ u) ⩽ 𝜑(𝜒(C) ∧ u). As a result, we obtain 𝜑(f ∧ u) − 𝜀/2 < 𝜑(𝜒(C) ∧ u). This gives 𝜑(𝜒(G) ∧ u) − 𝜑(𝜒(C) ∧ u) = 𝜑(𝜒(G) ∧ u) − 𝜑(f ∧ u) + 𝜑(f ∧ u) − 𝜑(𝜒(C) ∧ u) < 𝜀/2 + 𝜀/2 = 𝜀. ̌ ̌ − 𝜒(C)) ∧ u) = 𝜑(𝜒(G) ∧u − Using Lemma 6, we obtain 𝜑(𝜒(G\C) ∧ u) = 𝜑((𝜒(G) 𝜒(C) ∧ u) = 𝜑(𝜒(G) ∧ u) − 𝜑(𝜒(C) ∧ u) < 𝜀. If f ∈ A and |f | ⩽ 𝜒(G\C) ∧ u, then |𝜑f | ⩽ 𝜑|f | = 𝜑|f | ⩽ 𝜑(𝜒(G\C) ∧ u) < 𝜀. This means that 𝜑 is locally tight. Finally, if 𝜑 ∈ A∼ , then by Corollary 3 to Statement 5 (2.2.8) 𝜑 = 𝜑+ + 𝜑− , where 𝜑+ , −𝜑− are bounded and positive. Therefore, 𝜑+ , −𝜑− are locally tight. Then, by Proposition 1 (3.6.1) the functional 𝜑 is also locally tight. Theorem 4. Let ⟮T, G⟯ be a locally compact topological space and 𝜑 be an (order) bounded linear functional on C c (T, G). Then, 𝜑 is exact. Proof. The functional 𝜑 is pointwise continuous by Lemma 7 and locally tight by Proposition 7. We should only prove that 𝜑 is quite locally tight. According to Proposition 1 (3.6.1) and Proposition 1 (2.2.8), 𝜑 = 𝜑1 − 𝜑2 , where 𝜑1 and 𝜑2 are positive, locally tight, and pointwise continuous. Since C c (T, G) is a truncatable lattice-ordered linear subspace in S(T, G) and, by Proposition 2, has property (E), Theorem 2 implies that the functionals 𝜑1 and 𝜑2 are quite locally tight. Therefore, the functional 𝜑 is also quite locally tight.

380 | 3.6 Representation of a functional by the Radon integral

Corollary 1. Let ⟮T, G⟯ be a locally compact space. Then, C c (T, G)∼ = C c (T, G)󳶋 = C c (T, G)󳵻 and C c (T, G) ∼ = C c (T, G) 󳶋 = C c (T, G) 󳵻 = ⟮C c (T, G), ‖ ⋅ ‖u ⟯󸀠 . Proof. The equality C c (T, G)󳶋 = C c (T, G)󳵻 is valid by virtue of Corollary 1 to Lemma 2 (3.6.1). The inclusion C c (T, G)∼ ⊂ C c (T, G)󳶋 is fulfilled by Theorem 4. Lemma 2 (2.2.8) provides the inverse inclusion. Since C c (T, G) is a lattice-ordered linear subspace of F b (T), Corollary 1 to Lemma 3 (2.2.8) implies that C c (T, G) 󳵻 = C c (T, G) 󳶋 = C c (T, G) ∼ = ⟮C c (T, G), ‖ ⋅ ‖u ⟯󸀠 . Exactness of tight functionals on the classical space C b (T , G) on a Tychonoff space For a Tychonoff topological space ⟮T, G⟯, tight linear functionals turn out to be exact on C b (T, G). Lemma 8. Let ⟮T, G⟯ be a Tychonoff space, 𝜑 be a tight linear functional on C b (T, G). Then 𝜑 is pointwise continuous. Proof. Denote C b (T, G) by A. First, suppose that 𝜑 is positive. Let (f m ∈ A | m ∈ M) ↓ 0 in F(T). If 𝜑f m = 0 for all m ∈ M, then it is trivial that lim (𝜑f m | m ∈ M) = 0. If there is a number m0 such that 𝜑f m0 > 0, then ‖f m0 ‖u > 0 and 𝜑1 > 0. Consider M0 ≡ {m ∈ M | m ⩾ m0 }. It is clear that (f m | m ∈ M0 ) ↓ 0. Put g m ≡ f m / ‖f m0 ‖u . We have 0 ⩽ g m ⩽ g m0 ⩽ 1 for every m ∈ M0 and (g m | m ∈ M0 ) ↓ 0. For 𝜀 > 0, consider a compact set K such that the conditions f ∈ A and |f | ⩽ 𝜒(T\K) imply |𝜑f | < 𝜀/2. Take 𝛿 ≡ 𝜀/(2𝜑1) and consider open sets G m ≡ {t ∈ T | g m (t) < 𝛿}. Since (g m | m ∈ M0 ) ↓ 0, we get ⋃ (G m | m ∈ M0 ) = T. Consider a finite cover (G n | n ∈ N ⊂ M0 ) of K. For this N, there is m such that m ⩾ n for all n ∈ N. If t ∈ K, then t ∈ G n for some n. Therefore, g m (t) ⩽ g n (t) < 𝛿. Consider the continuous function h m ≡ (g m − 𝛿1) ∨ 0 ⩽ 1. It follows from 0 ⩽ h m ⩽ 𝜒(T\K) that 𝜑g m − 𝛿𝜑1 ⩽ 𝜑h m < 𝜀/2. Hence, 𝜑g m < 𝜀/2 + 𝜀/2 = 𝜀. Therefore, (𝜑g m | m ∈ M0 ) ↓ 0. This implies that (𝜑f m | m ∈ M0 ) ↓ 0 and, finally, (𝜑f m | m ∈ M) ↓ 0. Using Proposition 1 (3.6.1) and Proposition 1 (2.2.8), we complete the proof. Proposition 8. Let ⟮T, G⟯ be a Tychonoff space and 𝜑 be a tight linear functional on C b (T, G). Then, 𝜑 is locally tight. Proof. Denote C b (T, G) by A. First, assume that 𝜑 is positive. Take any G ∈ G, u ∈ A + , and 𝜀 > 0. Then, there is a compact set C0 such that f ∈ A and |f | ⩽ 𝜒(T\C0 ) implies |𝜑f | < 𝜀/2. According to Proposition 2, A envelopes from below the function 𝜒(T\C0 ) = 𝜒(T\C0 ) ∧ 1, therefore 𝜒(T\C0 ) ∈ S𝜏 (T, C b ). This implies 𝜑(𝜒(T\C0 )) ⩽ 𝜀/2. Consider the set M ≡ {f ∈ C b | 0 ⩽ f ⩽ 𝜒(G)}. Then, (f | f ∈ M) ↑ 𝜒(G) implies (f ∧ u | f ∈ M) ↑ 𝜒(G) ∧ u. Hence, 𝜒(G) ∧ u ∈ S𝜏 (T, C b ). Consequently, (𝜑(f ∧ u) | f ∈ M) ↑ 𝜑(𝜒(G) ∧ u). Therefore, there is f ∈ M such that 𝜑(𝜒(G) ∧ u) − 𝜀/4 < 𝜑(f ∧ u).

3.6.3 Characterizations of Radon integrals with respect to positive Radon measures | 381

Consider the functions f k ≡ (f − k1 1) ∨ 0. Since (f k ∧ u | k ∈ N) ↑ f ∧ u and by Lemma 8 𝜑 is pointwise continuous, we infer that (𝜑(f k ∧ u) | k ∈ N) ↑ 𝜑(f ∧ u). Then, there is k ∈ N such that 𝜑(f ∧ u) − 𝜀/4 < 𝜑(f k ∧ u). Consider the closed set F k ≡ supp f k . It follows from f k ⩽ 1 and F k ⊂ {t ∈ T | f (t) ⩾ 1/k} ⊂ coz f that f k ⩽ 𝜒(F k ) and F k ⊂ G. Since A envelopes from above the function 𝜒(F k ), we infer that 𝜒(F k ) ∈ I𝜏 (T, A), so 𝜒(F k ) ∧ u ∈ I𝜏 (T, A). Therefore, 𝜑(f k ∧ u) ⩽ 𝜑(𝜒(F k ) ∧ u). As a result, 𝜑(f ∧ u) − 𝜀/4 < 𝜑(𝜒(F k ) ∧ u). This gives 𝜑(𝜒(G) ∧ u) − 𝜑(𝜒(F k ) ∧ u) = 𝜑(𝜒(G) ∧ u) − 𝜑(f ∧ u) + 𝜑(f ∧ u) − 𝜑(𝜒(F k ) ∧ u) < 𝜀/4 + 𝜀/4 = 𝜀/2. Consider the compact set C ≡ C0 ∩ F k . Then, 𝜒(F k ) ∧ u = (𝜒(C) + 𝜒(F k \C0 )) ∧ u ⩽ ̌ 𝜒(C) ∧ u+𝜒(F k \C0 ) ∧ u ⩽ 𝜒(C) ∧ u+𝜒(T\C0 ). This implies 𝜑(𝜒(F k ) ∧ u) = 𝜑(𝜒(F k ) ∧ u) ⩽ ̌ ̌ )) = 𝜑(𝜒(C) ∧ u) + 𝜑(𝜒(T\C )) < 𝜑(𝜒(C) ∧ u) + 𝜀/2. As a conse𝜑(𝜒(C) ∧ u) + 𝜑(𝜒(T\C 0 0 quence, we obtain 𝜑(𝜒(G) ∧ u)−𝜑(𝜒(C) ∧ u) = 𝜑(𝜒(G) ∧ u)−𝜑(𝜒(F k ) ∧ u)+𝜑(𝜒(F k ) ∧ u)− 𝜑(𝜒(C) ∧ u) < 𝜀/2 + 𝜀/2 = 𝜀. Using Lemma 6 as in the proof of Proposition 7, we conclude that 𝜑 is locally tight. Using Proposition 1 (3.6.1), we complete the proof. Theorem 5. Let ⟮T, G⟯ be a Tychonoff space and 𝜑 be a tight linear functional on C b (T, G). Then, 𝜑 is exact. Proof. This proof is quite analogous to the proof of Theorem 4. We should only use Lemma 8 and Proposition 8 instead of Lemma 7 and Proposition 7. Corollary 1. Let ⟮T, G⟯ be a Tychonoff space. Then, C b (T, G)𝜋 = C b (T, G)󳶋 = C b (T, G)󳵻 = C b (T, G) 𝜋 = C b (T, G) 󳶋 = C b (T, G) 󳵻 . Proof. The equality C c (T, G)󳶋 = C c (T, G)󳵻 is valid by virtue of Corollary 1 to Lemma 2 (3.6.1). The imbedding C b (T, G)𝜋 ⊂ C b (T, G)󳶋 fulfilled by Theorem 5. Show that exact functionals are tight. If 𝜑 is exact, then for G ≡ T, u ≡ 1, and 𝜀 > 0 there is a compact set C ⊂ G such that the conditions f ∈ C b and |f | ⩽ 𝜒(T\C) = 𝜒(G\C) ∧ u imply |𝜑f | < 𝜀. Thus, 𝜑 is tight. Prove now that all functionals belonging to C b (T, G)𝜋 = C b (T, G)󳶋 = C b (T, G)󳵻 are uniformly bounded. If 𝜑 ∈ (C b (T, G)󳵻 )+ , then for every function f ∈ C b (T, G) such that |f | ⩽ 1 ∈ C b (T, G), we have |𝜑f | ⩽ 𝜑|f | ⩽ 𝜑1 < ∞, i. e. 𝜑 is uniformly bounded. Using Corollary 3 to Statement 5 (2.2.8), we conclude that an arbitrary functional 𝜑 ∈ C b (T, G)󳵻 is uniformly bounded. 3.6.3 Characterizations of Radon integrals with respect to positive Radon measures on a Hausdorff space as linear functionals Integral representations of a 𝜎-exact linear functional Now, we can prove the first result about representations of functionals by Radon integrals.

382 | 3.6 Representation of a functional by the Radon integral

Let ⟮T, G⟯ be a Hausdorff topological space, A(T) be a truncatable latticeordered linear subspace in the lattice-ordered linear space of symmetrizable functions S(T, G) possessing property (E 𝜎 ) or property (E)&(D). Consider the truncatable lattice-ordered linear subspace B(T) ≡ Sm (T, G, A(T)) in S(T, G), the mapping V ∧ : (B(T)∧ )+ → Mw (T)0 (see 3.4.2), and its restriction V 󳵻 ≡ V ∧ |(B(T)󳵻 )+ . Remember also the bijection P : (A(T)󳵻 )+ (B(T)󳵻 )+ from Proposition 6 (3.6.2) and define the 󳵻 󳵻 mapping R ≡ V ∘ P : (A(T) )+ → Mw (T)0 . Theorem 1. Let ⟮T, G⟯ be a Hausdorff space, A(T) be a truncatable lattice-ordered linear subspace in S(T, G) with property (E𝜎 ) or property (E)&(D). Let 𝜑 ∈ (A(T)󳵻 )+ and 𝜇 ≡ U𝜑 : M → R+ . Then, 1) I𝜎 (B(T)) ⊂ Am (T, G, A(T))𝛿 ; 2) 𝜇 ∈ RMwe (T, G)0 ; 3) 𝜇C = inf{𝜑f | f ∈ A(T) ∧ f ⩾ 𝜒(C)} for every C ∈ C; 4) the functional 𝜑 is representable by the Lebesgue integral over the measure space ⟮T, M, 𝜇⟯; 5) if 𝜑 is uniformly bounded, then 𝜇 is bounded; 6) the measure 𝜇 is unique in the following sense: if there is another measure 𝜈 : N → R+ in RMwe (T, G) such that 𝜈C = inf{𝜑f | f ∈ A(T) ∧ f ⩾ 𝜒(C)} for every C ∈ C and 𝜑 is representable by the Lebesgue integral over ⟮T, N, 𝜈⟯, then N = M and 𝜈 = 𝜇. Proof. According to Proposition 6 (3.6.2) there exists the functional 𝜑S ≡ P𝜑 ∈ (B(T)󳵻 )+ extending the functional 𝜑. Consider the ensemble R ≡ I𝜎 (B(T)) of all sets R ⊂ T such that the family B(T) 𝜎-envelope from above the function 𝜒(R). By Theorem 1 (3.4.2) the measure 𝜇 ≡ V ∧ 𝜑S ≡ R𝜑 is positive, wide, complete, strongly saturated, inner R-regular, and upper R-𝜎-continuous; besides, R ⊂ Mf (𝜇) and the functional 𝜑S is representable by the Lebesgue integral over ⟮T, M, 𝜇⟯. Note that 𝜇R ≡ inf{𝜑S f | f ∈ B(T)+ ∧ f ⩾ 𝜒(R)} according to the definition of the measure 𝜇 and 𝜇M = sup{𝜇R | R ∈ R ∧ R ⊂ M} By virtue of inner R-regularity of 𝜇. Since 𝜒(C) ∈ B(T) for every compact set C, we conclude that C ⊂ R. Therefore, 𝜇 is finite on compact sets. 1. Notice that for every f ∈ B(T) and every F ∈ F, we have f ∧ 𝜒(F) ∈ B(T). Indeed, first, 𝜒(F) ∈ S(T, G), where f ∧ 𝜒(F) ∈ S(T, G), secondly, |f | ⩽ u ∈ A(T)+ , where f ∧ 𝜒(F) = f ∧ u ∧ 𝜒(F), but the family A(T) envelopes from above any functions of the form u ∧ 𝜒(F); therefore, u ∧ 𝜒(F) ⩽ v ∈ A(T)+ , as a result, |f ∧ 𝜒(F)| = |f | ∧ |u ∧ 𝜒(F)| ⩽ u ∧ v ∈ A(T)+ , i. e. f ∧ 𝜒(F) ∈ B(T). Let F ∈ F and R ∈ R. Then, there is a sequence (f n ∈ B(T)+ | n ∈ 𝜔) such that (f n | n ∈ 𝜔) ↓ 𝜒(R). Consider the functions g n ≡ f n ∧ 𝜒(F) ∈ B(T)+ . It follows from (g n | n ∈ 𝜔) ↓ 𝜒(R∩F) and the definition of R that R∩F ∈ R ⊂ M. Since the measure 𝜇 is R-saturated, we obtain that F ∈ M. Thus, F ⊂ M. Therefore, G ⊂ M, because M is a 𝜎-algebra.

3.6.3 Characterizations of Radon integrals with respect to positive Radon measures | 383

Let R ∈ R and (f n ∈ B(T)+ | n ∈ 𝜔) ↓ 𝜒(R). Consider the sets Λ n ≡ {t ∈ T|f n (t) ⩾ 1}. Then, (Λ n | n ∈ 𝜔) ↓ R in P(T). Since f n is symmetrizable, there is a finite cover (B nj ∈ A(T, G) | j ∈ J n ) of the set T such that 𝜔(f n , B nj ) < 1/(n + 1). Consider the sets J n󸀠 ≡ {j ∈ J n | B nj ∩ Λ n = / ⌀} and B n ≡ ⋃ ⟮B nj | j ∈ J n󸀠 ⟯ ∈ A(T, G). If t ∈ B n , then t ∈ B nj for some j ∈ J n󸀠 . Take a point s ∈ B nj ∩Λ n . Then, |f n (t)−f n (s)| < 1/(n + 1) implies that f n (t) > f n (s) − 1/(n + 1) ⩾ n/(n + 1). Hence, Λ n ⊂ B n ⊂ Λ󸀠n ≡ {t ∈ T|(1 + 1/n)f n (t) > 1}. By the definition, of the family B(T), we have f n ⩽ u n for some u n ∈ A(T). Therefore, 𝜒(B n ) ⩽ (1 + 1/n)f n ⩽ (1 + 1/n)u n ∈ A(T) implies B n ∈ Am (T, G, A(T)) = Am (T, G, B(T)). If t ∈ ̸ R, then f n (t) < 1/2 for some n ⩾ 1. Therefore, t ∈ ̸ Λ󸀠n , where ⋂ ⟮Λ󸀠n | n ∈ 𝜔⟯ ⊂ R. As a result, from R = ⋂ ⟮Λ n | n ∈ 𝜔⟯ ⊂ ⋂ ⟮B n | n ∈ 𝜔⟯ ⊂ ⋂ ⟮Λ󸀠n | n ∈ 𝜔⟯ ⊂ R we conclude that R = ⋂ ⟮B n | n ∈ 𝜔⟯ ∈ Am (T, G, A(T))𝛿 . 2. Consider the sets A n ≡ ⋂ ⟮B m | m ∈ n + 1⟯ ∈ Am (T, G, A(T)); then we have (A n | n ∈ 𝜔) ↓ R. By Lemma 10 (2.2.4), B(T) = Sm (T, G, B(T)); therefore, by irtue of Proposition 4 (3.6.1) for the sequence (A n ∈ Am (T, G, A(T)) | n ∈ 𝜔) and for every number 𝜀 > 0 there is n0 ∈ 𝜔 and a compact set C ⊂ ⋂ ⟮A n | n ∈ 𝜔⟯ such that it follows from f ∈ B(T) and |f | ⩽ 𝜒(A n0 \C) ≡ g that |𝜑S f | < 𝜀. Since g ∈ B(T), we infer that 𝜑S g < 𝜀. Then, 𝜇R − 𝜇C ⩽ 𝜇A n0 − 𝜇C = 𝜇(A n0 \C) = ∫ g d𝜇 = 𝜑S g < 𝜀. Therefore, 𝜇R = sup{𝜇C | C ∈ C ∧ C ⊂ R}. Suppose M ∈ M and x is a number such that x < 𝜇M. Then, by virtue of inner R-regularity of the measure 𝜇, there is R ∈ R such that R ⊂ M and x < 𝜇R. Further, there is C ∈ C such that C ⊂ R and x < 𝜇C. Thus, x ⩽ sup{𝜇C | C ∈ C ∧ C ⊂ M} ⩽ 𝜇M. Since x is arbitrary, we infer that 𝜇M = sup{𝜇C | C ∈ C ∧ C ⊂ M}. This means that 𝜇 is compactly regular. Thus, we have established that 𝜇 is a wide positive Radon measure. 3. Let A(T) has property (E). Take an arbitrary compact set C. By virtue of property (E) of the space A(T), there exists a decreasing net (f n ∈ A(T) | n ∈ N) such that (f n | n ∈ N) ↓ 𝜒(C) in F(T). Then, 𝜒(C) ∈ I𝜏 (T, A(T)) ∩ B(T) and 𝜇C = ∫ 𝜒(C) d𝜇 = 𝜑Š 𝜒(C) = 𝜑𝜒(C) = inf(𝜑f n | n ∈ N). This implies that for every 𝜀 > 0 there is n such that 𝜇C + 𝜀 > 𝜑f n . If f ∈ A(T) and f ⩾ 𝜒(C), then 𝜇C = 𝜑Š 𝜒(C) ⩽ 𝜑Š f = 𝜑f . Therefore, 𝜇C = inf{𝜑f | f ∈ A(T) ∧ f ⩾ 𝜒(C)}. When A(T) has property (E𝜎 ) all the arguments are quite similar. 4. If f ∈ A(T), then 𝜑f = 𝜑S f = ∫ f d𝜇. Therefore, the functional 𝜑 is representable by Lebesgue integral over the measurable space ⟮T, M, 𝜇⟯. 5. Since 𝜑 is uniform bounded and 𝜇 is compactly regular we get 𝜇T = sup{𝜑𝜒(C) | C ∈ C} ⩽ sup{𝜑f | f ∈ A(T) ∧ 0 ⩽ f ⩽ 1} < ∞, i. e. 𝜇 is bounded. 6. Now, we should prove the uniqueness of 𝜇. Take any C ∈ C. Then, 𝜇C = inf{𝜑f | f ∈ A(T) ∧ f ⩾ 𝜒(C)} = 𝜈C. Let N ∈ N and C ∈ C. By the property of compact regularity of 𝜈 we can find some sequences (K n ⊂ C ∩ N | n ∈ N) ↑ and (L n ⊂ C\N | n ∈ N) ↑ of compact sets such

384 | 3.6 Representation of a functional by the Radon integral

that 𝜈(C∩N) = sup{𝜈K n | n ∈ N} and 𝜈(C\N) = sup{𝜈L n | n ∈ N}. Consider disjoint sets K ≡ ⋃ ⟮K n | n ∈ N⟯ ⊂ C ∩ N and L ≡ ⋃ ⟮L n | n ∈ N⟯ ⊂ C\N from M. We obtain 𝜈 ((C\K m )\L n ) = 𝜈(C\K m ) − 𝜈 ((C\K m ) ∩ L n ) = 𝜈C − 𝜈(C ∩ K m ) − 𝜈L n = 𝜈C − 𝜈K m − 𝜈L n . Whence 𝜇K = sup (𝜇K n | n ∈ N) = sup (𝜈K n | n ∈ N) = 𝜈(C ∩ N) and 𝜇L = sup (𝜇L n | n ∈ N) = sup (𝜈L n | n ∈ N) = 𝜈(C\N). Therefore, 𝜇(K ∪ L) = 𝜇K + 𝜇L = 𝜈(C ∩ N) + 𝜈(C\N) = 𝜈C. As a result, we get 𝜈C = 𝜇C = 𝜇 (C\(K ∪ L)) + 𝜇(K ∪ L) = 𝜇 (C\(K ∪ L)) + 𝜈C, what implies 𝜇 (C\(K ∪ L)) = 0. From (C ∩ N)\K ⊂ C\(K ∪ L) by completeness of 𝜇 we conclude that (C ∩ N)\K ∈ M. Therefore, N ∩ C = ((C ∩ N)\K) ∪ K ∈ M for every C ∈ C. Consider a set M ∈ Mf (𝜇). Since 𝜇 is compactly regular, we have 𝜇M = sup{𝜇D | D ∈ C ∧ D ⊂ M}. Then, there is a sequence (D n ∈ C | n ∈ N) ↑ such that ⋃ ⟮D n | n ∈ N⟯ ⊂ M and 𝜇M = sup{𝜇D n | n ∈ N}. Consider the sets E ≡ ⋃ ⟮D n | n ∈ N⟯ and H ≡ M\E from M. Since 𝜇D n ⩽ 𝜇E = 𝜇M − 𝜇H for every n, we get 𝜇M ⩽ 𝜇M − 𝜇H. Hence, 𝜇H = 0. By virtue of completeness of 𝜇, it follows from (N ∩ M)\E ⊂ H that (N ∩ M)\E ∈ M. On the other hand, the facts proven above imply (N ∩ M)∩ E = ⋃ ⟮(N ∩ D n )∩ M | n ∈ N⟯ ∈ M. As a result, N ∩ M ∈ M. The measure 𝜇 is strong saturated; therefore, N ∈ M. Thus, N ⊂ M. In the same way, we show that M ⊂ N. Now, compact regularity of 𝜇 and 𝜈 implies 𝜇 = 𝜈. For every subfamily A(T) in S(T, G), we shall consider the sets RMwe (T, G, A(T))0 ≡ {𝜇 ∈ RMwe (T, G)0 | A(T) ⊂ MI(T, dom 𝜇, 𝜇)} and RMw⋆ (T, G, A(T))0 ≡ {𝜇 ∈ RMw⋆ (T, G)0 | A(T) ⊂ MI(T, B(T, G), 𝜇)} of all positive Radon and Borel – Radon measures 𝜇 such that all functions from A(T) are 𝜇-integrable. Denote the mapping R|(A(T) 󳵻 )+ by R b . Corollary 1. Let ⟮T, G⟯ be a Hausdorff space and A(T) be a truncatable lattice-ordered linear subspace in S(T, G). 1) vA(T) has property (E𝜎 ), then the mapping R is a bijection of (A(T)󳵻 )+ onto RMwe (T, G, A(T))0 and R b is a bijection of (A(T) 󳵻 )+ onto RMwe b (T, G)+ . 2) If A(T) has property (E)&(D), then the mapping R is a bijection of (A(T)󳶋 )+ = (A(T)󳵻 )+ onto RMwe (T, G, A(T))0 and R b is a bijection of (A(T) 󳶋 )+ = (A(T) 󳵻 )+ onto RMwe b (T, G)+ . Proof. Since every function from A(T) is integrable with respect to every measure from RMwe (T, G, A(T))0 , the mappings R and R b are injective directly by assertion 4 of Theorem 1. By Corollary 1 to Lemma 2 (3.6.1), property (D) of the family A(T) possesses the equality A(T)󳶋 = A(T)󳵻 . Thus, we only should prove that R : (A(T)󳵻 )+ 󴀚󴀠 RMwe (T, G, A(T))0 is surjective.

3.6.3 Characterizations of Radon integrals with respect to positive Radon measures | 385

For each measure 𝜇 ∈ RMwe (T, G, A(T))0 , the integral functional i𝜇 is 𝜎-exact on A(T) ⊂ SI(T, M, 𝜇) by virtue of Proposition 5 (3.6.1). According to Lemma 1 (3.5.2), we see that S c (T, G) ⊂ SI(T, M, 𝜇). Therefore, 𝜒(C) ∈ SI(T, M, 𝜇) for every C ∈ C. Since A(T) envelopes [𝜎-envelopes] from above 𝜒(C), by Theorem 4 (3.3.3) we obtain that inf{i𝜇 f | f ∈ A(T) ∧ f ⩾ 𝜒(C)} = ∫ 𝜒(C) d𝜇 = 𝜇C. Now, by assertion 6 of Theorem 1, we get 𝜇 = Ui 𝜇 . Lemma 1 (3.5.2) implies that every function from A(T) ⊂ S(T, G) is intewe grable with respect to every measure from RMwe b (T, G), where RMb (T, G, A(T)) = we RMb (T, G). If one takes two different functionals from (A(T)󳵻 )+ , then by Theorem 1, there is an own measurable space for every functional such that every functional is representable by the Lebesgue integral over its space. Therefore, it is important to consider for all these functionals an integral representation over one and the same measurable space. For this purpose, consider the Borel 𝜎-algebra B ≡ B(T, G) and the mappingsS : (A(T)󳵻 )+ → Meas(T, B)0 and S b : (A(T) 󳵻 )+ → Measb (T, B)+ defined by S𝜑 ≡ (R𝜑)|B and S b 𝜑 ≡ (R b 𝜑)|B. Theorem 2. Let ⟮T, G⟯ be a Hausdorff space, A(T) be a truncatable lattice-ordered linear subspace in S(T, G) with property (E𝜎 ) or property (E)&(D), let 𝜑 ∈ (A(T)󳵻 )+ , 𝜇 ≡ R𝜑 : M → R+ , and 𝜇0 ≡ S𝜑 : B → R+ . Then, 1) 𝜇0 ∈ RMw⋆ (T, G)0 , 𝜇0 = 𝜇|B, and 𝜇 = (𝜇0 )#; 2) 𝜇0 C = inf{𝜑f | f ∈ A(T) ∧ f ⩾ 𝜒(C)} for every C ∈ C; 3) the functional 𝜑 is representable by the Lebesgue integral over the measure space ⟮T, B, 𝜇0 ⟯; 4) if 𝜑 is uniformly bounded, then 𝜇0 is bounded; 5) the measure 𝜇0 is unique in the following sense: if there is another measure 𝜘 ∈ RMw⋆ (T, G)0 such that 𝜘C = inf{𝜑f | f ∈ A(T) ∧ f ⩾ 𝜒(C)} for every C ∈ C and 𝜑 is representable by the Lebesgue integral over the measure space ⟮T, B, 𝜘⟯, then 𝜘 = 𝜇0 . Proof. Denote Sm (T, G, A(T)) by B(T) and V ∧ |(B(T)󳵻 )+ by V 󳵻 . It follows from the definitions that 𝜇0 ≡ S𝜑 ≡ (R𝜑)|B = 𝜇|B. By Theorem 1 𝜇 is a complete strongly saturated Radon measure. Therefore, by virtue of Lemma 4 (3.5.3) 𝜇0 is a Radon measure and 𝜇 = (𝜇0 )#. Besides, 𝜇0 C = 𝜇C = inf{𝜑f | f ∈ A(T) ∧ f ⩾ 𝜒(C)} for every C ∈ C. According to Theorem 1, we have R ≡ I𝜎 (B(T)) ⊂ Am (T, G, A(T))𝛿 ⊂ B(T, G). Consequently, B(T, R) ⊂ B(T, G) ⊂ M = dom 𝜇 ≡ dom R𝜑. As a result, since P is a bijection of (A(T)󳵻 )+ onto (B(T)󳵻 )+ , we obtain B(T, R) ⊂ B(T, G) ⊂ ⋂ ⟮dom R𝜑 | 𝜑 ∈ (A(T)󳵻 )+ ⟯ = ⋂ ⟮dom V ∧ (P𝜑) | 𝜑 ∈ (A(T)󳵻 )+ ⟯ = ⋂ ⟮dom V ∧ 𝜓 | 𝜓 ∈ (B(T)󳵻 )+ ⟯. Thus, we can apply Theorem 2 (3.4.2) to A(T) ≡ A(T)󳵻 , M0 ≡ B, and V ≡ V 󳵻 |M0 , and this theorem implies assertions 3 and 4.

386 | 3.6 Representation of a functional by the Radon integral

For every C ∈ C, we get 𝜇0 C = inf{𝜑f | f ∈ A(T) ∧ f ⩾ 𝜒(C)} = 𝜘C. Therefore, the property of compact regularity implies that 𝜇0 = 𝜘 on B. Corollary 1. Let ⟮T, G⟯ be a Hausdorff space, A(T) be a truncatable lattice-ordered linear subspace in S(T, G). 1) If A(T) has property (E𝜎 ), then the mapping S is a bijection of (A(T)󳵻 )+ onto RMw⋆ (T, G, A(T))0 and S b is a bijection of (A(T) 󳵻 )+ onto RMw⋆ b (T, G)+ . 2) If A(T) has property (E)&(D), then the mapping S is a bijection of (A(T)󳶋 )+ = (A(T)󳵻 )+ onto RMw⋆ (T, G, A(T))0 and S b is a bijection of (A(T) 󳶋 )+ = (A(T) 󳵻 )+ onto RMw⋆ b (T, G)+ . The proof of this Corollary is quite analogous to the proof of Corollary 1 to Theorem 1.

Parametric theorems on characterization of Radon integrals with respect to positive Radon measures and their corollaries Having the previous results we can prove the theorem on characterization of Radon integrals with respect to positive Radon measures. First, we shall formulate it for positive extended Radon measures. Theorem 3 (the Zakharov parametric theorem on characterization of Radon integrals with respect to positive Radon measures). Let ⟮T, G⟯ be a Hausdorff space, A(T) be a truncatable lattice-ordered linear subspace in S(T, G). If A(T) has either property (E𝜎 ) or property (E)&(D), then 1) for every positive 𝜎-exact linear functional 𝜑 there is a unique positive extended Radon measure 𝜇 : M → R+ such that 𝜑 is representable by the Lebesgue integral over the measurable space ⟮T, M, 𝜇⟯ and 𝜇C = inf{𝜑f | f ∈ A(T) ∧ f ⩾ 𝜒(C)} for every compact set C; 2) if, besides, 𝜑 is uniformly bounded, then 𝜇 is bounded; 3) the mapping R : 𝜑 󳨃→ 𝜇 is a bijection of (A(T)󳵻 )+ onto RMwe (T, G, A(T))0 ; 4) the mapping R b ≡ R|(A(T) 󳵻 )+ is a bijection of (A(T) 󳵻 )+ onto RMwe b (T, G)+ . Proof. All assertions follows immediately from Theorem 1 and its Corollary. Now, we shall formulate the theorem about characterization of Radon integral for positive Borel – Radon measures. Theorem 4 (the parametric theorem on characterization of Radon integrals with respect to positive Borel – Radon measures). Let ⟮T, G⟯ be a Hausdorff space, A(T) be a truncatable lattice-ordered linear subspace in S(T, G). If A(T) has either property (E𝜎 ) or property (E)&(D), then

3.6.3 Characterizations of Radon integrals with respect to positive Radon measures | 387

1)

for every positive 𝜎-exact linear functional 𝜑 there is a unique positive Borel – Radon measure 𝜇0 such that 𝜑 is representable by the Lebesgue integral over the measure space ⟮T, B, 𝜇0 ⟯ and 𝜇0 C = inf{𝜑f | f ∈ A(T) ∧ f ⩾ 𝜒(C)} for every compact set C; 2) if, besides, 𝜑 is uniformly bounded, then 𝜇0 is bounded; 3) the mapping S : 𝜑 󳨃→ 𝜇0 is a bijection of (A(T)󳵻 )+ onto RMw⋆ (T, G, A(T))0 ; 4) the mapping S b ≡ S|(A(T) 󳵻 )+ is a bijection of (A(T) 󳵻 )+ onto RMw⋆ b (T, G)+ . Proof. All assertions follows immediately from Theorem 2 and its corollary. Now, we show that these general theorems imply as partial cases some classical theorems on integral representation from Appendix (the other theorems will follow from Theorem 2 (3.6.4)) and derive some new characterizations. First, consider the case of a locally compact space. Corollary 1 (the Halmos – Hewitt – Edwards theorem). Let ⟮T, G⟯ be a locally compact space and A(T) ≡ C c (T, G). Then, 1) the mapping R is a bijection of (C c (T, G)∼ )+ onto RMwe (T, G)0 , the mapping S is a bijection of (C c (T, G)∼ )+ onto RMw⋆ (T, G)0 , and, in addition, 𝜑f = ∫ f dR𝜑 = ∫ f dS𝜑; 2) the mapping R b is a bijection of (C c (T, G) ∼ )+ onto RMwe b (T, G)+ , the mapping S b is a bijection of (C c (T, G) ∼ )+ onto RMw⋆ (T, G) , and, in addition, 𝜑f = ∫ f dR b 𝜑 = + b ∫ f dS b 𝜑. Proof. By Proposition 2 (3.6.2) the space A(T) ≡ C c (T, G) has property (E). Theorem 1 (2.3.4) provides that A(T) has property (D). By Corollary 1 to Theorem 4 (3.6.2) A(T)∼ = A(T)󳵻 . Since A(T) ⊂ MI(T, M, 𝜇) for every 𝜇 ∈ RMwe (T, G) by virtue of Lemma 1 (3.5.2), we obtain that RMwe (T, G, A(T))0 = RMwe (T, G)0 and RMw⋆ (T, G, A(T))0 = RMw⋆ (T, G)0 . Hence, Theorems 3 and 4 imply the desired assertions. Further, we consider the case of a Tychonoff space. Corollary 2 (the Bourbaki theorem for positive measures). Let ⟮T, G⟯ be a Tychonoff topological space and A(T) ≡ C b (T, G). Then, the mapping R = R b is a bijection of (C b (T, G)𝜋 )+ = (C b (T, G) 𝜋 )+ onto RMwe b (T, G)+ , the mapping S = S b is a bijection of (C b (T, G)𝜋 )+ = (C b (T, G) 𝜋 )+ onto RMw⋆ b (T, G)+ and, in addition, 𝜑f = ∫ f dR𝜑 = ∫ f dS𝜑. Proof. By Proposition 2 (3.6.2) the space A(T) has property (E). Theorem 1 (2.3.4) provides that A(T) has property (D). By virtue of Corollary 1 to Theorem 5 (3.6.2), A(T)𝜋 = A(T) 𝜋 = A(T)󳵻 = A(T) 󳵻 . Hence, Theorems 3 and 4 imply the desired assertions.

388 | 3.6 Representation of a functional by the Radon integral

Finally, consider the case of an arbitrary Hausdorff space. Corollary 3. Let ⟮T, G⟯ be a Hausdorff space and A(T) ≡ S c (T, G). Then, 1) the mapping R is a bijection of (S c (T, G)󳵻 )+ onto RMwe (T, G)0 , the mapping S is a bijection of (S c (T, G)󳵻 )+ onto RMw⋆ (T, G)0 , and, in addition, 𝜑f = ∫ f dR𝜑 = ∫ f dS𝜑; 2) the mapping R b is a bijection of (S c (T, G) 󳵻 )+ onto RMwe b (T, G)+ , the mapping S b is a bijection of (S c (T, G) 󳵻 )+ onto RMw⋆ (T, G) , and, in addition, 𝜑f = ∫ f dR b 𝜑 = + b ∫ f dS b 𝜑. Proof. By Proposition 2 (3.6.2), A(T) has property (E𝜎 ). According to Lemma 1 (3.5.2), A(T) ⊂ MI(T, M, 𝜇) for every 𝜇 ∈ RMwe (T, G); therefore, RMwe (T, G, A(T))0 = RMwe (T, G)0 and RMw⋆ (T, G, A(T))0 = RMw⋆ (T, G)0 . Now, Theorems 3 and 4 imply the desired assertions. Corollary 4. Let ⟮T, G⟯ be a Hausdorff space and A(T) ≡ S(T, G). Then, the mapping R = R b is a bijection of (S(T, G)󳵻 )+ = (S(T, G) 󳵻 )+ onto RMwe b (T, G)+ , the mapping 󳵻 󳵻 w⋆ S = S b is a bijection of (S(T, G) )+ = (S(T, G) )+ onto RMb (T, G)+ , and, in addition, 𝜑f = ∫ f dR𝜑 = ∫ f dS𝜑. Proof. By Proposition 2 (3.6.2), A(T) has property (E𝜎 ). Since 1 ∈ A(T), for every functional 𝜑 ∈ (A(T)󳵻 )+ and every function f ∈ A(T) such that |f | ⩽ 1, we have |𝜑f | ⩽ 𝜑|f | ⩽ 𝜑1 < ∞. Therefore, (A(T)󳵻 )+ = (A(T) 󳵻 )+ . Then, Theorems 3 and 4 imply the desired assertions.

3.6.4 The solution of the problem of characterization of Radon integrals as linear functionals The parametric theorem on characterization of Radon integrals with respect to Radon bimeasures Our purpose now is to extend bijections S : (A(T)󳵻 )+ RMw⋆ (T, G, A(T))0 and RMw⋆ S b : (A(T) 󳵻 )+ b (T, G)+ (see Theorem 4 (3.6.3)) to isomorphisms of latticeordered linear spaces. To extend the bounded bijection S b , we can use ordinary means of measure theory. Since the families RMw⋆ (T, G) and RMw⋆ (T, G, A(T)) are not linear spaces, the extension of the (unbounded) bijection S is a difficult problem, and Radon bimeasures were invented in 3.5.5 to solve it. Proposition 1. Let ⟮T, G⟯ be a Hausdorff space, A(T) be a truncatable lattice-ordered linear subspace in S(T, G). If A(T) has either property (E𝜎 ) or property (E)&(D), then the mappings S and S b are conic, isotone, and preserving any exact bounds.

3.6.4 The solution of the problem of characterization of Radon integrals | 389

Proof. Let 𝜑, 𝜓 ∈ (A(T)󳵻 )+ , 𝜇 ≡ S𝜑, 𝜈 ≡ S𝜓, x, y ∈ R+ . By Corollary 1 to Proposition 3 (3.5.3), we get 𝜆 ≡ x𝜇 + y𝜈 ∈ RMw⋆ (T, G)0 . It follows from assertion 3 of Theorem 4 (3.6.3) that A(T) ⊂ MI(T, B, 𝜇) ∩ MI(T, B, 𝜈), where B ≡ B(T, G). According to Proposition 3 (3.3.5) MI(T, B, 𝜆) = MI(T, B, 𝜇) ∩ MI(T, B, 𝜈) and ∫ f d𝜆 = x ∫ f d𝜇 + y ∫ f d𝜈 for every function f ∈ MI(T, B, 𝜆). Therefore, A(T) ⊂ MI(T, B, 𝜆) and for every function f ∈ A(T) we get (x𝜑 + y𝜓)f = x ∫ f d𝜇 + y ∫ f d𝜈 = ∫ f d𝜆 By virtue of Theorem 4 (3.6.3). Take a compact set C and consider X ≡ {f ∈ A(T) | f ⩾ 𝜒(C)}. By assertion 1 of Theorem 4 (3.6.3) we have 𝜇C = inf⟮𝜑f | f ∈ X⟯ and 𝜈C = inf⟮𝜓f | f ∈ X⟯. Put a ≡ inf⟮x𝜑f | f ∈ X⟯, b ≡ inf⟮y𝜓f | f ∈ X⟯, and c ≡ inf⟮(x𝜑f + y𝜓f )f | f ∈ X⟯. It is clear that 𝜆C = x𝜇C + y𝜈C = a + b ⩽ c. From c ⩽ x𝜑f + y𝜓f we obtain c − y𝜓f ⩽ a, c − a ⩽ b, and therefore, c ⩽ a + b. Then, 𝜆C = c. By virtue of Theorem 4 (3.6.3), we conclude that 𝜆 = S(x𝜑 + y𝜓). As a consequence, we get the same property for S b . Now, let 𝜑 ⩽ 𝜓. If C is a compact set, then 𝜇C = inf⟮𝜑f | f ∈ X⟯ ⩽ inf⟮𝜓f | f ∈ X⟯ = 𝜈C. Then, using the property of compact regularity, we establish that 𝜇 ⩽ 𝜈. Conversely, let 𝜇 ⩽ 𝜈. Therefore, 𝜑f = ∫ f d𝜇 ⩽ ∫ f d𝜈 = 𝜓f for every f ∈ A(T). This means that 𝜑 ⩽ 𝜓. Thus, V is isotone. Hence, Lemma 2 (1.1.15) guarantees that S preserves any exact bounds. The same arguments are convenient for S b . To construct the necessary extension of the mapping S, we shall use the embedding E0 : RMw⋆ (T, G)0 RB(T, G)+ from 3.5.5. According to Corollary 2 to RBb (T, G)+ has a unique Theorem 3 (3.5.5), the mapping E0b : RMw⋆ b (T, G)+ w⋆ extension to an isomorphism E b : RMb (T, G) RBb (T, G) of the given latticeordered linear spaces such that E b 𝜇 = E0b (𝜇+ ) − E0b (−𝜇− ). Consider the family RB(T, G, A(T)) ≡ {m ∈ RB(T, G) | A(T) ⊂ BI(T, G, m)} of all Radon bimeasures m such that all functions from A(T) are m-integrable. Since all functions from A(T) ⊂ S(T, G) are integrable with respect to every bounded measure (Lemma 1 (3.5.2)), we have A(T) ⊂ BI(T, G, p󸀠 (m)) ∩ BI(T, G, p󸀠󸀠 (m)) = BI(T, G, m) for every m ∈ RBb (T, G), i. e. RBb (T, G, A(T)) = RBb (T, G). If 𝜇 ∈ RMw⋆ (T, G, A(T))0 , then by Proposition 1 (3.5.6), ∫ f d𝜇 = ∫ f dE0 𝜇 for every function f ∈ BI(T, G, 𝜇) = BI(T, G, E0 𝜇). Consequently, E0 𝜇 ∈ RB(T, G, A(T))+ and ∫ f d𝜇 = ∫ f dE0 𝜇 for every f ∈ A(T). This implies that we can consider the composite mapping E0 ∘ S : (A(T)󳵻 )+ 󴀚󴀠 RB(T, G, A(T))+ . It is easily proven that E0b ∘ S b : (A(T) 󳵻 )+ 󴀚󴀠 RBb (T, G)+ . Theorem 1 (the Zakharov parametric theorem on characterization of Radon integrals with respect to Radon bimeasures). Let ⟮T, G⟯ be a Hausdorff space, A(T) be a truncatable lattice-ordered linear subspace in S(T, G). If A(T) has either property (E𝜎 ) or property (E)&(D), then 1) the mapping E0 ∘ S has a unique extension to an isomorphism J : A(T)󳵻 RB(T, G, A(T)) of the given lattice-ordered linear spaces and J𝜑 = E0 (S(𝜑+ )) − E0 (S(−𝜑− )) for every 𝜑 ∈ A(T)󳵻 ;

390 | 3.6 Representation of a functional by the Radon integral

2) 𝜑f = ∫ f dJ𝜑 = ∫ f dS(𝜑+ ) − ∫ f dS(−𝜑− ) for every f ∈ A(T) and every 𝜑 ∈ A(T)󳵻 ;

3) the mapping S b has a unique extension to an isomorphism I b : A(T) 󳵻 RMw⋆ b (T, G) of the given lattice-ordered linear spaces and I b 𝜑 = S b (𝜑+ ) − S b (−𝜑− ) for every 𝜑 ∈ A(T) 󳵻 ; 4) 𝜑f = ∫ f dI b 𝜑 = ∫ f dS b (𝜑+ )−∫ f dS b (−𝜑− ) for every f ∈ A(T) and every 𝜑 ∈ A(T) 󳵻 ;

5) the mapping E0b ∘ S b has a unique extension to an isomorphism J b : A(T) 󳵻 RBb (T, G) of the given lattice-ordered linear spaces and J b 𝜑 = E0b (S b (𝜑+ )) − E0b (S b (−𝜑− )) = E b (I b 𝜑) for every 𝜑 ∈ A(T) 󳵻 ; 6) 𝜑f = ∫ f dJ b 𝜑 = ∫ f dS b (𝜑+ )−∫ f dS b (−𝜑− ) for every f ∈ A(T) and every 𝜑 ∈ A(T) 󳵻 .

Proof. Put X ≡ A(T)󳵻 , Y0 ≡ RMw⋆ (T, G, A(T))0 , Z ≡ RB(T, G, A(T)). 1. By Theorem 4 (3.6.3) the mapping S is a bijection from X+ onto Y0 . By Theorem 2 (3.5.5) and Proposition 1 (3.5.6) E0 |Y0 is a bijection from Y0 onto Z+ . Thus, E0 ∘ S is a one-to-one mapping from X+ onto Z+ . According to Theorem 4 (3.6.3) and Proposition 1, if 𝜑, 𝜓 ∈ X+ and x, y ∈ R+ , then (E0 ∘ V)(x𝜑 + y𝜓) = E0 (xS𝜑 + yS𝜓) = x(E0 ∘ S)𝜑 + y(E0 ∘ S)𝜓. Consequently, using Statement 1 (3.4.2), we can uniquely extend the mapping E0 ∘ S to an injective linear operator J : X 󴀚󴀠 Z such that J𝜑 = E0 (S(𝜑+ )) − E0 (S(−𝜑− )) for all 𝜑 ∈ X. By Theorem 2 (3.5.5). the mapping E0 preserves any exact bounds. By virtue of Proposition 1, the mapping S also preserves any exact bounds. Consequently, if 𝜑1 , 𝜑2 ∈ X and 𝜑1 ∧ 𝜑2 = 0, then 𝜑1 , 𝜑2 ∈ X+ and J𝜑1 ∧ J𝜑2 = E0 (S𝜑1 ) ∧ E0 (S𝜑2 ) = E0 (S𝜑1 ∧ S𝜑2 ) = E0 (S(𝜑1 ∧ 𝜑2 )) = 0. Therefore, according to Statement 2 (3.2.2), the operator J is a homomorphism of lattice-ordered linear spaces. We shall obtain that J is bijective when assertion 2 will be proven. 2. Fix 𝜑 and consider m ≡ J𝜑, 𝜇1 ≡ S(𝜑+ ), 𝜇2 ≡ S(−𝜑− ). Since m = E0 𝜇1 − E0 𝜇2 and E0 𝜇1 ∧ E0 𝜇2 = 0, we have E𝜇1 = m+ and E𝜇2 = −m− . Then, it follows from Theorem 2 (3.5.5) that 𝜇1 and 𝜇2 are parent Borel – Radon measures of the bimeasure m, i. e. 𝜇1 = p󸀠 (m) and 𝜇2 = p󸀠󸀠 (m). By definition, BI(T, G, m) ≡ BI(T, G, 𝜇1 ) ∩ BI(T, G, 𝜇2 ). By virtue of Theorem 4 (3.6.3), the functional 𝜑+ is represented by measure 𝜇1 , −𝜑− is represented by 𝜇2 , and 𝜇1 , 𝜇2 ∈ Y0 . Therefore, A(T) ⊂ BI(T, G, 𝜇1 ) ∩ BI(T, G, 𝜇2 ) = BI(T, G, m). By definition, ∫ f dm ≡ ∫ f dp 󸀠 (m)−∫ f dp󸀠󸀠 (m) = ∫ f d𝜇1 −∫ f d𝜇2 = ∫ f dS(𝜑+ )−∫ f dS(−𝜑− ) for every f ∈ A(T). By Theorem 4 (3.6.3), ∫ f dS(𝜑+ ) = 𝜑+ f and ∫ f dS(−𝜑− ) = (−𝜑− )f . Then, ∫ f dJ𝜑 = ∫ f dS(𝜑+ ) − ∫ f dS(−𝜑− ) = 𝜑+ f + 𝜑− f = 𝜑f . Now, let m is a Radon bimeasure. By Theorem 2 (3.5.5) m = E0 𝜇󸀠 − E0 𝜇󸀠󸀠 for some positive Borel – Radon measures 𝜇󸀠 and 𝜇󸀠󸀠 . According to Theorem 4 (3.6.3), we have 𝜇󸀠 = S𝜑󸀠 and 𝜇󸀠󸀠 = S𝜑󸀠󸀠 for some functionals 𝜑󸀠 , 𝜑󸀠󸀠 ∈ X+ . Consider the functional 𝜑 ≡ 𝜑󸀠 − 𝜑󸀠󸀠 ∈ X. Then, m = E0 (S𝜑󸀠 ) − E0 (S𝜑󸀠󸀠 ) = J𝜑󸀠 − J𝜑󸀠󸀠 = J𝜑. This implies that J is surjective. Thus, J is the necessary isomorphism of the lattice-ordered linear spaces X and Z. 3. Consider the case of uniformly bounded functionals. By virtue of Theorem 4 (3.6.3) and Proposition 1, the mapping S b is a conic bijection from

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(A(T) 󳵻 )+ onto RMw⋆ b (T, G)+ . Then, according to Statement 1 (3.4.2), S b has the unique extension to the linear bijection I b : A(T) 󳵻 RMw⋆ b (T, G). In addition, 󳵻 I b 𝜑 = S b (𝜑+ ) − S b (−𝜑− ) for every 𝜑 ∈ A(T) . By Proposition 1, the mapping S b preserve any exact bounds. Consequently, if 𝜑1 ∧ 𝜑2 = 0, then I b 𝜑1 ∧ I b 𝜑2 = S b 𝜑1 ∧ S b 𝜑2 = S b (𝜑1 ∧ 𝜑2 ) = 0. Then, by Statement 2 (3.2.2), the mapping I b is a homomorphism of lattice-ordered linear spaces. So, I b is the necessary isomorphism of the given lattice-ordered linear spaces. 4. Suppose f ∈ A(T) and 𝜑 ∈ X ∘ ≡ A(T) 󳵻 . By Theorem 4 (3.6.3), the function f is integrable with respect to the measures 𝜇󸀠 ≡ S b (𝜑+ ) and 𝜇󸀠󸀠 ≡ S b (−𝜑− ), 𝜑+ f = ∫ f d𝜇󸀠 and −𝜑− f = ∫ f d𝜇󸀠󸀠 . Therefore, f ∈ BI(T, G, 𝜇󸀠 − 𝜇󸀠󸀠 ) = BI(T, G, I b 𝜑) and 𝜑f = 𝜑+ f − (−𝜑− )f = ∫ f d𝜇󸀠 − ∫ f d𝜇󸀠󸀠 = ∫ f dI b 𝜑. 5. According to Theorem 1 (3.5.5), Z b ≡ RBb (T, G, A(T)) = RBb (T, G) is a latticeordered linear space. Owing to Theorem 4 (3.6.3) and Proposition 1 the mapping S b is a w⋆ conic bijection from (X ∘ )+ onto (Y b )+ , where Y b ≡ RMw⋆ b (T, G, A(T)) = RMb (T, G). 0 The mapping E b is a conic bijection from (Y b )+ onto (Z b )+ . Then, the mapping E0b ∘ S b is a conic bijection from (X ∘ )+ onto (Z b )+ . Therefore, by Statement 1 (3.4.2) it is uniquely extended to the linear bijection J b : X ∘ Z b . In addition, J b 𝜑 = E0b (S b (𝜑+ )) − E0b (S b (−𝜑− )). Using Corollary 2 to Theorem 3 (3.5.5), we obtain J b 𝜑 = E b (S b 𝜑+ − S b (−𝜑− )) = E b (I b 𝜑) for every 𝜑 ∈ X ∘ . By virtue of (3) and Corollary 2 to Theorem 3 (3.5.5), the mappings I b and E b are isomorphisms of the given lattice-ordered linear spaces. Consequently, J b = E b ∘ I b is also an isomorphism. Since E0b = E0 |(Y b )+ and S b = S|(X ∘ )+ , we have J𝜑 = E0 (V𝜑+ ) − E0 (V(−𝜑− )) = E0b (S b 𝜑+ ) − E0b (S b (−𝜑− )) = J b 𝜑 for every 𝜑 ∈ X ∘ , i. e. J b = J|X ∘ . 6. Finally, since J b = J|X ∘ and S b = S|(X ∘ )+ , we conclude from (2) that 𝜑f = ∫ f dJ b 𝜑 = ∫ f dS b (𝜑+ ) − ∫ f dS b (−𝜑− ) for every f ∈ A(T) and every 𝜑 ∈ X ∘ .

The parametric theorem on characterization of Radon integrals with respect to arbitrary Radon measures and its corollaries Let A(T) be a linear space of functions on a set T and 𝜑 : A(T) → R be a linear functional on A(T). Define the lower and the upper boundedness indices of the functional 𝜑: b(𝜑) ≡ inf{𝜑f | f ∈ A(T)+ ∧ f ⩽ 1} and b(𝜑) ≡ sup{𝜑f | f ∈ A(T)+ ∧ f ⩽ 1}. It is clear that −∞ ⩽ b(𝜑) ⩽ 𝜑(0) = 0 and 0 = 𝜑(0) ⩽ b(𝜑) ⩽ ∞. If both the indicies are finite, then 𝜑 is uniformly (order) bounded and vice versa. The functional 𝜑 is said to be natural if at least one of its boundedness indices is finite (compare with the definition of a natural evaluation in 3.1.1). The family of all natural 𝜎-exact linear functionals on a space A(T) will be denoted by (A(T)󳵻 )nat . Clearly, A(T) 󳵻 ⊂ (A(T)󳵻 )nat .

392 | 3.6 Representation of a functional by the Radon integral

Theorem 2 (the Zakharov – Rodionov parametric theorem on characterization of Radon integrals with respect to arbitrary Radon measures). Let ⟮T, G⟯ be a Hausdorff space and A(T) be a truncatable lattice-ordered linear subspace in the space S(T, G) possessing property (E𝜎 ) or property (E)&(D). Suppose 𝜑 ∈ A(T)󳵻 . Then, the equality E𝜇 = J𝜑 defines the bijective mapping I : 𝜑 󳨃→ 𝜇 from the set U ≡ (A(T)󳵻 )nat onto the set V ≡ RMw⋆ (T, G, A(T)). Moreover, 𝜑f = ∫ f d𝜇 for all f ∈ A(T) and the bijection I is an isomorphism from the system U onto the system V in the following sense (see 5∘ (2.2.7)): 1) I[{0U }] = {0V }, 2) (I ∗m I)[−U ] = −V , 3) ((I ∗m I) ∗m I)[+U ] = +V , 4) (I ∗m I)[⩽U ] = ⩽V , 5) ((I ∗m I) ∗m I)[∨U ] = ∨V , 6) ((I ∗m I) ∗m I)[∧U ] = ∧V , 7) ((idR ∗m I) ∗m I)[⋅⟮U,R⟯ ] = ⋅⟮V ,R⟯ . Proof. We use here the mappings E : RMw⋆ (T, G) 󴀚󴀠 RB(T, G) from Theorem 3 (3.5.5), J : A(T)󳵻 W ≡ RB(T, G, A(T)) from Theorem 1, and S : (A(T)󳵻 )+ V0 from Theorem 4 (3.6.3). ii) ⊢ i). Consider the variations v+ (𝜇) and v− (𝜇) of measure 𝜇. According to Lemma 1 (3.5.3), 𝜇1 ≡ v+ (𝜇) and 𝜇2 ≡ −v− (𝜇) are positive Radon measures and at least one of them is finite. By definition, 𝜑f = ∫ f d𝜇 ≡ ∫ f d𝜇1 − ∫ f d𝜇2 for all f ∈ A(T). Suppose 𝜇1 is finite. Whence if f ∈ A(T)+ and f ⩽ 1, then 𝜑f ⩽ ∫ f d𝜇1 ⩽ ∫ 1 d𝜇1 = 𝜇1 (T) < ∞, i. e. b(𝜑) < ∞. Suppose 𝜇2 is finite. Whence if f ∈ A(T)+ and f ⩽ 1, then 𝜑f ⩾ − ∫ f d𝜇2 ⩾ − ∫ 1 d𝜇2 = −𝜇2 (T) = v− (T) > −∞, i. e. b(𝜑) > −∞. i) ⊢ ii). Consider functionals 𝜑+ and 𝜑− such that 𝜑 = 𝜑+ + 𝜑− from Corollary 3 to Statement 5 (2.2.8). According to Proposition 1 (3.6.1), 𝜑+ , −𝜑− ∈ (A(T)󳵻 )+ . Define positive Radon measures 𝜇1 ≡ S(𝜑+ ) and 𝜇2 ≡ S(−𝜑− ). Suppose b(𝜑) < ∞. Check that 𝜇1 is finite. By Corollary 3 to Statement 5 (2.2.8) 𝜑+ f = sup{𝜑g | g ∈ A(T)+ ∧ g ⩽ f }. Therefore, for every f ∈ A(T)+ such that f ⩽ 1 we have 𝜑+ f ⩽ sup{𝜑g | g ∈ A(T)+ ∧ g ⩽ 1} = b(𝜑) < ∞. Properties (E) or (E𝜎 ) guarantee that for every C ∈ C there is f ∈ A(T) such that 𝜒(C) ⩽ f ⩽ 𝜒(T) = 1. Then, by Theorem 4 (3.6.3) 𝜇1 C = ∫ 𝜒(C) d𝜇1 ⩽ ∫ f d𝜇1 = 𝜑+ f ⩽ b(𝜑) < ∞. Consequently, 𝜇1 B = sup{𝜇1 C | C ∈ C ∧ C ⊂ B} ⩽ b(𝜑) < ∞ for every B ∈ B by virtue of the property of compact regularity of 𝜇1 . Suppose now b(𝜑) > −∞. Check that 𝜇2 is finite. Again by Corollary 3 to Statement 5 (2.2.8) 𝜑− f = inf{𝜑h | h ∈ A(T)+ ∧ h ⩽ f }. Therefore, for every f ∈ A(T)+ it follows from f ⩽ 1 that 𝜑− f ⩾ inf{𝜑h | h ∈ A(T)+ ∧ h ⩽ 1} = b(𝜑) > −∞. As in the preceding case for any C ∈ C we take f ∈ A(T) such that 𝜒(C) ⩽ f ⩽ 1 and have

3.6.4 The solution of the problem of characterization of Radon integrals

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𝜇2 C = ∫ 𝜒(C) d𝜇2 ⩽ ∫ f d𝜇2 = −𝜑− f ⩽ −b(𝜑) < ∞. Hence, 𝜇2 B = sup{𝜇2 C | C ∈ C ∧ C ⊂ B} ⩽ −b(𝜑) < ∞ for every B ∈ B. Thus, at least one of the measures 𝜇1 or 𝜇2 is finite. Therefore, the measure 𝜇 ≡ 𝜇1 − 𝜇2 is well defined and 𝜇 ∈ RMw⋆ (T, G) by Lemma 1 (3.5.3). Then, Theorem 3 (3.5.5) and Theorem 1 imply that E𝜇 = E0 𝜇1 −E0 𝜇2 = J𝜑. Further, 𝜇 is the unique measure such that E𝜇 = J𝜑, because E is injective. Finally, taking into account Proposition 1 (3.5.6) we get 𝜑f = ∫ f dJ𝜑 = ∫ f dE𝜇 = ∫ f d𝜇. According to Theorem 3 (3.5.5) the mapping E is injective and isotone and is a homomorphism from the system V into the lattice-ordered linear space W in sense of 5∘ (2.2.7). By Proposition 1 (3.5.6) the mapping E preserves integral. By the above, I is a “constriction” of the isomorphism J from Theorem 1 in the sense that E ∘ I = J. Since J is an isomorphism of the lattice-ordered linear spaces A(T)󳵻 and W, the mapping I quite preserves all the indicated structures (see 5∘ (2.2.7)). Now, we show that this general theorem implies as partial cases the classical theorems on integral representation from Appendix not mentioned in 3.6.3 and derive some new characterizations. Corollary 1 (the Bourbaki theorem). Let ⟮T, G⟯ be a Tychonoff space and A(T) ≡ C b (T, G). Then, the mapping I is a bijection from C b (T, G)𝜋 onto RMw⋆ b (T, G). Moreover, 𝜑f = ∫ f dI𝜑 for all f ∈ C b (T, G) and I is an isomorphism of the given latticeordered linear spaces. Proof. By Proposition 2 (3.6.2) the family A(T) has property (E). Theorem 1 (2.3.4) provides that A(T) has property (D). By virtue of Corollary 1 to Theorem 5 (3.6.2), A(T)𝜋 = A(T) 𝜋 = A(T)󳵻 = A(T) 󳵻 . Hence, A(T)𝜋 = A(T) 󳵻 ⊂ (A(T)󳵻 )nat ⊂ A(T)󳵻 = A(T)𝜋 . Then, the desired assertion follows from Theorem 2, Corollary 2 to Theorem 3 (3.5.5), and assertion 5 of Theorem 1. Corollary 2. Let ⟮T, G⟯ be a locally compact space and A(T) ≡ C c (T, G). Then, the mapping I is a bijection from (C c (T, G)∼ )nat onto RMw⋆ (T, G). Moreover, 𝜑f = ∫ f dI𝜑 for all f ∈ C c (T, G) and I is an isomorphism in sense of Theorem 2. Proof. By Proposition 2 (3.6.2) the family A(T) ≡ C c (T, G) has property (E). Theorem 1 (2.3.4) provides that A(T) has property (D). By Corollary 1 to Theorem 4 (3.6.2), A(T)∼ = A(T)󳵻 . According to Lemma 1 (3.5.2), A(T) ⊂ MI(T, M, 𝜇) for every 𝜇 ∈ RMw⋆ (T, G); therefore, RMw⋆ (T, G, A(T)) = RMw⋆ (T, G). Now, Theorem 2 implies the desired assertions. Corollary 3 (the Radon – Saks – Kakutani theorem). Let ⟮T, G⟯ be a compact space and A(T) ≡ C(T, G). Then, the mapping I is a bijection from C(T, G)∼ onto RMw⋆ (T, G).

394 | 3.7 The Riemann integral

Moreover, 𝜑f = ∫ f dI𝜑 and I is an isomorphism of the given lattice-ordered linear spaces. Proof. Since ⟮T, G⟯ is compact, we have T ∈ C. Therefore, A(T) ≡ C(T, G) = C c (T, G) ∼ and RMw⋆ (T, G) = RMw⋆ b (T, G). Corollary 1 to Lemma 3 (2.2.8) provides that A(T) = ∼ ∼ ∼ ∼ ∼ A(T) . Hence, A(T) = A(T) ⊂ (A(T) )nat ⊂ A(T) . By Proposition 2 (3.5.3), the family ∼ RMw⋆ b (T, G) is a lattice-ordered linear space; by Statement 5 (2.2.8), A(T) is a lattice-ordered linear space as well. Now, Corollary 2 implies the desired assertion. Corollary 4. Let ⟮T, G⟯ be a Hausdorff space and A(T) ≡ S c (T, G). Then, the mapping I is a bijection from (S c (T, G)󳵻 )nat onto RMw⋆ (T, G). Moreover, 𝜑f = ∫ f dI𝜑 and I is an isomorphism in sense of Theorem 2. Proof. By Proposition 2 (3.6.2), the family A(T) has property (E𝜎 ). According to Lemma 1 (3.5.2), A(T) ⊂ MI(T, M, 𝜇) for every 𝜇 ∈ RMw⋆ (T, G); therefore, RMw⋆ (T, G, A(T)) = RMw⋆ (T, G). Now, Theorem 2 implies the desired assertions. Corollary 5. Let ⟮T, G⟯ be a Hausdorff space and A(T) ≡ S(T, G). Then, the mapping I is a bijection from S(T, G)󳵻 onto RMw⋆ b (T, G). Moreover, 𝜑f = ∫ f dI𝜑 and I is an isomorphism of the given lattice-ordered linear spaces. Proof. By Proposition 2 (3.6.2), the family A(T) has property (E𝜎 ). Since 1 ∈ A(T), for every functional 𝜑 ∈ A(T)󳵻 and every function f ∈ A(T) such that |f | ⩽ 1, we have |𝜑f | ⩽ 𝜑|f | ⩽ 𝜑1 < ∞. Hence, A(T)󳵻 = A(T) 󳵻 ⊂ (A(T)󳵻 )nat ⊂ A(T)󳵻 . Then, the desired assertion follows from Theorem 2, Corollary 2 to Theorem 3 (3.5.5), and assertion 5 of Theorem 1.

3.7 The Riemann integral After the appearance of the Lebesgue integral, the celebrated Riemann integral did not lose his value and continued developing (see e. g. [Marcus, 1959; Johnson, 1962; Henstock, 1988; Kurzweil, 2000; Lee, 2011; Lukashenko et al., 2011; Skvortsov and Tulone, 2013]). Further, we consider the notion of the Riemann integral suitable for the work in general measurable topological spaces. It is remarkable that many attainments in the previous sections of the book have come together in this section at focus. 3.7.1 The Riemann integral over a topological space with a positive bounded Radon measure Let ⟮T, G⟯ be a topological space, M be a 𝜎-algebra on T containing G, and 𝜇 be a positive bounded (quite topological) measure on M. A set P ⊂ T will be called 𝜇-Jordan

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if fr P ∈ M0 (𝜇). Remind that according to Lemma 9 (3.1.1), the ensemble M0 (𝜇) of all 𝜇-negligible sets is a 𝜎-ideal in M. Denote the ensemble of all 𝜇-Jordan sets by J(T, G, M, 𝜇). Lemma 1. Let ⟮T, G⟯ be a topological space, M be a 𝜎-algebra on T containing G, and 𝜇 be a positive bounded measure on M. Then, the ensemble J(T, G, M, 𝜇) is an algebra. Proof. Since M0 (𝜇) is a 𝜎-ideal, the inclusion fr(P1 ∩ P2 ) ⊂ fr P1 ∪ fr P2 ∈ M0 (𝜇) means that fr(P1 ∩ P2 ) ∈ M0 (𝜇). Similarly, fr(P1 ∪ P2 ) ∈ M0 (𝜇). In addition, the equality fr(T\P) = fr P implies that T\P ∈ J(T, G, M, 𝜇). Evidently, fr T = ⌀ means that T ∈ J(T, G, M, 𝜇). Consider the set Γ ≡ Γ(T, G, M, 𝜇) ≡ Parf (J(T, G, M, 𝜇), T) of all finite 𝜇-Jordan partitions of the set T (see also 3.1.3). Consider also the set Δ ≡ Δ(T, G, M, 𝜇) ≡ Parf (G ∪ M0 (𝜇), T) of all finite partitions of T consisting of open and 𝜇-negligible sets. Lemma 2. Let ⟮T, G⟯ be a topological space, M be a 𝜎-algebra on T containing G, and 𝜇 be a positive bounded measure on M. Then Δ(T, G, M, 𝜇) ⊂ Γ(T, G, M, 𝜇). Proof. Let 𝜘 ≡ (Q k ∈ G ∪ M0 (𝜇) | k ∈ K) ∈ Δ. Consider the sets K 󸀠 ≡ {k ∈ K | Q k ∈ G ∧ Q k ∈ ̸ M0 (𝜇)} and K 󸀠󸀠 ≡ {k ∈ K | Q k ∈ M0 (𝜇)}. If k ∈ K 󸀠 , then by Statement 3 (3.5.1), we have cl Q k ∩ Q k󸀠 = ⌀ for every k󸀠 ∈ K 󸀠 \{k}, and therefore, fr Q k = cl Q k \Q k ⊂ T\ ⋃ ⟮Q k󸀠 | k󸀠 ∈ K 󸀠 ⟯ = ⋃ ⟮Q k󸀠󸀠 | k󸀠󸀠 ∈ K 󸀠󸀠 ⟯ ∈ M0 (𝜇). If k ∈ K 󸀠󸀠 , then by the same reason, fr Q k ≡ cl Q k \ int Q k ⊂ cl Q k ⊂ T\ ⋃ ⟮Q k󸀠 | k󸀠 ∈ K 󸀠 ⟯ ∈ M0 (𝜇). Thus, every Q k is a 𝜇-Jordan set, and therefore, 𝜘 ∈ Γ. Every 𝜇-Jordan partition from Δ(T, G, M, 𝜇) will be called prime. With each 𝜇-Jordan partition 𝜋 ≡ (P k | k ∈ K) ∈ Γ, we shall associate the prime 𝜇-Jordan partition 𝜘 ≡ (Q ki | (k, i) ∈ K × 2), where Q k0 ≡ G k ≡ int P k ∈ G and Q k1 ≡ N k ≡ P k \G k ∈ M0 (𝜇). Thus, to define the Riemann integral on the measurable topological space ⟮T, G, M, 𝜇⟯, we need not the complicated ensemble J(T, G, M, 𝜇) and the set Γ(T, G, M, 𝜇) of all 𝜇-Jordan partitions 𝜋 but only its subset Δ(T, G, M, 𝜇) of prime 𝜇-Jordan partitions. Define an order ⩽ on Δ setting 𝜘 ≡ (Q k | k ∈ K) ⩽ (R l | l ∈ L) ≡ 𝜆 if for every k ∈ K there is L 󸀠 ⊂ L such that Q k = ⋃ ⟮R l | l ∈ L󸀠 ⟯. In this case, the partition 𝜆 is called finer than the partition 𝜘. Obviously, 𝜆 is a refinement of 𝜘 in the sense of 1.1.10. Lemma 3. Let ⟮T, G⟯ be a topological space, M be a 𝜎-algebra on T containing G, and 𝜇 be a positive bounded (quite topological) measure on M. Then, the preordered class ⟮Δ, ⩽⟯ is upward directed. Proof. Let 𝜘 ≡ (Q k | k ∈ K) ∈ Δ and 𝜆 ≡ (R l | l ∈ L) ∈ Δ. Consider the set N ≡ {(k, l) ∈ K × L | Q k ∩ R l ≠ ⌀} and the partition 𝜎 ≡ (S n | n ∈ N), where S n ≡ Q k ∩ R l . If Q k ,

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R l ∈ G, then S n ∈ G; if Q k ∈ M0 (𝜇) or R l ∈ M0 (𝜇), then S n ∈ M0 (𝜇). Hence, 𝜎 ∈ Δ. Since Q k = ⋃ ⟮S n | n ∈ N ∧ n = (k, l)⟯, we get 𝜎 ⩾ 𝜘. Similarly, we see that 𝜎 ⩾ 𝜆. Let 𝜘 ∈ Δ and f ∈ F b (T). Consider the lower Darboux sum s(f , 𝜘) ≡ ∑ (inf f [Q k ]𝜇Q k | k ∈ K) and the upper Darboux sum S(f , 𝜘) ≡ ∑ (sup f [Q k ]𝜇Q k | k ∈ K) . Clearly, the net (s(f , 𝜘) | 𝜘 ∈ Δ) is increasing, the net (S(f , 𝜘) | 𝜘 ∈ Δ) is decreasing, and s(f , 𝜘) ⩽ S(f , 𝜘). A bounded function f : T → R is called Riemann integrable on the measurable topological space ⟮T, G, M, 𝜇⟯ or Riemann 𝜇-integrable if sup (s(f , 𝜘) | 𝜘 ∈ Δ) = inf (S(f , 𝜘) | 𝜘 ∈ Δ) ≡ r𝜇 f . If f is Riemann 𝜇-integrable, then the number r𝜇 f is called the Riemann integral of the function f over the space ⟮T, G, M, 𝜇⟯ or Riemann 𝜇-integral of f and is also denoted by P(𝜇)f or by (R) ∫ f d𝜇. The set of all bounded Riemann 𝜇-integrable functions will be denoted by RI(T, G, M, 𝜇). The remarkable properties of the family RI(T, G, M, 𝜇) and of the Riemann integral r𝜇 : RI(T, G, M, 𝜇) → R will be established in the following subsection. The notion of uniform function will play the key role in the corresponding proofs.

3.7.2 Description of the family of Riemann integrable functions on a Tychonoff space and its consequences The 𝜎-ideal M0 (𝜇) of all 𝜇-negligible sets is too large for the family RI(T, G, M, 𝜇) of all bounded Riemann 𝜇-integrable functions. Introduce a more narrow ideal which is “native” for Riemann 𝜇-integrable functions. The Lebesgue – Vitali characterization for the family U(T , SP𝜇 ) Let ⟮T, G, M, 𝜇⟯ be a measurable topological space with a positive quite topological (i. e. G ⊂ M) measure. The ensemble Coz C(T, G) of all cozero-sets of the family C(T, G) ≡ M(T, G) of all continuous functions on ⟮T, G⟯ will be denoted also by G0 . The ensemble Zer C(T, G) = co-G0 will be denoted also by F 0 . Consider the ensemble U0𝜇 ≡ G0 ∩ co-M0 (𝜇) = {U ∈ G0 | T\U ∈ M0 (𝜇)}. Every element of U0𝜇 is called a cozero-set of full measure. Further, consider the ensemble N𝜇 ≡ {N ∈ P(T) | ∃ U ∈ U0𝜇 (N ⊂ T\U)}. Note that N𝜇 = Nl (T, F 0 , 𝜇|F 0 ) (see 3.1.4). Consider the ensemble SP𝜇 ≡ SP(G0 , N𝜇 ) ≡ {P ∈ P(T) | ∃ U ∈ G0 ∃ N ∈ N𝜇 (P = U ∪ N)} (see 2.1.4).

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Lemma 1. Let ⟮T, G⟯ be a topological space, M be a 𝜎-algebra on T containing G, and 𝜇 be a positive measure on M. Then, the following assertions hold: 1) the ensemble U0𝜇 is a lattice; 2) the ensemble N𝜇 is an ideal and its subensemble N𝜇 ∩F 0 is cofinal to N𝜇 (see 2.1.4); 3) the ensemble SP𝜇 is an additive foundation (≡ a lattice containing ⌀ and T). Proof. 1. By Proposition 1 (2.2.5), G0 is a 𝜎-foundation, by Lemma 9 (3.1.1) M0 (𝜇) is a 𝜎-ideal in M. This means, in particular, that G0 and M0 (𝜇) are additive and multiplicative. Hence, U0𝜇 has these properties. 2. Since U0𝜇 is multiplicative, N𝜇 is additive. Let N ∈ N𝜇 . By definition, N ⊂ T\U ≡ N 󸀠 for some U ∈ U0𝜇 . If M ⊂ N, then M ⊂ T\U. Consequently, N𝜇 is an ideal ensemble. Further, N ⊂ N 󸀠 , N 󸀠 ∈ N𝜇 , and N 󸀠 ∈ co-U0𝜇 ⊂ co-G0 = F 0 . Thus, N𝜇 ∩ F 0 is cofinal to N𝜇 . 3. Since G0 is an additive foundation and N𝜇 is an ideal, Lemma 5 (2.1.4) implies that SP𝜇 is an additive foundation. Consider the family U(T, SP𝜇 ) of all SP𝜇 -uniform functions on T. By Corollary 4 to Proposition 1 (2.4.3), this family is boundedly normal, in particular, it is a latticeordered linear space. The connection between uniform and quasi-uniform functions established in 2.5.2 allows to give the following useful characterization of this family. Lemma 2. Let ⟮T, G⟯ be a topological space, M be a 𝜎-algebra on T containing G, 𝜇 be a positive measure on M, and f ∈ F b (T). Then, the following conclusions are equivalent: 1) f ∈ U(T, SP𝜇 ); 2) for every number 𝜀 > 0 there are a set U ∈ U0𝜇 and its finite cover 𝜘 ≡ (G k ∈ (G0 )U | k ∈ K) such that 𝜔(f , 𝜘) < 𝜀. Proof. By Proposition 1 (2.2.5), G0 is a 𝜑-foundation, by Lemma 1 N𝜇 is an ideal and N𝜇 ∩F 0 is cofinal to N𝜇 . Then, Proposition 5 (2.5.2) guarantees that QU b (T, G0 , N𝜇 ) = U(T, SP((G0 )𝜑 , N𝜇 )) = U(T, SP𝜇 ). Check that a bounded function f : T → R belongs to QU b (T, G0 , N𝜇 ) iff assertion 2 holds. On the one hand, if f ∈ QU b (T, G0 , N𝜇 ), then, by definition, for every 𝜀 > 0, there are a set V ∈ co-N𝜇 and its finite cover 𝜋 ≡ (G k ∈ (G0 )V | k ∈ K) such that 𝜔(f , 𝜋) < 𝜀. By definition of N𝜇 , there is U ⊂ V such that U ∈ U0𝜇 . Therefore, 𝜘 ≡ (G k ∩ U ∈ (G0 )U | k ∈ K) is a cover of U and 𝜔(f , 𝜋) < 𝜀. On the other hand, if assertion 2 holds, then f ∈ QU b (T, G0 , N𝜇 ) because U0𝜇 ⊂ co-N𝜇 . Corollary 1. Let ⟮T, G⟯ be a topological space, M be a 𝜎-algebra on T containing G, and 𝜇 be a locally complete positive measure on M. Then, the following conclusions are equivalent: 1) f ∼ g mod N𝜇 (see 2.2.6); 2) f ∼ g mod M0 (𝜇).

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Proof. (1) ⊢ (2). This deduction is evident because the local completeness of 𝜇 (see 3.1.4) implies that N𝜇 = Nl (T, F 0 , 𝜇|F 0 ) ⊂ Nl (T, M, 𝜇) ⊂ M0 (𝜇). (2) ⊢ (1). Consider the function h ≡ f − g ∈ U(T, SP𝜇 ). Suppose that |h(t)| < 1/n for every t ∈ ̸ I n ∈ M0 (𝜇). By Lemma 2, there are a set U n ∈ U0𝜇 and its finite cover 𝜘n ≡ (G nk ∈ (G0 )U n | k ∈ K n ) such that 𝜔(h, 𝜘n ) < 𝜀. Let t ∈ U n . Then, t ∈ G nk for some k ∈ K n . Take some point s ∈ G nk \I n . Since |h(s)| < 1/n, we have |h(t)| < 2/n. Hence, f ∼ g mod N𝜇 . Corollary 2. Let f , g ∈ F b (T), f ∼ g mod N𝜇 , and f ∈ U(T, SP𝜇 ). Then, g ∈ U(T, SP𝜇 ). Proof. Let 𝜀 > 0. By Lemma 2, there are a set U ∈ U0𝜇 and its finite cover 𝜘 ≡ (G k ∈ (G0 )U | k ∈ K) such that 𝜔(f , 𝜘) < 𝜀/2. Since f ∼ g mod N𝜇 , I ≡ {t ∈ T | |f (t) − g(t)| ⩾ 𝜀/2} ∈ N𝜇 . By definition of N𝜇 , there is a set V ∈ U0𝜇 such that I ⊂ T\V. By Lemma 1, U0𝜇 is a lattice. Therefore, W ≡ U ∩ V ∈ U0𝜇 . Besides, 𝜋 ≡ (G k ∩ V ∈ (G0 )W | k ∈ K) is a finite cover of W such that 𝜔(g, 𝜋) ⩽ 𝜔(f , 𝜘) + 𝜀/2 < 𝜀. Hence, g ∈ U(T, SP𝜇 ). Lemma 3. Let ⟮T, G⟯ be a topological space, M be a 𝜎-algebra on T containing G, and 𝜇 be a positive bounded measure on M. Then, the set Y ≡ {y ∈ R | f −1 [{y}] ∈ ̸ M0 (𝜇)} is countable. Proof. Since M is a 𝜎-algebra, Corollary 2 to Lemma 1 (2.3.1) guarantees that f −1 [{y}] ∈ M for every y ∈ R. Consider the ensemble X ≡ {f −1 [{y}] | y ∈ R ∧ f −1 [{y}] ∈ ̸ M0 (𝜇)}. For every n ∈ N consider the ensemble Xn ≡ {X ∈ X | 𝜇X > 1/n}. Then, X = ⋃ ⟮Xn | n ∈ N⟯. Suppose that at least one Xn is infinite; then, by Corollary 4 to Proposition 1 (1.3.2), there is an injective mapping u : 𝜔 → Xn . Let X i ≡ u(i) ∈ Xn , m ∈ N, and L m ≡ ⋃ ⟮X i | i ∈ (m + 1)n⟯. If i ≠ j, then X i ≠ X j , and therefore, X i ∩ X j = ⌀. Since 𝜇 is additive, we get 𝜇L m = ∑ (𝜇X i | i ∈ (m + 1)n) ⩾ (1/n)(m + 1)n = m + 1 > m. Since m is arbitrary, this contradicts the property of boundedness of 𝜇. Therefore, every Xn is finite and Theorem 1 (1.3.9) implies that X is countable. Then, the set Y is also countable. Proposition 1. Let ⟮T, G⟯ be a topological space, M be a 𝜎-algebra on T containing G, and 𝜇 be a locally complete positive bounded measure on M. Then, U(T, SP𝜇 ) ⊂ RI(T, G, M, 𝜇). Proof. Let f ∈ U(T, SP𝜇 ), x, y ∈ R, and x < y. Consider the set S ≡ f −1 []x, y[]. By Lemma 1, N𝜇 ∩ F 0 is cofinal to N𝜇 . Consequently, Lemma 3 (2.5.2) implies that S ∈ (SP𝜇 )𝜎 . By Lemma 1 (2.3.5) G0 ⊂ G, where (G0 )𝜎 ⊂ G𝜎 = G. According to Lemma 9 (3.1.1), M0 (𝜇) is a 𝜎-ideal in M. The local completeness of 𝜇 implies that N𝜇 ⊂ M0 (𝜇), where (N𝜇 )𝜎 ⊂ M0 (𝜇)𝜎 = M0 (𝜇). Therefore, (SP𝜇 )𝜎 = SP(G0 , N𝜇 )𝜎

3.7.2 Description of the family of Riemann integrable functions

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= SP((G0 )𝜎 , (N𝜇 )𝜎 ) ⊂ SP(G, M0 (𝜇)). Thus, S = G ∪ J for some G ∈ G and J ∈ M0 (𝜇). Moreover, S = G ∪ I for disjoint sets G ∈ G and I = J\G ∈ M0 (𝜇). Since G ⊂ M, this implies S ∈ M. Therefore, f ∈ M(T, M). By Lemma 3, the set Y ≡ {y ∈ R | f −1 [{y}] ∈ ̸ M0 (𝜇)} is countable. Theorem 3 (1.4.4) and Corollary 1 to Lemma 2 (1.4.9) guarantee that for any n ∈ N and k ∈ Z, the set ]k/(2n), (k + 1)/(2n)[ is uncountable. Hence, Y nk ≡ (R\Y) ∩ ]k/(2n), (k + 1)/(2n)[≠ ⌀. By the choice axiom from 1.1.12, there exists a mapping c : P(R)\{⌀} → R such that p(P) ∈ P. Consider y nk ≡ c(Y nk ). Then, y nk ∈ ̸ Y and y n,k+1 − y nk < 1/n. Since f is bounded, the sets K n ≡ {k ∈ Z | [y nk , y n,k+1 ] ∩ rng f ≠ ⌀} are finite. As was shown above, X nk ≡ f −1 []y nk , y n,k+1 [] = G nk ∪ I nk for some disjoint G nk ∈ G and I nk ∈ M0 (𝜇). Consider the partition 𝜘n ∈ Δ(T, G, M, 𝜇) ≡ Parf (G ∪ M0 (𝜇), T) such that 𝜘n ≡ (Q nki | (k, i) ∈ K n × 3), where Q nk0 ≡ G nk ∈ G, Q nk1 ≡ I nk ∈ M0 (𝜇), and Q nk2 ≡ f −1 [{y nk }] ∈ M0 (𝜇). Finally, we have S(f , 𝜘n ) − s(f , 𝜘n ) < 𝜇T/n. Prove now an analogue of the Lebesgue – Vitali characterization for the family U(T, SP𝜇 ). Theorem 1. Let ⟮T, G⟯ be a topological space, M be a 𝜎-algebra on T containing G, and 𝜇 be a locally complete positive bounded Radon measure on M. Then, the following conclusions are equivalent: 1) f ∈ U(T, SP𝜇 ); 2) the set of all points t ∈ T such that f is not continuous at t is 𝜇-negligible. Proof. (1) ⊢ (2). By Proposition 1 (2.2.5), G0 is a 𝜎-foundation, by Lemma 1 N𝜇 is an ideal and N𝜇 ∩ F 0 is cofinal to N𝜇 . Then, Proposition 5 (2.5.2) guarantees that U(T, SP𝜇 ) = U(T, SP((G0 )𝜑 , N𝜇 )) = QU b (T, G0 , N𝜇 ). Further, by virtue of Theorem 1 (2.5.2), there are functions g ∈ SM bl (T, G0 ) ∩ U(T, SP𝜇 ) and h ∈ SM bu (T, G0 ) ∩ U(T, SP𝜇 ) such that g ⩽ f ⩽ h and g ∼ h mod N𝜇 . Consider the sets E ≡ {t ∈ T | g(t) = h(t)} and I n ≡ {t ∈ T | |g(t) − h(t)| ⩾ 1/n}. Then, T\E = ⋃ ⟮I n | n ∈ N⟯. Corollary 1 to Lemma 2 implies that g ∼ h mod M0 (𝜇), where I n ∈ M0 (𝜇). Therefore, 𝜇(T\E) = 0. Let 𝜀 > 0. For t0 ∈ E, we have g(t0 ) = f (t0 ) = h(t0 ) ≡ y. Consider the sets G ≡ g−1 []y − 𝜀, ∞[] ∈ G0 and H ≡ h−1 [] − ∞, y + 𝜀[] ∈ G0 . Then, t0 ∈ F ≡ G ∩ H ∈ G0 . Hence, E ⊂ F. In addition, by Lemma 1 (2.3.5), F ∈ G. For every t ∈ F we get y − 𝜀 < g(t) ⩽ f (t) ⩽ h(t) < y + 𝜀, where f [F] ⊂]y − 𝜀, y + 𝜀[. This means that the function f is continuous at the point t0 . Thus, the set D of all points t ∈ T such that f is not continuous at t contains in the 𝜇-negligible set T\E. Since 𝜇 is locally complete, we conclude that D is also 𝜇-negligible. (2) ⊢ (1). Let D be the set of all points t ∈ T such that f is not continuous at t and E ≡ T\D. Let m ∈ N and 𝜀 ≡ 1/m. If t ∈ E, then f is continuous at t, and therefore, there is H t ∈ G such that t ∈ H t and f [H t ] ⊂]f (t) − 𝜀/4, f (t) + 𝜀/4[. Consider the set

400 | 3.7 The Riemann integral

Φ ≡ {f ∈ C b (T, G) | 0 ⩽ f ⩽ 1 ∧ ∃ s ∈ H t (f (s) = 1) ∧ f [T\H t ] = {0} } . It is non-empty because ⟮T, G⟯ is a Tychonoff space. Moreover, H t = ⋃ ⟮coz f | f ∈ Φ⟯. Consequently, there is G t ∈ G0 such that t ∈ G t and f [G t ] ⊂]f (t) − 𝜀/4, f (t) + 𝜀/4[. Consider the set V ≡ ⋃ ⟮G t | t ∈ E⟯ ∈ G. Since 𝜇 is a bounded Radon measure, for every n ∈ N, there is a compact set K n ⊂ V m such that 𝜇(V m \K n ) ⩽ 1/n. It follows from K n ⊂ ⋃ ⟮G t | t ∈ E⟯, K n ∈ C, and G t ∈ G0 ⊂ G that there is a finite collection (G ni ∈ G0 | i ∈ I n ) such that K n ⊂ ⋃ ⟮G ni | i ∈ I n ⟯. Then, L ≡ ⋃ ⟮K n | n ∈ N⟯ ⊂ U m ≡ ⋃ ⟮⋃ ⟮G ni | i ∈ I n ⟯ | n ∈ N⟯ ∈ G0 ⊂ G. Since L ⊂ U m ⊂ V m and 𝜇L = 𝜇V m , we get 𝜇U m = 𝜇V m . Further, 0 ⩽ 𝜇(T\U m ) = 𝜇(T\V m ) ⩽ 𝜇(T\E) = 𝜇D = 0. Hence, U m ∈ U0𝜇 . For every n ∈ N, i ∈ I n , and t ∈ G ni , we obtain f [G ni ] ⊂ f [H t ] ⊂] f (t) − 𝜀/4, f (t) + 𝜀/4[. Consider the numbers x ni ≡ inf (f (t) | t ∈ G ni ) and y ni ≡ sup (f (t) | t ∈ G ni ). For these numbers, we get f [G ni ] ⊂]x ni , y ni [ and |y ni − x ni | < 𝜀/2. Consider the set A ≡ ⋃ ⟮{n} × I n | n ∈ N⟯ and indices 𝛼 ≡ (n, i) ∈ A such that i ∈ I n . By Theorem 1 (1.3.9), the set A is countable. Thus, we have the countable cover 𝜋 ≡ (G𝛼 ∩ U m ∈ G0 | 𝛼 ∈ A) of the set U m ∈ U0𝜇 ⊂ co-N𝜇 such that 𝜔(f , 𝜋) ⩽ 𝜀/2 < 𝜀. This means that f ∈ QD b (T, G0 , N𝜇 ). Finally, Propositions 1 and 5 (2.5.2) guarantee that f ∈ U(T, SP𝜇 ). The Zakharov characterization of the family RI(T , G, M, 𝜇) Denote by Gt the set of all open neighborhoods of a point t ∈ T. Let f ∈ F b (T). Define its lower regularization l(f ) and upper regularization u(f ) setting l(f )(t) ≡ sup (inf (f (s) | s ∈ G) | G ∈ Gt ) and u(f )(t) ≡ inf (sup (f (s) | s ∈ G) | G ∈ Gt ) for every t ∈ T. Lemma 4. Let ⟮T, G⟯ be a topological space and f ∈ F b (T). Then, its lower and upper regularizations have the following properties: 1) l(f ) ⩽ f ⩽ u(f ); 2) l(f ) ∈ SC lb (T, G) and u(f ) ∈ SC ub (T, G); 3) if f ∈ SC lb (T, G) [f ∈ SC ub (T, G)], then l(f ) = f [u(f ) = f ]; 4) if g ∈ SC lb (T, G) [h ∈ SC ub (T, G)] and g ⩽ f [f ⩽ h], then g ⩽ l(f ) [u(f ) ⩽ h]. Proof. 1. Let t ∈ T and G ∈ Gt . Then, inf (f (s) | s ∈ G) ⩽ f (t) ⩽ sup (f (s) | s ∈ G). Hence, l(f )(t) ⩽ f (t) ⩽ u(f )(t). −1 2. Let x ∈ R and A ≡ (u(f )) [] − ∞, x[]. If t ∈ A, then there is G t ∈ Gt such that sup (f (s) | s ∈ G t ) < x. Therefore, u(f )(s) < x for every s ∈ G t , where G t ⊂ A. Thus, A = ⋃ ⟮G t | t ∈ A⟯ ∈ G. This means that u(f ) ∈ SC u (T, G). It follows directly from the definition that u(f ) is bounded. The case of l(f ) is considered in the completely similar way.

3.7.2 Description of the family of Riemann integrable functions

| 401

3. Let f ∈ SC ub (T, G). Then, for every t ∈ T, we have S t ≡ f −1 [] − ∞, f (t)[] ∈ G. This implies sup (f (s) | s ∈ S t ) ⩽ f (t), where u(f )(t) ⩽ f (t). According to (1), this means that u(f ) = f . For f ∈ SC lb (T, G), all the arguments are analogous. 4. Let h ∈ SC ub (T, G) and f ⩽ h. Then, using (3), we get u(f ) ⩽ u(h) = h. Similarly, g ∈ SC lb (T, G) and g ⩽ f implies g = l(g) ⩽ l(f ). Lemma 5. Let ⟮T, G⟯ be a topological space, M be a 𝜎-algebra on T containing G, 𝜇 be a positive measure on M, and f ∈ F b (T). If 𝜘 ∈ Δ(T, G, M, 𝜇) (see 3.7.1), then s(f , 𝜘) = s(l(f ), 𝜘) and S(f , 𝜘) = S(u(f ), 𝜘). Proof. Let 𝜘 ≡ (Q k ∈ G ∪ M0 (𝜇) | k ∈ K). Consider the set K 󸀠 ≡ {k ∈ K | Q k ∈ G ∧ Q k ∈ ̸ M0 (𝜇)}. Let k ∈ K 󸀠 . Then, x k ≡ inf (f (t) | t ∈ Q k ) ⩾ inf (l(f )(t) | t ∈ Q k ) ⩾ ⩾ inf (inf (f (s) | s ∈ Q k ) | t ∈ Q k ) = inf (x k | t ∈ Q k ) = x k implies x k = inf (l(f )(t) | t ∈ Q k ). Similarly, y k ≡ sup (f (t) | t ∈ Q k ) ⩾ sup (u(f )(t) | t ∈ Q k ) ⩽ ⩽ sup (sup (f (s) | s ∈ Q k ) | t ∈ Q k ) = sup (y k | t ∈ Q k ) = y k implies y k = sup (u(f )(t) | t ∈ Q k ). Therefore, S(f , 𝜘) = ∑ (y k 𝜇Q k | k ∈ K) = ∑ (sup (u(f )(t) | t ∈ Q k ) 𝜇Q k | k ∈ K 󸀠 ) = S(u(f ), 𝜘) and s(f , 𝜘) = ∑ (x k 𝜇Q k | k ∈ K) = ∑ (inf (l(f )(t) | t ∈ Q k ) 𝜇Q k | k ∈ K 󸀠 ) = s(l(f ), 𝜘). Theorem 2. Let ⟮T, G⟯ be a topological space, M be a 𝜎-algebra on T containing G, and 𝜇 be a locally complete positive bounded Radon measure on M. Then, RI(T, G, M, 𝜇) ⊂ U(T, SP𝜇 ). Proof. Let f ∈ RI(T, G, M, 𝜇). For every a > 0, consider the set P a ≡ {t ∈ T | u(f )(t) − l(f )(t) ⩾ a}. Lemma 1 (3.5.2) implies that l(f ), u(f ) ∈ M(T, M). Hence, P a ∈ M. Let 𝜇P a > 0 for some a. Consider 𝜀 ≡ a𝜇P a > 0.

402 | 3.7 The Riemann integral

Suppose that S(f , 𝜋) − s(f , 𝜋) ⩾ 𝜀 for every 𝜋 ∈ Δ ≡ Δ(T, G, M, 𝜇); then, for every 𝜔, 𝜎 ∈ Δ, we have S(f , 𝜎) − s(f , 𝜔) ⩾ S(f , 𝜔 ∧ 𝜎) − s(f , 𝜔 ∧ 𝜎) ⩾ 𝜀, where 𝜎 ∧ 𝜔 ∈ Δ (see Lemma 3 (2.1.5)). It follows from S(f , 𝜎) ⩾ s(f , 𝜔) + 𝜀 that r𝜇 f ≡ inf (S(f , 𝜎) | 𝜎 ∈ Δ) ⩾ s(f , 𝜔) + 𝜀, where r𝜇 f ≡ sup (s(f , 𝜔) | 𝜎 ∈ Δ) ⩽ r𝜇 f − 𝜀, which is impossible. This contradiction shows that there is 𝜘 ≡ (Q k | k ∈ K) ∈ Δ such that x ≡ S(f , 𝜘) − s(f , 𝜘) < 𝜀. Lemma 5 implies that S(u(f ), 𝜘) − s(l(f ), 𝜘) = x < 𝜀. Consider the non-empty set K a ≡ {k ∈ K | 𝜇(Q k ∩ P a ) > 0}. For every k ∈ K a take some t k ∈ Q k ∩ P a . Then, x ⩾ ∑ ((sup (u(f )(t) | t ∈ Q k ) − inf (l(f )(t) | t ∈ Q k ))𝜇Q k | k ∈ K a ) ⩾ ⩾ ∑ ((u(f )(t k ) − l(f )(t k ))𝜇(Q k ∩ P a ) | k ∈ K a ) ⩾ ⩾ ∑ (a𝜇(Q k ∩ P a ) | k ∈ K a ) = a𝜇P a = 𝜀. This contradicts the inequality x < 𝜀. The last contradiction proves that 𝜇P a = 0 for every a > 0. Consider the set P ≡ {t ∈ T | u(f )(t) − l(f )(t) > 0}. Since P = ⋃ ⟮P1/n | n ∈ N⟯, we get 𝜇P = 0. Denote by D the set of all points t ∈ T such that f is not continuous at t. If t ∈ D, then Statement 1 implies that there is 𝛿 > 0 such that sup (f (s) | s ∈ G) − inf (f (s) | s ∈ G) ⩾ 𝛿 for every G ∈ Gt . Therefore, u(f )(t) − l(f )(t) ⩾ 𝛿, i. e. t ∈ P. Since 𝜇 is locally complete, we conclude that 𝜇D = 0. Then, Theorem 1 implies that f ∈ U(T, SP𝜇 ). Proposition 2. Let ⟮T, G⟯ be a topological space, M be a 𝜎-algebra on T containing G, and 𝜇 be a positive measure on M. Then, the following conclusions are equivalent: 1) f ∈ U(T, SP𝜇 ); 2) for every n ∈ N there are a set U n ∈ U0𝜇 and a function f n : T → R such that f n |U n ∈ C b (U n , GU n ) and |f (t) − f n (t)| < 1/n for every t ∈ U n . Proof. By Proposition 1 (2.2.5), G0 is a separable perfect 𝜎-foundation; by Lemma 1, N𝜇 is an ideal and N𝜇 ∩ F 0 is cofinal to N𝜇 . Then, Proposition 5 (2.5.2) guarantees that QU b (T, G0 , N𝜇 ) = U(T, SP((G0 )𝜑 , N𝜇 )) = U(T, SP𝜇 ). Further, by Corollary 1 to Theorem 4 (2.5.2), the family U(T, SP𝜇 ) is the uniform closure of the family AU b (T, G0 , N𝜇 ) in F b (T). Hence, there exists a sequence (f n ∈ AU b (T, G0 , N𝜇 ) | n ∈ N) such that ‖f − f n ‖u < 1/n. By Lemma 1 (2.5.1), f n ∈ AU b (T, G0 , N𝜇 ) iff there is E n ∈ co-N𝜇 such that f n |E n ∈ U(E n , (G0 )E n ). By definition of N𝜇 , there is U n ⊂ E n such that U n ∈ U0𝜇 . Therefore, Lemma 3 (2.5.1) implies that f n |U n ∈ U(U n , (G0 )U n ). Since (G0 )U n is a 𝜎-foundation, we get U(U n , (G0 )U n ) = M b (U n , (G0 )U n ). Finally, by Lemma 2 (2.3.5), we conclude that M(U n , (G0 )U n ) = M(U n , GU n ) ≡ C(U n , GU n ).

3.7.2 Description of the family of Riemann integrable functions

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Theorem 3 (the Zakharov theorem on characterization of the family of Riemann integrable functions). Let ⟮T, G⟯ be a Tychonoff space, M be a 𝜎-algebra on T containing G, 𝜇 be a locally complete positive bounded Radon measure on M, and f ∈ F b (T). Then, the following conclusions are equivalent: 1) f ∈ RI(T, G, M, 𝜇); 2) f ∈ U(T, SP𝜇 ); 3) f ∈ QU b (T, G0 , N𝜇 ); 4) the set of all points t ∈ T such that f is not continuous at t is 𝜇-negligible; 5) for every n ∈ N, there are a set U n ∈ U0𝜇 and a function f n : T → R such that f n |U n ∈ C b (U n , GU n ) and |f (t) − f n (t)| < 1/n for every t ∈ U n ; 6) there are functions g ∈ SC lb (T, G) and h ∈ SC ub (T, G) such that g ⩽ f ⩽ h and g ∼ h mod N𝜇 ; 7) there are functions g ∈ SC lb (T, G) ∩ RI(T, G, M, 𝜇) and h ∈ SC ub (T, G) ∩ RI(T, G, M, 𝜇) such that g ⩽ f ⩽ h and g ∼ h mod N𝜇 ; 8) there are countable collections u ≡ (g i ∈ C b (T, G) | i ∈ I) and v ≡ (h j ∈ C b (T, G) | j ∈ J) such that g i ⩽ f ⩽ h j for all i ∈ I and j ∈ J and g ∼ h mod N𝜇 , where g ≡ sup u and h ≡ inf v in F(T); 9) there are countable collections (g i ∈ C b (T, G) | i ∈ I) and (h j ∈ C b (T, G) | j ∈ J) and a sequence (U n ∈ U0𝜇 | n ∈ N) such that g i ⩽ f ⩽ h j for all i ∈ I and j ∈ J, and for every n ∈ N and t ∈ U n , there exist i and j such that h j (t) − g i (t) < 1/n. Proof. The equivalence of (1) and (2) follows from Proposition 1 and Theorem 2. By Proposition 1 (2.2.5), G0 is a 𝜑-foundation; by Lemma 1, N𝜇 is an ideal and N𝜇 ∩ F 0 is cofinal to N𝜇 . Then, Proposition 5 (2.5.2) guarantees that U(T, SP𝜇 ) = U(T, SP((G0 )𝜑 , N𝜇 )) = QU b (T, G0 , N𝜇 ). This proves the equivalence of (2) and (3). The equivalence of (2) and (4) follows from Theorem 1. The equivalence of (2) and (5) follows from Proposition 2. The equivalence of (3), (6), and (7) follows from the proved equivalence of (1) and (3) and Theorem 1 (2.5.2). The equivalence of (3) and (8) follows from Theorem 2 (2.5.2). (8) ⊢ (9). Since g ∼ h mod N𝜇 , for every n ∈ N, the set N n ≡ {t ∈ T | h(t) − g(t) ⩾ 1/(2n)} belongs to the ideal N𝜇 . By definition, N n ⊂ T\U n for some U n ∈ U0𝜇 . Let t ∈ U n . Then, there is i ∈ I such that g(t) < g i (t) + 1/(4n) and there is j ∈ J such that h(t) > h j (t) − 1/(4n). Consequently, h j (t) − g i (t) < h(t) + 1/(4n) − (g(t) − 1/(4n)) = h(t) − g(t) + 1/(2n) < 1/n. (9) ⊢ (8). Since U0𝜇 ⊂ co-N𝜇 , this deduction follows from the definition of equivalence of functions with respect to an ideal ensemble. Corollary 1. The family RI(T, G, M, 𝜇) is boundedly normal, i. e. it is a lattice-ordered linear space containing 1 and closed with respect to uniform convergence of sequences and bounded inversion.

404 | 3.7 The Riemann integral

Proof. The assertion follows from the equivalence of conclusions 1 and 2 of Theorem 3, Lemma 1, and Corollary 4 to Proposition 1 (2.4.3). Thus, Theorem 3 gives the comprehensive characterization of the lattice-ordered linear space RI(T, G, M, 𝜇) of Riemann integrable functions. Moreover, the equivalence of (1), (8), and (9) clarifies the countable pointwise nature of relations between this space and its subspace C b (T, G) of bounded continuous functions. The relation between the Lebesgue and the Riemann integrals Proposition 3. Let ⟮T, G⟯ be a topological space, M be a 𝜎-algebra on T containing G, and 𝜇 be a locally complete positive bounded Radon measure on M. Then, 1) RI(T, G, M, 𝜇) ⊂ MI(T, M, 𝜇), i. e. every Riemann integrable function is Mmeasurable and Lebesgue integrable; 2) r𝜇 f = i𝜇 f for every f ∈ RI(T, G, M, 𝜇). Proof. 1. By Theorem 2 and Lemma 4 (2.4.1), we get RI(T, G, M, 𝜇) ⊂ U(T, SP𝜇 ) ⊂ M b (T, (SP𝜇 )𝜎 ). Since SP𝜇 ⊂ M and M is a 𝜎-algebra, we infer that (SP𝜇 )𝜎 ⊂ M. Hence, RI(T, G, M, 𝜇) ⊂ M b (T, M). According to the beginning of 3.3.2, RI(T, G, M, 𝜇) ⊂ F b (T) ⊂ LI(T, M, 𝜇). As a result, we obtain RI(T, G, M, 𝜇) ⊂ MI(T, M, 𝜇). 2. Let f ∈ RI(T, G, M, 𝜇). Take some partition 𝜘 ≡ (Q k ∈ Δ(T, G, M, 𝜇) | k ∈ K) of T. Then, for the Lebesgue integral of f , we get i𝜇 f = ∫(f 1) d𝜇 = ∫(∑ (f 𝜒(Q k ) | k ∈ K)) d𝜇 = = ∑ (∫(f 𝜒(Q k )) d𝜇 | k ∈ K) ⩽ ∑ (∫(sup f [Q k ]𝜒(Q k )) d𝜇 | k ∈ K) . By Lemma 1 (3.3.2), the last sum is equal to ∑ (sup f [Q k ]𝜇Q k | k ∈ K) = S(f , 𝜘). Thus, i𝜇 f ⩽ S(f , 𝜘). Similarly, s(f , 𝜘) ⩽ i𝜇 f . These inequalities imply the final inequality r𝜇 f ⩽ i𝜇 f ⩽ r𝜇 f . Corollary 1. The Riemann integral r𝜇 is a positive linear functional on the latticeordered linear space RI(T, G, M, 𝜇). Proof. The assertion follows from Corollary 1 to Theorem 3, Proposition 3, and Theorem 2 (3.3.2).

Auxiliary definitions and statements 1∘ Let ⟮T, G⟯ be a topological space. A function f : T → R is called continuous at a point t ∈ T if for every 𝜀 > 0, there is a neighborhood V of t such that f [V] ⊂]f (t) − 𝜀, f (t) + 𝜀[.

3.7.3 The Riemann integral for Rn

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Statement 1. Let ⟮T, G⟯ be a topological space, f ∈ F(T), and t ∈ T. Then, the following conclusions are equivalent: 1) f is continuous at the point t; 2) there exists a neighborhood U of t such that f |U ∈ C(U, GU ); 3) for every interval ]x, y[⊂ R containing f (t), there is a neighborhood V of t such that f [V] ⊂]x, y[. 3.7.3 The Riemann integral for ℝn Show now that the definition of the Riemann integral given in 3.7.1 generalizes the classical definition for Rn . Consider the topological space ⟮Rn , Ost ⟯ with the standard topology Ost (see 1∘ (3.1.6)). Consider also the initial Borel – Lebesgue measure 𝜆 : R → R+ on the ring R ≡ R(Rn , Spar ) generated by the semiring Spar of all half-open parallelepipeds P[s, t[ (see 3.1.4) and the extended Borel – Lebesgue measure 𝜆× : L𝜆 → R+ on the 𝜎-algebra L𝜆 ≡ L(Rn , R(Rn , Spar ), 𝜆) of all Borel – Lebesgue measurable sets (see 3.1.6). The Jordan measure in ℝn Let E ⊂ Rn be a bounded set. Consider the ensembles Re (E) ≡ {R ∈ R | E ⊂ R} and Ri (E) ≡ {R ∈ R | R ⊂ E}. The set E is called Jordan measurable if, for every 𝜀 > 0, there are sets R e ∈ Re (E) and R i ∈ Ri (E) such that 𝜆R e − 𝜆R i < 𝜀. The ensemble of all Jordan measurable sets in Rn will be denoted by J𝜆 . For E ∈ J𝜆 , the equality inf{𝜆R e | R e ∈ Re (E)} = sup{𝜆R i | R i ∈ Ri (E)} holds. So we can define the Jordan evaluation m : J𝜆 → R+ setting mE ≡ sup{𝜆R i | R i ∈ Ri (E)}. Since E is bounded, i. e. E is contained in some parallelepiped, the last supremum is finite. The boundedness of E implies also the inclusion J𝜆 ⊂ H𝜎 (Rn , R(Rn , Spar )) (see 2.1.1). Hence, the outer evaluation 𝜆∗ induced by the measure 𝜆 (see 3.1.5) is defined on J𝜆 . Lemma 1. Let E ∈ J𝜆 . Then, 𝜆∗ E ⩽ mE. Proof. Take some R e ∈ Re (E). By Theorem 4 (2.1.1), there is some finite collection 𝜋 ≡ (P k ∈ Spar | k ∈ K) of pairwise disjoint parallelepipeds such that R e = ⋃ ⟮P k | k ∈ K⟯. 𝜎

Since 𝜋 ∈ Par (R, E), we infer that 𝜆∗ E ⩽ ∑ (𝜆P k | k ∈ K) = 𝜆R e . This implies that 𝜆∗ E ⩽ mE. Lemma 2. Let E ∈ J𝜆 . Then, mE ⩽ 𝜆∗ E. 𝜎

Proof. Take some 𝜋 ∈ Par (R, E) and some R i ∈ Ri (E). By definition, 𝜋 = (R k ∈ R | k ∈ K) for some countable set K. Since R i ⊂ E ⊂ rng 𝜋 = ⋃ ⟮R k | k ∈ K⟯, we infer

406 | 3.7 The Riemann integral

that 𝜆R i ⩽ ∑ (𝜆R k | k ∈ K) ≡ v(𝜆, 𝜋). Therefore, mE ⩽ v(𝜆, 𝜋). According to the definition of outer evaluation 𝜆∗ , the previous inequality implies the final inequality mE ⩽ 𝜆∗ E. Lemma 3. Every Jordan measurable set in Rn is Borel – Lebesgue measurable, i. e. J𝜆 ⊂ L𝜆 ≡ L(Rn , R(Rn , Spar ), 𝜆). In addition, the equality 𝜆× E = mE holds for every E ∈ J𝜆 . Proof. Lemmas 1 and 2 imply the equality 𝜆∗ E = mE. By Statement 1, fr E ∈ J𝜆 and m(fr E) = 0. Since fr E is closed, we infer that fr E ∈ L𝜆 . Therefore, 𝜆× (fr E) = 𝜆∗ (fr E) = m(fr E) = 0. Consider the sets G ≡ int E and N ≡ E\G. Since N ⊂ fr E and 𝜆× (fr E) = 0, we conclude that N ∈ L𝜆 and 𝜆× N = 0 (see 3.1.6). The representation E = G ∪ N shows that E ∈ L𝜆 . This means that 𝜆× E = 𝜆∗ E = mE. Now we can prove that the definition of a 𝜇-Jordan set given in 3.7.1 generalizes the classical definition of a Jordan measurable set in Rn considered in this subsection. Theorem 1. Let E be a bounded set in Rn . Then, the following conclusions are equivalent: 1) E is Jordan measurable, i. e. E ∈ J𝜆 ; 2) E is 𝜆× -Jordan, i. e. E ∈ J(Rn , Ost , L𝜆 , 𝜆× ). Proof. (1) ⊢ (2). By Statement 1, fr E ∈ J𝜆 and m(fr E) = 0. By Lemma 3, fr E ∈ L𝜆 and 𝜆× E = mE = 0. Hence, fr E ∈ L0𝜆 (𝜆× ). (2) ⊢ (1). Denote fr E by A. Take some 𝜀 > 0. By definition 𝜆× A = 0, i. e. there is some countable collection (R j ∈ R | j ∈ J) such that A ⊂ ⋃ ⟮R j | j ∈ J⟯ and ∑net (𝜆R j | j ∈ J) < 𝜀/2. By virtue of Theorem 4 (2.1.1) for every j ∈ J, there is some finite collection 𝜋j ≡ (P jl ∈ Spar | l ∈ L j ) of pairwise disjoint parallelepipeds such that R j = ⋃ ⟮P jl | l ∈ L j ⟯. Consider the countable set K ≡ ⋃ ⟮{j}× L j | j ∈ L⟯. Applying Proposition 1 (1.1.13), we get the inclusion A ⊂ ⋃ ⟮⋃ ⟮P jl | l ∈ L j ⟯ | j ∈ J⟯ = ⋃ ⟮P k | k ∈ K⟯. K. By assertion 1 of Proposition 1 (1.1.10), we have Take some bijection u : 𝜔 ⋃ ⟮P k | k ∈ K⟯ = ⋃ ⟮P u(m) | m ∈ 𝜔⟯. Denote P u(m) by Q m . Consider the numbers a j ≡ ∑ (𝜆P jl | (j, l j ) ∈ {j} × L j ) ∈ R+ . Assertion 2 of Theorem 1 (1.4.8) implies the equality ∑net (𝜆P k | k ∈ K) = ∑net (a j | j ∈ J) = = ∑net (∑ (𝜆P jl | (j, l j ) ∈ {j} × L j ) | j ∈ J) = = ∑net (∑ (𝜆P jl | l j ∈ L j ) | j ∈ J) = ∑net (𝜆R j | j ∈ J) < 𝜀/2.

3.7.3 The Riemann integral for Rn

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407

Applying assertion 1 of Theorem 1 (1.4.8), we get ∑net (𝜆Q m | m ∈ 𝜔) = ∑net (𝜆P u(m) | m ∈ 𝜔) = ∑net (𝜆P k | k ∈ K) < 𝜀/2. By virtue of Lemma 1 (3.1.4) for every m ∈ 𝜔, there is the non-empty ensemble Gm of all open parallelepipeds G such that P m ⊂ G and 𝜆G − 𝜆Q m < 𝜀/2m+1 . According to the axiom of choice from 1.1.12, there exists a mapping p : P(P(Rn ))\{⌀} → P(Rn ) such that p(E) ∈ E. Consider the set G m ≡ p(Gm ) ∈ Gm . It now follows that ∑net (𝜆Q m | m ∈ 𝜔) ⩽ ∑net (𝜆G m | m ∈ 𝜔) ⩽ ∑net (𝜆Q m + 𝜀/2m+1 | m ∈ 𝜔) = ∑ (𝜆Q m | m ∈ 𝜔) + ∑ (𝜀/2m+1 | m ∈ 𝜔) < 𝜀/2 + 𝜀/2 = 𝜀. By Statement 1 (3.1.4), the set A is Ost -compact. Hence, there is a finite set M ⊂ 𝜔 such that A ⊂ ⋃ ⟮G m | m ∈ M⟯ ≡ R e . This gives the inequality 𝜆R e ⩽ ∑ (𝜆G m | m ∈ M) ⩽ ∑ (𝜆G m | m ∈ 𝜔) < 𝜀. Therefore, y ≡ inf (𝜆R | R ∈ Re (A)) < 𝜀. Take some R ∈ Ri (A). From R ⊂ R e , we deduce 𝜆R < 𝜀, where x ≡ sup (𝜆R | R ∈ Ri (A)) ⩽ 𝜀. Since 𝜀 is arbitrary, we conclude that x = 0 = y. This means that A ∈ J𝜆 and mA = 0. Now Statement 1 provides that E ∈ J𝜆 . The Riemann integral in the classical sense. The equivalence of two definitions of the Riemann integral for ℝn Let T ⊂ Rn be a Jordan measurable set, G ≡ Ost |T, M ≡ L𝜆 |T, 𝜇 ≡ 𝜆× |T, and J ≡ J𝜆 |T. Consider the set Ω ≡ Ω(T, G, J, m) of all finite covers 𝜔 ≡ (J k ∈ J | k ∈ K ⊂ N) of the set T such that int J k ∩ int J l = ⌀ for every k, l ∈ K, k ≠ l. Let 𝜔 ∈ Ω and f ∈ F b (T). Consider the lower Darboux – Jordan sum 𝜎(f , 𝜔) ≡ ∑ (inf f [J k ]mJ k | k ∈ K) and the upper Darboux – Jordan sum Σ(f , 𝜔) ≡ ∑ (sup f [J k ]mJ k | k ∈ K) . Clearly, the net (𝜎(f , 𝜔) | 𝜔 ∈ Ω) is increasing, the net (Σ(f , 𝜔) | 𝜔 ∈ Ω) is decreasing, and 𝜎(f , 𝜔) ⩽ Σ(f , 𝜔). A bounded function f : T → R is called Riemann integrable in the classical sense on the Jordan measurable topological space ⟮T, G, J, m⟯ if sup (𝜎(f , 𝜔) | 𝜔 ∈ Ω) = inf (Σ(f , 𝜔) | 𝜔 ∈ Ω) ≡ R m f . If f is Riemann integrable in the classical sense, then the number R m f is called the Riemann integral in the classical sense of the function f over the space ⟮T, G, J, m⟯. It is also denoted by ∫ . . . ∫ f (x1 , . . . , x n ) dx1 . . . dx n . T

Finally, show that for Rn , the definition of the Riemann integral from 3.7.1 is equivalent to the given classical definition.

408 | 3.7 The Riemann integral

The classical definition uses covers of T consisted of Jordan measurable sets J ∈ J𝜆 . At the same time, the definition from 3.7.1 uses prime 𝜇-Jordan partitions of T. Therefore, we need in some delicate topological reasoning to prove the equivalence of these definitions. Theorem 2. Let T be a Jordan measurable subset in Rn and f ∈ F b (T). Then, the following conclusions are equivalent: 1) f is Riemann integrable in the sense of 3.7.1; 2) f is Riemann integrable in the classical sense. Besides, r𝜇 f = R m f for every Riemann integrable function f . Proof. (2) ⊢ (1). Let 𝜀 > 0. Then, there is a finite cover 𝜔 ≡ (J k | k ∈ K ⊂ N) ∈ Ω of the set T such that Σ(f , 𝜔) − 𝜎(f , 𝜔) < 𝜀. Consider the sets G k ≡ int J k and M k ≡ J k \G k ⊂ fr J k . Clearly, G k ∈ G and J k = G k ∪ M k . Since F k ≡ fr G k = cl G k \G k ⊂ fr J k and m(fr J k ) = 0, the set F k is Jordan measurable and mF k = 0. Therefore, Statement 1 guarantees that the set G k is Jordan measurable. The set M k is also Jordan measurable and mM k = 0. Define the sets N0 ≡ M0 , N1 ≡ M1 \J0 , and N k ≡ M k \ ⋃ ⟮J l | l ∈ k⟯, k ∈ K\{0}. Theorem 1 and Lemma 1 (3.7.1) guarantee that they are Jordan measurable. Let k, l ∈ K and k < l. Then, N k ∩ N l ⊂ J k ∩ (M l \ ⋃ ⟮J i | i ∈ l⟯) ⊂ J k ∩ (M l \J k ) = ⌀. Consider the sets G ≡ ⋃ ⟮G k | k ∈ K⟯ ∈ G and M ≡ ⋃ ⟮M k | k ∈ K⟯. Lemma 3 implies that M ∈ M0 (𝜇). It is clear that G ∩ M = ⌀ and T = G ∪ M, where N k = M k \ ⋃ ⟮M l | l ∈ k⟯. This implies that M = ⋃ ⟮N k | k ∈ K⟯. Consider the partition 𝜘 ≡ (Q ki | (k, i) ∈ K × 2) ∈ Δ(T, G, M, 𝜇), where Q k0 ≡ G k and Q k1 ≡ N k . It is clear that 𝜘 is finer than 𝜔, i. e. 𝜔 ⩽ 𝜘, where 𝜎(f , 𝜘) ⩾ 𝜎(f , 𝜔) and Σ(f , 𝜘) ⩽ Σ(f , 𝜔). Since G k and N k are Jordan measurable, Lemma 3 implies that mG k = 𝜆× G k = 𝜇G k and mN k = 𝜆× N k = 𝜇N k . Therefore, S(f , 𝜘) = Σ(f , 𝜘) and s(f , 𝜘) = 𝜎(f , 𝜘), where S(f , 𝜘) − s(f , 𝜘) ⩽ Σ(f , 𝜔) − 𝜎(f , 𝜔) < 𝜀. This means that f is Riemann integrable in sense of 3.7.1. Moreover, r𝜇 f ⩾ s(f , 𝜘) = 𝜎(f , 𝜘) ⩾ 𝜎(f , 𝜔) = R m f − (R m f − 𝜎(f , 𝜔)) ⩾ R m f − (Σ (f , 𝜔) − 𝜎(f , 𝜔)) > R m f − 𝜀 and r𝜇 f ⩽ S(f , 𝜘) = Σ(f , 𝜘) ⩽ Σ(f , 𝜔) = R m f + (Σ(f , 𝜔)− R m f ) ⩽ R m f + (Σ(f , 𝜔) − 𝜎(f , 𝜔)) < R m f + 𝜀. Since 𝜀 is arbitrary, we conclude that r𝜇 f = R m f . (1) ⊢ (2). Let 𝜀 > 0. According to 1) there is a finite partition 𝜘 ≡ (Q k | k ∈ K) ∈ Δ(T, G, M, 𝜇) of the set T such that S(f , 𝜘) − s(f , 𝜘) < 𝜀. Consider the sets K 󸀠 ≡ {k ∈ K | Q k ∈ G ∧ Q k ∉ M0 (𝜇)}, K 󸀠󸀠 ≡ {k ∈ K | Q k ∈ M0 (𝜇)}, G ≡ ⋃ ⟮Q k | k ∈ K 󸀠 ⟯ ∈ G, and N ≡ ⋃ ⟮Q k | k ∈ K 󸀠󸀠 ⟯ ∈ M0 (𝜇). Since 𝜘 is a partition, T = G ∪ N and G ∩ N = ⌀. Denote by T 󸀠 the closure of the set T in the topological space ⟮Rn , Ost ⟯. Consider the topology G󸀠 ≡ Ost |T 󸀠 on the set T 󸀠 . Since fr T 󸀠 = T 󸀠 \ int T 󸀠 ⊂ fr T and m(fr T) = 0, according to Statement 1, we see that m(fr T 󸀠 ) = 0. Hence, the set T 󸀠 is Jordan measurable. It follows from X ≡ T 󸀠 \T ⊂ fr T that X ∈ J and mX = 0, where 𝜆× X = 0.

3.7.3 The Riemann integral for Rn

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Let k ∈ K 󸀠 . Since Q k ∈ G, Q k = T ∩ G k for some G k ∈ Ost . Consider the set N k ≡ X ∩ G k . It follows from N k ⊂ X that N k ∈ J and mN k = 0. By Statement 1, m(fr N k ) = 0. Since N k󸀠 ≡ cl N k \N k ⊂ fr N k , we have N k󸀠 ∈ J and mN k󸀠 = 0. Therefore, M k ≡ cl N k = N k ∪ N k󸀠 ∈ J and mM k = 0, where M = ⋃ ⟮M k | k ∈ K 󸀠 ⟯ ∈ J and mM = 0. Consider the sets H k ≡ G k \M ∈ Ost and R k ≡ H k ∩ T 󸀠 ∈ G󸀠 for every k ∈ K 󸀠 . Since G k = Q k ∪ (G k \T 󸀠 ) ∪ N k , we get G k \M = (Q k \M) ∪ ((G k \T 󸀠 )\M) ∪ (N k \M) ⊂ Q k ∪ (G k \T 󸀠 ). This implies that R k = (G k \M) ∩ (T 󸀠 \M) ⊂ (Q k ∪ (G k \T 󸀠 )) ∩ (T 󸀠 \M) ⊂ (Q k ∪ (G k \T 󸀠 )) ∩ T 󸀠 = Q k . Thus, R k ∩ R l ⊂ Q k ∩ Q l = ⌀ for all k, l ∈ K 󸀠 , k ≠ l. Besides, X k ≡ Q k \R k = (G k ∩ T) ∩ ((G k ∩ T 󸀠 )\M) ⊂ (((G k ∩ T 󸀠 ) ∩ M) ∪ ((G k ∩ T 󸀠 )\ 󸀠 M)) ∩ ((G k ∩ T )\M) ⊂ M. Therefore, X k ∈ J and mX k = 0, where 𝜆× X k = 0. Denote by F k the closure of the set R k in the topological space ⟮T 󸀠 , G󸀠 ⟯. It follows from R k ⊂ T 󸀠 ⊂ Rn that F k = T 󸀠 ∩ cl R k . Since the set T 󸀠 is closed, we have cl R k ⊂ cl T 󸀠 = T 󸀠 . Hence, F k = cl R k . For all l ∈ K 󸀠 , the sets R l are open in the topological space ⟮T 󸀠 , G󸀠 ⟯ and R l ∩ R k = ⌀ for every l ≠ k. Therefore, R l ∩ F k = ⌀ for every l ∈ K 󸀠 \{k}, where R l ∩ (F k \R k ) = ⌀ for all l ∈ K 󸀠 . Consider the set Y ≡ (T 󸀠 \T) ∪ (T\ ⋃ ⟮Q l | l ∈ K 󸀠 ⟯) ∪ (⋃ ⟮Q l | l ∈ K 󸀠 ⟯\ ⋃ ⟮R l | l ∈ K 󸀠 ⟯) = X ∪ (T\G) ∪ ⋃ ⟮Q l \R l | l ∈ K 󸀠 ⟯ = X ∪ N ∪ ⋃ ⟮X l | l ∈ K 󸀠 ⟯. Since 𝜇N = 0, we get 𝜆× N = 0. Therefore, 𝜆× Y ⩽ 𝜆× X + 𝜆× N + ∑ (𝜆× X l | l ∈ K 󸀠 ) = 0. By virtue of F k \R k ⊂ Y, this implies that 𝜆× (F k \R k ) = 0 for every l ∈ K 󸀠 . It follows from R k ≡ H k ∩ T 󸀠 that int R k = int(H k ∩ T 󸀠 ) = H k ∩ int T 󸀠 = R k ∩ int T 󸀠 in the topological space ⟮Rn , Ost ⟯. Hence, R k \ int R k = R k \ int T 󸀠 ⊂ fr T 󸀠 . Since m(fr T 󸀠 ) = 0, we get R k \ int R k ∈ J and m(R k \ int R k ) = 0. Therefore, 𝜆× (R k \ int R k ) = 0. Thus, 𝜆× (F k \R k ) = 𝜆× (R k \ int R k ) = 0 for every k ∈ K 󸀠 . Using the equality fr R k = cl R k \ int R k = (F k \R k ) ∪ (R k \ int R k ), we get 𝜆× (fr R k ) = 0. Since the set fr R k is closed, it is 𝜆× -Jordan. Then, according to Theorem 1, it is Jordan measurable and, according to Lemma 3, m(fr R k ) = 0. Therefore, R k ∈ J, where Q k = R k ∪ X k ∈ J for every k ∈ K 󸀠 . As was shown above, 𝜆× Y = 0. The equality Y = T 󸀠 \ ⋃ ⟮H k | k ∈ K 󸀠 ⟯ implies that the set Y is closed in Rn . Hence, Y ∈ J and mY = 0. Since Q k ⊂ N = T\G ⊂ Y, this implies that Q k ∈ J and mQ k = 0 for every k ∈ K 󸀠󸀠 . Thus, the partition 𝜘 consists of Jordan measurable sets. Therefore, mQ k = 𝜇Q k for every k ∈ K, where S(f , 𝜘) = Σ(f , 𝜘) and s(f , 𝜘) = 𝜎(f , 𝜘). This provides that Σ(f , 𝜘) − 𝜎(f , 𝜘) < 𝜀. This means that f is Riemann integrable in the classical sense. Moreover, R m f ⩾ 𝜎(f , 𝜘) = s(f , 𝜘) ⩾ r𝜇 f − (r𝜇 f − s(f , 𝜘)) ⩾ r𝜇 f − (S(f , 𝜘) − s(f , 𝜘)) > r𝜇 f − 𝜀 and R m f ⩽ Σ(f , 𝜘) = S(f , 𝜘) = r𝜇 f + (S(f , 𝜘) − r𝜇 f ) ⩽ r𝜇 f + (S(f , 𝜘) − s(f , 𝜘)) < r𝜇 f + 𝜀. Since 𝜀 is arbitrary, we conclude that R m f = r𝜇 f . Auxiliary statements Statement 1. A set E ⊂ Rn is Jordan measurable iff fr E ∈ J𝜆 and m(fr E) = 0.

D Historical notes on the Riesz – Radon – Fréchet problem of characterization of Radon integrals as linear functionals This appendix is devoted to a survey of results connected with the problem of characterization of integrals as linear functionals, which has a long and rich history. It goes back to a famous result of F. Riesz (1909) about the descriptive characterization of Riemann – Stiltjes integrals as linear functionals. By virtue of the significance of this result, we present it in more details.

D.1 The original Riesz representation theorem The interest in characterization of integrals appeared simultaneously with the appearance of the Riemann – Stiltjes integrals (1894).

D.1.1 Functions of bounded variation Let T ≡ [a, b] be a closed interval in R. According to 1.1.15, |s, t| denotes an interval of a general kind in R determined by the end-points s ⩽ t. Denote by R the ensemble of all intervals |s, t| ⊂ T. Consider the set Parf (R, T) of all finite partitions (R j ∈ R | j ∈ J) of T (see also 3.1.3). For a function v : T → R and an interval R ≡ |s, t|, put v|R| ≡ |v(t) − v(s)|. For a finite partition 𝜌 ≡ (R j ∈ R | j ∈ J) of T, put v(v, 𝜌) ≡ ∑ (v|R j | | j ∈ J). A real-valued function v : T → R is called a function of bounded variation on T if var(v) ≡ sup(v(v, 𝜌) | 𝜌 ∈ Parf (R, T)) < ∞. The set of all functions of bounded variation on T is denoted by BV(T). Collect some basic properties of functions of bounded variation in the following proposition. Proposition 1. Let T ≡ [a, b] be a closed interval in R. Then, 1) if v ∈ BV(T), then v is bounded; 2) every bounded increasing function on T belongs to BV(T); 3) for every v ∈ BV(T) there are bounded positive increasing functions v1 and v2 such that v = v1 − v2 ; 4) the family BV(T) is a lattice-ordered linear space.

https://doi.org/10.1515/9783110550962-003

412 | D.1 The original Riesz representation theorem

D.1.2 The Riemann – Stiltjes integral For an interval |s, t| ⊂ R, define its length l|s, t| ≡ t − s. Take 𝜌 ∈ Parf (R, T) and define the mesh l(𝜌) ≡ gr (lR j | j ∈ J) of the partition 𝜌. Every collection 𝜌∗ ≡ (r j ∈ R j | j ∈ J) is called a selection from the partition 𝜌. Denote by M the set of all triplets 𝜇 ≡ (𝜌, 𝜌∗ , l(𝜌)). Define a preorder on M setting 𝜇 ≡ (𝜌, 𝜌∗ , l(𝜌)) ⩽ (𝜎, 𝜎∗ , l(𝜎)) ≡ 𝜈 if l(𝜎) ⩽ l(𝜌). The preordered class ⟮M, ⩽⟯ is upward directed (see 1.1.15). Let f ∈ F(T), v ∈ BV(T), 𝜇 ≡ (𝜌, 𝜌∗ , l(𝜌)) ∈ M, 𝜌 ≡ (R j | j ∈ J), R j ≡ |r󸀠j , r󸀠󸀠j |, and ∗ 𝜌 ≡ (r j ∈ R j | j ∈ J). Consider the Riemann – Stiltjes integral sum S(f , v, 𝜇) ≡ ∑ (f (r j )(v(r󸀠󸀠j ) − v(r󸀠j )) | j ∈ J) . The function f is called Riemann – Stiltjes integrable on T with respect to v if there is a number S ∈ R such that for every 𝜀 > 0 there exists 𝜆 ≡ (𝜘, 𝜘∗ , l(𝜘)) ∈ M such that for every 𝜇 ⩾ 𝜆 the inequality |S − S(f , v, 𝜇)| < 𝜀 holds. This number is unique, and it is denoted by S ≡ lim (S(f , v, 𝜇) | 𝜇 ∈ M) ≡ i v f ≡ ∫T f dv and is called the Riemann – Stiltjes integral of the function f with respect to the function v. The set of all Riemann – Stiltjes integrable functions on T with respect to v will be denoted by RI(T, v). Proposition 1. Let v ∈ BV(T). Then, 1) |i v f | ⩽ var(v)‖f ‖u for every f ∈ F b (T); 2) the family RI(T, v) is a lattice-ordered linear space; 3) the mapping i v : RI(T, v) → R is a linear functional; 4) if v = v1 − v2 for some bounded increasing v1 and v2 , then i v = i v1 − i v2 and the functionals i v1 and i v2 are positive.

D.1.3 The Hadamard – Fréchet problem Since there are many Riemann – Stiltjes integrals i v with respect to different functions v and they are defined on different function families RI(T, v), it is natural to consider them on some good linear subspace A(T) of the linear space ⋂⟮RI(T, v) | v ∈ BV(T)⟯ of all universally Riemann – Stiltjes integrable functions on T. Such a good subspace exists, this is the space C(T) ≡ C(T, G) ≡ M(T, G) of all continuous functions on T. Thus, we get the family I(C(T), BV(T)) ≡ {i v |C(T) | v ∈ BV(T)} of all Riemann – Stiltjes integrals on C(T) considered as linear functionals on C(T). Proposition 1. The family I(C(T), BV(T)) is a proper linear subspace of the linear space C(T)× of all linear functionals on C(T), i. e. I(C(T), BV(T)) ⫋ C(T)× .

D.1.4 The Riesz theorem

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In connection with this result, J. Hadamard [1903] and M. Fréchet [1904] had raised the following problem: to characterize the linear space I(C(T), BV(T)) of all Riemann – Stiltjes integrals on C(T) in the linear space C(T)× of all linear functionals on C(T) (see Historical notes in [Bourbaki, 1965; 1956)].

D.1.4 The Riesz theorem Consider the family C(T)∼ of all order bounded (see 1.1.15) or bipositive (≡ order regular, see 4∘ (2.2.7) and 2∘ (2.2.8)) linear functionals on C(T). It follows from Corollary 1 to Lemma 3 (2.2.8) that this family coincides with the family ⟮C(T), ‖ ⋅ ‖u ⟯󸀠 of all norm bounded (see 4∘ (2.2.7)) or continuous linear functionals on C(T). Proposition 1. Every Riemann – Stiltjes integral on C(T) is an (order) bounded functional, i. e. I(C(T), BV(T)) ⊂ C(T)∼ . In attempt to prove the inverse inclusion, J. Hadamard [1903] proved that every funcb tional 𝜑 ∈ C(T)∼ had the representation 𝜑f = lim ∫a f (x)p n (x) dx, where p n ∈ C(T). n→∞

This result was refined in [Fréchet, 1904]. F. Riesz announced in [Riesz, 1909] the following result. Theorem 1 (the Riesz representation theorem). The family of all Riemann – Stiltjes integrals on C(T) coincides with the family of all (order) bounded linear functionals on C(T), i. e. I(C(T), BV(T)) = C(T)∼ . Some proofs of this famous theorem were given in papers [Riesz, 1911; Helly, 1912]. The most transparent proof was obtained by F. Riesz himself [Riesz, 1914], who used W. H. Young’s method [Young, 1911] of extension of a positive linear functional on the lattice-ordered linear space of semicontinuous functions (see 2.3.8). Below, we expound a scheme of this proof.

D.1.5 Extension of positive functionals by Young’s method According to Lemma 7 (3.6.2), any functional 𝜑 from C(T)∼ is pointwise 𝜎-continuous. Using this property, we can extend 𝜑 on a wider family of functions. This method was first used by W. H. Young [1911]. Consider the conic spaces SC lb (T) ≡ SC lb (T, G) and SC ub (T) ≡ SC ub (T, G) of all bounded lower semicontinuous and bounded upper semicontinuous functions on T ≡ [a, b] ⊂ R (see 3.5.2 and 2.3.8). Consider also the family SC(T) ≡ BSM b (T, G) ≡ {g1 − g2 | g1 , g2 ∈ SC lb (T)+ } = {h1 − h2 | h1 , h2 ∈ SC ub (T)+ }. It is clear

414 | D.1 The original Riesz representation theorem

that SC(T) = {g1 + h1 | g1 ∈ SC lb (T) ∧ h1 ∈ SC ub (T)}. By Lemma 2 (2.3.8), the family SC(T) is a lattice-ordered linear space. It follows directly from the definitions that 𝜒([a, y[), 𝜒(]x, y[), 𝜒(]x, b]) ∈ SC lb (T) ⊂ SC(T) and 𝜒([x, y]) ∈ SC ub (T) ⊂ SC(T) for any x, y ∈ T, x < y. Let 𝜑 ∈ (C(T)∼ )0 . Define the mapping 𝜑∗ : SC lb (T) → R setting 𝜑∗ g ≡ sup{𝜑f | f ∈ C(T) ∧ f ⩽ g} for every g ∈ SC lb (T) (compare with the definition of 𝜑 in 3.6.2). Since |g| ⩽ x1 for some x ∈ R+ and 1 ∈ C(T), we have the functional 𝜑∗ : SC lb (T) → R. Clearly, 𝜑∗ is an extension of 𝜑. Lemma 1. Let g ∈ SC lb (T) and (f n | n ∈ N) be a sequence of continuous functions on T such that (f n | n ∈ N) ↑ g. Then, 𝜑∗ g = sup (𝜑f n | n ∈ N) (cf. Lemma 3 (3.6.2)). Proof. Let f ∈ C(T) and f ⩽ g. Then, ((f − f n ) ∨ 0 | n ∈ N) ↓ (f − g) ∨ 0 = 0. Since 𝜑 is pointwise 𝜎-continuous, we infer that lim (𝜑((f − f n ) ∨ 0) | n ∈ N) = 0. By virtue of positivity of 𝜑, we conclude that 𝜑f = 𝜑f − 0 = 𝜑f − lim (𝜑((f − f n ) ∨ 0) | n ∈ N) = = lim (𝜑f − 𝜑((f − f n ) ∨ 0) | n ∈ N) = lim (𝜑(f − (f − f n ) ∨ 0) | n ∈ N) = = lim (𝜑(f + (f − f n ) ∧ 0) | n ∈ N) = lim (𝜑((f + f n − f ) ∧ f ) | n ∈ N) = = lim (𝜑(f n ∧ f ) | n ∈ N) = sup (𝜑(f n ∧ f ) | n ∈ N) ⩽ sup (𝜑f n | n ∈ N) . Therefore, 𝜑∗ g ⩽ sup (𝜑f n | n ∈ N) ⩽ 𝜑∗ g, where 𝜑∗ g = sup (𝜑f n | n ∈ N). Lemma 2. The functional 𝜑∗ : SCl b (T) → R is conic. Lemma 3. Let g1 , g2 , g1󸀠 , g2󸀠 ∈ SC lb (T) and g1 − g2 = g1󸀠 − g2󸀠 . Then, 𝜑∗ g1 − 𝜑∗ g2 = 𝜑∗ g1󸀠 − 𝜑∗ g2󸀠 . Proof. Since g1 + g2󸀠 = g1󸀠 + g2 ∈ SC lb , Lemma 2 guarantees that 𝜑∗ g1 + 𝜑∗ g2󸀠 = 𝜑∗ g1󸀠 + 𝜑∗ g2 . This implies the desired equality. This lemma allows to define the functional 𝜑 : SC(T) → R setting 𝜑(g1 − g2 ) ≡ 𝜑∗ g1 − 𝜑∗ g2 . Lemma 4. The functional 𝜑 is linear and positive.

D.1.6 The property of norm preserving The idea of construction of the functional 𝜑 is from W. H. Young. But F. Riesz proved that 𝜑 has same norm as the functional 𝜑. It is a key property of the Young extension.

D.1.6 The property of norm preserving

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To prove the property of norm preserving, F. Riesz used his refined method of alternation of two increasing sequences of continuous functions. Let 𝜉 ∈ SC(T) and 𝜉 = g󸀠 − g󸀠󸀠 for some g󸀠 , g󸀠󸀠 ∈ SC lb (T). Since the standard open topology G on T ≡ [a, b] ⊂ R is perfect and separable, Theorem 2 (2.3.5) and Corollary 1 to Theorem 1 (2.3.8) provides that there exist (initial) increasing sequences (f n󸀠 ∈ C(T) | n ∈ N) ↑ g 󸀠 and (f n󸀠󸀠 ∈ C(T) | n ∈ N) ↑ g 󸀠󸀠 . Denote ‖𝜉‖u ≡ sup (𝜉(t) | t ∈ T) by e. Consider the disjoint sets F n ≡ {s ∈ T | 󸀠 |f n (s) − f n󸀠󸀠 (s)| ⩽ e}, G n ≡ {s ∈ T | f n󸀠 (s) − f n󸀠󸀠 (s) > e}, and H n ≡ {s ∈ T | f n󸀠 (s) − f n󸀠󸀠 (s) < −e}. It is clear that T = F n ∪ G n ∪ H n for every n ∈ N. Define the new sequence (f n ∈ F(T) | n ∈ N) setting f n (t) ≡ f n󸀠 (t) for every t ∈ F n , f n (t) ≡ f n󸀠󸀠 (t) + e for every t ∈ G n , and f n (t) ≡ f n󸀠󸀠 (t) − e for every t ∈ H n . Obviously, f n ∈ C(T). This new sequence has the property of uniform proximity to the sequence (f n󸀠󸀠 | n ∈ N), which the initial sequence (f n󸀠 | n ∈ N) has not. Lemma 1. ‖f n − f n󸀠󸀠 ‖u ⩽ e ≡ ‖𝜉‖u for every n ∈ N. Lemma 2. f n ⩽ g󸀠 for every n ∈ N. Lemma 3. The sequence (f n | n ∈ N) increases. Lemma 4. p-lim (f n | n ∈ N) = g󸀠 . Proof. Fix t ∈ T and 𝜀 > 0. Then, there is m ∈ N such that |g 󸀠 (t) − f n󸀠 (t)| < 𝜀 and |g󸀠󸀠 (t) − f n󸀠󸀠 (t)| < 𝜀 for every n ⩾ m. Remind that by construction T = F n ∪ G n ∪ H n . Let t ∈ F n . Then, f n (t) ≡ f n󸀠 (t) implies |g 󸀠 (t) − f n (t)| < 𝜀. Let t ∈ G n . Using Lemma 2, we get 0 ⩽ g󸀠 (t) − f n (t) = (g󸀠 (t) − g󸀠󸀠 (t)) + (g󸀠󸀠 (t) − 󸀠󸀠 f n (t)) − e ⩽ e + (g 󸀠󸀠 (t) − f n󸀠󸀠 (t)) − e < 𝜀. Therefore, |g󸀠 (t) − f n (t)| < 𝜀. Let t ∈ H n . Since f n󸀠 (t)− f n󸀠󸀠 (t) < −e, using Lemma 2, we get 0 ⩽ g󸀠 (t)− f n (t) = g󸀠 (t) − 󸀠󸀠 f n (t) + e = g 󸀠 (t) − f n󸀠 (t) + (f n󸀠 (t) − f n󸀠󸀠 (t)) + e < 𝜀 − e + e = 𝜀. Therefore, |g󸀠 (t) − f n (t)| < 𝜀. Thus, we conclude that g󸀠 (t) = lim (f n (t) | n ∈ N). Corollary 1. g󸀠 = sup (f n | n ∈ N). Proof. The assertion follows from Lemmas 3 and 4 and Lemma 7 (1.4.7). Corollary 2. 𝜑∗ g󸀠 = sup (𝜑f n | n ∈ N) = lim (𝜑f n | n ∈ N). Proof. The assertion follows from Lemmas 3 and 4 and Lemma 1 (D.1.5). Using all the properties proven above, we obtain the property of norm preserving. Proposition 1. Let 𝜉 ∈ SC(T). Then, |𝜑𝜉| ⩽ ‖𝜑‖󸀠 ‖𝜉‖u .

416 | D.1 The original Riesz representation theorem

Proof. Take g󸀠 , g󸀠󸀠 ∈ SC lb (T) such that 𝜉 = g󸀠 − g󸀠󸀠 and the sequences (f n ∈ C(T) | n ∈ N) ↑ g 󸀠 and (f n󸀠󸀠 ∈ C(T) | n ∈ N) ↑ g 󸀠󸀠 constructed above. Then, using Corollary 2 to Lemma 4, we get 𝜑𝜉 ≡ 𝜑∗ g󸀠 − 𝜑∗ g󸀠󸀠 = lim (𝜑f n | n ∈ N) − lim (𝜑f n󸀠󸀠 | n ∈ N) = lim (𝜑(f n − f n󸀠󸀠 ) | n ∈ N). According to Lemma 1, 𝜑(f n − f n󸀠󸀠 ) ⩽ |𝜑(f n − f n󸀠󸀠 )| ⩽ ‖𝜑‖󸀠 ‖f n − f n󸀠󸀠 ‖u ⩽ ‖𝜑‖󸀠 ‖𝜉‖. Therefore, lim (𝜑(f n − f n󸀠󸀠 ) | n ∈ N) ⩽ ‖𝜑‖󸀠 ‖𝜉‖. Similarly, 𝜑(f n − f n󸀠󸀠 ) ⩾ −|𝜑(f n − f n󸀠󸀠 )| ⩾ −‖𝜑‖󸀠 ‖f n − f n󸀠󸀠 ‖u ⩾ −‖𝜑‖󸀠 ‖𝜉‖ implies lim (𝜑(f n − f n󸀠󸀠 ) | n ∈ N) ⩾ −‖𝜑‖󸀠 ‖𝜉‖. Thus, |𝜑𝜉| =

| lim (𝜑(f n − f n󸀠󸀠 ) | n ∈ N) | ⩽ ‖𝜑‖󸀠 ‖𝜉‖. This implies that ‖𝜑‖󸀠 = ‖𝜑‖󸀠 .

D.1.7 Construction of the integral corresponding to a given functional Let 𝜑 ∈ (C(T)∼ )0 and 𝜑 be its extension constructed above. Define the function v : T → R setting v(a) = 0 and v(t) ≡ 𝜑(𝜒([a, t])) for every t ∈]a, b]. Lemma 1. The function v is increasing and bounded, and therefore, v ∈ BV(T). Proof. By Lemma 4 (D.1.5), the functional 𝜑 is linear and positive, and therefore, the function v is increasing. Since 0 ⩽ v ⩽ v(b)1 = (𝜑1)1, v is bounded. Then, assertion 2) of Lemma 1 (D.1.1) guarantees that v ∈ BV(T). Theorem 1. Let f ∈ C(T). Then, 𝜑f = i v f . Proof. Fix 𝜀 > 0. By the definition of the Riemann – Stiltjes integral there is 𝜆 ≡ (𝜘, 𝜘∗ , l(𝜘)) ∈ M such that |i v f − S(f , v, 𝜇)| < 𝜀/3 for every 𝜇 ⩾ 𝜆. Since the function f is continuous and the closed interval T ≡ [a, b] is compact, Proposition 2 (2.3.1) guarantees that f is uniformly continuous, i. e. there is 𝛿 > 0 such that the conditions t, s ∈ T and |t − s| < 𝛿 imply that |f (t) − f (s)| < 3𝜀/(8‖𝜑‖󸀠 ). Consider a partition 𝜌 ≡ (R j | j ∈ J) ∈ Parf (R, T) with l(𝜌) = l(𝜘) ⊼ (𝛿/2), the selection 𝜌∗ ≡ (r j ∈ R j | j ∈ J) with r j ≡ (r󸀠j + r󸀠󸀠j )/2 for R j ≡ |r󸀠j , r󸀠󸀠j |, and 𝜇 ≡ (𝜌, 𝜌∗ , l(𝜌)). Since l(𝜌) ⩽ l(𝜘), we have 𝜇 ⩾ 𝜆. Hence, |i v f − S(f , v, 𝜇)| < 𝜀/3. Denote the index j ∈ J such that r󸀠j = a by j0 and denote the set J\{j0 } by J0 . Using the definition of the function v and linearity of the functional 𝜑, we get S(f , v, 𝜇) ≡ ∑ (f (r j )(v(r󸀠󸀠j ) − v(r󸀠j )) | j ∈ J) = 𝜑(∑ (f (r j )𝜒(]r󸀠j , r󸀠󸀠j ]) | j ∈ J0 ) + f (r j0 )𝜒([a, r󸀠󸀠j0 ])) ≡ x. Then, by Proposition 1 (D.1.6), we obtain |𝜑f − i v f | ⩽ |𝜑f − S(f , v, 𝜇)| + |S(f , v, 𝜇) − i v f | < |𝜑f − x| + 𝜀/3 ⩽ ‖𝜑‖󸀠 ‖f − ∑ (f (r j )𝜒(]r󸀠j , r󸀠󸀠j ]) | j ∈ J0 ) − f (r j0 )𝜒([a, r󸀠󸀠j0 ])‖u + 𝜀/3. For every t ∈ T, either t = a or t ∈]r󸀠i , r󸀠󸀠i ] for some i ∈ J. For t = a, put i = j0 . In either case, |t − r i | ⩽ l(𝜌) < 𝛿, and therefore, |f (t) − ∑ (f (r j )𝜒(]r󸀠j , r󸀠󸀠j ]) | j ∈ J0 ) (t) −

D.1.7 Construction of the integral corresponding to a given functional

| 417

f (r j0 )𝜒([a, r󸀠󸀠j0 ])(t)| = |f (t) − f (r i )| < 3𝜀/(8‖𝜑‖󸀠 ). Thus, |𝜑f − i v f | < 3‖𝜑‖󸀠 𝜀/(8‖𝜑‖󸀠 ) + 𝜀/3 < 7𝜀/8 < 𝜀. Since 𝜀 > 0 is arbitrary, we infer that 𝜑f = i v f . A sophisticated generalization of the Riesz theorem for a parallelepiped P[s, t] in Rn (see 2.1.1 and 3.1.4) was given by C. A. Fisher [1915], who generalized the twodimensional Riemann – Stiltjes integral introduced in [Fréchet, 1910]. But unfortunately, this elegant method is absolutely not suitable for any domain T ⊂ Rn different from the parallelepiped P[s, t]. Therefore, the proper way to generalize the Riemann – Stiltjes integrals onto other domains, T ⊂ Rn is to abandon functions of bounded variation and to conceive the notion of a measure. This transition was made by J. Radon.

D.2 Transition from functions of bounded variation to bounded measures on compact spaces When the Lebesgue integral (more precisely, Lebesgue – Radon – Fréchet integral) had appeared the problem of characterization of integrals as linear functionals obtained further development. Combining the ideas of H. Lebesgue and T. Stiltjes, J. Radon introduced in paper [Radon, 1913] the notion of quite additive set functions 𝜇 : M(T) → R defined on some 𝜎-algebras M(T) on a compact set T in Rn . He essentially used the following topological property of a quite additive set function 𝜇: for every M ∈ M(T) and every 𝜀 > 0, there are a closed-set F and an open-set G such that F ⊂ M ⊂ G and |𝜇|(G\F) < 𝜀. Later on, this property was called the topological regularity, and 𝜎-additive set functions (i. e. measures) possessing this property were called Radon measures (see 3.5.3 and 3.5.1). Denote the family of these set functions by RF(T). Using the Lebesgue integral construction for Rn , he defined the integral i𝜇 f ≡ ∫T f d𝜇 giving a linear functional on the space C(T) of all continuous functions on T. The family I(C(T), RF(T)) ≡ {i𝜇 |C(T) | 𝜇 ∈ RF(T)} of all Lebesgue – Radon integrals on C(T) is as above a proper linear subspace of the linear space C(T)× . J. Radon proved the following theorem: Theorem 1. The space of all Lebesgue – Radon integrals on C(T) for closed bounded (compact) subspace T in Rn coincides with the space of all bounded linear functionals on C(T), i. e. I(C(T), RF(T)) = C(T)∼ ⫋ C(T)× . “Almost at once after issuing Radon’s memoir Fréchet notices that almost all results of this paper may be extended to the case when ‘quite additive set function’ is not considered only for measurable subsets of the space Rn but is considered for some subsets of an arbitrary set E (subsets having the following property: a function is also defined on every countable union and every relative complement of these subsets)” (see Historical notes in [Bourbaki, 1965; 1956]).

418 | D.3 Stage of unbounded positive measures on locally compact spaces

“. . . in 1915 Fréchet defined in (IIa) (see [Fréchet, 1915a] — authors) ‘abstract’ measures on a set equipped with a fixed 𝜎-algebra of subsets and integrals with respect to these measures;. . . ” [Bourbaki, 1969, Note hist.]. The above-mentioned Fréchet’s remark did not concern the Radon theorem because it talked about not the abstract but the topologically regular measures. After these papers of J. Radon and M. Fréchet, the problem of characterization of integrals as linear functionals began to be understood as the problem of extension of Radon’s theorem from T ⊂ Rn to more general topological spaces with regular measures. This problem turned out to be difficult and its solution has a long and abundant history. Therefore, it may be naturally called the Riesz – Radon – Fréchet problem of characterization of integrals. The extension of Radon’s theorem proceeded in several qualitatively different stages. First essential advance in the solution of the problem was made by S. Saks in 1938. Using the quite additive set functions (abstract measures) introduced by M. Fréchet and the topological property of regularity, S. Saks generalized Radon’s theorem to the case of a compact metric space [Saks, 1938] (see also [Dunford and Schwartz, 1958, IV.16]). In 1941, S. Kakutani generalized the result of Saks to an arbitrary compact space [Kakutani, 1941]. Theorem 2. For each compact topological space ⟮T, G⟯, the space of all bounded Radon integrals on C(T, G) coincides with the space of all bounded linear functionals on C(T, G), i. e. I(C(T, G), RMb (T, G)) = C(T, G)∼ ⫋ C(T, G)× . Moreover, the mapping 𝜇 󳨃→ i𝜇 |C(T, G) is an isomorphism of lattice linear spaces RMb (T, G) and C(T, G)∼ . This theorem finished the stage of bounded Radon measures on compact spaces. Note that in the theorems of Riesz, Radon, and Saks – Kakutani, the hardest part is integral representation of a functional (more accurately speaking, reconstruction of a measure), i. e. constructing of a function v and a measure 𝜇 according to the functional 𝜑 such that 𝜑 = i v |C[a, b] and 𝜑 = i𝜇 |C(T, G).

D.3 Stage of unbounded positive measures on locally compact spaces Unfortunately, the Kakutani theorem did not concern the most famous integral functional i𝜆 (f ) ≡ ∫R f d𝜆 with respect to the Lebesgue measure 𝜆 on the real line R. The space R is not compact, and the Lebesgue measure 𝜆 : B(R) → R+ ≡ [0, ∞] is not bounded. Properties of the Lebesgue measure 𝜆 was the starting point for the definition of a positive not necessary bounded Radon measure 𝜇 : B ≡ B(T, G) → R+ on a topological space ⟮T, G⟯.

D.1.7 Construction of the integral corresponding to a given functional

| 419

In 1950–53, P. Halmos, E. Hewitt, and R. Edwards obtained the following result (see [Halmos, 1950; Hewitt, 1952; Edwards, 1953]). Theorem 1. Let (T, G) be a locally compact space. Then, the cone of all integrals i𝜇 with respect to positive (not necessary bounded) Radon measures 𝜇 on (T, G) coincides with the cone of all positive linear functionals on the space A(T) = C c (T, G) of all continuous functions on T with compact supports, i. e. I(C c (T, G), RM(T, G)0 ) = (C c (T, G)∼ )+ . Moreover, the mapping 𝜇 󳨃→ i𝜇 |C c (T, G) is an isomorphism of the lattice cones RM(T, G)0 and (C(T, G)∼ )+ . In some publications, this theorem is called the Riesz representation theorem (see e. g. [Hewitt and Stromberg, 1965, 12.36]). Since the authors considered only the case of a locally compact space and positive measures on it, their proofs were not appropriate for more general spaces and measures. The boundedness of extending to locally compact spaces and positive unbounded measures inspired the natural interest to generalization of the Riesz – Radon – Saks – Kakutani – Halmos – Hewitt – Edwards theorem to (1) a more general class of topological spaces and (2) a more general class of “Radon” measures. The further essential rupture in solving the problem was made in the case of bounded Radon measures on Tychonoff spaces.

D.4 Stage of bounded measures on non-compact spaces. Tight functionals If ⟮T, G⟯ is a Tychonoff (completely regular) space, then the space C c (T, G) is not usable because the equality C c (T, G) = {0} is possible, and therefore, C c (T, G) does not separate Radon measures (i. e. only one trivial zero integral functional corresponds to all Radon measures). In this connection, the space C b (T, G) of all bounded continuous functions may be taken for the basis. But for this space, it has the sense to consider only the linear space I(C b (T, G), RMb (T, G)) of all integrals with respect to bounded Radon measures. In this case, the Riesz – Radon equality does not hold because I(C b (T, G), RMb (T, G)) ⫋ C b (T, G)∼ . Therefore, it is necessary to find a new characteristic property of Radon integrals. In 1969, N. Bourbaki, basing on Prokhorov’s ideas from [Prokhorov, 1956] arising from studying the weak compactness of families of bounded Radon measures (see also [Bogachev, 2007, 8.6]), singled out a new property of tightness of a functional (see 3.6.1) and proved the following theorem [Bourbaki, 1969, ch. IX, 5, no. 2].

420 | D.5 Transition to unbounded measures on arbitrary Hausdorff spaces

Theorem 1. The space of all Radon integrals with respect to bounded Radon measures on a Tychonoff space ⟮T, G⟯ coincides with the space of all bounded tight functionals on the space C b (T, G) of all bounded continuous functions on T, i. e. I(C b (T, G), RMb (T, G)) = C b (T, G)𝜋 . Moreover, the mapping 𝜇 󳨃→ i𝜇 |C b (T, G) is an isomorphism of the lattice-ordered linear spaces RMb (T, G) and C b (T, G)𝜋 . In Fremlin’s monograph, this theorem is called the Prokhorov theorem [Fremlin, 1974, 73G(e)]. Note that for a compact space ⟮T, G⟯ the equality C b (T, G)𝜋 = C b (T, G)∼ is valid. Therefore, the Bourbaki theorem is a natural generalization of the Kakutani theorem. The Bourbaki theorem has completely closed the Riesz – Radon – Saks – Kakutani line in description of bounded Radon integrals as functionals on the space C b (T, G).

D.5 Transition to unbounded measures on arbitrary Hausdorff spaces Despite the essential progress in the case of bounded Radon integrals on a Tychonoff space, any characterization remained unknown for unbounded integrals on a Tychonoff space, for bounded integrals on a Hausdorff space, and all the more for unbounded integrals on a Hausdorff space. First, it was unclear what unbounded “Radon” measures should be considered. Second, it was unclear what linear function space A(T) all “Radon” integrals i𝜇 should be considered on. If ⟮T, G⟯ is not a Tychonoff space, then the space C b (T, G) is not appropriate because it may contain only constant functions [Engelking, 1977, 2.7.17], and therefore, does not separate Radon measures. If ⟮T, G⟯ is a Tychonoff space but measures are unbounded, then C b (T, G) is not suitable because i𝜇 (1) = ∞, and the space C c (T, G) is not suitable because it may contain only the null function 0. Third, it was unclear what particular properties of Radon integrals as linear functionals on an unknown space A(T) should be considered. Thus, it was necessary to find the following: first, a suitable definition of the general Radon measure; second, a suitable space A(T) of not necessary continuous functions to replace the space C(T, G) of continuous functions, which had played its rôle in this problem; third, a new characteristic property of Radon integrals i𝜇 on this space. This was done by V. K. Zakharov, A. V. Mikhalev, and T. V. Rodionov in papers [Zakharov and Mikhalev, 1997a; 1997b; 1999; 2002; Zakharov, 2002b; 2005a; Zakharov et al., 2010a; 2010b; 2012a; 2012b] and is expounded in subsections 3.5.3, 2.4.5, and section 3.6 of this book.

Index of terms absolutely continuous semimeasure [measure] 3.2.4 absolutely homogeneous mapping 4∘ (2.2.7) abstract B.5.1 – infinite sequence B.5.2 – interpretation B.5.2 – model B.5.2 – , universal B.5.1 additive – ensemble 2.1.1 – evaluation 3.1.1 – mapping 3∘ (2.2.7) – series of collection¹ [sequence] of real numbers 1.4.8 admissible substitution A.1.1 a-envelope 2.4.4 a-foundation 2.1.1 Alexandrov – algebra 2.1.1 – set 2.1.1 – space 2.1.1 – theorem 2.3.5 Alexandrov – Stone – Fremlin integral representation theorem 3.4.2 algebra – , Banach 8∘ (2.2.7) – , lattice-ordered linear 3∘ (2.2.4) – , seminormed 8∘ (2.2.7) – , linear 2.2.4 – , seminormed 8∘ (2.2.7) algebra (of sets) 2.1.1 – , Alexandrov 2.1.1 – generated by ensemble 2.1.1 almost – distributable function 2.5.1 – everywhere 3.3.1 – measurable function 2.5.1 – uniform function 2.5.1 antiisotone mapping 1.1.15 antimonotone mapping 1.1.15 antisymmetric relation 1.1.14

Archimedean ordered linear space 9∘ (2.2.7) Archimedes principle 1.4.1, 1.4.2, 1.4.3 arrow B.2.1 – , 𝛼-functorial B.2.2 – , 𝛼-transformational B.2.2 a-space 2.1.1 assembly – defined by formula B.1.1 – universal B.1.1 – ordered B.3.3 – well-ordered B.3.3 associativity of – cardinal product 1.3.5 – cardinal sum 1.3.5 – infimum 1.1.15 – intersection of classes 1.1.5 – intersection of collection 1.1.10 – netful product 1.4.8 – netful sum 1.4.8 – product 1.3.6 (N), 1.4.1 (Z), 1.4.2 (Q), 1.4.3 (R) – product of classes 1.1.8 – product of collection 1.1.12 – sum 1.3.6 (N), 1.4.1 (Z), 1.4.2 (Q), 1.4.3 (R) – supremum 1.1.15 – union of classes 1.1.5 – union of collection 1.1.10 auxiliary – carrier of mathematical system 1∘ (2.2.2) – sets 2.2.2, C.1.2 – type C.1.1 axiom 1.1.3 – of binary union 1.1.6 (NBG), B.1.1 (LTS)² – of choice 1.1.12 (NBG), A.2.1 (ZF), B.1.1 (LTS) – , empty class B.1.1 – , empty set A.2.1 – , equiuniversality B.1.1 – , explicit 1.1.3 – , extensionality 1.1.5, A.2.1, B.1.1 – of full ensemble 1.1.6 (NBG), A.2.1 (ZF), B.1.1 (LTS)

1 In this index, any articles (a, an, the) are omitted. 2 In this index, for terms having different values in different set theories, the abbreviations in brackets indicate subsections where the term is explained within the framework of the corresponding theory. https://doi.org/10.1515/9783110550962-004

422 | Index of terms

– , full union B.1.1 – of foundation 1.1.11 – , galacticity A.7.3 – of general union 1.1.11 – , implicit 1.1.3 – , inaccessibility A.4.3 – , 𝜔-inaccessibility A.5.1 – , 1-inaccessibility A.5.2 – of inaccessible cardinal A.5.2 – of infinity 1.1.11, A.2.1 – , infra-infinity B.1.1 – , infra-universality B.1.1 – , local pair B.7.1 – , pair A.2.1 (ZF), B.7.1 (LTS) – , power set A.2.1, A.6.2 – , quasitransitivity B.1.1 – of regularity 1.1.11 (NBG), A.2.1 (ZF), B.1.1 (LTS) – scheme 1.1.3 – , full comprehension 1.1.5, B.1.1 – , replacement A.2.1 – , separation A.2.1 – , subset B.1.1 – , Tarski A.7.1 – text 1.1.3 – , transitivity B.1.1 – of union 1.1.11 (NBG), A.2.1 (ZF) – of universal set A.5.2 – , universality A.4.3 (ZF), B.1.1 (LTS) – , 𝜔-universality A.5.1 – , transitive A.5.2 – , unordered pair B.7.3 – , 1-universality A.5.2 – of values 1.1.11 – axioms of Peano 1.2.6 axioms of second-order generalized Peano – Landau arithmetic C.3.4 axioms of specification B.7.1 (LTS), B.7.3 (NBG) Baire – envelope of functional family 2.2.4 – extension of C b (T , G) 2.5.2 – , bounded 2.2.4 – collection for functional family 2.2.4, 2.3.7 – functional hierarchy 2.3.7 – functions of class 𝛼 2.2.4 – , bounded 2.2.4 Baire – Borel correlation 2.3.7, 2.4.7 Banach – algebra 8∘ (2.2.7)

– lattice-ordered space 8∘ (2.2.7) – local convergence classification theorem 2.3.7 – space 6∘ (2.2.7) balanced mathematical system C.2.2 band disjoint to set 3.2.2 band generated by set 3.2.2 base of topology 1∘ (3.5.1) belonging 1.1.1, A.1.3 – , generalized C.1.3 – type subdomain C.1.3 Beppo Levi theorem 3.3.3 Bernays theorem on finite axiomatization of NBG B.7.3 bijection 1.1.7, A.2.1, B.1.1 bimeasure 3.5.5 Birkhoff – identity 1.4.5, 2.2.2 – inequalities 1.4.5, 2.2.2 binary – additive ensemble 2.1.1 – additive evaluation 3.1.1 – intersection 1.1.5, A.2.1 – multiplicative ensemble 2.1.1 – relation 1.1.14 – superadditive evaluation 3.1.1 – union 1.1.5, A.2.1 bipositive mapping 4∘ (2.2.7) Borel – envelope of ensemble 2.1.3 – function 2.3.7 – of class 𝛼 2.3.7 – integrable function 3.5.6 – well-integrable function 3.5.6 – functional hierarchy 2.3.7 – measure on Rn 3.1.6 – set 2.1.1, 2.1.3 Borel – Lebesgue – initial evaluation on Rn 3.1.4 – measurable subsets of Rn 3.1.6 – measure, extended 3.1.6 Borel – Lebesgue – Hausdorff theorem on normal families 2.3.6 Borel – Radon measure 3.5.3 – , narrow 3.5.4 – , wide 3.5.3 Borel – Radon triplet 3.5.5 boundary of parallelepiped 3.1.6 boundary of set in topological space 1∘ (3.5.1)

Index of terms |

bounded – collection 1.1.15 – function 2.2.1 – mapping 1.1.15 (of preordered classes), 7∘ (2.2.7) (seminorm bounded) – set 1.1.15 (in preordered class), 1∘ (3.1.4) (in Rn ) boundedly normal – envelope of functional family 2.2.4, 2.4.6 – family of functions 2.2.4, 2.4.6 boundedness indicies of functional 3.6.4 Bourbaki theorem 3.6.3, 3.6.4 Bunina theorem on galactic sets A.7.2 Cantor completeness of R 1.4.4 Cantor theorem on cardinality of set of all subsets 1.3.2 cardinal degree 1.3.5 cardinal number 1.3.1, A.2.2, B.3.3 – , first denumerable 1.3.1 – , first uncountable 1.3.4 – , inaccessible A.2.2 – , U-inaccessible B.3.3 – , regular A.2.2 – , U-regular B.3.3 U-cardinal number B.3.3 cardinal product – of simple collection 1.3.5 – of sequential suit [pair, . . . ] 1.3.5 cardinal sum of simple collection 1.3.5 cardinality of set 1.3.2 carrier of mathematical system 1∘ (2.2.2) carrier of semimeasure [measure] 3.5.1 𝛼-category (category of class 𝛼) B.2.1 – , local B.2.1 – , locally-small B.2.1 – , small B.2.1 Cauchy criterion 1.4.4 – for uniform convergence 2.2.3 Cauchy sequence 1.4.3, 1.4.4 chain w. r. t.³ relation 1.1.14 – , maximal 1.2.11 chain of subelements of set A.7.2 choice – axiom 1.1.12 (NBG), A.2.1 (ZF), B.1.1 (LTS) – mapping 1.1.12, A.2.1

423

class 1.1.5, B.1.1 – of assignment of correspondence 1.1.7 – of class 𝛼 B.1.1 – , completely order closed 1.1.15 – completely determined by formula A.2.1 – , Dedekind complete 1.1.15 – , 𝛼-Dedekind complete 1∘ (2.2.8) – defined by formula A.2.1 – , downward directed 1.1.15 – , empty 1.1.5 – , latticed 1.1.15 – , lattice-ordered 1.1.15 – , order 𝛼-closed 1∘ (2.2.8) – , order complete 1.1.15 – , ordered 1.1.15, A.2.2 – , preordered 1.1.15 – , proper 1.1.5, A.2.1 – , quasitransitive A.2.2 – selected by property 1.1.5 – , solitary 1.1.6 – , supertransitive A.2.2 – , transitive 1.2.2, A.2.2 – , universal 1.1.5, B.1.1 – , upward directed 1.1.15 – of values of correspondence 1.1.7 – well-defined by formula A.2.1 – , well-ordered 1.2.1, A.2.2 𝛼-class B.1.1 closed – final interval 1.1.15 – initial interval 1.1.15 – set 2.1.1 closedly regular evaluation 3.5.1 closure of set in topological space 1∘ (3.5.1) closure (in F (T )) – , pointwise 2.2.4 – , bounded 2.2.4 – , uniform 2.2.4 – , bounded 2.2.4 co-ensemble 2.1.1 cofinal subclass 1.1.15 cofinality of ordinal number A.2.2 co-foundation 2.1.1 coinitial subclass 1.1.15 collection 1.1.9, A.2.1 – , bounded 1.1.15

3 In this index, the words “with respect to” are abbreviated to “w. r. t.”.

424 | Index of terms

– , commutatively summarized [multiplied] 1.4.8 – , countable 1.2.6 – of extensions of ensemble of types Σ, Δ, Λ, and Γ 2.1.2 – , finite 1.2.6 – , identical 1.1.9 – , multivalued 1.1.9, A.2.1 – , pairwise disjoint 1.1.10 – , simple 1.1.9, A.2.1 – , total 1.1.9 – , unconditionally summarized [multiplied] 1.4.8 – , uniformly bounded 2.2.2 – , well-summarized [well-multiplied] 1.4.8 – with graduation 2.4.4 𝛼-collection B.1.1 commutatively summarized [multiplied] collection 1.4.8 commutativity of – cardinal product 1.3.5 – cardinal sum 1.3.5 – infimum 1.1.15 – intersection of classes 1.1.5 – intersection of collection 1.1.10 – netful product 1.4.8 – netful sum 1.4.8 – product 1.3.6 (N), 1.4.1 (Z), 1.4.2 (Q), 1.4.3 (R) – product of classes 1.1.8 – product of collection 1.1.12 – sum 1.3.6 (N), 1.4.1 (Z), 1.4.2 (Q), 1.4.3 (R) – supremum 1.1.15 – union of classes 1.1.5 – union of collection 1.1.10 compact – descriptive space 2.1.1 – set 2.1.1 compactly regular evaluation 3.5.1 complement of class 1.1.5 complete – ensemble 1.1.5, 2.1.1 – evaluation 3.1.4 – pseudometric space 2∘ (2.2.7) – saturated [strongly saturated] extension 3.1.4 – seminormed linear space 6∘ (2.2.7) completely – additive ensemble 2.1.1 – additive evaluation 3.1.1

– normal – envelope of functional family 2.2.4, 2.3.6 – family of functions 2.2.4, 2.3.6 – multiplicative ensemble 2.1.1 – order closed class 1.1.15 completion of semimeasurable [measurable] space 3.1.4 completion of semimeasure [measure] 3.1.4 composition of – correspondences 1.1.7 – function and measure 3.3.7 – 𝛼-functors B.2.2 – 𝛼-transformations B.2.2 conclusion of theorem 1.1.3 H-concordant mathematical systems C.2.2 H-concordant evaluated mathematical systems C.2.3 conditionally summarized [multiplied] sequence 1.4.8 congruence 1∘ (2.2.6) conic space 1∘ (3.4.2) conic operator 1∘ (3.4.2) connected occurrence 1.1.2 connecting relation 1.1.14 constant function 2.2.1 constants C.1.3 – individual C.1.3 – predicate C.1.3 continuous function 2.3.1 consistent theory [totality] A.1.2 contradictory theory [totality] A.1.2 convergent – net (in R) 1.4.7 – net in topological space 2∘ (2.2.7) – sequence (in R) 1.4.4 – series of collection [sequence] of real numbers 1.4.8 convex mapping 4∘ (2.2.7) convex subclass 1.1.15 coordinate – pair 1.1.6 – 𝛼-pair B.1.1 – product 1.1.6, A.2.1 – 𝛼-product B.1.1 co-perfect ensemble 2.1.1 correspondence 1.1.7, A.2.1 – , bijective 1.1.7, B.1.1 – , identical 1.1.7 – , injective 1.1.7, B.1.1

Index of terms |

– , inverse to 1.1.7 – , simple 1.1.8 – , single-valued 1.1.7, B.1.1 – , surjective 1.1.7, B.1.1 – , total 1.1.7, B.1.1 – of several arguments 1.1.12 𝛼-correspondence B.1.1 countable set 1.2.6, A.2.2 countably – additive ensemble 2.1.1 – additive evaluation 3.1.1 – additive foundation 2.1.1 – multiplicative ensemble 2.1.1 – step function 2.2.4 cover – of class 1.1.10 – of set 2.1.5 – , one-member 2.1.5 covering 2.1.5 – multiplicative 2.1.5 cozero-set 2.2.5 – of level n 2.2.5 crossing of collection of evaluations C.3.1 crossing of evaluation C.3.2 cumulative classes B.3.4 cumulative sets A.3.1 – , inaccessible A.3.3 – , scheme-inaccessible A.8.1 Daniell functions 3.6.2 Darboux sums 3.7.1 Darboux – Jordan sums 3.7.3 decomposition of functional – , Riesz 2.2.8 decomposition of semimeasure [measure] – , Jordan 3.2.1 – , Lebesgue 3.2.4 – , Riesz 3.2.2 – w. r. t. semimeasure [measure] 3.2.4 decreasing – mapping 1.1.15 – net 1.1.15 – in F (T ) 2.2.3 – sequence 1.4.4 Dedekind – complete class 1.1.15 – completeness of R 1.4.5 – cut 1.4.5 𝛼-Dedekind complete class 1∘ (2.2.8)

425

deduction – , D-bounded A.1.3 – from axiom text 1.1.3 – from condition and axiom text 1.1.3 – from totality A.1.2 – , symbol of 1.1.3 – theorem 1.1.3 deducible formula A.1.2 degree of – cardinal 1.3.5 – class 1.1.8 – integer 1.4.1 – natural number 1.3.6 – rational number 1.4.2 – real number – , integer 1.4.3 – , rational 1.4.6 dense set w. r. t. ensemble 2.1.4 density of measure w. r. t. measure 3.3.7 denumerable set 1.2.6, A.2.2 derivative ensembles 2.1.1 derivative mapping of mapping(s) w. r. t. – degrees 1.1.8 – ensembles 1.1.8 – coordinate products 1.1.8 derivative mapping of collection of mappings w. r. t. products 1.1.12 designation 1.1.1 descriptive space 2.1.1 – , compact 2.1.1 – with covering 2.1.5 – with negligence 2.1.4 difference – of classes 1.1.5, A.2.1 – of integers 1.4.1 – of natural numbers 1.3.6 – of real numbers 1.4.3 – , symmetrical 2.1.4 Dini property 3.6.1 Dini theorem 2.3.4 directed class – upward [downward] 1.1.15 disjoint – classes 1.1.5 – collection 1.1.10 – elements (in lattice-ordered linear space) 3.2.2 – union 1.1.10 – union of sequential suit [pair, . . . ] 1.1.11

426 | Index of terms

dissection of class 1.1.10 distance between points (in metric space) 1∘ (2.2.7) distributable function 2.3.1 distributivity of – cardinal product 1.3.5 – cardinal sum 1.3.5 – infimum 1.1.15 – in R 1.4.5 – intersection of classes – w. r. t. union 1.1.5 – w. r. t. multiplication 1.1.6 – intersection of collection 1.1.13 – product 1.3.6 (N), 1.4.1 (Z), 1.4.2 (Q), 1.4.3 (R) – product of collection 1.1.12 – supremum 1.1.15 – in R 1.4.5 – union of classes – w. r. t. intersection 1.1.5 – w. r. t. multiplication 1.1.6 – union of collection 1.1.13 division of natural numbers 1.3.6 divisor of number 1.3.6 domain of definition of correspondence 1.1.7 dominant set A.7.2 dominated convergence theorem 3.3.3 dual to ordered linear space 2∘ (2.2.8) dual to seminormed linear space 7∘ (2.2.7) edge ensemble 2.1.1 Egorov theorem 3.3.1 empty class 1.1.5, B.1.1 – axiom B.1.1 empty set axiom A.2.1 ensemble 1.2.11, 2.1.1 – , additive 2.1.1 – closed under – complement 2.1.1 – difference 2.1.1 – , complete 1.1.5, 2.1.1, A.2.1 – , co-perfect 2.1.1 – , derivative 2.1.1 – , disjointly additive 2.1.1 – , disjointly multiplicative 2.1.1 – , edge 2.1.1 – , extended derivative 2.1.1 – , complete 1.1.5, 2.1.1, A.2.1 – , hereditary 2.1.4

– , ideal 2.1.4 – , initial derivative 2.1.1 – , latticed 2.1.1 – , multiplicative 2.1.1 – , perfect 2.1.1 – , reducible 2.1.1 – , saturated 2.1.1 – , separable 2.1.1 – of sets of finite character 1.2.11 – , Young – Hausdorff 2.1.2 – , Zakharov – Koldunov 2.1.2 envelope function from above [below] 2.2.9 envelope of ensemble – , Borel 2.1.3 envelope of functional family – , Baire convergence 2.2.4, 2.3.7 – , normal 2.2.4, 2.3.6 – , boundedly 2.2.4, 2.4.6 – , completely 2.2.4, 2.3.7 envelopment properties 3.6.2 essentially – bounded function 2.2.7 – bounded part of functional family 2.2.7 – positive [negative] set for measure 3.2.1 – uniform convergence 2.2.7 equality axioms C.1.3 equiuniversality axiom B.1.1 equivalent classes 1.1.8, A.2.2, B.3.3 U-equivalent classes B.3.3 equivalent formulas 1.1.3 equivalence relation 1.1.14 equivalence of functions w. r. t. ideal 2.1.4 equivalence class of element 1.1.14 Euclidean division 1.3.6 evaluable topological space 3.5.1 evaluated mathematical system C.2.2 evaluation 3.1.1 – , additive 3.1.1 – , bounded 3.1.1 – , complete 3.1.4 – , decreasing 3.1.1 – defined on topological space 3.5.1 – , finite [𝜎-finite, 𝜏-finite] 3.1.1 – , increasing 3.1.1 – , initial Borel – Lebesgue 3.1.4 – , internally finite 3.1.1 – , inner regular 3.4.1 – , Jordan 3.7.3 – , locally bounded 3.5.1

Index of terms |

– , locally complete 3.1.4 – , lower continuous 3.4.1 – , natural 3.1.1 – , negative 3.1.1 – , outer regular 3.4.1 – , positive 3.1.1 – , regular 3.4.1 – , saturated 3.1.4 – , strongly saturated 3.1.4 – , subadditive 3.1.1 – , superadditive 3.1.1 – , topological 3.5.1 – , topologically internally finite 3.5.1 – , upper continuous 3.4.1 evaluation on mathematical system C.2.3 even natural number 1.3.6 exact bounds 1.1.15 exact functional 3.6.1 explicit axiom 1.1.3 exponential function 1.4.7 exponential set A.7.2 exponentiality property A.7.1 extended – Borel – Lebesgue measure 3.1.6 – Radon measure 3.5.3 – Radon triplet 3.5.5 – real number(s) 1.4.3 extendedly Lebesgue integrable function 3.3.2, 3.3.6 extending mapping 3.4.1 extension of – correspondence 1.1.7 – collection 1.1.9 – ensemble of types Σ, Δ, Λ, and Γ 2.1.2 – initial family C b (T cG) 2.5.2 – measure – , complete saturated [strongly saturated] 3.1.4 – , large complete saturated [strongly saturated] 3.1.5 – , Lebesgue – Caratheodory 3.1.5 – of narrow bounded Borel – Radon measure 3.5.4 extensional mathematical system C.2.2 extensionality axiom 1.1.5, A.2.1, B.1.1 factor-class of class w. r. t. equivalence 1.1.14 factor-correspondence w. r. t. equivalence 1.1.14

427

factor-ensemble of ensemble w. r. t. ideal 2.1.4 factor-mapping w. r. t. equivalence 1.1.14 family of functions 2.2.4 – closed under – bounded inversion 2.2.4 – finite addition 2.2.4 – finite exact bounds 2.2.4 – finite multiplication 2.2.4 – inversion 2.2.4 – multiplication by real numbers 2.2.4 – pointwise convergence 2.2.4 – Stone truncation 2.2.4 – uniform convergence 2.2.4 – , normal 2.2.4 – , boundedly 2.2.4 – , completely 2.2.4, 2.3.7 – separating points and closed sets 3.5.2 – truncatable 2.2.9 Fatou lemma 3.3.3 Fine – Gillman – Lambek problem 2.5.2 finer partition 3.7.1 filter (of sets) C.3.1, 2.1.4 final – interval 1.1.15 – subclass 1.1.15 finally constant [non-constant] sequence 1.2.7 finite – collection 1.2.6 – evaluation 3.1.1 – set 1.2.6, A.2.2 finitely – additive ensemble 2.1.1 – additive evaluation 3.1.1 – closed set theory B.5.2 – multiplicative ensemble 2.1.1 – superadditive [subadditive] evaluation 3.1.1 formation C.1.2 – , auxiliary C.1.1 formula 1.1.2, A.1.1 – , atomic C.1.3 – , closed 1.1.2 – , elementary A.1.1 – , false 1.1.3 – , infrafiltrated C.3.2 – , normalizable C.2.4 – , predicative 1.1.5 – , 𝛼-predicative B.1.1 – scheme 1.1.2, A.1.2 – , true 1.1.3

428 | Index of terms

foundation 2.1.1 foundation axiom 1.1.11 free occurrence 1.1.2, A.1.1 frontier of set in topological space 1∘ (3.5.1) full comprehension axiom scheme 1.1.5, B.1.1 full ensemble 1.1.5, 2.1.1 – axiom 1.1.6 (NBG), B.1.1 (LTS) full 𝛼-ensemble B.1.1 full union axiom B.1.1 function 1.1.8, A.2.1 – , Baire – of class 𝛼 2.2.4 – , Borel measurable 2.3.7 – of class 𝛼 2.3.7 – , bounded 2.2.1 – of bounded variation D.1.1 – , constant 2.2.1 – , continuous 2.3.1 – continuous at point 1∘ (3.7.2) – , Daniell 3.6.2 – , distributable 2.3.1 – , almost 2.5.1 – , essentially bounded 2.2.7 – , exponential 1.4.7 – , extendedly Lebesgue integrable 3.3.2, 3.3.6 – , infimal 3.6.2 – , Lebesgue integrable 3.3.2, 3.3.6 – , Lebesgue 𝜎-integrable 3.3.2, 3.3.6 – , majorized by function 2.2.4 – , measurable 2.3.1 – , almost 2.5.1 – , positive [negative] 2.2.1 – , Riemann integrable 3.7.1 – in classical sense (for Rn ) 3.7.3 – , Riemann – Stiltjes integrable D.1.2 – , quasidistributable 2.5.2 – , quasimeasurable 2.5.2 – , quasiuniform 2.5.2 – , real-valued 2.2.1 – , semicontinuous 2.3.8, 3.5.2 – , semimeasurable 2.3.8 – separating points and closed sets 3.5.2 – , set 3.1.1 – , signed 2.2.4 – , step 2.2.4 – , countably 2.2.4 – , quite 2.2.4 – , quite countably 2.2.4 – , supremal 3.6.2

– , symmetrizable 2.4.5 – , uniform 2.4.1 – , almost 2.5.1 – , uniformly continuous 2.3.1 – , universally integrable 3.3.6 – with compact support 3.5.2 functional 6∘ (2.2.7) – , exact [𝜎-exact] 3.6.1 – , locally tight 3.6.1 – , natural 3.6.4 – , pointwise continuous [𝜎-continuous] 2.2.8 – representable by Lebesgue integral over measurable space 3.4.2 – , quite locally tight 3.6.1 – , tight 3.6.1 – , uniformly order bounded 2.2.8 𝛼-functor (functor of class 𝛼) B.2.2 – , identity B.2.2 𝛼-functorial arrows B.2.2 fundamental sequence – in R 1.4.3, 1.4.4 – in pseudometric space 2∘ (2.2.7) galactic set A.7.2 galacticity axiom A.7.3 general structure of mathematical system 1∘ (2.2.2) generalization rule 1.1.3 geometric progression 1.4.8 graduation of collection 2.4.4 graph of mapping 1.1.8 greatest – element of subclass 1.1.15 – member of collection 1.1.15 – lower bound of collection 1.1.15 Hahn decomposition 3.2.1 Halmos – Hewitt – Edwards theorem 3.6.3 Hausdorff – maximality principle 1.2.11 – space 2∘ (3.5.1) – theorem on normal envelope 2.3.6 homogeneous mapping 3∘ (2.2.7) homomorphism – of linear spaces 3∘ (2.2.7) – of lattice-ordered linear spaces 3∘ (2.2.7) hull of covering 2.1.5 hulls of ensemble 2.1.1

Index of terms |

ideal negligence 2.1.4 ideal of linear algebra 3∘ (2.2.6) ideal ensemble (of sets) 2.1.4 identical – collection 1.1.9 – correspondence 1.1.7 identity – , Birkhoff 1.4.5, 2.2.2 – 𝛼-functor B.2.2 – 𝛼-transformation B.2.2 – , Neumann A.3.3 image of subclass under correspondence 1.1.7, A.2.1 implication rule 1.1.3 implicit axiom 1.1.3 improvement property 2.1.1 inaccessibility axiom A.4.3 inaccessible cardinal number A.2.2 U-inaccessible cardinal number B.3.3 inaccessible cumulative sets A.3.3 increasing – mapping 1.1.15 – net 1.1.15 – in F (T ) 2.2.3 – sequence 1.4.4 individual constants C.1.3 induction – , natural 1.2.6, B.3.3 – , general principle of 1.3.6 – , construction of mappings by 1.2.7 – , principle of 1.2.1 – , transfinite 1.2.8, A.2.2, B.3.3 – , construction of mappings by 1.2.7, A.2.2 inductive set A.2.1 inequality – , Bernoulli 1.4.3 – , Birkhoff 1.4.5 inferior limit of net 1.1.15 infimal functions 3.6.2 infimum – of simple collection 1.1.15 – of set 1.1.15 – of simple sequential pair [triplet, . . . ] 1.1.15 infinite – geometric progression 1.4.8 – sequence A.2.2 – set 1.2.6, A.2.2 infinity – axiom 1.1.11, A.2.1

429

– in R 1.4.3 infra-infinity axiom B.1.1 infrafiltrated formula C.3.2 infra-D-power of system C.3.2 infra-D-product of collection of mathematical systems C.3.1 infra-universal class B.1.1 infra-universality axiom B.1.1 initial – alphabet 1.1.1, A.1.1 – interval 1.1.15 – subclass 1.1.15 – Borel – Lebesgue measure 3.1.4 injection 1.1.7, A.2.1 inner convergent sequence – in R 1.4.3, 1.4.4 – in pseudometric space 2∘ (2.2.7) inner regular evaluation 3.4.1 inner uniformly convergent sequence 2.2.3 inscribe set in functional family 2.2.9 integer(s) 1.4.1 integrable function 3.3.2, 3.3.6 integral of function 3.3.2, 3.3.6 interior of set in topological space 1∘ (3.5.1) internally finite evaluation 3.1.1 interpretation of first order theory A.1.3 – , abstract B.5.2 – , standard A.6.1 interpretation of signature on support C.2.1 intersection of classes 1.1.5 𝛼-intersection of 𝛼-classes B.1.1 intersection of collection 1.1.10, A.2.1 𝛼-intersection of 𝛼-collection B.1.1 intersection of sequential suit [pair, . . . ] 1.1.11 interval in preordered class – , closed 1.1.15 – , final 1.1.15 – of general kind 1.1.15 – , half-open 1.1.15 – , initial 1.1.15 – , open 1.1.15 inverse image of subclass under correspondence 1.1.7, A.2.1 irrational number(s) 1.4.3 isomorphism – of lattice-ordered linear spaces [algebras] 5∘ (2.2.7) – of seminormed lattice-ordered linear spaces [algebras] 6∘ (2.2.7)

430 | Index of terms

– of seminormed linear spaces 6∘ (2.2.7) isotone mapping 1.1.15 J0 -extension of C b (T , G) 2.5.2 Jordan – decomposition 3.2.1 – partition 3.7.1 – , prime 3.7.1 – measurable set 3.7.3 – measure 3.7.3 Kuratowski pair 1.1.6 Kuratowski – Zorn lemma 1.2.11 L-space 8∘ (2.2.7) language of generalized signature C.1.3 large complete saturated extension of measure 3.1.5 large complete strongly saturated extension of measure 3.1.5 lattice (of sets) 2.1.1 – generated by ensemble 2.1.1 lattice envelope 2.1.1 latticed ensemble 2.1.1 l-ideal – of lattice-ordered class 1.1.15 – of lattice-ordered linear space 3∘ (2.2.6) – of lattice-ordered linear algebra 3∘ (2.2.6) lattice-ordered – class 1.1.15 – linear algebra 5∘ (2.2.4) – linear space 4∘ (2.2.4) Lebesgue – decomposition of measure w. r. t. measure 3.2.4 – dominated convergence theorem 3.3.3 – integrable function 3.3.2, 3.3.6 – integral 3.3.2, 3.3.6 – theorem on countable additivity of integral 3.3.3 Lebesgue – Caratheodory extension of measure 3.1.5 Lebesgue – Hausdorff local convergence classification theorem 2.3.7 Lebesgue – Radon – Nikodym theorems 3.3.8 Lebesgue – Urysohn theorem on Cantor staircase 2.3.5 lemma – , Fatou 3.3.3

– , Tukey 1.2.11 – , Kuratowski – Zorn 1.2.11 limit of net 1.1.15, 1.4.7 – in R 1.4.7 – in topological space 2∘ (2.2.7) – inferior 1.1.15 – , order- 1.1.15 – , pointwise (in F (T )) 2.2.3 – superior 1.1.15 – , uniform (in F (T )) 2.2.3 limit of sequence in R 1.4.4 limit ordinal 1.2.3, A.2.2, B.3.3 linear – algebra 3∘ (2.2.4) – mapping 3∘ (2.2.7) – order 1.1.14, A.2.2 – space 2∘ (2.2.4) – subalgebra 2∘ (2.2.6) – subspace 2∘ (2.2.6) local completion of semimeasurable [measurable] space 3.1.4 local completion of semimeasure [measure] 3.1.4 locally bounded evaluation 3.5.1 locally compact space 2∘ (3.5.1) locally tight functional 3.6.1 logical axiom schemes 1.1.4, A.1.2, C.1.3 lower – bound of collection 1.1.15 – , greatest 1.1.15 – boundedness index of functional 3.6.4 – 𝜎-continuous [continuous] evaluation 3.4.1 – Darboux sum 3.7.1 – regularization of function 3.7.2 – semimeasurable function 2.3.8 Lusin theorem 3.5.2 M-algebra 8∘ (2.2.7) M-space 8∘ (2.2.7) main convergence classification theorem 2.3.7 main fine convergence classification theorem 2.4.7 main open sets 2.5.2 main part of ordered class 1.1.15 majorized – function 2.2.4 – set 2.2.4 mapping 1.1.8, A.2.1 – , additive 3∘ (2.2.7)

Index of terms | 431

– , antiisotone 1.1.15 – , antimonotone 1.1.15 – , bipositive 4∘ (2.2.7) – , choice 1.1.12 – compatible with subclass 1.2.7 – , conic 1∘ (3.4.2) – , convex 4∘ (2.2.7) – , decreasing 1.1.15 – , derivative – w. r. t. degrees 1.1.8 – w. r. t. ensembles 1.1.8 – w. r. t. coordinate products 1.1.8 – of collection of mappings w. r. t. products 1.1.12 – extending 3.4.1 – , homogeneous 3∘ (2.2.7) – , absolutely 4∘ (2.2.7) – , positively 4∘ (2.2.7) – , increasing 1.1.15, A.2.2 – , strictly 1.1.15, A.2.2 – , isotone 1.1.15, A.2.2 – , linear 3∘ (2.2.7) – , modulusly monotone 4∘ (2.2.7) – , monotone 1.1.15, A.2.2 – , multiplicative 4∘ (2.2.7) – , multivalued 1.1.7 – , order bounded 1.1.15 – , order changing 1.1.15 – , order preserving 1.1.15 – , order regular 4∘ (2.2.7) – , preserving exact bounds 1.1.15 – , positive 4∘ (2.2.7) – , seminorm [norm] bounded 7∘ (2.2.7) – , subadditive 4∘ (2.2.7) – , sublinear 4∘ (2.2.7) – , submultiplicative 4∘ (2.2.7) – thinning net out 1.1.15 – thinning the sequence out 1.2.6 𝛼-mapping B.1.1 mathematical system 1∘ (2.2.2), C.2.1 – , balanced C.2.2 – , evaluated C.2.3 – , extensional C.2.2 – , normal C.2.2 – , regular C.2.2 – with true generalized equalities and belongings C.2.3 maximal – element w. r. t. relation 1.1.14

– chain 1.2.11 – member of collection 1.1.9 member of collection 1.1.9 – , greatest [smallest] 1.1.15 – , maximal [minimal] 1.1.15 measure 3.1.1 – absolutely continuous w. r. t. measure 3.2.4 – concentrated on set 3.2.4 – , extended Borel – Lebesgue 3.1.6 – , initial Borel – Lebesgue 3.1.4 – , Jordan 3.7.3 – , narrow 3.1.1 – , overfinite 3.2.1 – singular to measure 3.2.4 – , wide 3.1.1 measurable function 2.3.1 – , almost 2.5.1 measurable set w.r.t. evaluation 3.1.1 measurable space 3.1.1 measurable topological space 3.5.1 mesh of partition D.1.2 metric 1∘ (2.2.7) – space 1∘ (2.2.7) – topology 1∘ (2.2.7) minimal – element w. r. t. relation 1.1.14 – member of collection 1.1.9 Mirimanov – von Neumann collection A.3.1, B.3.4 Mirimanov – von Neumann sets A.3.1 model of axiomatic theory A.1.3 – , abstract B.5.2 – , standard A.6.1 model for totality of formulas C.2.3 – , balanced C.2.3 – , extensional C.2.3 – , normal C.2.3 – , regular C.2.3 – , second-order C.2.3 modulus of – element in lattice-ordered space 4∘ (2.2.4) – integer 1.4.1 – overfinite semimeasure [measure] 3.2.2 – rational number 1.4.2 – real number 1.4.3 – real-valued function 2.2.2 modulusly monotone mapping 4∘ (2.2.7) modus ponens 1.1.3, A.1.2 monotone mapping 1.1.15

432 | Index of terms

multiple of number 1.3.6 multiplicative – ensemble 2.1.1 – covering 2.1.5 – mapping 4∘ (2.2.7) – series of collection [sequence] of real numbers 1.4.8 – unit 4∘ (2.2.4) multivalued – collection 1.1.9 – mapping 1.1.7 – sequence 1.2.6 mutually deducible formulas 1.1.3 Nakano – Shimogaki problem 2.5.2 narrow measure 3.1.1 narrow Radon measure 3.5.4 natural evaluation 3.1.1 natural functional 3.6.4 natural induction 1.2.6 natural number(s) 1.2.6, A.2.2 – , even 1.3.6 – , odd 1.3.6 natural series of Peano – Landau of the second order C.3.4 negative part of element (in lattice-ordered class) 1.1.15 negative part of lattice-ordered class 1.1.15 negligence (on descriptive space) 2.1.4 – , ideal 2.1.4 negligible sequence 1.4.3 negligible set 3.1.1 neighborhood of point 1∘ (3.5.1) net 1.1.15 – , convergent to number 1.4.7 – , increasing [decreasing] in F (T ) 2.2.3 – , strictly 2.2.3 – , increasing [decreasing] to element 1.1.15 – , order-convergent to element 1.1.15 – , pointwise convergent 2.2.3 – , uniform convergent 2.2.3 netful series of collection of real numbers 1.4.8 netful sum [product] of collection of real numbers 1.4.8 Neumann relation 1.2.2 Neumann identity A.3.3 neutral subclass 1.1.14 Newton binomial theorem 1.4.6 non-consistent theory [totality] A.1.2

non-contradictory theory [totality] A.1.2 non-ordered – pair 1.1.6, A.2.1 – 𝛼-pair B.1.1 – suit 1.1.11 norm 2.2.7 – of convergence in mean 3.3.4 – of uniform convergence 2.2.7 norm bounded functional [mapping] 7∘ (2.2.7) norm dual to seminormed linear space 7∘ (2.2.7) normed – metric 6∘ (2.2.7) – lattice-ordered linear algebra 8∘ (2.2.7) – lattice-ordered linear space 8∘ (2.2.7) – linear algebra 8∘ (2.2.7) – linear space 6∘ (2.2.7) – topology 7∘ (2.2.7) normal – envelope of family of functions 2.2.4, 2.3.6 – family of functions 2.2.4, 2.3.6 – mathematical system C.2.2 – topological space 2.1.1 normalizable formula C.2.4 nowhere dense set 2.1.4 null sequence 1.4.3 null set 3.1.1 number – , cardinal 1.3.1, A.2.2 – , first denumerable 1.3.1 – , first uncountable 1.3.4 – , natural 1.2.6, A.2.2 – , ordinal 1.2.2 – , limit 1.2.3, A.2.2, B.3.3 – , U-ordinal B.3.3 – , successive A.2.2 – , rational 1.4.2 – , real 1.4.3 – objective constants C.1.3 occurrence 1.1.1, A.1.1 – , connected 1.1.2, A.1.1 – , free 1.1.2, A.1.1 odd natural number 1.3.6 one-member cover of set 2.1.5 one-valued correspondence 1.1.7 open – ball (in pseudometric space) 1∘ (2.2.7) – final interval 1.1.15

Index of terms |

– initial interval 1.1.15 – neighborhood of point 1∘ (3.5.1) – set 2.1.1 – in pseudometric space 1∘ (2.2.7) – in seminormed space 7∘ (2.2.7) – topology 2.1.1 – unit ball (in seminormed space) 7∘ (2.2.7) openly regular evaluation 3.5.1 operator 1.1.8 – linear 3∘ (2.2.7) order 1.1.14 – changing mapping 1.1.15 – complete class 1.1.15 – dual to ordered linear space 2∘ (2.2.8) – equivalent classes 1.1.15, A.2.2 – by inclusion 1.2.11 – , linear 1.1.14, A.2.2 – preserving mapping 1.1.15 – , natural (on a subset of 𝜔) 1.3.7 – relation 1.1.14 order bounded – collection 1.1.15 – function 2.2.1 – mapping 1.1.15 – set 1.1.15 order-convergent net 1.1.15 order dual to ordered linear space 2∘ (2.2.8) order-limit of net 1.1.15 order regular mapping 4∘ (2.2.7) order type of well-ordered set 1.2.5 ordered – assembly B.3.3 – class 1.1.15, A.2.2 – disjoint union 1.2.9 – linear space 4∘ (2.2.4) – , Archimedean 9∘ (2.2.7) – pair 1.1.6, A.2.1 U-ordered U-class B.3.3 orderly summarized [multiplied] sequence 1.4.8 ordinal 1.2.2, A.2.2 – , limit 1.2.3, A.2.2, B.3.3 U-ordinal B.3.3 ordinal number 1.2.2, A.2.2 – , even 1.3.6 – , odd 1.3.6 – , scheme-inaccessible A.8.1 – , scheme-regular A.8.1 – , successive A.2.2 U-ordinal number B.3.3

433

ordinal sum 1.2.9 oscillation of function – on set 2.2.1 – on cover 2.3.1 outer regular evaluation 3.4.1 overfinite semimeasure [measure] 3.2.1 pair – , non-ordered 1.1.6, A.2.1 – , ordered 1.1.6 – , sequential 1.1.11, A.2.2 𝛼-pair B.1.1 pairwise disjoint collection 1.1.10 parallelepiped in Rn 2.1.1 – , half-bounded 1∘ 3.1.6 parent measures of Radon bimeasure 3.5.5 parents of type C.1.1 partial product [sum] of collection of real numbers 1.4.8 partial structures of mathematical system 2.2.2 partition of class 1.1.10 partition of well-ordered set 1.2.9 Peano axioms (for natural numbers) 1.2.6 Peano – Landau postulates C.3.4 perfect ensemble 2.1.1 n-placed correspondence on class A.2.2 n-placed operation on class A.2.2 pointwise – continuous [𝜎-continuous] functional 2.2.8 – closure 2.2.4 – convergent net 2.2.3 – limit in F (T ) 2.2.3 pointwisely dense family 2.2.4 positive mapping 4∘ (2.2.7) positive part of element (in lattice-ordered class) 1.1.15 positive part of lattice-ordered class 1.1.15 positively homogeneous mapping 4∘ (2.2.7) power of – class B.3.3 – continuum 1.4.4 – set 1.3.2, A.2.2, B.3.3 precompact set 2∘ (3.5.1) predicative formula 1.1.5 𝛼-predicative formula B.1.1 preimage of subclass under correspondence 1.1.7, A.2.1 pre-L-space 8∘ (2.2.7) pre-M-space 8∘ (2.2.7)

434 | Index of terms

preorder 1.1.14 – relation 1.1.14 – , opposite 1.1.14 preordered class 1.1.15 prescriptive space 2.1.5 preserve exact bounds 1.1.15 prime Jordan partition 3.7.1 principal carrier of mathematical system 1∘ (2.2.2) principle – , Archimedes 1.4.1, 1.4.2, 1.4.3 – of induction 1.2.1 – of natural induction 1.2.6 – of transfinite induction 1.2.6, A.2.2 – , Hausdorff maximality 1.2.11 – , Zermelo 1.2.11 – of ∈-induction A.2.2 product of classes 1.1.6, A.2.1 𝛼-product of 𝛼-classes B.1.2 product of – collection 1.1.12, A.2.2 – collection of preordered classes 1.1.15 – correspondences 1.1.7 – function and measure 3.3.7 – integers 1.4.1 – natural numbers 1.3.6 – rational numbers 1.4.2 – real numbers 1.4.3 – sequential pair [triplet, . . . ] 1.1.12, A.2.2 𝛼-product of 𝛼-collection B.1.2 product – , cardinal 1.3.5 – , netful (in R) 1.4.8 – , partial (in R) 1.4.8 projections onto factors 1.1.8 projection – into factor 1.1.12 – into subproduct 1.1.12 Prokhorov property of functional 3.6.1 proof of formula A.1.2 proof of theorem 1.1.3 property – (D) 3.6.1 – , Dini 3.6.1 – (E) 3.6.2 – (E𝜎 ) 3.6.2 – , exponentiality A.7.1 – , improvement 2.1.1 – , reduction 2.1.1

– , separation 2.1.1 – , Stone – of functional family 2.2.9 – of set 2.1.4 – , strong substitution A.4.1, A.6.1 – , Tarski A.7.1 – , transitivity A.4.1, A.7.1, A.8.2, A.8.4 pseudometric 1∘ (2.2.7) – space 1∘ (2.2.7) – topology 1∘ (2.2.7) quasidistributable function 2.5.2 quasimeasurable function 2.5.2 quasitransitive class A.2.2 quasitransitivity axiom B.1.1 quasiuniform function 2.5.2 quite – countably step function 2.2.4 – locally tight functional 3.6.1 – step function 2.2.4 – topological evaluation 3.5.1 quadruplet – , sequential 1.1.11 quotient at division 1.3.6 (N), 1.4.2 (Q), 1.4.3 (R) Radon bimeasure 3.5.5 – , bounded 3.5.5 Radon integral 3.5.3 Radon measure 3.5.3 – , extended 3.5.3 – , narrow 3.5.4 Radon triplet 3.5.5 Radon – Saks – Kakutani theorem 3.6.4 raising of real number to – integer degree 1.4.3 – rational degree 1.4.6 rational number(s) 1.4.2 real number(s) 1.4.3 real-valued function 2.2.1 reducible ensemble 2.1.1 reduction property 2.1.1 refinement of cover 1.1.10 reflexive relation 1.1.14 Regoli theorem on completely normal families 2.3.6 regular cardinal A.2.2 U-regular cardinal B.3.3 regular mathematical system C.2.2

Index of terms |

regularity axiom 1.1.11 (NBG), B.1.1 (LTS) regularization of function – , lower 3.7.2 – , upper 3.7.2 relation (binary relation) 1.1.14 – , antisymmetric 1.1.14 – , connecting 1.1.14 – , Neumann 1.2.2 – , order 1.1.14 – , preorder 1.1.14 – , reflexive 1.1.14 – , symmetric 1.1.14 – , total 1.1.14 – , transitive 1.1.14 – with property of minimality [maximality] 1.2.1 relativization of formula to class B.3.1 relativization of formula to set A.6.1 remainder at division 1.3.6 restriction of – collection 1.1.9 – correspondence 1.1.7, A.2.1 – relation 1.1.14 Riemann – extension of C b (T , G) 2.5.2 – integrable function 3.7.1 – in classical sense (for Rn ) 3.7.3 – integral 3.7.1 – in classical sense (for Rn ) 3.7.3 Riemann – Stiltjes – integrable function D.1.2 – integral D.1.2 – integral sum D.1.2 Riesz decomposition – of functional 2.2.8 – of semimeasure 3.2.2 – in Riesz space 2.2.4 – w. r. t. semimeasure [measure] 3.2.4 Riesz – Kantorovich theorem 2.2.8 Riesz – Radon – Fréchet problem 3.5.3 Riesz representation theorem D.1.4 Riesz space 4∘ (2.2.4) ring (of sets) 2.1.1 – generated by ensemble 2.1.1 root of real number with natural exponent 1.4.6 rules of deduction 1.1.3 – , derivative 1.1.4 rule of generalization 1.1.3, A.1.2 rule of implication 1.1.3, A.1.2

435

satisfaction of formula w. r. t. evaluation on system C.2.3 saturated – ensemble 2.1.1 – evaluation 3.1.4 – extension – , large complete 3.1.5 – , complete 3.1.4 saturation – of ensemble 2.1.1 – of semimeasurable space [measure] 3.1.4 – of semimeasure [measure] 3.1.4 scheme set A.8.1 scheme Tarski set A.8.4 scheme-regular ordinal number A.8.1 scheme-inaccessible ordinal number A.8.1 scheme-inaccessible cumulative set A.8.1 scheme-universal set A.8.2 selection from partition D.1.2 semimeasurable functions 2.3.8 semimeasurable space 3.1.1 semimeasurable topological space 3.5.1 semimeasure 3.1.1 – absolutely continuous w. r. t. semimeasure 3.2.4 – concentrated on set 3.2.4 – , overfinite 3.2.1 – singular to semimeasure 3.2.4 seminorm 6∘ (2.2.7) – bounded mapping [functional] 7∘ (2.2.7) – dual to seminormed linear space 7∘ (2.2.7) – of essentially uniform convergence 2.2.7 – of integral convergence 3.3.4 seminormed – pseudometric 6∘ (2.2.7) – lattice-ordered linear algebra 8∘ (2.2.7) – lattice-ordered linear space 8∘ (2.2.7) – linear algebra 8∘ (2.2.7) – linear space 6∘ (2.2.7) – topology 7∘ (2.2.7) semiring 2.1.1 semitype C.1.1 – of type C.1.1 sentence A.1.1 separable ensemble 2.1.1 separation axiom scheme A.2.1 separation property 2.1.1 sequence 1.2.6, A.2.2 – , bounded 1.4.3, 1.4.4

436 | Index of terms

– , Cauchy 1.4.3, 1.4.4 – , decreasing 1.4.4 – , strictly 1.4.4 – , commutatively summarized [multiplied] 1.4.8 – , conditionally summarized [multiplied] 1.4.8 – , convergent to number 1.4.4 – , finally constant 1.2.7 – , finally non-constant 1.2.7 – , finite A.2.2 – , fundamental – in R 1.4.3, 1.4.4 – in pseudometric space 2∘ (2.2.7) – , increasing 1.4.4 – , strictly 1.4.4 – , infinite A.2.2 – , abstract B.5.2 – , inner convergent – in R 1.4.3, 1.4.4 – in pseudometric space 2∘ (2.2.7) – , inner uniformly convergent 2.2.3 – , multivalued 1.2.6 – , null 1.4.3 – , sequentially summarized [multiplied] 1.4.8 – , simple 1.2.6 – , unconditionally summarized [multiplied] 1.4.8 – , uniformly fundamental 2.2.3 – , uniformly upper [lower] unbounded 1.4.3 – , well-summarized [well-multiplied] 1.4.8 sequential – pair 1.1.11, A.2.2 – product of sequential pair [triplet, . . . ] 1.1.12 – quadruplet 1.1.11 – triplet 1.1.11, A.2.2 – suit 1.1.11 – sum [product] of sequence of real numbers 1.4.8 sequentially additive evaluation 3.1.1 sequentially summarized [multiplied] sequence 1.4.8 series of collection [sequence] of real numbers 1.4.8 – , netful [sequential] 1.4.8 – , additive [multiplicative] 1.4.8 set 1.1.5 (NBG), A.2.1 (ZF) – , Alexandrov 2.1.1 – of all subsets of set 1.1.6 – of all parents of type C.1.1

– , Borel 2.1.1, 2.1.3 – , cardinality of 1.3.2 – of class 𝛼 (𝛼-set) B.1.1 – , compact 2.1.1 – , countable 1.2.6, A.2.2 – , cumulative A.3.1 – , inaccessible A.3.3 – , scheme-inaccessible A.8.1 – , denumerable 1.2.6, A.2.2 – , dense w. r. t. ensemble 2.1.4 – , dominant A.7.2 – , essentially positive [negative] 3.2.1 – , exponential A.7.2 – , finite 1.2.6, A.2.2 – function 3.1.1 – , galactic A.7.2 – , infinite 1.2.6, A.2.2 – , inductive A.2.1 – , inscribed [𝜎-inscribed] in functional family 2.2.9 – , Jordan measurable 3.7.3 – , main open 2.5.2 – , majorized by function 2.2.4 – , measurable w.r.t. evaluation 3.1.1 – , negligible w.r.t. evaluation 3.1.1 – , nowhere dense w. r. t. ensemble 2.1.4 – , order bounded 1.1.15 – , precompact 2∘ (3.5.1) – , scheme A.8.1 – , scheme-universal A.8.2 – , solitary A.2.1 – , squarable w.r.t. evaluation 3.1.1 – with Stone property 2.1.4 – , symmetrizable 2.1.1, 2.4.5 – , Tarski A.7.1 – , scheme A.8.4 – theory 1.1.1, A.1.3 – , topologically bounded (in Rn ) 1∘ 3.1.4 – , uncountable 1.2.6 – , universal A.4.1 𝛼-set B.1.1 signature – of generalized belongings C.1.3 – of category B.2.1 – of constants C.1.3 – of generalized equalities C.1.3 – , generalized C.1.3 – of variables C.1.3 signed function 2.2.4

Index of terms |

simple – collection 1.1.9 – correspondence 1.1.8 – sequence 1.2.6 single-valued correspondence 1.1.7 singular semimeasure [measure] 3.2.4 smallest – element of subclass 1.1.15 – member of collection 1.1.15 – upper bound of – collection 1.1.15 – set 1.1.15 – subclass A.2.2 solitary – class 1.1.6 – 𝛼-class B.1.1 – set A.2.1 space – , Alexandrov 2.1.1 – , Banach 6∘ (2.2.7) – , lattice-ordered 8∘ (2.2.7) – , conic 1∘ (3.4.2) – , descriptive 2.1.1 – with covering 2.1.5 – , compact 2.1.1 – with evaluation 3.1.1 – , Hausdorff 2∘ (3.5.1) – , lattice-ordered linear 4∘ (2.2.4) – , locally compact 2∘ (3.5.1) – , linear 4∘ (2.2.4) – with measure 3.1.1 – , metric 1∘ (2.2.7) – , normed linear 6∘ (2.2.7) – , complete 2.2.7 – , ordered linear 4∘ (2.2.4) – , prescriptive 2.1.5 – , pseudometric 1∘ (2.2.7) – , complete 2∘ (2.2.7) – with semimeasure 3.1.1 – , seminormed linear 6∘ (2.2.7) – , complete 6∘ (2.2.7) – , topological 2.1.1 squarable set 3.1.1 standard – interpretation A.6.1 – model A.6.1 – topology – on R 1∘ (2.3.1) – on Rn 1∘ (3.1.6)

437

statement of theorem 1.1.3 step function 2.2.4 – , countably 2.2.4 – , quite 2.2.4 Stone – truncation 2.2.4 – property of functional family 2.2.9 – property of set 2.1.4 strong order unit 4∘ (2.2.4) strong saturation – of semimeasurable [measurable] space 3.1.4 – of semimeasure [measure] 3.1.4 strong substitution property A.4.1, A.6.1 strongly saturated – evaluation 3.1.4 – extension – , large complete 3.1.5 – , complete 3.1.4 strongly scheme-inaccessible ordinal number A.8.1 strongly inaccessible cardinal A.2.2 subadditive evaluation 3.1.1 subadditive mapping 4∘ (2.2.7) subalgebra of linear algebra 2∘ (2.2.6) subassembly B.1.1 subclass 1.1.5, A.2.1 – , cofinal 1.1.15 – , coinitial 1.1.15 – , convex 1.1.15 – , final 1.1.15 – , initial 1.1.15 subcover of cover 1.1.10 sublinear mapping 4∘ (2.2.7) submultiplicative mapping 4∘ (2.2.7) subnet of net 1.1.15 subordinated to element (in lattice-ordered linear space) 3.2.2 subsequence of sequence 1.2.6 subset 1.1.5 – axiom B.1.1 – , uniformly order bounded (in F b (T )) 2.2.2 subspace of linear space 2∘ (2.2.6) succession relation of Peano C.3.4 successive ordinal number A.2.2 successor of – natural number C.3.4 – ordinal number B.3.3 suit – , non-ordered 1.1.11

438 | Index of terms

– , sequential 1.1.11 sum – , cardinal 1.3.5 – , netful (in R) 1.4.8 – , partial (in R) 1.4.8 – , ordinal 1.2.9 sum of – integers 1.4.1 – natural numbers 1.3.6 – rational numbers 1.4.2 – real numbers 1.4.3 superadditive evaluation 3.1.1 superstructure of signature over formation C.2.1 supertransitive class A.2.2 superior limit of net 1.1.15 support of – function 3.5.2 – mathematical system 1∘ (2.2.2), C.2.1 – semimeasure [measure] 3.5.1 supremal functions 3.6.2 supremum – of simple collection 1.1.15 – of set 1.1.15 – of subclass A.2.2 – of simple sequential pair [triplet, . . . ] 1.1.15 surface of parallelepiped 3.1.6 surjection 1.1.7, A.2.1 symbol-string 1.1.1, A.1.1 symmetric relation 1.1.14 symmetrizable – function 2.4.5 – set 2.1.1, 2.4.5 Tarski axiom A.7.1 Tarski property A.7.1 Tarski set A.7.1 term 1.1.2, A.1.1, C.1.3 – free for variable 1.1.2 terminal C.1.2 theorem 1.1.3 – , Alexandrov 2.3.5 – , Alexandrov – Stone – Fremlin integral representation 3.4.2 – , Banach local convergence classification 2.3.7 – , Beppo Levi 3.3.3 – , Bernays (on finite axiomatization of NBG) B.7.3 – , Borel – Lebesgue – Hausdorff 2.3.6

– , Bourbaki 3.6.3, 3.6.4 – , Bunina (on galactic sets) A.7.2 – , Cantor (on cardinality of set of all subsets) 1.3.2 – on characterization of Radon integrals with respect to positive Borel – Radon measures 3.6.3 – of deduction 1.1.3 – , Dini 2.3.4 – , Egorov 3.3.1 – , Halmos – Hewitt – Edwards 3.6.3 – , Hausdorff 2.3.6 – , improvement 2.1.1, 2.1.3 – , Lebesgue – Hausdorff local convergence classification 2.3.7 – , Lebesgue – Radon – Nikodym 3.3.8 – , Lebesgue – Urysohn 2.3.5 – , Lebesgue dominated convergence 3.3.3 – , Lusin 3.5.2 – , main convergence classification 2.3.7 – , main fine convergence classification 2.4.7 – , Newton binomial 1.4.6 – on one-step pointwise limits 2.3.7 – , Radon – Saks – Kakutani 3.6.4 – , reduction 2.1.1 – , Regoli 2.3.6 – , Riesz – Kantorovich 2∘ (2.2.8) – , Riesz representation D.1.4 – , Russell 1.1.12 – , Schröder – Cantor – Bernstein 1.3.2 – , reduction 2.1.1, 2.1.3 – , separation 2.1.1, 2.1.3 – , untwining 2.1.3 – , Young – Daniell 3.6.2 – , Zakharov – on boundedly normal families 2.4.6 – on characterization of family of Riemann integrable functions 3.7.2 – on characterization of Lebesgue integrals as linear functionals 3.4.2 – on characterization of natural models of ZF A.8.4 – on characterization of Radon integrals w. r. t. positive Radon measures 3.6.2 – on characterization of Radon integrals w. r. t. Radon bimeasures 3.6.4 – on compactness for generalized second-order language C.3.3 – on finite axiomatization of LTS B.7.2

Index of terms |

– on initial synchronization of powers of cumulative sets A.3.2 – on Lebesgue integral with respect to linear combinations of finite measures 3.3.8 – , Zakharov – Rodionov – on boundedly normal envelope 2.4.6 – on characterization of Radon integrals w. r. t. arbitrary Radon measures 3.6.4 – on completely normal envelope in descriptive form 2.3.6 – on completely normal envelope in constructive form 2.3.7 – on completely normal families 2.3.7 – , Zermelo well-ordering 1.2.11 – , Zermelo – Shepherdson A.6.2 thinning – net out (mapping) 1.1.15 – sequence out (mapping) 1.2.6 tight functional 3.6.1 topological evaluation 3.5.1 topological space 2.1.1 – with evaluation 3.5.1 – with measure 3.5.1 – , normal 2.1.1 – , perfect 2.1.3 – with semimeasure 3.5.1 topologically bounded set in Rn 1∘ 3.1.4 topologically internally finite evaluation 3.5.1 topology 2.1.1 – , metric 1∘ (2.2.7) – , pseudometric 1∘ (2.2.7) – , standard – on R 1∘ (2.3.1) – on Rn 1∘ (3.1.6) total – collection 1.1.9 – correspondence 1.1.7 – relation 1.1.14 – variation of evaluation 3.1.3 trace of ensemble 2.1.1 transfinite induction 1.2.8, A.2.2 transformation (mapping) 1.1.8 𝛼-transformation from functor to functor B.2.2 𝛼-transformational arrows B.2.2 transitive – class 1.2.2, A.2.2 – relation 1.1.14 – 𝜔-universality axiom A.5.2 transitivity axiom B.1.1

439

transitivity property A.4.1, A.7.1, A.8.2, A.8.4 translation on R 1.4.9 translation of formula on sequence under interpretation A.1.3 translation of formula on sequence under abstract interpretation B.5.2 triplet – , non-ordered 1.1.11 – , sequential 1.1.11, A.2.2 truncatable family of functions 2.2.9 truncation 2.2.4 Tukey lemma 1.2.11 type C.1.1 – , auxiliary C.1.1 – , basic C.1.1 – , first-order C.1.1 – domain C.1.3 – , second-order C.1.1 u-fundamental sequence 2.2.3 ultrafilter C.3.1 unconditionally summarized [multiplied] collection 1.4.8 uncountable set 1.2.6 uniform – closure 2.2.4 – function 2.4.1 – , almost 2.5.1 – limit in F (T ) 2.2.3 uniformly – order bounded functional 2.2.8 – order bounded collection [set] in F b (T ) 2.2.2 – continuous function 2.3.1 – convergent net 2.2.3 – dense family 2.2.4 – fundamental sequence 2.2.3 union of classes 1.1.5 𝛼-union of 𝛼-classes B.1.1 union of collection 1.1.10, A.2.1 – , disjoint 1.1.10 – , ordered disjoint 1.2.9 𝛼-union of 𝛼-collection B.1.1 union of sequential suit [pair, . . . ] 1.1.11 – , disjoint 1.1.11 – , ordered disjoint 1.2.9 unit – , multiplicative 3∘ (2.2.4) – , strong order 4∘ (2.2.4) – , weak order 4∘ (2.2.4)

440 | Index of terms

unit function 2.2.1 unity element 1.4.2 (Q), 1.4.3 (R) universal – abstract B.5 – assembly B.1.1 – class 1.1.5, A.2.1, B.1.1 – set A.4.1 universality axiom A.4.3 (ZF), B.1.1 (LTS) universally integrable functions 3.3.6 unordered – pair 1.1.6, A.2.1 – 𝛼-pair B.1.1 – suit 1.1.11 upper bound of collection 1.1.15 – , smallest 1.1.15 upper boundedness index of functional 3.6.4 upper continuous [𝜎-continuous] evaluation 3.4.1 upper Darboux sum 3.7.1 upper regularization of function 3.7.2 upper semimeasurable function 2.3.8 value of term on sequence under interpretation A.1.3 value of term on sequence under abstract interpretation B.5.2 value of term w. r. t. evaluation on system C.2.3 variable 1.1.2 – , connected 1.1.2, A.1.1 – , free 1.1.2, A.1.1 – , objective C.1.3 – , predicate C.1.3 variation of evaluation 3.1.3 variation of measure – , positive [negative] 3.2.1 weak order unit 4∘ (2.2.4) well-multiplied collection 1.4.8 well-ordered assembly B.3.3 well-ordered class 1.2.1, A.2.2 well-U-ordered U-class B.3.3 well-summarized collection 1.4.8 wide – Borel – Radon extension of narrow bounded Borel – Radon measure 3.5.4 – measure 3.1.1 – Radon measure 3.5.3 Young – Daniell theorem 3.6.2

Young – Hausdorff – ensembles 2.1.2 – classification theorems 2.1.3 Z-extension of C b (T , G) 2.5.2 Z 0 -extension of C b (T , G) 2.5.2 Zakharov theorem – on boundedly normal families 2.4.6 – on characterization of – family of Riemann integrable functions 3.7.2 – Lebesgue integrals as linear functionals 3.4.2 – natural models of ZF A.8.4 – Radon integrals w. r. t. positive Radon measures 3.6.2 – Radon integrals w. r. t. Radon bimeasures 3.6.4 – on finite axiomatization of LTS B.7.2 – on compactness for generalized second-order language C.3.3 – on initial synchronization of powers of cumulative sets A.3.2 – on Lebesgue integral with respect to linear combinations of finite measures 3.3.8 Zakharov – Koldunov – ensembles 2.1.2 – classification theorems 2.1.3 Zakharov – Rodionov theorem – on boundedly normal envelope 2.4.6 – on characterization of Radon integrals w. r. t. arbitrary Radon measures 3.6.4 – on completely normal families 2.3.7 – on completely normal envelope – in descriptive form 2.3.6 – in constructive form 2.3.7 Zermelo principle 1.2.11 Zermelo – Shepherdson theorem A.6.2 zero element 1.4.1 (Z), 1.4.2 (Q), 1.4.3 (R), 2.2.4 (in linear space) zero evaluation 3.1.2 zero function 2.2.1 zero-set 2.2.5 𝛾-hull of ensemble 2.1.1 𝛿-co-foundation 2.1.1 𝛿-hull of ensemble 2.1.1 𝛿-multiplicative ensemble 2.1.1 𝛿-ring (of sets) 2.1.1

Index of terms | 441

𝜀-co-foundation 2.1.1 𝜀-hull of ensemble 2.1.1 𝜀-multiplicative ensemble 2.1.1 𝜂-co-foundation 2.1.1 𝜂-hull of ensemble 2.1.1 𝜂-hull of covering 2.1.5 𝜂-multiplicative ensemble 2.1.1 𝜆-hull of ensemble 2.1.1 𝜎-additive ensemble 2.1.1 𝜎-additive evaluation 3.1.1 𝜎-algebra (of sets) 2.1.1 𝜎-continuous evaluation 3.4.1 pointwise 𝜎-continuous functional 2.2.8 𝜎-Daniell functions 3.6.2 𝜎-exact functional 3.6.1 𝜎-finite evaluation 3.1.1 𝜎-hull of ensemble 2.1.1

𝜎-ideal of sets 2.1.4 𝜎-infimal functions 3.6.2 𝜎-integrable function 3.3.2 𝜎-inscribe set in functional family 2.2.9 𝜎-lattice (of sets) 2.1.1 𝜎-ring (of sets) 2.1.1 𝜎-supremal functions 3.6.2 𝜏-additive ensemble 2.1.1 𝜏-finite evaluation 3.1.1 𝜏-hull of ensemble 2.1.1 𝜑-additive ensemble 2.1.1 𝜑-hull of ensemble 2.1.1 𝜔-inaccessibility axiom A.5.1 𝜔-universality axiom A.5.1 – , transitive A.5.2 ∈-induction principle A.2.2

Index of notations Latin alphabet A(S) 2.4.4 A(T , S) 2.1.1 𝜈𝜇ac 3.2.4 AD(T , S, I), AD(T , S) 2.5.1 AG A.7.3 AI A.4.3, B.4.1 AI(𝜔) A.5.1, B.4.1 AI(𝜔 + 𝜔) B.4.1 AI(1) A.5.2 AIC A.5.2, B.4.1 AM(T , S, I), AM(T , S) 2.5.1 Arr B.2.1 𝛽 𝛽 f Arr𝛼 , c Arr𝛼 B.2.2 Ar2 C.3.4 ASR B.6.1 AS2 1.1.5, B.1.1 AS3 A.2.1 AS6 A.2.1 AT A.7.1 ATU(𝜔) A.5.2, B.4.2 AU(T , S, I), AU(T , S) 2.5.1 AU A.4.3 AU(𝜔) A.5.1, B.4.2 AU(1) A.5.2 AUS A.5.2 A1 1.1.5, A.2.1, B.1.1 A1, . . . , A4 (Peano – Landau arithmetic) C.3.4 A2 A.2.1 A2.1, . . . , A2.9 B.7.3 (NBG)⁴ A2.1, . . . , A2.15 B.7.1 (LTS) A3 1.1.6 (NBG), B.1.1 (LTS) A3󸀠 , A3󸀠󸀠 A.6.2 A4 1.1.6 (NBG), A.2.1 (ZF), B.1.1 (LTS) A5 1.1.11 (NBG), A.2.1 (ZF), B.1.1 (LTS) A5󸀠 , A5󸀠󸀠 1.1.11 A6 1.1.11 (NBG), B.1.1 (LTS) A7 1.1.11 (NBG), A.2.1 (ZF), B.1.1 (LTS) A8 1.1.12 (NBG), A.2.1 (ZF), B.1.1 (LTS)

A9, A10 A.2.1 (ZF), B.1.1 (LTS) A11, A12, A13, A14 B.1.1 B(T , S) 2.1.1, 2.1.3 B(A(T )), B b (A(T )) 2.2.4 F b (T ), A b (T ) 2.2.1 b(𝜑), b(𝜑) 3.6.4 Bc (T , G) 3.5.4 F bc (T , G), A bc (T , G) 3.5.2 BI(T , G, 𝜇), BI(T , G, m) 3.5.6 BM(T , S) 2.3.7 BN(A(T )) 2.2.4 BSM(T , S), BSM b (T , S) 2.3.8 bu-Lim A(T ), bp-Lim A(T ) 2.2.4 BV(T ) D.1.1 C(T , G) 2.3.1 C x f 2.2.7 𝛽

C𝛼 B.2.2 c 1.4.4 A c 1.1.5 +c 1.3.5 k Cm 1.4.6 F c (T , G), A c (T , G) 3.5.2 car 𝜇 3.5.1 Card 1.3.1 card A 1.3.2, A.2.2 𝛽 Cat𝛼 B.2.2 cf (𝛼) A.2.2 cl S 1∘ (3.5.1) CN(A(T )) 2.2.4 Cn(𝛼) A.2.2, B.4.1 Cn A.2.2 Cn B.3.3 co-S 2.1.1 cons(T , Φa ), cons(T ) A.1.2 Cor(A, B), Cors (A, B) 1.1.7 Cov S 2.1.5 Cov A(T ) 2.4.6 coz, cozn 2.2.5 Coz, CozN 2.2.5

4 In this index, for notations having different values in different set theories, the abbreviations in brackets indicate subsections where the notation is explained within the framework of the corresponding theory. https://doi.org/10.1515/9783110550962-005

444 | Index of notations

(D) (property (D)) 3.6.1 D(T , S) 2.3.1 D(S) 2.1.4 ⋃d ⟮A i | i ∈ I⟯ 1.1.10 ⋃d ⟮A, A󸀠 ⟯, ⋃d ⟮A, A󸀠 , A󸀠󸀠 ⟯,. . . 1.1.11 D𝜏 (T , A(T ), 𝜑), D𝜎 (T , A(T ), 𝜑) 3.6.2 Dc 1, Dc 2, Dc 3, Dc 4 1.1.3 Dc 5, Dc 6 1.1.4 ⋃dm ⟮u i | i ∈ I⟯ 1.1.10 ⋃dm ⟮u, u󸀠 ⟯, ⋃dm ⟮u, u󸀠 , u󸀠󸀠 ⟯,. . . 1.1.11 ∪d , ∪dm 1.1.11 ⋃do , ∪do 1.2.9 dom u 1.1.7, A.2.1 domf 𝜀 3.1.1 1.1.11 D1, D2, D3, D4 1.1.3 D5, D6 1.1.4 (E) (property (E)) 3.6.2 SE 2.5.1 e(K) 3.4.1 ‖ ⋅ ‖ 6∘ (2.2.7) ‖ ⋅ ‖e , ‖ ⋅ ‖eu , ‖ ⋅ ‖eu,A(T ) 2.2.7 Eval(T , S), Evalb (T , S), Evalf (T , S), Eval(T , S)0 3.1.2 Eval(T , S, ] − ∞, ∞]), Eval(T , S, [−∞, ∞[) 3.1.2 exp 1.4.7 (E𝜎 ) (property (E𝜎 )) 3.6.2 E1, E2, E3, E4 C.1.3 F𝛼 2.1.3 𝛽

F𝛼 B.2.2 domf 𝜀, Sf (𝜀), S𝜎f (𝜀), S𝜏f (𝜀) 3.1.1 𝜇f 3.1.4 Pf (I) 1.4.8 F |s, t| 3.1.6 F (T ), F b (T ) 2.2.1 F+ (T ), F− (T ) 2.2.2 F (T , I), F b (T , I) 2.2.7 ̂ F b (T , I), F∞ (T , I) 2.2.7 fr S 1∘ (3.5.1) func(f ) A.2.1, B.4.1 𝛽

Funct𝛼 B.2.2 F0 3.7.2 G𝛼 2.1.3 Gen 1.1.3, A.1.2, C.1.3 Gen D A.1.3

gr X 1.1.15 gr (a i | i ∈ I) 1.1.15 gr(a, a󸀠 ), gr(a, a󸀠 , a󸀠󸀠 ), . . . 1.1.15 G0 3.7.2 H𝛿 (T , S), H𝜎 (T , S) 2.1.1 f

H𝜎0 (T , R, 𝜇), H𝜎 (T , R, 𝜇), 𝜎f H𝜎 (T , R, 𝜇) 3.1.5

I(T , M(T )), I(T , A(T ), M(T )) 3.3.6 I(C(T ), BV(T )) D.1.3 I(N𝜑 ) 2.1.4 I(A(T )), I𝜎 (A(T )) 2.2.9 ‖ ⋅ ‖i 3.3.4 Icn(𝜘) A.2.2, B.4.1 IdA , IdX ,A 1.1.7 iff 1.1.3 In A.2.2, B.4.1 In B.3.4 inf⟮a i | i ∈ I⟯ 1.1.15 inf I 1.1.15 inf(a, a󸀠 ), inf(a, a󸀠 , a󸀠󸀠 ), . . . 1.1.15 infra-D-power(U0 , F ) C.3.2 infra-D-prod⟮U f | f ∈ F ⟯ C.3.1 int S 1∘ (3.5.1) i v f D.1.2 i𝜇 f 3.3.2 I𝜏 (T , A(T )), I𝜎 (T , A(T )) 3.6.2 J(T , G, M, 𝜇) 3.7.1 K(T , S) 2.1.1 K𝛼 2.1.3 K(T , S, 𝜀), Kl (T , S, 𝜀) 3.1.4 l(f ) 3.7.2 g g L(Σ2 ), L(Σ2 ) C.1.3 L(T , S) 2.1.1 L𝛼 2.1.3 L(T , K) 3.4.1 L(T , R, 𝜇), L0 (T , R, 𝜇), Lf (T , R, 𝜇), L𝜎f (T , R, 𝜇) 3.1.5 L(T , M, 𝜇), L b (T , M, 𝜇), L∞ (T , M, 𝜇) 3.3.1 L1 (T , M, 𝜇) 3.3.4, 3.3.6 LAS1, . . . , LAS14 1.1.4, A.1.2 LI(T , M, 𝜇), LI e (T , M, 𝜇), LI𝜎 (T , M, 𝜇) 3.3.2 (𝜇 ⩾ 0), 3.3.6 lim(a𝜇 | 𝜇 ∈ M), lim(a𝜇 | 𝜇 ∈ M) 1.1.15

Index of notations | 445

lim(x n | n ∈ N) 1.4.4 Lim𝛼 A(T ), Lim𝛼b A(T ) 2.2.4 Lon(𝛼) A.2.2 LTS*, LTSf B.7.1 M(𝜏) C.1.1 M(T , S), M b (T , S) 2.3.1

u ∗m u󸀠 , Pm (u), (u)Im 1.1.8 ∏m ⟮u i | i ∈ I⟯ 1.1.12 u ×m u󸀠 , u ×m u󸀠 ×m u󸀠󸀠 1.1.12 Em (T , A(T )), Em (A(T )) 2.2.4 M(T ), Mn (T ), Mw (T ) 3.1.2 M(T , I𝜎 (A(T )))0 , Me (T , I𝜎 (A(T )))0 3.4.2 M(T , I(A(T )))0 , Me (T , I(A(T )))0 3.4.3 M#(T , R, 𝜇), M≠ (T , R, 𝜇) 3.1.4 ̂ ̂ , R, 𝜇), M(T ̂ , R, 𝜇) 3.1.5 M(T M∞ (T , M, 𝜇) 3.3.1 Map(A, B) 1.1.8, A.2.1 Meas(T , R), Measb (T , R), Measf (T , R), Meas(T , R)0 3.1.2 Meas(T , R, ] − ∞, ∞]), Meas(T , R, [−∞, ∞[) 3.1.2 Measof (T , R) 3.2.1 Meas(T , M0 , I𝜎 (A(T )))0 , Measb (T , M0 , I𝜎 (A(T ))) 3.4.2 Meas(T , M0 , I(A(T )))0 , Measb (T , M0 , I(A(T ))) 3.4.3 MI(T , M, 𝜇), MI e (T , M, 𝜇), MI𝜎 (T , M, 𝜇) 3.3.2 (𝜇 ⩾ 0), 3.3.6 a = a󸀠 (mod 𝜀) 1.1.14 A ∼ B mod I, S mod I 2.1.4 f ∼ g mod I 2.2.6 MP 1.1.3, A.1.2, C.1.3 N 1.2.6 N0 C.3.4 N(A(T )) 2.2.4 N(𝜏) C.1.1 A n (main part of A) 1.1.15 N(S) 2.1.4 N(T , S, 𝜀), N0 (T , S, 𝜀), Nl (T , S, 𝜀) 3.1.4 (A(T )󳵻 )nat 3.6.4 NBG*, NBGf B.7.3 ∑o , +o 1.2.9 Oint 2.3.1 Opar 3.1.6 Ost 2.3.1 (on R), 3.1.6 (on Rn )

O0 2.5.2 Obj B.2.1 o-lim(a𝜇 | 𝜇 ∈ M) 1.1.15 On(𝛼) A.2.2, B.4.1 On A.2.2 On B.3.3 on(x) B.5.2 Ord 1.2.3 ord⟮A, ⩽⟯ 1.2.5 P(A) (power of set A) 1.3.2 P(A) 1.1.5, A.2.1, B.3.2 P(𝜏) C.1.1 P(𝛼i | i ∈ I) (cardinal product) 1.3.5 P(x i | i ∈ I) 1.4.1 (in Z), 1.4.2 (in Q), 1.4.3 (in R) P(x i | i ∈ N) (product of sequence) 1.4.8 P(f i | i ∈ I) 2.2.1 (in F (T )) p󸀠 m, p󸀠󸀠 m 3.5.5 Pf (I) 1.4.8 Pm (u) 1.1.8 P c 1, . . . , P c 7 B.2.1 P1, P2, P3, P4 C.3.4 P|s, t|, Ppar 2.1.1 Parf (S, E), Par𝜎 (S, E) 3.1.3 𝜎 Parf (S, E), Par𝜎 (S, E), Par (S, E) 3.1.3 PE1, . . . , PE4 C.3.4 p-lim (f n | n ∈ N) 2.2.3 p-Lim A(T ) 2.2.4 Pnet (x i | i ∈ I) 1.4.8 prA 1.1.8 pJ , prA j , prj 1.1.12 Q, Q1 , Q+ , Q− 1.4.2 QD(T , S, I), QD(T , S) 2.5.2 QM(T , S, I), QM(T , S) 2.5.2 QU(T , S, I), QU(T , S) 2.5.2 q St(T , S), qStc (T , S) 2.2.4 R, R+ , R− , R, R+ , R− 1.4.3 r𝜇 f 3.7.1 R(T , S), R𝛿 (T , S), R𝜎 (T , S) 2.1.1 f

𝜎f

R0 𝜎 (T , R, 𝜇), R𝜎 (T , R, 𝜇), R𝜎 (T , R, 𝜇) 3.1.5 R(T , G), Rb (T , G), Re (T , G), R⋆ (T , G) 3.5.5 RB(T , G), RBb (T , G) 3.5.5 RB(T , G, A(T )) 3.6.4 Rcn(𝜘) B.4.1 restX u 1.1.7 RI(T , G, M, 𝜇) 3.7.1

446 | Index of notations

RI(T , v) D.1 R m f 3.7.3 RMn (T , G), RMn⋆ (T , G) 3.5.4 RMw (T , G), RMwe (T , G), RMw⋆ (T , G) 3.5.3 RMw (T , G, M) 3.5.3 RMwe (T , G, A(T ))0 , RMw⋆ (T , G, A(T ))0 3.6.3 rng u 1.1.7, A.2.1 ⟮R, S⟯ 1.4.4 S(T , S) 2.4.5 S(f , 𝜘), s(f , 𝜘) 3.7.1 S(f , v, 𝜇) D.1.2 SE 2.5.1 S|s, t| 3.1.6 𝜈𝜇s 3.2.4 ̌ , 𝜑̂S 3.6.2 𝜑S S ≡ ⟮S c , S e , S b , S v ⟯ C.2.1

SC l (T , S), SC u (T , S) 2.3.8, 3.5.2 SI(T , M, 𝜇) 3.6.1 sign f 2.2.4 Slb (T , M, 𝜇) 3.3.1 sm X 1.1.15 sm (a i | i ∈ I) 1.1.15 sm(a, a󸀠 ), sm(a, a󸀠 , a󸀠󸀠 ), . . . 1.1.15 a m Snet (x i | i ∈ I), Snet (x i | i ∈ I) a m S (x i | i ∈ N), S (x i | i ∈ N) 1.4.8 SMeas(T , R), SMeasb (T , R), SMeasf (T , R), SMeas(T , R)0 3.1.2 SMeas(T , R, ] − ∞, ∞]), SMeas(T , R, [−∞, ∞[) 3.1.2 SMeasof (T , R) 3.2.1 SMeasof (T , R)+ , SMeasof (T , R)− 3.2.2 SM l (T , S), SM u (T , S) 2.3.8 Son(𝛼) A.2.2 SP(S, I) 2.1.4 Spar 2.1.1 St(T , S), Stc (T , S) 2.2.4 sup⟮a i | i ∈ I⟯ 1.1.15 sup I 1.1.15, A.2.2 sup(a, a󸀠 ), sup(a, a󸀠 , a󸀠󸀠 ), . . . 1.1.15 S𝜏 (T , A(T )), S𝜎 (T , A(T )) 3.6.2 supp f 3.5.2 supp 𝜇 3.5.1 U 1.1.5, B.5.1 U A.4.1 U B.7.3

U(T , C), U(T , S) 2.4.1 u(f ) 3.7.2 ‖ ⋅ ‖u , ‖ ⋅ ‖u,A(T ) 2.2.7 U(A(T )), U b (A(T )) 2.2.4 U(T , S), UE (T , S) 2.1.1 U0 (𝜇) 3.7.2 Us (T , S, 𝜀) 3.1.4 UI(T , M(T )) 3.3.6 u-lim (f n | n ∈ N) 2.2.3 u-Lim A(T ) 2.2.4 V B.1.1 V𝛼 A.3.1 V𝛼 B.3.4 v(𝜀, 𝜋) 3.1.3 v(v, 𝜌) D.1.1 var 𝜀, v(𝜀) 3.1.3 var(v) D.1.1 v+ (𝜇), v− (𝜇) 3.2.1 WI(T , G, m) 3.5.6 Z, Z+ , Z− , Z∗ 1.4.1 zer, Zer 2.2.5 Greek alphabet Γ𝛼 (T , S) 2.1.2 Γ(T , G, M, 𝜇) 3.7.1 S𝛾 2.1.1 Δ 𝛼 (T , S) 2.1.2 Δ(T , G, M, 𝜇) 3.7.1 S𝛿 2.1.1 S𝜀 2.1.1 S𝜂 (𝜂-hull of ensemble) 2.1.1 C𝜂 (𝜂-hull of covering) 2.1.5 𝜃I , 𝜃I,A(T ) , 𝜃A0 (T ,I) 2.2.6 Θ𝛼 (T , R) 2.1.2 Θ2Ar2 C.3.4 𝜆 (Borel – Lebesgue measure) 3.1.4 ̂ 𝜆× , 𝜆,̂ 𝜆,̂ 𝜆̃ 3.1.6 ̂ ̂ ,M ̂ 3.1.6 R ,L ,M 𝜆

𝜆

𝜆

Λ 𝛼 (T , S) 2.1.2

𝜆

Index of notations | 447

S𝜆 2.1.1 Λ(𝜇) (integral) 3.3.2 (for 𝜇 ⩾ 0), 3.3.6 Π A.3.3 A(T )𝜋 , A(T ) 𝜋 3.6.1 ∏⟮A i | i ∈ I⟯ 1.1.12, A.2.2 ∏⟮A, A󸀠 ⟯, ∏⟮A, A󸀠 , A󸀠󸀠 ⟯, . . . 1.1.12, A.2.2 ∏m ⟮u i | i ∈ I⟯ 1.1.12 ∏0 ⟮U i | i ∈ I⟯ 1.1.15 ∏0 ⟮U, U󸀠 ⟯, ∏0 ⟮U, U󸀠 , U󸀠󸀠 ⟯ 1.1.15 Σ𝛼 (T , S) 2.1.2 ∑(𝛼i | i ∈ I) (cardinal sum) 1.3.5 ∑(x i | i ∈ I) 1.4.1 (in Z), 1.4.2 (in Q), 1.4.3 (in R) ∑(x i | i ∈ N) (sum of sequence) 1.4.8 ∑(f i | i ∈ I) 2.2.1 (in F (T )) 𝜎(f , 𝜋) 3.3.2 𝜎(f , 𝜔), Σ(f , 𝜔) 3.7.3 ∑net (x i | i ∈ I) 1.4.8 ∑o 1.2.9 S𝜎 2.1.1 S𝜎f (𝜀) 3.1.1 Σc , Σe , Σb , Σv C.1.3 g Σg , Σ2 C.1.3 g

ΣAr2 C.3.4 S𝜏 2.1.1 S𝜏f (𝜀) 3.1.1 Y𝛼 (T , S) 2.1.2 S𝜑 2.1.1 𝜒(R) 2.2.4 𝜔 1.2.6, A.2.2, B.1.1, B.5.2 Ω, 𝜔1 1.3.4 𝜔(f , E) 2.2.1 𝜔(f , 𝜋) 2.3.1 Ω(T , G, J, m) 3.7.3 Digits 0 1.2.2, 1.2.6 0A 2∘ (2.2.4) A0 (T , I) 2.2.7 A0 (main part of A) 1.1.15 Eval(T , R)0 3.1.2

SMeas(T , R)0 , Meas(T , R)0 3.1.2 S0 (𝜀) 3.1.1 ∏0 ⟮U i | i ∈ I⟯ 1.1.15 ∏0 ⟮U, U󸀠 ⟯, ∏0 ⟮U, U󸀠 , U󸀠󸀠 ⟯ 1.1.15 0, 1 2.2.1 1,̄ ‖ ⋅ ‖1̄ 2.2.7 1, 2, 3, 4, . . . 1.1.11, 1.2.6 1A 3∘ (2.2.4) k ∈ 2, n \ 1, (n + 1) \ 2, 𝜔 \ 3 . . . 1.2.6 A2 , A3 , . . . 1.1.12 u−1 [Y], u−1 ⟨b⟩ 1.1.7, A.2.1 x −1 (inverse number) 1.4.2, 1.4.3 Arrows u:A B 1.1.7, A.2.1 u : x 󳨃→ y 1.1.8, A.2.1 u : A → B 1.1.8, A.2.1 u:A B 1.1.8, A.2.1 u:A B 1.1.8, A.2.1 u:A B 1.1.8, A.2.1 a𝜇 ↑ a, (a𝜇 | 𝜇 ∈ M) ↑ 1.1.15 a𝜇 ↓ a, (a𝜇 | 𝜇 ∈ M) ↓ 1.1.15 ]←, b[, ]←, b], ]a, →[, [a, →[ 1.1.15 F 󴀔󴀭 A → B A.2.1, B.4.1 Other symbols of operations and relations a ∈ A 1.1.5 a ∈ 𝛼 (ordinals) 1.2.2 k ∈ n, k ⊂ n (natural numbers) 1.2.6 < −𝜏 C.2.1 A ∼ B 1.1.8, A.2.2, B.3.3 a ∼ a󸀠 1.1.14 A ∼ B mod I 2.1.4 f ∼ g mod I 2.2.6 A∼ 2∘ (2.2.8) ⟮A, ⩽⟯ ≈ ⟮B, ⩽⟯ 1.1.15, A.2.2 ≈𝜏 , ≈𝜏̌ C.2.1 A ∪ B 1.1.5, A.2.1, B.3.2 A ∪ A󸀠 ∪ A󸀠󸀠 , . . . 1.1.11 a0 ∪ . . . ∪ a n−1 1.2.6 ∪C A.2.1, B.3.2 ⋃⟮A i | i ∈ I⟯ 1.1.10, A.2.1 ⋃⟮A, A󸀠 ⟯, ⋃⟮A, A󸀠 , A󸀠󸀠 ⟯, . . . 1.1.11 ⋃d ⟮A i | i ∈ I⟯, ⋃dm ⟮u i | i ∈ I⟯ 1.1.10

448 | Index of notations

⋃d ⟮A, A󸀠 ⟯, ⋃d ⟮A, A󸀠 , A󸀠󸀠 ⟯,. . . 1.1.11 ⋃dm ⟮u, u󸀠 ⟯, ⋃dm ⟮u, u󸀠 , u󸀠󸀠 ⟯,. . . 1.1.11 ∪d , ∪dm 1.1.11 ⋃do , ∪do 1.2.9 A ∩ B 1.1.5, A.2.1, B.3.2 A ∩ A󸀠 ∩ A󸀠󸀠 , . . . 1.1.11 a0 ∩ . . . ∩ a n−1 1.2.6 ⋂⟮A i | i ∈ I⟯ 1.1.10, A.2.1 ⋂⟮A, A󸀠 ⟯, ⋂⟮A, A󸀠 , A󸀠󸀠 ⟯, . . . 1.1.11 B\A 1.1.5, A.2.1 n\k, 𝜔\k (natural numbers) 1.2.6 P\Q 2.1.4 . . . ∨ . . . (. . . or . . . ) 1.1.1 a ∨ a󸀠 , a ∨ a󸀠 ∨ a󸀠󸀠 , . . . 1.1.15 a ⊻ a󸀠 , a ⊻ a󸀠 ⊻ a󸀠󸀠 , . . . 1.1.15 P ∨ Q 2.1.4 S ∨ R 2.1.1 A(T )∨ , A(T ) ∨ 2.2.8 ϕ, 𝜑∨ 3.6.2 󳶋

A(T ) , A(T )

󳶋

3.6.1

. . . ∧ . . . (. . . and . . . ) 1.1.1 a ∧ a󸀠 , a ∧ a󸀠 ∧ a󸀠󸀠 , . . . 1.1.15 a ⊼ a󸀠 , a ⊼ a󸀠 ⊼ a󸀠󸀠 , . . . 1.1.15 P ∧ Q 2.1.4 𝜋1 ∧ 𝜋2 (coverings) 2.1.5 ⋀ (𝜋𝛼 | 𝛼 ∈ A) 2.1.5 S ∧ R 2.1.1 A(T )∧ , A(T ) ∧ 2.2.8 𝜑∧ , 𝜑∧ 3.6.2 A(T )󳵻 , A(T ) 󳵻 3.6.1 (A(T )󳵻 )nat 3.6.4 a ⩽ a󸀠 , a󸀠 ⩾ a, a < b, b > a 1.1.14 a ⩽ b, 𝛼 < 𝛽 (ordinals) 1.2.2 ⟮A, ⩽⟯ 1.1.15 f < g, f ⩽ g, f ≪ g 2.2.2 ⩽A 1∘ (2.2.6) 𝜈 ≪ 𝜇 3.2.4 +o 1.2.9 +c 1.3.5 +A (in linear space A) 2∘ (2.2.4)

𝛼 + 1, 𝛼+ 1.2.3, B.3.3 x + x 󸀠 , x + x 󸀠 + x 󸀠󸀠 , . . . 1.4.1 (in Z), 1.4.2 (in Q), 1.4.3 (in R) f + g 2.2.1 (in F (T )) a+ , A+ 1.1.15 f+ 2.2.2 𝜇+ , SMeasof (T , R)+ , Measof (T , R)+ 3.2.2 F (T )+ 2.2.2 v+ (𝜇) 3.2.1 −x 1.4.1, 1.4.2, 1.4.3 n − m 1.3.6 −f , f − g 2.2.1 −A 2∘ (2.2.4) a− , A− 1.1.15 f− 2.2.2 𝜇− , SMeasof (T , R)− , Measof (T , R)− 3.2.2 F (T )− 2.2.2 v− (𝜇) 3.2.1 u−1 [Y], u−1 ⟨b⟩ 1.1.7 x −1 (inverse number) 1.4.2, 1.4.3 A(T )∗ 2.3.6 A ∗ B 1.1.6, A.2.1, B.3.2 u ∗m u󸀠 1.1.8 𝜇∗ (for semimeasure 𝜇) 3.1.5 𝜑∗ (for formula 𝜑) C.2.4 v ∘ u 1.1.7 A(T ) ⃝ 2.2.8 A × A󸀠 , A × A󸀠 × A󸀠󸀠 , . . . 1.1.12 a0 × . . . × a n−1 1.2.6 u ×m u󸀠 , u ×m u󸀠 ×m u󸀠󸀠 1.1.12 U ×0 U󸀠 , U ×0 U󸀠 ×0 U󸀠󸀠 1.1.15 A× 6∘ (2.2.7) 𝜇× 3.1.5 𝜆× 3.1.6 xx 󸀠 , xx 󸀠 x 󸀠󸀠 , . . . 1.4.1 (in Z), 1.4.2 (in Q), 1.4.3 (in R) fg 2.2.1 (in F (T )) f ⋅ 𝜇 (product of function and measure) 3.3.7 m/n 1.3.6 (in N) m/p 1.4.2 (in Q) x/y 1.4.2 (in Q), 1.4.3 (in R) A/𝜀 1.1.14

Index of notations | 449

S/I 2.1.4 1/f , f /g 2.2.1 A(T )/𝜃, A(T )/A0 (T , I) 2.2.7 𝜇 ⊥ 𝜈, B⊥ , B⊥⊥ , 𝜇⊥⊥ 3.2.2 Brackets u(a) 1.1.8 (u)Im 1.1.8 (a i ∈ A | i ∈ I), (a i | i ∈ I) 1.1.9, A.2.1 (A, A󸀠 ), (A, A󸀠 , A󸀠󸀠 ), . . . 1.1.11, A.2.2 ⃗ 𝜑(x,⃗ p)⃗ 1.1.2, A.8.1, B.1.1 𝜑(x, y), 𝜑(p), u[X] 1.1.7, A.2.1 u−1 [Y] 1.1.7, A.2.1 ]a, b[, [a, b], ]a, b], [a, b[ 1.1.15 ]←, b[, ]←, b], ]a, →[, [a, →[ 1.1.15 ] − ∞, x[ , ] − ∞, x], ]x, ∞[ , [x, ∞[ 1.4.3 t M [s] A.1.3 tM [s] A.1.3 q[𝛾] C.2.3 M ⊨ 𝜑[s] A.1.3 M ⊨ 𝜑[s] B.5.2 U ⊨ 𝜑[𝛾] C.2.3 ⃗ 𝜑[x,⃗ p]⃗ B.1.1 𝜑[x], ⃗ [𝜑(x, y, p)|A] A.8.1

⟮R, S⟯ 1.4.4 |a, b| 1.1.15 𝜑|x, y| 1.1.2 |x| 1.4.1 (in Z), 1.4.2 (in Q), 1.4.3 (in R) |f | 2.2.2 (in F (T )) |a| 4∘ (2.2.4) (in lattice-ordered space) |𝜇| 3.2.2 (in SMeasof (T , R)) |A| (≡ card A) 1.3.2, A.2.2, B.3.3 ‖ ⋅ ‖∞ 2.2.7 ‖ ⋅ ‖u , ‖ ⋅ ‖u,A(T ,I) 2.2.7 ‖ ⋅ ‖e , ‖ ⋅ ‖eu , ‖ ⋅ ‖eu,A(T ) 2.2.7 ‖ ⋅ ‖1̄ 2.2.7 ‖ ⋅ ‖i 3.3.4 ‖ ⋅ ‖󸀠 7∘ (2.2.7) Infinities ∞, −∞ 1.4.3 A∞ (T ) 2.2.7 F∞ (T ) 2.2.7 ‖ ⋅ ‖∞ 2.2.7 L∞ (T , M, 𝜇), M∞ (T , M, 𝜇) 3.3.1 ] − ∞, x[ , ] − ∞, x], ]x, ∞[ , [x, ∞[ 1.4.3 Ascenders

⟨A, B⟩ 1.1.6, A.2.1, B.3.2 u⟨a⟩ 1.1.7 u−1 ⟨b⟩ 1.1.7 ⃗ ⟨𝜎(x; u)|A⟩ A.8.1 {A}, {A, B} 1.1.6, A.2.1, B.3.2 {A, A󸀠 , A󸀠󸀠 }, . . . 1.1.11 {a i | i ∈ I} 1.1.9 {x x ∈ I | x ∈ X} 1.1.9 {a0 , . . . , a n−1 } 1.2.6 ⃗ 1.1.5, A.2.1 {x | 𝜑(x)}, {x | 𝜑(x, p)} {x ∈ A | 𝜑(x)} 1.1.5, A.2.1 ⟮T , S⟯, ⟮T , G⟯ 2.1.1 ⟮T , C⟯ 2.1.5 ⟮T , S, I⟯ 2.1.4 ⟮T , S, 𝜀⟯ 3.1.1 ⟮A i ⊂ A | i ∈ I⟯, ⟮A i | i ∈ I⟯ 1.1.9, A.2.1 ⟮A, A󸀠 ⟯, ⟮A, A󸀠 , A󸀠󸀠 ⟯, . . . 1.1.11, A.2.2 ⟮A xx 󸀠 | x ∈ X , x 󸀠 ∈ X 󸀠 ⟯, . . . 1.1.12 ⟮{x}x ⊂ I | x ∈ X⟯ 1.1.9

a,̄ Ā 1.1.14 1̄ 2.2.7 A(T , I), F (T , I), F b (T , I) 2.2.6 󸀠

P 2.1.4 𝜘(k), 𝜘(k) 2.4.4 𝜀̄̄ 3.1.3 𝜑, 𝜑 3.6.2 b(𝜑), b(𝜑) 3.6.4 ∀x,⃗ ∃x⃗ A.8.1, B.1.1 𝜑∨ , 𝜑∨ , 𝜑∧ , 𝜑∧ 3.6.2 ̂ (for ensemble S) 2.1.1 S ̂ F b (T , I) 2.2.7 ̂ 𝜑,̌ 𝜑̌S , 𝜑̂S 3.6.2 𝜑, 𝜏̌ (for type 𝜏) C.1.1 𝜇,̌ 𝜇,̄ 𝜇󸀠 , 𝜇󸀠󸀠 , 𝜇#, 𝜇≠ (for semimeasure 𝜇) 3.1.4 A† 2∘ (2.2.8) ‖ ⋅ ‖󸀠 7∘ (2.2.7) ⟮A, ‖ ⋅ ‖A ⟯󸀠 , A󸀠 7∘ (2.2.7)

450 | Index of notations

𝜑∗ (for formula 𝜑) C.2.4 𝜇∗ , 𝜇× , 𝜇,̂ 𝜇̃ (for semimeasure 𝜇) 3.1.5 ̂ 𝜆× , 𝜆,̂ 𝜆,̂ 𝜆̃ 3.1.6 Special cases of using upper and lower indices B A 1.1.8, A.2.1 𝛼𝛽 1.3.5 𝜑U (relativity of formula 𝜑 to class U) A.6.1, B.3.1 A n (= A × . . . × A) 1.2.6 x n (= x ⋅ ⋅ ⋅ x) 1.4.1, 1.4.2, 1.4.3 m √x, x p/m 1.4.6 r √ m f , f 2.2.1 P𝛼 , ∪𝛼 , ∩𝛼 , ⋃𝛼 , ⋂𝛼 , {. . .}𝛼 , ⟨. . .⟩𝛼 , ⟮. . .⟯𝛼 , ∗𝛼 , 𝛼 , →𝛼 , 𝛼, 𝛼, 𝛼 , dom𝛼 , A rng𝛼 , B(𝛼) , Map𝛼 (A, B) B.1.1 ∏𝛼 , ×𝛼 B.1.2 ∼U , cardU B.3.3 Miscellaneous ¬, ∧, ∨, ⇒, ∀, ∃, ≡ 1.1.1, A.1.1

∈, =, ⊂, ⊃, = / , ∉ 1.1.5, A.2.1, B.1.1 #, ∘, ↔ B.2.1 ⇔ 1.1.2, A.1 ⊢ 1.1.3, A.1.2, B.5.2 ⌀ 1.1.5, B.1.1 ℵ 1.3.4 a B.1.1 U⋈ A.4.1, B.1.1 ⋈⟮𝛾f | f ∈ F ⟯ C.3.1 ⋈⟮𝛾0 , F ⟯ C.3.2 𝜁(v ‖ 𝜏) (substituted in formula) 1.1.2, A.1.1 u|X, u‖X (restriction of correspondence) 1.1.7 m + 1|n (rank of formation, etc.) C.1.2 ⃗ [𝜑(x, y, p)|A] A.8.1 ⃗ ⟨𝜎(x; u)|A⟩ A.8.1 n!, (m k ) 1.4.6 ∫ f d𝜇 3.3.2, 3.3.6 (R) ∫ f d𝜇 3.7.1 M ⊨ 𝜑[s] A.1.3 M ⊨ 𝜑[s] B.5.2 U ⊨ 𝜑[𝛾] C.2.3 ⃗ 1.1.5, A.2.1 {x | 𝜑(x)}, {x | 𝜑(x, p)} ⃗ B.1.1 {x | 𝜑[x, p]} {x ∈ A | 𝜑(x)} 1.1.5, A.2.1, B.1.1

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