Measure and integration 9783030187460, 9783030187477

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Springer Undergraduate Mathematics Series

Satish Shirali Harkrishan Lal Vasudeva

Measure and Integration

Springer Undergraduate Mathematics Series Advisory Editors M. A. J. Chaplain, St. Andrews, UK Angus MacIntyre, Edinburgh, UK Simon Scott, London, UK Nicole Snashall, Leicester, UK Endre Süli, Oxford, UK Michael R. Tehranchi, Cambridge, UK John F. Toland, Bath, UK

The Springer Undergraduate Mathematics Series (SUMS) is a series designed for undergraduates in mathematics and the sciences worldwide. From core foundational material to final year topics, SUMS books take a fresh and modern approach. Textual explanations are supported by a wealth of examples, problems and fully-worked solutions, with particular attention paid to universal areas of difficulty. These practical and concise texts are designed for a one- or two-semester course but the self-study approach makes them ideal for independent use.

More information about this series at http://www.springer.com/series/3423

Satish Shirali Harkrishan Lal Vasudeva •

Measure and Integration

123

Satish Shirali Formerly of Panjab University Chandigarh, India

Harkrishan Lal Vasudeva Formerly of Panjab University Chandigarh, India

University of Bahrain Zallaq, Bahrain

IISER Mohali Mohali, India

ISSN 1615-2085 ISSN 2197-4144 (electronic) Springer Undergraduate Mathematics Series ISBN 978-3-030-18746-0 ISBN 978-3-030-18747-7 (eBook) https://doi.org/10.1007/978-3-030-18747-7 Mathematics Subject Classification (2010): 28-01, 28A05, 28A10, 28A12, 28A20, 28A25, 28A35 © Springer Nature Switzerland AG 2019 This work is subject to copyright. All rights are reserved by the Publisher, whether the whole or part of the material is concerned, specifically the rights of translation, reprinting, reuse of illustrations, recitation, broadcasting, reproduction on microfilms or in any other physical way, and transmission or information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed. The use of general descriptive names, registered names, trademarks, service marks, etc. in this publication does not imply, even in the absence of a specific statement, that such names are exempt from the relevant protective laws and regulations and therefore free for general use. The publisher, the authors and the editors are safe to assume that the advice and information in this book are believed to be true and accurate at the date of publication. Neither the publisher nor the authors or the editors give a warranty, expressed or implied, with respect to the material contained herein or for any errors or omissions that may have been made. The publisher remains neutral with regard to jurisdictional claims in published maps and institutional affiliations. This Springer imprint is published by the registered company Springer Nature Switzerland AG The registered company address is: Gewerbestrasse 11, 6330 Cham, Switzerland

Preface

A good part of the subject matter of the present book is concerned with Measure Theory and Integration. It provides a substantial part of the necessary background for several branches of modern mathematics, applied mathematics and probability. The Riemann theory of integration for bounded functions f:[a, b]! R, where [a, b] is a bounded closed interval of R, serves all the needs of elementary calculus. However, it has some limitations that are inherent in its definition: (i) It is defined for bounded functions only. The function x1=2 is not Riemann integrable on [0,1] because it is not bounded on the interval. However, it is integrable in an extended sense and the value of the integral is 2. (ii) It is defined on bounded intervals only. On the interval [1,∞), which is not bounded, the function 1=x2 is not Riemann integrable; however, it is well known that the value of its integral over [1,∞) is 1 in an extended sense. On the other hand, the following limitations cannot be overcome by using the same kind of extension: (iii) Functions having finite range, even if defined on a bounded interval, need not be Riemann integrable. For example, the Dirichlet function f ð xÞ ¼ 1 if x is rational and 0 otherwise is not Riemann integrable. Indeed, R1 R1 0 f ðxÞdx = 1 but 0 f ðxÞdx = 0. (iv) The interchange of the notions of limit and integral is not always possible in the Riemann theory of integration. The following example will substantiate the foregoing statement: Let {r1,r2,…} be an enumeration of the rationals in [0,1]. Define for every n 2 N,

v

vi

Preface

 fn ðxÞ ¼

if x 2 fr1 ; r2 ;    ; rn g otherwise:

1 0

Then lim fn ðxÞ ¼ f ðxÞ, the Dirichlet function of (iii) above. It is not even Riemann n!1

integrable, leave alone the possibility that Z1

Z1 fn ðxÞdx ¼

lim

n!1 0

f ðxÞdx: 0

In modern mathematics the need was felt for a theory of integration which is free of the drawbacks (iii) and (iv) above. Mathematicians of the time made several attempts to enlarge the class of functions which could be brought under a satisfactory theory of integration. É. Borel (1898) had presented a theory of measure for a large class of sets that included unions of countably many disjoint intervals and differences of such sets. The decisive advance was made by H. Lebesgue (1902), who formulated a meaningful theory of integration based on an improvement and generalisation of the work of Borel. He generalised the concept of length of an interval to the measure of an even larger class of sets of real numbers than Borel did. The definition of the Lebesgue integral permits a much more general class of functions, including certain unbounded functions, as integrands and replaces the interval [a,b] by more general kinds of sets called “measurable” sets. It also facilitates the interchange of the notions of limit and integral. More specifically, the equality Z Z lim fn dm ¼ f dm; n!1

X

X

where lim fn ðxÞ ¼ f ðxÞ, X is a measurable subset of R and “dm” indicates n!1

Lebesgue integration, holds under mild conditions on the functions fn and f. While the Riemann theory begins by partitioning the domain of the function sought to be integrated, Lebesgue’s approach starts by partitioning the range, or more precisely, partitioning an interval containing the range. It then becomes necessary to have some sort of length for the inverse image of each subinterval of the partition, although the inverse image may not be an interval. In other words, one needs to assign a size, or “measure”, to a larger class of sets than just the intervals, and that too, in a manner that is consistent with expectations about the measure of a union of disjoint sets. The procedure for developing the theory of measure and integration follows a well established procedure: outer measure and measure, measurable functions, integrable functions, convergence theorems for sequences of integrable functions are introduced successively. This is carried out in Chaps. 2 and 3. The Lebesgue theory of integration developed in the latter enables us to define certain spaces such as Lp spaces 1  p  ∞, which are Banach spaces. These were introduced by F. Riesz. The primary purpose of Sect. 3.3 is the construction of these spaces and

Preface

vii

the derivation of their properties. The fact that the vital property of completeness can be achieved is a direct consequence of the “congenial” behavior of the integral when limits are taken. The class of square integrable functions is obtained for p = 2. In this case p = q = 2 and Hölder’s Inequality takes the form Z

Z

1=2 Z

jfgjdl  X

1=2

f 2 dl X

g2 dl X

and is referred to as the Cauchy–Schwarz Inequality, where “dl” indicates integration with respect to a general measure l. This inequality leads to the introduction of inner product spaces and orthogonality of two elements of the space L2. In Sect. 3.4, dense subsets of Lp, 1  p < ∞, for the case when l is Lebesgue measure are pointed out. In Analysis, it is often desirable and fruitful to approximate objects by simpler objects, e.g., irrational numbers are approximated by rational numbers and continuous functions by polynomials. We approximate measurable sets by open sets, measurable functions by simple functions as well as by continuous functions. That every convergent sequence of measurable functions is nearly uniformly convergent is proved. It is shown that each of the classes of simple functions, continuous functions of compact support and step functions is dense in the space of Lp-functions. The class of square integrable functions and the concept of mean square convergence of a sequence of functions of this class are included in Chap. 3. An arbitrary integrable 2p-periodic function can be expressed formally as 1 X 1 f ð x Þ ¼ a0 þ ðan cos nx þ bn sin nxÞ; 2 n¼1

with coefficients an and bn given by the formulae an ¼

1 p

bn ¼

1 p

Z

p p

f ðxÞ cos nx dx; n ¼ 0; 1; 2; . . .

and Z

p p

f ðxÞ sin nx dx; n ¼ 1; 2; 3; . . .:

The above series is called the “Fourier series” of the function f. In Chap. 4, the representation of a square integrable 2p-periodic function by Fourier series and its convergence in the mean are studied. The important result connecting measure theory and Fourier series, namely the Riesz–Fischer Theorem, constitutes a part of the chapter. The pointwise convergence (Dirichlet’s Theorem), L2 convergence and Cesàro summability of Fourier series (Fejér’s Theorem) are proved under appropriate hypotheses. The Weierstrass Polynomial Approximation Theorem is derived as an

viii

Preface

application of Fejér’s Theorem. The chapter concludes with Fejér’s example of a continuous function whose Fourier series diverges at 0. In Chap. 5, we study monotone functions, functions of bounded variation and absolutely continuous functions. As is well known, a continuous function need not be differentiable. K. Weierstrass (1861) constructed a continuous function having no derivative at any point whatsoever. B. Bolzano (1831) knew the result but was debarred from publishing by an imperial decree. However, if we additionally suppose that the continuous function in question is monotone, it possesses a derivative almost everywhere. This result was proved by H. Lebesgue in 1904. W. H. Young (1911) gave a proof of the Lebesgue Theorem without the continuity assumption. These and other proofs of the seminal result, including the one by L. A. Rubel (1963), are provided. Equally important and related to the class of monotone functions is the class of functions of bounded variation. The important properties of the two classes and the relationship between them are obtained. After proving some of the important properties of the special class of ‘absolutely continuous’ functions, we provide a characterisation of an absolutely continuous function f on a domain [a,b]: Z f ðxÞ ¼ c þ g; ½a;x

where c is a constant and g is in L1[a,b]. The relationship between rectifiable curves and functions of bounded variation is indicated in the problems. Also included is the change of variables formula in Lebesgue integration. Signed measures, which are a useful generalisation of measures, the Hahn Decomposition Theorem and the Jordan Decomposition Theorem constitute Sect. 5.9. In Sect. 5.10, the Radon– Nikodým Theorem and Lebesgue decomposition of a measure into an absolutely continuous part and singular part are discussed. The connection between bounded increasing left continuous real-valued functions on the real line and finite measures on Borel subsets of the real line are discussed in Sect. 5.11. Chapter 6 begins with a quick overview of real normed linear spaces and bounded linear functionals on them, the focus being on the Lp spaces. Hölder’s Inequality suggests a way of defining bounded linear functionals on these spaces. The Riesz Representation Theorem, which asserts in the r-finite case that all bounded linear functionals on Lp are indeed as suggested by Hölder’s Inequality when 1  p < ∞, is proved for Lebesgue measure on [0,1]. A proof of the Riesz Representation Theorem for a general finite measure is also provided and is based on the Radon–Nikodým Theorem. Up to this point in the book, various types of convergence of sequences of measurable functions have been encountered, such as pointwise, almost everywhere, uniform and in the Lp norm. Other modes of convergence of sequences of measurable functions that also play a significant role in Analysis are almost uniform convergence and convergence in measure. The interplay between them is thoroughly dwelt upon in Chap. 6.

Preface

ix

Let ðX; F ; lÞ and ðY; G; mÞ be two r-finite measure spaces and ðX  Y; F  G; l  mÞ be the product measure space, where F  G denotes the product r-algebra generated by the measurable rectangles A  B; A 2 F ; B 2 G and l  m the product measure. If f is an F  G-measurable function defined on X  Y, the integral of f on X  Y is denoted by Z f ðx; yÞdðl  mÞðx; yÞ; XY

or some variant thereof, and is called a double integral. We may integrate with respect to y and obtain a function of x, namely, Z f ðx; yÞdmðyÞ: Y

Integrating it with respect to x, we obtain Z

Z f ðx; yÞdmðyÞ:

dlðxÞ X

Y

In Chap. 7, we seek conditions which ensure the following equality: Z

Z

X

Z f ðx; yÞdmðyÞ ¼

dlðxÞ Y

Z f ðx; yÞdðl  mÞðx; yÞ ¼

XY

Z f ðx; yÞdlðxÞ:

dmðyÞ Y

X

The application of the above equality to convergence of Fourier series in Lp spaces is indicated. A reader with a modest background in Mathematical Analysis in one variable, including rudiments of Metric Spaces, which are included in Chap. 1, will find the material covered in the book well within reach. The treatment of the Cantor set and the Cantor function, though not new, is well worth the reader’s attention. The authors are grateful to the referees for useful suggestions. Gurugram, India Chandigarh, India March 2019

Satish Shirali Harkrishan Lal Vasudeva

Contents

1 Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . 1.1 Sets and Functions . . . . . . . . . . . . . . . 1.2 Countable and Uncountable Sets . . . . . 1.3 Metric Spaces . . . . . . . . . . . . . . . . . . . 1.4 Sequences and Series of Extended Real 1.5 Sequences and Series of Functions . . . . 1.6 Derivatives . . . . . . . . . . . . . . . . . . . . . 1.7 Riemann Integration . . . . . . . . . . . . . . 1.8 Decimals and the Cantor Set . . . . . . . .

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. . . . Numbers . ........ ........ ........ ........

2 Measure in Euclidean Space . . . . . . . . . . . . . . . . . 2.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . 2.2 Lebesgue Outer Measure . . . . . . . . . . . . . . . . 2.3 Measurable Sets and Lebesgue Measure . . . . . 2.4 Measurable Functions . . . . . . . . . . . . . . . . . . 2.5 Extended Real-Valued Functions . . . . . . . . . . 2.6 Egorov’s and Luzin’s Theorems . . . . . . . . . . 2.7 Lebesgue Outer Measure in Rn . . . . . . . . . . . 2.8 Measurable Sets and Lebesgue Measure in Rn

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3 Measure Spaces and Integration . . . . . 3.1 Integrals of Simple Functions . . . 3.2 Integrals of Measurable Functions 3.3 Lp Spaces . . . . . . . . . . . . . . . . . . 3.4 Dense Subsets of Lp . . . . . . . . . .

1 1 7 8 18 20 22 24 28

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. 43 . 43 . 46 . 56 . 71 . 77 . 87 . 93 . 100

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109 109 119 138 157

4 Fourier Series . . . . . . . . . . . . . . . . . . . . . . 4.1 Introduction . . . . . . . . . . . . . . . . . . . 4.2 The Convergence Problem . . . . . . . . 4.3 Cesàro Summability of Fourier Series

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163 163 170 185

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xi

xii

Contents

4.4 Even and Odd Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 191 4.5 Orthonormal Expansions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 197 Appendix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 205 5 Differentiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.1 Background . . . . . . . . . . . . . . . . . . . . . . . . . 5.2 Monotone Functions and Continuity . . . . . . . 5.3 Monotone Functions and Differentiability (A) . 5.4 Monotone Functions and Differentiability (B) . 5.5 Integral of the Derivative . . . . . . . . . . . . . . . 5.6 Total Variation . . . . . . . . . . . . . . . . . . . . . . . 5.7 Absolute Continuity . . . . . . . . . . . . . . . . . . . 5.8 Differentiation and the Integral . . . . . . . . . . . 5.9 Signed Measures . . . . . . . . . . . . . . . . . . . . . . 5.10 The Radon–Nikodým Theorem . . . . . . . . . . . 5.11 The Lebesgue–Stieljes Measure . . . . . . . . . . .

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211 211 212 217 228 240 252 266 277 294 305 316

6 Lebesgue Spaces and Modes of Convergence . . . . . . . . . . . . . . . . . . 337 6.1 The Spaces Lp as Normed Linear Spaces . . . . . . . . . . . . . . . . . . 337 6.2 Modes of Convergence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 355 7 Product Measure and Completion . . . . . . . . . . . . . . . . . . . . . . . . . . 375 7.1 Product Measure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 375 7.2 The Completion of a Measure . . . . . . . . . . . . . . . . . . . . . . . . . . 392 8 Hints . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 405 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 591 Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 593

Chapter 1

Preliminaries

We shall find it convenient to use logical symbols such as 8, 9, ∋, ) and ,. These are listed below with their meanings. A brief summary of Set Algebra, Functions, Elementary Real Analysis, which will be used throughout this book, is included in this chapter. Our purpose is descriptive and no attempt has been made to give proofs of the results stated, except in Sect. 1.8. The reader is expected to be familiar with the material up to Sect. 1.7. The words ‘set’, ‘class’, ‘collection’ and ‘family’ are regarded as synonymous and we do not define these terms.

1.1

Sets and Functions

The following commonly used symbols will be employed in this book: 8 means “for all” or “for every” 9 means “there exists” ∋ means “such that” ) means “implies that” or simply “implies” , means “if and only if”. The concept of set plays an important role in every branch of modern mathematics. Although it is easy and natural to define a set as a collection of objects, it has been shown that this definition leads to a contradiction. The notion of set is, therefore, left undefined and a set is described by simply listing its elements or by its properties or properties of its elements. Thus {x1, x2, …, xn} is the set whose elements are x1, x2, …, xn; and {x} is the set whose only element is x. If X is the set of all elements x such that some property P(x) is true, we shall write X ¼ fx : PðxÞg:

© Springer Nature Switzerland AG 2019 S. Shirali and H. L. Vasudeva, Measure and Integration, Springer Undergraduate Mathematics Series, https://doi.org/10.1007/978-3-030-18747-7_1

1

2

1 Preliminaries

The symbol £ denotes the empty set. We write x 2 X if x is a member of the set X; otherwise x 62 X. If Y is a subset of X, that is, if x 2 Y implies x 2 X, we write Y  X. If Y  X and X  Y, then X = Y. If Y  X and Y 6¼ X, then Y is proper subset of X. Observe that £  X for every set X. Given a set X, the collection of all its subsets is called the power set of X and will be denoted by PðXÞ. We list below the standard notations for the most important sets of numbers: N the set of all natural numbers Z the set of all integers Q the set of all rational numbers R the set of all real numbers C the set of all complex numbers. Given two sets X and Y, we can form the following new sets from them: X [ Y ¼ fx : x 2 X or x 2 Y g; X \ Y ¼ fx : x 2 X and x 2 Y g: X [ Y and X \ Y are the union and intersection respectively of X and Y. More generally, given a nonempty family F of sets, we define their union to be [ F2F

F ¼ fx : x 2 F for at least one F 2 F g:

S It is also denoted by F or [ F . Given a nonempty family F of sets, we define their intersection to be \ F2F

It is also denoted by The symbol

T

F ¼ fx : x 2 F for every F 2 F g:

F or \ F . f : X!Y

means that f is a function (or mapping or map) from the nonempty set X into the set Y; that is, f assigns to each x 2 X an element f(x) 2 Y. The elements assigned to members of X by f are often called values of f. If A  X and B  Y, the image of A and inverse image of B are respectively f ðAÞ ¼ ff ðxÞ : x 2 Ag; f 1 ðBÞ ¼ fx : f ðxÞ 2 Bg: Note that f−1(B) may be empty even when B 6¼ £. The domain of f is X and the range is f(X); the range space is Y. If f(X) = Y, the function f is said to map X onto Y (or the function is said to be surjective). We write

1.1 Sets and Functions

3

f−1(y) instead of f−1({y}) for every y 2 Y. When f−1(y) consists of at most one element for each y 2 Y, f is said to be one-to-one (or injective). If f is one-to-one, then f−1, which is actually a map from PðYÞ to PðXÞ, gives rise to a function with domain f(X) and range X. The latter is also denoted by f−1, as it does not cause any confusion. A function which is both injective and surjective is said to be bijective and is often spoken of as a bijection or one-to-one correspondence. If f is bijective, then f−1 is a function with domain Y and range X, and is called the inverse of f. A map is said to be invertible if it has an inverse. It is obvious that being invertible is equivalent to being bijective and that the inverse is unique if it exists. There can never exist a bijection between any set X and its power set PðXÞ. It is sometimes necessary to consider a function f only on a nonempty subset S of its domain X. Technically that makes it a different function and it is called the restriction of f to S. Introducing a new symbol to denote a restriction can clutter the notation and we shall avoid it as far as possible. Let g: U ! V and f: X ! Y be maps, where X has a nonempty intersection with the range g(U). Then the inverse image Z ¼ g1 ðX \ gðUÞÞ  U is nonempty and the function f  g : Z ! Y such that ðf  gÞðzÞ ¼ f ðgðzÞÞ is called the composition of f and g. For most theoretical purposes, it is sufficient to work with the case X  g(U), because this ensures that X \ g(U) = g(U) and hence that Z = U. Among the important examples of functions are those defined on subsets of the set N of natural numbers. By a finite sequence of n terms we understand a function f whose domain is the set {1, 2, …, n}. The range of f is the set {f(1), …, f(n)}. The function value ak = f(k) is called the kth term of the sequence. By an infinite sequence we shall mean a function f whose domain is N. The range of f is the set {f(1), f(2), …}, which is also written as {f(k): k  1}. Here again, the function value ak = f(k) is called the kth term of the sequence. For brevity, we shall use the notation ff ðkÞgk  1 or ff ðkÞg1 k¼1 , or simply {f(k)}, to denote the infinite sequence whose kth term is f(k). It is sometimes more convenient to denote the sequence by fak gk  1 or fak g1 k¼1 , or simply {ak}, especially when a letter such as f may not have been introduced to denote the function. Sometimes the range space X of a sequence is specified by speaking of a sequence in X. Ordinarily, a sequence will be understood to be an infinite sequence unless the context indicates the contrary. A sequence of n terms may be denoted by ff ðkÞgnk¼1 or by fak gnk¼1 . Given a sequence fak g1 k¼1 , consider a sequence {kn} of positive integers such that k1 \k2 \k3 \   . Then the sequence {akn } is called a subsequence of fak g1 k¼1 . When we have a map from a set K into the power set of some set X, we speak of an indexed family or indexed collection of subsets of X. Here K is often referred to as the indexing set. One can create an indexed family from any nonempty family F by taking F itself as the indexing set and selecting the map to be the identity map. This often makes it possible to carry over results and definitions about an indexed family of sets with a general type of indexing set to an arbitrary nonempty family of subsets.

4

1 Preliminaries

If {Xa} is an indexed collection of sets, where a runs through some indexing set K, we write \ [ Xa and Xa a2K

a2K

for the union and intersection respectively of Xa: [

Xa ¼ fx : x 2 Xa for at least one a 2 Kg;

a2K

\

Xa ¼ fx : x 2 Xa for every a 2 Kg:

a2K

If K ¼ N, the set of all natural numbers, the customary notations are 1 [

Xn and

n¼1

1 \

Xn :

n¼1

Two sets A and B are said to be disjoint if they have no element in common, i.e., A \ B = ∅. If no two members of a collection of sets, indexed or otherwise, have any element in common, then the sets or the collection is said to be pairwise disjoint. We shall often omit the word “pairwise”, because no confusion will arise. A union of disjoint sets is often called a disjoint union. If Y  X, the complement of Y in X is the set of elements that are in X but not in Y, that is XnY ¼ fx : x 2 X; x 62 Y g: The complement of Y is denoted by Yc whenever it is clear from the context with respect to which larger set the complement is taken. If {Xa} is a collection of subsets of X, then the following De Morgan’s laws hold: ð

[ a2K

Xa Þc ¼

\

ðXa Þc

a2K

\

and ð

a2K

Xa Þc ¼

[

ðXa Þc :

a2K

Given finitely many sets X1, X2, …, Xn, by an ordered n-tuple (x1, x2, …, xn), we understand a finite sequence fxk gnk¼1 , where xk 2 Xk for k = 1, 2, …, n. The Cartesian product X1  X2      Xn of the sets X1, X2, …, Xn is the set of all ordered n-tuples (x1, x2, …, xn). The element xk is called the kth coordinate of (x1, x2, …, xn) and is the same as the kth term of the sequence fxk gnk¼1 . If {Xa: a 2 K} is any family of subsets of X, then [ [ fð Xa Þ ¼ f ðXa Þ a2K

a2K

1.1 Sets and Functions

5

and

\

fð

Xa Þ 

a2K

\

f ðXa Þ:

a2K

Also, if {Ya: a 2 K} is a family of subsets of Y, then f 1 ð

[

Ya Þ ¼

a2K

and f 1 ð

\ a2K

[

f 1 ðYa Þ

a2K

Ya Þ ¼

\

f 1 ðYa Þ:

a2K

If Y1 and Y2 are subsets of Y, then f 1 ðY1 nY2 Þ ¼ f 1 ðY1 Þnf 1 ðY2 Þ: One of the methods of producing subsets from a given family of sets is via the Axiom of Choice. Axiom of Choice 1.1.1 Given a nonempty family {Aa: a 2 K} of nonempty disjoint subsets of a set X, there exists an S  X containing exactly one element from each Aa. A binary relation R on a set X is a subset of X  X. The statement that (x, y) 2 R is written as xRy. Let X be a set with a binary relation R. The relation R is said to be an equivalence relation if, for all x, y, z 2 X, (i) xRx, (ii) xRy ) yRx, (iii) xRy, yRz ) xRz. The properties (i), (ii) and (iii) are, respectively, known as reflexivity, symmetry and transitivity. Let R be an equivalence relation on X. For each x 2 X, the subset Rx = {y 2 X: yRx} is called the equivalence class of x and often denoted by ½½ x . The fundamental result in this direction is the following: (i) [ {Rx: x 2 X} = X, (ii) if xRy then Rx = Ry, (iii) if x6 Ry, i.e., x is not related to y, then Rx \ Ry = ∅. Thus equivalence classes constitute a partition of X, by which we mean a family of disjoint subsets whose union is X. Conversely, a partition of X yields an equivalence relation on X, whose equivalence classes are precisely the disjoint sets of the partition. An element of an equivalence class is called a representative of that class.

6

1 Preliminaries

Let R be the set which is the union of R with two elements, written ∞ and −∞ (points at infinity). The enlarged set, when endowed with the structure described in (a)–(i) below, in addition to that already available on its subset R, is called the extended real numbers: (a) (b) (c) (d) (e) (f) (g) (h) (i)

−∞ < x < ∞ for every x 2 R; x + ∞ = ∞ + x = ∞ and x + (−∞) = (−∞) + x = −∞ for every x 2 R; x∞ = ∞x = ∞ and x(−∞) = (−∞)x = −∞ for every positive x 2 R; x(∞) = (∞)x = −∞ and x(−∞) = (−∞)x = ∞ for every negative x 2 R; ±∞0 = 0(±∞) = 0; ∞ + ∞ = ∞ and (−∞) + (−∞) = −∞; ∞(±∞) = (±∞)∞ = ±∞ and (±∞)(−∞) = (−∞)(±∞) = ∞; −(−∞) = ∞ and −(∞) = −∞; (∞)p = ∞ for real p > 0.

It can be verified on a case by case basis that the associative property of addition and the distributive property continue to be valid as long as ∞ and −∞ do not both appear in a sum. This fact is essential for working with infinite series in R . The question of whether multiplication is associative will not arise anywhere. In view of the restricted validity of the associative and distributive properties, the extended real numbers do not form a number “system” in the sense usually understood. The reader is cautioned that there is no unanimity among authors about the properties (a)–(i). For instance [11, p. 1] and [19, p. 12] include a property to the effect that (x/±∞) = 0 for real x but [12, pp. 54–55] does not; also, [11] includes our (g) but [12] specifically excludes all parts of it except for ∞∞ = ∞. The work [20, pp. 18–19] does not define any addition or multiplication involving −∞. While [20] defines 0∞ to be 0, the same author leaves it undefined in [19, p. 12]. The absolute value is defined in the obvious manner and the Triangle Inequality can be verified, whenever the smaller side is well defined. Let A  R . The smallest extended real number which is greater than or equal to each x 2 A is called the supremum of A and is denoted by supA or by supx2Ax, or by some variant thereof, depending on how the set A has been described. If A = ∅, then every extended real number is vacuously greater than or equal to each x 2 A and hence sup∅ is −∞. Analogously for the infimum, infA and inf∅. It may be noted in passing that ∅ is the only set for which the infimum is greater than the supremum. Let {xn}n  1 be a sequence in R . Then supn  1xn and infn  1xn are respectively the supremum and infimum of the range of the sequence. One of the advantages of the extended real numbers is that every subset has a supremum and an infimum. Another is that results about the integral of a sum or product of two functions can be stated without recourse to separate cases.

1.2 Countable and Uncountable Sets

1.2

7

Countable and Uncountable Sets

Two sets A and B are said to be equivalent, and we write A * B, if there exists a one-to-one mapping, i.e., a bijection of A onto B. (Two finite sets are equivalent if and only if they have the same number of elements.) This relation clearly has the following properties: It is reflexive: A * A for all A, it is symmetric: if A * B, then B * A, it is transitive: if A * B and B * C, then A * C. The transitivity is a consequence of the fact that the composition of two bijections is a bijection. Here are some examples of equivalent sets: 1. The set X ¼ f2n : n 2 Ng and the set N are equivalent. Indeed, n ! 2n is a bijection of N onto X. 2. It is easy to see that if I is an interval (open, closed, semi-open) with more than one point, then there is a bijection between I and R. Indeed, the function f : ð0; 1Þ ! R given by f ðxÞ ¼

 2x1

x ; 2x1 1x

0\x\ 12 x\1

1 2

defines a bijection between (0, 1) and R. Another such function f is given by x f ðxÞ ¼ ln 1x . Each of these functions is continuous along with its inverse. If we extend the domain to [0, 1] by setting f(0) = −∞ and f(1) = ∞, we obtain a bijection between [0, 1] and R . On the other hand, the function g: [0, 1] ! (0, 1) given by gðxÞ ¼

8 < :

1 2 x 2x þ 1

x

x¼0 x ¼ 1n ; n 2 N x 6¼ 0; 1n ; n 2 N

defines a bijection between [0, 1] and (0, 1). The latter function, suitably modified, shows that [0, 1) * (0, 1). Thus R ð0; 1Þ ½0; 1Þ ½0; 1 R . However, we emphasise that this equivalence is only a set-theoretic one in the sense defined at the beginning of this section. For example, the elements n þn 2 and 1 of [0, 1], where n 2 N, would be considered as being “arbitrarily” close but not gðn þn 2Þ ¼ n þn 2 and gð1Þ ¼ 13. This means some of the mathematical features of the set [0, 1] are not “preserved” by g. In more precise terminology, g is not continuous. The knowledgeable reader will observe that the bijection g is far from being what is called a “homeomorphism”.

8

1 Preliminaries

Theorem 1.2.1 (Schröder–Bernstein) Let A and B be two sets. If each of them is equivalent to a subset of the other, then A * B. For any positive integer n, let Nn be the set whose elements are 1, 2, …, n. A set A is finite if A Nn for some n. A set is infinite if it is not finite. A set A is said to be countable if A N. It is said to be uncountable if it is neither finite nor countable and is said to be at most countable if it is either finite or countable. An infinite subset of a countable set is countable. When a set A is countable, a bijection that maps N onto A is often called an enumeration of the elements of A and we speak of elements a1, a2, … of A. An important result that will often be used in the sequel is that the Cartesian product of N with itself finitely many times is countable. The set Q of rational numbers is countable, whereas the set R is uncountable. Another important result that will S often be used in the sequel is that if {An}n  1 is a sequence of countable sets, then n  1 An is countable. Also, if each An is at most S countable, then so is n  1 An . Let X be any nonempty set. The set of all maps X ! {0, 1}, where {0, 1} is the set consisting of the two integers 0 and 1, is denoted by 2X. For any set A  X 6¼ £, the function vA : X ! R such that vA(x) is 0 if x 62 A and 1 if x 2 A is called the characteristic function of A. Observe that vA, where A  X, is a member of the set 2X. The following simple result has important consequences: 2X PðXÞ. Indeed the mapping f : PðXÞ ! 2X defined by f ðAÞ ¼ vA ; A  X, is clearly a bijection between PðXÞ and 2X. In subsequent chapters, we shall need to compare the size of two sets. It turns out that with every set A is associated a unique set called its cardinal number, denoted by card(A), such that A and card(A) are equivalent. If card(A) and card(B) are two distinct cardinal numbers, then A and B are not equivalent but one of the sets, say A, and a proper subset of B are equivalent. In this case, card(A) is said to be a smaller cardinal number and we write card(A) < card(B). We shall not go into the rigorous treatment of cardinal numbers. For a rigorous treatment of this topic, the reader may refer to [10]. The cardinal number of R occurs frequently and we shall denote it by c when necessary. The cardinal number of N is denoted by @0 . The following results, which we state without proof, will be needed in the sequel: 1. For any set A, cardðAÞ\cardðPðAÞÞ; 2. cardðPðNÞÞ ¼ card(2N Þ ¼ card(RÞ ¼ card([0, 1]) ¼ c.

1.3

Metric Spaces

The ideas of convergence and open sets in a variety of spaces encountered in the present text can be best described in a metric space. The general reference for Sects. 1.3 to 1.5 is [26].

1.3 Metric Spaces

9

Definition 1.3.1 Let X be a nonempty set. A nonnegative function d defined on the set of ordered pairs {(x, y): x 2 X, y 2 X} is a metric on X if it satisfies the following properties: (i) d(x, y) = 0 if and only if x = y, (ii) d(x, y) = d(y, x), (iii) d(x, z) d(x, y) + d(y, z), for all x, y, z 2 X. The pair (X, d) is called a metric space. Customarily, one calls X the metric space when d is understood from the context. ðR; dÞ, where d(x, y) = |x − y|, x; y 2 R, is the real line with the usual metric. The n-fold Cartesian product R  R      R is denoted by Rn and is called Euclidean n-space. There are three useful metrics on it, given by d1 ðx; yÞ ¼

n X

jxi  yi j;

i¼1

d2 ðx; yÞ ¼

n pX

ðxi  yi Þ2

i¼1

and

d1 ðx; yÞ ¼ max1 i n jxi  yi j;

where x ¼ ðx1 ; x2 ; . . .; xn Þ 2 Rn and y ¼ ðy1 ; y2 ; . . .; yn Þ 2 Rn . The inequality p d1 ðx; yÞ d1 ðx; yÞ nd2 ðx; yÞ nd1 ðx; yÞ

ð1:1Þ

holds. Definition 1.3.2 Any nonempty subset Y  X together with the induced metric defined by dY(x, y) = d(x, y), x, y 2 Y, is a metric subspace of (X, d). It is denoted by (Y, dY). The subscript Y of dY is often dropped when there is no scope for confusion. In fact, the subset Y itself is referred to as a metric subspace when d is understood. The subset ð0; 1Þ ¼ fx 2 R : x [ 0g of R, with the induced metric, denoted yet again by d, is a metric subspace of R. If we relax the condition (i) above to read as d(x, y) = 0 if x = y, we obtain a pseudometric space. More precisely, we have the following definition. Definition 1.3.3 Let X be a nonempty set. A nonnegative function d defined on the set of ordered pairs {(x, y): x 2 X, y 2 X} is a pseudometric on X if it satisfies the following properties: (i) d(x, y) = 0 if x = y, (ii) d(x, y) = d(y, x), (iii) d(x, z) d(x, y) + d(y, z), for all x, y, z 2 X.

10

1 Preliminaries

The pair (X, d) is called a pseudometric space. Customarily, one calls X the pseudometric space when d is understood from the context. The notion of pseudometric space will prove useful in Chaps. 3 and 6. In a pseudometric space, it is standard procedure to generate a related metric space by “identifying” x and y whenever d(x, y) = 0, i.e. by forming equivalence classes corresponding to the binary relation defined by d(x, y) = 0—quite obviously an equivalence relation. One can verify for equivalence classes ½½ x ; ½½ y with respective representatives x, y that d ð½½ x ; ½½ y Þ ¼ dðx; yÞ is unambiguous (it is independent of the choice of representatives x, y) and that it gives rise to a metric on the set of equivalence classes. We next consider the notion of convergence of sequences. Definition 1.3.4 Let (X, d) be a metric space and {xn}n  1 a sequence in X. An element x 2 X is said to be a limit of the sequence {xn}n  1 if, for every e > 0, there exists a natural number n0 such that d(x, xn) < e whenever n  n0. If this is the case, we say that the sequence converges to x and write it in symbols as xn ! x as n ! ∞, or alternatively, as limnxn = x. If there is no such x, we say that the sequence {xn}n  1 diverges. The symbol limn is often written as lim or simply as lim. n!1

When a limit exists in a metric space, it is unique. Let xn ¼ 1n, n = 1, 2, …. Then {xn}n  1 is a sequence in R and limnxn = 0. The sequence {xn}n  1, where xn = (−1)n, n = 1, 2, …, is divergent. Suppose a sequence in Rn converges to a limit in the sense of any one of the three metrics d1, d2, d∞ described above. In view of the inequality (1.1) between the three metrics, it follows that it converges to the same limit in the sense of the other two metrics. To enable us to operate in the extended real number system, we shall enlarge the scope of the above definition. We say that a sequence {xn}n  1 of extended real numbers diverges to ∞, and write xn ! ∞ as n ! ∞ if for any M > 0, we can find an n0 such that xn > M for any n  n0. In this case, limxn is the extended real number ∞. Similarly, a sequence {xn}n  1 diverges to −∞ and we write xn ! −∞ as n ! ∞ if for any M > 0, we can find an n0 such that xn < −M for any n  n0. In this case, limxn is the extended real number −∞. Whether limxn is an extended real number or a real number, it is unique whenever it exists. Definition 1.3.5 A sequence {xn}n  1 in a metric space X is said to be Cauchy if, for every e > 0, there exists a natural number n0 = n0(e) such that d(xn, xm) < e whenever n  n0 and m  n0. A metric space is said to be complete if every Cauchy sequence in it converges. It is trivial that every sequence that converges is a Cauchy sequence. The space ðR; dÞ is a complete metric space, where d(x, y) = |x − y|, x; y 2 R, is the usual metric. This is a consequence of the following well-known result in Analysis:

1.3 Metric Spaces

11

Theorem 1.3.6 (Cauchy’s Principle of Convergence) Let {xn}n  1 be a sequence in R satisfying the condition that, for every e > 0, there exists a natural number n0 = n0(e) such that d(xn, xm) < e whenever n  n0 and m  n0. Then the sequence converges to a limit in R. The subspace ðQ; dÞ is incomplete. p It is well known that there is a sequence of rational numbers that converges to 2. It therefore converges to a point that does not belong to the space. Suppose a sequence in Rn is Cauchy in the sense of any one of the three metrics d1, d2, d∞ described above. In view of the inequality (1.1) between the three metrics, it follows that it is Cauchy in the sense of the other two metrics. Moreover, Rn is complete in the sense of each of the metrics. Proposition 1.3.7 If a subsequence of a Cauchy sequence in a metric space converges, then the Cauchy sequence converges to the same limit as the subsequence. The open and closed sets, defined below, will subsequently play a distinguished role. Definition 1.3.8 Let (X, d) be a metric space. The set S(x0, r) = {x 2 X: d(x, x0) < r}, where r > 0 and x0 2 X, is called the open ball with centre x0 and radius r. The set Sðx0 ; r Þ ¼ fx 2 X : d ðx; x0 Þ r g, where r  0 and x0 2 X, is called the closed ball with centre x0 and radius r. The open ball S(x0, r) on the real line is the bounded open interval (x0 − r, x0 + r) with midpoint x0 and total length 2r. Conversely, any nonempty bounded open interval on the real line is an open ball. So the open balls on the real line are precisely the nonempty bounded open intervals. The closed balls on the real line are precisely the nonempty bounded closed intervals. In Rn , n > 1, the open ball of radius r with centre (x1, x2, …, xn) depends on which metric is being used. When the metric is d∞, it is the Cartesian product of the n open intervals of length 2r with midpoints at x1, x2, …, xn, respectively. It is an example of what will later be called a “cuboid”. Definition 1.3.9 A set U in a metric space (X, d) is an open set if, given any x0 2 U, there exists an r > 0 such that Sðx0 ; r Þ  U. That is, U contains an open ball around each of its points. So, X, ∅ are open and it also follows that any union of open sets is open and that any finite intersection of open sets is open. intersection of open sets need not be open. Indeed, the collection  An1 arbitrary    n ; 1n : n 2 N has intersection {0}, which is certainly not open. The inequalities (1.1) have the consequence that a subset of Rn which is open in the sense of any one of the three metrics d1, d2, d∞ is open in the sense of the other two. Therefore we need not specify the metric when we mention an open subset. Let Y be a nonempty subset of the metric space (X, d) together with the induced metric. A subset in Y is open if and only if it is the intersection of an open set in X with Y.

12

1 Preliminaries

By a neighbourhood of a point x we mean an open set containing x. Thus a sequence {xn}n  1 converges to x if and only if, for every neighbourhood of x, there exists a natural number n0 such that xn lies in the neighbourhood whenever n  n0. Definition 1.3.10 A set F in a metric space (X, d) is a closed set if its complement Fc is open. A set in a metric space (X, d) is bounded if it is contained in some ball. For example, [0, ∞) = (−∞, 0)c is closed in ðR; dÞ, where d is the usual metric. A subset of a metric space may be neither open nor closed: the semi-open interval (0, 1] is neither open nor closed in R. Definition 1.3.11 The closure F of a set F  X, where (X, d) is a metric space, is the intersection of all closed subsets of X that contain F. Definition 1.3.12 A point a is said to be a limit point of F if, given e > 0, there exists an x 2 F, x 6¼ a, such that d(x, a) < e. In other words, a limit point of a set is characterised by the property that every neighbourhood of it contains a point of the set other than the point in question. The limit points of an interval are those numbers that belong to it and also any endpoints it may have, regardless of whether the endpoints belong to the interval or not. A point of a set F is said to be isolated if it is not a limit point of F, which is to say, some neighbourhood of it contains no other point of the set. A set is closed if and only if it is the same as its closure. The closure is the union of the set with the set of its limits points (if any). A set is closed if and only if its limit points, if any, belong to it; in particular, a set having no limit points is always closed. Let f be an extended real-valued function defined on a subset F of a metric space X and a be a limit point of F. We say that f(x) tends to l 2 R as x tends to a if, for every e > 0, there exists some d > 0 such that jf ðxÞ  lj\e

8x 2 F

satisfying

0\dðx; aÞ\d:

Also, we say that f(x) tends to l ¼ 1 2 R as x tends to a if, for every real M, there exists some d > 0 such that f ðxÞ [ M

8x 2 F

satisfying

0\dðx; aÞ\d:

Similarly for l = −∞. The value l 2 R is uniquely determined when it exists. The extended real number l is said to be the limit of f(x) as x tends to a and we write lim f ðxÞ ¼ l or

x!a

limx!a f ðxÞ ¼ l

or

f ðxÞ ! l as x ! a:

Note that one among f(a) and limx!a f ðxÞ may make sense but not the other. More precisely, if a is a limit point not belonging to the domain F, only limx!a f ðxÞ makes sense, while if a belongs to the domain F but is not a limit point (is an isolated point), only f(a) makes sense.

1.3 Metric Spaces

13

If lim f ðxÞ and lim gðxÞ both exist and are real, then so do the limits x!a

x!a

limðf ðxÞ þ gðxÞÞ and limðf ðxÞgðxÞÞ; moreover,

x!a

x!a

limðf ðxÞ þ gðxÞÞ ¼ lim f ðxÞ þ lim gðxÞ

x!a

x!a

x!a

and limðf ðxÞgðxÞÞ ¼ ðlim f ðxÞÞðlim gðxÞÞ:

x!a

x!a

x!a

If lim f ðxÞ and lim gðxÞ both exist and one or both are ±∞, the above assertion x!a

x!a

about the sum holds unless the sum lim f ðxÞ þ lim gðxÞ is undefined, i.e., unless x!a

x!a

one of the limits is ∞ and the other is −∞. The assertion regarding the product holds unless one limit is 0 and the other is ±∞. If a is any real number, then lim ðaf ðxÞÞ ¼ aðlim f ðxÞÞ. This holds even if x!a

a = ±∞ but with the proviso that lim f ðxÞ 6¼ 0.

x!a

x!a

If there exists an η > 0 such that f(x) h(x) g(x) whenever 0 < d(x, a) < η and if lim f ðxÞ and lim gðxÞ both exist and are equal, then lim hðxÞ also exists and x!a

x!a

lim hðxÞ ¼ lim f ðxÞ ¼ lim gðxÞ.

x!a

x!a

x!a

x!a

The following special aspect of limits when the domain is an interval is crucial in Chap. 5. When we speak of the limit of a function defined on a subset F of R, it is understood that R has the usual metric, which is given by d(x, y) = |x − y|. So the meaning of “limx!a f ðxÞ ¼ l 2 R”, is that, for every e > 0, there exists some d > 0 such that jf ðxÞ  lj\e 8x 2 F satisfying 0\jx  aj\d: By definition of absolute value, this is equivalent to the conjunction of the two statements: jf ðxÞ  lj\e 8x 2 F satisfying 0\x  a\d; jf ðxÞ  lj\e 8x 2 F satisfying 0\a  x\d: Suppose the domain F is an interval. (Since a is understood to be a limit point of F, it is either a point of F or an endpoint.) If for every e > 0, there exists some d > 0 such that the first of these statements holds and a is not the right endpoint of F, the number l must be unique and we speak of l as the right-hand the limit of f(x) as x tends to a or as the limit of f(x) as x tends to a+. Similarly, if for every e > 0, there exists some d > 0 such that the second of these statements holds and a is not the left endpoint of F, the number l must be unique and we speak of l as the left-hand limit of f(x) as x tends to a or as the limit of f(x) as x tends to a−. There is no left[resp. right-] hand limit at the left [resp. right] endpoint of an interval. The modifications required when l = ±∞ are obvious.

14

1 Preliminaries

The left-hand and right-hand limits are denoted by f ðaÞ

or

f ða þ Þ

or

lim f ðxÞ

x!a

and lim f ðxÞ

x!a þ

respectively. Together, they are referred to as one-sided limits. At the left [resp. right] endpoint of F, the right- [resp. left-] hand limit is the same as the limit. At an interior point, a limit exists if and only if both one-sided limits exist and are equal, in which case, their common value is the limit, sometimes called the two-sided limit for emphasis. It will be proved in Proposition 5.2.2 that for an increasing function, the left [resp. right] hand limit always exists except at the left- [resp. right-] endpoint of the domain. It is trivial that f ða þ Þ ¼ lim f ða þ hÞ and f ðaÞ ¼ lim f ða þ hÞ. h!0 þ

h!0

Suppose f is an extended real-valued function defined on a subset F of R, and supF = ∞. We say that f(x) tends to l 2 R as x tends to ∞ if, for every e > 0, there exists some K 2 R such that jf ðxÞ  lj\e 8x 2 F satisfying x [ K: Also, we say that f(x) tends to l ¼ 1 2 R as x tends to ∞ if, for every real M, there exists some K 2 R such that f ðxÞ [ M 8x 2 F satisfying x [ K: Similarly for l = −∞. The value l 2 R is uniquely determined when it exists. The extended real number l is said to be the limit of f(x) as x tends to ∞ and we write lim f ðxÞ ¼ l or

x!1

limx!1 f ðxÞ ¼ l or

f ðxÞ ! l as x ! 1:

The meaning of lim f ðxÞ when the domain F satisfies infF = −∞ is clear. x!1

The observations made earlier about sums and products of limits are of course valid for one-sided limits and also for lim f ðxÞ and lim f ðxÞ. x!1

x!1

The following important formulation of the limit of a function is phrased in terms of limits of sequences. Proposition 1.3.13 Let f : F ! R, where F is a subset of some metric space and let a be a limit point of F. Then lim f ðxÞ ¼ l 2 R if and only if, for every sequence x!a

{xn}n  1 in F that converges to a, and which satisfies xn 6¼ a for every n, the sequence {f(xn)}n  1 converges/diverges to l.

1.3 Metric Spaces

15

There is an analogous formulation of one-sided limits of functions in terms of limits of sequences. Suppose the domain F is an interval and let a be a limit point of F but not be its right endpoint. Then f ða þ Þ ¼ l 2 R if and only if, for every sequence {xn}n  1 in F that converges to a and satisfies xn > a for every n, the sequence {f(xn)}n  1 converges/diverges to l. Analogously for f(a−), where a is not the left endpoint. Let f be a real-valued function whose domain of definition is a subset F of a metric space. We say that f is continuous at the point a 2 F if, given e > 0, there exists a d > 0 such that for all x 2 X with d(x, a) < d, we have |f(x) − f(a)| < e. The function is said to be continuous on F if it is continuous at every point of F. If we merely say that a function is continuous, we mean that it is continuous on its domain. It may be checked that f is continuous at a limit point a 2 F if and only if lim f ðxÞ ¼ f ðaÞ. The following criterion of continuity follows from the preceding x!a

paragraph and Proposition 1.3.13. Proposition 1.3.14 Let f be a real-valued function defined on a subset F of a metric space and a 2 F. Then f is continuous at a if and only if, for every sequence {xn}n  1 in F that converges to a, we have lim f(xn) = f(lim xn) = f(a). This result shows that continuous functions are precisely those which send every convergent sequence and its limit into a convergent sequence and its limit, in other words, ‘preserve’ limits of sequences. As an aside, we note that if a function ‘preserves’ convergence of sequences, it necessarily preserves their limits as well. Suppose once again that the domain F is an interval, and let a 2 F but not be its right endpoint. Then f is said to be right continuous at the point a if, given e > 0, there exists a d > 0 such that for all x 2 F with 0 < x − a < d, we have |f(x) − f(a)| < e. It is easy to verify that right continuity at a means the same thing as f(a+) = f(a). The formulation in terms of sequences is that f is right continuous at a if and only if, for every sequence {xn}n  1 in F that converges to a and satisfies xn > a for every n, the sequence {f(xn)}n  1 converges to f(a). Analogous observations apply to left continuity, where a 2 F is not the left endpoint. If a 2 F is neither the left endpoint nor the right endpoint, then f is continuous at a if and only if it is left as well as right continuous at a. On the other hand, if a 2 F is the left [resp. right] endpoint, then f is continuous at a if and only if it is right [resp. left] continuous at a. Let f be an extended real-valued function whose domain of definition is a metric space X and Y  X. We define sup(f, Y) and inf(f, Y) as follows: supðf ; YÞ ¼ supff ðyÞ : y 2 Y g

and

infðf ; YÞ ¼ inf ff ðyÞ : y 2 Y g:

A set consisting of only isolated points is said to be discrete; it is always closed. A set F is said to be dense in X, where (X, d) is a metric space, if F ¼ X. The set Q of rational numbers in R is dense in ðR; dÞ, where d is the usual metric. The set Qn is dense in Rn (regardless of which metric).

16

1 Preliminaries

An open cover (covering) of a subset A of a metric space is a collection C ¼ fGa : a 2 Kg of open sets in the metric space whose union contains A, that is, [ A Ga : a

If C0 is a subcollection of C such that the union of sets in C0 also contains A, then C0 is called a subcover (or subcovering) from C of A. If C0 consists of finitely many sets, then we say that C0 is a finite subcover (or finite subcovering). Definition 1.3.15 A nonempty subset K of a metric space is said to be compact if every open cover of K contains a finite subcover. Obviously any nonempty finite subset of a metric space is compact. A nonempty subset K is compact if and only if every sequence in it has a subsequence which converges to a limit belonging to K. This in turn is equivalent to every infinite subset of K having a limit point belonging to K. Nontrivial examples in Rn are given by the following result. Theorem 1.3.16 (Heine–Borel) A nonempty subset of Rn is compact if and only if it is closed and bounded. An intersection of a sequence of sets belonging to a class A of sets is called a countable intersection of sets of A. The intersection of a finite sequence of sets belonging to A can obviously be represented as the intersection of an infinite sequence of sets belonging to A and is therefore spoken of as a countable intersection. Similarly, one speaks of a countable union. Sometimes the sets in a countable union are required to be (pairwise) disjoint and it is often convenient to use the abbreviation countable disjoint union. If £ 2 A, as happens often, the union of a finite sequence of disjoint sets in A is also a countable disjoint union of sets in A. We shall also use the following results. Theorem 1.3.17 Let U be an open subset of R. Then U can be written as a countable disjoint union of open intervals. Theorem 1.3.18 An open set V contained in Rn can be written as a countable union of Cartesian products of open intervals. Definition 1.3.19 Let (X, dX) and (Y, dY) be metric spaces and A  X. A function f: A ! Y is said to be continuous at a 2 A if, for e > 0, there exists some d > 0 such that dY ðf ðxÞ; f ðaÞÞ\e whenever x 2 A and dX ðx; aÞ\d: If f is continuous at every point of A, then f is said to be continuous on A. The analogue of Proposition 1.3.14 is valid for a function from a subset of a metric space to a metric space.

1.3 Metric Spaces

17

Theorem 1.3.20 (Extreme Value Theorem) A continuous real-valued function on a compact set is bounded and has a maximum value as well as a minimum value. Suppose a function defined on a subset of Rn is continuous at a point of its domain in the sense of any one of the three metrics d1, d2, d∞ described above. In view of the inequality (1.1) between the three metrics, it follows that it is continuous at that point in the sense of the other two metrics. This allows us to speak of continuity without specifying which metric we have in mind. Theorem 1.3.21 Let (X, dX) and (Y, dY) be metric spaces and A  X. A function f: A ! Y is continuous on A if and only if for every open set V  Y, there exists an open set U  X such that f−1(V) = U \ A. Definition 1.3.22 Let (X, dX) and (Y, dY) be metric spaces and A  X. A function f: A ! Y is said to be uniformly continuous on A if, for e > 0, there exists some d > 0 such that dY ðf ðxÞ; f ðaÞÞ\e whenever x; a 2 A and dX ðx; aÞ\d: Theorem 1.3.23 Let (X, dX) and (Y, dY) be metric spaces and A  X be compact. If the function f: A ! Y is continuous on A, then f is uniformly continuous on A. Proposition 1.3.24 Let A be a nonempty subset of a set M with metric q. The distance of a point x 2 M from the subset A is defined as d(x, A) = inf{q(x, a): a 2 A}. The map / : M ! R defined by /(x) = d(x, A) is continuous. Proof Let x1, x2 2 M and p 2 A be arbitrary. Then qðx1 ; pÞ qðx1 ; x2 Þ þ qðx2 ; pÞ; this implies d(x1, A) q(x1, x2) + q(x2, p). Choosing p appropriately, we have q(x2, p) d(x2, A) + e, where e > 0 is arbitrary. Then d(x1, A) q(x1, x2) + d(x2, A) + e. Therefore d(x1, A) − d(x2, A) q(x1, x2), since e > 0 is arbitrary. Similarly, d(x2, A) − d(x1, A) q(x1, x2). Consequently, |d(x1, A) − d(x2, h A)| q(x1, x2), which, in fact, implies uniform continuity. Proposition 1.3.25 (needed in Example 5.5.2) Let (X, dX) and (Y, dY) be metric spaces. If f: X ! Y and g: X ! Y are both continuous and agree on a dense subset of X, then they agree everywhere on X. The next result is relevant only in the special metric space R. Known as the Bolzano Intermediate Value Theorem, it guarantees that a continuous function on an interval assumes (at least once) every value that lies between any two of its values. Theorem 1.3.26 (Intermediate Value Theorem) Let I be an interval and f : I ! R be a continuous mapping on I. If a, b 2 I and a 2 R satisfy f(a) < a < f(b) or f(a) > a > f(b), then there exists a point c 2 I between a and b such that f(c) = a.

18

1.4

1 Preliminaries

Sequences and Series of Extended Real Numbers

In Sect. 1.1, sequences with values in an arbitrary set were defined, and in Definition 1.3.4, convergence of sequences with values in a metric space was introduced. Sequences in R , often called extended real sequences, enjoy additional properties, which will be needed in the sequel. We record them below for convenience. It is important to distinguish between the extended real sequence {xn}n  1 and its range fxn : n 2 Ng, which is a subset of R . In accordance with Definition 1.3.4, a real number l 2 R is a limit of the sequence {xn}n  1 if for each e > 0, there is a positive integer n0 such that for all n  n0, we have |xn − l| < e. As in any metric space, a sequence has at most one limit. When {xn}n  1 does have a limit, we denote it by lim xn. In symbols, l = lim xn if 8 e [ 0; 9 n0 3 n  n0 ) jxn  lj\e: A real sequence that has a real limit is said to converge (or to be convergent). Recall that the concept of limit of a sequence was enlarged to include the case of extended real numbers. If lim xn and lim yn both exist and both are real, then so do lim(xn + yn) and lim(xnyn); moreover, limðxn þ yn Þ ¼ lim xn þ lim yn

and

limðxn yn Þ ¼ ðlim xn Þðlim yn Þ:

If a is any real number, then lim (axn) = a(lim xn). Suppose one or both limits are ±∞. Then the assertion about the sum holds provided that the sum lim xn + lim yn is defined, and the assertion regarding the product holds unless one limit is 0 and the other is ±∞. The assertion regarding axn holds even if a = ±∞ but with the proviso that lim xn 6¼ 0. The sequence xn ¼ ð1 þ 1nÞn has a limit denoted by e; this number is irrational and lies between 2 and 3. Also, lim n1/n = 1. A sequence {xn}n  1 of extended real numbers is said to be increasing if it satisfies the inequalities xn xn+1, n = 1, 2, …, and decreasing if it satisfies the inequalities xn  xn+1, n = 1, 2, …. We say that the sequence is monotone if it is either increasing or it is decreasing. A sequence {xn}n  1 of real numbers is said to be bounded if there exists a real number M > 0 such that |xn| M for all n 2 N. The following simple criterion for the convergence of a monotone sequence is very useful. Proposition 1.4.1 A monotone sequence of real numbers is convergent if and only if it is bounded. Bolzano–Weierstrass Theorem 1.4.2 A bounded sequence of real numbers has a convergent subsequence.

1.4 Sequences and Series of Extended Real Numbers

19

The convergence criterion described in Proposition 1.4.1 is restricted to monotone sequences. It is important to have a condition implying the convergence of a sequence of real numbers that is applicable to a larger class and preferably does not require knowledge of the value of the limit. The Cauchy Principle of Convergence stated in Theorem 1.3.6 provides such a condition. Definition 1.4.3 A sequence of intervals is said to be nested if each interval contains the next. Nested Interval Theorem 1.4.4 Given a nested sequence of closed intervals whose lengths tend to 0, there exists a unique real number that belongs to all of them. Definition 1.4.5 Let {xn}n  1 be a sequence in R . We define the limit superior of {xn}n  1 to be lim sup xn ¼ inf n  1 ðsupk  n xk Þ ¼ limn ðsupfxn ; xn þ 1 ; . . .gÞ and the limit inferior to be lim inf xn ¼ supn  1 ðinf k  n xk Þ ¼ limn ðinf fxn ; xn þ 1 ; . . .gÞ: It can be checked that the limit exists if and only if the limits superior and inferior both exist, in which case they agree with the limit. For the sequence {xn}n  1, where xn = (−1)n, n = 1, 2, …, lim sup xn = 1 and lim inf xn = −1. In particular, limnxn does not exist. Definition 1.4.6 Suppose {fn}n  1 is a sequence of extended real-valued functions defined on a nonempty set X. Then for x 2 X, we define ðsupn fn ÞðxÞ ¼ supn fn ðxÞ ðinf n fn ÞðxÞ ¼ inf n fn ðxÞ ðlim supn fn ÞðxÞ ¼ lim supn fn ðxÞ ðlim inf n fn ÞðxÞ ¼ lim inf n fn ðxÞ: For the sequence {fn}n  1, where fn(x) = xn, n = 1, 2, … and x 2 [0, 1), supn fn ¼ f1 and inf n fn ¼ 0. Let fak gk  1 be a sequence of extended real numbers, in which ∞ and −∞ do n P not both appear as terms. Let {sn}n  1, where sn ¼ ak , be the sequence of partial k¼1

sums associated with fak gk  1 . A sequence of partial sums is called a series. The limit lim sn, if it exists, is called the sum of the series and the series is said to converge or be convergent if the sum is a real number. If either the sum does not 1 P exist or is ±∞, the series is said to be divergent. The symbol ak , or its k¼1 P abbreviated form ak , denotes the series as well as the sum, if any. The context determines which of the two is intended. By the kth term of the series we mean ak. 1 P The series k p is convergent if and only if p > 1. k¼1

20

1 Preliminaries

Proposition 1.4.7 (Comparison Test for Series) If 0 ak bk for each k and the 1 1 P P series bk is convergent, then the series ak is also convergent and k¼1 1 P

1 P

k¼1

ak

k¼1 1 P

ak \

k¼1

bk . Suppose further that there exists a j such that aj < bj. Then

k¼1 1 P k¼1

A series

bk . 1 P

ak of real numbers is said to converge absolutely if the related series

k¼1

1 P

jak j converges.

k¼1

Proposition 1.4.8 If a series converges absolutely then it converges. Proposition 1.4.9 If the terms of an absolutely convergent series are rearranged in any manner whatsoever, the resulting series also converges and has the same sum.

1.5

Sequences and Series of Functions

As elsewhere in Mathematical Analysis, the distinction between pointwise and uniform convergence of a sequence or series of functions is crucial in Measure Theory. Here we summarise some of the basic facts concerning the two kinds of convergence. Definition 1.5.1 Let {fn}n  1 and f be extended real-valued functions defined on some set X. We say that {fn}n  1 converges pointwise to f if and only if limnfn(x) = f(x) for each x 2 X. If fn converges to f pointwise, we write (according to convenience) limn fn ¼ f pointwise or

fn ! f

or

fn ðxÞ ! f ðxÞ; x 2 X:

The sequence {fn}n  1, where fn(x) = xn, x 2 [0, 1], converges pointwise to the function f defined by f(x) = 0, 0 x < 1, f(1) = 1. Definition 1.5.2 The sequence {fn}n  1 of real-valued functions converges uniformly to a real-valued function f if and only if for each e > 0, there exists an integer n0(e) such that jf ðxÞ  fn ðxÞj\e for all n  n0 ðeÞ and all x 2 X: If fn converges to f uniformly, we write (according to convenience) unif

limn fn ¼ f ðunifÞ or fn ! f :

1.5 Sequences and Series of Functions

21

The sequence {fn}n  1, where fn(x) = xn, x 2 [0, a], and 0 < a < 1, converges uniformly to the identically zero function. It is immediate that uniform convergence implies pointwise convergence. The reverse implication is, however, not true. The sequence {fn}n  1, where fn(x) = xn, x 2 [0, 1], is not uniformly convergent. That it is pointwise convergent has been observed above. Proposition 1.5.3 Let {fn}n  1 be sequence of real-valued functions defined on a metric space X and x0 2 X. Suppose that fn ! f uniformly on X. If each fn is continuous at x0, then f is continuous at x0. Briefly put, a uniform limit of functions continuous at a point is continuous at that point. Remark It follows from Proposition 1.5.3 that the sequence {fn}n  1, where fn(x) = xn, x 2 [0, 1], does not converge uniformly to the function f defined by f(x) = 0, 0 x < 1, f(1) = 1, as the limit function is discontinuous at x = 1. The Cauchy criterion of uniform convergence of sequences of functions is as follows. Theorem 1.5.4 (Cauchy Criterion of Uniform Convergence) The sequence of realvalued functions {fn}n  1 defined on X converges uniformly on X if and only if, given e > 0, there exists an integer n0 such that, for all x 2 X and all n  n0, m  n0, we have |fn(x) − fm(x)| < e. The above necessary and sufficient condition for uniform convergence is called the uniform Cauchy property and a sequence satisfying it is said to be uniformly Cauchy. A finite sum n X a0 þ ðak cos kx þ bk sin kxÞ k¼1

is called a trigonometric polynomial. For a proof of the result below, the reader may consult Problem 6.3.P8 of [28]; it will be used in Proposition 3.4.4 and an independent proof based on Fourier series will later be given in Sect. 4.3. Theorem 1.5.5 (Weierstrass Approximation Theorem) Let f be a continuous function on [−p, p] satisfying f(−p) = f(p). Then for any e > 0, there exists a trigonometric polynomial n X PðxÞ ¼ a0 þ ðak cos kx þ bk sin kxÞ k¼1

such that jf ðxÞ  PðxÞj\e for all x 2 ½p; p : 1 P Recall that a series xk of real numbers converges to x 2 R if the sequence k¼1

{sn}n  1, where sn ¼

n P

k¼1

xk (the nth partial sum), converges to x. We write

22

1 Preliminaries

x ¼ lim sn ¼

1 P

xk and x is called the sum of the series. If

k¼1

every x 2 X, and if we define f ðxÞ ¼ of the series

1 P

1 P

fn ðxÞ, x 2 X, the function f is called the sum 1 P

fn . We say that the series

We say that the series

fn converges uniformly on X if the sequence {sn}n  1

n¼1

n P

fk ðxÞ, x 2 X, converges uniformly on X. As with

k¼1

series of numbers, we speak of the series sequence 1 P

n P

fn converges pointwise.

n¼1

1 P

of functions, where sn ðxÞ ¼

fn ðxÞ converges for

n¼1

n¼1

n¼1

1 P

1 P

fn converging absolutely when the

n¼1

jfk j of partial sums converges pointwise. It is easy to check that when

k¼1

jfn j converges uniformly, so does

n¼1

1 P

fn . For an example when

n¼1 1 P

uniformly as well as absolutely but

1 P

fn converges

n¼1

jfn j does not converge uniformly, see

n¼1

Problem 5.2.P9 on p. 160 of [28]. The following test for uniform convergence of series of functions is due to Weierstrass. Proposition 1.5.6 (Weierstrass M-test) Suppose {fn}n  1 is a sequence of functions 1 P defined on X and suppose that |fn(x)| Mn, x 2 X and n = 1, 2, …. Then fn converges uniformly on X if

1 P

n¼1

Mn converges.

n¼1

Proposition 1.5.7 (needed in Proposition 5.2.11) Suppose {fn}n  1 is a sequence of functions defined on an interval, converging uniformly to a limit function f and c is a point in the interval such that fn(c−) exists for every n. Then limnfn(c−) exists and equals f(c−). Similarly for right-hand limits. The gist of the proof is the same as that of Proposition 6.2.6 on p. 178 of [28]. Uniform convergence cannot be dropped. Take fn(x) = xn on [0, 1] and c = 1. Then fn(c−) = 1 for every n but f = lim fn satisfies f(c−) = 0.

1.6

Derivatives

Let S  R and x 2 S be a limit point of the set. A function f : S ! R is differentiable at x if

1.6 Derivatives

23

lim

h!0

f ðx þ hÞ  f ðxÞ h

exists, in which case, the limit is called the derivative of f at x and is denoted by d f 0 ðxÞ. It is often more convenient to write dx f ðxÞ for f 0 ðxÞ. The derivative function f 0 is the one that maps each point of differentiability into the derivative at that point and is called simply the derivative of f. If f 0 ðxÞ and g0 ðxÞ both exist; then so do ðf þ gÞ0 ðxÞ and ðfgÞ0 ; moreover; ðf þ gÞ0 ðxÞ ¼ f 0 ðxÞ þ g0 ðxÞ and ðfgÞ0 ðxÞ ¼ f 0 ðxÞgðxÞ þ f ðxÞg0 ðxÞ: If a is any real number, then ðaf Þ0 ðxÞ ¼ aðf 0 ðxÞÞ. If x is a limit point of the set on which g 6¼ 0 and also belongs to the set, then  0 f f 0 ðxÞgðxÞ  f ðxÞg0 ðxÞ ðxÞ ¼ : g gðxÞ2 We assume that the reader is aware of trigonometric functions, the exponential and natural logarithm functions, and also of their limit and differentiation properties, such as d sin x ¼ cos x; dx

d 1 tan1 x ¼ ; dx 1 þ x2

d 1 ln x ¼ dx x

and so forth:

The functions can be defined variously via limit processes and all their properties learned in calculus can be derived from there. The manner in which this is done will be of no consequence for the material in this book. Proposition 1.6.1 (Chain Rule) Suppose f : S ! R is differentiable at x 2 S and g maps a set containing f(S) into R. If g is differentiable at f(x) 2 f(S), then the composition g  f is differentiable at x and ðg  f Þ0 ðxÞ ¼ g0 ðf ðxÞÞ  f 0 ðxÞ: Let I denote an interval. A function f : I ! R is said to have a local maximum at c 2 I if there exists a d > 0 such that x 2 I, |x − c| < d ) f(x) f(c). Similarly for a local minimum. A function f : D ! R, where D is a subset of R, is said to be increasing if, for all x1, x2 2 D, x1 \x2 ) f ðx1 Þ f ðx2 Þ and strictly increasing if x1 \x2 ) f ðx1 Þ\f ðx2 Þ:

24

1 Preliminaries

Correspondingly for decreasing and strictly decreasing. A monotone function on an interval is one which is either increasing or decreasing. Proposition 1.6.2 If f : ½a; b ! R satisfies f 0 ðxÞ  0 for every x 2 [a, b], then f is increasing. Similarly, if f : ½a; b ! R satisfies f 0 ðxÞ 0 for every x 2 [a, b], then f is decreasing. Proposition 1.6.3 If f : ½a; b ! R has a local maximum or a local minimum at c 2 (a, b) and is differentiable at c, then f 0 ðcÞ ¼ 0. Proposition 1.6.4 Suppose f : ½a; b ! R satisfies f 0 ðcÞ ¼ 0, where c 2 (a, b). If f 00 ðcÞ\0 then f has a local maximum at c and if f 00 ðcÞ [ 0 then f has a local minimum at c. Mean Value Theorem 1.6.5 Suppose the continuous function f : ½a; b ! R is differentiable on (a, b). Then there exists some n 2 (a, b) such that f ðbÞ  f ðaÞ ¼ f 0 ðnÞðb  aÞ: Proposition 1.6.6 Suppose f : ½a; b ! R has derivative zero at every point of its domain. Then the function is a constant. Proposition 1.6.7 (needed in Proposition 5.2.11) Suppose {fn}n  1 is a pointwise convergent sequence of functions defined on an interval. If each fn is increasing, then the limit function is also increasing.

1.7

Riemann Integration

By a partition P of an interval [a, b] we mean a finite sequence of points xk in the interval such that P : a ¼ x0 \x1 \    \xn ¼ b: For a bounded function f : ½a; b ! R and any partition, the nonempty set {f(x): xk−1 x xk} is bounded above as well as below for each k. Consequently it has supremum Mk and an infimum mk. The upper and lower sums of f over the partition P are respectively n X k¼1

Mk ðxk  xk1 Þ and

n X

mk ðxk  xk1 Þ:

k¼1

Their respective infimum and supremum are called the upper and lower integrals Rb Rb respectively of f and are denoted by a f and f. a

1.7 Riemann Integration

25

Thus Z

(

b

f ¼ inf

n X

a

) Mk ðxk  xk1 Þ : all partitions P

k¼1

and Z

b

( f ¼ sup

n X

) mk ðxk  xk1 Þ : all partitions P :

k¼1

a

Rb Rb It turns out that a f  f for every bounded function f. If equality holds, then the a function f is said to be Riemann integrable, and the Riemann integral of f from a to b is the common value of the upper and lower integrals, denoted by Z

Z

b

f

b

or

a

f ðxÞ dx:

a

Sometimes it is convenient to speak of f being Riemann integrable on [a, b]. The Riemann integral exists, for instance, if f is continuous or monotone. In the present section, we shall abbreviate “Riemann integrable” to simply “integrable”, although in the context of Measure Theory later, the word will have a technically different meaning. If the restriction of f to ½a; b  ½a; b is integrable, we say that f is integrable on [a, b]. If f : ½a; b ! R and g : ½a; b ! R are both integrable on [a, b], then so are f + g, fg and af (a a real number); moreover, Z a

b

Z

b

ðf þ gÞ ¼

Z fþ

Z

b

g

a

and

a

a

b

Z ðaf Þ ¼ a

b

f:

a

Suppose f : ½a; b ! R is bounded and a < c < b. If f is integrable on [a, b], then it is integrable on [a, c] as well as [c, b], and the equality Z

Z

b

f ¼

a

c

Z fþ

b

f

a

c

holds. Conversely, if f is integrable on [a, c] as well as [c, b], then it is integrable on [a, b] and the foregoing equality holds. It holds without the proviso that a < c < b if we agree that Z a

b

Z

a

f ¼ b

f:

26

1 Preliminaries

If f : ½a; b ! R and g : ½a; b ! R are both integrable and f(x) g(x) for each Rb Rb x 2 [a, b], then a f a g. If f is integrable on [a, b], and ½a; b  ½a; b , then f is integrable on [a, b]. If also Rb Rb f  0 on [a, b] then a f a f . Proposition 1.7.1 Let f : ½a; b ! R be integrable. Then j f j : ½a; b ! R is also integrable and Z

b

a

Z f

b

j f j:

a

Theorem 1.7.2 (a) First Fundamental Theorem of Calculus Let f : ½a; b ! R be integrable and let Z

x

FðxÞ ¼

f;

a x b:

a

Then F is continuous on [a, b]. Moreover, if f is continuous at a point c 2 [a, b], then F is differentiable at c and F 0 ðcÞ ¼ f ðcÞ: (b) Second Fundamental Theorem of Calculus Let f be Riemann integrable on [a, b] and suppose that there exists a function F on the same interval such that F 0 ðxÞ ¼ f ðxÞ everywhere. Then Z

b

f ðxÞ dx ¼ FðbÞ  FðaÞ:

a

There are two versions of the Substitution Rule or Change of Variables Formula. Proposition 1.7.3 (Version 1) Suppose u : ½a; b ! [a; b] is a bijection having a continuous derivative that vanishes nowhere. If (f  uÞju0 j is integrable on [a, b] and f is integrable on the image uð½a; b Þ ¼ ½a; b], then Z

b a

Z f ¼

b

ðf  uÞju0 j:

a

The reason for the absolute value on the right-hand side is that if u0 \0 everywhere, then we have uðaÞ ¼ b and uðbÞ ¼ a. It is possible to deduce the integrability of (f  uÞju0 j from the remaining hypotheses. See [15, pp. 170–171]. Proposition 1.7.3 (Version 2) Let F : ½a; b ! R and u : ½a; b ! ½a; b both be differentiable. If F 0 and (F 0  uÞu0 are both Riemann integrable, then

1.7 Riemann Integration

27

Z

uðbÞ

F0 ¼

uðaÞ

Z

b

ðF 0  uÞu0 :

a

It is not presumed in either version of the above proposition that uðaÞ\uðbÞ. The next result is the Formula of Integration by Parts. Proposition 1.7.4 Let f and g be differentiable functions on [a, b] having integrable derivatives f 0 and g0 . Then the products fg0 and f 0 g are integrable, and Z

b

Z

0

fg ¼ f ðbÞgðbÞ  f ðaÞgðaÞ 

a

b

f 0 g:

a

Proposition 1.7.5 Suppose {fn}n  1 is a sequence of Riemann integrable functions on [a, b] with uniform limit f. Then f is Riemann integrable on [a, b] and Z lim a

b

Z fn ¼

b

f:

a

If a function f : ½a; 1Þ ! R is integrable on [a, b] whenever a < b, the symbol Rb R1 lim a f , even if the limit does not exist. If it does not, we say that a f means b!1 R1 R1 a f is divergent; otherwise convergent. In either case, a f is called the improper integral of f over [a, ∞). If itR is convergent, we speak of f being Rimproper Riemann 1 1 integrable over [a, ∞). If a j f j is convergent, we say that a f is absolutely R1 R1 convergent. If a j f j is divergent but a f is convergent, then we say that the latter R1 is conditionally convergent. If a f is absolutely convergent, then it is convergent. Corresponding remarks apply to a function f : ð1; b ! R and the improper Rb integral 1 f . As an illustration, Z

1 1

1 dx ¼ lim b!1 x R1 1

Z 1

b

1 dx ¼ lim ðln b  ln 1Þ ¼ 1 b!1 x

and the improper integral 1 x dx is therefore divergent. If a function f : ða; b ! R is integrable on [c, b] whenever a < c < b, but is Rb Rb unbounded on (a, b], the symbol a f means lim c f , even if the limit does not c!a þ Rb exist. If it does not, we say that a f is divergent; otherwise convergent. In either Rb case, a f is called the improper integral of f over [a, b]. If it is convergent, we speak of f being improper Riemann integrable over [a, b]. Corresponding remarks Rb apply to a function f : ½a; bÞ ! R and the improper integral a f . The meaning of absolute convergence in this context is clear.

28

1 Preliminaries

As an illustration, Z

1

x

12

Z dx ¼ lim

c!0 þ

0

and the improper integral

R1 0

1

x2 dx ¼ lim ð2  2c2 Þ ¼ 2 1

c

1

c!0 þ

x2 dx is therefore convergent. 1

Proposition 1.7.6 (Comparison Test for Improper Integrals) Suppose the functions f : ½a; 1Þ ! R and g : ½a; 1Þ ! R are integrable R 1 on [a, b] for every bR>1a. If 0 f(x) g(x) for every x 2 [a, ∞) and if a g is convergent, then a f is R1 R1 convergent; moreover, a f a g. Similar assertions apply to improper integrals over bounded intervals.

1.8

Decimals and the Cantor Set

Let b 2 N and n 2 N, where b > 1. Every integer j such that 0 j < bn−1 then has nP 1 a representation as j ¼ ak bnk1 , where a1, a2, …, an−1 are integers satisfying k¼1

0 ak b − 1. For n = 1, this is trivially true, because j = 0 in this case and the sum is empty, which is understood to have value 0. For n > 1, the assertion is essentially Theorem 1.11.7 of [28, p. 51]. In the context of such representations, the integer b is called the base and is tacitly taken to be greater than 1. In much of our discussion later, there will be particular emphasis on the bases b = 2 and b = 3. Consider an arbitrary x 2 [0, 1]. In what follows, we shall first exhibit an infinite sequence fak : k 2 Ng of integers such that x ¼ a1 b1 þ    þ ak bk þ    ¼

1 X

ak bk ;

k¼1

0 ak b  1: Divide the interval [0, 1] into b equal parts. Then x lies in some interval where a1 is an integer in {0, 1, …, b − 1}. If x ðx 6¼ 0; x 6¼ 1Þ is one of the points of subdivision, then two values of a1 are possible and either may be chosen. Regardless of the choice, we have the inequality

½ab1 ; a1 bþ 1 ,

a1 a1 þ 1 ; x b b where a1 2 {0, 1, …, b − 1}. The chosen subinterval is again divided into b equal subintervals and the process is continued indefinitely. In this way, we obtain a sequence of nested intervals

1.8 Decimals and the Cantor Set

29

a1 a2 an a1 a2 an 1 þ 2 þ  þ n x þ 2 þ  þ n þ n ; b b b b b b b

ð1:2Þ

where a1 ; a2 ; . . .; ak ; . . . is a sequence of integers such that 0 ak b − 1, k 2 N. By virtue of (1.2), we 1 P have x ¼ ak bk and we have thus exhibited the kind of sequence that was k¼1

promised. We call any representation of x by a series

1 P

ak bk a decimal representation of

k¼1

x in the base b and write x ¼ :a1 a2 . . .ak . . .. The matter of its uniqueness or otherwise will be taken up further below, culminating in Theorem 1.8.3. The 1 P expression :a1 a2 . . .ak . . . represents the series ak bk in an abbreviated form that k¼1

does not mention b explicitly. Here the integers a1, a2, …, ak, … are called the 1st digit, 2nd digit and so on, of the decimal representation. They may be spoken of as digits “of x” when a representation is understood to be given. We shall speak of the cases b = 2 and 3 as binary and ternary respectively. Conversely, given a sequence a1, a2, …, ak, … of integers such that 0 ak b − 1, k 2 N, we claim that there exists a unique x 2 [0, 1] satisfying 1 P (1.2) for all n and hence also satisfying x ¼ ak bk . To obtain such an x, first k¼1

construct intervals defined by (1.2). They have the following properties: (i) each succeeding interval is a subinterval of the preceding one (keeping in mind that an þ 1 1 1 bn þ 1 þ bn þ 1 bn ); (ii) the lengths of the intervals tend to 0 as n ! ∞. It follows from the Nested Interval Theorem 1.4.4 that there exists a unique x 2 [0, 1] as claimed. The reader will recall from previous knowledge that the number of sequences of b terms, with each term belonging to a given set consisting of b elements (such as {0, 1, …, b − 1}), is bn. Remark 1.8.1 The endpoints of the interval described in (1.2) have a useful alternative description that we proceed to derive. Given an, the bn−1 possibilities for the left-hand side in (1.2) can be described without explicit reference to a1, a2, …, an−1. Indeed, the left endpoint of the interval can be recast as follows (including n n1 n1 P P P when n = 1): b1n ð ak bnk Þ ¼ b1n ð ak bnk þ an Þ ¼ b1n ðb ak bnk1 þ an Þ k¼1

¼

1 bn

n1 P k¼1

ðbj þ an Þ, where j ¼

n1 P

k¼1

ak b

nk1

k¼1

is an integer satisfying 0 j ðb  1Þ

k¼1

bnk1 ¼ bn1  1. Conversely, if j is an integer satisfying 0 j bn−1 − 1,

30

1 Preliminaries

then it follows from the representation of j mentioned in the opening paragraph of nP 1 nP 1 this section that b1n ðbj þ an Þ ¼ b1n ðb ak bnk1 þ an Þ ¼ b1n ð ak bnk þ an Þ ¼ 1 bn

ð

n P

k¼1

ak b

nk

k¼1

Þ. Thus, for a given an, the possible left-hand sides in (1.2) can be

k¼1

alternatively described as precisely the bn−1 numbers of the form b1n ðbj þ an Þ with 0 j bn−1 − 1. This alternative description yields the consequence that, for a given an, a number lies in an interval of the form (1.2) if and only if it lies in some interval an bj þ an þ 1 ½bj þ , where 0 j bn−1 − 1. The case when an = 1 is of special bn bn ; 1 P interest: a number x 2 [0, 1] has a representation x ¼ ak bk with nth digit k¼1

an = 1 if and only if it lies in some interval ½bjbþn 1 ; bjbþn 2 with 0 j bn−1 − 1. For b = 3, we thus find that a number x 2 [0, 1] has a ternary representation 1 P x¼ ak 3k with nth digit an = 1 if and only if it lies in an interval of the form k¼1

½3j3þn 1 ; 3j3þn 2 , where 0 j 3n−1 − 1. Since the “if” part and “only if” part will be used separately, we record them separately for ease of reference when needed later. (A) If x 2 ½3j3þn 1 ; 3j3þn 2 for some j satisfying 0 j 3n−1 − 1, then x has a ternary representation with 1 as the nth digit. (B) If x 2 [0, 1] does not belong to an interval ½3j3þn 1 ; 3j3þn 2 for any j satisfying 0 j 3n−1 − 1, then x cannot have a ternary representation with 1 as the nth digit. The interval can also be described as ½33jn þ 31n ; 33jn þ 32n . The endpoints of any such interval will be seen to have more than one ternary representation, the details of which are discussed below in the third paragraph of Remark 1.8.6. Before proceeding, we remind the reader of the elementary equalities ðb  1Þ

n X

bnk ¼ bn  1

and

ðb  1Þ

1 X

bk ¼ bn :

k¼n þ 1

k¼1

The latter of these has the consequence that ðb  1Þ

m X

bk \bn for all m [ n:

k¼n þ 1

Consider x ¼

1 P k¼1

ak bk and y ¼

1 P k¼1

bk bk , where 0 ak b − 1 and

0 bk b − 1 for all k 2 N. Then ak − bk  −(b − 1) for all k 2 N and equality obtains if and only if ak = 0, bk = b − 1, a fact that will be used presently.

1.8 Decimals and the Cantor Set

31

Suppose the sequences fak g and fbk g are distinct, without prejudice to whether x and y are distinct or not. Then there must exist a smallest k such that ak 6¼ bk ; denote it by n. This means an 6¼ bn and k < n implies ak = bk (without ruling out the possibility that n = 1). We have xy¼

1 X

ðak  bk Þbk ¼

k¼1

¼ ðan  bn Þb

1 X

ðak  bk Þbk

k¼n n

1 X

þ

ð1:3Þ k

ðak  bk Þb :

k¼n þ 1

Using the fact that ak − bk  −(b − 1) for all k 2 N, we deduce from (1.3) that x  y  ðan  bn Þbn  ðb  1Þ

1 X

bk ¼ ðan  bn Þbn  bn

k¼n þ 1

ð1:4Þ

n

¼ b ððan  bn Þ  1Þ: This shows that an > bn implies x  y. Remembering that an 6¼ bn , the contrapositive is that x < y implies an < bn, or equivalently, x > y implies an > bn. Remark 1.8.2 In the event that there exists a positive integer m > n such that ak = bk for all k > m, i.e., the digits agree beyond a certain stage, we deduce from (1.3) the stronger inequality that x  y  ðan  bn Þbn  ðb  1Þ

m X

bk [ ðan  bn Þbn  bn

k¼n þ 1

¼ bn ððan  bn Þ  1Þ: Therefore an > bn implies x > y. Combining this with the conclusion of the paragraph preceding this remark, we infer that, if the digits agree beyond a certain stage, then the inequality x > y is equivalent to the condition that an > bn, where n denotes the smallest positive integer k for which ak 6¼ bk. It will be convenient to rephrase the condition without explicit mention of the symbols ak, bk and n as the earliest digit of x that differs from the corresponding digit of y is the greater one.

The equivalence of the condition with the inequality x > y when digits agree beyond a certain stage will be invoked in Remark 1.8.14(e) below. To consider the matter of uniqueness of a decimal representation, we specialise to x = y in the foregoing discussion up to and including (1.4). Since an 6¼ bn , we may assume without loss of generality that an > bn. Then it follows from (1.4) that 0  bn ððan  bn Þ  1Þ  0: Consequently, b−n((an − bn) − 1) = 0 and hence an − bn = 1. Using this in (1.3), we find that

32

1 Preliminaries

0 ¼ bn þ

1 X

ðak  bk Þbk :

k¼n þ 1

We can deduce from here that k > n implies ak = 0, bk = b − 1. Indeed, if this were not so, there would exist some k > n such that ak − bk > −(b − 1) and the pre1 P ceding equality would lead to 0 [ bn  ðb  1Þ bk ¼ bn  bn ¼ 0, a k¼n þ 1

contradiction. Since k > n implies ak = 0, we can write x¼

n X

ak bk ¼ bn

k¼1

n X

n X m ; where m ¼ ak bnk : bn k¼1

ak bnk ¼

k¼1

Here, 0\bn ðbn þ 1Þbn ¼ an bn an bnn

n X

ak bnk ¼ m ðb  1Þ

k¼1

¼ bn  1;

n X

bnk

k¼1

which is to say, 0 < m < bn. As b − 1  an = bn + 1 > 0, the integer an is not divisible by b, and hence the integer m¼

n X

ak bnk ¼ a1 bn1 þ a2 bn2 þ    þ an

k¼1

is also not divisible by b. Let us summarise what has been proved about uniqueness in the preceding three paragraphs. Theorem 1.8.3 Suppose x 2 [0, 1] has two distinct representations x¼

1 X k¼1

ak bk ¼

1 X

bk bk ;

k¼1

where 0 ak b − 1, 0 bk b − 1, k 2 N, and n is the smallest value of k such that ak 6¼ bk . If an > bn, then (a) an − bn = 1, (b) k > n implies ak = 0, bk = b − 1, (c) x ¼ bmn , where 0 < m < bn (in particular, x 6¼ 0 and x 6¼ 1) and m is not divisible by b. It follows from (b) that there cannot be a third representation distinct from the two described above. So, there can be at most two distinct representations.

1.8 Decimals and the Cantor Set

33

Although (c) tells us that numbers in [0, 1] having at least two distinct representations must be of a certain form, it does not tell us that numbers of that form must conversely have at least two distinct representations. We prove the converse by proceeding as below. Let x ¼ bmn , where 0 < m < bn and m is not divisible by b. By Theorem 1.11.7 of [28, p. 51], there exists a finite sequence c1, …, cn of integers such that n P 0 ck b − 1 and m ¼ ck bnk . Therefore k¼1 n m X ¼ c bk : bn k¼1 k

This can be rewritten as

m bn

¼

1 P

ak bk , where ak = ck for k n and ak = 0 for

k¼1

k > n. In other words, m ¼ :c1 c2 . . .cn 00. . .: bn n P

Since m is not divisible by b, the representation m ¼

ck bnk shows that cn 6¼ 0

k¼1

and therefore cn  1. But an = cn and therefore an  1. If we define the sequence fbk : k 2 Ng by setting bk ¼ ak ¼ ck

for

k\n;

bn ¼ an  1 ¼ cn  1;

bk ¼ b  1

for

k [ n;

obviously the smallest value of k such that ak 6¼ bk is n, an > bn, and 0 bk b − 1 for k 6¼ n. Since an  1, the inequality 0 bk b − 1 holds for k = n as well and hence for all k. Furthermore, 1 X

bk bk ¼

k¼1

n1 X

¼

bk bk

k¼n þ 1

k¼1 n1 X

1 X

bk bk þ bn bn þ

ak bk þ ðan  1Þbn þ ðb  1Þ

k¼n þ 1

k¼1

¼

n X

ak bk  bn þ ðb  1Þ

¼

1 X k¼n þ 1

k¼1 n X

ak bk  bn þ bn

k¼1

¼

n X k¼1

1 X

ak bk ¼

n X k¼1

ck bk ¼

m : bn

bk

bk

34

1 Preliminaries

In other words, m ¼ :b1 b2 . . .bk . . . ¼ :c1 c2 . . .cn1 ðcn  1Þðb  1Þðb  1Þ. . .: bn This completes the converse proof that numbers of the form mentioned in (c) have 1 P ak bk , where a1, a2, … are at least two distinct representations of the form k¼1

integers satisfying 0 ak b − 1, k 2 N. Taking into account Theorem 1.8.3, we have actually proved something more, which we now formulate: Corollary 1.8.4 Any x 2 [0, 1] either has unique representation as

1 P

ak bk , i.e.,

k¼1

as .a1a2…ak…, or has precisely two such representations; in the latter case, there exists an integer n  1 such that one of the representations is :c1 c2 . . .cn 00. . . with cn  1 and the other one is :c1 c2 . . .cn1 ðcn  1Þðb  1Þðb  1Þ. . .: (It is of course understood that when n = 1, there are no c1, c2, …, cn−1.) For instance, 1 ¼ :0111    ¼ :1000    2 1 ¼ :0222    ¼ :1000    3 2 ¼ :1222    ¼ :2000    3

ðbase 2Þ; ðbase 3Þ; ðbase 3Þ:

If m is divisible by b, then the number x ¼ bmn still has two distinct representations 1 1 P P x¼ ak bk ¼ bk bk but the smallest k such that ak 6¼ bk is less than n. k¼1

k¼1

The hypothesis of the next corollary may seem a bit contrived, but we shall encounter such a situation later. Corollary 1.8.5 Let x 2 [0, 1] and b > 2. Suppose that, for every n 2 N, x has a representation :a1 a2 . . .ak . . . in which the nth digit satisfies an 6¼ 1. Then it has a representation :a1 a2 . . .ak . . . in which, for every n 2 N, the nth digit satisfies an 6¼ 1, which is to say, none of the digits is 1. Proof If x has a unique representation as :a1 a2 . . .ak . . ., there is nothing to prove. So, consider the case when the representation x ¼ :a1 a2 . . .ak . . . is not unique. By Corollary 1.8.4, there exists an integer k  1 such that x has precisely two representations

1.8 Decimals and the Cantor Set

x ¼ :c1 c2 . . .ck 00. . .

and

35

x ¼ :c1 c2 . . .ck1 ðck  1Þðb  1Þðb  1Þ. . .;

where 0 cj b − 1 for 1 j k and ck  1. We have to show that in at least one of these representations, none of the digits is 1. Assume that one of the digits in the first representation is 1. We need only argue that none of the digits in the second representation is 1. Let the rth digit in the first representation be 1. Then 1 r k and cr = 1. If r < k, then the rth digit in the second representation is also 1, whereby the hypothesis is violated for n = r. So we must have r = k and cj 6¼ 1 for 1 j k − 1. What this implies regarding the second representation is that none of the first k − 1 digits is 1, the kth digit is 0 and the remaining ones are b − 1. As b − 1 > 1, none of the digits in the second representation can be 1. h Remark 1.8.6 Let n 2 N be given. When the numerator m ¼

n P

ck bnk of the

k¼1

number x ¼ bmn exceeds a multiple of b by 1, we have cn = 1 and the two representations are x ¼ :c1 c2 . . .cn1 100. . . and x ¼ :c1 c2 . . .cn1 0ðb1Þðb1Þ. . .. In the rest of this remark, we consider ternary representations, i.e., b = 3. Let j be an integer such that 0 j 3n−1 − 1. The integer 3j + 1 satisfies 3j + 1 < 3n and exceeds a multiple of 3 by 1, and therefore we have 3j þ 1 It follows that 3j3þn 2 ¼ 3n ¼ :c1 c2 . . .cn1 1000. . . ¼ :c1 c2 . . .cn1 0222. . .. :c1 c2 . . .cn1 2000. . . ¼ :c1 c2 . . .cn1 1222. . .. This shows that the endpoints of the interval ð3j3þn 1 ; 3j3þn 2Þ have ternary representations without 1 as the nth digit. We shall argue that interior points of the interval do not have this property. We have recorded in (A) of Remark 1.8.1 that any x in the interval has a ternary representation with 1 as the nth digit. If it also has a representation without 1 as the nth digit, then the smallest k for which the two representations differ is some integer q n. It follows by (c) of Theorem 1.8.3 that the number must be of the form 3pq , where p is some integer. If it is in the interior of the interval in question, we must have 3j + 1 < 3n−qp < 3j + 2, which means 3n−qp is not an integer even though q n. This contradiction shows that the number in the interior can have no ternary representation without 1 as the nth digit. What we have seen about the intervals ð

3j þ 1 3j þ 2 ; n Þ; 3n 3

0 j 3n1  1

so far is that (A) Any number in the interior of such an interval must have 1 as the nth digit in each of its ternary representations. (B) Any endpoint of such an interval has at least one ternary representation without 1 as the nth digit. (C) A number that is neither of the above two kinds cannot have a ternary representation with 1 as the nth digit (see (B) of Remark 1.8.1); a fortiori, such a number has at least one ternary representation without 1 as the nth digit.

36

1 Preliminaries

In other words, the points of [0, 1] lying in the complement of the union of all these intervals, 0 j 3n−1 − 1, are precisely those that have at least one ternary representation without 1 as the nth digit. What we have said in this paragraph up to this point applies to any given n 2 N. We finally consider the consequence for all n 2 N: The intersection of the complements in [0, 1] of all these intervals, n = 1, 2, …, consists of all those numbers in [0, 1] which, for any given n, have at least one ternary representation without 1 as the nth digit. By Corollary 1.8.5 and its converse (which is trivial), these are precisely all those numbers in [0, 1] which have a ternary representation in which none of the digits is 1. What is known as the Cantor set will be obtained by removing open intervals from the closed interval C0 = [0, 1] in stages, finitely many  at a time. At the first stage, we remove the open middle third I1;1 ¼ 13 ; 23 of C0 to obtain

  C1 ¼ 0; 13 [ 23 ; 1 , a union of 2 disjoint closed intervals of length 13 each. At the     second stage, we remove the open middle thirds I2;1 ¼ 19 ; 29 , I2;2 ¼ 79 ; 89 of the



 closed intervals 0; 13 and 23 ; 1 to obtain the set





1 2 1 2 7 8 C2 ¼ 0; [ ; [ ; [ ; 1 : 9 9 3 3 9 9 of 22 = 4 disjoint closed intervals, each of the form

Observe  that C2 consists i iþ1 1 32 ; 32 , of length 32 each. At the third stage, we remove the open middle thirds of each of the closed intervals in C2 to obtain 23 = 8 disjoint closed intervals of length 1 33 each. We continue in this way. In general, if Cn has been constructed and is a union of 2n disjoint closed intervals of length 31n each, then we obtain Cn+1 by removing the open middle thirds of each of the closed intervals in Cn. Note that Cn+1 is a union of 2n+1 disjoint closed intervals of length 3n1þ 1 each. We shall denote the 2n disjoint closed intervals, of which Cn is the union, by Jn,k and call them the component intervals of Cn. Here, k ranges from 1 to 2n and the intervals Jn,k will be numbered from left to right. The middle third of any Jn,k is flanked on either side by component intervals of Cn+1 and all three have length 3n1þ 1 . The Cantor set C is what remains after the process of removing the open middle third of each component interval of Cn has been carried out for every n 2 N. Definition 1.8.7 The Cantor set C is the intersection of the sets Cn, n 2 N, obtained by removing the middle thirds, starting with C0 = [0, 1]: C¼

1 \

Cn :

n¼1

For each n, the Cantor set C is a subset of Cn, which is a union of intervals of  n total length 23 .   The interval I1;1 ¼ 13 ; 23 , which is removed from C0 = [0, 1] at the first stage to obtain C1, will be called the removed (or complementary) interval “of C1”, notwithstanding the fact that it is disjoint from C1. It is of the same length as the 2

1.8 Decimals and the Cantor Set

37

component intervals of C1, which is 13. In general, the 2n−1 intervals that are removed from Cn–1 at the nth stage in obtaining Cn will be called the removed (or complementary) intervals “of Cn” notwithstanding the fact that they are disjoint from Cn. They are of the same length as the component intervals of Cn, which is 31n . They will be numbered from left to right as In,k, where k ranges from 1 to 2n−1. If we by Enthe union of the open intervals that are removed at the nth  denote S2n1 stage En ¼ k¼1 In;k , then C ¼ ½0; 1 n

1 [ n¼1

En ¼ ½0; 1 n

n1 1 2[ [

! In;k :

n¼1 k¼1

The first removed interval In,1 has only one component interval of Cn to its left, namely, ½0; 31n . Between any pair of consecutive removed intervals of Cn, there are two component intervals of Cn. Thus, the kth removed interval In,k has 2k − 1 component intervals of Cn to its left. Remark 1.8.8 One can show by induction that In;k ¼ ð3j3þn 1 ; 3j3þn 2Þ for some j, 0 j 3n−1 − 1. This is obvious for n = 1, because in this case, j can only be 0   and ð3j3þn 1 ; 3j3þn 2Þ works out to be 13 ; 23 , which is indeed the interval removed at the first stage. Assume for some n that every interval removed at the nth stage is of the form ð3j3þn 1 ; 3j3þn 2Þ for some j, 0 j 3n−1 − 1. Then the intervals that remain at the nth stage are ½3j3þn 0 ; 3j3þn 1 and ½3j3þn 2 ; 3j3þn 3 . It is these intervals whose middle thirds are removed at the (n + 1)th stage. However, the middle thirds of these þ 1 3ð3jÞ þ 2 3ð3j þ 2Þ þ 1 3ð3j þ 2Þ þ 2 intervals are ð3ð3jÞ ; 3n þ 1 Þ, both of which have the 3n þ 1 ; 3n þ 1 Þ and ð 3n þ 1 n required form because 0 3j 3 − 3 < 3n − 1 and 0 < 3j + 2 3n − 3 + 2 = 3n − 1. This completes the induction argument that In;k ¼ ð3j3þn 1 ; 3j3þn 2Þ for some j, 0 j 3n−1 − 1. In the light of Remark 1.8.6, it now follows that the right endpoint of the removed interval In,k has a ternary representation of the form :c1 c2 . . .cn1 2000. . . and the other ternary representation contains the digit 1. (We shall see in Remark 1.8.14(d) later that none of the digits in :c1 c2 . . .cn1 2000. . . can be 1.) nP 1 aj 2 Another way to express this representation is 3 j þ 3n , with each aj even. j¼1

Remark 1.8.6 also shows that the left endpoint of In,k has the representation nP 1 1 P aj 2 3 j. 3j þ j¼1

j¼n þ 1

Having shown that an interval removed at the nth stage is indeed of the form 3j þ 1 3j þ 2 ð 3n ; 3n Þ, we can now assert that the intervals that remain at the nth stage are of the form ½3j3þn 0 ; 3j3þn 1 and ½3j3þn 2 ; 3j3þn 3 . It may be noted that not every interval ð3j3þn 1 ; 3j3þn 2Þ with 0 j 3n−1 − 1 is removed at the nth stage; in fact, these intervals are 3n−1 in number while only 2n−1

38

1 Preliminaries

intervals are removed at the nth stage. However, it is important to take into account that all the 3n−1 intervals of this form, whether removed at the nth stage or not, are disjoint from Cn, as we shall now demonstrate. In the light of what was recorded in the preceding paragraph, it is sufficient to show that an interval of the form ð3j3þn 1 ; 3j3þn 2Þ is disjoint from all intervals of the form either ½3j3þn 0 ; 3j3þn 1 or ½3j3þn 2 ; 3j3þn 3 , even with a different j. To show this, it turns out that we need not restrict ourselves only to 0 j 3n−1 − 1. Suppose 3p þ 0 3p þ 1 3j þ 1 3j þ 2 ð 3n ; 3n Þ \ ½ 3n ; 3n 6¼ £, where j; p 2 Z. Then there exists some x such that 3j þ 1\x\3j þ 2 as well as 3p þ 0 x 3p þ 1: If either 3j + 2 3p + 0 or 3p + 1 3j + 1, we immediately obtain a contradiction. Therefore we have 3j + 2 > 3p + 0 as well as 3p + 1 > 3j + 1. The second of these inequalities implies p > j, which leads to p  j + 1 and hence to 3p  3j + 3, contradicting the first inequality. This contradiction establishes that no such x can exist, which implies that ð3j3þn 1 ; 3j3þn 2Þ \ ½3p3þn 0 ; 3p3þn 1 ¼ £. Next consider the possibility that ð3j3þn 1 ; 3j3þn 2Þ \ ½3p3þn 2 ; 3p3þn 3 6¼ £, where j; p 2 Z. Then there exists some x such that 3j þ 1\x\3j þ 2 as well as 3p þ 2 x 3p þ 3: If either 3j + 2 3p + 2 or 3p + 3 3j + 1, we immediately obtain a contradiction. Therefore we have 3j + 2 > 3p + 2 as well as 3p + 3 > 3j + 1. The first of these inequalities implies j > p, which leads to j  p + 1 and hence to 3j + 1  3p + 4, contradicting the second inequality. This contradiction establishes that no such x can exist, which implies that ð3j3þn 1 ; 3j3þn 2Þ \ ½3p3þn 2 ; 3p3þn 3 ¼ £. Thus all the 3n−1 intervals of the form ð3j3þn 1 ; 3j3þn 2Þ are disjoint from Cn. It follows from (B) and (C) of Remark 1.8.6 that a number in Cn has one ternary representation without 1 as its nth digit, and may or may not have another ternary representation. Since a number in the Cantor set is in Cn for every n, we conclude from Corollary 1.8.5 that it has a ternary representation with no digit equal to 1. We claim that the converse also holds: Suppose x 2 [0, 1] has a ternary representation with no digit equal to 1. By (A) of Remark 1.8.6, it cannot be in the interval ð3j3þn 1 ; 3j3þn 2Þ for any j and for any n. Since all the removed intervals are of this form, x cannot be in any of the removed intervals at any stage. This means it is in the Cantor set. We summarise the discussion as the following theorem. Theorem 1.8.9 A point x 2 [0, 1] is in the Cantor set if and only if it has a ternary representation which consists of the digits 0 and 2 only (i.e., every digit is even). Remark 1.8.10 Moreover, this representation of a point in C is unique, as is easily deduced from (a) of Theorem 1.8.3. However, we can give an independent proof as below.

1.8 Decimals and the Cantor Set

If

1 P an n¼1

3n

¼

1 P bn n¼1

3n ,

39

where each of an [resp. bn] is either 0 or 2, we shall conclude

that an = bn for every n. Suppose that there exists an n such that an 6¼ bn . Let m be the smallest integer such that am 6¼ bm . Then |am − bm| = 2 and |an − bn| 2 for every n, so that X 1 1 1 an X bn X an  bn 0¼  ¼ n¼m 3n n¼m 3n n¼m 3n a  b 1 X an  bn m m ¼ þ 3m 3n n¼m þ 1 1 jam  bm j X an  bn   n¼m þ 1 3n 3m 

1 X jam  bm j jan  bn j  m 3 3n n¼m þ 1



1 X 1 2  a  b j j m m n 3m 3 n¼m þ 1

¼

1 ; 3m

a contradiction:

Hence an = bn for every n. Terminology If in a decimal representation in any arbitrary base b, every digit from a certain point onward is 0, we shall call it a representation with ‘0 recurring’. The term ‘b − 1 recurring’ has the corresponding meaning. In the rest of this section, ‘two representations’ will be understood to mean at least two distinct representations. Remark 1.8.11 According to Theorem 1.8.3 and the discussion thereafter, the following are equivalent for any number x 2 (0, 1), the base b being arbitrary: (1) (2) (3) (4) (5)

x x x x x

has two decimal representations; has a decimal representation with 0 recurring; has a decimal representation with b − 1 recurring; has no decimal representation with neither 0 recurring nor b − 1 recurring; is of the form bmn , where 0 < m < bn.

Proposition 1.8.12 card(C) ¼ 2N ¼ c. Proof Define g : C ! R as follows: By Theorem 1.8.9 and Remark 1.8.10, any x 2 C has a unique representation with every digit even:

40

1 Preliminaries

x¼

1 X ak k¼1

Put gðxÞ ¼

1 P bk k¼1

2k ,

3k

;

each ak being either 0 or 2:

where bk ¼ 12 ak .

Observe that g(0) = 0, g(1) = 1. Clearly, the range of g is contained in [0, 1]. In fact, the range of g can be shown to be [0, 1]. Consider y 2 [0, 1] with binary representation y ¼ :d1 d2 . . ., where dn 2 {0, 1} for all n 2 N. Let x ¼ :ð2d1 Þð2d2 Þ. . . in base 3. Then by Theorem 1.8.9, have x 2 C and the definition of g yields 1 2 we 1 g(x) = y. [Thus, e.g., g 3 ¼ g 3 ¼ 2, illustrating that g is not one-to-one.] Suppose g(x) = 0. Since 0 has a unique representation in any base (see Theorem 1.8.3), it follows that bk = 0 for each k. Hence ak = 0 for each 0 and x = 0. Consequently, x > 0 implies g(x) > 0. Since 1 also has unique representation in any base (see Theorem 1.8.3), a similar argument shows that x < 1 implies g(x) < 1. Thus, x 2 (0, 1) if and only if g(x) 2 (0, 1). Let A  C consist of numbers having two ternary representations and B  ½0; 1 consist of numbers having two binary representations. Since 0 and 1 have unique representations in any base, we have A; B  ð0; 1Þ. We shall prove that x 2 A if and only if g(x) 2 B. First suppose x 2 A. Then x 2 (0, 1) and hence g(x) 2 (0, 1). By Remark 1.8.11, any ternary representation of x has either 0 recurring or 2 recurring. It follows from the definition of g that one binary representation of g(x) has either 0 recurring or 1 recurring. Since g(x) 2 (0, 1), we conclude on the basis of Remark 1.8.11 that g(x) has two binary representations, i.e., g(x) 2 B. For the converse, suppose g(x) 2 B. Then g(x) 2 (0, 1) and hence x 2 (0, 1). By Remark 1.8.11, any binary representation of g(x) has either 0 recurring or 1 recurring. It follows from the definition of g that one ternary representation of x has either 0 recurring or 2 recurring. Since x 2 (0, 1), we conclude on the basis of Remark 1.8.11 that x has two ternary representations, i.e., x 2 A. This proves the converse and hence completes the proof that x 2 A if and only if g(x) 2 B. The fact established earlier that g maps C onto [0, 1], when combined with the fact that x 2 A only if g(x) 2 B, implies that g maps C\A onto [0, 1]\B. We shall now argue that it does so in a one-to-one manner. Let x; x0 2 C\A and x 6¼ x0 . Since it has been proved that x 2 A if g(x) 2 B, we 1 1 0 P P ak ak 0 know that g(x), gðx0 Þ 62 B. We write x ¼ 3k , and x ¼ 3k (each ak as well as k¼1

k¼1

each a0k being even). Then there exists some k such that ak 6¼ a0k and hence 1 1 0 0 2 ak 6¼ 2 ak . Therefore the binary representations by means of which g(x) and gðx Þ are defined are distinct from each other. However, since g(x), gðx0 Þ 62 B, neither can have two binary representations. This implies gðxÞ 6¼ gðx0 Þ. Thus g has been shown to be one-to-one on C\A.

1.8 Decimals and the Cantor Set

41

Consequently, the mapping g sets up a one-to-one correspondence between C\A and [0, 1]\B. By Remark 1.8.11, the set A consists of all numbers of the form m n m n 3n ; 0\m\3 and the set B consist of all numbers of the form 2n ; 0\m\2 . Therefore both A and B are countable. By making the countable subsets A and B correspond to each other, we obtain a one-to-one correspondence between the Cantor set and [0, 1]. Consequently, cardðCÞ ¼ 2N ¼ c. h Remark 1.8.13 We have noted in paragraph 3 of the above proof that x > 0 implies g(x) > 0. Combining this with the consequence of Theorem 1.8.9 that C contains arbitrarily small positive numbers, we infer that there exist arbitrarily small numbers at which g takes positive values. Another property of the function g that will be useful in Sect. 5.5 is that the value of 2ng at the right endpoint of any removed interval equals the number of component intervals to the left of that removed interval. To arrive at this conclusion, we need to avail of some facts about removed and component intervals that were noted earlier and also some fresh ones. We present the deduction in a sequence of remarks. Remarks 1.8.14 (a) The set Cn has 2n component intervals, each being closed and having length 3−n. There are half as many removed intervals In,k, each being open with length 3−n and flanked on either side by a component interval. These 2n−1 open intervals are the ones that were removed from Cn−1 to obtain Cn, and they are disjoint from Cn. (b) The kth removed interval In,k (numbered from left to right) of Cn has 2k − 1 component intervals of Cn to its left. Therefore, if we arrange the removed intervals from left to right and count the number of component intervals to the left of the respective intervals, we get the arithmetic progression 1, 3, …, 2n − 1. (c) The endpoints of a removed interval of Cn do not belong to the removed interval but do belong to the component intervals of Cn as well as to the component intervals of Cm with m > n, and hence to C. (d) As recorded in Remark 1.8.8, the right endpoint of a removed interval has a ternary representation of the form :c1 c2 . . .cn1 2000. . . and its other ternary representation contains the digit 1. Since it belongs to C (as observed in (c)), it must have a ternary representation with every digit even (Theorem 1.8.9), which must then be the one having the form :c1 c2 . . .cn1 2000. . .. Consequently, every digit in this representation of the right endpoint of a removed interval must be even. (e) Another consequence is that for right endpoints x and y of any two removed intervals, the ternary representations with every digit even have the property that the digits agree beyond a certain stage. Suppose x > y. Then by Remark 1.8.2, the earliest digit of x that differs from the corresponding digit of y is the greater one. It is transparent that the same must also be true of the binary representations of g(x) and g(y) in the definition of g : C ! R in

42

1 Preliminaries

Proposition 1.8.12. Moreover, they also have the property that the digits agree beyond a certain stage. It follows by Remark 1.8.2 again that g(x) > g(y). Thus we have shown that when x and y are right endpoints of removed intervals, x > y implies g(x) > g(y). (f) Consider again the 2n−1 removed intervals In,k of Cn arranged from left to right as in (b); their right endpoints are in strictly increasing order and have the form noted in (d). Therefore, upon applying g to them, we get numbers in strictly increasing order in view of (e) and having binary representations : c1  c2 . . . cn1 1000. . . with every  cp either 0 or 1. There are precisely 2n−1 such binary representations. If we multiply each one by 2n, we get 2n−1 odd positive integers in strictly increasing order, none exceeding 2n. But there are only 2n−1 odd positive integers not exceeding 2n. It follows that we get all of them and hence the odd positive integers in strictly increasing order that we get are 1, 3, …, 2n − 1. This is the same as the arithmetic progression in (b) above. Thus, upon multiplying the value of g at the right endpoint of a removed interval of Cn by 2n, we obtain the number of component intervals of Cn to the left of that removed interval.

Chapter 2

Measure in Euclidean Space

2.1

Introduction

The theory of integration has its roots in the method of “exhaustion” which was developed by Archimedes for the purpose of calculating the areas and volumes of certain geometric figures. The work of Newton and Leibniz made this method into a systematic tool for such calculations but it was Riemann who gave a formal definition of the integral. The definition due to Riemann may be summarised as follows: The Riemann integral of a bounded real-valued function f defined on [a, b] is the limit (when it exists) of the Riemann sums n X j¼1

f ðtj Þðxj  xj1 Þ ¼

n X

f ðtj Þ‘ðIj Þ;

j¼1

  where Ij = [xj−1, xj] are contiguous intervals with union [a, b], ‘ Ij denotes the length xj − xj − 1 of Ij and tj belongs to Ij for 1  j  n. The integral exists for a large class of functions, which includes, among others, continuous functions and monotone functions. The reader will discern that this definition and the one given in Sect. 1.7 seem to be at variance. However, they are equivalent (see Darboux’s Theorem 10.1.7, [28]). It is well known that if a sequence fn of integrable functions converges uniformly to a limit function f, then the latter has an integral, which is equal to the limit of the integrals of fn. In other words, the Riemann integral “behaves well” with respect to uniform convergence. Unfortunately, it does not behave well with respect to what is called “monotone convergence”, because a sequence fn of integrable functions such that fn  fn+1 can converge pointwise to a limit function f that is not even integrable. We give an illustration.

© Springer Nature Switzerland AG 2019 S. Shirali and H. L. Vasudeva, Measure and Integration, Springer Undergraduate Mathematics Series, https://doi.org/10.1007/978-3-030-18747-7_2

43

44

2 Measure in Euclidean Space

Example 2.1.1 The reader is undoubtedly familiar with the standard example of the function f : ½0; 1 ! R defined by  f ðxÞ ¼

1 if x is rational 0 if x is irrational;

which is not integrable in the Riemann theory. Now take any enumeration of the rationals in [0, 1] and let En denote the set of the first n rationals in the enumeration. Then En  En+1 and hence the characteristic functions fn of En satisfy 0  fn  fn+1 and clearly converge pointwise to f. Each fn differs from the identically zero function at only a finite number of points and therefore has Riemann integral 0, but the limit function f is not Riemann integrable. (The fact that each set En+1\En contains only one point, and is therefore an interval of length 0, will be needed shortly.) Thus problems which involve integration with a limiting process are often awkward with this integral. However, if we extend the concept of the length of an interval to a notion of size for more general sets, technically called the “measure” of the set, then the integral of the function f in Example 2.1.1 may be taken as representing the area of a figure of height 1 and base A = {x 2 [0, 1]: f(x) = 1} and should therefore be just the measure of the set A. On the other hand, since each fn has integral 0, the limit of the sequence of integrals is also 0, and hence we would like the integral of f to be 0 as well. Thus the measure of A should be taken as 0. Note that A is the countable disjoint union of singleton sets En+1\En, which are intervals of length 0. This observation already provides an inkling of what is involved in extending the concept of length of an interval to that of a measure of more general sets. More precisely, for a set which is a countable disjoint union of intervals, the measure should be the sum of the lengths of those intervals. Unfortunately, there are two reasons why this is unsatisfactory as a definition. One is that the representation of the set as a union of disjoint intervals may not be unique; another is that we may need to include sets more general than such unions. Towards the end of the nineteenth century, many mathematicians considered it desirable to replace the Riemann integral by some other type of integral which is more general and better suited for dealing with limiting processes. Amongst the attempts made in this direction, it was Lebesgue’s contribution, dating back to the beginning of the last century, that turned out to be the most successful. His definition enables us to integrate some, but not all, functions for which the Riemann integral does not exist, in particular the one mentioned in Example 2.1.1. Although the enlargement is useful in itself, its main virtue is that theorems relating to the interchange of limit and integral are valid under less stringent assumptions than are required for the Riemann integral. In order to achieve the objective outlined above, Lebesgue approached the matter by partitioning an interval containing the range rather than partitioning the domain. For a controversy about whether it makes any difference which integration theory (Riemann or Lebesgue) is used in aircraft design, see the article [7].

2.1 Introduction

45

The length ‘ðIÞ of an interval I is defined, as usual, to be the difference of its endpoints. If I = (a, b) (a, b], [a, b) or [a, b], then ‘ðIÞ ¼ b  a. This also makes sense for the empty set ∅ = (a, a) and for unbounded intervals; recall that that ∞ − a = ∞ and so on (see Chap. 1, Sect. 1.1). Thus ‘ðIÞ is a nonnegative extended real number. It will be convenient to include the empty set as a special case of an interval. Length is an example of a “set function” in the following sense. Definition 2.1.2 A set function is a function whose domain is some class of subsets of a given set. In the case of length, the given set is R and the class of subsets that constitute the domain is the class of intervals; the range is the set of nonnegative extended reals. We remind the reader that some of the properties satisfied by ‘ are as follows: (i) ‘ðIÞ  0 for all intervals I; S (ii) if {Ij}j  1 is a sequence of disjoint intervals such that 1 j¼1 Ij is an interval, 1   S1 P then ‘ð j¼1 Ij Þ ¼ ‘ Ij ; j¼1

(iii) if x is any fixed real number and I is any interval (so that I + x = {y + x: y 2 I} is also an interval), then ‘ðI þ xÞ ¼ ‘ðIÞ. When we extend the concept of the length ‘ of an interval to a notion of size or measure m for more general subsets of R, we would like the analogues of the above properties to continue to hold for the extension. To recap, we now list some of the desired properties in terms of m. (i) m is an nonnegative extended real-valued set function; (ii) mðEÞ ¼ ‘ðEÞ if E is an interval; (iii) m is “countably additive”, meaning thereby that if {Ej}j  1 is a sequence of disjoint subsets in the domain of m and [ j  1 Ej is also in the domain of m, then mð

1 [ j¼1

(iv)

Ej Þ ¼

1 X

mðEj Þ;

j¼1

for E in the domain of m and x a fixed real number, m(E + x) = m(E), provided E + x is in the domain of m, where E + x = {y + x: y 2 E}; this is called “translation invariance” of m.

It can be shown that it is impossible to have such a measure m with domain consisting of all subsets of R (see [30]). Later we shall construct one which is defined on a much larger class of subsets of R than just intervals; in particular, the set A of Example 2.1.1 will satisfy m(A) = 0. It turns out that in order to obtain satisfactory theorems relating to the interchange of limit and integral, property (iii) of m needs to be strengthened to read as:

46

2 Measure in Euclidean Space

(iii) if {Ej}j  1 is a sequence of disjoint subsets in the domain of m, then first, [ j  1 Ej is also in the domain of m, and second, mð

1 [

Ej Þ ¼

j¼1

1 X

mðEj Þ:

j¼1

Lebesgue was the first to establish the existence of such a measure, but by a construction that differs from the one we shall present. Problem Set 2.1 2.1.P1. From properties (i) and (iii) of m, show that m(∅) = 0, provided that at least one set C in the domain of m satisfies m(C) < ∞. 2.1.P2. From properties (i) and (iii) of m, show that, if A \ B = ∅, where A, B, A [ B are in the domain of m, then m(A [ B) = m(A) + m(B). 2.1.P3. From properties (i) and (iii) of m, show that, if A\B, B\A, A, B, A [ B and A \ B are in the domain of m, then m(A [ B) + m(A \ B) = m(A) + m(B).

2.2

Lebesgue Outer Measure

As a first attempt at extending the concept of the length ‘ of an interval (the empty set being regarded as an open interval of length 0) to a notion of size or measure m for more general subsets of R, we introduce the following concept. Let A be any subset of R. Consider a sequence {In}n  1 of open intervals that covers A, that is, 1 P A  [ n  1 In . For each such sequence, the sum ‘ðIn Þ is independent of the order n¼1

of the terms, since ‘ðIn Þ  0 for each n.

Definition 2.2.1 For any subset A of R, we define the Lebesgue outer measure m (A) to be the infimum of all numbers 1 X

‘ðIn Þ;

n¼1

where {In}n  1 is a sequence of open intervals that covers A and the infimum is taken over all such sequences. For most of this chapter, we shall refer to Lebesgue outer measure as simply “outer measure”, as there will be no other outer measure under discussion, except in some problems, where the matter is clarified. Remarks 2.2.2 (a) Some or all of the In may be empty. It follows in particular that m (∅) = 0. (b) The outer measure m is a set function defined on all subsets of R and has values in the set of nonnegative extended real numbers.

2.2 Lebesgue Outer Measure

47

The following properties of the outer measure are almost immediate from the definition. The property stated in (c) is called monotonicity. Proposition 2.2.3 (a) (b) (c) (d)

0  m (A)  ∞ for all subsets of R; m (∅) = 0; If A  B  R, then m (A)  m (B); For x 2 R, we have m ({x}) = 0.

Proof (a) and (b) are obvious. If {In}n  1 is a sequence of open intervals that covers B, then it also covers A and hence m (A) is the infimum taken over a larger or as large a set of numbers as compared to m (B). So, m (A)  m (B). This proves (c). For the proof  of (d), note that, for every e > 0, the sequence {In}n  1 of intervals, where I1 ¼ x  2e ; x þ 2e and In = ∅ for n  2, covers {x}, while 1 P ‘ðIn Þ ¼ e. h n¼1

It is immediate from Proposition 2.2.3(a) and (c) that if B  A and m (A) = 0, then m (B) = 0. Suppose that A is any subset of R and x is any real number. The translate of A by x is the set A + x = {y + x: y 2 A}. The assertion of the next proposition is called translation invariance of the Lebesgue outer measure m . Proposition 2.2.4 For any A  R and x 2 R, we have m (A + x) = m (A). Proof Given any open interval I, bounded or unbounded, its translate I + x is also an open interval and ‘ðI þ xÞ ¼ ‘ðIÞ. Now let any e > 0 be given. Suppose m (A)S< ∞. Then there is a sequence {In}n  1 of open intervals such that A  n  1 In and satisfies 1 X

‘ðIn Þ\m ðAÞ þ e:

n¼1

Clearly, A + x  [ n  1(In + x). Therefore m ðA þ xÞ 

1 X n¼1

‘ðIn þ xÞ ¼

1 X

‘ðIn Þ\m ðAÞ þ e:

n¼1

Since e > 0 is arbitrary, we have m (A + x)  m (A). As A is the translate of A + x by −x, it follows (on replacing A by A + x and x by −x in the above inequality) that m (A)  m (A + x). Since the reverse inequality has already been proved, it follows that m (A + x) = m (A) when m (A) < ∞. The same argument with obvious modifications works when m (A) = ∞. h Example 2.2.5 Suppose that A has at most countably many points. Then m (A) = 0. In view of the monotonicity of m , we need consider only countable A. So, let

48

2 Measure in Euclidean Space

{xk}k  1 be an enumeration of the points of A. Given e > 0, let Ik be an open interval of length e/2k that contains the point xk. The sequence {Ik}k  1 then covers 1 1 P P A and satisfies ‘ðIk Þ ¼ e=2k ¼ e. So, m (A)  e. In particular k¼1

k¼1

m ðQÞ ¼ m ðNÞ ¼ m ðZÞ ¼ 0. It follows from what has just been noted that a set of positive outer measure must be uncountable. Proposition 2.2.6 If A is any interval, then its outer measure is the same as its length: m ðAÞ ¼ ‘ðAÞ. Proof The case when A is empty is trivial. So we shall consider only the case of nonempty A. First consider the case when A is aclosed bounded interval [a, b]. Given e > 0, we can choose I1 ¼ a  2e ; b þ 2e and Ik ¼ £ for k  2 to see that m (A)  b − a + e. This being true for each e > 0, we have m ðAÞ  b  a ¼ ‘ðAÞ. We shall show that m (A)  b − a. Let {Ik}k  1 be a sequence of open intervals that covers A. Then the family of intervals {Ik: k  1} is an open cover of A and, by the Heine–Borel Theorem 1.3.16, contains a finite subcover, which we shall name as F. We may assume that each interval in F has a nonempty intersection with A, so that the latter 1 P ‘ðIk Þ [ b  a if we is an interval with well-defined endpoints. It will follow that k¼1

show that the total length of the intervals in F is greater than or equal to b − a. To show the latter, we proceed as follows. By renumbering the intervals if necessary, we may assume that the finite subcover F consists of intervals Ik, k = 1, …, p. Let the intervals A \ Ik have endpoints c(k), d(k). Arrange all the numbers c(k) and d(k), k = 1, …, p, as a single increasing sequence a(0) < a(1) <  < a(m). Note that we must necessarily have a = a(0) and b = a(m) because the union of the intervals A \ Ik is precisely equal to A. Consequently, the single increasing sequence provides a partition T of the interval [a, b] and also gives rise to a partition of each interval A \ Ik. Let T and Tk denote respectively the systems of subintervals generated by the partitions. Since the length of an interval equals the total length of all the subintervals of any partition of it, we have on the one hand ‘ðAÞ ¼

X

‘ðJÞ

ð2:1Þ

J2T

and on the other hand ‘ðIk Þ  ‘ðA \ Ik Þ ¼

X J2Tk

‘ðJÞ;

2.2 Lebesgue Outer Measure

49

so that p X

‘ðIk Þ 

p X X

‘ðJÞ:

ð2:2Þ

k¼1 J2Tk

k¼1

Note that each interval belonging to T is a constituent of at least one collection Tk but could belong to several Tk. Therefore p X X

‘ðJÞ 

X

k¼1 J2Tk

‘ðJÞ:

J2T

Together with (2.1) and (2.2), this implies the required conclusion.  is a Next, consider the case when A is any bounded nonempty interval. Then A  ¼ ‘ðAÞ, but also, according closed bounded interval and not only do we have ‘ðAÞ  ¼ ‘ðAÞ,  so that to the case already proved, m ðAÞ m ðAÞ ¼ ‘ðAÞ: Now, given e > 0, there exists a closed bounded interval I such that I  A and ‘ðIÞ [ ‘ðAÞ  e: According to what has already been proved, ‘ðIÞ ¼ m ðIÞ:  we know from Proposition 2.2.3(c) that Since I  A  A, m ðIÞ  m ðAÞ  m ðAÞ: It follows from the four statements displayed above that ‘ðAÞ  e\m ðAÞ  ‘ðAÞ: Since this is true for an arbitrary e > 0, we have m ðAÞ ¼ ‘ðAÞ. Finally, consider the case when A is an unbounded interval. Then for any M > 0, there exists a closed bounded interval I  A such that ‘ðIÞ  M. Hence m ðAÞ  m ðIÞ ¼ ‘ðIÞ  M: Thus m (A)  M for every positive real number M. It follows that m ðAÞ ¼ 1 ¼ ‘ðAÞ: h

50

2 Measure in Euclidean Space

A part of the proof above consisted in showing that the total length of a finite family F of open intervals covering a closed bounded interval [a, b] is no less than that of the latter. An alternative argument, which actually proves that the total length exceeds the latter, is as follows. Clearly, we may assume that the intervals of F are bounded. Since F covers [a, b], the number a belongs to some interval ðu1 ; v1 Þ 2 F Then u1 \a\v1 :

ð2:3Þ

  Suppose the n intervals uj ; vj 2 F, 1  j  n, have the property that u1 \a\v1

and

uj þ 1 \vj \vj þ 1 whenever 1  j  n  1:

ð2:4Þ

When n = 1, (2.4) asserts the same thing as (2.3), because there can be no j satisfying 1  j  n − 1. If vn < b, then we have a < vn < b, so that vn 2 [a, b] and, in view of the hypothesis that F covers [a, b], there must exist ðun þ 1 ; vn þ 1 Þ 2F for which un+1 < vn < vn+1. This means we now have n + 1 intervals uj ; vj 2 F having the property that u1 \a\v1

and

uj þ 1 \vj \vj þ 1

whenever 1  j  n:

This has been deduced from the assumption that vn < b, and says that (2.4) holds with n + 1 in place of n. We have already noted that (2.4) holds when n = 1. It follows that, if vn < b for every n, then there is a sequence of intervals {(un, vn)} of F satisfying (2.4). However, (2.4) also forces the intervals to be different from each other, thus we get an infinite subset of F, which contradicts the finiteness of the latter. Therefore vn  b for some n. So, we get only finitely many intervals uj ; vj 2 F, 1  j  n, and they satisfy u1 \a\v1 ; uj þ 1 \vj \vj þ 1

for

1jn  1

and

b  vn :

ð2:5Þ

For any pair of finite sequences {uj} and {vj} of real numbers, we shall prove by induction that (2.5) implies n X

ðvj  uj Þ [ b  a:

ð2:6Þ

j¼1

If n = 1, then the last inequality in (2.5) implies b  v1 and the first implies n   P vj  uj ¼ v1  u1  b  u1 [ b  a. Thus (2.5) implies (2.6) when n = 1. j¼1

Assume (induction hypothesis) for some n that (2.5) implies (2.6), and consider a pair of finite sequences {uj} and {vj}, 1  j  n + 1, which satisfy (2.5) with n + 1 in place of n:

2.2 Lebesgue Outer Measure

u1 \a\v1 ;

51

uj þ 1 \vj \vj þ 1

for

1jn

and

b  vn þ 1 :

Taking j = n, we see that un+1 < vn, and so the finite sequences {uj} and {vj}, 1  j  n, consisting of the first n terms, satisfy (2.5) with b replaced by un+1. The induction hypothesis yields n X

ðvj  uj Þ [ un þ 1  a;

j¼1

which leads to

nP þ1 

 vj  uj [ vn þ 1  un þ 1 þ un þ 1  a ¼ vn þ 1  a  b  a.

j¼1

This completes the induction argument that (2.5) always implies (2.6). Since we have shown above that F contains a subfamily {(uj,vj): 1  j  n} satisfying (2.5), it follows that the total length of the intervals in F is greater than b − a. For the next important property of outer measure, the following definition will turn out to be useful. Definition 2.2.7 Let A be a nonempty collection of subsets of a set X and m be an extended real-valued set function with domain A. The set function m is said to be finitely additive if, for every finite sequence of sets {Aj}1  j  n such that each Aj 2 A, [ 1  j  n Aj 2 A, and Aj \ Ak ¼ £ whenever j 6¼ k (i.e. disjoint), we have m

n [ j¼1

n  X Aj ¼ mðAj Þ: j¼1

It is said to be finitely subadditive if, for every finite sequence of sets {Aj}1  j  n such that each Aj 2 A and [ 1  j  n Aj 2 A, we have m

n [ j¼1

n  X Aj  mðAj Þ; j¼1

and countably subadditive if for every sequence of sets {Aj}j  1 such that each Aj 2 A and [ j  1 Aj 2 A, we have m

1 [ j¼1

1  X Aj  mðAj Þ: j¼1

It is said to be countably additive if for every sequence of sets {Aj}j  1 such that each Aj 2 A and [ j  1 Aj 2 A, and Aj \ Ak ¼ £ whenever j 6¼ k (i.e. disjoint), we have

52

2 Measure in Euclidean Space

mð

1 [

Aj Þ ¼

j¼1

1 X

mðAj Þ:

j¼1

Before proving that the Lebesgue outer measure is countably subadditive, we give some simple instances of finitely and countably subadditive set functions in a general set X. Examples 2.2.8 (a) Let X be any infinite set and suppose A consists of all nonempty finite subsets of X. Take m(A) to be the number of elements in A. Then m is finitely as well as 1   P countably subadditive. In fact, m Aj will always be ∞, because each term j¼1

will be 1 or greater; this will cease to be so if we enlarge A to include £. But m will nevertheless be finitely as well as countably subadditive. Indeed, m is finitely as well as countably additive regardless of whether we enlarge A to include £. This example works even when X is finite but nonempty, although for slightly different reasons. (b) Let X be any set containing 4 or more elements and suppose A consists of all finite subsets of X containing 2 or more elements. Take m(A) to be 1 less than 1   P the number of elements in A. Then m Aj will always be ∞, and therefore m j¼1

is countably subadditive. However, it is not finitely subadditive, as can be seen by considering two disjoint subsets containing 2 elements each. Proposition 2.2.9 The outer measure m is countably subadditive; that is, if {Aj}j  1 is a sequence of subsets of R, then 1 1 [ X m ð Aj Þ  m ðAj Þ: j¼1

ð2:7Þ

j¼1

Proof Assume that e > 0 is given. For each j 2 N, choose a sequence {Ij,k}k  1 of open intervals that covers Aj and satisfies 1 X

‘ðIj;k Þ  m ðAj Þ þ e=2 j :

k¼1

Then the sequence {Ij,k}j,k  1 covers [ j  1 Aj and has total length less than or equal 1   P to m Aj þ e. Hence j¼1

m ð

1 [ j¼1

Aj Þ 

1 X

m ðAj Þ þ e:

j¼1

Since e > 0 is arbitrary, we conclude that (2.7) holds.

h

2.2 Lebesgue Outer Measure

53

Corollary 2.2.10 The outer measure m is finitely subadditive; that is, if {Aj}1  j  n is a finite sequence of subsets of R, then n n [ X m ð Aj Þ  m ðAj Þ: j¼1

j¼1

Proof Set Aj = ∅ for j > n and use Propositions 2.2.9 and 2.2.3(b).

h

Remarks 2.2.11 (a) If m (A) = 0, then m (A [ B) = m (B). In fact, by Proposition 2.2.3(c) and Corollary 2.2.10, m ðBÞ  m ðA [ BÞ  m ðAÞ þ m ðBÞ ¼ m ðBÞ: (b) Let A be any subset of R and AI denote the set of irrationals in A, then m (A) = m (AI). This follows from (a) above and the result noted in Example 2.2.5. (c) The set C, known as the Cantor set (see Sect. 1.8 for details), has outer measure 0. Recall that for each natural number n 2 N, the set C is a subset of a union Cn of intervals whose total length is (2/3)n. It follows by Propositions 2.2.6 and 2.2.9 that m (Cn)  (2/3)n. Since C  Cn , it further follows that m (C)  (2/3)n ! 0 as n ! ∞. In the rest of this section, we explore more closely the role of open sets in determining the Lebesgue outer measure of an arbitrary subset of R. The material can be omitted for now without loss of continuity and the reader may wish to take it up only when it is needed later. Proposition 2.2.12 Let A  R be arbitrary and e > 0 be given. Then there exists an open set O A such that m (O)  m (A) + e. Moreover, in the case when m (A) < ∞, the set O can be so chosen that the inequality is strict. Proof If m (A) = ∞, the assertion is trivial, because we can choose O ¼ R. So, suppose m (A) < ∞. By definition, m (A) is the infimum of all possible sums 1 P ‘ðIn Þ, where {In}n  1 is a sequence of open intervals that covers A. Therefore, n¼1

there exists such a covering sequence {In}n  1 such that 1 X

‘ðIn Þ\m ðAÞ þ e:

n¼1

Now O ¼ [ n  1 In is a union of open intervals (hence open sets) and is therefore an open set. As {In}n  1 is a sequence of open intervals that covers A, we have A  O. Also, by countable subadditivity of m (Proposition 2.2.9) and the fact that the outer measure of an interval is its length (Proposition 2.2.6),

54

2 Measure in Euclidean Space

m ðOÞ ¼ m ð

1 [ n¼1

In Þ 

1 X n¼1

m ðIn Þ ¼

1 X

‘ðIn Þ\m ðAÞ þ e:

h

n¼1

It is natural to ask whether one can sharpen the preceding result to eliminate the e. In other words, can one always find an open set O A such that m (O) = m (A)? Of course, one cannot: if A contains a single point, then its outer measure is 0 but any open set containing A has to be a nonempty open set and must therefore contain an open interval of positive length, which makes its outer measure greater than that of A. The next natural question is what happens if we take a sequence of open sets On such that m ðOn Þ \m ðAÞ þ 1n and then set G ¼ \ n  1 On . Of course, we get  an intersection of a sequence of open sets need not G A and m (G) = Tm (A), 1but be open (example: n  1 ðn; 1nÞ ¼ f0g, which is not open). Here is the best we can do: Proposition 2.2.13 Let A  R be arbitrary. Then there exists a set G A such that m (G) = m (A) and G is an intersection of a sequence of open sets. Proof If m (A) = ∞, we may take G ¼ R. So, suppose m (A) < ∞. Then by Proposition 2.2.12, for each n 2 N, there exists an open set On A such that m ðOn Þ\m ðAÞ þ 1n. Let G ¼ \ n  1 On . Then G is an intersection of a sequence of open sets, G A, and m ðAÞ  m ðGÞ  m ðOn Þ  m ðAÞ þ

1 n

for each n 2 N:

Hence m (G) = m (A). h The above proposition can be rephrased as saying that a given subset of R is a subset of a set which is a countable intersection of open sets and has the same outer measure as the given set. This can be shortened further by introducing the following terminology. Definition 2.2.14 The class of those subsets of a metric space that are countable intersections of open sets is called Gd . Sets in this class are called Gd -sets or said to be of type Gd . The class of those subsets of a metric space that are countable unions of closed sets is called F r . Sets in this class are called F r -sets or said to be of type F r. Thus every subset of R is contained in a Gd -set of the same outer measure as itself. Recall that if we take the union of a countable number of families, each of which is itself countable, the resulting family is again countable. It is a consequence that a countable intersection [resp. union] of Gd -sets [resp. F r -sets] is a Gd -set [F r -set].

2.2 Lebesgue Outer Measure

55

Examples 2.2.15 (a) Consider the subset Q of all rational numbers in R. Let x1, x2,… be an enuS meration of Q. We can now write Q ¼ fxi g. Since each of the single point i1

sets {xi} is closed, it follows that Q is an F r -set. (b) If E 2 F r , then by using the definition of F r and taking complements, we conclude via De Morgan’s Laws that Ec 2 Gd . In particular, it follows that the set of irrationals in R is a set of type Gd . Similar considerations show that, if E 2 Gd then EcS2 F r . T (c) Since ða; bÞ ¼ n  1 ½a þ 1n; b  1n and f0g ¼ n  1 ð1n; 1nÞ, it follows that (a, b) is a set of type F r and that {0} is of type Gd . (d) It is clear that every open set is of type Gd and every closed set is of type F r . It is easily seen that every open set is also of type F r , because it is a countable union of open intervals, each of which is an F r -set by (c). On taking complements, we see that every closed set is of type Gd . Problem Set 2.2 2.2.P1. Use the results of this section to show that [0, 1] is uncountable. 2.2.P2. If the domain of a countably subadditive set function m includes ∅ and if m(∅) = 0, show that m is finitely subadditive. 2.2.P3. Show that for any two sets A and B with union [0, 1], the outer measure satisfies m ðAÞ  1  m ðBÞ: 2.2.P4. Let {Ij}1  j  n be a finite sequence of open intervals covering the rationals n   P in [0, 1]. Show that ‘ Ij  1. (Note: With a little extra effort, it can be shown j¼1

that the sum of lengths is strictly greater than 1; however, we shall not need this fact.) 2.2.P5. Show that if we were to define outer measure as approximation by finitely many open intervals, i.e. if m (A) were to be inff

n X

‘ðIi Þ : A 

i¼1

n [

Ii ;

each Ii an open intervalg;

i¼1

then it would not be countably subadditive. 2.2.P6. Show that if we were to define outer measure as approximation from within, i.e. if m (A) were to be supf

1 X i¼1

‘ðIi Þ : A

1 [

Ii ; each Ii an open interval and i 6¼ j ) Ii \ Ij ¼ £g;

i¼1

then it would not be finitely subadditive.

56

2 Measure in Euclidean Space

2.2.P7. Prove that, if the open set A is the union of a sequence {In}n  1 of disjoint open intervals, then m ðAÞ ¼

1 X

‘ðIn Þ:

n¼1

2.2.P8. Show that using closed intervals instead of open intervals in the definition of outer measure does not change the evaluation of m (A). 2.2.P9. Show that using intervals closed only on the left (or on the right, or a mixture of various types of intervals) rather than open intervals does not change the evaluation of m (A). 2.2.P10. (a) For k > 0 and A  R, let kA denote {x: k−1x 2 A}. Show that m (kA) = k  m (A). (b) For A  R let −A denote {x: −x 2 A}. Show that m (−A) = m (A). 2.2.P11. Let A = {x 2 [0, 1]: x has a decimal expansion not containing the digit 5}. Note that 0.5 2 A, because 0.5 = 0.4999…, but 0.51 62 A. Then show that m (A) = 0. 2.2.P12. Suppose m ðA \ IÞ  12 m ðIÞ for every interval I. Prove that m (A) = 0. 2.2.P13. For nonempty E  R, the diameter is diam E ¼ supfjx  yj: x; y 2 Eg. Show that m (E)  diamE.

2.3

Measurable Sets and Lebesgue Measure

The Lebesgue outer measure m of Definition 2.2.1 is an example of a nonnegative, monotone, countably subadditive set function, which is defined on all subsets of R and satisfies m (∅) = 0. (A set function l is said to be monotone if A  B implies lðAÞ  lðBÞ.) There is one further essential property that one would like to have, namely, countable additivity. In fact, m is not even finitely additive. For an example of this phenomenon, see Problem 2.3.P16. So to speak, in our attempt to extend the concept of length of an interval to a notion of measure to more general sets, we have gone too far, and the problem now is to restrict m to a subclass on which m is countably additive, as will unfold in this section. Such a subclass exists and is indeed a “r-algebra” (formally described in Definition 2.3.10 below) containing open as well as closed sets. It turns out to be useful to select precisely those subsets E  R for which m ðAÞ ¼ m ðA \ EÞ þ m ðA \ Ec Þ

for every A  R;

where Ec denotes the complement of E in R. This property of E can be regarded as saying informally that E “splits” every subset A “additively”.

2.3 Measurable Sets and Lebesgue Measure

57

Definition 2.3.1 A subset E  R is said to be measurable if m ðAÞ ¼ m ðA \ EÞ þ m ðA \ E c Þ

for every A  R:

Remarks 2.3.2 (a) Observe that the finite subadditivity property (see Corollary 2.2.10) of m implies m ðAÞ  m ðA \ EÞ þ m ðA \ Ec Þ

for every A  R:

Thus in testing the measurability of E, it is enough to show the reverse of this inequality. This condition in Definition 2.3.1 is known as the Carathéodory condition. (b) If E is measurable, so is its complement Ec. (c) For any subset A of R, we have m ðAÞ ¼ m ðAÞ þ 0 ¼ m ðA \ RÞ þ m ðA \ £Þ. Thus R is measurable, and hence by (b), the same is true of £. (d) If m (E) = 0, where E  R, then E is measurable; in fact, every subset of E is also measurable. Indeed, for A  R, we have m (A \ E)  m (E) = 0 and m ðA \ E c Þ  m ðAÞ. So, m ðAÞ  m ðA \ Ec Þ þ 0  m ðA \ EÞ þ m ðA \ Ec Þ. In particular, the Cantor set (see Remark 2.2.11(c)) is measurable and so is every subset of it. (e) A countable subset of R is measurable. This follows from Example 2.2.5 and (d) above. Consequently, the set Q of rationals in R is measurable. The set of irrationals in R, being the complement of Q, is measurable in view of (b) above. Notation 2.3.3 We denote the collection of all measurable subsets of R by M. In what follows, we prove some of the properties of the collection M. Lemma 2.3.4 If E1 and E2 are in M, then so is E1 [ E2. Proof Let A be any subset of R. From the measurability of E1, we have m ðAÞ ¼ m ðA \ E1 Þ þ m ðA \ E1c Þ:

ð2:8Þ

Since A \ E1 and A \ E1c are subsets of R, it follows from the measurability of E2 that m ðA \ E1 Þ ¼ m ðA \ E1 \ E2 Þ þ m ðA \ E1 \ E2c Þ

ð2:9Þ

m ðA \ E1c Þ ¼ m ðA \ E1c \ E2 Þ þ m ðA \ E1c \ E2c Þ:

ð2:10Þ

and

58

2 Measure in Euclidean Space

Now, ðE1 \ E2 Þ [ ðE1 \ E2c Þ [ ðE1c \ E2 Þ ¼ E1 [ E2 and ðA \ E1 \ E2 Þ [ ðA \ E1 \ E2c Þ [ ðA \ E1c \ E2 Þ ¼ A \ ðE1 [ E2 Þ:

ð2:11Þ

Also, ðA \ E1c \ E2c Þ ¼ A \ ðE1 [ E2 Þc :

ð2:12Þ

On using the finite subadditivity of outer measure (Corollary 2.2.10), it follows from (2.11) and (2.12) that m ðA \ ðE1 [ E2 ÞÞ þ m ðA \ ðE1 [ E2 Þc Þ  m ðA \ E1 \ E2 Þ þ m ðA \ E1 \ E2c Þ þ m ðA \ E1c \ E2 Þ þ m ðA \ E1c \ E2c Þ: Using (2.9) and (2.10), and then (2.8), we further obtain from here that m ðA \ ðE1 [ E2 ÞÞ þ m ðA \ ðE1 [ E2 Þc Þ  m ðA \ E1 Þ þ m ðA \ E1c Þ ¼ m ðAÞ: By Remark 2.3.2(a), this completes the proof. h We introduce some terminology that will make it possible to summarise the important properties of M conveniently. Definition 2.3.5 Let X be a nonempty set. A family F of subsets of X is called an algebra if ðaÞ X 2 F

ðbÞ A 2 F ; B 2 F ) A [ B 2 F

ðcÞ A 2 F ) Ac 2 F :

Remarks 2.3.6 (a) (b) (c) (d) (e)

£ 2 F. S A1 ; A2 ; . . .; An 2 F ) nj¼1 Aj 2 F : If A 2 F ; B 2 F ) A \ TB 2 F : A1 ; A2 ; . . .; An 2 F ) nj¼1 Aj 2 F : A 2 F ; B 2 F ) AnB 2 F :

Proposition 2.3.7 The collection M of measurable subsets of R is an algebra. In particular, if E1, E2, …, En are measurable subsets, then so are [ 1  k  n Ek and \ 1  k  n Ek . Proof This is merely a summary of Remarks 2.3.2(b) and 2.3.2(c), Lemma 2.3.4 and Remarks 2.3.6(b) and 2.3.6(d). h We now show that the outer measure restricted to M is finitely additive.

2.3 Measurable Sets and Lebesgue Measure

59

Proposition 2.3.8 If E1, E2, …, En is a finite sequence of disjoint measurable subsets of R and A  R is arbitrary, then m ðA \

n [

Ej Þ ¼

j¼1

n X

m ðA \ Ej Þ:

ð2:13Þ

j¼1

Taking A ¼ R in particular, we have m ð

n [

Ej Þ ¼

j¼1

n X

m ðEj Þ:

j¼1

Proof We shall prove this by induction on n. For n = 1, the equality (2.13) is obvious. Suppose it holds for n = k, that is (induction hypothesis), m ðA \

k [ j¼1

Ej Þ ¼

k X

m ðA \ Ej Þ

ð2:14Þ

j¼1

whenever E1, E2, …, Ek is a finite sequence of disjoint measurable subsets of R and A  R is arbitrary. In order to deduce the validity of (2.13) for n = k + 1, observe that, in view of the measurability of Ek+1, m ðA \

k[ þ1

Ej Þ ¼ m ððA \

j¼1

k[ þ1

Ej Þ \ Ek þ 1 Þ þ m ððA \

j¼1

k[ þ1

Ej Þ \ Ekc þ 1 Þ: ð2:15Þ

j¼1

Using the disjointness, we have ðA \

k[ þ1

Ej Þ \ Ek þ 1 ¼ A \ Ek þ 1 and ðA \

j¼1

k[ þ1

Ej Þ \ Ekc þ 1 ¼ A \

j¼1

k [

Ej :

j¼1

Therefore (2.15) becomes m ðA \

k[ þ1 j¼1

Ej Þ ¼ m ðA \ Ek þ 1 Þ þ m ðA \

k [

Ej Þ:

j¼1

The induction hypothesis (2.14) now yields the desired equality (2.13). h The outer measure m restricted to M is in fact countably additive, as we shall show below in Theorem 2.3.13. The first step in this direction is to show that M is closed under countable unions. We begin with a lemma. Lemma 2.3.9 Let F be an algebra of subsets of X. If {Aj}j  1 is a sequence of sets in F , then there exists a sequence {Bj}j  1 of disjoint sets in F such that

60

2 Measure in Euclidean Space 1 [

Bj ¼

j¼1

1 [

Aj :

j¼1

Proof Set B1 ¼ A1 ; B2 ¼ A2 \ Ac1 ; B3 ¼ A3 \ ðA1 [ A2 Þc ; . . .; Bj ¼ Aj \ ðA1 [ A2 [    [ Aj1 Þc ; . . .: Clearly, each Bj 2 F because the latter is an algebra. Moreover, each Bj  Aj . Suppose m < n; then Bm \ Bn  Am \ An \ ðAc1 \ Ac2 \    \ Acn1 Þ  Am \ Acm ¼ £: Since each Bj  Aj , the inclusion 1 [

Bj 

j¼1

1 [

Aj

j¼1

holds. On the other hand, suppose that x 2 [ j  1 Aj and let j be the smallest positive  c integer such that x 2 Aj. Then x 2 Aj and x 2 A1 [ A2 [    [ Aj1 . So x 2 Bj and h hence x 2 [ j  1 Bj . Thus the reverse inclusion also holds. Definition 2.3.10 An algebra F of subsets of X is called a r-algebra (or a r-field) if every union of a sequence of sets in F is again in F . That is, if {Aj}j  1 is a sequence of sets in F , then [ j  1 Aj must be in F . Remarks 2.3.11 (a) If {Aj}j  1 is a sequence of sets in a r-algebra F , then \ j  1 Aj 2 F . In fact, if Aj 2 F , then so is Acj ; since F is a r-algebra, [ j  1 Acj 2 F and hence  c \ j  1 Aj ¼ [ j  1 Acj 2 F . (b) A class F of subsets of X is a r-algebra if and only if ðiÞ X 2 F

ðiiÞ Aj 2 F

8j 2 N )

[ j1

Aj 2 F

ðiiiÞ A 2 F ) Ac 2 F :

(c) Trivially, a finite union or intersection of sets in a r-algebra is again in the r-algebra. (d) A knowlegable reader will discern the difference between the postulates of a r-algebra and those of a “topology” on a set X.

2.3 Measurable Sets and Lebesgue Measure

61

Theorem 2.3.12 The collection M of measurable subsets of R is a r-algebra. Proof We have shown in Proposition 2.3.7 that the collection M of measurable sets is an algebra. All we need to show now is that if Ej  Rðj 2 NÞ are measurable, then so is their union [ j  1 Ej . Since M is an algebra, by Lemma 2.3.9, there exists a sequence {Fj}j  1 of disjoint measurable sets such that [ j  1 Ej ¼ [ j  1 Fj . To complete the proof, we show 1 1 [ [ m ðAÞ  m ðA \ ð Fj ÞÞ þ m ðA \ ð Fj Þc Þ j¼1

j¼1

for any set A  R. Observe that, for any positive integer p, [ 1  j  p Fj is measurable, because M is an algebra. Therefore m ðAÞ ¼ m ðA \ ð

p [

Fj ÞÞ þ m ðA \ ð

j¼1



p X

p [

F j Þc Þ

j¼1

1 [ m ðA \ Fj Þ þ m ðA \ ð Fj Þc Þ; j¼1

j¼1

 c  c where we have used Proposition 2.3.8, the fact that [ 1  j  p Fj [ j  1 Fj and Proposition 2.2.3(c). Since this inequality is valid for all p and the sequence ( ) p 1    P P  m A \ Fj has supremum m A \ Fj , we have j¼1

j¼1

p1

m ðAÞ 

1 X

m ðA \ Fj Þ þ m ðA \ ð

j¼1

 m ð

1 [

Fj Þc Þ

j¼1 1 [

1 [ ðA \ Fj ÞÞ þ m ðA \ ð Fj Þc Þ by Propostion 2:2:9

j¼1

j¼1

1 1 [ [  m ðA \ ð Fj ÞÞ þ m ðA \ ð Fj Þc Þ: j¼1

h

j¼1

Theorem 2.3.13 The outer measure m is countably additive on the r-algebra M of measurable subsets of R. Proof Let {Ej}j  1 be a sequence of disjoint measurable subsets. From Proposition 2.2.3(c) and Proposition 2.3.8, it follows for an arbitrary positive integer p that m ð

1 [ j¼1

Ej Þ  m ð

p [ j¼1

Ej Þ ¼

p X j¼1

m ðEj Þ:

62

2 Measure in Euclidean Space

Therefore 1 1 [ X m ð Ej Þ  m ðEj Þ: j¼1

j¼1

The reverse inequality is a consequence of countable subadditivity (see Proposition 2.2.9). h Definition 2.3.14 The restriction of m to M is called (Lebesgue) measure and will be denoted by m. For any E 2 M, the (extended) real number m(E) = m (E) is called the (Lebesgue) measure of the set E. Since m is countably subadditive according to Proposition 2.2.9, it follows trivially that Lebesgue measure is also countably subadditive. We shall have occasion to use this fact in Chap. 5. Proposition 2.3.15 The interval (a, ∞) is measurable. Proof Let A be any subset of R and let A1 ¼ A \ ða; 1Þ and A2 ¼ A \ ða; 1Þc ¼ A \ ð1; a. We must show that m (A)  m (A1) + m (A2). If m (A) = ∞, then there is nothing to prove. Suppose that m (A) < ∞. By the definition of outer measure, for every e > 0, there exists a sequence {In}n  1 of open intervals such that A  [ n  1 In and 1 X

‘ðIn Þ  m ðAÞ þ e:

n¼1

Let In0 ¼ In \ ða; 1Þ and In00 ¼ In \ ð1; a. Then In′ and In″ are intervals (some of them possibly empty) and ‘ðIn Þ ¼ ‘ðIn0 Þ þ ‘ðIn00 Þ ¼ m ðIn0 Þ þ m ðIn00 Þ: Since A1  [ n  1 In0 and A2  [ n  1 In00 , we have m ðA1 Þ  m ð

1 [

In0 Þ 

n¼1

1 X

m ðIn0 Þ

n¼1

and m ðA2 Þ  m ð

1 [

n¼1

In00 Þ 

1 X n¼1

m ðIn00 Þ:

2.3 Measurable Sets and Lebesgue Measure

63

Hence m ðA1 Þ þ m ðA2 Þ  ¼

1 X n¼1 1 X n¼1

m ðIn0 Þ þ

1 X

m ðIn00 Þ

n¼1

½m ðIn0 Þ þ m ðIn00 Þ ¼

1 X

‘ðIn Þ  m ðAÞ þ e:

n¼1

Since e > 0 is arbitrary, we have m (A)  m (A1) + m (A2), as required.

h

Proposition 2.3.16 A subset of R that is either open or closed must be measurable. Proof Since the collection M of measurable sets is a r-algebra, it follows from Proposition S 2.3.15 that (−∞, a] = (a,∞)c is measurable for every real a. Since ð1; bÞ ¼ n  1 ð1; b  1n, it further follows that (−∞, b) is measurable and hence each open interval ða; bÞ ¼ ða; 1Þ \ ð1; bÞ is measurable. We can now conclude that each open set, being a countable union of open intervals (see Theorem 1.3.17), must be measurable. Since M is a r-algebra, the complement of an open set, i.e. a closed set, must be measurable. h Proposition 2.3.17 Let A be a class of subsets of X. Then there is a smallest r-algebra S of subsets of X that contains A. Proof Let fS a g be the collection of all those r-algebras of subsets of X that contain A. Since the collection of all subsets of X is a r-algebra, the collection fS a g is nonempty. It is easily verified that \ a S a is a r-algebra and contains A. It is necessarily the smallest r-algebra of subsets of X that contains A. h The r-algebra in the above proposition is called the r-algebra generated by A (or by the sets of A). Definition 2.3.18 The r-algebra generated by all the open subsets of R is called the Borel algebra and will be denoted by B. A set in the Borel algebra is called a Borel measurable set or simply a Borel set. A set in the algebra M (hitherto called simply “measurable”) will be called a Lebesgue measurable set when a distinction needs to be made. Remarks 2.3.19 (a) Since every open set is Lebesgue measurable (Proposition 2.3.16) and Lebesgue measurable sets constitute a r-algebra, it follows by Proposition 2.3.17 and Definition 2.3.18 that B  M. In other words, every Borel set is Lebesgue measurable. The restriction of m to B is called Borel measure and will again be denoted by m. Thus, for E 2 B, m(E) = m (E) is called the Borel measure of E and is the same as its Lebesgue measure.

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2 Measure in Euclidean Space

(b) The cardinality of the Borel algebra B is c and that of M is strictly larger. The proof of the former may be found in [12, p. 134]. For the latter, see Problem 2.3.P20. The triples (R, B, m) and (R, M, m) are called the Borel measure space and Lebesgue measure space respectively. We have verified the following properties of Lebesgue measurable sets and Lebesgue measure: (i) Complements, countable unions and countable intersections of measurable sets are measurable. (ii) Any interval is measurable and its measure is its length. (iii) If {En}n  1 is a sequence of disjoint measurable sets, then mð

1 [

En Þ ¼

n¼1

1 X

mðEn Þ:

n¼1

The above properties are also valid for Borel measurable sets and Borel measure: (i) is a direct consequence of the definition of B as being a r-algebra, while (ii) and (iii) follow from their Lebesgue counterparts taken with Remark 2.3.19. Some additional properties of measure will now be verified. We begin with what is known as “continuity” of measure. The following definition will be needed. Definition 2.3.20 For any sequence {En}n  1 of sets, the limit superior and limit inferior are respectively lim sup En ¼

1 [ \

En

and

lim inf En ¼

i¼1 n  i

1 \ [

En :

i¼1 n  i

If limsupEn = liminfEn, then we call this set the limit of the sequence {En}n  1 of sets and denote it by limEn. It is easily seen from the definition that limsupEn is the set of all points belonging to infinitely many of the sets En and liminfEn is the set of all points belonging to all but finitely many. Consequently, liminfEn  limsupEn. If E1  E2  , then we have lim En ¼ [ n  1 En ; if E1 E2   , then we have lim En ¼ \ n  1 En . Proposition 2.3.21 (Continuity) Let {En}n  1 be a sequence of Lebesgue [or Borel] measurable sets. Then (a) E1  E2     implies mðlim En Þ ¼ lim mðEn Þ (inner continuity), n!1

(b) E1 E2    with m(E1) < ∞ implies mðlim En Þ ¼ lim mðEn Þ (outer n!1

continuity), (c) m(liminfEn)  liminfm(En), (d) m(limsupEn)  limsupm(En) provided mð

1 S k¼i

Ek Þ\1 for some i.

2.3 Measurable Sets and Lebesgue Measure

65

Proof It is sufficient to argue the case of Lebesgue measurable sets only, because the result will then follow for Borel measurable sets by Remark 2.3.19. c (a) Write F1 = E1, and Fk ¼ Ek \ Ek1 for k > 1. Then En ¼ F1 [ F2 [    [ Fn and also, [ k  1 Ek ¼ [ k  1 Fk ; besides, the sets Fk are measurable and disjoint. Hence, Theorem 2.3.13 yields

mðlim En Þ ¼ mð

1 [

Ek Þ ¼ mð

k¼1

¼ lim mð n!1

1 [

Fk Þ ¼

k¼1 n [

1 X

mðFk Þ ¼ lim

n X

n!1

k¼1

mðFk Þ

k¼1

Fk Þ ¼ lim mðEn Þ: n!1

k¼1

(b) Write Fk ¼ E1 \ Ekc . Since Ek  E1 , therefore E1 is the disjoint union of Ek and Fk, where Fk is measurable. Before proceeding, note that F1  F2     and consequently, lim Fn ¼ [ k  1 Fk . Now, by Theorem 2.3.13, we know that m(E1) = m(Fk) + m(Ek). Since m(E1) < ∞, we can rewrite this as mðFk Þ ¼ mðE1 Þ  mðEk Þ:

ð2:16Þ

Moreover, lim Fn ¼

1 [ k¼1

Fk ¼

1 [

ðE1 \ Ekc Þ ¼ E1 \

k¼1

1 [

Ekc ¼ E1 \ ð

k¼1

¼ E1 \ ðlim En Þc ;

because lim En ¼

\

1 \

E k Þc

k¼1

E: k1 k

Since lim En  E1 , a similar argument as for (2.16) now leads to mðlim Fn Þ ¼ mðE1 Þ  mðlim En Þ:

ð2:17Þ

Also, F1  F2     and hence by (a) and (2.16), we have mðlim Fn Þ ¼ lim mðFn Þ ¼ mðE1 Þ  lim mðEn Þ: n!1

n!1

ð2:18Þ

On comparing (2.17) and (2.18), and once again using the fact that m(E1) < ∞, we obtain the required equality. (c) Set Fn ¼ \ k  n Ek . Then by definition, we have lim inf En ¼

1 [

Fn :

n¼1

Observe that {Fn} is an increasing sequence of measurable sets. Therefore by part (a),

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2 Measure in Euclidean Space

mð

1 [

Fn Þ ¼ lim mðFn Þ; n!1

n¼1

so that mðlim inf En Þ ¼ lim mðFn Þ: n!1

ð2:19Þ

Let n 2 N be fixed. Then Fn  En þ k and so mðFn Þ  mðEn þ k Þ for all k 2 N: Thus we have mðFn Þ  lim inf mðEn þ k Þ ¼ lim inf mðEk Þ: k

This is true for all n 2 N. Hence lim mðFn Þ  lim inf mðEk Þ:

n!1

ð2:20Þ

The result now follows from (2.19) and (2.20). (d) Similar to (c).

h

Remark 2.3.22 Part (b) of the above proposition does not hold without the hypothesis that m(E1) < ∞. Indeed, if En = (n, ∞) for each n 2 N, then [ n  1 En ¼ £; however, lim mðEn Þ ¼ 1 and mðlim En Þ ¼ mð£Þ ¼ 0. n!1

Proposition 2.3.23 (Translation Invariance of Lebesgue measure on R) Let E 2 M [resp. B], x 2 R. Then (a) E þ x 2 M [resp. B], (b) mðE þ xÞ ¼ mðEÞ: That is, the Lebesgue [resp. Borel] measure on M [resp. B] is translation invariant. Proof (a) First we prove this for M. Let A  R be arbitrary and E 2 M. Note that

A \ ðE þ xÞ ¼ ½ðA  xÞ \ E þ x

and

A \ ðE þ xÞc ¼ ½ðA  xÞ \ Ec  þ x:

It follows immediately from the measurability of E and Proposition 2.2.4 (translation invariance of m ) that

2.3 Measurable Sets and Lebesgue Measure

67

m ðA \ ðE þ xÞÞ þ m ðA \ ðE þ xÞc Þ ¼ m ð½ðA  xÞ \ E þ xÞ þ m ð½ðA  xÞ \ E c  þ xÞ ¼ m ð½ðA  xÞ \ EÞ þ m ð½ðA  xÞ \ Ec Þ ¼ m ðA  xÞ ¼ m ðAÞ: This proves that E þ x 2 M, thus establishing (a) for M. To prove the same for B, consider the class F ¼ fF  R : F þ x 2 Bg. First, since R þ x ¼ R 2 B, we have R 2 F . Second, Aj 2 F 8j 2 N ) [ j  1 Aj 2 F . To see why, suppose Aj 2 F 8j 2 N. Then every Aj þ x 2 B. Now, y 2 Aj + x ,     y − x 2 Aj. Therefore y 2 [ j  1 Aj þ x , y  x 2 [ j  1 Aj , y 2 [ j  1 Aj þ x.     In other words, [ j  1 Aj þ x ¼ [ j  1 Aj þ x. But since B is a r-algebra, we   have [ j  1 Aj þ x 2 B. Therefore ð [ j  1 Aj Þ þ x 2 B, so that [ j  1 Aj 2 F . And third, A 2 F ) Ac 2 F . This is because (A + x)c can be shown to be the same as Ac + x by arguing that y 2 (A + x)c , y 62 A + x , y − x 62 A , y − x 2 Ac. By Remark 2.3.11(b), it follows that F is a r-algebra. Surely, O + x is open, and hence belongs to B whenever O  R is open. Therefore the r-algebra F contains all open sets and hence contains B. Thus F 2 B ) F 2 F ) F þ x 2 B. (b) Since E þ x 2 M (or B) and m is the restriction of m to M (or B), by Proposition 2.2.4 (translation invariance of m ), we have mðE þ xÞ ¼ m ðE þ xÞ ¼ m ðEÞ ¼ mðEÞ:

h

Finally, we prove the following characterisation of Lebesgue measurable subsets of R. Proposition 2.3.24 Let E be a given subset of R. The following five statements are equivalent: (a) E is Lebesgue measurable; (b) For every e > 0, there exists an open set O E such that m (O\E) < e; (c) There exists a Gd -set G E such that m (G\E) = 0; (d) For every e > 0, there exists a closed set F  E such that m (E\F) < e; (η) There exists an F r -set F  E such that m (E\F) = 0. Proof We shall prove (a) ) (b) ) (c) ) (a) and then (d) , (a) , (η). (a) ) (b). For any set E and e > 0, there exists (see Proposition 2.2.12) an open set O E such that m ðOÞ  m ðEÞ þ

e : 2

Since O E, we have O \ E = E. From the Lebesgue measurability of E, we have

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2 Measure in Euclidean Space

m ðOÞ ¼ m ðO \ EÞ þ m ðO \ E c Þ ¼ m ðEÞ þ m ðOnEÞ: Thus m ðEÞ þ

e  m ðOÞ ¼ m ðEÞ þ m ðOnEÞ: 2

In the case when m (E) < ∞, this shows that m ðOnEÞ  2e \e. Now let m (E) = ∞ and set En = E \ [−n, n] for n 2 N. Then En 2 M and  m (En)  m ([−n, n]) < ∞. Moreover, E ¼ [ n2N En . Since m (En) < ∞, it follows on applying the case of finite measure proved above that, for any e > 0, there exists an open set On En such that m ðOn nEn Þ\

e : 2n

Note that [

O¼

On

n2N

[

En ¼ E

n2N

and OnE ¼

[

On n

n2N

[ n2N

En 

[

ðOn nEn Þ:

n2N

So, m ðOnEÞ 

1 X n¼1

m ðOn nEn Þ\

1 X e ¼ e: n 2 n¼1

(b) ) (c). It follows from (b) that, for any n 2 N, there exists an open set On E such that 1 m ðOn nEÞ\ : n Let G = \ n  1On; then G is a Gd -set containing E. Moreover, m ðGnEÞ ¼ m ð

1 \ n¼1

On nEÞ  m ðOn nEÞ\

1 n

for every n 2 N:

Hence m (G\E) = 0. (c) ) (a). Note that E = G\(G\E) and that the set G, being a countable intersection of Lebesgue measurable sets, is Lebesgue measurable by Theorem 2.3.12; also, G\E is Lebesgue measurable by Remark 2.3.2(d). Use Theorem 2.3.12 once again.

2.3 Measurable Sets and Lebesgue Measure

69

We have shown so far that (a) ) (b) ) (c) ) (a). The consequence that (b) , (a) will shortly be used in proving that (a) , (d). (a) , (d). Assume (a), i.e. E 2 M. Then E c 2 M. Since (a) ) (b), it follows that, for any e > 0, there exists an open set O Ec such that m (O\Ec) < e. So the closed set F = Oc satisfies F  E and also m (E\F) = m (E\Oc) = m (E \ O) = m (O\Ec) < e. Thus (d) holds. Conversely, assume (d). Then the open set O = Fc satisfies O Ec and  m (O\Ec) < e. In other words, (b) holds with Ec in place of E. Since (b) ) (a), it follows that E c 2 M and hence E 2 M. Thus (a) holds. (a) , (η). This is analogous to the argument that (a) , (d). h Remark 2.3.25 Let E  R be Lebesgue measurable, i.e. E 2 M. Then it follows from the preceding proposition that there exist F, G 2 B such that F  E  G with m(G\F) = 0. Conversely, the existence of such F and G can be proved to imply that E 2 M by arguing in the following manner: Suppose such F, G exist. Then F 2 M because B  M; besides, E\F  G\F, which implies m (E\F)  m (G\F) = 0 and hence EnF 2 M by Remark 2.3.2(d). Therefore E ¼ ðEnFÞ [ F 2 M. Note that in this situation, m(E) = m(F) = m(G). The foregoing equivalence is related to the fact that Lebesgue measure is the “completion” of Borel measure (see Definition 7.2.3 and the paragraph preceding Theorem 7.2.2). Problem Set 2.3 2.3.P1. Let A, B be subsets of R and A  B. Show that (a) If m (B\A) = 0, then m (B) = m (A). (b) If m (B) = m (A) < ∞ and A 2 M, then m (B\A) = 0. 2.3.P2. Suppose that A is a subset of R with the property that, for e > 0, there exist measurable sets B and C such that B  A  C and m(C \ Bc) < e. Show that A is measurable. 2.3.P3. Let {En}n  1 be a sequence of sets such that E1  E2  . Show that m ðlim En Þ ¼ lim m ðEn Þ. n!1

2.3.P4. Given a subset A of R, let An = A \ [−n, n] for n 2 N. Show that m ðAÞ ¼ lim m ðAn Þ. n!1

2.3.P5. The symmetric difference of sets A, B is defined to be ADB = (A\B) [ (B\A), consisting of points belonging to one of the two sets but not to both. Show that if A 2 M and m (ADB) = 0, then B 2 M. 2.3.P6. Show that every nonempty open subset O has positive measure. 2.3.P7. Let O ¼ [ n  1 ðxn  n12 ; xn þ n12 Þ, where x1, x2, … is an enumeration of all the rationals. Prove that m(ODF) > 0 for any closed set F. 2.3.P8. The number of elements in a r-algebra generated by n given sets cannot n exceed 22 . 2.3.P9. Show that a r-algebra consisting of infinitely many distinct sets contains an uncountable number of sets.

70

2 Measure in Euclidean Space

2.3.P10. Show that, if E1 and E2 are measurable, then mðE1 [ E2 Þ þ mðE1 \ E2 Þ ¼ mðE1 Þ þ mðE2 Þ: 2.3.P11. If E1 2 M and E2 2 M are such that E1 E2 and m(E2) < ∞, then m (E1\E2) = m(E1) – m(E2). (Since B  M, it follows trivially that this is true also when E1 2 B and E2 2 B.) 2.3.P12. Let 0 < a < 1. Construct a measurable set E  [0, 1] of measure 1 − a and containing no interval of positive length. 2.3.P13. Show that, if E is such that 0 < m (E) < ∞ and 0 < a < 1, then there exists an open interval I such that m ðI \ EÞ [ a‘ðIÞ. 2.3.P14. (a) Suppose m (E) < ∞ and, for every e > 0, there exists an open set U such that m (EDU) < e. Show that E 2 M. (b) If E is measurable and m(E) < ∞, then show that, for every e > 0, there exists a finite union U of disjoint open intervals such that m (EDU) < e. 2.3.P15. Give an example of a nonmeasurable set. 2.3.P16. Show that the outer measure m is not finitely additive. 2.3.P17. Let f be defined on [0, 1] by f(0) = 0 and f ðxÞ ¼ x sin 1x for x > 0. Show that the measure of the set {x: f(x) > 0} is 1 − (ln2)/p. 2.3.P18. Let E be Lebesgue measurable with 0 < m(E) < ∞ and let e > 0 be given. Then there exists a compact set K  E such that m(E\K) = m(E) − m(K) < e. In what follows, we provide a characterisation of the r-algebra generated by an algebra A in terms of what are called “monotone classes”. A family M of subsets of a nonempty set X is called a monotone class if it satisfies the following two conditions: (i) if A1  A2   and each Aj 2 M, then [ j  1 Aj 2 M; (ii) if B1 B2  and each Bj 2 M, then \ j  1 Bj 2 M. Any r-algebra is a monotone class. Let A  X, where X is any nonempty set and M ¼ fAg. Then M is a monotone class which is not a r-algebra. 2.3.P19. (a) If Y is any class of subsets of a nonempty set X, then show that there exists a smallest monotone class containing Y. We shall denote it by M0 ðYÞ. (b) If A is an algebra, show that SðAÞ ¼ M0 ðAÞ; that is, the r-algebra generated by an algebra is also the smallest monotone class containing the algebra. 2.3.P20. Show that (a) the cardinality of M is greater than c and (b) the Cantor set has a subset which is not a Borel set. 2.3.P21. Let F be an algebra of subsets of a set X. If A and B are subsets of X such that B, B \ Ac, A \ Bc all belong to F , show that A also belongs to F . 2.3.P22. Let f be a real-valued function on [a, b]. Suppose E  [a, b], f 0 exists and satisfies jf 0 ðxÞj  M for all x 2 [a, b]. Prove that m (f(E))  M m (E). 2.3.P23. (a) Let F and Y be subsets of a set X. Then show that Fc \ Y is the complement in Y of F \ Y.

2.3 Measurable Sets and Lebesgue Measure

71

(b) Let Y  X and G be a r-algebra of subsets of Y. Show that the family F 0 ¼ fF  X: F \ Y 2 Gg of subsets of X is a r-algebra. 2.3.P24. Let A be a family of subsets of a set X and F the r-algebra generated by it. Suppose Y  X and AY ¼ fA \ Y : A 2 Ag;

F Y ¼ fF \ Y: F 2 F g:

Show that F Y is the r-algebra of subsets of Y generated by AY .

2.4

Measurable Functions

The class of what are called measurable functions includes continuous and monotone functions among others, and plays a vital role in the theory of Lebesgue integration, which will be discussed in the sequel. The concept of measurability for functions depends only on a prescribed r-algebra of subsets of the domain set (in our case MX ¼ fX \ Y: Y 2 Mg or BX ¼ fX \ Y: Y 2 Bg, where X  R is measurable and M is the r-algebra of Lebesgue measurable subsets of R and B is the r-algebra of Borel measurable subsets of R). It is analogous to the concept of continuous functions, with which the reader is undoubtedly familiar. We recall the appropriate characterisation of continuous functions to emphasise the analogy between the definition of measurable functions and that of continuous functions: A real-valued function f with domain X is continuous if and only if for every open set V  R, the inverse image f−1(V) = {x 2 X: f(x) 2 V} is open in X, that is, there is an open set U  R such that f−1(V) = X \ U (see Theorem 1.3.21). Let X be a subset of R and a be an arbitrary real number. Suppose that f is a real-valued function defined on X. The following notations will be employed: Xðf  aÞ ¼ fx 2 X : f ðxÞ  ag; Xðf ¼ aÞ ¼ fx 2 X : f ðxÞ ¼ ag; Xðf  aÞ ¼ fx 2 X : f ðxÞ  ag; Xðf [ aÞ ¼ fx 2 X : f ðxÞ [ ag; Xðf \aÞ ¼ fx 2 X : f ðxÞ\ag: Before stating the next definition, we note the simple fact that, if X 2 M [resp. B], then the class fA  X : A 2 M ½resp:Bg of those subsets of X that are Lebesgue [resp. Borel] measurable is the same as the class of intersections MX ½resp:BX  ¼ fX \ Y : Y 2 M ½resp:Bg and is a r-algebra. MX [resp. BX ] is called the r-algebra of Lebesgue [resp. Borel] measurable subsets of X.

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2 Measure in Euclidean Space

Definition 2.4.1 Let F be a r-algebra of subsets of X  R. A real-valued function f defined on X is said to be F -measurable if, for every real number a, the set X(f > a) is in F . In particular, if X 2 M and if F is the r-algebra MX ¼ fX \ Y : Y 2 Mg of Lebesgue measurable subsets of X, then f is said to be Lebesgue measurable. Similarly, if X 2 B and if F is the r-algebra BX ¼ fX \ Y : Y 2 Bg of Borel measurable subsets of X, then f is said to be Borel measurable. From here onwards, both will be referred to as simply “measurable” when it is clear from the context which r-algebra is intended. It may be noted that every Borel measurable function is Lebesgue measurable. The converse does not hold. Indeed, let A be a Lebesgue measurable set which is not Borel measurable (see 2.3.P20). Then it follows from Example 2.4.4(b) below that vA is Lebesgue measurable but not Borel measurable. We now prove a proposition that shows that we could have modified the form of the set in defining measurability. Proposition 2.4.2 Let f be a real-valued function defined on a measurable subset X  R. Then the following statements about the function f are equivalent: (a) For every a 2 R, the set X(f > a) is measurable; (b) For every a 2 R the set X(f  a) is measurable; (c) For every a 2 R, the set X(f  a) is measurable; (d) For every a 2 R, the set X(f < a) is measurable. Proof Since the sets X(f > a) and X(f  a) are complements of each other, and the collection MX [resp. BX ] is closed under complementation (being a r-algebra), it follows that (a) and (b) are equivalent. Similarly, the statements (c) and (d) are equivalent. If (a) holds, then Xðf [ a  1nÞ is measurable for each n and 1 \

Xðf  aÞ ¼

1 n

Xðf [ a  Þ;

n¼1

so, the set X(f  a), being the intersection of a sequence of measurable sets, is measurable. Thus (a) implies (c). Again, since Xðf [ aÞ ¼

1 [ n¼1

1 n

Xðf  a þ Þ;

it follows that X(f > a), being the union of a sequence of measurable sets, is measurable. Thus (c) implies (a). h Corollary 2.4.3 A real-valued function f defined on a measurable subset X  R is measurable if and only if any one of the statements (a), (b), (c) or (d) in Proposition 2.4.2 holds. Proof An immediate consequence of the preceding definition and proposition. h We give below examples of measurable functions.

2.4 Measurable Functions

73

Examples 2.4.4 (a) Any constant function defined on a measurable set X  R is measurable. For, if f(x) = k for all x 2 X and a  k, then X(f > a) = £, whereas, if a < k, then X(f > a) = X. (b) Suppose X  R is measurable, A  X and vA is defined on X, that is, vA ðxÞ ¼ 1 if x 2 A and 0 otherwise. Then vA is a measurable function if and only if A is measurable. In fact, for an arbitrary real number a, the set {x 2 X: vA(x) > a} is X, A or ∅, depending on whether a < 0, 0  a < 1 or a  1 respectively. (c) If X  R is measurable and f : X ! R is continuous, then X(f > a) is open and is therefore a Borel set (see Definitions 2.3.18 and 2.4.1); this means f is Borel measurable and hence also Lebesgue measurable. (d) Let I be an interval and f : I ! R be monotonically increasing, that is, x < x′ ) f(x)  f(x′). Then f is Borel measurable, because I(f > a) is an interval; indeed, if x1 and x2 are any two points in it, then so is any y lying between x1 and x2. Similarly, a monotonically decreasing function is always Borel measurable. (e) Suppose X  R is a countable union [ k  1Xk of sets Xk, not necessarily disjoint. Then for any function f : X ! R, we obviously have Xðf [ aÞ ¼

1 [

Xk ðf [ aÞ:

k¼1

Therefore if each Xk is measurable and the restriction of f to each Xk is measurable, then f is measurable. As this applies when the sets Xk are not disjoint, it follows that the result remains true when they are finite in number. (f) Suppose X  R is measurable and let f : X ! R be measurable. Then the truncated function 8 < f ðxÞ jf ðxÞj  A fA ðxÞ ¼ A f ðxÞ [ A : A f ðxÞ\A can be seen to be measurable. Since its restriction to each of the three measurable sets (whichever ones may be nonempty) fx 2 X : f ðxÞ [ Ag;

fx 2 X : A  f ðxÞ  Ag

and

fx 2 X : f ðxÞ\  Ag;

having union X, is measurable, it follows from part (e) above that fA is measurable. Alternatively, fA = min{A, max{− A, f}} and Theorem 2.4.6(e) below implies that fA is measurable. Proposition 2.4.5 If f is a measurable function defined on a measurable subset X  R, then the set X(f = a) is measurable for each real number a.

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2 Measure in Euclidean Space

Proof For a real number a, the sets X(f  a) and X(f  a) are both measurable, as f is measurable. Therefore, so is their intersection X(f  a) \ X(f  a). But the intersection is precisely X(f = a). h Remark The converse of Proposition 2.4.5 is, however, not true: Let A  [0, 1) be a nonmeasurable set described in the solution to Problem 2.3.P15. Consider the function f defined on [0, 1) by  f ðxÞ ¼

x2 x2

x2A x 2 ½0; 1ÞnA

Then for each real number a, the set {x 2 [0, 1): f(x) = a} consists of at most two elements and is therefore measurable. However, the set {x 2 [0, 1): f(x)  0} is A, which is not measurable. Consequently, the function f is not measurable. We shall next show that certain simple algebraic combinations of measurable functions are measurable (“algebra of measurable functions”). Their limits are deferred to a later section. Theorem 2.4.6 Let X be a measurable subset of R and suppose that f, g: X!R are measurable. Then the following are also measurable: (a) (b) (c) (d) (e)

af ða 2 RÞ; j f j; f + g; fg; min{f, g} and max{f, g}.

Proof (a) If a = 0, the statement is trivial. If a > 0, then a fx 2 X : af ðxÞ [ ag ¼ fx 2 X : f ðxÞ [ g a and the set on the right-hand side is measurable. The case when a < 0 is similar. (b) If a < 0, then the set fx 2 X : j f ðxÞj [ ag ¼ X, whereas, if a  0, then fx 2 X : j f ðxÞj [ ag ¼ fx 2 X : f ðxÞ [ ag [ fx 2 X : f ðxÞ\  ag and since each of the sets on the right is measurable, so is their union. (c) By hypothesis, if r is a rational number, then Ar ¼ fx 2 X : f ðxÞ [ rg \ fx 2 X : gðxÞ [ a  rg is measurable. Since

2.4 Measurable Functions

75

fx 2 X : ðf þ gÞðxÞ [ ag ¼

[ r2Q

Ar

and a countable union of measurable sets is measurable, it follows that the set on the left-hand side is measurable. (d) We shall first show that, if f is measurable, then so is f 2 . If a < 0, then fx 2 X : f ðxÞ2 [ ag ¼ X; which is a measurable set; if a  0, then pffiffiffi pffiffiffi fx 2 X : f ðxÞ2 [ ag ¼ fx 2 X : f ðxÞ [ ag [ fx 2 X : f ðxÞ\  ag and since each of the sets on the right is measurable, the one on the left is measurable. It follows that f 2 is measurable. This fact, together with (a), (c) and the equality 1 fg ¼ ½ðf þ gÞ2  ðf  gÞ2 ; 2 implies that fg is measurable. (e) For an arbitrary real number a, fx 2 X : maxff ; ggðxÞ [ ag ¼ fx 2 X : f ðxÞ [ a or gðxÞ [ ag ¼ fx 2 X : f ðxÞ [ ag [ fx 2 X : gðxÞ [ ag: Since each of the sets on the right is measurable, the one on the left is measurable. It follows that max{f, g} is measurable. The argument for min{f, g} is similar. h Remarks 2.4.7 (a) If f is any real-valued function defined on X, let f + and f − be the nonnegative functions on X defined as f þ ðxÞ ¼ maxff ðxÞ; 0g and

f  ðxÞ ¼ maxff ðxÞ; 0g:

The function f+ is called the positive part of f and f − is called the negative part of f. The following identities are obvious: f ¼ f þ  f

and

j f j ¼ f þ þ f :

It is a consequence of these identities that

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2 Measure in Euclidean Space

1 f þ ¼ ðj f j þ f Þ and 2

1 f  ¼ ðj f j  f Þ: 2

From Theorem 2.4.6, we infer that f is measurable if and only if f + and f − are. (b) The converse of (b) in Theorem 2.4.6 is not true. If A  [0, 1) is a nonmeasurable subset, define  f ðxÞ ¼

1 1

x2A x 2 ½0; 1ÞnA:

Then f is not measurable, because the set {x 2 [0, 1): f(x) > 0} = A is nonmeasurable. But j f j is the constant function 1 and is therefore measurable. (c) The above example also shows that f 2 can be measurable when f is not. Problem Set 2.4 2.4.P1. Let the function f : ½0; 1 ! R be defined by f ðxÞ ¼ 1x if 0 < x  1, f(0) = 0. Show that f is measurable. 2.4.P2. Show that, if f is a real-valued function on a measurable subset X  R such that X(f  r) is measurable for every rational number r, then f is measurable. 2.4.P3. Let X  R be measurable. Without using Theorem 2.4.6, show that if f and g are measurable functions defined on X, then the set {x 2 X: f(x) > g(x)} is measurable. 2.4.P4. Let f be a real-valued function defined on a measurable subset X  R. Then f is measurable if and only if, for every open set V  R, f 1 ðVÞ ¼ fx 2 X : f ðxÞ 2 Vg is measurable. 2.4.P5. Let f be a measurable function defined on a measurable subset X  R and / be defined and continuous on the range of f. Then / f is a measurable function on X. 2.4.P6. Show that any function f defined on a set X of Lebesgue measure zero is Lebesgue measurable. 2.4.P7. Let X  R and  f : X ! R be any function. Show that the family of subsets of X given by F ¼ f 1 ðVÞ: V 2 B is a r-algebra. (The same is true if B is replaced by M.) 2.4.P8. Let X  R and f : X ! R be any function. If F is a r-algebra of subsets of X, show that the family of subsets of R given by G ¼ fV  R : f 1 ðVÞ 2 F g is a r-algebra. Hence show that, if f is Borel measurable, then V 2 B ) f 1 ðVÞ 2 fX \ U : U 2 Bg. Is it true that, if f is Lebesgue measurable, then V 2 B ) f 1 ðVÞ 2 fX \ U : U 2 Mg?

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77

2.4.P9. Let f be a real-valued measurable function defined on a measurable set X  R. Prove that (a) if a > 0, then the function j f ja is measurable; (b) if f(x) 6¼ 0 on X and a < 0, then j f ja is measurable. 2.4.P10. A complex-valued function f with domain a measurable set X  R is said to be measurable if its real and imaginary parts, which are real-valued functions on X, are measurable. Prove that a complex-valued function is measurable if and only if f −1(V) is measurable for every open set V  C (the complex plane).

2.5

Extended Real-Valued Functions

The preceding discussion of measurable functions pertained to real-valued functions. However, in dealing with a sequence of measurable functions, we need to form its supremum, infimum, limit superior, limit inferior or limit. It is then convenient to work with the extended real number system R [ f1g [ f1g as the range space. With this in mind, we proceed to define measurability for extended real valued functions. Much of what we say is intended for Borel as well as Lebesgue measurability, except when we explicitly mention only one of them. Definition 2.5.1 Let f be an extended real-valued function defined on a measurable subset X  R. Then f is said to be measurable if, for each real number a, the set X(f > a) is measurable. Proposition 2.5.2 An extended real-valued function f on a measurable set X  R is measurable if and only if (a) the sets A = X(f = ∞), B = X(f = −∞) are measurable and (b) the restriction of f to X\(A [ B), which is real-valued, is measurable. Proof We take up the “only if” part first. Observe that, for each positive integer n, the T1 set X(f > n) is measurable by definition; hence, so is their intersection The set n¼1 Xðf [ nÞ, which is A = X(f = ∞). This shows that A is measurable. S B = X(f = −∞) is measurable because it is the complement of n2Z X(f > n). Moreover, for each real number a, the set {x 2 X\(A [ B): f(x) > a} is measurable, because it equals the intersection {x 2 X: f(x) > a} \ {X\(A [ B)}. We have thus proved that if an extended real-valued function f is measurable, then the sets A = X(f = ∞), B = X(f = −∞) and {x 2 X\(A [ B): f(x) > a}, where a is an arbitrary real number, are measurable. This proves the “only if” part. The converse is immediate from the fact that, for any real number a, we have Xðf [ aÞ ¼ fx 2 XnðA [ BÞ : f ðxÞ [ ag [ A:

h

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2 Measure in Euclidean Space

Remarks 2.5.3 (a) It is a consequence of Theorem 2.4.6 and Proposition 2.5.2 that, if f and g are extended real-valued measurable functions, then so are cf ðc a real numberÞ;

j f j;

minff ; gg

and

maxff ; gg;

keeping in mind the conventions (see Chap. 1) that 0  (±∞) = 0. (b) If f and g are extended real-valued measurable functions, then so are f + g and fg. In connection with the measurability of f + g, we proceed as follows. Consider the sets A = X(f = ∞), B = X(f = − ∞), C = X(g = ∞) and D = X(g = −∞). The domain of definition of f + g is Y = X\((A \ D) [ (B \ C)). By Proposition 2.5.2, the sets A, B, C and D are all measurable and hence the domain of definition of f + g is also measurable. To prove the measurability of f + g by using Proposition 2.5.2, we first note that the sets, Yðf þ g ¼ 1Þ ¼ ðAnDÞ [ ðCnBÞ ¼ E;

say,

and Yðf þ g ¼ 1Þ ¼ ðBnCÞ [ ðDnAÞ ¼ F;

say,

are both measurable. On the complement of their union, i.e., Y\(E [ F), the function f + g is real-valued and therefore f and g are also real-valued, so that Theorem 2.4.6 is applicable to them on the domain Y\(E [ F). We thereby obtain the measurability of the restriction of f + g to Y\(E [ F). Since E and F have already been seen to be measurable, an application of Proposition 2.5.2 to f + g on the measurable set Y leads to the measurability of the function. We next consider the measurability of the product. For any real number a < 0, fx : f ðxÞgðxÞ [ ag 8 fx : f ðxÞ; gðxÞ are finite and f ðxÞgðxÞ [ ag < ¼ [ fx : f ðxÞ; gðxÞ are of the same sign and at least one is infiniteg : [ fx : f ðxÞ ¼ 0 or gðxÞ ¼ 0 g: By Proposition 2.5.2 and Theorem 2.4.6, all the sets on the right-hand side are measurable. If a  0, then fx : f ðxÞgðxÞ [ ag  fx : f ðxÞ; gðxÞare finite and f ðxÞgðxÞ [ ag ¼ [ fx : f ðxÞ; gðxÞare of the same sign and at least one is infiniteg: Again by Proposition 2.5.2 and Theorem 2.4.6, all the sets on the right-hand side are measurable.

2.5 Extended Real-Valued Functions

79

If f is any extended real-valued function defined on X, the positive and negative parts f + and f − respectively are defined exactly as for real-valued functions: f þ ðxÞ ¼ maxff ðxÞ; 0g

and

f  ðxÞ ¼ maxff ðxÞ; 0g:

The following identities continue to be valid: f ¼ f þ  f ;

j f j ¼ f þ þ f ;

1 f þ ¼ ðj f j þ f Þ 2

1 and f  ¼ ðj f j  f Þ: 2

As before, when f is measurable, the associated functions f + and f − are also measurable. We recall the definitions of the limit superior and limit inferior of a sequence {an}n  1 in the extended real number system R ¼ R [ f1g [ f1g: lim sup an ¼ inf ðsup ak Þ ¼ limðsupfan ; an þ 1 ; . . .gÞ; n1 kn

n

lim inf an ¼ sup ð inf ak Þ ¼ limðinffan ; an þ 1 ; . . .gÞ: n1 kn

n

If {an}n  1 converges or diverges to ±∞, then evidently, lim an ¼ lim sup an ¼ lim inf an : n

From the definition we easily get lim sup an ¼  lim infðan Þ and an  bn 8n 2 N ) lim inf an  lim inf bn ) lim sup an  lim sup bn : Suppose {fn}n  1 is a sequence of extended real-valued functions on a domain X  R. Then lim sup fn and lim inf fn are functions defined on the domain X as ðlim sup fn ÞðxÞ ¼ lim supðfn ðxÞÞ and

ðlim inf fn ÞðxÞ ¼ lim infðfn ðxÞÞ:

If f(x) = limnfn(x), the limit being assumed to exist at every x 2 X, then we call f the (pointwise) limit of the sequence {fn}n  1 and denote it by limnfn or limfn. We also define sup1  i  nfi and inf1  i  nfi respectively as ð sup fi ÞðxÞ ¼ sup ðfi ðxÞÞ 1in

1in

and ð inf fi ÞðxÞ ¼ inf ðfi ðxÞÞ; 1in

1in

80

2 Measure in Euclidean Space

and sup fn and inf fn respectively as ðsup fn ÞðxÞ ¼ supðfn ðxÞÞ and ðinf fn ÞðxÞ ¼ inf ð fn ðxÞÞ: Theorem 2.5.4 If {fn}n  1 is a sequence of extended real-valued measurable functions on the same measurable set X  R, then the following hold: (a) (b) (c) (d) (e) (f)

sup1  i  n fi is measurable for each n; inf1  i  n fi is measurable for each n; sup fn is measurable; inf fn is measurable; lim sup fn is measurable; lim inf fn is measurable.

Proof (a) For each real number a, Xð sup fi [ aÞ ¼ 1in

n [

Xðfi [ aÞ;

i¼1

the right-hand side, being a countable (actually finite) union of measurable sets, is measurable. It follows that the left-hand side is measurable. Since this is true for each real number a, the function sup1  i  n fi is measurable. (b) By Remark 2.5.3, each −fn is measurable. Moreover, inf1  i  nfi = −sup1  i  n(−fi). Therefore it follows from (a) that inf1  i  nfi is also measurable. (c) For each real number a, 1 [ Xðsup fn [ aÞ ¼ Xðfn [ aÞ; n¼1

the right-hand side, being a countable union of measurable sets, is measurable. It follows that the left-hand side is measurable. Since this is true for each real number a, the function supfn is measurable. (d) Follows from (c) and Remark 2.5.3, because inffn = −sup(−fn). (e) limsupfn = infn  1(supk  nfk) and is therefore measurable by (c) and (d). (f) lim inf fn = −lim sup (−fn) and is therefore measurable by (e) and Remark 2.5.3. h Corollary 2.5.5 Let {fn}n  1 be a sequence of extended real-valued measurable functions on a measurable set X  R. If limfn exists, then it is measurable. Proof An immediate consequence of the theorem above. h The definition of the Lebesgue integral is based on what are called “simple” functions. They are a generalisation of step functions, which take only finitely many values and each value is taken on an interval; thus step functions are linear

2.5 Extended Real-Valued Functions

81

combinations of characteristic functions of intervals. For example, if s:[1, 3]!R is the step function which equals 5 on [1, 2), 8 on (2, 3] and s(2) = 6, then s = 5v[1,2) + 6v[2,2] + 8v(2,3]. The generalisation is achieved by replacing intervals by measurable sets, so that a simple function is a linear combination of characteristic functions of measurable sets. However, we require the values to be nonnegative and finite. The formal description is as follows. Definition 2.5.6 A measurable function defined on a measurable subset of R is called a simple function if its range consists of finitely many points of [0, ∞). Examples 2.5.7 (a) The characteristic function vA of a nonempty measurable set A  R is a simple function with range consisting of the two numbers 0 and 1. If A is empty or equals R, then the range consists of a single number, namely, 0 or 1. (b) Suppose A and B are nonempty measurable subsets of R, both distinct from R. Then each of vA and vB has range consisting of two numbers; but the range of the sum vA + vB consists of the single number 1 if A \ B ¼ £ and A [ B ¼ R two numbers 0 and 1 if A \ B ¼ £ and A [ B 6¼ R two numbers 0 and 2 two numbers 1 and 2

if A ¼ Bðin which case A [ B 6¼ RÞ if A [ B ¼ R and A \ B 6¼ £

three numbers 0; 1; 2

if A [ B 6¼ R; A 6¼ B and A \ B 6¼ £:

(c) The sum of two functions with finite range has a finite range. Since it is also true that the sum of two measurable functions is measurable, we conclude that the sum of two simple functions is simple. Therefore a finite sum R1  j  najvAj, where A1, …, An are measurable and a1, a2, …, an are nonnegative real numbers, is always a simple function. As illustrated in the preceding example, the values of the simple function need not be among the numbers aj. Remarks 2.5.8 (a) An extended real-valued measurable function on a measurable set with finite range is also sometimes called a simple function. The above situation (with real and nonnegative values) is the one that will mainly interest us in the sequel. (b) The value ∞ is explicitly excluded from the range of a simple function. (c) If a1, a2, …, an are the distinct values assumed by a simple function s and Aj denotes the set X(s = aj), then the sets Aj are nonempty, disjoint with [ 1  j  nAj = X, and are measurable (by Proposition 2.4.5); moreover, sðxÞ ¼

n X j¼1

aj vAj ðxÞ:

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2 Measure in Euclidean Space

Conversely, if the sets Aj are nonempty, disjoint with [ 1  j  nAj = X and are measurable, and if a1, a2, …, an are distinct, then the simple function s ¼ P 1  j  n aj vAj has the coefficients a1, a2, …, an as its distinct values and satisfies X(s = aj) = Aj. When these conditions are fulfilled, we shall refer to the sum as the canonical representation of s. The range s(X) of the function is {aj: 1  j  n}. (d) In Example 2.5.7(b), the canonical representations of vA + vB in the five cases displayed there are respectively 1  vR ; 0  vðA [ BÞc þ 1  vA [ B ; 2  vA þ 0  vAc ; 0  vðA [ BÞc þ 1  vADB þ 2  vA \ B :

1  vðA \ BÞc þ 2  vA \ B ;

This shows that simple functions have simple representations as linear combinations of characteristic functions, and yet they are not so simple to add! We have seen in Corollary 2.5.5 that the limit of a sequence of measurable functions is measurable. We shall next show that a measurable nonnegative extended real-valued function is the limit of an increasing sequence of nonnegative simple (measurable) functions. Theorem 2.5.9 Let f be a measurable extended nonnegative real-valued function on a measurable subset X of R. Then there exists a sequence {sn}n  1 of simple functions such that for every x 2 X, (a) 0  s1(x)  s2(x)    f(x), (b) sn(x) ! f(x) as n ! ∞. Moreover, if f is bounded, then {sn}n  1 converges to f uniformly on X. Proof Since f is an extended nonnegative real-valued function, its range is contained in [0, ∞]. For each positive integer n, split [0, ∞] by the sequence of points 0;

1 2 n2n  1 ; n; 1; ; n ; . . .; n 2 2 n

so that X ¼ f 1 ð½0; 1Þ ¼

n2n [

k1 k f 1 ð½ n ; n ÞÞ [ f 1 ð½n; 1Þ: 2 2 k¼1

For 1  k  n2n, set En;k ¼ f 1 ð½

k1 k ; ÞÞ and 2n 2n

Fn ¼ f 1 ð½n; 1Þ:

Note that, for n 2 N and 1  k  n2n, the sets En,k and Fn are measurable by hypothesis and are also disjoint. Put

2.5 Extended Real-Valued Functions

sn ¼

83

n2n X k1 k¼1

2n

vEn;k þ nvFn

for n 2 N:

It is clear that 0  sn(x)  f(x) everywhere. We shall now show for any x 2 X that sn ðxÞ  sn þ 1 ðxÞ;

ð2:21Þ

which will prove (a). To this end, first consider any x 2 X such that f(x)  n + 1. Then x 2 Fn+1  Fn, whence sn(x) = n and sn+1(x) = n + 1, so that (2.21) holds. Next, consider an x 2 X such that n  f(x) < n + 1. Then x 2 Fn and therefore sn(x) = n. Also, n2n+1  f(x)2n+1 < (n + 1)2n+1, so that k − 1  f(x)2n+1 < k for some integer k satisfying n2n+1 < k  (n + 1)2n+1. (Take k to be 1 plus the integer part of f(x)2n+1.)This implies that x 2 En+1,k, where n2n+1 < k  (n + 1)2n+1. Consequently, sn+1(x) = (k − 1)/2n+1  n. Since sn(x) = n as already noted, we conclude that (2.21) holds if n  f(x) < n + 1. Now consider the remaining case, i.e., f(x) < n. Here, 0  f(x)2n < n2n so that k − 1  f(x)2n < k for some integer k (again, 1 plus the integer part of f(x)2n) satisfying 0 < k  n2n. This implies that x 2 En,k, where 0 < k  n2n. Consequently, sn(x) = (k − 1)/2n. Note that the inequality k − 1  f(x)2n < k holds if and only if either

2ðk  1Þ  f ðxÞ2n þ 1 \2k  1

or

2k  1  f ðxÞ2n þ 1 \2k:

This is equivalent to either

x 2 En þ 1;2k1

or

x 2 En þ 1;2k ;

bearing in mind that 1  2k − 1 < 2k  (n + 1)2n+1. If x 2 En+1,2k-1, then sn+1(x) = (2k − 2)/2n+1 = (k − 1)/2n. Since sn(x) = (k − 1)/2n as already noted, we find that (2.21) holds if x 2 En+1,2k−1. On the other hand, if x 2 En+1,2k, then sn+1(x) = (2k − 1)/2n+1 > (k − 1)/2n = sn(x). We can therefore conclude that (2.21) always holds even in this remaining case. If x is such that f(x) < ∞, then x 2 Fcn for all sufficiently large n and therefore sn(x)  f(x) − 2−n. (Since x 2 En,k for some k, it follows that sn(x) = (k − 1)/2n. As f(x) < k/2n, we have f(x) − 2−n < (k − 1)/2n = sn(x).) So, sn(x) increases monotonically to f(x). If f(x) = ∞, then x 2 \ n  1Fn, so that sn(x) = n for all n and again sn(x) increases monotonically to f(x). Finally, if 0  f(x)  M for every x 2 X, then sn(x)  f(x) − 2−n for every n  n0, where n0  M, that is, 0  f(x) − sn(x)  2−n for every x 2 X and for every n  n0, where n0  M. This implies that {sn}n  1 converges to f uniformly on X. h

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2 Measure in Euclidean Space

We now prepare to prove that, if two functions differ on a set of Lebesgue measure zero and one of them is Lebesgue measurable, then so is the other. This is best phrased in terminology (formally introduced later) that is useful in other contexts as well. We shall begin by illustrating the contexts. The function f defined on R by f(x) = 1 for rational x and 0 for irrational x has the property that f(x) = 0 except on the set of rationals, which has Lebesgue measure zero. The phenomenon is described by saying that f is zero “almost everywhere”. The greatest integer function f(x) = [x] is continuous except on the set Z of integers, which has Lebesgue measure zero. We say that this function is continuous “almost everywhere”. The two properties considered in the preceding paragraph, namely vanishing at x and being continuous at x, are “hereditary” properties in the sense that: if either of them holds on some set, then it holds on every subset thereof. The same is true of uniform continuity. More generally, consider a hereditary property that holds everywhere on a measurable set X  R except on a subset E 2 BX with m(E) = 0. Since B  M, it is trivial that the property holds everywhere on a measurable set X  R except on a set E1 2 MX with m(E1) = 0. The converse is also true: suppose a hereditary property holds everywhere on a measurable set X  R except on a subset E1 2 MX with m (E1) = 0. Then E1 2 MX and by the equivalence (a) , (c) in Proposition 2.3.24, there exists a Gd -set G, which has to be in B by virtue of being Gd , such that G E1 and m(G) = m(E1) = 0. Now, E1  X \ G 2 BX and m(X \ G) = 0; by heredity, the property holds everywhere on X except X \ G. Thus E = X \ G serves as the subset of X such that the hereditary property holds everywhere on the measurable set X except on the subset E 2 BX with m(E) = 0. The upshot of this discussion is that, in the forthcoming definition of “almost everywhere”, it makes no difference whether the exceptional set is in MX or in BX . Definition 2.5.10 If a hereditary property holds everywhere on a measurable set X except on a subset E in either MX or BX having m(E) = 0, we say that it holds almost everywhere. The phrase is abbreviated as a.e. When the property is described in terms of an explicit x 2 X, we can say that it holds for almost all x. We note that convergence of a sequence of functions, whether pointwise or uniform, is also a hereditary property. Therefore it makes sense to speak of pointwise or uniform convergence almost everywhere of a sequence of functions. The exceptional set referred to in the definition is not required to be nonempty; in other words, a property that holds everywhere also holds a.e. Examples 2.5.11 (a) Let X = [0, 1] and suppose f : X ! R is 0 everywhere, while g : X ! R is 1 on Q \ X but 0 on the rest of X. Then the property that f(x) = g(x) holds everywhere on X except on the subset Q \ X. Since this subset has measure zero, we have f = g a.e.

2.5 Extended Real-Valued Functions

85

(b) Let X = [0, 1] and {fn}n  1 be the sequence of functions on X given by fn(x) = xn. This sequence converges to 0 except at 1. Since the subset {1} consisting of the exceptional point 1 has measure 0, we have lim fn ¼ 0 a.e. on X. n!1

(c) Let X = [−1, 1] and {fn}n  1 be the sequence of functions on X given by fn(x) = 1/(1 + x2n). Then lim fn ðxÞ ¼ 1 if x 2 (−1, 1). Since the complement of n!1

(−1, 1) in X, which consists of the two numbers ±1, has measure zero, we have lim fn ¼ 1 a.e. on X. It is a fact that l lim fn ðxÞ ¼ 12 when x = ±1, but this fact

n!1

n!1

has no bearing on the assertion that lim fn ¼ 1 a.e. on X. n!1

Proposition 2.5.12 Let f and g be continuous functions on an interval I  R of positive length. Suppose f = g almost everywhere on X. Then f = g everywhere on I. Proof Suppose x0 2 X is such that f(x0) 6¼ g(x0). Then {x 2 X: f(x) 6¼ g(x)} is nonempty and the continuity of f − g at x0 implies that it is an open subset of I. Now, a nonempty open subset of any interval of positive length must contain an interval of positive length. So, mðfx 2 X : f ðxÞ 6¼ gðxÞgÞ [ 0; which contradicts the hypothesis that f = g a.e.

h

Proposition 2.5.13 Let f and g be two functions on a Lebesgue measurable subset X  R such that f = g a.e. on X and f is Lebesgue measurable. Then g is Lebesgue measurable. Proof Let A = X(f > a) and B = X(g > a). Note that ADB  Xðf 6¼ gÞ: Since f is measurable, it follows that A is Lebesgue measurable. Also, m ðADBÞ  m ðXðf 6¼ gÞÞ ¼ 0: In view of Problem 2.3.P5, this implies that the set B is Lebesgue measurable, i.e. the function g is Lebesgue measurable. h Proposition 2.5.14 If a real-valued function f defined on a Lebesgue measurable subset X  R is continuous a.e., then f is Lebesgue measurable on X. Proof Let Y = {x 2 X: f is not continuous at x}. Note that m (Y) = 0 and so Y is measurable. Let a be an arbitrary real number. Since f is continuous on the measurable set X\Y, it follows that Z = {x 2 X\Y: f(x) > a} is measurable [analogously to Example 2.4.4(c)]. Also, the set W = {x 2 Y: f(x) > a}, being a subset of the set Y of measure zero, is measurable. Consequently, the union Z [ W is measurable; but this union is the same as {x 2 X: f(x) > a}. h

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2 Measure in Euclidean Space

Problem Set 2.5 2.5.P1. Let g(x) = f 0 (x) exist for every x 2 [a, b]. Prove that g is Lebesgue measurable. 2.5.P2. Prove that the set of points at which a sequence of measurable functions converges or diverges to ±∞ is a measurable set. 2.5.P3. Show that sup{fa: a 2 K} is not necessarily measurable even if each fa is. 2.5.P4. If f is a real-valued Lebesgue measurable function defined on a Borel set X  R, then there exists a Borel measurable function g on X such that f(x) = g(x) a.e. 2.5.P5. Let f and g be extended real-valued measurable functions defined on a measurable subset X  R. Then their product fg is also measurable. 2.5.P6. P Let s be a simple function on a measurable set X such that s ¼ 1  i  p ai vAi , where i 6¼ i0 ) Ai \ Ai0 ¼ £

and

p [

Ai ¼ X:

i¼1

(a) If c1, c2, …, cn are the distinct elements of the range of s, show that fcj : 1  j  ng ¼ fai : 1  i  p; Ai 6¼ £g: (b) Show that, for any j, 1  j  n, there must exist some index i (1  i  p) for which the coefficient ai equals cj and, at the same time, Ai 6¼ ∅. If for each j (1  j  n), Nj is the set of all such indices i corresponding to a given j, namely, Nj ¼ fi : 1  i  p; ai ¼ cj

and

Ai 6¼ £g;

  show that [ i2Ni Ai ¼ X s ¼ cj . P (c) Finally, show that the canonical representation of s is 1  j  n cj vBj , where Bj = [ i2NjAi. P 2.5.P7. Let 1  i  p ai vAi be the canonical representation of a simple function s. If P 0 < a < ∞, show that 1  i  p ðaai ÞvAj is the canonical representation of as. 2.5.P8. Give an example when (f + g)+ is not the same as f + + g+. 2.5.P9. If f and g are extended real-valued functions and g  0, show that (fg)+ = f +g and (fg)− = f −g. 2.5.P10. Let f be a bounded measurable function defined on a bounded closed interval [a, b] and let e > 0 be arbitrary. Show that there exists a step function g on [a, b] such that mðfx 2 ½a; b : j f ðxÞ  gðxÞj  egÞ\e:

2.6 Egorov’s and Luzin’s Theorems

2.6

87

Egorov’s and Luzin’s Theorems

The theorems mentioned in the title of this section are important in their own right. Egorov’s Theorem will be used subsequently. We begin with a preliminary result which is of independent interest. Proposition 2.6.1 Suppose {fn}n  1 is a sequence of functions on a measurable set X and lim fn ¼ f a:e: n!1

(a) If each fn is Lebesgue measurable, then so is f. (b) If X is a Borel set and each fn is Borel measurable, then there exists a Borel measurable function g on X such that f = g a.e. Proof (a) Indeed, f = limsupfn a.e. and the result follows from Theorem 2.5.4 and Proposition 2.5.13. (b) By part (a), f is Lebesgue measurable. Now use Problem 2.5.P4. h In the rest of this section, “measure” and “measurable” will be understood exclusively in the sense of Lebesgue, as there will be no occasion to work with Borel measure or measurability. The first main result here relates the notions of convergence almost everywhere and uniform convergence. Theorem 2.6.2 (Egorov). Suppose that a sequence {fn}n  1 of measurable functions converges a.e. to f on E, where m(E) < ∞. Then, for any e > 0, there exists a measurable set Ee  E such that m(E\Ee) < e and the sequence {fn}n  1 converges to f uniformly on Ee. Proof According to Proposition 2.6.1, f is measurable. Set En;p ¼

\

1 fx 2 E : jfi ðxÞ  f ðxÞj\ g; p in

that is, En,p is the set of all x for which jfi ðxÞ  f ðxÞj\

1 p

for all i  n:

Let Ep ¼ [ n  1 En;p . It is clear from the definition of En,p that E1;p  E2;p     for fixed p. Therefore, by Proposition 2.3.21 and Problem 2.3.P11, for arbitrary p and e > 0, there exists an integer n(p) depending on p such that

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2 Measure in Euclidean Space

mðEp nEnðpÞ;p Þ\

e : 2p

We set Ee ¼

1 \

EnðpÞ;p

p¼1

and show that Ee is the required set. For an arbitrary η > 0, choose p so large that 1 p < η. For x 2 Ee, we have 1 jfi ðxÞ  f ðxÞj\ \g p for i > n(p). This proves the uniform convergence of {fn}n  1 to f on Ee. It remains to show that m(E\Ee) < e. In order to show this, we begin by noting that m(E\Ep) = 0 for every p. Indeed, if x0 2 E\Ep, then jfi ðx0 Þ  f ðx0 Þj 

1 p

for infinitely many values of i, that is, the sequence {fn}n  1 does not converge to f at x0. Since fn ! f a.e. by hypothesis, it follows that m(E\Ep) = 0. Consequently, mðEnEnðpÞ;p Þ ¼ mðEp nEnðpÞ;p Þ\

e : 2p

Therefore mðEnEe Þ ¼ mðEn

1 \

EnðpÞ;p Þ ¼ mð

p¼1



1 X p¼1

mðEnEnðpÞ;p Þ\

1 [

ðEnEnðpÞ;p ÞÞ

p¼1 1 X e ¼ e: 2p p¼1

This completes the proof. h The requirement that m(E) < ∞ cannot be dropped in Theorem 2.6.2. In other words, there is a measurable set E with m(E) = ∞ and a sequence {fn}n  1 of measurable functions converging a.e. to f on E such that for some η > 0 and every measurable set Eη  E satisfying m(E\Eη) < η, the sequence {fn}n  1 fails to converge uniformly on Eη. To demonstrate this, we take E ¼ R þ ¼ fx 2 R : x  0g, fn ¼ v½n1;n and g ¼ 12. Then {fn}n  1 converges everywhere to the identically zero function. Note

2.6 Egorov’s and Luzin’s Theorems

89

that when n > m + 1, the functions fn and fm are characteristic functions of disjoint sets; this disjointness has the consequence that if x 2 ½n  1; n and n [ m þ 1; then jfn ðxÞ  fm ðxÞj ¼ 1:

ð2:22Þ

Now consider an arbitrary measurable set Eη  E satisfying m(E\Eη) < η. We have to show that the sequence {fn}n  1 fails to converge uniformly on Eη. In order to prove this, we first note that Eη \ [n − 1, n] 6¼ ∅ for every n (for otherwise, [n − 1, n] would be contained in E\Eη, which has measure less than 1n). Let n0 be an arbitrary positive integer. Choose n > m + 1 > n0 and x 2 Eη \ [n − 1, n]. Then by (2.22), we have jfn ðxÞ  fm ðxÞj ¼ 1, ruling out uniform convergence on Eη. The definition of a measurable function is fashioned after a characterisation of a continuous function. In what follows, we shall prove a theorem due to Luzin, which shows that a measurable function can in a certain sense be approximated by a continuous function. The reader who chooses to skip the rest of this section will experience no loss of continuity, because the results will not be used later in this book. We begin with a lemma which says that every measurable function that is finite a.e. becomes bounded when we disregard a set of arbitrarily small measure. Lemma 2.6.3 Let a measurable function f defined on a set E of finite measure be finite a.e. Then for any e > 0, there exists a bounded measurable function g such that m(E(f 6¼ g)) < e. Proof For each n, let En ¼ Eðj f j [ nÞ and E1 ¼ Eðj f j ¼ 1Þ. By hypothesis, m(E∞) = 0. In view of the obvious relations E1 E2 E3   

and

E1 ¼

1 \

En

n¼1

and the fact that m(E) < ∞, Proposition 2.3.21(b) shows that mðEn Þ ! mðE1 Þ

as n ! 1:

Hence there exists an n0 such that m(En0 ) < e. Define a function g on E by  gðxÞ ¼

f ðxÞ 0

if x 2 EnEn0 if x 2 En0 :

The function g is then measurable and E(f 6¼ g)  En0 . So, m(E(f 6¼ g))  m(En0 ) < e. Also, the definition of En0 ensures that g is bounded above in absolute value by n0. h Lemma 2.6.4 Let the sets {Fi}1  i  n be closed and disjoint. If the function f defined on their union F is constant on each of the sets Fi, then it is continuous.

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2 Measure in Euclidean Space

Proof Let {xk}k  1 be a sequence of points in F such that xk ! x. Since F is closed, x 2 F and hence x 2 Fm for some m, 1  m  n. Since Fi \ Fj = ∅, i 6¼ j, it follows that x 62 Fm′ for m′ 6¼ m. Since [ i6¼mFi is closed and does not contain x, no subsequence of {xk}k  1 can lie in it. Thus all terms of the sequence from some index k0 onwards lie in Fm. Since f is constant on Fm, we have f(x) = f(xk) for k  k0, whence f(x) = limf(xk). h Lemma 2.6.5 Let F be a closed set contained in the closed interval [a, b]. If f is a function defined and continuous on the set F, then there exists a continuous function g on [a, b] such that g(x) = f(x) for all x 2 F and max jgðxÞj ¼ maxjf ðxÞj. Proof Let [a, b] be the smallest closed interval containing F. If a function g of the required kind has been obtained on [a, b], then define g on [a, b]\[a, b] by  gðxÞ ¼

f ðaÞ f ðbÞ

if x 2 ½a; aÞ if x 2 ðb; b:

Then g is the required function on [a, b]. So, without loss of generality, assume [a, b] is the smallest closed interval containing F; in particular, a, b 2 F. If F = [a, b], then there is nothing to prove. Suppose that F  [a, b]. Then, [a, b]\F = (a, b)\F, being open in R, is a countable union of disjoint open intervals (c, d) with endpoints in F. For the rest of this proof, we shall call them component intervals. On F, define g(x) = f(x). Then for each component interval (c, d), we have g(c) = f(c) and g(d) = f(d); define g on (c, d) by a straight line graph. Note that f is bounded because F is compact (see Theorem 1.3.20) and hence that g is bounded (with the same bounds as f); moreover, it is defined on the entire interval [a, b]. We shall show that g is continuous from the right on [a, b). Continuity from the left on (a, b] is proved similarly. Within each component interval (c, d), the function, being linear, is continuous. If x0 2 F is the left endpoint c of one of the intervals (c, d), then the function g, being linear on (c, d), is continuous there. So, let x0 2 F be different from the left endpoint of any of the component intervals, and x0 6¼ b. Let {xm}m  1 be a sequence of points in [a, b] converging from the right to x0, i.e. xm ! x+0 . First consider the case that xm 2 F for all m. Then g(xm) = f(xm) ! f(x0) = g(x0), since f is continuous on F. Next, consider the case that xm 62 F for all m. Suppose, if possible, that the sequence {g(xm)}m  1 does not converge to g(x0). It is nonetheless bounded and therefore {xm}m  1 has a subsequence, which we shall denote by {yk}k  1, such that the corresponding subsequence {g(yk)}k  1 converges to a limit other than g(x0). Denote the component interval of [a, b]\F that contains yk by (ck, dk). Since yk > x0 2 F and F \ (ck, dk) = ∅, we have x0 < ck < yk < dk, so that it is possible to extract a subsequence {yk(j)}j  1 of {yk}k  1 such that x0 < yk(j+1) < ck(j) < yk(j). This property of the subsequence ensures that the intervals (ck(j), dk(j)) are distinct and hence also disjoint. Since the sequence {xm}m  1 converges to x0, its subsequence {yk(j)}j  1 does the same. It follows that ck(j) ! x0. Now the sums of the lengths of the disjoint intervals (ck(j), dk(j)) cannot exceed that of [a, b] and therefore,

2.6 Egorov’s and Luzin’s Theorems

91

dk(j) − ck(j) ! 0, which implies dk(j) ! x0. But ck(j), dk(j) and x0 all lie in F and hence f(ck(j)) ! f(x0), f(dk(j)) ! f(x0). Since g is linear on the interval [ck(j), dk(j)], each g(yk(j)) lies between f(ck(j)) and f(dk(j)) and therefore also converges to f(x0) = g(x0). But this is a contradiction, because {g(yk(j))}j  1 is a subsequence of {g(yk)}k  1, which converges to a limit other than g(x0). The contradiction shows that {g(xm)}m  1 indeed converges to g(x0). Finally, consider the remaining case that xm 2 F for some m but not others. Again suppose, if possible, that the sequence {g(xm)}m  1 does not converge to g(x0). As before, {xm}m  1 has a subsequence, which we shall denote by {yk}k  1, such that the corresponding subsequence {g(yk)}k  1 converges to a limit other than g(x0). Now {yk}k  1 must either have a subsequence {yk(j)}j  1 that lies in F or have a subsequence {yk(j)}j  1 that lies outside F. In either event, one of the above cases applies to {yk(j)}j  1, whereby {g(yk(j))}j  1 is seen to converge to g(x0). Like before, this is a contradiction, showing that {g(xm)}m  1 indeed converges to g(x0). h Theorem 2.6.6 (Luzin). If f is a Lebesgue measurable function finite a.e. on E = [a, b], then, given any e > 0, there is a continuous function g on [a, b] such that m(E(f 6¼ g)) < e. Proof To begin with, we shall show that, for an arbitrary e > 0, there exists a closed set F  [a, b] such that m([a, b]\F) < e and the restriction of the function f to the set F is continuous. We shall assume that f is bounded on [a, b], i.e. jf ðxÞj  M. Let e > 0 be arbitrary. For any k 2 N, divide the interval [−M, M] into 2k parts by means of the points yj ¼ M þ

jM ; k

j ¼ 0; 1; 2;    ; 2k

and construct the sets Ek;1 ¼ Eðy0  f  y1 Þ and Ek;j ¼ Eðyj1 \f  yj Þ;

j ¼ 2; 3; . . .; 2k:

These sets are measurable and disjoint, and ½a; b ¼

2k [

Ek;j :

j¼1

By Proposition 2.3.24, there exists a closed set Fk,j  Ek,j such that

92

2 Measure in Euclidean Space

mðEk;j nFk;j Þ\

e 2k þ 1 k

:

Let Fk = [ 1  j  2kFk,j. Then [a, b]\Fk = [ 1  j  2k(Ek,j\Fk,j) and therefore m([a, b]\Fk) < e/2k. Now we define on Fk the function fk by fk(x) = yi for x 2 Fk,i, i = 1, 2, …, 2k. From Lemma 2.6.4, it follows that fk is continuous on Fk. The definition of Ek,j implies 0  fk ðxÞ  f ðxÞ 

M k

ð2:23Þ

for all x 2 Fk. Put F = \ k  1Fk. Then ½a; bnF ¼

1 [

ð½a; bnFk Þ

k¼1

and mð½a; bnFÞ 

1 X k¼1

mð½a; bnFk Þ\

1 X e ¼ e: 2k k¼1

Moreover, for x 2 F, the inequality (2.23) holds for any k. Therefore fn ! f uniformly on F. Since all fk are continuous, the aforementioned uniform convergence implies that f is also continuous on F. Suppose now that the function is finite a.e. on [a, b] but not bounded. Using Lemma 2.6.3, we get a bounded measurable function ~g such that mðfx 2 ½a; b : f ðxÞ 6¼ ~ gðxÞgÞ\e: By applying the result proved in the preceding paragraph to ~g, it follows that this function is continuous on some closed set F satisfying m([a, b]\F) < e. Using Lemma 2.6.5, we obtain a continuous function g defined on [a, b] such that g(x) = ~ g(x) for all x 2 F. Since fx 2 ½a; b : f ðxÞ 6¼ gðxÞg  fx 2 ½a; b : f ðxÞ 6¼ ~ gðxÞg [ fx 2 ½a; b : ~gðxÞ 6¼ gðxÞg; it follows that mðfx 2 ½a; b : f ðxÞ 6¼ gðxÞgÞ  mðfx 2 ½a; b : f ðxÞ 6¼ ~gðxÞgÞ þ mðfx 2 ½a; b : ~gðxÞ 6¼ gðxÞgÞ\2e:

h

2.7 Lebesgue Outer Measure in Rn

2.7

93

Lebesgue Outer Measure in Rn

In defining a measure on R, we extended the concept of length of an interval to the notion of outer measure of an arbitrary subset of R. Although outer measure m was defined on all subsets of R, it failed to satisfy the desirable properties of a measure described following Definition 2.1.2. (see Problem 2.3.P16) The collection M of Lebesgue measurable subsets of R was defined to consist of those subsets E  R that satisfy the Carathéodory condition of measurability, that is, m ðAÞ ¼ m ðA \ EÞ þ m ðA \ E c Þ for all A  R. It turned out that M was a r-algebra containing all intervals and that the set function m ¼ m jM satisfied all the desirable properties. In this section, we propose to replicate the first part of the above procedure in Rn , n > 1. We shall use the concept of the volume of a “cuboid” to define a nonnegative set function mn on PðRn Þ, called Lebesgue outer measure in Rn . In the next section, we shall select those subsets of Rn that satisfy the analogue of the above Carathéodory condition. These selected subsets will be shown to constitute a r-algebra Mn , called the r-algebra of measurable subsets of Rn . The Lebesgue outer measure mn restricted to Mn will turn out to have the desirable properties of a measure. It is known as Lebesgue measure in Rn and is denoted by mn (Definition 2.8.10 below). Several proofs are exactly analogous to those of corresponding results in Sect. 2.2 and will therefore be omitted. A straightforward analogue of a bounded interval in higher dimensions is a cartesian product of bounded intervals. Although many authors call them intervals, we shall refer to them as “cuboids”. They are best visualised as rectangles in R2 and as “boxes” in R3 . It is clear from our formal definition below that a nonempty cuboid is a Cartesian product of bounded intervals of positive length. Definition 2.7.1 By a cuboid in Rn we mean either the empty set or a set of points I ¼ fðx1 ; x2 ; . . .; xn Þ 2 Rn : ai  xi  bi for 1  i  ng where ai ; bi 2 R and ai < bi. A set of points characterised in a similar manner but with any or all of the signs  replaced by strict inequalities < will also be called a cuboid, and will be denoted by ha1 ; b1 ; a2 ; b2 ; . . .; an ; bn i: By the volume of any such cuboid I we mean the product Y vðIÞ ¼ ðbi  ai Þ 1in and by the volume v(∅) of the empty cuboid, we mean 0.

94

2 Measure in Euclidean Space

We consider the empty set as a cuboid with volume zero for the same reason that we consider the empty set as an interval with length zero in the one-dimensional case. In view of the requirement that ai < bi, the volume of a nonempty cuboid is always positive. We note that the symbol 〈a1, b1; a2, b2; …; an, bn〉 can denote any of 22n cuboids. The interior of a set A will be denoted by A°. Two sets A and B are said to overlap if A° \ B° is nonempty. It is clear that two intervals of positive length are nonoverlapping if they are disjoint or have only an endpoint in common. Suppose that each of the intervals 〈ai, bi〉, which may be open, closed or neither but nonempty, is split into a finite number of nonoverlapping intervals with the help of a partition ð0Þ

ð1Þ

ðki Þ

ai ¼ ai \ai \    \ai

¼ bi

and form the cuboids D E ðj Þ ðj þ 1Þ Ij1 ;j2 ;...;jn ¼ a1 1 ; a1 1 ; . . .; aðjn n Þ ; aðjn n þ 1Þ : Here the indices satisfy 0  ji  ki – 1. The total number of these cuboids must be k1 k1    kn . The cuboids Ij1 ;j2 ...;jn form a paving of the (nonempty) cuboid I ¼ ha1 ; b1 ; a2 ; b2 ; . . .; an ; bn i: The fact that the union of all cuboids in a paving may be a proper subset of I is of no consequence. It should be noted that the cuboids in a paving are nonoverlapping, as this will play a role in the proof of Lemma 2.7.2(i) below. It follows from the above definition of the volume of a cuboid that the total volume of all the cuboids in the paving agrees with the volume of the cuboid. This is clear in R, R2 and R3 . It can be verified by induction that the result holds in Rn when n > 3. In the notation of the preceding paragraph, this means: vðIÞ ¼

kX 1 1 k 2 1 X j1 ¼0 j2 ¼0



kX n 1

vðIj1 ;j2 ;...;jn Þ:

jn ¼0

The following lemma will be used in the sequel. Lemma 2.7.2 (i) If a cuboid I is represented as a union of pairwise nonoverlapping cuboids Ik, k = 1, …, p, that is, I = [ 1  k  pIk, where Ik° \ Ik′° = ∅ whenever k 6¼ k′, then

2.7 Lebesgue Outer Measure in Rn

95

vðIÞ ¼

p X

vðIk Þ:

ð2:24Þ

k¼1

(ii) If the cuboids I and Ik, k = 1, …, p, are such that I  [ 1  k  pIk, where the cuboids are allowed to overlap, then vðIÞ 

p X

vðIk Þ:

ð2:25Þ

k¼1

Proof (i) If I = ∅, then each Ik = ∅ and there is nothing to prove. Let I = 〈a1, b1; a2, b2; …; an, bn〉. We may assume that each Ik is nonempty because the empty ones, if any, can be discarded without affecting (2.24). So, let D E ðkÞ ðkÞ ðkÞ ðkÞ ðkÞ Ik ¼ c1 ; d1 ; c2 ; d2 ; . . .; cðkÞ ; d ; n n

k ¼ 1; . . .; p:

(k) For each i (1  i  n), arrange all the numbers c(k) i and di , k = 1,   , p, as ki (0) (1) a single increasing sequence ai < ai <    < a1 : Note that we must ki necessarily have ai = a(0) i and bi = a1 because the union of these cuboids is precisely equal to I. Consequently, the single increasing sequence partitions the interval [ai, bi] into a finite number of nonoverlapping subintervals. The cuboids

D E ðj Þ ðj þ 1Þ Ij1 ;j2 ;...;jn ¼ a1 1 ; a1 1 ; . . .; aðjn n Þ ; aðjn n þ 1Þ constructed with the help of these partitions constitute a paving of I and also give rise to a paving of each cuboid Ik. Let Tk denote the system of those cuboids Ij1 ;J2 ;...;jn which form a paving of a cuboid Ik. Since the volume of a cuboid equals the total volume of all the cuboids of any paving of it, we have on the one handbelongs to one and only vðIÞ ¼

X

vðIj1 ;j2 ;...;jn Þ

j1 ;...;jn

and on the other hand vðIk Þ ¼

X Ij1 ;...;jn 2Tk

vðIj1 ;...;jn Þ;

ð2:26Þ

96

2 Measure in Euclidean Space

so that p X

vðIk Þ ¼

k¼1

p X k¼1

0

1

X

@

vðIj1 ;...;jn ÞA:

ð2:27Þ

Ij1 ;...;jn 2Tk

However, since the cuboids Ik are nonoverlapping, so are the cuboids Ij1 ;J2 ;...;jn ; and hence we know that each Ij1 ;J2 ;...;jn belongs to one and only one Tk and therefore, the sum on the right-hand side of (2.27) is equal to the sum on the right-hand side of (2.26). Thus (2.24) is seen to hold. (ii) Once again, we may assume that I and Ik are all nonempty. Since v(I) = v(I°), the left-hand side of (2.25) remains unaffected if I is replaced by I°. Also, I  [ 1  k  pIk implies I°  [ 1  k  pIk. We may therefore assume without loss of generality that I is open. We can also suppose that the Ik’s intersect I because the insertion of some additional summands in the right-hand side of (2.25) only strengthens the inequality. Let Ik′ = Ik \ I, k = 1, …, p. We have I = [ 1  k  pIk′. As in the proof of (i), we construct a paving of the cuboid I which generates a paving of all the cuboids Ik′. It may be noted that the construction of such a paving is valid even when the cuboids Ik′ are allowed to overlap. As before, we denote the cuboids of the paving by Ij1 ;J2 ;...;jn and let Tk denote the system of those cuboids which form a paving of Ik′. Since the total volume of all the cuboids in a paving of any cuboid agrees with the volume of that cuboid, we have p X k¼1

0

X

@

1 vðIj1 ;...;jn ÞA ¼

Ij1 ;...;jn 2Tk

p X

vðIk0 Þ 

k¼1

p X

vðIk Þ

k¼1

and vðIÞ ¼

X

vðIj1 ;...;jn Þ:

j1 ;...;jn

Note that each cuboid Ij1 ;J2 ;...;jn is a constituent of at least one collection Tk but could belong to several Tk. Therefore p X k¼1

0 @

X Ij1 ;...;jn 2Tk

1 vðIj1 ;...;jn ÞA 

X

vðIj1 ;...;jn Þ:

j1 ;...;jn

Together with the preceding two displayed statements, this implies the required conclusion. h

2.7 Lebesgue Outer Measure in Rn

97

Since cuboids are the only sets we know how to measure at this stage, it is natural that the definition of outer measure involves coverings with cuboids. Definition.2.7.3 For any subset A of Rn , we define the Lebesgue outer measure mn (A) to be the infimum of all numbers 1 X

vðIk Þ;

k¼1

where {Ik}k  1 is a sequence of open cuboids that covers A and the infimum is taken over all such sequences. As in the case when n = 1, we shall refer to Lebesgue outer measure as simply “outer measure”, as there will be no other outer measure under discussion except in Problems 2.7.P2 and 2.8.P4. Remarks 2.7.4 (a) Some or all of the Ik may be empty. It follows in particular that mn (∅) = 0. (b) The outer measure mn is a set function defined on all subsets of A  R and has values in the set of nonnegative extended real numbers. The following properties of the outer measure are almost immediate from the definition. The property stated in (c) is called monotonicity. Proposition 2.7.5 (a) (b) (c) (d)

0  mn (A)  ∞ for all subsets of Rn ; mn (∅) = 0; If A  B  Rn , then mn (A)  mn (B); For x 2 Rn , we have mn ({x}) = 0.

Suppose that A is any subset of Rn and x is any element of E  R. As in the case of R, the translate of A by x is the set A + x = {y + x: y 2 A}. The assertion of the next proposition is called translation invariance of the Lebesgue outer measure mn . Proposition 2.7.6 For any A  Rn and x 2 Rn , we have mn (A + x) = mn (A). Example 2.7.7 Suppose that A has at most countably many points. Then mn (A) = 0. In view of the monotonicity of mn , we need consider only countable A. So, let {xk}k  1 be an enumeration of the points of A. Given e > 0, let Ik be an open cuboid of volume e/2k that contains the point xk. The sequence {Ik}k  1 then 1 1 P P covers A and satisfies vð I k Þ ¼ e=2k ¼ e. So, mn (A)  e. In particular, k¼1

k¼1

mn ðQ Q    QÞ ¼ mn ðN N    NÞ ¼ mn ðZ Z    ZÞ ¼ 0: It follows from what has just been noted that a set of positive outer measure must be uncountable.

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2 Measure in Euclidean Space

Proposition 2.7.8 If I is any cuboid, then its outer measure is the same as its volume: mn (I) = v(I). Proof If I is empty, there is nothing to prove; so assume I nonempty. It is easily seen that, given any η > 0, there exists an open cuboid I′ I such that v(I′)  (1 + η)v(I). Then {I′, ∅, ∅, …} is a sequence of open cuboids that covers I and therefore mn (I)  v(I′)  (1 + η)v(I). But η is an arbitrary positive number; therefore mn (I)  v(I). To prove the reverse inequality, we proceed as follows: First suppose I is closed and let η > 0 be given. There is a sequence of open 1 P cuboids {Ik}k  1 such that I  [ k  1Ik and mðIk Þ  mn  ðIÞ þ g. By the Heine– k¼1

Borel Theorem, any collection of open cuboids covering I contains a finite subcollection, which we again denote by {Ik}1  k  p, such that I  [ 1  k  pIk. By p P Lemma 2.7.2(ii), we have vðIÞ  vðIk Þ. Thus k¼1

vðIÞ 

p X k¼1

vðIk Þ 

1 X

vðIk Þ  mn  ðIÞ þ g:

k¼1

Since η > 0 is arbitrary, it follows that v(I)  mn (I). This completes the proof in the case when I is a closed cuboid. If I is any nonempty cuboid, then, given η > 0, there is a closed cuboid I″  I such that vðI 00 Þ [ ð1  gÞvðIÞ: Hence ð1  gÞvðIÞ\vðI 00 Þ ¼ mn  ðI 00 Þ  mn  ðIÞ  mn  ðIÞ ¼ vðIÞ ¼ vðIÞ: Thus for each η > 0, ð1  gÞvðIÞ\mn  ðIÞ  vðIÞ; and so, vðIÞ ¼ mn  ðIÞ:

h

The reader is reminded of the concepts of finitely additive, countably subadditive and countably additive as laid down in Definition 2.2.7. Illustrations of these concepts in a general context were given in Examples 2.2.8.

2.7 Lebesgue Outer Measure in Rn

99

Proposition 2.7.9 The outer measure mn is countably subadditive; that is, if {Aj}j  1 is a sequence of subsets of Rn , then 1 1 [ X mn  ð Aj Þ  mn  ðAj Þ: j¼1

j¼1

Corollary 2.7.10 The outer measure mn is finitely subadditive; that is, if {Aj}1  j  p is a finite sequence of subsets of Rn , then p p [ X mn  ð Aj Þ  mn  ðAj Þ: j¼1

j¼1

Remarks 2.7.11 (a) If mn (A) = 0, then mn (A [ B) = mn (B). In fact, by Proposition 2.7.5(c) and Corollary 2.7.10, mn  ðBÞ  mn  ðA [ BÞ  mn  ðAÞ þ mn  ðBÞ ¼ mn  ðBÞ: (b) Let A be any subset of Rn and AI denote the set of points in A with at least one irrational coordinate. Then mn (A) = mn (AI). This follows from (a) above and the result noted in Example 2.7.7. In the rest of this section, we explore more closely the role of open sets in determining the Lebesgue outer measure of an arbitrary subset of Rn . The material can be omitted for now without loss of continuity and the reader may wish to take it up only when it is needed later. Proposition 2.7.12 Let A  Rn be arbitrary and e > 0 be given. Then there exists an open set O A such that mn (O)  mn (A) + e. Proposition 2.7.13 Let A  Rn be arbitrary. Then there exists a set G A such that mn (G) = mn (A) and G is an intersection of a sequence of open sets. Thus every subset of Rn is contained in a Gd -set of the same outer measure as itself. Proposition 2.7.14 Any nonempty open set O is a countable union of open cuboids and also a countable union of closed cuboids. Proof Since O is nonempty, each point of it must belong to an open cuboid I contained in O. It is obvious how to “shrink” the open cuboid I slightly in such a manner that the point continues to belong to it while the corresponding closed cuboid remains within I. Moreover, one can choose the smaller cuboid to have rational “corners”. Then O is the union of all these smaller cuboids as well as of the corresponding closed cuboids, both of which must necessarily be countable in number. h

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Examples 2.7.15 (a) Since the subset Q Q    Q of Rn is countable and a single point set is closed, it follows that the aforementioned set is an F r -set. (b) If E 2 F r , then by using the definition of F r and taking complements, we conclude that E c 2 Gd . In particular, it follows that the set of points in Rn with at least one coordinate irrational is a set of type Gd . (c) Since ða; bÞ ¼ [ p  1 ½a þ 1p; b  1p and fag ¼ \ p  1 ða  1p; a þ 1pÞ, it follows that an open cuboid ða1 ; b1 ; a2 ; b2 ; . . .; an ; bn Þ ¼

1 [ p¼1

½a1 þ

1 1 1 1 ; b1  ; . . .; an þ ; bn   p p p p

is a set of type F r and that a single point set is Qof type Gd . Here the left-hand side is a Cartesian product of open intervals ni¼1 ðai ; bi Þ and the right-hand Q side is a Cartesian product of closed intervals ni¼1 ½ai þ 1p; bi  1p. (d) It is clear that every open set including ∅ is of type Gd and every closed set including ∅ is of type F r . It is easily seen that every nonempty open set is also of type F r , because it is a countable union of closed cuboids by Proposition 2.7.14. On taking complements, we see that every closed set is of type Gd . Problem Set 2.7 2.7.P1. Prove that jmn  ðAÞ  mn  ðBÞj  maxfmn  ðAnBÞ, mn (B\A)}  mn (ADB) provided mn (A) and mn (B) are finite. 2.7.P2. An outer measure on Rn is an extended real-valued, nonnegative, monotone and countably subadditive set function l defined on all subsets of Rn and satisfying l (∅) = 0. If {lk }k  1 is a sequence of outer measures and {ak} k  1 a sequence of positive real numbers, then show that the set function defined by 1 P l ðEÞ ¼ ak lk ðEÞ is an outer measure. k¼1

2.8

Measurable Sets and Lebesgue Measure in Rn

The outer measure has the advantage that it is defined on P ðRn Þ; however, it is not countably additive (see Definition 2.2.7), which is a desirable property for a measure. The outer measure becomes countably additive if the family of sets on which it is defined is suitably restricted. We pursue this matter in this section. Proofs that are similar to their analogues in Sect. 2.3 will be omitted. Our first definition is wholly analogous to the one in R.

2.8 Measurable Sets and Lebesgue Measure in Rn

101

Definition 2.8.1 A subset E  Rn is said to be measurable if mn  ðAÞ ¼ mn  ðA \ EÞ þ mn  ðA \ E c Þ for every A  Rn : Remarks 2.8.2 (a) Observe that the finite subadditivity property (see Corollary 2.7.10) of mn implies mn  ðAÞ  mn  ðA \ EÞ þ mn  ðA \ E c Þ for every A  Rn :

(b) (c) (d)

(e)

Thus in testing the measurability of E, it is enough to show the reverse of this inequality. This condition of measurability in Definition 2.8.1 is known as the Carathéodory condition. If E is measurable, so is its complement Ec. For A  Rn , we have mn  ðAÞ ¼ mn  ðAÞ þ 0 ¼ mn  ðA \ Rn Þ þ mn  ðA \ £Þ. Thus Rn is measurable, and hence by (b), so is £. If mn (E) = 0, where E  Rn , then E is measurable; in fact, every subset of E is also measurable. This is because, for A  Rn , we have mn (A \ E)  mn (E) = 0 and mn (A \ Ec)  mn (A). So, c mn (A)  mn (A \ E ) + 0  mn (A \ E) + mn (A \ Ec). A countable subset of Rn is measurable. This follows from Example 2.7.7 and (d) above. Consequently, the set Q Q    Q in Rn is measurable. The set of points in Rn with at least one coordinate irrational, being the complement of Q Q    Q, is measurable in view of (b) above.

Notation 2.8.3 We denote the collection of all measurable subsets of Rn by Mn . In what follows, we record some of the properties of the collection Mn . Lemma 2.8.4 If E1 and E2 are in Mn , then so is E1 [ E2. Remarks 2.8.5 (a) (b) (c) (d) (e)

£ 2 Mn . S A1 ; A2 ; . . .; Ap 2 F ) pj¼1 Aj 2 Mn . If A 2 F ; B 2 F ) A \ TB 2 Mn . A1 ; A2 ; . . .; Ap 2 F ) pj¼1 Aj 2 Mn . A 2 F ; B 2 F ) AnB 2 Mn .

Proposition 2.8.6 The collection Mn of measurable subsets of Rn is an algebra. In particular, if E1, E2, …, Ep are measurable subsets, then so are [ 1  k  pEk and \ 1  k  pEk. Proof This is merely a summary of Remarks 2.8.2(b) and 2.8.2(c), Lemma 2.8.4 and Remarks 2.8.5(b) and 2.8.5(d). h We now note that the outer measure restricted to Mn is finitely additive (Cf. Proposition 2.3.8).

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2 Measure in Euclidean Space

Proposition 2.8.7 If E1, E2, …, Ep is a finite sequence of disjoint measurable subsets of Rn and A  Rn is arbitrary, then mn ðA \

p [ j¼1

Ej Þ ¼

p X

mn ðA \ Ej Þ:

j¼1

In particular, if A ¼ Rn , then mn

p [ j¼1

p  X Ej ¼ mn ðEj Þ: j¼1

The outer measure mn restricted to Mn is in fact countably additive (Theorem 2.8.9). Theorem 2.8.8 (Cf. Theorem 2.3.12) The collection Mn of measurable subsets of Rn constitute a r-algebra. Theorem 2.8.9 (Cf. Theorem 2.3.13) The outer measure mn is countably additive on the r-algebra Mn of measurable subsets of Rn . Definition 2.8.10 The restriction of mn to Mn is called (Lebesgue) measure and will be denoted by mn. For any E 2 Mn , the (extended) real number mn(E) = mn (E) is called the (Lebesgue) measure of the set E. Since mn is countably subadditive according to Proposition 2.7.9, it follows trivially that Lebesgue measure is also countably subadditive. Definition 2.8.11 By a closed upper half-space we mean a subset H of Rn for which there exists j (1  j  n) and a real number cj such that H ¼ fðx1 ; x2 ; . . .; xn Þ 2 Rn : cj  xj g: If the inequality is replaced by < , we speak of an open upper half-space. If the inequalities are reversed, we have what are called closed and open lower half-spaces. A half-space is the n-dimensional analogue of an interval with one finite endpoint. Remarks 2.8.12 (a) A cuboid is an intersection of 2n half-spaces. (b) An open [resp. closed] half-space is an open [resp. closed] set. (c) The complement of an open [resp. closed] upper half space is a closed [resp. open] lower half-space, and vice versa. (d) If H is an open upper half-space (so that its complement Hc is also a half-space) and I is any open cuboid, then I \ H and I \ Hc are both cuboids, possibly empty, and v(I) = v(I \ H) + v(I \ Hc). This is obvious in the one-dimensional case. Suppose

2.8 Measurable Sets and Lebesgue Measure in Rn

I ¼ ða1 ; b1 ; a2 ; b2 ; . . .; an ; bn Þ

and

103

H ¼ fðx1 ; x2 ; . . .; xn Þ 2 Rn : cj \xj g:

Then (x1, x2, …, xn) 2 I \ H must satisfy aj < xj < bj as well as cj < xj. This is not possible if cj  bj, and therefore I \ H = ∅, so that I \ Hc = I and there is nothing left to prove in this case. Also, (x1, x2, …, xn) 2 I \ Hc must satisfy aj < xj < bj as well as cj  xj. This is not possible if cj  aj, and therefore I \ Hc = ∅, so that I \ H = I and there is nothing left to prove in this case either. So, suppose aj < cj < bj. Then I \ H is given by the inequalities ai \xi \bi for i 6¼ j and

cj \xj \bj :

Thus I \ H is a cuboid with volume Y

vðI \ HÞ ¼ ðbj  cj Þ 

ðbi  ai Þ:

i6¼j

At the same time, I \ Hc is given by the inequalities ai \xi \bi for i 6¼ j

aj \xj  cj :

and

Thus I \ Hc is a cuboid with volume vðI \ H c Þ ¼ ðcj  aj Þ 

Y i6¼j

ðbi  ai Þ:

It is now immediate that v(I) = v(I \ H) + v(I \ Hc). Proposition 2.8.13 An open upper half-space H is measurable. Proof Let A be any subset of Rn and let A1 = A \ H and A2 = A \ Hc. We must show that mn (A)  mn (A1) + mn (A2). If mn (A) = ∞, then there is nothing to prove. Suppose that mn (A) < ∞. By the definition of outer measure, for every e > 0, there exists a sequence {In}n  1 of open cuboids such that A  [ k  1Ik and 1 X

vðIk Þ  mn  ðAÞ þ e:

k¼1

Let I′k = Ik \ H and I″k = Ik \ Hc. By Remark 2.8.12(d), I′k and I″k are cuboids (some of them possibly empty) and vðIk Þ ¼ vðIk0 Þ þ vðIk00 Þ

¼ mn  ðIk0 Þ þ mn  ðIk00 Þ

by Proposition 2.7.8. Since A1  [ k  1I′k and A2  [ k  1I″k, we have

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2 Measure in Euclidean Space

mn  ðA1 Þ  mn  ð

1 [

Ik0 Þ 

k¼1

and

mn  ðA2 Þ  mn  ð

1 [

mn  ðIk0 Þ

k¼1

Ik00 Þ 

k¼1

1 X

1 X

mn  ðIk00 Þ:

k¼1

Hence mn  ðA1 Þ þ mn  ðA2 Þ 

1 X

mn  ðIk0 Þ þ

k¼1

¼

1 X k¼1

1 X

mn  ðIk00 Þ

k¼1

½mn  ðIk0 Þ þ mn  ðIk00 Þ ¼

1 X

vðIk Þ  mn  ðAÞ þ e:

k¼1

Since e > 0 is arbitrary, we have mn (A)  mn (A1) + mn (A2), as required.

h

Proposition 2.8.14 A subset of Rn that is either open or closed must be measurable. Every cuboid is measurable. Moreover, Gd -sets and F r -sets are always measurable. Proof Since the collection Mn of measurable sets is a r-algebra, it follows from Proposition 2.8.13 that a closed lower half-space is measurable. Since an open lower half-space given by xj < cj is the countable union of the closed lower half-spaces given by xj  cj  1n, it further follows that an open lower half-space is measurable. Hence a closed upper half-space is also always measurable. By Remark 2.8.12(a), any cuboid is measurable. Now any nonempty open set O is a countable union of open cuboids by Proposition 2.7.14. Since Mn is a r-algebra, it follows that O is measurable. Thus, every open set is measurable. Once again, since Mn is a r-algebra, the complement of an open set, i.e. a closed set, must be measurable. The final assertion about Gd and F r -sets now follows immediately from the fact that Mn is a r-algebra. h Recall from Proposition 2.3.17 that there exists a smallest r-algebra generated by any collection of subsets. Definition 2.8.15 The r-algebra generated by all the open subsets of Rn is called the Borel algebra (in Rn ) and will be denoted by Bn . A set in the Borel algebra is called a Borel measurable set (in Rn ) or simply a Borel set (in Rn ). A set in the algebra Mn (hitherto called simply “measurable”) will be called a Lebesgue measurable set when a distinction needs to be made. Remark 2.8.16 Since every open set is Lebesgue measurable by Proposition 2.8.14 and Lebesgue measurable sets constitute a r-algebra, it follows by Definition 2.8.15 that B  M. In other words, every Borel set is Lebesgue measurable. The restriction of mn to Bn is called Borel measure (in Rn ) and will again be denoted by mn. Thus, for E 2 Bn , the extended real number mn(E) = mn (E) is called the Borel measure of E and is the same as its Lebesgue measure.

2.8 Measurable Sets and Lebesgue Measure in Rn

105

The triples ðRn ; Bn ; mn Þ and ðRn ; Mn ; mn Þ are called the Borel measure space and Lebesgue measure space respectively. We have verified the following properties of Lebesgue measurable sets and Lebesgue measure: (i) Complements, countable unions and countable intersections of measurable sets are measurable. (ii) Any cuboid is measurable and its measure is its volume (see Proposition 2.7.8). (iii) If {Ek}k  1 is a sequence of disjoint measurable sets, then mn ð

1 [ k¼1

Ek Þ ¼

1 X

mn ðEk Þ:

k¼1

The above properties are also valid for Borel measurable sets and Borel measure: (i) is a direct consequence of the definition of Bn as being a r-algebra, while (iii) follows from its Lebesgue counterpart taken with Remark 2.8.16. Assertion (ii) about Borel sets follows from Remark 2.8.12(a) and the fact that every open half space, being an open set, is a Borel set and hence so is every half-space. Proposition 2.8.17 (Translation Invariance of Lebesgue measure on Rn ) Let E 2 Mn ½resp: Bn ; x 2 Rn . Then (a) E þ x 2 Mn ½resp:Bn , (b) mn(E + x) = mn(E). That is, the Lebesgue [resp. Borel] measure on Mn ½resp:Bn  is translation invariant. Proof Analogous to the proof of Proposition 2.3.23. h Finally, we prove the following characterisation of Lebesgue measurable subsets of Rn . Proposition 2.8.18 Let E be a given subset of Rn . The following five statements are equivalent: (a) E is Lebesgue measurable; (b) For every e > 0, there exists an open set O E such that mn (O\E) < e; (c) There exists a Gd -set G E such that mn (G\E) = 0; (d) For every e > 0, there exists a closed set F  E such that mn (E\F) < e; (η) There exists an F r -set F  E such that mn (E\F) = 0. Proof We shall prove (a) ) (b) ) (c) ) (a) and then (d) , (a) , (η). (a) ) (b). For any set E and e > 0, there exists [see Proposition 2.7.12] an open set O E such that

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2 Measure in Euclidean Space

mn  ðOÞ  mn  ðEÞ þ

e : 2

Since O E, we have O \ E = E. From the Lebesgue measurability of E, we have mn  ðOÞ ¼ mn  ðO \ EÞ þ mn  ðO \ Ec Þ ¼ mn  ðEÞ þ mn  ðOnEÞ: Thus mn  ðEÞ þ

e  mn  ðOÞ ¼ mn  ðEÞ þ mn  ðOnEÞ: 2

In the case when mn (E) < ∞, this shows that mn  ðOnEÞ  2e \e. Now let mn (E) = ∞. Observe that for each k 2 N, the set Ik ¼



 ðx1 ; x2 ; . . .; xn Þ 2 Rn : xj \k; 1  j  n is a cuboid of volume (2k)n, which is finite. For each k 2 N, the set Ek ¼ E \ Ik 2 Mn and mn (Ek)  mn (Ik) < ∞. Moreover E = [ k  1Ek. Since mn (Ek) < ∞, it follows on applying the case of finite measure proved above that, for any e > 0, there exists an open set Ok Ek such that mn  ðOk nEk Þ\

e : 2k

Note that 1 [

O¼

Ok

k¼1

1 [

Ek ¼ E

k¼1

and 1 [

OnE ¼

Ok n

k¼1

1 [ k¼1

Ek 

1 [

ðOk nEk Þ:

k¼1

So, mn  ðOnEÞ 

1 X

mn  ðOk nEk Þ\

k¼1

1 X e ¼ e: kþ1 2 k¼1

(b) ) (c). It follows from (b) that, for any k 2 N, there exists an open set Ok E such that 1 mn  ðOk nEÞ\ : k Let G = \ k  1Ok; then G is a Gd -set containing E. Moreover,

2.8 Measurable Sets and Lebesgue Measure in Rn

mn  ðGnEÞ ¼ mn  ð

1 \

Ok nEÞ  mn  ðOk nEÞ\

k¼1

107

1 k

for every k 2 N:

Hence mn (G\E) = 0. (c) ) (a). Note that E = G\(G\E) and that the set G, being a countable intersection of Lebesgue measurable sets, is Lebesgue measurable by Theorem 2.8.8; also, G\E is Lebesgue measurable by Remark 2.8.2(d). Use Theorem 2.8.8 once again. We have shown so far that (a) ) (b) ) (c) ) (a). The consequence that (b) , (a) will shortly be used in proving that (a) , (d). (a) , (d). Assume (a), i.e. E 2 Mn . Then E c 2 Mn . Since (a) ) (b), it follows that, for any e > 0, there exists an open set O Ec such that mn (O\Ec) < e. So the closed set F = Oc satisfies F  E and also mn (E\F) = mn (E\Oc) = mn (E \ O) = mn (O\Ec) < e. Thus (d) holds. Conversely, assume (d). Then the open set O = Fc satisfies O Ec and  mn (O\Ec) < e. In other words, (b) holds with Ec in place of E. Since (b) ) (a), it follows that E c 2 Mn and hence E 2 Mn . Thus (a) holds. (a) , (η). This is analogous to the argument that (a) , (d). h Remark 2.8.19 Let E  Rn be Lebesgue measurable, i.e. E 2 Mn . Then it follows from the preceding proposition that there exist F; G 2 Bn such that F  E  G with mn(G\F) = 0. Conversely, the existence of such F and G can be proved to imply E 2 M in the following manner: Suppose such F, G exist. Then F 2 Mn because Bn  Mn ; besides, E\F  G\F, which implies mn (E\F)  mn (G\F) = 0 and hence E  R by Remark 2.8.2(d). Therefore E ¼ ðEnFÞ [ F 2 Mn . Note that in this situation, mn(E) = mn(F) = mn(G). The foregoing equivalence is related to the fact that Lebesgue measure is the “completion” of Borel measure (see Definition 7.2.3 and Theorem 7.2.5). Problem Set 2.8 2.8.P1. Let {Ei}i  1 be a sequence of measurable sets. Then (a) if Ei  Ei+1, then mn ðlim Ek Þ ¼ lim mn ðEk Þ; k!1

(b) if Ei Ei+1 and mn(E1) < ∞, then mn ðlim Ek Þ ¼ lim mn ðEk Þ. k!1

2.8.P2. Prove that for subsets A, B, C of any nonempty set X, ADC  ðADBÞ [ ðBDCÞ: 2.8.P3. Let A, B be subsets of Rn . Show that mn  ðADBÞ  mn  ðAÞ þ mn  ðBÞ: 2.8.P4. If l is an outer measure on Rn and if A and B are subsets of Rn , of which at least one is l -measurable, then show that

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2 Measure in Euclidean Space

l ðAÞ þ l ðBÞ ¼ l ðA [ BÞ þ l ðA \ BÞ: Remark: Under the additional hypothesis that at least one among A and B has finite outer measure, the above equality can be written as l ðA [ BÞ ¼ l ðAÞ þ l ðBÞ  l ðA \ BÞ: 2.8.P5. Let A  E, where E is measurable and mn(E) < ∞. Show that A is measurable provided mn  ðEÞ ¼ mn  ðAÞ þ mn  ðEnAÞ: 2.8.P6. If F 2 Mn and mn (FDG) = 0, then show that G is measurable. 2.8.P7. Show that every nonempty open set has positive measure. 2.8.P8. Let q1, q2, … be an enumeration of points in Rn with rational coordinates and let G = [ k  1Ik, where Ik is an open cuboid centred at qk with volume 1/k2. Prove that for any closed set F, mn(GDF) > 0. 2.8.P9. Let E be Lebesgue measurable with 0 < mn(E) < ∞ and let e > 0 be given. Then there exists a compact set K  E such that mn(E\K) = mn(E) − mn(K) < e.

Chapter 3

Measure Spaces and Integration

3.1

Integrals of Simple Functions

The Riemann integral of a function over a real interval is defined via upper and lower sums or via Riemann sums, all of which have finitely many terms. What is often not emphasised is that they are in fact integrals of certain step functions. Thus, step functions are used as building blocks for Riemann integration, which is possible because their integrals are finite sums. In the development of the Lebesgue theory, we prefer to use as building blocks those functions that are constant on measurable sets and have only finitely many nonnegative real values, in other words, simple functions as in Definition 2.5.6. The reason for not using step functions will become clear soon. All intervals are instances of measurable sets and therefore nonnegative-valued step functions are instances of simple functions. However, they are far from being the only ones, because there exist measurable sets other than intervals. For example, the set of irrational numbers in [0,1] is measurable and therefore its characteristic function is a simple function, although it is not a step function. A simple function can be expressed as a linear combination of characteristic functions in several ways. We shall define the Lebesgue integral of a simple function in terms of its canonical representation so as to avoid ambiguity. The awkwardness of expressing the canonical representation of the sum of two simple functions in terms of their respective canonical representations [see Remark 2.5.8 (d) and Problem 2.5.P6] creates a slight difficulty, which will be surmounted in Proposition 3.1.5. The context in which we shall be working is that we have a set function l having a r-algebra F as its domain and satisfying the following properties: (M1) l(∅) = 0 and l(E)  0 for every E 2 F ; (M2) l is countably additive [as in Definition 2.2.7], that is, for every disjoint sequence of sets fAj gj  1 such that each Aj 2 F , we have © Springer Nature Switzerland AG 2019 S. Shirali and H. L. Vasudeva, Measure and Integration, Springer Undergraduate Mathematics Series, https://doi.org/10.1007/978-3-030-18747-7_3

109

110

3 Measure Spaces and Integration

l

1 [ j¼1

! Aj

¼

1 X

lðAj Þ:

j¼1

There will be no provision for special elements of F called “intervals”, thereby taking away all scope for step functions; in particular, l will not be presumed to have been built up from the concept of length or volume of simpler kinds of sets. The nature of the set X of which the elements of F are subsets will not matter, as it will not enter into our considerations. In such a general set up, the set function involved is sometimes called a general or an abstract measure for emphasis. Definition 3.1.1 A measure space (X, F , l) is a nonempty set X with a r-algebra F of subsets of X and an extended real-valued function l on F such that (M1) and (M2) hold. The set function l is called a measure on F . Sometimes it will be convenient to speak of a “measure on X (or in X)”, meaning a measure on some r-algebra of subsets of X. A subset that belongs to the r-algebra is called a measurable subset without presuming any outer measure. To recap, a measure on a set X is a countably additive nonnegative extended real-valued function on a r-algebra of subsets of X that vanishes on the empty set. The reader may note that the continuity property in Proposition 2.3.21 was proved by using only properties (M1) and (M2) of Lebesgue and Borel measures while the manner in which these measures were constructed from lengths of intervals did not play any role. The property is therefore valid for abstract measures too. Examples 3.1.2 The only examples we have seen so far are the Borel and Lebesgue measures on measurable subsets of Rn . Here are some more. (a) For a set E  X, where X is any nonempty set, define l(E) = ∞ if E is an infinite set and let l(E) be the number of points in E if E is finite. Then l is called the counting measure on PðXÞ, the collection of all subsets of X and (X, PðXÞ, l) is a measure space. In the particular case when X ¼ N, the counting measure on PðNÞ is often denoted by c. (b) Fix x0 2 X, where X is any nonempty set whatsoever. Define l(E) = 1 if x0 2 E and l(E) = 0 otherwise. This l may be called the unit mass concentrated at x0. It is easy to verify that l is a measure defined on the r-algebra PðXÞ. (c) Let (X, F , l) be a measure space and Y be a measurable set (which means simply that Y 2 F ). The triple (Y, F Y , ljF Y ), where F Y ¼ fY \ A : A 2 F g and ljF Y is the restriction of l to the r-algebra F Y , is a measure space. In particular, if X ¼ R, F ¼ M, the r-algebra of Lebesgue measurable subsets of R and l the Lebesgue measure defined on M, then for any Y 2 F ¼ M, F Y is the collection of Lebesgue measurable subsets of Y and ljF Y is the Lebesgue measure restricted to the measurable subsets of Y. The corresponding statement with B in place of M is also true.

3.1 Integrals of Simple Functions

111

The definitions and results of Sect. 2.5 up to Theorem 2.5.9 carry over when the set X is understood to be any set with a given r-algebra and measure l. Sometimes a property of a function or of a sequence of functions, such as vanishing at x or converging to a limit function, is a “hereditary” property in the sense that if it holds on some set, then it holds on every subset thereof. The same is true of uniform continuity. As in the case of Lebesgue and Borel measure, we have a concept of “almost everywhere” as below [cf. Definition 2.5.10]. Definition If a hereditary property holds everywhere on X except on a subset E in F having l(E) = 0, we say that it holds almost everywhere. The phrase is abbreviated as a.e. When the property is described in terms of an explicit x 2 X, we can say that it holds for almost all x. In the rest of this chapter, X will denote a set with a r-algebra F and a measure l on it. However, when X is specified as a (Lebesgue) measurable subset of Rn (n  1), it will be understood that l is Lebesgue measure mn on the r-algebra of Lebesgue or Borel measurable subsets of X, unless the context calls for some other r-algebra or measure. We note in passing that if we were not to work with an abstract measure but were instead to stay with Lebesgue or Borel measure on an interval, then we could have used step functions as building blocks and avoided simple functions altogether. For treatments that proceed in this manner, see [1] or [17]. Also, we would have the option of proceeding via the Henstock–Kurzweil “gauge integral”, which is rather more complicated to define than the Riemann integral, but leads directly to the Lebesgue as well as Riemann integrals (including improper) without depending on the concept of measure. For an introduction to the gauge integral, see [6] or the Internet article [22]. P Before proceeding, we remind the reader [see Remark 2.5.8c] that 1  j  n aj vAj is a canonical representation of a simple function on X if and only if (a) the measurable sets Aj are disjoint with union X, while (b) the nonnegative real numbers aj are distinct and the sets Aj nonempty. We shall say that the sets Aj form a partition of X when they satisfy (a); in particular, the sets must be measurable. Definition P 3.1.3 Let s be a simple function on X, having the canonical representation s ¼ 1  j  n aj vAj . The integral of s is defined to be Z s¼

n X

aj lðAj Þ:

j¼1

It is also denoted by

R X

s,

R X

s dl or

R

s dl according to convenience.

Thus the integral of a simple function is a sum of products, each of which corresponds to a term in the sum that represents the function canonically.

112

3 Measure Spaces and Integration

P A simple function s with canonical representation s ¼ 1  j  n aj vAj takes the value 0 (vanishes) a.e. ifRand only if l(Aj) = 0 for every j such that aj 6¼ 0. When this is the case, clearly, X s ¼ 0. When l is Lebesgue measure m on Rn (n  1) or on a measurable subset thereof, we speak of the Lebesgue integral. If it is necessary to emphasise that we are working with an abstract measure, we speak of the measure space integral. Examples 3.1.4 (a) Suppose sR is zero everywhere on X. Its canonical representation is 0vX and R therefore X s dl ¼ X 0 dl ¼ 0  lðXÞ ¼ 0, even if l(X) = ∞. The function has another representation as 1v∅ (not canonical) and the corresponding sum of products is 1l(∅), which also turns out to be 0. (b) The rationals in [0,1] form a subset of Lebesgue measure zero. Therefore the characteristic function of this set is a simple function with Lebesgue integral 0. This is in contrast to the Riemann theory, in which the function has no Riemann integral at all on [0,1]. (c) Consider s = avA, where a is a nonnegative real number and A is a measurable subset of X, that is, A 2 F . If a > 0, then the canonical representation of s is c c av R A+ 0vA provided Ac and A are both cnonempty, and therefore ) = ∞. If a = 0, then the X s dl = al(A) + 0l(A ) = al(A) even if l(A R canonical representation is 0vX and therefore X s dl ¼ 0 as in (a) above; but R now al(A) is also 0 and so it is still true that X s dl ¼ a  lðAÞ. Thus R X ðavA Þdl ¼ a  lðAÞ when a  0. Note that this holds also when A = ∅ or X, because av∅ has canonical representation 0vX and avX is already in canonical form. (d) Let X = [0,3) P and s have the values 1,4,5 on [0,1), [1,1], (1,3) respectively. Then s ¼ 1  j  3 aj vAj , where a1 = 1, a2 = 4, a3 = 5 and A1 = [0,1), A2 = [1,1], A3 = (1,3). According to the above definition, Z s dl ¼ 1  ð1  0Þ þ 4  ð1  1Þ þ 5  ð3  1Þ ¼ 11: X

If Pwe split A3 as the disjoint union (1,2] [ (2,3), then we can represent s as 1  i  4 bi vBi , where b1 = 1, b2 = 4, b3 = b4 = 5 and B1 = [0,1), B2 = [1,1], B3 = (1,2], B4 = (2,3). This representation is not canonical, but we nevertheless have X 1i4

bi lðvBi Þ ¼ 1  ð1  0Þ þ 4  ð1  1Þ þ 5  ð2  1Þ þ 5  ð3  2Þ ¼ 11:

In this sum, the last two terms can be “grouped” together to form the last term in the previous sum. The purpose of the next proposition is to encapsulate this phenomenon in general.

3.1 Integrals of Simple Functions

113

Proposition 3.1.5 Let s be a simple function on X such that s¼

p X

ai vAi ¼

n X

bj vBj ;

j¼1

i¼1

where the sets Ai (1  i  p) as well as the sets Bj (1  j  n) both form partitions of X, i.e. i 6¼ i0 ) Ai \ Ai0 ¼ £;

j 6¼ j0 ) Bj \ Bj0 ¼ £

ð3:1Þ

Bj ¼ X:

ð3:2Þ

and p [

Ai ¼

n [ j¼1

i¼1

Then p X

ai lðAi Þ ¼

n X

Z bj lðBj Þ ¼

s dl:

ð3:3Þ

X

j¼1

i¼1

Proof By (3.1), the sets Ai \ Bj, 1  j  n, are disjoint for each fixed i and, in view of (3.2), their union is n [

ðAi \ Bj Þ ¼ Ai \

j¼1

n [

Bj ¼ Ai \ X ¼ Ai :

j¼1

It follows that lðAi Þ ¼

n X

lðAi \ Bj Þ for each i;

j¼1

and hence p X

ai lðAi Þ ¼

p X n X

ai lðAi \ Bj Þ:

i¼1 j¼1

i¼1

Similarly, lðBj Þ ¼

p X i¼1

lðAi \ Bj Þ for each j

ð3:4Þ

114

3 Measure Spaces and Integration

and hence n X

bj lðBj Þ ¼

p n X X

j¼1

p X n X

bj lðAi \ Bj Þ ¼

j¼1 i¼1

bj lðAi \ Bj Þ:

ð3:5Þ

i¼1 j¼1

Consider a fixed i and any j (1  j  n). If Ai \ Bj 6¼ ∅, then s takes the value ai as well as bj on this nonempty set and therefore ai = bj, so that ai lðAi \ Bj Þ ¼ bj lðAi \ Bj Þ: On the other hand, if Ai \ Bj = ∅, then the above equality still holds, because both sides are zero. Thus it holds for all i and j, which implies the first equality in (3.3) in view of (3.4) and (3.5). Since (3.1) and (3.2) are certainly fulfilled when P 1  i  p ai vAi is a canonical representation of s, it is immediate from Definition 3.1.3 that the second equality in (3.3) also holds. h Suppose as in Proposition 3.1.5 that the sets Ai (1  i  p) as well as the sets p n P P Bj (1  j  n) form partitions of X. Then s ¼ ai vAi , t ¼ bj vBj are simple i¼1

j¼1

functions (not necessarily equal to each other of course). Now, s + t takes the value ai + bj on Ai \ Bj and the sets ! Ai \ Bj form a partition of X, so that p n P P ððai þ bj ÞvAi \ Bj Þ . This means that the proposition above covers sþt ¼ i¼1

j¼1

the foregoing representations of all the three functions s, t and s + t. This is how it helps overcome the difficulty arising from the lack of a conveniently expressible canonical representation of s + t in terms of the respective canonical representations of s and t. [But see Problem 3.1.P8.] We shall now prove what is called the additivity of the integral for simple functions. Proposition 3.1.6 Let s and t be simple functions on X. Then Z

Z ðs þ tÞdl ¼

X

Proof Let s ¼

p P i¼1

s dl þ X

ai vAi and t ¼

p P j¼1

Z t dl: X

bj vBj , where the sets Ai (1  i  p) as well as

the sets Bj (1  j  n) form partitions of X. Then sþt ¼

p n X X i¼1

! ððai þ bj ÞvAi \ Bj Þ

j¼1

and the sets Ai \ Bj also form a partition of X. Therefore by Proposition 3.1.5,

3.1 Integrals of Simple Functions

Z s dl ¼ X

p X

115

Z ai lðAi Þ and

t dl ¼ X

i¼1

n X

bj lðBj Þ

ð3:6Þ

j¼1

and Z ðs þ tÞdl ¼ X

p n X X

p n X X

¼

! ðai lðAi \ Bj ÞÞ þ

j¼1

i¼1 p X

ððai þ bj ÞlðAi \ Bj ÞÞ

j¼1

i¼1

¼

!

ai

n X

j¼1

! lðAi \ Bj Þ þ

j¼1

i¼1

p n X X

n X

bj

j¼1

! ðbj lðAi \ Bj ÞÞ

i¼1 p X

ð3:7Þ

! lðAi \ Bj Þ :

i¼1

For each i, the set Ai is the disjoint union [ 1  j  n (Ai \ Bj) and therefore n X

lðAi \ Bj Þ ¼ lðAi Þ:

j¼1

Similarly, for each j, p X

lðAi \ Bj Þ ¼ lðBj Þ:

i¼1

Using these equalities in (3.7) and then appealing to (3.6), we get Z ðs þ tÞdl ¼ X

p X

n X

ai lðAi Þ þ

Z bj lðBj Þ ¼

s dl þ X

j¼1

i¼1

Z t dl:

h

X

The two properties proved in the next proposition will sometimes be referred to as linearity and monotonicity respectively of the integral for simple functions. Proposition 3.1.7 Let s and t be simple functions on X and a,b  0 be real numbers. Then Z Z Z ðas þ btÞdl ¼ a s dl þ b t dl: X

If s  t everywhere, then

X

R X

s dl 

R X

X

t dl:

will follow from Proposition 3.1.6 as soon as we show that R RProof The equality as dl ¼ a s dl. [Homogeneity of the integral for simple functions.] This X X equality holds when a = 0 because then as = 0vX (canonically) and therefore

116

R

3 Measure Spaces and Integration

XRas dl

¼ 0l(X) = 0, even if l(X) = ∞, while at the same time, R R a X s dl ¼ 0  X s dl ¼ 0, even if X s dl ¼ 1. Now consider the case when P 0 < a < ∞. Let s ¼ 1  i  p ai vAi , where the sets Ai (1  i  p) form a partition P of X. Then as ¼ 1  i  p ða ai ÞvAi and, by Proposition 3.1.5, Z

p X

s dl ¼ X

Z ai lðAi Þ;

as dl ¼ X

i¼1

p X

ða ai ÞlðAi Þ:

i¼1

R R Hence X as dl ¼ a X s dl, as we wished to show. P P Let s ¼ 1  i  p ai vAi and t ¼ 1  j  n bj vBj , where the sets Ai (1  i  p) as well as the sets Bj (1  j  n) form partitions of X. Then the sets Ai \ Bj (1  i  p, 1  j  n) form a partition of X. Moreover, s takes the value ai on the disjoint sets Ai \ Bj (1  j  n), so that [see Problem 3.1.P4(b) for details] s¼

p n X X i¼1

! ai vAi \ Bj :

j¼1

Therefore by Proposition 3.1.5, Z s dl ¼ X

p n X X i¼1

! ai lðAi \ Bj Þ :

ð3:8Þ

j¼1

A similar argument shows that Z t dl ¼ X

p n X X j¼1

! bj lðAi \ Bj Þ :

ð3:9Þ

i¼1

Next, we claim that ai lðAi \ Bj Þ  bj lðAi \ Bj Þ

for all i; j:

ð3:10Þ

If Ai \ Bj = ∅, then there is nothing to prove because l(Ai \ Bj) = 0. Consider the case when Ai \ Bj 6¼ ∅. Then some x belongs to the set. For this x, we have s(x) = ai because x 2 Ai and also t(x) = bj because x 2 Bj. Since s  t everywhere, we must have ai  bj. This inequality and the fact that l(Ai \ Bj)  0 together show that (3.10) holds in the present case as well, thereby completing the proof of the claim. The inequality resulting from taking the summation in (3.10) over all i and j, when seen in the light of (3.8) and (3.9), is precisely what we sought to prove. h Proposition 2.3.21 carries over easily to general measures [cf. Problem 2.8.P1].

3.1 Integrals of Simple Functions

117

Proposition 3.1.8 (Continuity) Let {En}n  1 be a sequence of sets belonging to F . Then (a) E1  E2   implies lðlim En Þ ¼ lim lðEn Þ [inner continuity], (b) E1  E2  

with

l(E1) < ∞

n!1

lðlim En Þ ¼ lim lðEn Þ

implies

n!1

[outer

continuity]. Proof The argument proceeds along the same lines as in Proposition 2.3.21, with (M2) serving the purpose that Proposition 2.3.13 did. h The following simple consequence of inner continuity will be superseded by the Monotone Convergence Theorem 3.2.4 later, but will be useful for establishing it. Proposition 3.1.9 Let {An}n  1 be a sequence of measurable subsets of X such that A1  A2  , and [ n  1 An = X. Then, for any simple function s, Z lim

n!1

X

Z ðsvAn Þdl ¼

s dl: X

Proof Any simple function is a finite sum of functions of the form avA. Therefore by the linearity of the integral for simple functions [Proposition 3.1.7], it is sufficient to prove for any nonnegative real number a that Z

Z

lim

n!1

X

ða vA vAn Þdl ¼

ða vA Þdl: X

Since vA vAn ¼ vA \ An , the integrals on the left- and right-hand sides here are al(A \ An) and al(A) respectively [see Example 3.1.4c]. Therefore we need only argue that lim lðAn \ AÞ ¼ lðAÞ. This equality follows from the inner continuity n!1

h

property of measures. Problem Set 3.1

R R 3.1.P1. Let X = [0,4], s = 2v[0,2] + 3v(2,4] and t = 6v(1,3]. Find X s dm and X t dm. R Also, find the canonical form of s + t and use it to compute X ðs þ tÞ dm from the definition.   3.1.P2. Let X = (0,4] and An ¼ 1n ; 4 , so that A1  A2   and [ n  1An = X. For R R R s = v(0,2], find X ðsvAn Þdm, X s dm: and lim X ðsvAn Þdm. n!1

3.1.P3. Let X be a measurable subset of R and {An}n  1 be a sequence of measurable subsets such that A1  A2  . If X\ [ n  1RAn has positiveR measure, show that there exists a simple function s such that lim X ðsvAn Þdl 6¼ X s dl: n!1

3.1.P4. Let X be a measurable set, Ai (1  i  p) disjoint measurable subsets of it and ai (1  i  p) nonnegative real numbers. (a) If the sets Ai do not form a partition of X, show that there are infinitely many simple functions taking the respective values ai on Ai.

118

3 Measure Spaces and Integration

(b) If the sets Ai form a partition of X, show that there is a unique simple function P s taking the respective values ai on Ai and that it is given by s ¼ 1  i  p ai vAi . 3.1.P5. Is Proposition 3.1.5 valid without the hypothesis that the Ai and the Bi form partitions of X? 3.1.P6. [Needed in 3.2.P13] Let s be a simple function on X. Define /s: [0,∞) ! [0,∞] as /s(u) = l(X(s > u)). If l(X) < ∞, then /s takes values in [0,∞). Show that (a) /s(u) = 0 if u  M = max{s(x) : x 2 X}. (b) /s is a step function. RM (c) The Riemann integral 0 /s ðuÞdu, which is equal to the improper integral R1 R 0 /s ðuÞdu in view of (a), is also equal to the measure space integral X s dl. (d) What happens if l(X) = ∞? 3.1.P7. Let {sn}n  1 be the sequence of simple functions given by snR(x) = 1/n for j xj  n and 0 for j xj > n. Show that sn ! 0 uniformly on X ¼ R, but X sn dm ¼ 2 for every n 2 N. [Uniform convergence on a set X of finite measure to a bounded limit function does imply convergence of integrals; this will be seen to be true in the next section even for more general functions in the light of the Dominated Convergence Theorem 3.2.16.] 3.1.P8. Optional. [The principle that permits grouping of terms in a sum according to Pany scheme whatsoever can be expressed in this manner: Suppose we have a sum 1  i  p bi to evaluate and that the set {i : 1  i  p} of indices has been partitioned into n subsets Nj, 1  j  n, meaning thereby that these subsets are nonempty and disjoint with union equal to {i : 1  i Pp}. Then the required sum can be evaluated as the grand total of the n subtotals i2Nj bi . In other words, ! X X X bi ¼ bi 1ip

provided

n S

1jn

i2Nj

Nj ¼ f1; 2; . . .; pg, every Nj 6¼ ∅ and j 6¼ j0 ) Nj \ Nj0 = ∅.

j¼1

This is valid even if one or more of the bi are ∞ as long as each bi is nonnegative. We shall refer to it as the grouping principle.] Let s be a simple function on a measurable set X such that s¼

p X

ai vAi ;

i¼1

where i 6¼ i0 ) Ai \ Ai0 ¼ £

and each Ai is measurable

3.1 Integrals of Simple Functions

119

and p [

Ai ¼ X:

i¼1

Using 2.5.P6 and the grouping principle, but not Proposition 3.1.5, show that p X

Z ai lðAi Þ ¼

i¼1

s dl:

ð3:11Þ

X

3.1.P9. If in a measure space, {An}n  1 is a sequence of sets of measure 0 and {Bn}n  1 is a descending sequence of sets of finite measure such that their measure tends to 0, show that \ n  1(An [ Bn) has measure 0. 3.1.P10. Let F be a r-algebra of subsets of a set X and f : X ! R be any function whatsoever. Show that the family G ¼ fAR : f 1 ðAÞ 2 F g of subsets of R is a r-algebra. Hence show that if f is F -measurable, then f 1 ðAÞ 2 F whenever AR is any Borel subset of R. 3.1.P11. Suppose f : R ! R and g : R  R ! R are both Borel measurable. Show that the composition f g : R  R ! R is Borel measurable. 3.1.P12. Let l be a measure on the Borel r-algebra B of subsets of R, satisfying the following properties: l(0,1) > 0; l(0,1) < ∞; for E 2 B and e > 0, there exists an open set O  E such that l(O\E) < e; l(E + x) = l(E) for an arbitrary E 2 B and arbitrary x 2 R (translation invariance).

(i) (ii) (iii) (iv)

Show that there exists a positive a 2 R such that l(E) = am(E) for all E 2 B.

3.2

Integrals of Measurable Functions

It is an immediate consequence of the monotonicity of the integral for simple functions [Proposition 3.1.7] that Z

Z s dl ¼ sup X

t dl : 0  t  s and

 t : X ! R a simple function :

X

The right-hand side here remains meaningful if we replace s by an arbitrary nonnegative extended real-valued function f. We take advantage of this to extend the concept of integral to a broader class of nonnegative functions, but restrict ourselves

120

3 Measure Spaces and Integration

to measurable ones (which means that Xð f [ aÞ 2 F for every real a); this is because we shall need the restriction for most of our proofs in the sequel: Definition 3.2.1 For any measurable extended nonnegative real-valued function f on X, we define the integral of f to be Z

Z



f ¼ sup

t dl : 0  t  f

and

t : X ! R a simple function :

X

It is also denoted by

R X

f,

R X

f dl or

R

f dl according to convenience.

The above notation for the integral can become inconvenient if we have an expression, such as f (x) = x2, 0  x < 1, available for the function but do not wish to introduce a letter f, or the like, to represent it. When this is so, we can use the R R R1 notation x2 dmðxÞ or ½0;1 x2 dx. The symbol 0 x2 dx will, as far as possible, be reserved for the Riemann integral. Remark 3.2.2 The set over which the sup is taken in the definition is nonempty, because the function t which has value 0 everywhere is simple and satisfies 0  t  f. Therefore the sup is always nonnegative. Note that existence of the sup poses no R difficulties, because we allow ∞ as a value. In this connection, we note that, if X f dl\1, then X(f = ∞) has measure zero, i.e. f has finite values almost everywhere. Indeed, if the aforementioned set, call it A, were to have positive measure, then for every n 2 N, the simple function t ¼ nvA would satisfy 0  tR  f, and hence by the above definition, we would have R 1 [ X f dl  X t dl ¼ n  lðA) for every n 2 N, a contradiction. It is easy to prove homogeneity and monotonicity of the integral for nonnegative extended real-valued functions. Proposition 3.2.3 Let f and g be measurable extended nonnegative real-valued functions on X and a  0 be a real number (not ∞). Then Z

Z ðaf Þdl ¼ a

X

If f  g everywhere, then

R X

f dl 

f dl: X

R X

g dl.

Proof If a = 0, both sides of the equality to be proved are 0. Suppose 0 < a < ∞. Then t is a simple function satisfying 0 R t  f if andRonly if at is a simple function satisfying 0  at  af. Also, X ðatÞ dl ¼ a X t dl by Proposition 3.1.7. Together with the Definition 3.2.1, this implies the required equality when 0 < a < ∞. The second part (i.e. monotonicity) follows from the observations that firstly, when f  g, any simple function t satisfying 0  t  g also satisfies 0  t  f, and that secondly, A  B ) supA  supB. h

3.2 Integrals of Measurable Functions

121

We are about to prove one of the fundamental results about Lebesgue integration. One straightforward consequence of it is that the infirmity in handling limiting processes with Riemann integration that was pointed out in Example 2.1.1 disappears in the present theory. Before stating the result, we point out that it has an alternative version in which the inequalities and convergence in (a) and (b) of the hypothesis are assumed to hold only almost everywhere; however, that complicates the precise statement of the conclusion because the limit function need not be measurable. These matters are relegated to Problem 3.2.P30. Theorem 3.2.4 (Monotone Convergence Theorem). Let {fn}n  1 be a sequence of measurable nonnegative extended real-valued functions on X such that, for every x belonging to the set, (a) 0  f1(x)  f2(x)  , (b) fn(x) ! f(x) as n ! ∞. Then f is measurable and Z

Z fn dl ¼

lim

n!1

f dl:

X

X

Proof From (a) and (b), we have fn(x)  f(x) everywhere on X. The measurability of f is a consequence of Corollary 2.5.5. R R R In view of monotonicity [Proposition R 3.2.3], we have X fn dl  X fn þ 1 dl  X f dl for all n. Therefore lim X fn dl n!1

exists as an extended real number and Z fn dl 

lim

n!1

Z

X

f dl:

ð3:12Þ

X

To arrive at the reverse inequality, consider any simple function s such that 0  s  f. For arbitrary b 2 (0,1), let An ¼ Xð fn  bsÞ for all n 2 N:

ð3:13Þ

Then each An is measurable. Moreover, our hypothesis (a) implies that A1  A2     :

ð3:14Þ

It is also true that x 2 X ) x 2 [ n  1An; indeed, for those x 2 X for which s(x) = 0, the implication follows from (a) and for those x for which s(x) > 0, it follows from (b) and the fact that b < 1 (because bs(x) < s(x)  f(x)). Thus 1 [ n¼1

An ¼ X:

ð3:15Þ

122

3 Measure Spaces and Integration

From the definition (3.13) of An, it follows that fn vAn  ðbsÞvAn everywhere on X. But fn  fn vAn and therefore fn  ðbsÞvAn . By monotonicity, we have Z

Z fn dl 

X

ððbsÞvAn Þdl:

X

ð3:16Þ

Since bs is a simple function, we conclude from R (3.14), (3.15) and Proposition 3.1.9 that the right-hand side of (3.16) has limit X ðbsÞdl as n ! ∞. We have already notedR that theR left-hand side R has a limit as n ! ∞. Therefore, we have lim X fn dl  X ðbsÞdl ¼ b  X s dl. Since this is true for all b 2 (0,1), it follows

n!1

that Z

Z fn dl 

lim

n!1

s dl:

X

X

But this has been established for an arbitrary simple function s such that 0  s  f. So, Z

Z fn dl 

lim

n!1

X

f dl: X

In conjunction with (3.12), this proves the required equality.

h

Remark If the hypothesis (a) above is replaced by f1  f2    0, then the conclusion need not hold. R R To see why, take fn = 1/n on X ¼ R. Then f = 0 on R and X f dl ¼ 0, but X fn dl ¼ 1 for every n. If we add the hypothesis that R X f1 \1, then the conclusion does indeed hold but the proof requires some results yet to come. A full discussion is therefore postponed to Remark 3.2.11(d). Theorem 3.2.5 Let f and g be measurable nonnegative extended real-valued functions on X and a,b  0 be real. Then Z

Z

Z

ðaf þ bgÞdl ¼ a

f dl þ b

X

X

g dl: X

Proof As homogeneity has already been proved in Proposition 3.2.3, it remains to prove only additivity, i.e., Z

Z ð f þ gÞdl ¼ X

Z f dl þ

X

g dl: X

By Proposition 2.5.9, there exist increasing sequences {sn}n  1 and {tn}n  1 of simple functions with respective limits f and g. It follows that the sequence

3.2 Integrals of Measurable Functions

123

{sn+ tn}n  1, which has limit f + g, is also an increasing sequence of measurable functions. Now, Z Z Z ðsn þ tn Þdl ¼ sn dl þ tn dl: X

X

X

Upon taking limits in this equality and appealing to the Monotone Convergence Theorem 3.2.4, we get additivity. h Corollary 3.2.6 Let {Ek}k  1 be a sequence of disjoint measurable subsets of X such that [ k  1Ek = X, and f be a measurable nonnegative extended real-valued function on X. Then Z f dl ¼ X

Proof Let fn ¼

n P k¼1

1 Z X k¼1

X

ð f vEk Þdl:

ð f vEk Þ. Since the measurable sets of the sequence {Ek}k  1 are

disjoint and [ k  1Ek = X, the sequence {fn}n  1 consists of measurable functions satisfying (a) and R (b) of the Monotone Convergence Theorem 3.2.4. Therefore R f dl ¼ lim fn dl. On the other hand, Theorem 3.2.5 implies that X n!1 X n R R P h X fn dl ¼ X ð f vEk Þdl. The required conclusion is now immediate. k¼1

Example 3.2.7 We illustrate the use of this Corollary. Let f : ð0; 1 ! R be defined by f(x) = [−logx], where [ ] denotes the integer part function and log means logarithm to base 10, and f(0) = 0. Then f is a step function on [e,1] whenever 0 < e < 1. In fact, for 10k1 \x  10k ;

f ðxÞ ¼ k

where k þ 1 2 N:

R It follows that ð0;1 f vEk = k(10−k − 10−k−1) = 9 k/10 k+1, where Ek = (10−k−1,10−k]. Now f  0, the measurable sets Ek are disjoint and [ k  0Ek = (0,1]. Therefore R P kþ1 , a convergent series by comparison Corollary 3.2.6 yields ð0;1 f ¼ 1 k¼1 9k=10 P p kþ2 [see Proposition 1.4.7] with the geometric series 1 , considering k¼1 1=ð 10Þ p p that k/10 k+1 = (k/( 10)k)(1/( 10)k+2). Theorem 3.2.8 (Fatou’s Lemma) Let {fn}n  1 be a sequence of measurable nonnegative extended real-valued functions on X. Then Z Z ðlim inf fn Þdl  lim inf fn dl: X

X

Proof Let {gn}n  1 be the sequence of functions on X defined by

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3 Measure Spaces and Integration

gn ðxÞ ¼ inf fk ðxÞ for all x 2 X and all n 2 N: kn

Then gn  fn everywhere on X, so that Z Z gn dl  fn dl: X

ð3:17Þ

X

Also, {gn}n  1 is an increasing sequence of measurable nonnegative extended real-valued functions such that, by definition of liminf [see Remark 2.5.3], lim gn ! lim inf fn everywhere. Therefore by the Monotone Convergence

n!1

Theorem 3.2.4, we get Z

Z

Z



ðlim inf fn Þ dl ¼ lim X

gn dl ¼ lim inf gn dl X Z  lim inf fn dl by ð3:17Þ and Remark 2:5:3: n!1

X

h

X

For an example when strict inequality holds in the above Theorem, see Problem 3.2.P2. So far we have dealt with nonnegative functions only. We now extend the concept of Lebesgue integral to measurable functions that take positive as well as negative values. Definition 3.2.9 For any measurable extended real-valued function f on X, the integral of f is defined to be Z Z Z þ f ¼ f  f ; R R provided that at least one among f þ and f  is finite. If both happen to be ∞, R R then f is taken as undefined. If f \1, the function is said to be integrable. The R R R integral is also denoted by X f , X f dl or f dl according to convenience. Proposition 3.2.10

R (a) Let A be a measurable subset of X such that l(Ac) = 0. If X f dl exists, then R R X f vA dl exists and equals X f dl. (b) Let f and g be measurable extended real-valued functions on X satisfying f = g a.e. Then f and g is integrable if and only if the other is, in which R one among R case X f dl ¼ X g dl. Proof (a) If B is any measurable subset, then so is B \ A; l(B \ Ac)  l(Ac) = 0. Therefore l(B \ Ac) = 0 and hence

moreover,

3.2 Integrals of Measurable Functions

125

lðBÞ ¼ lðB \ ðA [ Ac ÞÞ ¼ lðB \ AÞ þ lðB \ Ac Þ ¼ lðB \ AÞ: For any simple function s ¼

n P j¼1

aj vAj , we have svA ¼

n P j¼1

aj vAj \ A and therefore,

by the equality proved above, Z Z n n X X ðsvA Þdl ¼ aj lðAj \ AÞ ¼ aj lðAj Þ ¼ s dl: X

j¼1

j¼1

X

Thus the result holds for simple functions. Consider any measurable nonnegative function f. For any simple function such that 0  s  f, the and satisfies 0  svA  fvA. Therefore R Rfunction RsvA is simple s dl ¼ ðsv Þdl  f v dl. for every simple function s such A A R This holds X X X R that 0  s  f. Consequently, X f dl  X f vA dl. But the reverse inequality also holds, because fvA  f. This proves the required equality for measurable nonnegative extended real-valued functions. The equality now follows for measurable extended real-valued functions from the fact that ( fvA)+ = f +vA  f + and (fvA)− = f −vA  f − [see 2.5.P9]. (b) Consider measurable extended real-valued functions f and g on X satisfying f = g a.e. Accordingly, let A be a measurable subset of X such that l(Ac) = 0 + + and f = g on A. It is trivial that f + = g+ and f − = g− on R A,þ so thatR f vþA = g vA − − and f vA = g vA on X. By part (a), we have X f dl ¼ X f vA dl ¼ R þ R þ − and g−. Together with X g vA dl ¼ X g dl and correspondingly for f Definition 3.2.9, this leads to the conclusion that R one among R f and g is integrable if and only if the other is, in which case X f dl ¼ X g dl. h Remarks 3.2.11

R (a) If f is nonnegative-valued, then f = f + and f as defined now is the same as what it would be under Definition. 3.2.1. In particular, it is still true that R f  0 ) X f dl  0. (b) If an extended real-valued function f defined on X is integrable, then each of the sets Xð f ¼ 1Þ and Xð f ¼ 1Þ has measure zero [see Remark 3.2.2]. In other words, an integrable function is finite-valued almost everywhere. (c) Suppose f is a measurable extended real-valued function that is defined a.e. This means there exists some measurable Y  X such that f is defined and measurable on Y, while l(X\Y) = 0. Then there are several measurable extended real-valued functions F on X that agree with f on Y, for example, the one obtained by extending f outside Y to be ∞ on X\Y. Clearly, the positive and negative parts of all such functions F agree on Y and thus agree a.e. It follows from Proposition 3.2.10(b) that one such F is integrable if and only if all of them are, in which case,R they all have the same integral. This integral is what we shall understand by X f dl when f is defined a.e. on X.

126

3 Measure Spaces and Integration

(d) We resume the discussion about replacing the hypothesis (a) of the Monotone Convergence Theorem 3.2.4 by f1  f2    0 and adding the hypothesis R that X f1 \1. We aim to show that the conclusion of the theorem still holds. The additional hypothesis and Remark 3.2.2 together show that the function f1 is finite a.e. and accordingly, there exists a measurable subset A of X such that l(Ac) = 0 and f1 is finite on A. From the hypothesis that f1  f2    0, it follows that all the functions fn and the limit f are finite on A. Upon multiplying all functions concerned by vA, we obtain functions that not only satisfy the same conditions but are also finite everywhere on X. Moreover, we find from Proposition 3.2.10(a) that their respective integrals are the same as before. Therefore we may assume that the functions fn and the limit f are all finite-valued on X, which permits subtractions like f1 − fn and f1 − f. Now, the sequence of functions f1 − fn satisfies the hypotheses of the Monotone Convergence Theorem 3.2.4 and has limit f1 − f. Hence R R R lim X ð f1  fn Þdl ¼ X ð f1  f Þdl  X f1 dl\1. Since ( f1 − fn) + fn = n!1

f1 = ( f1 − f ) + f, it follows by Theorem 3.2.5 that Z Z Z Z Z ð f1  fn Þ dl þ fn dl ¼ f1 dl ¼ ð f1  f Þ dl þ f dl: X

X

Since lim

R

fn n!1 X

X

X

X

dl exists and, as already noted, Z Z ð f1  fn Þdl ¼ ð f1  f Þ dl; lim n!1

X

X

it follows from the above equality that Z

Z ð f1  f Þ dl þ lim X

n!1

Z fn dl ¼

X

Z ð f1  f Þ dl þ

X

f dl: X

All four terms involved in this equality are finite and hence the required conclusion follows. Proposition 3.2.12 (a) The function f is integrable if and only if its absolute value j f j is integrable. R  R R (b) If X f dl exists (perhaps ± ∞), then  X f dl  X j f j dl. Proof (a) In accordance withR the above definition, for f to be integrable, it is necessary R and sufficient that X f þ dl and X f  dl both be finite. This is the case if and

3.2 Integrals of Measurable Functions

127

R R only if X j f j dl < ∞, because j f j = f + + f −, which implies X j f j dl ¼ R R þ dl þ X f  dl by Theorem 3.2.5. Xf R (b) We need consider only the case when X j f j dl < ∞. Since j f j = f + + f −, we R R R have X j f j dl ¼ X f þ dl þ X f  dl by Theorem 3.2.5. Hence the finiteness R R R of X j f j dl implies the finiteness of both X f þ dl and X f  dl. Applying the above definition and using the finiteness just proved, we have Z  Z  Z  Z  Z Z Z         þ  f dl ¼  f þ dl  f  dl   f þ dl þ  f  dl ¼ f dl þ f  dl         X X X X X X X Z ¼ h j f jdl: X

As the reader can easily verify, ( f + g)+ is not the same as f + + g+. Therefore we make a slight detour to establish the additivity of the integral on the basis the above definition. Proposition 3.2.13 Suppose f = g − h, where f, g, h are extended real-valued functions on X, and g, h are nonnegative. Then f +  g and f −  h. [No measurability involved.] Proof It is sufficient to prove the first inequality, because f ¼ ðf Þ þ and − f = − g + h. Let x 2 X. If either g(x) = ∞ or f +(x) = 0, there is nothing to prove. So, let g(x) < ∞ and f +(x) > 0; since f +(x) = max{f(x),0}, we have f(x) > 0, and since f − (x) = max{ − f(x),0}, we further have f −(x) = 0. Therefore f +(x) = f +(x) − f −(x) = f(x) = g(x) − h(x). Since h(x)  0, it follows from here that f +(x)  g(x). h Proposition 3.2.14 Let f = g − h, where f, g, h are R measurable Rextended realvalued functions on X and g, h are nonnegative. Then X g dl < ∞) X f þ dl < ∞ R R R R and X h dl < ∞) X f  dl < ∞. If at least one among X g dl and X h dl is R R finite, then the same is true of X f þ dl and X f  dl, and furthermore, R R R X f dl ¼ X g dl  X h dl. Proof The two implications are immediate from Proposition 3.2.13 and they lead directly to the first part of the last statement. Only the equality needs an argument. To begin with, observe that the equality f + − f − = f = g − h leads to + − R þif not all the numbers involved are finite. Suppose Rf + h = f + g even g dl < ∞. Then dl < ∞ and X Xf Z

f þ dl þ

X

Since

R X

g dl and

R X

Z

Z

f  dl þ

h dl ¼ X

X

Z g dl: X

f þ dl are both finite, we deduce from here that

128

3 Measure Spaces and Integration

Z

Z

þ

Z



f dl 

Z

f dl ¼

X

g dl 

X

h dl:

X

X

In view of Definition 3.2.9, this R is precisely the equality we wished to prove. Similar considerations apply if X g dl < ∞ instead. h Theorem 3.2.15 Let f and g be integrable extended real-valued functions on X and a, b be real. Then af + bg is integrable and Z

Z

Z

ðaf þ bgÞdl ¼ a X

f dl þ b

g dl:

X

If f  g everywhere, then

R X

f dl 

R X

X

g dl.

Proof We first prove the result only for finite-valued functions. As observed in Remark 2.5.3(b), af + bg is measurable. By Proposition 3.2.3 and Theorem 3.2.5, Z

Z

Z

jaf þ bgj dl  X

Z

ðjajj f j þ jbjjgjÞ dl ¼ jaj X

j f j dl þ jbj X

jgj dl\1: X

Thus af + bg is integrable by Proposition 3.2.12. To establish the equality (called linearity of the Lebesgue integral) it is sufficient to prove additivity: Z

Z

Z

ð f þ gÞdl ¼ X

f dl þ X

g dl X

and homogeneity: Z

Z ðaf Þdl ¼ a

X

f dl: X

As f and g are both integrable and real-valued, so are f +, f −, g+, g− as well as f + g+ and f − + g−. Since all functions involved are real-valued, we have f + g = (f + + g+) − (f − + g−); besides, both the functions f + + g+ and f − + g− are nonnegative. Therefore it follows by Proposition 3.2.14 that Z Z Z ð f þ gÞdl ¼ ð f þ þ g þ Þdl  ð f  þ g Þdl: +

X

X

X

Since all integrals here are finite, we can rearrange terms on the right-hand side to get

3.2 Integrals of Measurable Functions

Z

Z ð f þ gÞdl ¼ X

ðf

þ



129

Z

 f Þdl þ

X

ðg

þ

Z



Z

 g Þdl ¼

X

f dl þ X

g dl: X

This proves additivity for real-valued Lebesgue integrable functions f and g. We next consider the case when the functions involved are extended real-valued. Let Y  X be the set of points where neither of the functions is ± ∞, i.e., both are real-valued. By Remark 3.2.11(b), l(X\Y) = 0 since the functions are integrable. Observe that f + g is defined and real-valued on Y. Thus f + g is defined a.e. By Remark 3.2.11(c), the desired conclusion about f + g will follow if we can prove the existence of some R R integrable R function F on X that agrees with f + g on Y and satisfies X F dl ¼ X f dl þ X g dl. Now, fvY and gvY are real-valued on all of X and are measurable by the second assertion of Remark 2.5.3(b). Also, f = fvY and g = gvY on Y, and hence F = (fR + g)vY agrees with f + gRon Y. Now, R R it follows from Proposition 3.2.10(a) that X f dl ¼ X f vY dl and X g dl ¼ X gvY dl. Hence, from the additivity proved above in the case, we find that ( f + g)vY is integrable and R real-valued R R ð f þ gÞv dl ¼ f dl þ g dl. Therefore F = ( f + g)vY is a function of the Y X X X kind we sought to prove the existence of. We next prove homogeneity. For a  0, the required equality is an immediate consequence of Definition 3.2.9 and the homogeneity of the integral for nonnegative functions [Proposition 3.2.3]. The case of negative a now follows by using the relations f − = ( − f )+ and f + = ( − f)−. R R R For monotonicity, we use the consequence X ð f  gÞ dl ¼ X f dl  X g dl of the linearity that has just been proved and argue that Z Z Z ð f  gÞdl  0 ) f dl  g dl: h f g ) f  g0 ) X

X

X

We close this section with the second of the two fundamental convergence theorems that highlight the advantages of the measure space integral over the Riemann integral. Before stating the result, we point out that it has an alternative version in which the inequality and convergence in the hypothesis are assumed to hold only almost everywhere; however, that complicates the precise statement of the conclusion because the limit function need not be measurable. These matters are relegated to Problem 3.2.P30. Theorem 3.2.16 (Dominated Convergence Theorem) Let {fn}n  1 be a sequence of measurable extended real-valued functions on X, converging everywhere to a function f and let g be an integrable function on X such that j fn ðxÞj  gðxÞ Then fn as well as f are integrable,

for every x 2 X:

130

3 Measure Spaces and Integration

Z j fn  f jdl ¼ 0

lim

n!1

Z

Z and

X

fn dl ¼

lim

n!1

X

f dl: X

Proof Since j fn j  g everywhere, it follows by Theorem 3.2.15 and Proposition 3.2.12(a) that all the fn are integrable. Since the hypotheses imply that j f j  g everywhere it also follows that f is integrable. The hypotheses also imply that lim inf ð2g  j fn  f jÞ ¼ 2g and that j fn  f j  2g so that the functions 2g  j fn  f j are nonnegative. Fatou’s Lemma 3.2.8 can therefore be applied to obtain via Theorem 3.2.15 that Z

Z ð2gÞdl  lim inf ð2g  j fn  f jÞ dl X X  Z  Z ð2gÞ dl þ lim inf  j fn  f jdl  ZX Z X ¼ ð2gÞdl  lim sup j fn  f j dl: X

Since

R X

X

ð2gÞ dl < ∞, we further obtain from here that Z j fn  f jdl  0:

lim sup X

This yields the first of the two equalities in question. We get the second equality from the first upon applying Proposition 3.2.12(a) to fn − f and using Theorem 3.2.15 yet again. h Remark The preceding theorem and the Monotone Convergence Theorem 3.2.4 R R both say that if fn ! f, then under an additional hypothesis, X fn dl ! X f dl. The need for an additional hypothesis is underscored by the following example: Let X = (0,1] and fn = nv(0,1/n]. Then each fn is a simple function with   1 fn dm ¼ n  0 ¼ 1 for every n 2 N: n X

Z

Also, fn ! f, where f is zero everywhere. Indeed, for 0 < x  1, Rthere exists an R n0 2 N such that fn(x) = 0 for n  n0. Now, X f dm ¼ 0 but lim X fn dm ¼ 1. n!1

Example 3.2.17 Let g:(0,1] ! R be defined by  gðxÞ ¼

0 ½ log x

if x is rational otherwise;

3.2 Integrals of Measurable Functions

131

where [ ] denotes the integer part function and log means logarithm to base 10. Let B be the complement in (0,1] of the rationals. Then Bc has measure 0. Since g = fvB, where f is as in Example 3.2.7, and the latter has been shown there to be integrable, it follows by Proposition 3.2.10(a) that g has the same integral. Remark 3.2.18 It will be seen in Problem 3.2.P13 that the Lebesgue integral of a nonnegative measurable function is actually the improper Riemann integral of a related function. In principle, the Lebesgue integral could therefore have been directly defined as being the latter, without bringing in simple functions. However, such an approach would have made it difficult to prove the additivity of the integral. In fact, the improper integral of the related function continues to make sense when (M2) of Sect. 3.1 is replaced by the weaker requirement of monotonicity: A  B ) l(A)  l(B). It is then known as the Choquet integral. If we add the further requirement that l be inner continuous, then we can even prove the monotone convergence property for the Choquet integral, but not additivity. This fact may be regarded as a partial explanation of why we needed to prove the monotone convergence property before establishing additivity: the latter is actually a deeper property, despite being more elementary in appearance! When (M2) is replaced by monotonicity, l is often called a fuzzy measure and the concept has found applications in the areas of multicriteria decision making and pattern recognition. In such a context, F is usually taken to consist of all subsets of a finite set X and l is assumed to take only finite nonnegative values. The interested reader may consult [9, 13, 14, 31, 32 or 33]. Proposition 3.2.19 For a bounded function f : ½a; b ! R, there exists an increasing sequence of step functions /n and a decreasing sequence of step functions wn such that (i) /n  f  wn, (ii) /n ! /, wn ! w, R R Rb Rb (iii) ½a;b / dm ¼ f dx, ½a;b w dm ¼ a f dx, a (iv) /(x) = w(x) if f is continuous at x, (v) f is continuous at x if x is not a point of any partition associated with any /n or wn and /(x) = w(x). Proof By definition of lower and upper integrals of any bounded function, there exist sequences fP0n gn  1 and fP00n gn  1 of partitions of [a, b] satisfying

Rb Rb f ðxÞdx and lim U f ; P00n ¼ a f ðxÞdx. Let P*n be the partition lim L ( f, P0n ) = a consisting of the points xk ¼ a þ k

ba ; 0  k  n: n

132

3 Measure Spaces and Integration

Now let P1 ¼ P1 and Pn þ 1 ¼ Pn [ Pn þ 1 [ ð [ 1  k  n þ 1 ðP0k [ P00k ÞÞ 8 n  1. Then each partition is a refinement of the preceding one. Therefore for n  2, Rb Rb f ðxÞdx and a f ðxÞdx  U(f, Pn)  U(f, P00n ). Since L(f, P0n )  L(f, Pn)  a Rb Rb f ðxÞdx and limU(f, P00n ) = a f ðxÞdx, it follows that limL(f, P0n ) = a Rb Rb f ðxÞdx and limU(f, Pn) = a f ðxÞdx. Note that the length of the limL(f, Pn) = a longest subinterval of Pn approaches zero. For each n, let /n be the step function that is constant on each open subinterval of Pn, the constant value being the infimum of f on the corresponding closed subinterval; we set /n(b) equal to its value on the subinterval extending to the left of b and at every other point of Pn, we set /n equal to its value on the subinterval extending to the right of that point. Then Rb Rb Rb Rb f ðxÞdx, a /n ðxÞ dx ! f ðxÞdx and we also have a /n dx = L(f, Pn)  a a /n  f everywhere on [a, b]. Since Pn+1 is a refinement of Pn, the bounded sequence {/n}n  1 of step functions is increasing. By a similar argument with suprema and upper sums, we get a bounded decreasing sequence of step functions Rb Rb wn  f such that lim a wn dx ¼ a f ðxÞdx. Thus there exists a pair of bounded increasing and decreasing sequences of step functions /n and wn respectively, which satisfy the requirement (i) and also satisfy Z

b

lim a

Z /n ðxÞdx ¼

b a

Z f ðxÞdx

and

b

lim

Z

b

wn ðxÞdx ¼

a

f ðxÞdx:

ð3:18Þ

a

Rb Rb Note that a /n ðxÞdx and a wn ðxÞdx are equal to the corresponding Lebesgue integrals, because /n and wn are step functions. Since both sequences of step functions are (pointwise) monotone, they have pointwise limits / and w, say. Then (ii) holds, as required. All functions concerned are bounded and therefore the Dominated Convergence Theorem 3.2.16 can be applied to deduce from (3.18) that the requirement (iii) is satisfied. To prove (iv), suppose f is continuous at x. Then for any e > 0, there exists a d > 0 such that sup f − inf f < e, where the sup and inf are taken over (x − d, x + d). Since the length of the longest subinterval of Pn approaches zero, for sufficiently large n, any interval of Pn containing x will lie in (x − d, x + d), and hence 0  wn(x) − /n(x) < e. Since e > 0 is arbitrary, it follows that /(x) = w(x). Finally, to prove (v), let x not be a point of any Pn and suppose /(x) = w(x). Then it follows from (i) and (ii) that /ðxÞ ¼ wðxÞ ¼ f ðxÞ:

ð3:19Þ

Let e > 0. Since {/n} is increasing and {wn} is decreasing, (ii) yields an n 2 N such that

3.2 Integrals of Measurable Functions

wðxÞ  wn ðxÞ\wðxÞ þ e

133

and

/ðxÞ  e\/n ðxÞ  /ðxÞ:

ð3:20Þ

In view of (i), we have f(x) − e < wn(x) and /n(x) < f(x) + e. If x0 belongs to the subinterval of Pn containing x, then by definition of /n and wn, the value of the function f at x0 satisfies /n ðxÞ  f ðx0 Þ  wn ðxÞ, so that (3.20) leads to /ðxÞ  e\/n ðxÞ  f ðx0 Þ  wn ðxÞ\wðxÞ þ e; whereupon (3.19) leads to f ðxÞ  e\f ðx0 Þ\f ðxÞ þ e: Since x must belong to the interior of the subinterval of Pn that contains it, an interval (x − d, x + d) is contained in the subinterval; this d then has the property that x0 2 (x − d, x + d) ) f(x) − e < f ðx0 Þ < f(x) + e. h Theorem 3.2.20 [Needed in Problems 3.2.P7-11 & 3.2.P23-24] If a bounded function f : [a, b] ! R is Riemann integrable, then it is Lebesgue measurable with Lebesgue integral equal to its Riemann integral. Proof Let / and w be as in Proposition 3.2.19. Then by (i) and (ii) therein, we have / R  f  w. In view of (iii), the Riemann integrability of f leads to ½a;b ðw  /Þdx = 0. Upon using Problem 3.2.P1(b) it follows that / = w a.e., which then implies / = f = w a.e. Therefore by Proposition 2.5.13, f must be Lebesgue measurable. Since /  f  w and all three are Lebesgue measurable, their Lebesgue integrals are in the same order. Together with (iii), this leads to Rb R h ½a;b f ðxÞdx ¼ a f ðxÞdx. Problem Set 3.2 3.2.P1. R (a) If X f dl > 0, where f is a measurable extended real-valued function on X, show that X(f > 0) has positive measure. (b) [Needed in Theorem 3.2.20, Problems 3.2.P6 & 3.2.P14 and Theorem 3.3.5] When f is nonnegative, show that if X(f > 0) has positive measure, then R f dl > 0. X R R (c) When f and g are measurable, f  g and X f dl ¼ X g dl, does it follow that f = g a.e.? on X = [0, 2] defined Rby fn = v[0,1] 3.2.P2. Let {fn}n  1 be the sequence of functions R if n is even and v(1, 2] if n is odd. Find X ðlim inf fn Þ dl and liminf X fn dl. 3.2.P3.R Let L be the class of all Lebesgue integrable functions on X. Define q(f, g) to be X j f  gj dl. Show that q(f, g) can be 0 when f 6¼ g but that q has all the other properties of a metric (i.e. q is a “pseudometric”). In what way is this different Rb if X = [a, b] and we set q(f, g) = a j f ðxÞ  gðxÞj dx in the class of Riemann integrable functions?

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3.2.P4. Give an example of a nonnegative real-valued function on [0,1] which is unbounded on every subinterval of positive length (so that it cannot have an improper Riemann integral) but has Lebesgue integral 0. 3.2.P5. (a) [Needed in Problems 3.2.P8(a) & 3.2.P9] Let Y be a measurable subset of X. If f is a measurable function on X, then its restriction to Y, denoted by f jY , is a measurable function on Y. Show that the product fvY (defined on X) has an R only if the restriction f jY has an integral, and that when this is so, Rintegral if and ð f v Þ ¼ YR Y ð f jY Þ. [Note: It is customary to denote this common value by RX f or f dl.] Y Y (b) Suppose f is a measurable R extended nonnegative real-valued function on R. Let E 2 M or B, and let E f ¼ 0. Show that f vanishes a.e. on E, i.e. {x 2 E : f(x) > 0} has measure zero. 3.2.P6. [Theorem due to Lebesgue] Show that a bounded function f:[a, b] ! R is Riemann integrable if and only if it is continuous a.e. 3.2.P7. Show that there does not exist a nonnegative Lebesgue integrable function g on [0,1] such that g(x)  n2xn(1 − x) everywhere. 3.2.P8. h i 1 (a) Let I denote the interval ð2k þ1 1Þp ; 2kp , where k 2 N. Using Problem 3.2.P5 and Theorem 3.2.20, show that the Lebesgue integral  Z  1 1 sin vI ðxÞ dmðxÞ x ½0;1 x is no less than

2 ð2k þ 1Þp.

(b) Hence show that, for f ðxÞ ¼ 1x sin 1x, we have R (c) Does ½0;1 f dm exist?

R ½0;1

f þ dm ¼ 1.

3.2.P9. Suppose that the function f:(0,1] ! R is Riemann integrable on [c,1] R1 R1 whenever 0 < c  1, and that lim c j f ðxÞjdx and lim c f ðxÞdx exist. Then by c!0 c!0 R1 R1 definition, these limits are respectively equal to 0 j f ðxÞjdx and 0 f ðxÞdx. Show R R1 that f is measurable and that its Lebesgue integral ½0;1 f is the same as 0 f ðxÞdx. [Remark: The corresponding assertion is true when f:[A,∞) ! R isR Riemann 1 integrable over [A, B] for any finite B > A and the improper integral A f ðxÞdx converges absolutely.] 3.2.P10. Suppose f:[0,1] ! R is Lebesgue integrable and is also Riemann inteR1 grable on [e,1] for every positive e < 1. Show that lim e f ðxÞdx exists and equals e!0 R1 R ½0;1 f dm. In other words, the Riemann integral 0 f ðxÞdx exists, possibly as an improper integral, and whether improper or not, equals the Lebesgue integral of f over [0,1]. [Remark: The corresponding assertion is true when f:[A,∞) ! R is Lebesgue integrable over [A,∞).]

3.2 Integrals of Measurable Functions

135

3.2.P11. Let fn:[0,1] ! R be defined by fn(x) = nx=ð1 þ n10 x10 Þ. Find the pointwise limit f of {fn}n  1 and show that the convergence is neither uniform nor monotone. Also, establish the convergence of Riemann integrals R1 R1 0 fn ðxÞdx ! 0 f ðxÞdx. 3.2.P12. Let f:[0,1]  [0,1] ! R satisfy | f(x, y)|  1 everywhere. Suppose that for each fixed x 2 [0,1], f(x, y) is a measurable function of y, and also that for each fixed y R2 [0,1], f(x, y) is a continuous function of x. If g:[0,1] ! R is defined as gðxÞ ¼ ½0;1 f ðx; yÞdmðyÞ, show that g is continuous. R1 Note: The improper Riemann integral 0 gðuÞdu is the sum of limits of Riemann integrals Z

c

lim

a!0

a

Z gðuÞ du þ lim

b!1

b

gðuÞ du; c

where c can be taken as any positive real number without affecting the value of the sum. The improper integral is finite if and only if each of the limits is finite. This will be needed in the next problem. 3.2.P13. Let f be a measurable extended nonnegative real-valued function on X. Define /f: [0,∞) ! [0,∞] as /f(u) = l(X(f > u)). Show that (a) /f is a decreasing function; (b) f  g ) /f  /g; (c) For an increasing sequence {hn} of measurable functions converging to h, we have /hn ! /h . R1 R (d) 0 /f ðuÞdu is equal to the integral X f dl. R R1 (e) 0 lðXð f [ uÞÞpup1 du ¼ X f p dl, where 0 < p < ∞. 3.2.P14. (a) Let f be an integrable function on a set X such R that, for every measurable R E  X, we have E f dl = 0 (which means X f vE dl = 0, as explained in Problem 3.2.P5). Show that f = 0 a.e. R (b) Let f be a Lebesgue integrable function on [a, b]. If ½a;X f dm = 0 for all x 2 [a, b], show that f = 0 a.e. on [a, b]. 3.2.P15. Let f be an integrable function on a set E. Show that for every e > 0, there R exists a d > 0 such that whenever A  E with l(A) < d, we have  A f dl < e R  (which means  E ð f vA Þ dl < e, as explained in Problem 3.2.P5). R 3.2.P16. Let f be an integrable function on [a, b]. If F(x) = ½a;x f dm, prove that F is continuous on [a, b]. If f is an integrable function on [a,∞), is it true that F is uniformly continuous on [a,∞)? 3.2.P17. Let {fn}n  1 be a sequence ofR measurableR nonnegative functions on R such that fn ! f a.e. and suppose that R fn dm ! R f dm < ∞. Show that for each R R measurable set E  R, E fn dm ! E f dm. [The result also holds on a general measure space.]

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3 Measure Spaces and Integration

3.2.P18. Let f be a measurable nonnegative function on R and E be a measurable subset of finite measure. Show that R R (a) E f ¼ limN!1 E fN , where fN(x) = min{f(x), N}; R R (b) R f ¼ limN!1 ½N;N f . 3.2.P19. Let E  R be measurable and g an integrable function on E. Suppose {fn}n  1 is a sequence of measurable functions on E such that | fn|  g a.e. Show that Z Z Z Z lim inf fn  lim inf fn  lim sup fn  lim sup fn : E

E

E

E

3.2.P20 (a) Let {fn}n  1 and {gn}n  1 be sequences of measurable functions on X such that | fn|  gn for every n. Let f and g be functions such that R R measurable limnfn = f a.e. and limngn = g a.e. If limn X gn ¼ X g\1, show that Z Z lim fn ¼ f: n

X

X

(b) Let {fn}n  1 be a sequence of integrable functions on X such that fn ! f a.e., R where f is also integrable. Show that X j fn  f jdl ! 0 if and only if R R X j fn jdl ! X j f jdl. 3.2.P21. Let the function f:I1  I2 ! R, where I1 and I2 are intervals, satisfy the following conditions: (i) The function f(x, y) is measurable for each fixed y 2 I2 and f(x, y0) is Lebesgue integrable on I1 for some y0 2 I2; @ (ii) @y f ðx; yÞ exists for each interior point (x, y) 2 I1  I2; (iii) There exists a nonnegative integrable function g on I1 such that   @  @y f ðx; yÞ  g(x) for each interior point (x, y) 2 I1  I2. R Show that the Lebesgue integral I1 f ðx; yÞdx exists for every y 2 I2 and that the function F on I2 given by Z FðyÞ ¼ f ðx; yÞdx I1

is differentiable at each interior point of I2, the derivative being Z @ f ðx; yÞdx: F 0 ðyÞ ¼ @y I1 3.2.P22. Use the result of Problem 3.2.P21 to prove that

R1 0

xn expðxÞdx ¼ n!

3.2 Integrals of Measurable Functions

137

R1 3.2.P23. Prove that the improper integral 0 exy sinx x dx, where it is understood that sin x x is to be replaced by 1 when x = 0, converges absolutely for y > 0, and evaluate it. R1 R1 3.2.P24. Assuming that 0 exy sinx x dx converges for y = 0, i.e. 0 sinx x dx converges, find its value by using the Dominated Convergence Theorem 3.2.16. It is understood that sinx x is to be replaced by 1 when x = 0. [In Problem 4.2.P10, we shall obtain the value without employing Lebesgue integration.] Show that sinx x is nevertheless not Lebesgue integrable over [1,∞). 3.2.P25. Let f be a nonnegative integrable function on [a, b]. For each n 2 N, let 1 P En = X(n − 1  f < n). Prove that ðk  1Þ  mðEk Þ\1. k¼1

3.2.P26. Let xcf(x) be integrable over (0,∞) Rfor c = a and c = b, where 0 < a < b. Show that for each c 2 (a,b), the integral ð0;1Þ xc f ðxÞdmðxÞ exists and is a continuous function of c. 3.2.P27. [Complement to Fatou’s Lemma] Let {fn}n  1 be a sequence of measurable functionsR on X such that | fn|  g,R where g is an integrable function on X. Prove that X lim sup fn dl  lim sup X fn dl and give an example to show that the condition about the function g cannot be dropped.

n R1 3.2.P28. Show that the improper Riemann integral 0 1 þ nx sin nx dx has limit 0 as n ! ∞.

n Rn 3.2.P29. Evaluate the limit as n ! ∞ of the Riemann integral 0 1  nx ex=2 dx. 3.2.P30. Let (X, F , l) be a measure space and {fk}k  1 be a sequence of functions on X which converges almost everywhere to a limit function f. In other words, there exists some N 2 F such that l(N) = 0 and fk(x) ! f(x) for every x 62 N. In ae symbols, fk ! f . (a) Give an example to show that f need not be measurable even if each fk is. ae (b) Show that fk ! g if and only if f = g a.e. (c) If each fk is measurable, show that there always exists a measurable g such that ae fk ! g. (d) State the Monotone and Dominated Convergence Theorems for the case when the inequalities and the convergence in the hypotheses are assumed to hold only almost everywhere. ae (e) If fk ! f and each fk is measurable, then under the additional hypothesis that (X, F , l) is complete, which is to say, FE; lðEÞ ¼ 0 implies F 2 F ; show that f has to be measurable. 3.2.P31. Let c be the counting measure [see Example 3.1.2(a)] on either N or {1,2,…, n}. Show for any nonnegative extended real-valued function f that the P integral is given by the sum k f ðkÞ.

138

3.3

3 Measure Spaces and Integration

Lp Spaces

It is well known that the uniform limit of a sequence of Riemann integrable functions (a) is Riemann integrable and (b) has integral equal to the limit of the integrals of the functions. It may be noted that the functions as well as the interval in the preceding statement are assumed to be bounded. This does not apply to improper Riemann integrals. For bounded functions, uniform convergence is equivalent to convergence with respect to the uniform (also called Chebychev) metric dð f ; gÞ ¼ supfj f ðsÞ  gðsÞj : s 2 Sg; where S is the common domain of the functions in question. In terms of this metric, the above assertions (a) and (b) about uniform convergence leads to completeness of the space of Riemann integrable functions and continuity of the Riemann integral as a real-valued function on it. Once the monotonicity and linearity of the Riemann integral are established, one can quickly derive the part about continuity, i.e. Z fn ! f uniformly

b

)

Z

b

fn ðxÞ dx !

a

f ðxÞ dx:

a

One can also prove the analogous statement that: Z

b

Z j fn ðxÞ  f ðxÞj dx ! 0

b

)

a

Z fn ðxÞ dx !

a

b

f ðxÞ dx:

a

This second assertion is broader than the previous one; indeed, if [a, b] = [0,1], Rb fn(x) = xn and f(x) = 0 for x < 1 and f(1) = 1, we have a j fn ðxÞ  f ðxÞjdx ! 0 but fn converges to f on [0,1] only pointwise, not uniformly. It is natural to ask whether the broader assertion can also be formulated as continuity of the integral with reference to some complete metric. The obvious candidate for the metric is Z

b

dð f ; gÞ ¼

j f ðxÞ  gðxÞjdx:

ð3:21Þ

a

However, this d will not quite do because it can happen that d(f, g) = 0 when f¼ 6 g (e.g. if f and g differ only at finitely many points), although it does have all the other properties of a metric [see Problem 3.2.P3]. In other words, d is only a pseudometric on the space of Riemann integrable functions in the sense of Definition. 1.3.3. As explained there, one can generate a metric space of equivalence classes using the equivalence relation d( f, g) = 0 and it turns out for equivalence classes ½½ f

; ½½g

with respective representatives f, g that dð½½ f

; ½½g

Þ = d( f, g) is

3.3 Lp Spaces

139

unambiguous (it is independent of the choice of representatives f, g) and it gives rise to a metric on the set of equivalence classes. The convergence ½½fn

! ½½ f

in the metric space generated in the above manner translates in terms of representatives as lim dð fn ; f Þ ¼ 0 and the Cauchy condition n!1

on f½½fn

gn  1 as

for every e [ 0; there exists an N 2 N such that dð fn ; fk Þ\e whenever n; k  N: Regarding the metric space so obtained from (3.21) above, one can ask whether it is complete. The answer is NO if we are working with the Riemann integral but YES if we switch to the Lebesgue integral or measure space integral, even if the domain is of infinite measure! The metric space obtained by switching to the measure space integral in (3.21), where f and g are now assumed to be real-valued and integrable in the sense of Sect. 3.2, is denoted by L1(X, F , l), or simply L1(X) or L1 if it is clear from the context what the intended measure space (X, F , l) is. If we are working in the general context of X, F and l as in the preceding two sections, then the metric space is denoted by any convenient abbreviation as above. Below we deal with a family of spaces of the L1(X, F , l) type, called Lp spaces, 1  p  ∞. We begin with p < ∞. Definition 3.3.1 Let 1  p < ∞. By Lp(X) or Lp(X, F , l) or simply Lp we mean the of real-valued measurable functions f defined on X for which R class p f dl\1, i.e., j j X  L ðXÞ ¼ p



Z p

f :

j f j dl\1 : X

We associate with each f 2 Lp(X) the real number Z kf kp ¼

1=p j f jp dl X

and call k f kp the Lp norm of f. Let f be a nonnegative extended real-valued measurable function defined on X and S be the set of all real M > 0 such that lðfx 2 X : f ðxÞ [ MgÞ ¼ 0: If S = ∅, set b = ∞. If S 6¼ ∅, put b = inf S. When this is so, lðfx 2 X : f ðxÞ [ bgÞ 

  1 X 1 l x 2 X : f ðxÞ [ b þ ¼ 0; n n¼1

ð3:22Þ

and hence b 2 S. We call b the essential supremum of f and denote it by esssup f.

140

3 Measure Spaces and Integration

When the only set of measure 0 is the empty set, as happens in the case of the counting measure, it is easily verified that the essential supremum is nothing but the ordinary supremum. Definition 3.3.2 If f is a real-valued measurable function, we define kf k1 to be the essential supremum of | f | and write k f k1 ¼ ess supj f j; and let L∞(X) or L∞ (X, F , l) consist of all f for which k f k1 < ∞, i.e.  L1 ðXÞ ¼ f : k f k1 \1 : The real number k f k1 is called the L∞ norm of f. Remark It is a consequence of (3.22) that | f(x)|  b for almost all x if and only if b  k f k1 . The space L∞(X) is also denoted by L∞ when convenient. Many of the important classes of spaces employed in Analysis consist of measurable functions. A large number of these spaces have norms defined on them via integrals. The Lp spaces, known as Lebesgue spaces, are among such important spaces of functions. These spaces were first introduced by F. Riesz. (D. Hilbert, in connection with his work on integral equations, introduced the L2 space. It is the natural infinite dimensional analogue of Euclidean spaces.) Aside from their intrinsic value in Analysis, they play a significant role in Fourier Analysis and Mathematical Physics, in particular, in Quantum Mechanics. The main tools in the study of these spaces are the inequalities attributed to Hölder and to Minkowski. Throughout this chapter, (X, F , l) will denote a measure space. However, sometimes X will be a Lebesgue measurable subset of R with Lebesgue measure m. Before proving the Hölder and Minkowski inequalities, we shall establish a lemma about real numbers, which is a generalisation of the inequality between arithmetic and geometric means. Lemma 3.3.3 Let a and b be nonnegative real numbers, and suppose 0 < k 6¼ 1. If k < 1, then ak b1k  ka þ ð1  kÞb with equality if and only if a = b. Proof Consider the continuous function u defined for nonnegative real numbers x by uðxÞ ¼ ð1  kÞ þ kx  xk : Then u0 ðxÞ = k(1 − xk−1) for x > 0, and so x = 1 is the only possible point for the extrema of u for x > 0. Since u00 ðxÞ = − k(k−1)xk−2 for all x > 0, it follows that

3.3 Lp Spaces

141

u00 ð1Þ > 0 because 0 < k < 1. Thus the function has a local minimum at x = 1. As uð0Þ > 0 and uð1Þ = 0, we infer that x = 1 is the point of absolute minimum of u. Consequently, for all x  0, we have ð1  kÞ þ kx  xk ¼ uðxÞ  uð1Þ ¼ 0; with equality if and only if x = 1. If b 6¼ 0, the lemma follows on substituting a/b for x, while if b = 0, the lemma is trivial. h Definition 3.3.4 If p and q are positive real numbers such that p + q = pq, or equivalently, 1 1 þ ¼ 1; p q then p and q are called a pair of conjugate exponents. It is clear that, if p and q are a pair of conjugate exponents, then 1 < p < ∞ and 1 < q < ∞. An important special case is p = q = 2. Observe that q ! ∞ as p ! 1. Consequently, 1 and ∞ are also regarded as a conjugate pair. Theorem 3.3.5 (Hölder’s Inequality) Let p and q be a pair of conjugate exponents, 1 < p < ∞. Then for nonnegative extended real-valued measurable functions f and g defined on X, Z

Z

1=p Z

fg dl  X

1=q

f p dl X

gq dl

:

ð3:23Þ

X

Furthermore, (a) if equality holds in (3.23) with both sides finite, then af p = bgq a.e. for some nonnegative constants a and b, not both zero (another way to say this is that either f p = cgq a.e. or gq = cf p a.e. for some nonnegative constant c); (b) if afp = bgq a.e. for some nonnegative constants a and b, not both zero, then equality holds in (3.23). Proof Let the factors on the right-hand side of (3.23) be denoted by A and B respectively. First suppose A = 0. Then the right-hand side of (3.23) is zero. Besides, f = 0 a.e. by Problem 3.2.P1(b) and hence the same is true of fg. Therefore the left-hand side of (3.23) is also zero. Hence (3.23) holds with equality. A similar argument works when B = 0. So, suppose A > 0 and B > 0. If either A or B is ∞, there is nothing to prove. It remains to consider the case when 0 < A < ∞ and 0 < B < ∞. So, suppose both are finite. In this remaining case, the functions f and g are finite a.e. and we may replace them by finite-valued functions without affecting any of the integrals involved. The finiteness allows us to apply the lemma with

142

3 Measure Spaces and Integration

 a¼

f ðxÞ A

p

  gðxÞ q 1 1 ;b ¼ ; k ¼ and 1  k ¼ : B p q

Thus,     f ðxÞ gðxÞ 1 f ðxÞ p 1 gðxÞ q  þ ; A B p A q B

ð3:24Þ

and integrating both sides, we get 1 AB

Z fg dl  X

1 pAp

Z f p dl þ X

1 qBq

Z gq dl ¼ 1;

ð3:25Þ

X

which immediately implies (3.23) in the remaining case. For the proof of (a), suppose equality holds in (3.23) and both sides are finite. Again, we first dispose of the cases A = 0 and B = 0. If A = 0, then f = 0 a.e. by Problem 3.2.P1(b) and hence the desired equality holds with a = 1 and b = 0. Similarly, if B = 0, then the desired equality holds with a = 0 and b = 1. In the remaining case that A 6¼ 0 6¼ B, we must have 0 < A < ∞ and 0 < B < ∞. As before, (3.24) must hold and (3.25) follows from it upon integrating both sides. Now equality in (3.23) implies equality in (3.25) and hence equality a.e. in (3.24). By Lemma 3.3.3, this further implies 

f ðxÞ A

p

  gðxÞ q ¼ a:e:; B

which leads to the desired equality with a = Bq and b = Ap, both of which are nonzero. This establishes (a). To prove (b), suppose that there exist a and b, not both zero, such that af p = bgq a.e. Plainly, we may assume that a 6¼ 0. This permits us to write f as cgq/p a.e., where c = (b/a)1/p. It follows that fg = cg1+(q/p) R= cgq a.e. It is now easy to verify that the two sides of (3.23) are both equal to c X gq . h Remark In (a) of the above theorem, the hypothesis that both sides are finite cannot be omitted. Indeed, we may have f(x) = 1 and g(x) = x everywhere on [0,∞) with Lebesgue measure, in which case, equality holds in (3.23) with both sides equal to ∞, but the kind of constants a and b claimed in (a) do not exist. In (b) of the above theorem, the conclusion that both sides are finite cannot be drawn. To wit, we may have f and g both equal to 1 everywhere on [0,∞) with Lebesgue measure, in which case both sides of (3.23) are ∞. If p = q = 2, Hölder’s Inequality is known as the Cauchy–Schwarz Inequality [Theorem 3.3.16].

3.3 Lp Spaces

143

Theorem 3.3.6 (Minkowski’s Inequality) Let 1 < p < ∞. Then for any nonnegative extended real-valued measurable functions f and g on X, the following inequality holds:

Z 1=p Z 1=p Z 1=p p p p ð f þ gÞ dl  f dl þ g dl : ð3:26Þ X

X

X

Furthermore, (a) if equality holds in (3.26) with both sides finite, then af = bg a.e. for some nonnegative constants a and b, not both zero (another way to say this is that either f = cg a.e. or g = cf a.e. for some constant c); (b) if af = bg a.e. for some nonnegative constants a and b, not both zero, then equality holds in (3.26). Proof To prove (3.26), we write ð f þ gÞp ¼ f ð f þ gÞp1 þ gð f þ gÞp1 : On applying Hölder’s Inequality to each of the products on the right, we get Z

Z f ð f þ gÞp1 dl 

f p dl

X

1p Z

X

1q

ð f þ gÞðp1Þq dl

ð3:27Þ

1q ð f þ gÞðp1Þq dl ;

ð3:28Þ

X

and Z

Z

1p Z

gð f þ gÞp1 dl 

gp dl

X

X

X

where 1/p + 1/q = 1. Since (p − 1)q = p, adding (3.27) and (3.28) gives 1q "Z

Z

Z p

p

ð f þ gÞ dl  X

ð f þ gÞ dl X

1p #

Z 1 p

f dlÞ þ p

X

p

g dl

:

ð3:29Þ

X

Clearly, it is enough to prove (3.26) in the case when the left-hand side is greater than zero and the right-hand side is less than infinity. Using the well-known inequality [found in [26, Theorem 1.1.7, p. 27]] ð f þ gÞp  2p1 ð f p þ gp Þ; we deduce that the left-hand side of (3.26) is finite. On dividing both sides of (3.29) R 1=q by X ð f þ gÞp dl , and keeping in view that 1/p + 1/q = 1, we get (3.26). For the proof of (a), suppose equality holds in (3.26) and both sides are finite. We first dispose of the case when both sides are 0. In this case, f = g = 0 a.e. and we take a = b = 1.

144

3 Measure Spaces and Integration

So, suppose both sides of (3.26) are positive. Since they have been assumed finite, the right-hand sides of (3.27) and (3.28) are finite. We claim that equality holds in (3.27) as well as (3.28). If not, then adding them and using the finiteness of their right-hand sides, we would obtain (3.29) with strict R 1=q inequality; upon dividing by X ð f þ gÞp dl , which is finite and positive in the present case, we would arrive at (3.26) with strict inequality. This justifies our claim that (3.27) and (3.28) both hold with equality. Since (3.27) and (3.28) both hold with equality, it follows upon using Theorem 3.3.5 that qf p ¼ rð f þ gÞðp1Þq a:e: and

cgp ¼ dð f þ gÞðp1Þq a:e:

for some nonnegative constants q and r, not both zero, and some nonnegative constants c and d, not both zero. It is easy to deduce from this that af = bg a.e., where a and b are some nonnegative constants, not both zero. This establishes (a). To prove (b), suppose that there exist a and b, not both zero, such that af = bg a.e. Plainly, we may assume that a 6¼ 0. This permits us to write f as cg a.e., where c = b/a. It follows that f + g = (1 + c)g a.e. and it is easy to verify that the two R

1=p sides of (3.26) are both equal to ð1 þ cÞ X gp . h Remark In (a) of the above theorem, the hypothesis that both sides are finite cannot be omitted. Indeed, we may have f(x) = 1 and g(x) = x everywhere on [0,∞) with Lebesgue measure, in which case, equality holds in (3.26) with both sides equal to ∞, but the kind of constants a and b claimed in (a) do not exist. In (b) of the above theorem, the conclusion that both sides are finite cannot be drawn. To wit, we may have f and g both equal 1 everywhere on [0,∞) with Lebesgue measure, in which case both sides of (3.26) are ∞. Theorem 3.3.7 Let p and q be a pair of conjugate exponents and let 1  p  ∞. If f 2 Lp(X) and g 2 Lq(X), then fg 2 L1 and k fgk1  k f kp kgkq :

ð3:30Þ

Proof For 1 < p < ∞, this is just Hölder’s Inequality applied to j f j and jgj. If p = ∞, observe that j f ðxÞgðxÞj  k f k1 jgðxÞj

ð3:31Þ

for almost all x. On integrating (3.31), we obtain (3.30). If p = 1, then q = ∞ and the same argument applies. h Theorem 3.3.8 Suppose 1  p  ∞, and f, g 2 Lp(X). Then f + g 2 Lp(X) and the inequality

3.3 Lp Spaces

145

kf þ gkp  k f kp þ kgk p

ð3:32Þ

holds. Also, af 2 Lp(X) when a is a real number, and kaf kp ¼ jajk f kp :

ð3:33Þ

Proof For 1 < p < ∞, (3.32) follows from Minkowski’s Inequality applied to j f j and jgj, since Z Z ðj f j þ jgjÞp dl: j f þ gjp dl  X

X

For p = 1, it is a trivial consequence of the inequality j f þ gj  j f j þ jgj. For p = ∞, the argument does not involve any integrals but is nevertheless simple enough to be left to the reader. As regards (3.33), when 1  p < ∞, Z

1=p p

kaf kp ¼

jaf j dl

Z ¼ jaj

X

1=p ¼ jajk f kp : j f j dl p

X

When p = ∞, the argument does not involve any integrals but, as before, is simple enough to be left to the reader. h Remarks 3.3.9 (a) When X = {1,2,…, n}, a function on X is nothing but an n-tuple x = (x1, x2,…, xn). It is real-valued if and only if x 2 Rn . With the counting measure on X, sums can be interpreted as integrals [see Problem 3.2.P31] and k x kp ¼

n X

!1=p p

jxi j

; where 1  p\1 and

k xk1 ¼ supfjxi j : 1  i  ng:

i¼1

The sum and sup are both taken over finitely many numbers and are therefore finite if and only if each xi is finite, i.e., if and only if x 2 Rn . Thus Rn and Lp(X) are the same set, 1  p  ∞. Also, Theorem 3.3.8 leads to !1=p !1=p !1=p n n n X X X  þ ; jxi þ yi jp jxi jp j yi j p i¼1 n X

i¼1

!1=p jaxi jp

i¼1

for 1  p < ∞ and to

¼ jaj

n X i¼1

!1=p jxi jp

i¼1

146

3 Measure Spaces and Integration

supfjxi þ yi j : 1  i  ng  supfjxi j : 1  i  ng þ supfjyi j : 1  i  ng; supfjaxi j : 1  i  ng ¼ jaj supfjxi j : 1  i  ng for p = ∞. It is left to the reader to formulate what Theorem 3.3.8 leads to when X = N with the counting measure. (b) Set

Z 1=p dð f ; gÞ ¼ j f  gjp dl ; f ; g 2 Lp ðXÞ; 1  p\1 X

and dð f ; gÞ ¼ kf  gk1 ;

f ; g 2 L1 ðXÞ:

Then 0  d(f, g) < ∞, d(f, f) = 0, d(f, g) = d(g, f) and it follows from Theorem 3.3.8 that the Triangle Inequality, namely, dð f ; gÞ  dð f ; hÞ þ dðg; hÞ;

f ; g; h 2 Lp ðXÞ

is satisfied. The only property of a metric that d fails to satisfy in general is that “d(f, g) = 0 implies f = g”. In other words, d is only a pseudometric on the space Lp(X) in the sense of Definition 1.3.3. As explained there, one can generate a metric space of equivalence classes using the equivalence relation d(f, g) = 0. It turns out for equivalence classes ½½ f

; ½½g

with respective representatives f, g that d ð½½ f

; ½½g

Þ = d(f, g) is unambiguous (it is independent of the choice of representatives f, g) and it gives rise to a metric on the set of equivalence classes. The resulting metric space whose elements are equivalence classes is again denoted by Lp(X), etc. In the space Lp of equivalence classes ½½ f

, one can introduce operations by setting ½½ f

þ ½½g

¼ ½½f þ g

and

a½½ f

¼ ½½af

:

It is left to the reader to verify that these equalities determine the sum of two equivalence classes and multiplication of an equivalence class by a scalar in an unambiguous manner, which is to say, independently of the choice of representatives f and g. After that, it is quite straightforward to check that the operations have the necessary properties to make Lp what is called a linear space [see Sect. 6.1]. In our present situation, d(f, g) = 0 precisely when f = g a.e. In the case of the counting measure, the only set of measure 0 is the empty set and equality a.e. amounts to equality everywhere. In particular, d is a metric. This applies to Rn discussed in part (a) above. The reader is reminded that the number k f kp is actually associated with the equivalence class ½½ f

of the function f. The resulting real-valued function on Lp(X), 1  p  ∞, has the following properties:

3.3 Lp Spaces

(i) (ii) (iii) (iv)

147

k f kp  0 8 f 2 Lp(X); k f kp = 0 , f = 0 (meaning ½½ f

= 0); kaf kp ¼ jajk f kp 8 f 2 Lp(X), 8 a 2 R; kf þ gkp  k f kp þ kgkp 8 f, g 2 Lp(X).

Here (i), (ii) and (iv) are essentially restatements of what we have already said; (iii) is immediate from the definition of k f kp . Since much of the work is done in terms of the representatives of the equivalence classes, we shall follow the customary practice by writing ½½ f

as simply f and speaking of the elements of Lp(X) as though they are functions. Thus for example, we may speak of ½½ f

as being continuous or nonnegative, meaning that at least one function in the equivalence class has the property. The notions of convergence and Cauchy sequence are defined with reference to the metric mentioned in the remark above. More specifically, we have the following: If {fn}n  1 is a sequence in Lp(X), if f 2 Lp(X) and lim kfn  f kp ¼ 0, we say that n!1

fn converges to f in Lp norm or fn converges to f in the sense of the mean of order p. If for every e > 0, there exists a positive n0 such that kfn  fk kp < e whenever n, k  n0, we say that the sequence {fn}n  1 is a Cauchy sequence in Lp(X). The next result shows that Lp(X), 1  p < ∞, is a complete metric space in the metric dð f ; gÞ ¼ kf  gkp ;

f ; g 2 Lp ðXÞ;

that is, every Cauchy sequence {fn}n  1 in Lp(X) converges to an element of Lp(X). We state and prove the result in terms of representatives rather than equivalence classes, that is, for the pseudometric space from which Lp(X) is constructed by taking equivalence classes. Theorem 3.3.10 If {fn}n  1 is a Cauchy sequence in Lp(X), where 1  p < ∞, then there exists an f 2 Lp(X) such that lim kfn  f kp ¼ 0. In other words, Lp(X) is a n!1

complete metric space when 1  p < ∞.

Proof Since the sequence {fn}n  1 is Cauchy in Lp(X), there exists a subsequence ffni gi  1 , n1 < n2 < n3 < , such that   fn  fn  \2i ; iþ1 i p

i  1:

ð3:34Þ

Put gk ¼

k  X  fn i¼1

iþ1

  fni ;

g¼

1  X  fn

iþ1

i¼1

  fni :

148

3 Measure Spaces and Integration

Since (3.34) holds, the Minkowski Inequality shows that kgk kp < 1, k = 1,2,…. By  Fatou’s Lemma applied to gpk , we have Z

Z lim gpk dl  lim inf

kgkpp ¼

X k!1

k!1

X

gpk dl  1:

Thus g 2 Lp(X) and hence is finite a.e. Consequently, 1 X

fn1 þ

ð fni þ 1  fni Þ

ð3:35Þ

i¼1

converges absolutely for almost every x 2 X. Denote the sum of (3.35) by f(x) for those x where (3.35) converges and set f(x) = 0 on the remaining set of measure zero. Since k1 X

fn1 þ

ð fni þ 1  fni Þ ¼ fnk ;

i¼1

we see that j fnk j  j fn1 j þ

k1  X  fn

iþ1

1  X    fn  fn  ¼ j fn j þ g ¼ h;  fni   j fn1 j þ iþ1 i 1

i¼1

say:

i¼1

Thus j fnk j  h for k = 1,2,…, where h 2 Lp(X), being the sum of Lp functions. We have proved that lim fni ¼ f a.e. Consider an e > 0 and choose N as in the i!1

Cauchy condition. Since ffni g is a subsequence of {fn}, we have Z    fn  fn p dl\e whenever i; j [ N: i

j

X

Then

Z X

   fn  f p dl  lim inf j i!1

Z

   fn  fn p dl  e whenever j [ N; i j

X

using Fatou’s Lemma. Thus, the subsequence ffni gi  1 converges in Lp(X) to the limit f. But {fn}n  1 is Cauchy and it therefore follows that lim kf  fn kp ¼ 0. h n!1

The argument for the following corollary is contained in the preceding proof. Corollary 3.3.11 If {fn}n  1 is a sequence in Lp(X) converging to f 2 Lp(X), where 1  p < ∞, then there exists a subsequence ffni gi  1 such that lim fni ¼ f

ni !1

a:e:

Any subsequence that converges a.e. must have f as its limit a.e.

3.3 Lp Spaces

149

In the case p = ∞, any Cauchy sequence {fn}n  1 converges uniformly to some limit function f 2 L∞(X) a.e., as will be seen from the proof below. Theorem 3.3.12 L∞(X) is a complete metric space. Proof Suppose {fn}n  1 is a Cauchy sequence in L∞(X). Let  Bm;n x 2 X : j fn ðxÞ  fm ðxÞj [ kfn  fm k1 and

E¼

[ m;n

Bm;n :

Since the absolute value of a function is greater than its essential supremum only on a set of measure zero, each of the sets Bm,n, m, n = 1,2,3,… has measure zero; it therefore follows that l(E) = 0. We assert that {fn}n  1 has the uniform Cauchy property of Theorem 1.5.4 on Ec. To prove this, consider any e > 0. We must show that there exists an integer n0 such that n, m  n0 implies j fn ðxÞ  fm ðxÞj\e for all x 2 Ec. By the given Cauchy property in L∞(X), we know that there exists an integer n0 such that n, m  n0 implies k fn  fm k1 \e. Now let x 2 Ec be arbitrary. By definition of E, we have x 2 Bcm;n for all n, m, i.e., j fn ðxÞ  fm ðxÞj  k fn  fm k1 for all n, m. Therefore n, m  n0 implies j fn ðxÞ  fm ðxÞj\e, which shows that n0 is indeed the kind of integer whose existence was to be shown. It now follows by Theorem 1.5.4 that {fn}n  1 converges uniformly on Ec to some real-valued limit function f. Upon defining f(x) = 0 for each x 2 E, we get a measurable function on X. By virtue of the uniform convergence, for every η > 0, there exists an n0 such that n  n0 implies j fn ðxÞ  f ðxÞj\g for all x 2 Ec, which further implies k fn  f k1  g because l(E) = 0. Thus fn − f 2 L∞(X) and h k fn  f k1 ! 0. Moreover, f = ( f − fn) + fn 2 L∞(X). Example 3.3.13 The distinction between the Cauchy condition of Theorem 3.3.10 and the condition that the sequence {fn}n  1 is Cauchy almost everywhere can be illustrated by taking fn(x) = 1 þn nx on [0,1]. Since fn(x) ! 1x when x 6¼ 0, this sequenceR {fn}n  1 isR Cauchy almost everywhere on [0,1]. Each fn is in L1[0,1] because ½0;1 j fn j ¼ ½0;1 1 þn nx dx = ln(1 + n) < ∞. However, ln(1 + n) ! ∞ and therefore even a subsequence of {fn}n  1 is unbounded in the metric space L1[0,1] and hence cannot satisfy the Cauchy condition. To see this without using a metric space argument, we begin by noting that, for n > k, we have   Z Z Z   n k  n k  dmðxÞ ¼   dmðxÞ j fn  fk j ¼  1 þ kx 1 þ kx ½0;1

½0;1 1 þ nx ½0;1 1 þ nx  Z 1 n k 1þn  : dx ¼ ln ¼ 1 þ nx 1 þ kx 1þk 0 When n > 2 k + 1, this exceeds ln 2, which shows that even a subsequence of {fn}n  1 cannot satisfy the Cauchy condition of Theorem 3.3.10.

150

3 Measure Spaces and Integration

The following proposition shows that Theorem 3.3.10 breaks down if we interpret the integral as the Riemann integral. Proposition 3.3.14 There exists a sequence {fn}n  1 of continuous functions on [0,1] which fulfils the Cauchy condition that for every e [ 0; there exists an N 2 N such that Z 1 j fn ðxÞ  fk ðxÞjdx\e whenever n; k [ N 0

but for which there exists no Riemann integrable function f satisfying Z

1

lim

n!1

j fn ðxÞ  f ðxÞjdx ¼ 0:

0

Proof Let r1, r2,… be an enumeration of the rationals in [0,1]. For each i 2 N, let gi be a continuous nonnegative function on [0,1] such that Z gi ðri Þ [ i

1

and

gi ðxÞdx ¼ 1=2i :

ð3:36Þ

0

Define fk ¼

P 1ik

gi . Then each fk is continuous and satisfies f k  gk

on ½0; 1 :

ð3:37Þ

Also, n > k ) fn  fk and Z

1

Z j fn ðxÞ  fk ðxÞjdx ¼

0

1

0

Z ¼

1

ð fn ðxÞ  fk ðxÞÞdx X

0 k\i  n

gi ðxÞdx ¼

X 1 \12k : i 2 k\i  n

It follows that {fn}n  1 is an increasing sequence of continuous functions that satisfies the Cauchy condition. Suppose, if possible, that f is a Riemann integrable function on [0,1] such that Z lim

n!1

1

j fn ðxÞ  f ðxÞjdx ¼ 0:

ð3:38Þ

0

Consider an arbitrary subinterval [a, b]  [0,1], where a < b; being Riemann integrable, j f j must have an upper bound M on [a, b]. Since a < b, the interval [a, b] must contain infinitely many rational numbers and hence must contain some rk with k > M + 1. By (3.36) and (3.37), fk(rk)  gk(rk) > k > M + 1. By continuity, [a, b] has some subinterval [a,b] of positive length on which fk > M + 1.

3.3 Lp Spaces

151

Now, j f j  M on [a, b]  [a,b] and therefore fk − f > (M + 1) − M = 1 on [a,b]. Since {fn}n  1 is an increasing sequence, we have fn  f [ 1

on ½a; b for all n  k:

Hence for any n  k, we have Z

1

Z j fn ðxÞ  f ðxÞjdx 

b

a

0

j fn ðxÞ  f ðxÞjdx  b  a [ 0;

in contradiction with (3.38). So, there can be no such Riemann integrable function on [0,1]. h Remark 3.3.15 It is worth noting that the above proof delivers something slightly more than promised. It shows that in order for f to be a limit of the sequence constructed, it has to be unbounded on every subinterval of [0,1] having positive length, which means that it cannot even be eligible to have an improper integral. An informal way to put it is that, in order to reach the completeness target, one has to overshoot improper integrals. Theorem 3.3.10 may be interpreted as saying that this, i.e., completeness is precisely what the Lebesgue integral achieves; besides, it does so without “sacrificing” any Riemann integrable functions in the process, as seen in Theorem 3.2.20. That it does so without “overkill” will be the main point of Proposition 3.4.2. Before embarking on the preliminaries for it, we remind the reader that a conditionally convergent improper (Riemann) integral is not a Lebesgue integral, as illustrated in Problem 3.2.P24. We have seen that for 1  p  ∞, the spaces Lp ðX; F ; lÞ equipped with a norm Z k f kp ¼

1=p j f jp dl

;

f 2 Lp ðXÞ;

1  p\1

X

k f k1 ¼ ess supj f j;

f 2 L1 ðXÞ

and metric R 1=p ; f ; g 2 Lp ðXÞ; 1  p\1 dð f ; gÞ ¼ X j f  gjp dl and dð f ; gÞ ¼ kf  gk1 ; f ; g 2 L1 ðXÞ; are complete with respect to the metric. The case when p = 2 plays an important role in the study of mathematical physics, probability theory and several other areas of mathematics. In the study of analytic geometry, we recall that orthogonality of two vectors is determined analytically by considering their inner (or dot) product. The space L2(X, F , l) admits an inner product that gives rise to a norm coinciding with the one described in the

152

3 Measure Spaces and Integration

paragraph above. This facilitates the study of the space. These are some of the reasons why we discuss L2(X, F , l) separately. Theorem 3.3.16 (Cauchy–Schwarz Inequality) Let f and g be measurable functions on X. If f 2 L2(X) and g 2 L2(X), then fg 2 L1(X) and Z

Z fg dl 

12 Z

2

2

f dl

X

12

g dl

X

:

X

Moreover, equality holds if and only if either j f j ¼ cjgj or jgj ¼ cj f j a.e. for some constant c. h

Proof This follows from Hölder’s Inequality of Theorem 3.3.5.

Theorem 3.3.17 (Minkowski’s Inequality) Let f and g be measurable functions on X. If f 2 L2(X) and g 2 L2(X), then f + g 2 L2(X) and

Z

1=2 2

j f þ gj dl X

Z

1=2



2

f dl X

Z

1=2

þ

2

g dl

:

X

Moreover, equality holds if and only if either j f j ¼ cjgj or jgj ¼ cj f j a.e. for some constant c. Proof This follows from Minkowski’s Inequality of Theorem 3.3.6.

h

If f 2 L2(X) and g 2 L2(X), then the inner product of f and g is defined to be Z h f ; gi ¼

fg dl: X

By the Cauchy–Schwarz Inequality [Theorem 3.3.16], 〈 f, g〉 is always a (finite) real number. The inner product is also known as the scalar product or as the dot product and is often denoted by fg. The number 〈 f, g〉 is associated with the pair of equivalence classes ½½ f

; ½½g

, as can be checked by using the Cauchy–Schwarz Inequality, and Theorem 3.3.8 with p = 1. The following properties of the inner product are used constantly without reference: Proposition 3.3.18 Let f 2 L2(X), g 2 L2(X) and h 2 L2(X). Then (a) (b) (c) (d)

〈f, g〉 = 〈g, f〉; 〈kf, g〉 = k〈f, g〉 8 k 2 R; 〈f + g, h〉 = 〈f, h〉 + 〈g, h〉; hf ; f i ¼ k f k2 ¼ k f k22 .

Proof The proof is straightforward and is left to the reader. h Elements f, g of L2(X) are said to be orthogonal if 〈 f, g〉 = 0. As the reader can verify by evaluating the relevant integrals, that when k and n are distinct

3.3 Lp Spaces

153

nonnegative integers, the functions in L2[−p,p] given by f(x) = sin kx and g(x) = sin nx are orthogonal, and so are those given by f(x) = cos kx and g(x) = cos nx. The functions given by f(x) = cos kx and g(x) = sin nx, where k and n need not be distinct, can also be verified to be orthogonal. An element f 2 L2(X) is said to be normalised if k f k = 1, i.e. if 〈f, f〉 = 1. It is left to the reader to verify independently that (a) when n is a positive integer, the function in L2[−p,p] given by cos nx f ðxÞ ¼ pffiffiffi p

or

is normalised; (b) the functions in L2[−1,1] given by f ðxÞ ¼

sin nx f ðxÞ ¼ pffiffiffi p qffiffi

1 2

and gðxÞ ¼

qffiffi 3 2x are orthogonal

and each one is normalised. The reader may compute the relevant integrals to confirm this. We have already encountered convergence pointwise, convergence pointwise a.e., uniform convergence and convergence in the Lp metric of a sequence of functions. The following implications are trivial: uniform convergence ) pointwise convergence ) pointwise convergence a:e: It is known from elementary analysis that the reverse of the first implication is not true. The reverse of the second implication is also not true, as demonstrated in Examples 2.5.11(b) and (c). In the case when l(X) < ∞, it is easy to see that k f kp  k f k1 lðXÞ1=p . It follows that L∞(X)  Lp(X) and that convergence in the L∞ metric implies convergence in the Lp metric. Convergence in the L2 metric is also called mean square convergence: R limn!1 X ½fn ðxÞ  f ðxÞ 2 dlðxÞ ¼ 0, or limn!1 kfn  f k2 ¼ 0. The notion applies to R functions g for which X gðxÞ2 dlðxÞ is finite, the so-called square integrable functions. In the next chapter, we shall study mean square convergence of Fourier series of square integrable functions in the context of the Riemann integral as well as the Lebesgue integral. The use of the Lebesgue integral is preferred because of the availability of a completeness property [Theorem 3.3.10]. Problem Set 3.3 3.3.P1. Give an example of a nonnegative extended real-valued function on [0,1] which has Lebesgue integral 1 and is (a) the limit of an increasing sequence of continuous functions (b) unbounded on every subinterval of positive length (so that it cannot have an improper Riemann integral).

154

3 Measure Spaces and Integration

R 3.3.P2. Give an example when X j fn  f jdl ! 0 but fn does not converge pointwise to f anywhere. 3.3.P3. Let {fk}k  1 be a sequence of integrable functions on a measurable set 1 R 1 P P X such that fk converges a.e. to an inteX j fk j\1. Show that the series k¼1

grable sum f and that

R X

f ¼

k¼1

1 R P

X fk :

k¼1

3.3.P4. For f 2 L1 ðRÞ, show that Z lim

h!0

R

j f ðx þ hÞ  f ðxÞjdx ¼ 0;

i.e., the mapping u : R ! L1 ðRÞ defined by uðhÞ = fh, where fh(x) = f(x + h), x 2 R, is continuous. 3.3.P5. Suppose that f 2 L1 ðRÞ. Show that Z

Z

lim

jhj!1

R

j f ðx þ hÞ þ f ðxÞjdx ¼ 2

R

j f ðxÞjdx:

3.3.P6. If f 2 L2[0,1], show that f 2 L1[0,1]. However, if f 2 L2[0,∞) then f need not belong to L1[0,∞). 3.3.P7. The functions in L2[−p,p] given by fn(x) = sin nx, n 2 N, form a bounded and closed subset which is not compact. 3.3.P8. Let f and g be positive measurable functions on [0,1] such that fg  1. Prove that ! Z

Z

!

f dm ½0;1

½0;1

g dm  1:

3.3.P9. Show that the following inequalities are inconsistent for a function f 2 L2[0,p]: 2 kf  cosk  ; 3

kf  sink 

1 : 3

3.3.P10. Let fn ðxÞ ¼ 1 þnnpffix for x 2 [0,1] and n 2 N. (a) Show that the sequence {fn}n  1 has a pointwise limit a.e.; (b) Does fn belong to L2[0,1] for each (or some) n 2 N? (c) Does the pointwise limit (a.e.) of fn belong to L2[0,1]? 3.3.P11. [Cf. Problem 3.3.P14] Let ðX; F ; lÞ be a measure space with l(X) < ∞. Show that L∞(X)  Lp(X) for all p, 0 < p < ∞. Also, show that, if f is measurable, then

3.3 Lp Spaces

155

lim k f kp ¼ k f k1 :

p!1

3.3.P12. (a) Let 0 < p < 1 and let q be such that 1p þ 1q ¼ 1. Then q < 0. If f 2 Lp(X) and the R R function g on X is such that g 6¼ 0 a.e. and X jgjq dl\1, then X jgjq dl 6¼ 0 unless l(X) = 0, and Z

Z

1=p Z p

j fgjdl  X

q

j f j dl X

1q

jgj dl

:

X

R (This is Hölder’s Inequality for 0 < p < 1.) If the hypothesis that X jgjq dl\1 is omitted, then the inequality is valid trivially, provided we interpret a negative power of ∞ to mean 0. (b) Let 0 < p < 1 and f, g be measurable functions on X such that f  0, g  0. Then kf þ gkp  k f kp þ kgkp : (This is Minkowski’s Inequality for 0 < p < 1.) (c) For a  complex-valued measurable function R R  / dl  j/j dl. X X

/

on

X,

show

that

3.3.P13. Let 0 < p < 1 and X = [0,1]. Then there exist f, g 2 Lp(X) such that kf þ gkp [ k f kp þ kgkp : 3.3.P14. [Cf. Problem 3.3.P11] Show that, if l(X) < ∞ and 0 < p < q  ∞, then Lq(X)  Lp(X). Also, show that when l(X) = ∞, the inclusion does not hold. 3.3.P15. Let X be a measurable subset of R and 0 < p < q < ∞. If f 2 Lp(X) \ Lq(X), then f 2 Lr(X) for all r such that p < r < q. 3.3.P16. (a) Let p  1 and let kfn  f kp ! 0 as n ! ∞. Show that kfn kp ! k f kp as n ! ∞. (b) Suppose {fn}n  1 is a sequence in Lp(X), f 2 Lp(X), fn ! f a.e. and kfn kp ! k f kp as n ! ∞. Prove that kfn  f kp ! 0 as n ! ∞. R 3.3.P17. Show that ½0;p x1=4 sin x dx  p3=4 . 3.3.P18. For each n, consider the functions fn : R ! R given by fn = v[n,n+1], n = 1,2,…. Show that fn(x) ! 0 as n ! ∞ for each x 2 R. Also, show that for all p such that 1  p  ∞, we have fn 2 Lp ðRÞ, kfn kp ¼ 1 for all n, so that kfn kp 90

156

3 Measure Spaces and Integration

as n ! ∞. (This example shows that pointwise convergence does not imply convergence in any Lp norm, 1  p  ∞.) 3.3.P19. Give an example to show that convergence in the pth norm does not imply convergence a.e. 3.3.P20. Let p and q be conjugate indices and lim kfn  f kp ¼ 0 ¼ lim kgn  gkq , n!1

n!1

where fn, f 2 Lp(X) and gn, g 2 Lq(X), n = 1,2,…. Show that lim kfn gn  fgk1 ¼ 0. n!1

3.3.P21. For 1  p < ∞, we denote by ‘p the space of all sequences {xm}m  1 such 1 P that jxm jp \1. m¼1

(a) Without interpreting sums as integrals with respect to the counting measure, show that, if fxm gm  1 2 ‘p and fym gm  1 2 ‘q , where 1 < p, q < ∞ and 1 1 p þ q ¼ 1, then 1 X

j xm ym j 

m¼1

1 X

!1=p j xm j

p

m¼1

1 X

!1=q jym j

q

:

m¼1

For p = 1, q = ∞, show that 1 X

j xm ym j 

m¼1

1 X

!  jxm j fym g

m¼1

 

m  1 1:

(b) Using the inequality of part (a), show also that, if fxm gm  1 2 ‘p and fym gm  1 2 ‘p , then

1 X

!1=p p

jxm þ ym j



m¼1

1 X

!1=p p

jxm j

þ

m¼1

1 X

!1=p p

jym j

:

m¼1

[The inequalities in (a) and (b) are known as Hölder and Minkowski Inequalities respectively for sequences.] 3.3.P22. For the space ‘p (1  p < ∞) of Problem 3.3.P21, show that dp ðx; yÞ ¼

1 X

!1=p j xm  ym j

p

m¼1

defines a complete metric, without using Theorem 3.3.10. P 3.3.P23. Show that, if k1, k2,…, kn > 1, ni¼1 k1i ¼ 1 and fi 2 Lki ðXÞ for each i, then

3.3 Lp Spaces

157

 k1 Z Y  n n Z Y i   ki j fi j dl :  fi dl    X i¼1 X i¼1 Equality occurs if and only if either fi = 0 a.e. for some i or there exist positive constants ci, 1  i  n, such that  k ci j fi jki ¼ cj  fj  j a:e:1  i; j  n: 3.3.P24. (a) Let ∞  p  1 and fi 2 Lp(X) for i = 1,2,…, n. Show that    X n n X   fi   kf i kp :   i¼1  i¼1 p

(b) Let ∞ > p > 1. If equality holds in (a), show that there exist nonnegative constants ci, 1  i  n, not all 0, such that cifi = cjfj a.e. for 1  i, j  n. (c) If either p = 1 or p = ∞, show that the analogue of (b) does not hold. 3.3.P25. Show that, if for some p, where 0 < p < ∞, f 2 Lp(X) \ L∞(X), then for all q such that p < q < ∞, we have f 2 Lq(X) and ð1p=qÞ : k f kq  k f kp=q p k f k1

3.4

Dense Subsets of Lp

According to Definition 2.5.6, a simple function is required to be nonnegative. If we drop this requirement, the resulting function f will be a difference of simple functions, because f +, f − will both be simple. The next result says essentially that the integrable functions among such differences of simple functions form a dense subset of the space of integrable functions. Proposition 3.4.1 Given any f 2 Lp(X), 1  p < ∞, Rthere exists a sequence {sn}n  1 of differences of simple functions such that lim X jsn  f jp dl ¼ 0. If f is n!1

nonnegative, then we can choose the sequence {sn}n  1 to consist of nonnegative functions. In particular, differences of simple functions are dense in Lp(X). Proof Suppose f is nonnegative. By Theorem 2.5.9, there exists a sequence {sn}n  1 of simple functions such that

158

3 Measure Spaces and Integration

ðaÞ

0  s1 ðxÞ  s2 ðxÞ      f ðxÞ;

ðbÞ

sn ðxÞ ! f ðxÞ as n ! 1:

R Observe that jsn ðxÞ  f ðxÞjp  j f ðxÞjp . Since f 2 Lp(X), i.e., X j f jp dl\1, using R the Dominated Convergence Theorem 3.2.16, we have lim X jsn  f jp dl ¼ 0. n!1

Observe that the sequence {sn}n  1 consists of nonnegative functions. We get the result for a general f 2 Lp(X) by applying what has been proved to f +, − f separately. h Most of our considerations in this chapter have been valid for a general measure l. In the rest of this section, we shall be considering only Lebesgue measure m. We begin by proving that continuous functions on a compact set X are dense among integrable functions on X, which is the precise version of what we called “without overkill” in Remark 3.3.15. Proposition 3.4.2 Given any function f 2 Lp(X), 1  p < ∞, where XR is a compact set, there exists a sequence {fn}n  1 of continuous functions with domain X such that lim j fn  f jp dm ¼ 0. If f is nonnegative, then we can choose the sequence n!1

{fn}n  1 to consist of nonnegative functions. In particular, continuous functions are dense in Lp(X). Proof It is sufficient to prove this when f is a characteristic function vE on X with E  X. For, it will first extend immediately to simple integrable functions and then, by Proposition 3.4.1, to all functions in Lp(X). The case when m(E) = 0 needs no argument, because we can take every fn to be zero everywhere. So we work with the hypothesis that 0 < m(E) < ∞. The first step is to reduce to the case when E is compact. From the hypothesis 0 < m(E) < ∞, it follows by Problem 2.3.P18 that for each e > 0, one can find a compact K  E such that 0  m(E) − m(K) < e. Then Z Z Z Z p 0 ðvE  vK Þdm ¼ vE dm  vK dm jvE  vK j dm ¼ X

X

X

X

¼ mðEÞ  mðKÞ\e: This means that if the required kind of sequence can be found for every vK with K compact, then the same can be done for every vE with E  X. Therefore we need only prove the result for f = vK with nonempty compact K (the case when K = ∅ being trivial). Accordingly, consider any nonempty compact K  X. Define the sequence of functions gn on R as gn ðxÞ ¼

1 ; 1 þ n  dðx; KÞ

ð3:39Þ

3.4 Dense Subsets of Lp

159

where d(x, K) denotes the distance of x from K as in Proposition 1.3.24. As shown there, it is a continuous function of x and therefore every gn is continuous. (K has to be assumed nonempty for Proposition 1.3.24 to be applicable.) It follows that the restriction of gn to X is continuous, as is obviously the case with a restriction of any continuous function. In the rest of this proof, gn will mean the restriction to X. Then each gn satisfies 0  gn  1 everywhere on X:

ð3:40Þ

The compactness of K guarantees that dist(x, K) = 0 , x 2 K. Together with (3.39), this implies gn ! vK everywhere on X: Now (3.40) yields jgn  vK j  1 everywhere on X: Since X is compact, it is bounded and hence its Lebesgue measure must be finite and hence the constant function 1 is integrable on it. Therefore the Dominated Convergence Theorem 3.2.16 can be used to conclude that Z jgn  vK jp dm ¼ 0;

lim

n!1

X

which proves the result for f = vK. As already noted, this is all that we needed to prove. The fact that the gn are nonnegative justifies the assertion of the theorem that the functions fn can be chosen to be nonnegative if f is nonnegative. h When X is an interval [a, b], the space Lp ðX; F ; mÞ with Lebesgue measure m and F the r-algebra of either Lebesgue measurable sets or Borel sets will be denoted by Lp[a, b]. Proposition 3.4.3 Suppose f 2 LRp[a, b], 1  p < ∞ and let e > 0. Then there exists a step function h such that ½a;b j f  hjp dm\e. If f is nonnegative, then we can choose the step function h to be nonnegative. In particular, step functions are dense in Lp(X). Proof As f = f + − f −, we may assume throughout that R f  0. By Proposition 3.4.1, there exists a simple function s  0 such that ½a;b j f  sjp dm\e=2. The simple function s may be expressed as s¼

n X i¼1

ai vEi

160

3 Measure Spaces and Integration

e with [ 1  i  nEi = [a, b]. Let e0 ¼ 2nM p , where M = sup s on [a, b], and M is supposed to be positive. For each of the measurable sets Ei, there exist a union Gi of finitely many disjoint open intervals such that m(EiDGi) < e0 [see Problem 2.3.P14]. This inequality continues to hold if we replace each open interval by its intersection with  [a, b]. Then vGi is a step function on [a, b] such that R 0  p ½a;b vEi  vGi dm ¼ mðEi DGÞ\e . Therefore

Z

p    n n X X e   ai vGi  dm\ api e0  nM p e0 ¼ : s    2 ½a;b

i¼1 i¼1

By Minkowski’s Inequality of Theorem 3.3.6, p #1=p "Z p #1=p  #1=p "Z      n n X X     p ai vGi  dm  þ ai vGi  dm j f  sj dm f  s      ½a;b

½a;b

½a;b

i¼1 i¼1  e1=p \2 ; 2

"Z

and

n P

ai vGi is step function.

h

i¼1

We note that a bounded measurable function on a closed bounded interval is integrable and therefore the above proposition is applicable. Proposition 3.4.4 Trigonometric polynomials form a dense subset of L2[−p,p] and of L1[−p,p]. Proof Consider any f 2 L2[− p,p] and an arbitrary e > 0. In view of Proposition 3.4.2, there exists a continuous / such that kf  /k\e=2:

ð3:41Þ

Define w by w(x) = /(x) in [−p, p−d], w(p) = /(−p) and w linear in [p−d, p], where d 2 (0,2p) will be chosen subsequently but is arbitrary for now. If K > 0 is such that j/ðxÞj  K, then jwðxÞj  K as well. Therefore Z k/  wk2 ¼

½pd;p

ð/  wÞ2 dm  4K 2 d\

e2 16

for sufficiently small d. Then by (3.41), kf  wk\

3e : 4

ð3:42Þ

3.4 Dense Subsets of Lp

161

Observe that w is continuous and w(p) = w(−p). From the Weierstrass Approximation Theorem 1.5.5, it follows that there exists a trigonometric polynomial T which uniformly approximates w arbitrarily closely: 1

supfjTðxÞ  wðxÞj : x 2 ½p; p g\e=4ð2pÞ2 : It follows that s ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi e2 e 2p ¼ : ðTðxÞ  wðxÞÞ dmðxÞ  4 32p ½p;p

sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi Z kT  wk ¼

2

Now (3.42) and (3.43) together imply that kf  T k\e. A similar argument works in L1[− p,p].

ð3:43Þ

h

In Proposition 3.4.2, the set X is assumed to be compact. Therefore, not only does it have finite Lebesgue measure, but also every continuous function on it is bounded. This ensures that every continuous function on it is in Lp. If X is to be replaced by R, it will no longer be the case that every continuous function is in Lp. However, every continuous function that vanishes outside some compact set will be. Such functions are said to be of “compact support” in accordance with the definition below. Definition 3.4.5 The support of a function f defined on a metric space X is the closure of the subset {x 2 X : f(x) 6¼ 0}. For example, if X = (0, 2] and f = X(0,1), then the support of f is the closure of (0,1) in (0,2], which is (0, 1]. This support is not compact. However, if we understand f to be the same characteristic function X(0,1) but on R, then the support is [0,1], which is compact. The support of any function f = X[1,a) on (0,2], where 1 < a  2, is [1,a], which is a compact set. An example of a continuous function on R with compact support is the product f(x) = v[0,p](x) sin x. It is trivial that sums and constant multiples of continuous functions of compact support are again continuous functions of compact support. Proposition 3.4.6 Given any function f 2 Lp ðRÞ, 1  p < ∞, there exists a sequence {fn}n  1 of continuous functions with compact support such that R lim R j fn  f jp dm ¼ 0. If f is nonnegative, then we can choose the sequence

n!1

{fn}n  1 to consist of nonnegative functions. In particular, continuous functions with compact support are dense in Lp ðRÞ. Proof It is sufficient to prove the result for a nonnegative f. There exists an increasing sequence {sn}n  1 of nonnegative simple functions converging to f on R [see Theorem 2.5.9]. The sequence of products snv[−n, n] then p     p p also converges to f and satisfies sn v½n;n  f   j f j and sn v½n;n  f  ! 0. Besides, these products are all in Lp ðRÞ. By the Dominated Convergence

162

3 Measure Spaces and Integration

Theorem 3.2.16, we get lim

n!1

p R   s v  f   dm ¼ 0. Note that snv[−n,n] vanishes R n ½n;n

outside a bounded interval. Let e > 0 be arbitrary. It follows from what has been established in the preceding paragraph that there exists a simple nonnegative function s such that R p s  f dm\e and s vanishes outside a bounded interval. Since s is a finite linear j j R combination of characteristic functions of bounded sets, we have only to prove the theorem in the case f = vE, where ER is a bounded measurable set. If m(E) = 0, then f = 0 a.e. and we may take fn = 0 for all n. So, assume m(E) > 0. By Proposition 2.3.24, there exists a closed set F and an open set O such that F  E  O and m(O\F) < min{e, m(E)}. Since E is bounded, it is contained in some bounded open interval and, by intersecting O with that interval if necessary, we may assume that O is bounded. In particular, Oc 6¼ ∅. Also, m(E) − m(F)  m(O\F) < m(E) < ∞, so that m(F) > 0 and hence F 6¼ ∅. Since Oc 6¼ ∅ 6¼ F, the distances d(x, Oc) and d(x, F) are well defined on R. It is clear from the definition of dist(x, A) as being inf{ jx  aj : a 2 A} that, when A is closed, we have d(x, A) = 0 , x 2 A. Since Oc and F are both closed, it follows that d(x, Oc) + d(x, F) = 0 , x 2 Oc \ F; however Oc \ F = ∅ because F  O and hence d(x, Oc) + d(x, F) > 0 for all x 2 R. So, there exists a real-valued function g : R ! R given by gðxÞ ¼

dðx; Oc Þ : dðx; FÞ þ dðx; Oc Þ

Clearly, we have (i) 0  g  1 everywhere (ii) g = 1 on F and (iii) g = 0 on Oc, so that {x 2 X : g(x) 6¼ 0}  O, a bounded set, which has the consequence that the support of g, which is closed by definition, is bounded and therefore compact. We also know from Proposition 1.3.24 that g is continuous. Thus, g is a nonnegative continuous function of compact support and satisfies 0  g  1 everywhere. From the fact that g = 1 on F and g = 0 on Oc, we conclude that g  vE can be on O\F. Also, we have jg  vE j  1 everywhere on R. It follows that Rnonzero only p j g  v j dm  mðOnFÞ\e. E R The fact that g is nonnegative justifies the assertion of the theorem that the h functions fn can be chosen to be nonnegative if f is nonnegative.

Chapter 4

Fourier Series

4.1

Introduction

Fourier series arose in connection with the study of two physical problems, namely, the problem of the vibrating string and that of heat conduction in solids. The new series became one of the most important tools in mathematical physics and had a far-reaching influence in mathematical research. In 1807, Fourier announced that an ‘arbitrary’ function f could be represented in the form 1 X 1 f ð x Þ ¼ a0 þ ðan cos nx þ bn sin nxÞ; 2 n¼1

with coefficients an and bn given by the formulae 1 an ¼ p

Z

p p

f ðxÞ cos nx dx;

n ¼ 0; 1; 2; . . .

f ðxÞ sin nx dx;

n ¼ 1; 2; 3; . . .:

and bn ¼

1 p

Z

p p

The above series is called the “Fourier series” of the function f. The reason for using 12a0 in the Fourier series is that for the constant function 1, it turns out that a0 = 2 while an = bn = 0 for n > 0. This announcement led to the discovery of mathematical theories such as Cantor’s theory of infinite sets, the Riemann and Lebesgue integrals and summability of series. The purpose of this chapter is to acquaint the reader with some of the most important aspects of Fourier series and the role of L2 ½p; p. Research in the last © Springer Nature Switzerland AG 2019 S. Shirali and H. L. Vasudeva, Measure and Integration, Springer Undergraduate Mathematics Series, https://doi.org/10.1007/978-3-030-18747-7_4

163

164

4

Fourier Series

fifty years or so has added enormously to our knowledge of Fourier series and has done much to improve the earlier treatments of the subject. The origins of Fourier series in mathematical physics and all its generalisations and ramifications have been scrupulously avoided here, and we shall be content to derive only some basic theorems. Definition 4.1.1 A function f with domain R is called periodic if there is some positive number l such that f ðx þ lÞ ¼ f ðxÞ

for all real x:

Any such number l is called a period of f. The graph of such a function is obtained by repetition of its graph over any interval of length l on both sides of that interval. The trigonometric functions sine and cosine are the simplest examples of nonconstant differentiable periodic functions; they have 2p as a period. For a function f with period l, we have (from the definition) f ðxÞ ¼ f ðx þ lÞ ¼ f ððx þ lÞ þ lÞ ¼ f ðx þ 2lÞ ¼ f ðx þ 3lÞ ¼    ¼ f ðx þ nlÞ for any positive integer n. Hence 2l, 3l, … are also periods of f. Furthermore, if f and g both have l as a period, then the function h defined by hðxÞ ¼ af ðxÞ þ bgðxÞ; where a and b are constants, also has l as a period. If a periodic function f has a smallest positive period, then this period is often called the fundamental period of f. Thus sine and cosine have 2p as the fundamental period while a constant function, though periodic, has no fundamental period. An example of a nonconstant periodic function f having no fundamental period is  f ðxÞ ¼

0 if x is rational 1 if x is irrational:

The pointwise limit of a sequence of functions with period l again has period l. We shall be discussing the possibility of representing real-valued integrable periodic functions as the sum of a series, in which the terms are constant multiples of the trigonometric functions 1, cos x, sin x, cos 2x, sin 2x, …, i.e., as a0 þ a1 cos x þ b1 sin x þ a2 cos 2x þ b2 sin 2x þ    ; where a0, a1, a2, … and b1, b2, … are real constants. Such a series is called a trigonometric series and the an and bn are called its coefficients. It may be compactly written as

4.1 Introduction

165

a0 þ

1 X

ðan cos nx þ bn sin nxÞ:

n¼1

In what follows, we shall need to evaluate some important integrals involving the trigonometric functions 1, cos x, sin x, cos 2x, sin 2x, …. Lemma 4.1.2 The following equalities hold: Rp (a) p cos kx cos nx dx ¼ 0 for k; n ¼ 0; 1; 2; . . . and k 6¼ n;  Rp p for n ¼ 1; 2; . . . (b) p cos2 nx dx ¼ 2p for n ¼ 0; Rp (c) p sinkx sin nx dx ¼ 0 for k; n ¼ 1; 2; . . . and k 6¼ n; Rp (d) p sin2 nx dx ¼ p for n ¼ 1; 2; . . .; Rp (e) p cos kx sin nx dx ¼ 0 for k; n ¼ 0; 1; 2; . . .: Proof Each of these results follows easily from one or the other of the identities cosðkx  nxÞ ¼ cos kx cos nx  sin kx sin nx sinðkx  nxÞ ¼ sin kx cos nx  cos kx sin nx; h

and elementary methods of integration. Proposition 4.1.3 If f ð x Þ ¼ a0 þ

1 X

ðan cos nx þ bn sin nxÞ;

ð4:1Þ

n¼1

where the trigonometric series converges uniformly on [− p,p], then its coefficients are given by the formulae R 1 p f ðxÞ dx a0 ¼ 2p Rp 1 p f ðxÞ cos nx dx; b ¼ n p p p f ðxÞ sin nx dx

R 1 p

an ¼ p

 for n ¼ 1; 2; . . .:

ð4:2Þ

Proof In view of the given uniform convergence in (4.1) the limit function f is continuous, so that all integrals in question exist. Moreover, we have Z

p p

Z f ðxÞdx ¼

p

Z ¼

p

p

p

½a0 þ

1 X

ðan cos nx þ bn sin nxÞdx

n¼1 1 X

a0 dx þ

n¼1

Z ½an

p p

¼ 2p a0 ; 1 using Lemma 4.1.2. Thus a0 ¼ 2p

Rp

p

f ðxÞdx.

Z cos nx dx þ bn

p

sin nx dx p

166

4

Fourier Series

From (4.1), it follows that f ðxÞ cos nx ¼ a0 cos nx þ

1 X

ðak cos kx cos nx þ bk sin kx cos nxÞ:

ð4:3Þ

k¼1

What is more, the convergence here is once again uniform. This is because the difference between any two partial sums of (4.3) is obtained by multiplying cos nx by the corresponding difference in (4.1), whereby the uniform Cauchy property is preserved because |cos nx|  1. In view of the uniform convergence in (4.3), we have by Proposition 1.7.5 Z

p

p

Z f ðxÞ cos nx dx ¼

p

a0 p 1 X

þ

¼ an

k¼1 Z p p

cos nx dx Z ½an

p

p

Z cos nx cos kx dx þ bn

cos2 nx dx ¼ p an

p

cos nx sin kx dx p

when n 6¼ 0;

Rp by Lemma 4.1.2. Thus an ¼ p1 p f ðxÞ cos nx dx for n = 1,2, …. A similar argument establishes the equality claimed for bn. h Formulae (4.2) proved above are known as Euler–Fourier formulae for the coefficients. It is easy to write down trigonometric series that diverge somewhere. For example, let a0 = an = bn = 1 for n = 1, 2, …; then the series becomes 1 + 1 + … when x = 0 and therefore diverges. Indeed, the series diverges when x is any rational multiple of p, because the terms do not approach 0. However, Proposition 4.1.3 shows that, if a trigonometric series converges to a sum function f uniformly, then the coefficients of the series are given by the Euler–Fourier Formulae (4.2). Therefore the approach we adopt is to start with a function that has period 2p and is integrable over a closed interval of length 2p (it must then be integrable over any closed bounded interval), define the coefficients as the right-hand sides of (4.2) and then study the relationship of the series so formed to the function. Definition 4.1.4 Let f : R ! R have period 2p, i.e. f ðxÞ ¼ f ðx þ 2pÞ

for all x 2 R:

Assume that f is Lebesgue integrable over an interval of length 2p. Then the Fourier series of f is the series 1 X 1 a0 þ ðan cos nx þ bn sin nxÞ; 2 n¼1

4.1 Introduction

167

where an ¼

1 p

bn ¼

1 p

and

Z

p p

Z

p p

f ðxÞ cos nx dx;

n ¼ 0; 1; 2; . . .

f ðxÞ sin nx dx;

n ¼ 1; 2; 3; . . .:

The an and bn are called the Fourier coefficients of f and we write 1 X 1 f ð x Þ  a0 þ ðan cos nx þ bn sin nxÞ: 2 n¼1

The Fourier series on the right-hand side here is said to ‘correspond to’ the function. Note that we have used the symbol ‘*’ and not ‘=’, because the sum of the series may not be equal to f(x) for some values of x, i.e. the Fourier series may not represent the function, as the first of the following examples shows. Examples 4.1.5 (a) The Fourier series corresponding to the function  f ðxÞ ¼

0 p  x\0 1 0xp

extended to R so as to be periodic, fails to converge to the function at x = 0. Here  Z Z 1 p 1 p 1 when n ¼ 0 an ¼ f ðxÞ cos nx dx ¼ cos nx dx ¼ 0 when n 6¼ 0 p p p 0 and 1 bn ¼ p

Z

p

1 f ðxÞ sin nx dx ¼ p p

Z

p

sin nx dx ¼

0

1  cos np : np

Thus bn ¼ Consequently,

2 for n ¼ 1; 3; 5; . . . and bn ¼ 0 for n ¼ 2; 4; 6; . . .: np f ðxÞ 

1 1 2X sinð2n þ 1Þx þ : 2 p n¼0 2n þ 1

When x = 0, the sum of this series is 12 whereas f(0) = 1.

168

4

Fourier Series

(b) Let f(x) = cos px, where p > 0 is an integer. Then by Lemma 4.1.2, the Fourier coefficients of f are all 0 except for ap, which is 1. Thus for example, when p = 3, f ðxÞ  0 þ ð0 þ 0Þ þ ð0 þ 0Þ þ ðcos 3x þ 0Þ þ ð0 þ 0Þ þ    : A corresponding statement is true when f(x) = sin px. Note the linear dependence of the Fourier coefficients on the function; this has the consequence, for instance, that the Fourier series of a sum of two functions is obtained by simply adding corresponding terms; similarly for constant multiples. Moreover, the Fourier series of a constant function consists of a single term equal to that constant, all terms in the summation being zero. The following lemma will be useful in a later section. Lemma 4.1.6 For any real x, ( 1 þ cos x þ cos 2x þ    þ cos nx ¼ 2

sinð2n 2þ 1Þx 2 sin 2x 1 2

þn

if x 6¼ 2p k; k 2 Z if x ¼ 2p k; k 2 Z:

Proof Using the identity sin A  sin B ¼ 2 sin

AB AþB cos ; 2 2

we have sin

2n þ 1 x 2

    1 3 1 5 3 ¼ sin x þ sin x  sin x þ sin x  sin x 2 2 2 2 2   2n þ 1 2n  1 þ    þ sin x  sin x 2 2 1 1 1 ¼ sin x þ 2 sin x cos x þ 2 sin x cos 2x 2 2 2 1 þ    þ 2 sin x cos nx   2 1 1 ¼ 2 sin x þ cos x þ cos 2x þ    þ 2 2

 cos nx :

The identity in question follows immediately from here for x 6¼ 2kp. The case when x = 2kp is straightforward. h Problem Set 4.1 4.1.P1. (a) [Needed in (b)] Prove Abel’s Lemma: If b1 b2    bn 0 and p P k ur  K for 1  p  n, where u1, u2, …, un are any n real numbers, then r¼1

4.1 Introduction

169

b1 k 

n X

br ur  b1 K:

r¼1

The next two parts together constitute what is called Bonnet’s form of the Second Mean Value Theorem for Integrals and will be needed in Theorem 4.2.8 below. Rb Rb (b) If a f and a g both exist and f is decreasing on [a, b] with f(b) 0, then there exists a n 2 [a, b] such that Z

b

Z

n

fg ¼ f ðaÞ

a

g: a

Rb Rb (c) If a f and a g both exist and f is increasing on [a, b] with f(a) 0, then there exists a n 2 [a, b] such that Z

b

Z

b

fg ¼ f ðbÞ

g: n

a

4.1.P2. Find the Fourier coefficients of the periodic function f for which  f ðxÞ ¼

a a

p  x\0 0  x\p:

4.1.P3. Using Bonnet’s form of the Second Mean Value Theorem for Integrals of 4.1.P1(c), prove that for a monotone function f on [− p, p], the Fourier coefficients an and bn (n 6¼ 0) satisfy the inequalities jan j 

1 1 jf ðpÞ  f ðpÞj; jbn j  jf ðpÞ  f ðpÞj: np np

4.1.P4. Suppose f is Lebesgue integrable with period 2p and “modulus of continuity” x, which means xðdÞ ¼ supfjf ðxÞ  f ðyÞj : jx  yj  dg: Show that its Fourier coefficients satisfy jan j  xðp=nÞ, jbn j  xðp=nÞ (n ¼ 6 0). 1 P 1 4.1.P5. For the trigonometric series 2 a0 þ ðan cos nx þ bn sin nxÞ, suppose 1 P

n¼1

ðjak j þ jbk jÞ converges. Show that the trigonometric series converges uniformly

k¼1

and that it is the Fourier series of its own sum.

170

4

Fourier Series

4.1.P6. A finite set of continuous functions g1, …, gn on [a, b] is said to be an orthonormal system with weight function w (Lebesgue integrable and nonnegative) if 

Z



b

gi ; g j ¼

 gi ðxÞgj ðxÞwðxÞdx ¼

a

0 1

i 6¼ j i ¼ j:

(Example: By Lemma 4.1.2, any finite number of functions among 12, cos kx, sin kx form an orthonormal set with domain [− p, p] and weight wðxÞ ¼ p1.) For the n P function ak gk ðxÞ, show that k¼1 n X k¼1

pffiffiffi jak j  n 

Z a

b

 ( )  X n   wðxÞdx max  a g ðxÞ : x 2 ½a; b :  k¼1 k k 

12

4.1.P7. Suppose the sequence of partial sums sn of the trigonometric series 1 P a0 þ ðan cos nx þ bn sin nxÞ has a subsequence SnðkÞ that converges uniformly n¼1

to a function f. Show that the trigonometric series is the Fourier series corresponding to f.

4.2

The Convergence Problem

Let us assume that f 2 L1[− p, p]. For later use, we define f on the entire real line R by periodicity: f(x + 2kp) = f(x) for x 2 [− p, p) and k 2 Z (the number f(p) has no importance for f 2 L1[− p, p]). The fundamental questions to be investigated are whether the Fourier series of f [see Definition 4.1.4] converges, and if it does, whether the sum is f(x). Briefly speaking, does the Fourier series of a 2p-periodic Lebesgue integrable function represent that function? Associated with the Fourier series is the sequence of its partial sums n X 1 sn ðxÞ ¼ a0 þ ðak cos kx þ bk sin kxÞ; 2 k¼1

n ¼ 1; 2; . . .:

We have seen in Example 4.1.5(a) that there are functions whose Fourier series fail to converge to f(x) at some point. In 1876, du Bois-Reymond gave an example of a continuous function whose Fourier series diverges at some point. It surprised many mathematicians of the era, as it was believed until then that the Fourier series of a continuous function should converge at every point. In the appendix to this chapter, we present Fejér’s construction of a continuous function whose Fourier series diverges at a point.

4.2 The Convergence Problem

171

If, however, the function has a continuous derivative everywhere and has period 2p, then its Fourier series can be shown to converge to the function everywhere. This is a consequence of Dirichlet’s Theorem, which will be proved later in this section. We shall also investigate to what extent the Fourier series determines the function. Actually, the class of functions that can be represented by their Fourier series is much larger. Example 4.2.1 For the Fourier series of the function f given by f(x) = cos px, where p > 0 is an integer, recall from Example 4.1.5(b) that every Fourier coefficient is 0 except for ap, which is 1. In other words, when p > 0, cos px  0 þ ð0 þ 0Þ þ ð0 þ 0Þ þ    þ ð0 þ 0Þ þ ðcos px þ 0Þ þ ð0 þ 0Þ þ    ; where the term (cos px + 0) corresponds to n = p in the Fourier series. It is understood of course that when p = 0, this becomes cos 0x  1 þ ð0 þ 0Þ þ ð0 þ 0Þ þ    : This shows that the above mentioned partial sums of the Fourier series are  sn ðxÞ ¼

0 cos px

n\p n p:

Similarly for sin px. Remark 4.2.2 To express the partial sum sn(x) of the Fourier series of f in a manageable form, we use the Euler–Fourier formulae to obtain n X 1 sn ðxÞ ¼ a0 þ ðak cos kx þ bk sin kxÞ 2 k¼1 Z Z p Z p n 1 p 1X f ðxÞdx þ ðcos kx f ðtÞ cos kt dt þ sin kx f ðtÞ sin kt dtÞ ¼ 2p p p k¼1 p p Z n 1 p 1 X f ðtÞ½ þ ðcos kx cos kt þ sin kx sin ktÞdt ¼ p p 2 k¼1 Z n 1 p 1 X f ðtÞ½ þ cos kðx  tÞdt: ¼ p p 2 k¼1

Using Lemma 4.1.6, we may rewrite this as sn ðxÞ ¼

1 p

Z

p p

f ðtÞDn ðx  tÞdt;

172

4

Fourier Series

where Dn : R ! R is the function for which ( Dn ðuÞ ¼

sinð2n 2þ 1Þu 2 sin u2 1 2 þn

if u 6¼ 2kp; k 2 Z if u ¼ 2kp; k 2 Z

for all u 2 R. This achieves the manageable form we set out to obtain; since Dn enters into it in a particularly simple manner, there is reason to expect that a study of this function may yield information about how f is related to its Fourier series. Accordingly, it deserves a special name: Dn is called the Dirichlet kernel. Proposition 4.2.3 The Dirichlet kernel has the following properties: (a) Dn is continuous; (b) Dn(–t) = Dn(t) for all real t; Rp (c) p2 0 Dn ðtÞdt ¼ 1. Proof This is immediate from Lemma 4.1.6, according to which Dn ðtÞ ¼

1 þ cos t þ cos 2t þ    þ cos nt 2

for all t 2 R:

h

Remark 4.2.4 An alternative form of the integral representation for sn(x) that was obtained above turns out to be useful. Substituting t = x − u in the integral, we get Z 1 xþp f ðx  uÞDn ðuÞdu p xp Z Z p Z xþp 1 p ¼ ½ f ðx  uÞDn ðuÞdu þ f ðx  uÞDn ðuÞdu þ f ðx  uÞDn ðuÞdu: p xp p p

sn ðxÞ ¼

Observe that 1 p

Z

p xp

1 f ðx  uÞDn ðuÞ du ¼ p

Z

p xþp

f ðx  uÞDn ðuÞ du;

using the periodicity of the integrand. Thus sn ðxÞ ¼

1 p

Z

p p

f ðx  uÞDn ðuÞ du:

4.2 The Convergence Problem

173

Consequently, Z Z 1 0 1 p sn ðxÞ ¼ f ðx  uÞDn ðuÞ du þ f ðx  uÞDn ðuÞ du p p p 0 Z p Z p 1 1 ¼ f ðx þ uÞDn ðuÞ du þ f ðx  uÞDn ðuÞ du p 0 p 0 Z p 1 ½f ðx þ uÞ þ f ðx  uÞDn ðuÞ du; ¼ p 0 since Dn(−u) = Dn(u) for all u. This is the alternative form which we shall be using below. In the proof of Dirichlet’s Theorem, we need a well-known and important result. It states that the Fourier coefficients ak, bk of a Lebesgue integrable function with period 2p must approach 0 as k ! ∞. It is often called the Riemann–Lebesgue Lemma. Theorem 4.2.5 (Riemann–Lebesgue) Let f be a Lebesgue integrable function on R. Then Z lim

k!1

R

Z f ðxÞ sin kx dx ¼ 0 ¼ lim

k!1

R

f ðxÞ cos kx dx:

Proof Let [a, b] be any closed interval. If f is the characteristic function of [a, b], then  Z        f ðxÞ sin kx dx ¼ cos ka  cos kb  2 :  jkj    k R

The right-hand side tends to 0 as k ! ∞. Thus the result holds for any step function, which is constant on a finite number of bounded intervals and vanishes outside their union. Let e > 0 be chosen arbitrarily. Then corresponding to e, there is a positive number A [see 3.2.P18(b)] such that Z

1 jf ðxÞjdx\ e: 2 j xj [ A

ð4:4Þ

By Proposition 3.4.3, there is a step function / vanishing outside [−A, A] such that Z

A

1 jf ðxÞ  /ðxÞjdx\ e: 2 A

ð4:5Þ

174

4

Fourier Series

It follows from (4.4) and (4.5) that Z

Z R

jf ðxÞ  /ðxÞjdx ¼ ¼

Z

A

jf ðxÞ  /ðxÞjdx þ

A Z A

j xj [ A

jf ðxÞ  /ðxÞjdx

Z jf ðxÞ  /ðxÞjdx þ

A

j xj [ A

jf ðxÞjdx

\e: Therefore Z Z Z j f ðxÞ sin kx dxj  j ðf ðxÞ  /ðxÞÞ sin kx dxj þ j /ðxÞ sin kx dxj R R ZR Z  jf ðxÞ  /ðxÞjdx þ j /ðxÞ sin kx dxj R R Z \e þ j /ðxÞ sin kx dxj: R

Since the result has already been shown to be true for step functions, it now follows R R that lim R f ðxÞ sin kx dx ¼ 0. The argument that lim R f ðxÞ cos kx dx ¼ 0 is k!1

k!1

h

similar. Corollary 4.2.6 For any fixed positive d < p, we have Z lim

n!1

p

Dn ðtÞdt ¼ 0:

d

Proof In fact, Z

p

Z

p

Dn ðtÞdt ¼

d

d





1 1 1 2 sin t sin n þ t dt ! 0 2 2

as

n ! 1;

by Theorem 4.2.5 applied to the Riemann integrable function ð2 sin 12tÞ1 on the interval [d, p], where d > 0. h The next result shows that the behaviour of the sequence {sn(x)} for large n depends only on the behaviour of f on the interval [x − d, x + d], where d > 0 is arbitrary but less than p. Proposition 4.2.7 (Localisation Theorem) Let f : R ! R have period 2p and be Lebesgue integrable on [− p, p] and 0 < d < p. Then for each x 2 R, Z lim sn ðxÞ ¼ lim

n!1

n!1

d

p

½f ðx þ tÞ þ f ðx  tÞDn ðtÞdt ¼ 0:

4.2 The Convergence Problem

175

Proof By definition of Dn, Z

p

Z ½f ðx þ tÞ þ f ðx  tÞDn ðtÞ dt ¼

d

p

d f ðx þ tÞ þ f ðxtÞ 2 sin2t

Since

f ðx þ tÞ þ f ðx  tÞ 1 sinðn þ Þt dt: 2 2 sin 12 t

ð4:6Þ

is Lebesgue integrable over [d, p], it follows from Theorem

4.2.5 that the integral on the right-hand side of (4.6) approaches 0 as n ! ∞. h Theorem 4.2.8 For every Riemann integrable function f of period 2p and for every fixed number d, 0 < d  p, the integral Z

p

½f ðx þ tÞ þ f ðx  tÞDn ðtÞ dt

d

tends to 0 as n ! ∞, uniformly for all real values of x. Proof By definition of Dn, Z

p

Z ½f ðx þ tÞ þ f ðx  tÞDn ðtÞ dt ¼

d

d

p

f ðx þ tÞ þ f ðx  tÞ 1 sinðn þ Þt dt: 2 2 sin 12 t

ð4:7Þ

Applying Bonnet’s form of the Mean Value Theorem for Integrals [see Problem 4.1.P1], the right-hand side of (4.7) becomes 1 2 sin d2

Z

n d

1 ½f ðx þ tÞ þ f ðx  tÞ sinðn þ Þt dt; 2

where d  n  p. We shall show that the last integral tends to 0 uniformly for all real values of x, as n ! ∞. It suffices to consider the integral Z d

p

1 f ðx þ tÞ sinðn þ Þt dt 2

as the other integral with f(x − t) in place of f(x + t) behaves analogously. The last integral equals Z

nþx dþx

1 1 1 1 f ðuÞ½sinðn þ Þu cosðn þ Þx  cosðn þ Þu sinðn þ Þxdu: 2 2 2 2

ð4:8Þ

176

4

Fourier Series

Since the function f has period 2p, it is enough to consider values of x lying in the interval [0, 2p]. Since 0\d  d þ x  d þ 2p;

n  n þ x  n þ 2p

and

0\d  n  p;

it that  follows

 the limits of the integral

 lie in the fixed interval [0, 3p]. Observe that sin n þ 1 x  1 and cos n þ 1 x  1 for all x. Using the Riemann–Lebesgue 2 2 Lemma [see Theorem 4.2.5], the integral (4.8) above is seen to tend to 0 as n ! ∞, uniformly for all real x. h The next result, which is of independent interest, will prove useful in establishing the main theorem. Proposition 4.2.9 The partial sums of the series

1 P sin kx k¼1

absolute value by 1 þ p on R.

k

are uniformly bounded in

Proof We have     X n n sin kx 1  1 X 1 1 1   sin x ¼  ½cosðk  Þx  cosðk þ Þx   k¼m k 2  2  k¼m k 2 2    

n1 X 1  1 1 1 1 1 1 1  ¼  cosðm  Þx þ  cosðk þ Þx cosðn þ Þx 2 m 2 kþ1 k 2 n 2  k¼m 

n1 1 1 X 1 1 1 1  ½   þ ¼ : 2 m k¼m k þ 1 k n m

ð4:9Þ Let Sn(x) denote the nth partial sum of the series, i.e. Sn ðxÞ ¼

n X sin kx ; k k¼1

x 2 ð0; p:

Assuming that x 6¼ 0, denote the integer part of 1x by m, so that m  1x \m þ 1; in particular, mx  1. Since j sin tj  t for t 0 [reason: the functions t ± sin t are increasing for t 0 and are 0 at t = 0], we get jSn ðxÞj 

n X kx k¼1

k

¼ nx:

For n  m, we obtain from here that jSn ðxÞj  mx  1. On the other hand, for m < n and x 2 (0, p] we have

4.2 The Convergence Problem

177

   X n sin kx  jSn ðxÞj  jSm ðxÞj þ   k¼m þ 1 k  1 ; upon using (4:9Þ mþ1 1 x 1þ2 2 1 sin 2 x 1þ

 1 þ p; using the fact that sin t 2tp for 0  t  p2. In fact, the function sint t is decreasing for 0  t  p2 and its value at t ¼ p2 is p2. This shows that for x 2 (0, p] and all natural numbers n, we have jSn ðxÞj  1 þ p. For x = 0, this inequality holds trivially, because Sn(0) = 0. Thus the upper bound 1 + p has been established for all partial sums and all x 2 [0, p]. Since Sn ðxÞ ¼ Sn ðxÞ, the bound also applies when x 2 [− p, 0]. In view of the periodicity, it applies on all of R. h A set of sufficient conditions for the convergence of a Fourier series, which is due to Dirichlet, possesses a degree of generality that makes it useful for most purposes. Before stating the theorem, we recall that f(x+) and f(x−) are respectively the right limit and left limit of f at x, and that when the function is monotonic on each of (a, x) and (x, b) and is also bounded, these limits necessarily exist. For, when f is increasing on (a, x), say, given any e > 0, there exists an x0 2 (a, x) such that f ðx0 Þ [ supða;xÞ f  e; then x0 < t < x implies supða;xÞ f f ðtÞ f ðx0 Þ, which,     together with the preceding inequality, implies f ðtÞ  supða;xÞ f \e . The function of Example 4.1.5(a) satisfies f(0−) = 0, f(0+) = 1 and it was noted that, when x = 0, its Fourier series converges to 12, a value that agrees with 12ðf ð0Þ þ f ð0 þ ÞÞ. Dirichlet’s Theorem states that this happens under broad conditions. Theorem 4.2.10 (Dirichlet) Let f : R ! R be Riemann integrable on [− p, p] with period 2p and let x 2 R be arbitrary but fixed. If (a, b) is an interval such that x 2 (a, b) and f is increasing on each of (a, x) and (x, b), then 1 sn ðxÞ ! ½f ðx þ Þ þ f ðx Þ: 2 Here sn(x) denotes the nth partial sum of the Fourier series of f. If, moreover, f is continuous on (a, b), then sn ðxÞ ! f ðxÞ

as n ! 1

uniformly on every closed subinterval [a + l, b − l].

178

4

Fourier Series

Proof Since f is increasing on each of (a, x) and (x, b) and is also bounded (because it is Riemann integrable), the limits f(x+) and f(x–) both exist. By Remark 4.2.4 and Proposition 4.2.3(c), we may write 1 1 sn ðxÞ  ½f ðx þ Þ þ f ðx Þ ¼ 2 p  ¼

1 p

Z

p

½f ðx þ tÞ þ f ðx  tÞDn ðtÞdt

Z

0

1 p Z

p

½f ðx þ Þ þ f ðx ÞDn ðtÞdt

0 p

½f ðx þ tÞ þ f ðx  tÞ  f ðx þ Þ  f ðx ÞDn ðtÞdt

0

Z Z 1 d 1 p þ Þ½f ðx þ tÞ þ f ðx  tÞ  f ðx þ Þ  f ðx ÞDn ðtÞdt; p 0 p d for all d 2 ð0; pÞ

¼ð

¼ I þ J;

say:

ð4:10Þ Now I can be written in the form 1 p

I¼

Z

d

½f ðx þ tÞ  f ðx þ ÞDn ðtÞdt 

0

1 p

Z

d

½f ðx Þ  f ðx  tÞDn ðtÞdt:

ð4:11Þ

0

Since x 2 (a, b), we can choose the number d so small that x ± d both belong to (a, b). Since f is increasing on (x, x + d), the difference f(x + t) − f(x+) is an increasing function of t on the interval (0, d) and its limit at 0 is 0. On applying Bonnet’s form of the Second Mean Value Theorem for Integrals [see Problem 4.1. P1(c)], we obtain Z

d

½f ðx þ tÞ  f ðx þ ÞDn ðtÞdt ¼ ½f ðx þ dÞ

0

 f ðx þ Þ

Z

d

Dn ðtÞdt

for some g 2 ½0; d:

g

By Lemma 4.1.6 and Proposition 4.2.9, we have Z   

d

g

Denoting

    dg n n X  d  g X sin kd sin kg  þ  þ 2ð1 þ pÞ: Dn ðtÞdt ¼     2 k k 2 k¼1 k¼1 dg 2

þ 2ð1 þ pÞ by M1, we therefore have

Z   

0

d

  ½f ðx þ tÞ  f ðx ÞDn ðtÞdt  ½f ðx þ dÞ  f ðx þ ÞM1 : þ

ð4:12Þ

4.2 The Convergence Problem

179

Similarly, Z   

d

0

  ½f ðx Þ  f ðx  tÞDn ðtÞdt  f ðf ðx Þ  f ðx  dÞM1 : 

ð4:13Þ

The right-hand sides of the above two inequalities approach 0 as d ! 0. It follows that the left-hand sides do so uniformly in n. Therefore by (4.11), for any e > 0, we can choose d so small that x ± d both belong to (a, b) and 1 jI j  e for every natural number n: 2

ð4:14Þ

Fixing such a d, we next consider the integral denoted above by J. This integral can be written as J¼

1 p

Z

p d

½f ðx þ tÞ þ f ðx  tÞDn ðtÞdt 

1 p

Z

p

½f ðx þ Þ þ f ðx ÞDn ðtÞdt:

d

From the Localisation Theorem 4.2.7 applied to f and to the constant function f(x+) + f(x–), we deduce that J ! 0 as n ! ∞. Therefore, we have 1 jJ j\ e for sufficiently large n: 2

ð4:15Þ

Since e > 0 is arbitrary, the first part of the required conclusion now follows from (4.10), (4.14) and (4.15). Now suppose f is also continuous on (a, b). Then on any closed subinterval [a + l, b − l], f is uniformly continuous and bounded. Consequently, the right-hand sides of the inequalities (4.12) and (4.13), and hence also |I|, are uniformly small for values of x lying in the interval [a + l, b − l], provided l is sufficiently small. Rp For fixed d, p1 d ½f ðx þ tÞ þ f ðx  tÞDn ðtÞdt is uniformly small in absolute value  Rp  in view of Theorem 4.2.8. Also,p1 d ½f ðx þ Þ þ f ðx ÞDn ðtÞdt is uniformly small Rp since f is bounded and d Dn ðtÞdt tends to 0 as n ! ∞. So J is uniformly small for all x in [a + l, b − l] as n ! ∞. Since I and J have both been shown to be uniformly small when n is sufficiently large, it follows from (4.10) that sn(x) ! f(x) as n ! ∞ uniformly on every closed subinterval [a + l, b − l]. h Corollary 4.2.11 Let f : R ! R be Riemann integrable with period 2p and monotonic on [a, b]. If f is continuous at a point x 2 (a, b), then sn ðxÞ ! f ðxÞ: Here sn(x) denotes the nth partial sum of the Fourier series of f.

180

4

Fourier Series

Proof Since f is continuous at x, we have f(x+) = f(x) = f(x–). The result now follows from Theorem 4.2.10. h The problem of uniform convergence on an interval of a Fourier series of a function satisfying what is called a “Lipschitz condition” is considered in the next theorem. Definition 4.2.12 A real-valued function f satisfies a Lipschitz condition of order a, where 0 < a  1, on an interval [a, b] contained in its domain if there exist M > 0 and d > 0 such that jf ðx þ tÞ  f ðxÞj  M jtja ;

a  x  b; jtj\d:

A Lipschitz condition is understood to be of order 1 unless specified otherwise. The constant M in the condition is not unique and is called a Lipschitz constant for the function. Theorem 4.2.13 Suppose f : R ! R is Riemann integrable with period 2p and satisfies a Lipschitz condition of order a on the interval [a, b]. Then the Fourier series of f converges to f uniformly on [a, b]. Proof Let d and M be as in Definition 4.2.12; we may suppose that d  p. Denote the partial sum of the Fourier series by sn(x), as usual. From Remark 4.2.4 and Proposition 4.2.3(c), we have Z 2 p f ðx þ tÞ þ f ðx  tÞ  f ðxÞDn ðtÞ dt ½ p 0 2 Z Z 2 g 2 p f ðx þ tÞ þ f ðx  tÞ  f ðxÞDn ðtÞ dt þ Þ½ ¼ð p 0 p g 2

sn ðxÞ  f ðxÞ ¼

¼ I þ J;

ð4:16Þ

say,

where η will be chosen appropriately later. For the moment suppose that 0 < η  d  p. Let x 2 [a, b]. Then  Z g  2  f ðx þ tÞ þ f ðx  tÞ   f ðxÞDn ðtÞ dt jI j ¼  ½ p 0 2   Z g sinðn þ 1Þt 1  2   jf ðx þ tÞ þ f ðx  tÞ  2f ðxÞj  dt  2 sin 12 t  p 0 Z 1 g 1  2Mta dt; an improper integral: p 0 2 sin 12 t 1

t

1

t

1 a1 2 As the function ta 2 sin is such that sin2 1t ! 1 as t ! 0 while a − 1 > −1, 1 ¼ t sin1t t 2

2

2

it follows that the improper integral converges and therefore approaches 0 as η ! 0, by Problem 3.2.P15 and the fact that the integrand is nonnegative. Hence, for a given e > 0, there exists an η > 0 such that, for any natural number n,

4.2 The Convergence Problem

181

jI j 

1 e; 2

ð4:17Þ

for all x 2 [a, b]. Now fix such an η and write J as J¼

1 p

Z

p g

2 ½f ðx þ tÞ þ f ðx  tÞDn ðtÞdt  f ðxÞ p

Z

p

Dn ðtÞdt:

g

By Theorem 4.2.8, the first integral on the right-hand side here tends to 0 uniformly for all real values of x as n ! ∞. Since f is Riemann integrable and hence bounded, it follows that the second integral also tends to 0 uniformly for all real values of x by Corollary 4.2.6. Thus, for large enough n, we have 1 jJ j\ e 2

ð4:18Þ

for all real values of x. From (4.16), (4.17) and (4.18), it follows that, for large enough n, jsn ðxÞ  f ðxÞj\e

for all x 2 ½a; b:

h

Examples 4.2.14 (a) Let f be the function defined by f ðxÞ ¼ cos ax

for  p  x  p;

where a is a real number that is not an integer. Since f is even and f(−p) = f(p), it can be extended to R so as to have period 2p and also be even. Since f satisfies a Lipschitz condition on [− p, p], it follows by Theorem 4.2.13 that its Fourier series converges uniformly on [− p, p]. Observe that the Fourier coefficients bn are all 0 while 1 1 a0 ¼ 2 p

Z

p

cos ax dx ¼

0

sin ap ; ap

and, for n > 0, 1 an ¼ p

Z

p

½cosða þ nÞx þ cosða  nÞxdx ¼ ð1Þn

0

a2

2a sin ap :  n2 ap

Therefore, for −p  x  p, we have cos ax ¼

1 sin ap sin ap X a cos nx þ2 ð1Þn 2 ; ap p n¼1 a  n2

182

4

Fourier Series

or equivalently 1 X p cos ax 1 a cos nx ¼ þ ð1Þn 2 : 2 sin ap 2a a  n2 n¼1

Setting x = 0 in the latter, we obtain 1 X p 1 a ¼ þ ð1Þn 2 ; 2 sin ap 2a a  n2 n¼1

which may be written as cosec ap ¼

1 1 2a X 1 þ ð1Þn 2 : ap p n¼1 a  n2

This formula is valid whenever a is not an integer. Setting x = p instead, we obtain cot ap ¼

1 1 2a X 1 þ ; ap p n¼1 a2  n2

which is also valid whenever a is not an integer. (b) Let f(x) = sin ax, for −p  x  p, where a is not an integer. Extend it to all of R so as to be periodic. We shall find its Fourier series by calculating its coefficients. Since the restriction to (−p, p) is odd, the Fourier coefficients an are all 0 and the remaining Fourier coefficients are given by

2 bn ¼ p

Z

p 0

1 sin ax sin nx dx ¼ p ¼2

Z

p

½cosða  nÞx  cosða þ nÞxdx

0

sin ap n ð1Þn 2 : p a  n2

The function satisfies a Lipschitz condition on any closed subinterval of the open interval (−p, p). Therefore by Theorem 4.2.13, its Fourier series converges uniformly on any closed subinterval of (−p, p). So sin ax ¼ 2

1 sin ap X n ð1Þn 2 sin nx: p n¼1 a  n2

4.2 The Convergence Problem

183

This series can also be obtained from the one for cosax above by term differentiation provided we can prove the uniform convergence of the differentiated series [19, Theorem 7.17] independently. Such a proof can be based on Dirichlet’s Test [28, Theorem 4.4.2] and we present it for the sake of those readers who may know the latter: nþ1 n Choose an ¼ n2 a sin nx. Observe that for n a, the sequence 2 ; bn ¼ ð1Þ {an} decreases to 0 as n ! ∞. Also, 1 3 1 5 3 2 cos x½sin x  sin 2x þ    þ ð1Þn þ 1 sin nx ¼ ðsin x þ sin xÞ  ðsin x þ sin xÞ 2 2 2 2 2 1 1 nþ1 þ    þ ð1Þ ðsinðn þ Þx þ sinðn  ÞxÞ 2 2 1 1 nþ1 ¼ ð1Þ sinðn þ Þx þ sin x: 2 2

So,   1 sin x  sin 2x þ    þ ð1Þn þ 1 sin nx   1     cos 1 x mincos 1 x ; 2 2   where the min is taken over x 2 [a, b] (−p, p). Here mincos 12 x is guaranteed to be positive because cos 12x is continuous and positive on (−p, p) and hence also on [a, b]. Thus the partial sums of Rbn are uniformly bounded on [a, b]. It follows by Dirichlet’s Test that Rn a an bn converges uniformly on [a, b]. (c) Consider the function f(t) = t2/4, t 2 [−p, p]. It satisfies f(−p) = f(p) and therefore can be extended to be periodic on R. Moreover, it will satisfy a Lipschitz condition on [− p, p]; in fact,  p 1 jf ðxÞ  f ðyÞj ¼ x2  y2   jx  yj: 4 2 Thus the Fourier series of the function converges uniformly to the function. Since the function is even, its Fourier coefficients bn all vanish, while Z Z 2 p x2 1 p 2 cos nx dx ¼ an ¼ x cos nx dx p 0 4 2p 0 

 cos nx 1 2 sin nx sin nx p ½x  ð2xÞ  ¼ þ 2  0 2p n n2 n3 1 2p ¼ ð1Þn 2p n2

184

4

Fourier Series

and 2 a0 ¼ p

Z

p 2 0

x p2 dx ¼ : 4 6

Thus the Fourier series is 1 x2 p2 X cos nx ¼  ð1Þn þ 1 : n2 4 12 n¼1

Setting x = 0 here yields 1 p2 X ð1Þn þ 1 ¼ : 12 n¼1 n2

Problem Set 4.2 4.2.P1. Suppose that f is a 2p-periodic function that is Lebesgue integrable on [−p, p] and is differentiable at a point x0. Prove that lim sn ðx0 Þ ¼ f ðx0 Þ;

n!1

where sn denotes the partial sum of the Fourier series of f. 4.2.P2. (Bessel’s Inequality) If f is any Riemann integrable periodic function, its Fourier coefficients satisfy the inequality Z 1 1 2 X 1 p a0 þ ða2n þ b2n Þ  f ðxÞ2 dx: 2 p p n¼1 Note that it is part of the assertion that the series on the left is convergent. [It follows from this result that the Fourier coefficients an and bn of a Riemann integrable function tend to zero as n ! ∞.] 1 P cos pffiffinx is uniformly convergent on every closed subinterval 4.2.P3. The series n n¼1

[a, 2p−a], 0 < a < p of (0, 2p). Show that it is not the Fourier series of a Riemann integrable function. 1 P sin nx 4.2.P4. Is the series lnðn þ 1Þ the Fourier series of a continuous function? n¼1 n R p sinðn þ 1tÞ P 1 4 1 4.2.P5. Show that p2 0  2 sin 1t2 dt p42 k p2 ðn þ ln nÞ: 2 k¼1 R p sinðn þ 1tÞ 1 þ ln p þ p2 : 4.2.P6. Show that p2 0  2 sin 1t2 dt  ln n þ np 2

4.2.P7. Let the Fourier series corresponding to a continuous periodic function f(x) 1 P be given by 12 a0 þ ðan cos nx þ bn sin nxÞ. Show that the series obtained by n¼1

4.2 The Convergence Problem

185

Rx integrating this Fourier series term by term converges to 0 f ðtÞdt. [The remarkable thing about this result is that the Fourier series of f is not assumed to converge to it. Note however that the integrated series is not a trigonometric series.] 4.2.P8. If f is continuously differentiable on [a, b], use integration by parts (but not the Riemann–Lebesgue Theorem 4.2.5) to show that Z lim

k!1

b

Z f ðtÞ sin kt dt ¼ 0 ¼ lim

k!1

a

b

f ðtÞ cos kt dt:

a

4.2.P9. Let f be continuous with period 2p and Fourier coefficients ak, bk. Show that n X

1

ðjak j þ jbk jÞ  2ð2n þ 1Þ2 supfjf ðxÞj : x 2 ½p; pg:

k¼0

R1 4.2.P10. [Cf. Problem 3.2.P24] Show that the improper integral 0 sinx xdx converges and evaluate it by applying Dirichlet’s Theorem 4.2.10. It is understood that sin x x is to be replaced by 1 when x = 0.

4.3

Cesàro Summability of Fourier Series

As observed in Sect. 4.2, although the Fourier series of even a continuous function need not converge to it, strengthening the hypothesis on the function ensures that the corresponding Fourier series does converge to it. In the other direction, we may weaken the notion of convergence and seek an answer to the question of whether the Fourier series corresponding to a Lebesgue integrable periodic function converges in the weakened sense, and if it does, whether it converges to the function. It is known [see the example following Proposition 3.1.8 in [28] or Ex. 10 on p. 39 of [4]] that the arithmetic means of the first n terms of a convergent sequence converge to the same limit as the latter. Even for a nonconvergent sequence, such as fð1Þn : n 2 Ng, the sequence of arithmetic means can be convergent. This leads us to consider the arithmetic means of the first n partial sums sn of the Fourier series of a Lebesgue integrable function f with period 2p: 1 rn ðxÞ ¼ ½s0 ðxÞ þ s1 ðxÞ þ    þ sn1 ðxÞ; n where n X 1 sn ðxÞ ¼ a0 þ ðak cos kx þ bk sin kxÞ 2 k¼1

186

4

Fourier Series

and an, bn are the Fourier coefficients of f. In what follows, we shall show that the rn(x), known as the Fejér sums, converge to f(x) for all those x at which f is continuous; the Fourier series is then said to be Cesàro summable to f(x). Example 4.3.1 For the Fourier series of the function f given by f(x) = cos px, where p > 0 is an integer, recall from Example 4.2.1 that the partial sums of the Fourier series are  sn ðxÞ ¼

0 if n\p cos px if n p:

It follows that the Fejér sums are  rn ðxÞ ¼

0

ðnpÞ cos px n

if n  p if n [ p:

Remark 4.3.2 To express rn(x) in a more manageable form, we use the definition of sn(x) to obtain 1 1 rn ðxÞ ¼ ½s0 ðxÞ þ s1 ðxÞ þ    þ sn1 ðxÞ ¼ n p

Z

p

½f ðx þ tÞ þ f ðx  tÞFn ðtÞdt;

0

where Fn : R ! R is the function defined as 1 Fn ðtÞ ¼ ½D0 ðtÞ þ D1 ðtÞ þ    þ Dn1 ðtÞ: n In view of the definition of Dn, this becomes Fn ðtÞ ¼

8 P sinðk þ 12Þt < 1 n1 :

n

k¼0

2 sin 12t

;

n 2;

t 6¼ 2kp; k 2 Z t ¼ 2kp; k 2 Z:

Fn is known as the Fejér kernel. When t is not an integral multiple of 2p, it can also be written in the “closed form” Fn ðtÞ ¼

sin2 nt2 : 2n sin2 2t

This is a consequence of the trigonometric relation cos A  cos B ¼ 2 sin whence one can obtain

AþB A  B sin ; 2 2

4.3 Cesàro Summability of Fourier Series

187

n1 X 1 1 nt 2 sinðk þ Þt sin t ¼ ½cos kt  cosðk þ 1Þt ¼ 1  cos nt ¼ 2 sin2 : 2 2 2 k¼0 k¼0

n1 X

The following properties of the Fejér kernel are important. Proposition 4.3.3 The Fejér kernel Fn has the following properties: (a) (b) (c) (d) (e)

Fn is continuous; Fn(–t) = Fn(t) for all real t; Fn(t) 0 for all real t; R 2 p p 0 Fn ðtÞdt ¼ 1 for all n; Fn ðtÞ  2n 1sin2 d for d\t\p; 2 Rp (f) lim d Fn ðtÞdt ¼ 0; where 0\d\p: n!1

Proof Parts (a) and (b) follow from the corresponding properties of Dn. Part nP 1 (c) follows from the closed form of Fn derived above. Since Fn ðtÞ ¼ 1n Dk ðtÞ, we k¼0

have by Proposition 4.2.3(c) 2 p

Z

p

Fn ðtÞdt ¼

0

2 np

Z

n1 pX

0

Dk ðtÞdt ¼

k¼0

Z n1 n1 X X 2 p 1 ¼ 1: Dk ðtÞdt ¼ np n 0 k¼1 k¼0

This proves part (d). To prove (e), note that sin2 2x increases as x increases from 0 to p. Finally, it follows from (e) that Z d

p

Fn ðtÞdt 

pd !0 2n sin2 d2

as

n ! 1;

provided 0\d\p: h

This completes the proof of (f).

Theorem 4.3.4 (Fejér) If f : R ! R is Lebesgue integrable with period 2p and continuous at x 2 [− p, p], then its Fourier series at x is Cesàro summable to f(x), i.e. the arithmetic means rn of the partial sums of the Fourier series of f converge to f at x. Moreover, if f is continuous on [− p, p], then the convergence is uniform on R. Proof We multiply both sides of the equality in Proposition 4.3.3(d) by f(x) and subtract from the expression for rn(x) derived in Remark 4.3.2 to obtain rn ðxÞ  f ðxÞ ¼

2 p

Z

p 0

f ðx þ tÞ þ f ðx  tÞ  f ðxÞFn ðtÞdt: ½ 2

Fix e > 0. Because of the above equality, in order to show that rn(x) ! f(x), it suffices to prove that, for sufficiently large n,

188

4

Fourier Series

 Z p  2  f ðx þ tÞ þ f ðx  tÞ   f ðxÞFn ðtÞdt\e: ½ p 2 0

ð4:19Þ

For uniform convergence, it suffices to prove (4.19) for sufficiently large n and all real x. Since f is continuous at x, we can find d satisfying 0 < d < p such that 1 jf ðx þ tÞ  f ðxÞj\ e 2

for jtj\d:

Thus, if 0  t < d, then   f ðx þ tÞ þ f ðx  tÞ  1    ½jf ðx þ tÞ  f ðxÞj þ jf ðx  tÞ  f ðxÞj  f ð x Þ   2 2 1 1 1 1 \ ð e þ eÞ ¼ e: 2 2 2 2 Consequently,  Z d  Z 2  1 2 d f ðx þ tÞ þ f ðx  tÞ     f ðxÞF e ½ ðtÞdt Fn ðtÞdt n p  2 p 2 0 0 Z 1 2 p 1  e Fn ðtÞdt ¼ e; 2 p 0 2

ð4:20Þ

using Proposition 4.3.3(c) & (d). Observe that, if f is continuous on [− p, p], then it is uniformly continuous on R by virtue of its periodicity, and this makes it possible to choose the number d independently of x, thereby ensuring that (4.20) holds for all real x. If t d, then Fn ðtÞ  2n 1sin2 d by Proposition 4.3.3(e), and thus 2

 Z p  2  f ðx þ tÞ þ f ðx  tÞ    f ðxÞF ½ ðtÞdt n p  2 d Z p 2  jf ðx þ tÞ þ f ðx  tÞ  2f ðxÞjdt 4pn sin2 d2 d Z p 2 ½2  jf ðtÞjdt þ 2pjf ðxÞj 4pn sin2 d2 p 1 \ e for sufficienly large n (depending on xÞ: 2

ð4:21Þ

Inequalities (4.20) and (4.21) prove (4.19). As already noted, this completes the proof that rn(x) ! f(x) when f is continuous at x.

4.3 Cesàro Summability of Fourier Series

189

If f is continuous on [− p, p], then it is bounded and therefore (4.21) holds for sufficiently large n and for all real x, and hence so does (4.19). This completes the proof of uniform convergence in this case. h Remark Lebesgue proved that for a Lebesgue integrable function f, the Fourier series has almost everywhere Cesàro sum f [see Problem 5.8.P8]. Kolmogorov succeeded in finding an integrable function whose Fourier series diverges everywhere [34]. When f is square integrable, we shall see that the partial sums of the Fourier series converge to f in the L2 norm [see Corollary 4.5.9]. For 1  p < ∞ and f 2 Lp[− p, p], the Cesàro means converge to f in the Lp norm [see Problem 7.2.P4]. Recall from Sect. 1.5 that a finite sum a0 þ

n X

ðak cos kx þ bk sin kxÞ

k¼1

is called a trigonometric polynomial. Clearly, sums and constant multiples of trigonometric polynomials are again trigonometric polynomials. Therefore so are partial sums of Fourier series and Fejér sums. It should be borne in mind however, that although every trigonometric polynomial is the partial sum of a trigonometric series, a sequence of such polynomials can consist of partial sums of different series. Therefore being the limit of a sequence of trigonometric polynomials is not the same as being the sum of a trigonometric series. In particular, the next result does not claim that every continuous periodic function is the sum of a trigonometric series, which would be false. It is the same as Theorem 1.5.5, which was used in Proposition 3.4.4, but we now give a fresh proof based on Fourier series, without using any consequences of Proposition 3.4.4. Theorem 4.3.5 (Weierstrass Approximation Theorem) Every continuous real-valued function on R with period 2p can be approximated uniformly by trigonometric polynomials. Proof This is an immediate consequence of Fejér’s Theorem 4.3.4 because the Fejér sums are trigonometric polynomials. h Corollary 4.3.6 If all Fourier coefficients of a continuous function of period 2p vanish, then the function vanishes everywhere. Proof If all Fourier coefficients of a continuous function of period 2p vanish, then its Fejér sums vanish as well and therefore f ðxÞ ¼ lim rn ðxÞ ¼ 0 for all real x. h n!1

Problem Set 4.3 4.3.P1. If f : ½a; b ! R is continuous and e > 0, then prove by using Fejér’s Theorem 4.3.4 that there exists a polynomial P(x) such that

190

4

Fourier Series

supfjf ðxÞ  PðxÞj : x 2 ½a; bg\e: [This is known as the Weierstrass Polynomial Approximation Theorem.] 4.3.P2. Use the Weierstrass Polynomial Approximation Theorem to prove that if f and g are continuous functions on [a, b] such that Z

b

Z xn f ðxÞdx ¼

a

b

xn gðxÞdx

for all n 0;

a

then f = g. 4.3.P3. If an and bn are the Fourier coefficients of f, show that the Fejér sums rn are

n1  X 1 k 1  ðak cos kx þ bk sin kxÞ: rn ðxÞ ¼ a0 þ 2 n k¼1 Use this fact to show that 1 p

Z

n1  1 2 X k 2 2 rn ðxÞ dx ¼ a0 þ 1 ðak þ b2k Þ: 2 n p k¼1 p

2

Deduce Parseval’s Theorem that if f has period 2p and is continuous with Fourier coefficients an, bn, then 1 p

Z

1 X 1 f ðxÞ2 dx ¼ a20 þ ða2k þ b2k Þ: 2 p k¼1 p

4.3.P4. Two different periodic functions can have the same Fourier series (e.g. if they differ only on a finite set of points). Can this happen if the functions are continuous? 4.3.P5. If f and g are continuous functions of period 2p with Fourier coefficients an, bn and an, bn respectively, show that 1 p

Z

1 X 1 f ðxÞgð xÞdx ¼ a0 a0 þ ðan an þ bn bn Þ: 2 p n¼1 p

4.3.P6. Suppose 0 < d < p and f ðxÞ ¼ 1 ifjxj  d, f ðxÞ ¼ 0 if d\jxj\p, and f(x + 2p) = f(x) for all x. (a) Compute the Fourier coefficients of f. 1 P sin nd pd (b) Conclude that n ¼ 2 when 0 < d < p. n¼1

4.3 Cesàro Summability of Fourier Series

191

(c) Deduce from Parseval’s Theorem [see Problem 4.3.P3] that

1 P sin2 nd n¼1

n2 d

¼ pd 2

when 0 < d < p. R 1 2 (d) Let d ! 0 and prove that 0 sinx x dx ¼ p2. 4.3.P7. Show that

1 P cos nx n¼1

n3=2

is the Fourier series of a continuous function and

converges to it uniformly.

4.4

Even and Odd Functions

The standard form of a Fourier series is the one we have considered in the preceding sections. In this form, the function has period 2p and it is sufficient to consider it on [− p, p], because its behaviour elsewhere is a repetition of its behaviour on this interval. In many applications, it is necessary to work with functions with some positive period 2l and it is therefore desirable to consider the function on [−l, l] and adopt the corresponding form of the Fourier series. This is done by a change of variables. Let f : R ! R have period 2l and be Lebesgue integrable on some interval of length 2l (and hence on any bounded closed interval). Introduce a new variable t so that t x ¼ p l and define FðtÞ ¼ f pl t ¼ f ðxÞ . Then F is defined on R, is Lebesgue integrable and has period 2p, so that we may form its Fourier series. In the Fourier series of F, we replace t by pl x. The resulting series is taken as the Fourier series of f, 1 X 1 npx npx þ bn sin Þ; f ð x Þ  a0 þ ðan cos 2 l l n¼1

where an ¼

1 p

Z

p p

FðtÞ cos nt dt ¼

1 l

Z

l l

f ðxÞ cos

npx dx; l

n ¼ 0; 1; 2; . . .

and 1 bn ¼ p

Z

p

1 FðtÞ sin nt dt ¼ l p

Z

l

l

f ðxÞ sin

npx dx; l

n ¼ 1; 2; 3; . . .;

192

4

Fourier Series

where the integrals have been transformed by using the change of variables made above and applying Remark 5.8.21 with pl as ‘c’. We emphasise that the remark is independent on any results from this chapter. A function f is said to be odd if f ðxÞ ¼ f ðxÞ

for every real x

and is said to be even if f ðxÞ ¼ f ðxÞ

for every real x:

For example, sine is an odd function and cosine is an even function. Also, odd powers of x are odd functions and even powers are even functions. We note further that sums and constant multiples of even functions are even, and likewise for odd functions. The product of a pair of even or odd functions is even while the product of an even function with an odd function is odd. The following facts may be familiar in the case of Riemann integrals from an elementary course on Calculus at least when the integrand f is a derivative. They involve breaking up the integral over [−l, l] as a sum of integrals over [−l, 0] and [0, l], and then effecting the change of variables t = −x in the integral over [−l, 0]. For this particular change of variables, the justification in the case of Lebesgue integration is given in Remark 5.8.21, which is independent of any results in this chapter. (a) If a function f is Lebesgue integrable on [−l, l] and satisfies f(−x) = f(x) for x 2 (−l, l), then Z

l

l

Z f ðxÞdx ¼ 2

l

f ðxÞdx:

0

(b) If a function f is Lebesgue integrable on [−l, l] and satisfies f(−x) = −f(x) for x 2 (−l, l), then Z

l

l

f ðxÞdx ¼ 0:

Proposition 4.4.1 The Fourier series of an even function f of period 2l is a “Fourier cosine series”, which means every bn = 0: f ð xÞ  with coefficients an ¼ 2l

Rl 0

1 X 1 npx a0 þ an cos 2 l n¼1

f ðxÞ cos npl x dx;

n ¼ 0; 1; 2; . . .:

4.4 Even and Odd Functions

193

The Fourier series of an odd function of period 2l is a “Fourier sine series”, which means every an = 0: f ðxÞ 

1 X

bn sin

n¼0

with coefficients bn ¼ 2l

Rl 0

f ðxÞ sin npl x dx;

npx l

n ¼ 1; 2; 3; . . .:

npx Proof If f is an even function, then f ðxÞ sin npx l is odd and f ðxÞ cos l is even; so

bn ¼

Z

1 l

l

l

f ðxÞ sin

npx dx ¼ 0; l

n ¼ 1; 2; . . .

and 2 an ¼ l

Z

l

f ðxÞ cos

0

npx dx; l

n ¼ 0; 1; 2; . . .:

npx If f is an odd function, then f ðxÞ cos npx l is odd and f ðxÞ sin l is even; so

an ¼

1 l

Z

l

f ðxÞ cos

l

npx dx ¼ 0; l

n ¼ 0; 1; 2; . . .

and bn ¼

2 l

Z

l 0

f ðxÞ sin

npx dx; l

n ¼ 1; 2; . . .:

The result now follows by substituting these values in the Fourier series of f as described in the second paragraph of this section. h Examples 4.4.2 (a) For the odd periodic function f shown in Fig. 4.1, [which is f(x) = x on (−p, p], extended by periodicity], we have

Fig. 4.1 Function in Example 4.4.2(a)

2 p

Z

p

2 1 x x sin nx dx ¼ ½ 2 sin nx  cos nxp0 p n n 0 2p nþ1 2 cos np ¼ ð1Þ : ¼ pn n

bn ¼

194

4

Fourier Series

Thus the sine series in accordance with Proposition 4.4.1 is 1 X

2 ð1Þn þ 1 sin nx: n n¼1

Since f is monotonic on [0, p] and continuous at p2, it follows from Corollary 4.2.11 that its Fourier series at p2 converges to f ðp2Þ. Therefore 1 p X 2 p 1 1 ¼ ð1Þn þ 1 sin n ¼ 2ð1  þ  þ   Þ; 2 n¼1 n 2 3 5

so that p4 ¼ 1  13 þ 15  þ    : Since the partial sums are continuous, if they were to converge uniformly on any neighbourhood of p, the limit would have to be continuous, which it obviously cannot be. Therefore they cannot converge uniformly on a neighbourhood of p. (b) For the even periodic function f shown in Fig. 4.2, [which is f(x) = |x| on (−p, p], extended by periodicity], we have

an ¼

2 p

Z

p 0

2 1 x 2 cos np  1 x cos nx dx ¼ ½ 2 cos nx þ sin nxp0 ¼ : p n n p n2



 pn4 2 if n is odd 0 if n is even, n 6¼ 0: For n = 0, we have

Thus an ¼

a0 ¼

2 p

Z

p

xdx ¼ p:

0

Therefore the cosine series in accordance with Proposition 4.4.1 is 1 p 4X cosð2n þ 1Þx ¼ ; 2 p n¼0 ð2n þ 1Þ2

which converges at each x 2 (0, p) in view of Corollary 4.2.11. In particular, this series converges when x ¼ p4 or p3 Fig. 4.2 Function in Example 4.4.2(b)

4.4 Even and Odd Functions

195

and therefore p p 4 1 1 1 1 1 1 1 ¼  pffiffiffi ð1  2  2 þ 2 þ 2  2  2 þ þ     Þ; 4 2 p 2 3 5 7 9 11 13 which leads to and

p2 16

pffiffiffi 2 ¼ 1  312  512 þ

1 72

þ

1 92

 1112  1312 þ þ      ;

p p 41 2 1 1 2 1 1 ¼  ð1  2 þ 2 þ 2  2 þ 2 þ 2  þ þ   Þ; 3 2 p2 3 5 7 9 11 13 which leads to

p2 12

¼ 1  322 þ

1 52

þ

1 72

 922 þ

1 112

þ

1 132

 þ þ :

(c) Let f : R ! R be the function with period 2p such that 1 f ðxÞ ¼ ðp  xÞ for 0  x\2p: 2 See Fig. 4.3. With some effort one can check that this function is odd, with jumps at the points 2pk and is continuous as well as decreasing on the intervals ½2p k; 2pðk þ 1ÞÞ, k ¼ 0; 1; 2; . . .. Since it is odd, its Fourier coefficients an are all 0 while the coefficients bn are given by 1 bn ¼ p

Z

p

2 f ðxÞ sin nx dx ¼ p p

Z

p 0

1 f ðxÞ sin nx dx ¼ p

Z

p 0

1 ðp  xÞ sin nx dx ¼ : n

Therefore lits Fourier series is 1 X 1 n¼1

n

sin nx:

By Corollary 4.2.11, this series converges to 12 ðp  xÞ at each point of ð2p k; 2pðk þ 1ÞÞ, k ¼ 0; 1; 2; . . ., and at any point 2pk, the sum of the series is 0. The latter can be verified directly. In applications, we often want to employ a Fourier series for a (Lebesgue integrable) function f that is defined on an interval of the form [0, l]. One way of doing this is to extend the function to [−l, l] to be even by setting

Fig. 4.3 Function in Example 4.4.2(c)

196

4

f ðxÞ ¼ f ðxÞ

Fourier Series

for  l  x\0

and then extend it further to all of R so as to have period 2l. The latter is possible since the extension to [−l, l] satisfies f(−l) = f(−l + 2l) = f(l). Moreover, the extension to R is even and Lebesgue integrable on [−l, l], and therefore has a cosine series. We speak of this as the “cosine series” of the original function given on [0, l]. Alternatively, we can first extend f to an odd function on [−l, l] by setting f ðxÞ ¼ f ðxÞ

for  l\x\0

and f ðlÞ ¼ f ðlÞ: A periodic extension to R is now possible. Moreover, the extension to R is odd and Lebesgue integrable on [−l, l], and therefore has a sine series. We speak of this as the “sine series” of the original function given on [0, l]. Example 4.4.3 For f(x) = x, 0  x  p, we shall determine the sine series and cosine series below. The odd and even periodic extensions are precisely the functions discussed in Examples 4.4.2(a) & (b) respectively. Therefore the sine and cosine series are none other than the Fourier series derived there, namely, 1 X

2 ð1Þn þ 1 sin nx n n¼1

and 1 p 4X cosð2n þ 1Þx  : 2 p n¼0 ð2n þ 1Þ2

Problem Set 4.4 4.4.P1. Find the Fourier cosine series for the function given on [0, p] by 1 P 1 sin x. Deduce from the series that 12 ¼ 4n2 1 . n¼1

4.4.P2. Find the Fourier series of the function 8 m, 2    X n n X   ck /k  ¼ c2 ; ks n  s m k ¼   k¼m þ 1 k k¼m þ 1 2

so that fsn gn 1 is a Cauchy sequence in L2[a, b]. By Theorem 3.3.10, there is a function f 2 L2[a, b] such that lim kf  sn k ¼ 0:

n!1

Again, for n > k, Z

Z ½a;b

ðf /k Þdm  ck ¼

½a;b

Z ðf /k Þdm 

½a;b

ðsn /k Þdm

in view of Definition 4.5.1. Hence  Z  Z         ðf /k Þdm  ck  ¼  ððf  sn Þ/k Þdm  kf  sn kk/k k ¼ kf  sn k;    ½a;b   ½a;b where we have used the Cauchy–Schwarz Inequality 3.3.16. Letting n ! ∞, we obtain

4.5 Orthonormal Expansions

201

Z ck ¼

½a;b

ðf /k Þdm ¼ h/k ; f i;

for k ¼ 1; 2; 3; . . .:

h

Definition 4.5.6 An orthogonal sequence f/k gk 1 of nonzero elements in L2[a, b] is said to be complete if the only element of L2[a, b] that is orthogonal to each /k is ½½0 ; in other words, if f/k gk 1 satisfies the following condition: whenever h/k ; f i ¼ 0 8 k 2 N, we have f = 0 almost everywhere (i.e. ½½ f  ¼ ½½0). We shall show below in Theorem 4.5.8 that the trigonometric sequence 1; cos x; sin x; cos 2x; sin 2x; . . .: is complete in L2[− p, p]. The sequence obtained by deleting any term from a complete orthogonal sequence is never complete, because the deleted term is a nonzero element orthogonal to each term in it. In Bessel’s Inequality, the sign of equality holds whenever the orthonormal sequence in question is complete. This is, in fact, the content of the last part of our next result. Theorem 4.5.7 Given a complete orthonormal sequence f/k gk 1 in L2[a, b], the orthonormal expansion of every element f 2 L2[a, b] converges to it in the mean: f ¼

1 X

ck /k ;

where ck ¼ h/k ; f i 8 k 2 N;

k¼1

i.e.     m X   ck /k  ! 0 as m ! 1: f    k¼1 Furthermore, we have Parseval’s Theorem: k f k2 ¼

1 X

c2k :

k¼1

[This equality was proved in 4.3.P3 for continuous 2p-periodic functions with reference to the orthonormal sequence 1 cos x sin x cos 2x sin 2x pffiffiffiffiffiffi ; pffiffiffi ; pffiffiffi ; pffiffiffi ; pffiffiffi ; . . .: p p p p 2p Proof For the complete orthonormal sequence f/k gk 1 in L2[a, b], and ck ¼ h/k ; f i 8 k 2 N, we have

202

4

2 *  +   X n n n n X X X   ck / k  ¼ ck /k ; ck / k ¼ c2k :   k¼m þ 1 k¼m þ 1 k¼m þ 1 k¼m þ 1 Since by Bessel’s Inequality, the series

m P k¼1

Fourier Series

ð4:22Þ

c2k is convergent, it follows that the

right-hand side of (4.22) tends to 0 as m, n ! ∞. On using Theorem 3.3.10, we obtain a function g 2 L2[a, b] such that     m X   ð4:23Þ ck /k  ! 0 as m ! 1: g    k¼1   We next show that /j ; g ¼ cj for every j. In fact, for any given j and any m j, we have    * +        m m m X X X           /j ; g  cj  ¼  /j ; g  ck /k   g  ck /k /j  ¼ g  ck /k :       k¼1 k¼1 k¼1 When m ! ∞, the right-hand side tends to 0 by virtue of (4.23) and so we get /j ; g ¼ cj .   It is a consequence of what has just been proved that /j ; g  f ¼ 0 for every j. The completeness of the orthonormal sequence f/k gk 1 implies g = f a.e. We may now write (4.23) as     m X   ck /k  ! 0 as m ! 1: f    k¼1 Also, 2 *  +   m m m m m X X X X X   c k /k  ¼ f  ck /k ; f  ck /k ¼ hf ; f i  2 ck h/k ; f i þ c2k f    k¼1 k¼1 k¼1 k¼1 k¼1 ¼ k f k2 2

m X k¼1

c2k þ

m X k¼1

c2k ¼ k f k2 

m X

c2k :

k¼1

So, 2    m X   c2k ¼ f  ck /k  ! 0 as m ! 1: k f k2    k¼1 k¼1 m X

4.5 Orthonormal Expansions

203

Consequently, k f k2 ¼

1 X

c2k :

h

k¼1

If for some orthonormal sequence, equality always holds in Bessel’s Inequality, i.e. if Parseval’s Theorem holds, then that orthonormal sequence is complete. This is because if an orthonormal sequence fails to be complete, then by definition of completeness, there must exist a nonzero f that is orthogonal to every term in the 1 P sequence, which implies c2k ¼ 0 but k f k [ 0 . k¼1

Finally, we show that the orthogonal trigonometric sequence is complete in L2[− p, p]. Theorem 4.5.8 If f 2 L2[− p, p] satisfies the conditions Z

Z

½p;p

f ðxÞ dmðxÞ ¼

8 n 2 N;

Z ½p;p

f ðxÞ cos nx dmðxÞ ¼

½p;p

f ðxÞ sin nx dmðxÞ ¼ 0

then f = 0 a.e. Proof From the hypotheses, it immediately follows that Z ½p;p

f ðxÞTðxÞdmðxÞ ¼ 0 for an arbitrary trigonometric polynomial T: ð4:24Þ

By Proposition 3.4.4, there exists a sequence fTn gn 1 of trigonometric polynomials such that lim kf  Tn k ¼ 0:

ð4:25Þ

n!1

But by Proposition 3.3.18, 0  k f k2 ¼ hf ; f i  hf ; f i þ hTn ; Tn i ¼ hf ; f i þ hTn ; Tn i  2hf ; Tn i

by (4:24Þ

¼ hf  Tn ; f  Tn i: From this and (4.25), it follows upon letting n ! ∞ that k f k ¼ 0, i.e. f = 0 a.e. h The following is an immediate corollary of Theorems 4.5.7 & 4.5.8.

204

4

Fourier Series

Corollary 4.5.9 With every f 2 L2[− p, p] we associate its “Fourier series” 1 X 1 f ð x Þ ¼ a0 þ ðan cos nx þ bn sin nxÞ; 2 n¼1

where an ¼

1 p

Z

p p

f ðxÞ cos nx d x;

n ¼ 0; 1; 2; . . .

f ðxÞ sin nx dx;

n ¼ 1; 2; 3; . . .:

and 1 bn ¼ p

Z

p p

Then lim kf  sn k ¼ 0;

n!1 n P

where sn ðxÞ ¼ 12 a0 þ

ðak cos kx þ bk sin kxÞ, n = 1, 2, … and Parseval’s

k¼1

Theorem holds: Z 1 1 2 X 1 a0 þ ða2k þ b2k Þ ¼ f 2 dm: 2 p ½p;p k¼1 Problem Set 4.5 4.5.P1. If A  [− p, p] is measurable, prove that Z

Z cos nx dmðxÞ ¼ lim

lim

n!1

A

n!1

sin nx dmðxÞ ¼ 0: A

4.5.P2. Let n1, n2, … be a sequence of positive integers and let E ¼ x 2 ½p; p : lim sin nk x exists: k!1

Prove that E has measure m(E) = 0. 4.5.P3. Show that each of the two sequences 1; cos x; cos 2x; cos 3x; . . . sin x; sin 2x; sin 3x; . . . is a complete orthogonal sequence in L2[0, p].

4.5 Orthonormal Expansions

205

4.5.P4. Suppose E is a subset of [− p, p] with positive measure and let d > 0 be given. Deduce from Bessel’s Inequality that the number of positive integers k such that sin kx > d for all x 2 E is finite.

Appendix The purpose of this appendix is to prove two results. One is a weaker version of Fejér’s Theorem 4.3.4, in which continuity is assumed everywhere and which can be deduced from a theorem due to P. P. Korovkin. The latter is of independent interest and has been the subject of much research and generalisation. The second is the existence of a continuous function with a Fourier series that does not converge at 0. The construction of the function that we present is due to Fejér but du Bois-Reymond was the first to construct such a function in 1876. Definition Let C[a, b] be the space of all continuous real-valued functions on the interval [a, b]. A map P:C[a, b] ! C[a, b] (also called an operator in C[a, b]) is said to be positive if P(f) 0 wherever f 0. It is said to be linear if Pða f þ b gÞ ¼ a Pðf Þ þ b PðgÞ for all a; b 2 R and all f, g 2 C[a, b]. It may be observed that a linear positive operator P is monotone, in the sense that f  g ) Pðf Þ  PðgÞ . Also, one can have linear positive operators in the space C2p[− p, p] of all continuous functions on [− p, p] with f(−p) = f(p). The functions f0 = 1, f1 = cos and f2 = sin belong to C2p ½p; p. Any f 2 C2p ½p; p can be uniquely extended to all of R by periodicity and the space of the extended functions is essentially the same as C2p ½p; p. In effect, we may assume every function in C2p ½p; p to be so extended and speak of its Fourier series. One often writes P(f) as simply Pf, especially when the operator is denoted by an upper case letter and the function to which it is applied is denoted by a lower case letter. Theorem (Korovkin’s Theorem) Let {Pn}n 1 denote a sequence of positive linear operators on C2p ½p; p. In order that Pnf ! f uniformly for every f 2 C2p ½p; p, it is necessary and sufficient that such convergence occur for f0 = 1, f1 = cos and f2 = sin. Proof We first prove sufficiency. Suppose the uniform convergence occurs for the three particular functions f0 = 1, f1 = cos and f2 = sin. Put uy ðxÞ ¼ sin2

yx 1 1 ¼ ð1  cosðy  xÞÞ ¼ ð1  cos y cos x  sin y sin xÞ: 2 2 2

206

4

Fourier Series

Then we have 1 uy ¼ ðf0  f1 cos y  f2 sin yÞ 2 and 1 Pn ðuy Þ ¼ ðPn ðf0 Þ  Pn ðf1 Þ cos y  Pn ðf2 Þ sin yÞ: 2 Thus 1 ðPn uy ÞðyÞ ¼ f½ðPn f0 ÞðyÞ  1  cos y½ðPn f1 ÞðyÞ  cos y  sin y½ðPn f2 ÞðyÞ  sin yg 2 1  fkPn f0  f0 k þ jcos yjkPn f1  f1 k þ jsin yjkPn f2  f2 kg; 2

where k f k means sup fjf ðzÞj : z 2 ½p; pg for any f 2 C2p ½p; p. Since |cos y| and |sin y| are bounded on [− p, p], it follows that Pn uy ðyÞ converges uniformly to 0 in y. Now let f be an arbitrary element of C2p ½p; p and let e > 0 be given. Since f 2 C2p ½p; p is uniformly continuous, there is a positive d < p such that jx  yj\d ) jf ðxÞ  f ðyÞj\e:

ð4:26Þ

Also, the function f 2 C2p ½p; p is bounded and therefore there exists an M > 0 such that M  f ðxÞ  M

8 x 2 ½p; p:

ð4:27Þ

Let y be an arbitrary but fixed point of [− p, p]. If y − d < x < y + d, i.e., x 2 (y − d, y + d), then jf ðxÞ  f ðyÞj\e by (4.26) above and if d  x − y  2p − d, i.e., x 2 [y + d, 2p + y − d] then jf ðxÞ  f ðyÞj  2M  2M

uy ðxÞ sin2 d2

:

by (4.27) above combined with the following argument: if d  x − y  2p − d, 2d d 1 d then and uy ðxÞ ¼ sin2 xy Thus, for 2  2 ðx  yÞ  p  2 2 sin 2. x 2 ðy  d; 2p þ y  d ¼ ðy  d; y þ dÞ [ ½y þ d; 2p þ y  d, we have

Appendix

207

e 

2M 2M uy ðxÞ  f ðxÞ  f ðyÞ  e þ 2 d uy ðxÞ; sin2 d2 sin 2

that is, ef0 

2M 2M uy  f  f ðyÞf0  ef0 þ 2 d uy sin2 d2 sin 2

on ðy  d; 2p þ y  d:

On using the periodicity of the functions involved, the above inequality is seen to hold on all of R and in particular on [−p, p]. Since the operators Pn are positive and linear, and hence monotone, we have eðPn f0 ÞðyÞ 

2M ðPn uy ÞðyÞ  ðPn f ÞðyÞ  f ðyÞðPn f0 ÞðyÞ sin2 d2 2M  eðPn f0 ÞðyÞ þ 2 d ðPn uy ÞðyÞ; sin 2

which implies jðPn f ÞðyÞ  f ðyÞðPn f0 ÞðyÞj  ekPn f0 k þ

2M ðPn uy ÞðyÞ: sin2 d2



Since Pnf0 ! f0 and Pn uy ðyÞ ! 0 uniformly in y, it follows that (Pnf) (y) ! f(y) uniformly in y. This completes the proof of sufficiency. The necessity is trivial. h Theorem (Fejér) The Cesàro means of the Fourier series of a continuous function of period 2p converge uniformly to the function. Proof It is clear from Remark 4.3.2 that the Fejér sums (operators) rn are positive and linear. We may complete the proof by verifying that rn f ! f uniformly when f = 1 or cos or sin. This requires only some simple computation: rn 1 ¼ 1n ð1 þ 1 þ    þ 1Þ ¼ 1 ! 1; ðrn cosÞðxÞ ¼ cos x þ cos x þ    þ cos xÞ ¼ n1 n cos x ! cos x; 1 ðrn sinÞðxÞ ¼ n ð0 þ sin x þ sin x þ    þ sin xÞ ¼ n1 sin x ! sin x: h n 1 n ð0 þ

Theorem There exists a continuous function with period 2p for which the Fourier series diverges to ∞ at 0. Proof For integers m > n > 0, set Tðx; m; nÞ ¼ 2 sin mx

n X sin kx : k k¼1

ð4:28Þ

208

4

Fourier Series

By Proposition 4.2.9, the sum on the right-hand side is bounded in absolute value (by 1 + p) on R and therefore so is T(x, m, n). Using elementary trigonometric identities, we can rewrite T(x, m, n) as the sum of the sums cosðm  nÞx cosðm  n þ 1Þx cosðm  1Þx þ þ  þ n n1 1

ð4:29Þ

  cosðm þ 1Þx cosðm þ 2Þx cosðm þ nÞx  þ þ  þ : 1 2 n

ð4:30Þ

and

It will be convenient to refer to the sum of (4.29) and (4.30) as the expansion of T(x, m, n). Observe that the multiples of x in (4.29) and in (4.30), taken together, are in increasing order, and consequently, the expansion of T(x, m, n) consists of terms of a trigonometric series in their proper order. Now P let {nk} and {mk} be sequences of positive integers such that nk < mk and let ak be a convergent series of positive terms. As T(x, m, n) is bounded in absolute value, the series 1 X

ak Tðx; mk ; nk Þ

ð4:31Þ

k¼1

converges uniformly and absolutely on R. Therefore its sum is a continuous function f(x) on R with period 2p. Each T is an even function and therefore the same is true of f. We can choose the sequences {nk} and {mk} such that mk þ nk \mk þ 1  nk þ 1

8 k:

To do so, all we need to arrange for is that nk+1 > 3nk and mk = 2nk, so that mk + nk = 3nk < nk+1 = 2nk+1 − nk+1 = mk+1 − nk+1. This ensures that, when k > j, all the multiples of x in the expansion of T(x, mk, nk) are higher multiples than all those in the expansion of T(x, mj, nj). Therefore in view of the observation above, upon writing out the expansion of each T(x, mk, nk) in the sum (4.31), we get a trigonometric series with terms in the proper order. Denote the series by P aj cos jx. The partial sums of (4.31) form a subsequence of the partial sums of the trigonometric series. Since the former converge uniformly to f(x), it follows [see Problem 4.1.P7] that the Fourier series corresponding to f is in fact Raj cos jx. Now consider the partial sum smp ðxÞ of the Fourier series. [This will not be a partial sum of (4.31) because it stops midway in T(x, mp, np).]

Appendix

209

smp ðxÞ ¼

mp X

aj cos jx

j¼1

¼

p1 X

 ak Tðx; mk ; nk Þ þ ap

k¼1

cosðmp  np Þx np

 cosðmp  np þ 1Þx cosðmp  1Þx þ  þ þ : np  1 1 Since T(0, m, n) = 0 by (4.28), we have smp ð0Þ ¼ ap

np X 1 k¼1

By Problem 4.2.P5,

np P 1 k¼1

k[

k

:

ln np and therefore smp ð0Þ\ap ln np. If we now

show that the sequences {nk}, {mk} and {ak} can be so chosen that ap ln np! ∞ as p ! ∞, then it will follow that the partial sums of the Fourier series of f diverge to ∞ when x = 0. To choose {nk}, {mk} and {ak} in the required manner, let ak = 1/k2 and arrange not only that mk = 2nk and nk+1 > 3nk as mentioned earlier, but also that nk+1 > exp(k + 1)3. h

Chapter 5

Differentiation

5.1

Background

In 1806, A.M. Ampère, a great scholar of his times, in a paper entitled ‘Sur la théorie des fonctions derivées’, tried without success to establish the differentiability of an arbitrary function except at isolated points. The lack of success can safely be attributed to the fact that the concept of function had not fully evolved then. During the nineteenth century, repeated attempts were made to establish the differentiability of arbitrary continuous functions. It was K. Weierstrass who, in the year 1861, put an end to these attempts by constructing a continuous function without a derivative at any point whatsoever. B. Bolzano had already obtained a function with this property in 1830. He was however debarred by an imperial decree from teaching and publishing. Later on, simpler examples of functions continuous everywhere but differentiable nowhere were constructed. Mathematicians of the era sought additional conditions under which a continuous function is at least differentiable at “very many points” of its domain. H. Lebesgue, in the year 1904, established as a consequence of his theory of integration (an equivalent version of which we have discussed in Chap. 3) that a continuous monotone function is differentiable except on a set of measure zero. Subsequently, in the year 1911, W. H. Young gave a proof of this striking and important theorem of Lebesgue without the assumption of continuity (Theorem 5.4.6). A simpler proof of the theorem, which is due to L. A. Rubel, as well as three other proofs, will be given in Sects. 5.3 and 5.4. In this chapter, we shall consider a class of real-valued functions that have a derivative almost everywhere and the derivative is a measurable function. The derivative is then “eligible” to have an integral and we shall discuss how the integral is related to the original function.

© Springer Nature Switzerland AG 2019 S. Shirali and H. L. Vasudeva, Measure and Integration, Springer Undergraduate Mathematics Series, https://doi.org/10.1007/978-3-030-18747-7_5

211

212

5.2

5 Differentiation

Monotone Functions and Continuity

In this section, we take up some matters about discontinuity of monotone functions (see Sect. 1.6) and one-sided limits (see Sect. 1.3). Remarks 5.2.1 (a) Since a decreasing function becomes an increasing function when multiplied by −1, we shall restrict our consideration of monotone functions to the class of increasing functions only. (b) If f :D ! R is a nonconstant monotone function, and D is compact, then f is bounded; for, if [a, b] is the smallest closed interval containing D, then the range of f is contained in an interval whose endpoints are f(a) and f(b). The result is of course trivial if the function is constant. Examples (a) The function f defined by f ðxÞ ¼ jxxj for x 6¼ 0 and f(0) = k, where −1  k  1, is increasing on R, and is discontinuous at 0, because f ð0 þ Þ ¼ lim f ðhÞ ¼ lim h!0 þ

h!0 þ

h ¼1 jhj

and h ¼ 1: h!0 jhj

f ð0Þ ¼ lim f ðhÞ ¼ lim h!0

(b) The integer part function on R is increasing and has a discontinuity at each of the integral points. (c) The function f defined on R by f(x) = x is continuous and increasing. We shall prove that functions which are monotone on compact intervals always have right-hand and left-hand limits at each interior point and a right [resp. left] limit at the left [resp. right] endpoint. Proposition 5.2.2 If f is increasing on [a, b], then f ðx þ Þ ¼ lim f ðx þ hÞ exists and f ðx þ Þ  f ðxÞ for every x 2 ½a; bÞ; h!0 þ

f ðxÞ ¼ lim f ðx þ hÞ exists and f ðxÞ  f ðxÞ for every x 2 ða; b: h!0

Furthermore, if a  x < y  b, then f(x+)  f(y −). Proof For the first two assertions, it is sufficient to prove for every x 2 (a, b] that f(x−) exists and does not exceed f(x), because the analogous statement for every x 2 [a, b) will follow in the same manner.

5.2 Monotone Functions and Continuity

213

Consider any x 2 (a, b]. For an arbitrary n2 [a, x), the set of numbers ff ðtÞ : n\t\xg is nonempty while being bounded above by the number f(x), and therefore has a least upper bound A, say. Evidently, A  f(x). We need only show that A = f(x−) in order to conclude that the one-sided limit f(x−) exists and f(x −)  f(x). It will also follow that f ðxÞ ¼ supff ðtÞ : n\t\xg

8 n 2 ½a; xÞ:

ð5:1Þ

Let e > 0 be given. By definition of the least upper bound, we have A  e\f ðyÞ  A

ð5:2Þ

for some y 2 (n, x). In view of the monotonicity of f, the inequality f ðyÞ  f ðtÞ  A holds for all t such that y < t < x. Together with (5.2), this implies that jf ðtÞ  Aj\e for all t satisfying the inequality y < t < x. Hence A = f(x−), which is all that was needed to be shown for the existence of the one-sided limit f(x−) and for the validity of the equality (5.1). By a similar argument, we can establish for any x 2 [a, b) that f(x+) exists, f(x+)  f(x) and f ðx þ Þ ¼ infff ðtÞ : x\t\ng

8 n 2 ðx; b:

ð5:3Þ

Now suppose that a  x < y  b. Then f(x+) = inf{f(t) : x < t < y} by (5.3) and f(y−) = sup{f(t) : x < t < y} by (5.1). It therefore follows that f(x+)  f(y −). h Remark 5.2.3 At a point in the interior of the interval of definition of a monotone function f, either f(x−) = f(x+), in which case the function is continuous at x, or f(x−) < f(x+) (see Proposition 5.2.2). In the latter case, there are two gaps in the range of the function if f(x−) < f(x) < f(x+) and one gap if f(x) = f(x−) or f(x+). Consequently, when the range of a monotone function is an interval, there cannot be any discontinuity at an interior point of the interval of definition. Moreover, there cannot be any discontinuity even at an endpoint. This is because the function can be extended beyond an endpoint to be constant, thereby preserving its monotonicity as well as its range but rendering that endpoint an interior point. Proposition 5.2.2 has an obvious analogue for decreasing functions.

214

5 Differentiation

Definition 5.2.4 If f:[a, b]!R is monotone and x is an interior point of [a, b], the number jf ðx þ Þ  f ðxÞj is called the jump or saltus of f at x. The jump of f at a [resp.b] is defined to be jf ða þ Þf ðaÞj [resp. jf ðbÞf ðbÞj]. Theorem 5.2.5 Let f:[a, b]!R be increasing. Then the set of points at which f is discontinuous is at most countable. Proof Let E be the set of points in (a, b) at which f is discontinuous. With every point x of E we associate an interval (f(x−), f(x+)). Let r(x) be a rational number such that f ðxÞ\rðxÞ\f ðx þ Þ: If x1 < x2 are points of discontinuity, then for any n such that x1 < n < x2, we obviously have f(x1+)  f(n)  f(x2−). Thus the intervals corresponding to different points of discontinuity do not overlap, although they may have an endpoint in common. We have thus shown that the set E is in one-to-one correspondence with a subset of the set of rational numbers, which is known to be countable. Adjoining the endpoints a and b, if they are points of discontinuity, does not alter its finiteness/ countability. This completes the proof. h Remark 5.2.6 The function f defined on R by f(x) = [x] has a discontinuity at each x 2 Z; which is a countable discrete set. However the points of discontinuity of a monotone function need not be isolated. Indeed, it is possible to construct a monotone function with discontinuities at precisely the points of a given countable subset of R; which may even be dense, as illustrated in the example below: Example 5.2.7 Let E be the range of a sequence {xn}n  1 of distinct terms in (a, b) and {an}n  1 be a sequence of positive real numbers such that Ran converges. Define fn ðxÞ ¼ 0 if x\xn and fn ðxÞ ¼ an if x  xn :

ð5:4Þ

Since jfn ðxÞj  an and Ran converges, it follows by the Weierstrass M-test (see Proposition 1.5.6) that the series Rfn converges uniformly to f, say. If x0 does not coincide with any of the xn, then it is a point of continuity for all the fn and hence a point of continuity of f by Proposition 1.5.3. On the other hand, if x0 = xm for some m, then precisely one of the functions, namely fm, is discontinuous at x0. Then Rn6¼m fn ¼ f  fm is continuous at xm. Hence the function f = (f − fm) + fm, which is the sum of two functions, one continuous at xm and the other discontinuous, is itself discontinuous at xm. Nevertheless, the function f is right continuous by the obvious analogue of Proposition 1.5.3. We next show that f is increasing on (a, b). Indeed, if a < b, then in view of (5.4),

5.2 Monotone Functions and Continuity

f ðaÞ ¼

X xm  a

fm ðaÞ ¼

X

215

am

X

and f ðbÞ ¼

xm  a

fm ðbÞ ¼

xm  b

X

am ;

xm  b

where the summations extend over all indices m for which xm  a and xm  b respectively. As the terms of the convergent series are nonnegative, the order of summation is immaterial. Since all the xm that are less than or equal to a are also less than or equal to b, it follows that f(a)  f(b). Thus the function f has all the desired properties mentioned in Remark 5.2.6. f ðxn þ Þ ¼ lim f ðxÞ ¼ lim ðf  fn ÞðxÞ þ lim fn ðxÞ ¼ lim ðf  fn ÞðxÞ þ an x!xn þ

x!xn þ

x!xn þ

x!xn þ

and f ðxn Þ ¼ lim f ðxÞ ¼ lim ðf  fn ÞðxÞ þ lim fn ðxÞ ¼ lim ðf  fn ÞðxÞ þ 0: x!xn

x!xn

x!xn

x!xn

Therefore, by continuity of f − fn at xn, we have f ðxn þ Þ  f ðxn Þ ¼ an : Our next result will lead to the conclusion that the sum of the jumps of a monotone function defined on a closed bounded interval is bounded. Proposition 5.2.8 Let f:[a, b]!R be an increasing function and suppose a < x1, x2, …, xn < b. The sum of the jumps of f at the points x1, x2, …, xn does not exceed the difference [f(b−)−f(a+)]. Proof Without loss of generality, we may assume that x1, x2,…,xn are in increasing order. Then by Proposition 5.2.2, f(a+)  f(x1−), f(xn+)  f(b−) and f(xi+)  f(xi+1−) for i = 1,…,n − 1. Hence we have n X i¼1

½f ðxi þ Þ  f ðxi Þ ¼

n1 X

½f ðxi þ Þ  f ðxi þ 1 Þ þ f ðxn þ Þ

i¼1

 f ðx1 Þ  0 þ f ðbÞ  f ða þ Þ;

as asserted. Theorem 5.2.5 can also be derived as a consequence of Proposition 5.2.8:

h

Corollary 5.2.9 The set of points of discontinuity of an increasing function on [a, b] is at most countable. Proof Let H be the set of all points in (a, b) at which f is discontinuous and Hn be the set of points x at which [f(x+)−f(x−)] > 1n, n = 1, 2, …. Then H ¼ [ n  1 Hn . We shall show that Hn is finite for each n and this implies that H is at most countable. If

216

5 Differentiation

Hn consists of infinitely many points, then for any finitely many x1, x2,…,xk amongst them, we shall have k

k 1 X  ½f ðxi þ Þ  f ðxi Þ  f ðbÞ  f ða þ Þ; n i¼1

using Proposition 5.2.8. This will lead to a contradiction on letting k tend to ∞. This shows that H must be at most countable; adjoining the endpoints a and b, if they are points of discontinuity, does not alter the finiteness/countability. h In view of Corollary 5.2.9, we may speak of the sum of (all) jumps on [a, b] meaning a sum f ða þ Þ  f ðaÞ þ

1   X    f xj þ  f xj  þ f ðbÞ  f ðbÞ j¼1

where the numbers in the sequence {xj}j  1 include all points of discontinuity in (a, b) but may also include other points. Proposition 5.2.10 Let f:[a, b]!R be an increasing function. Then the sum of its jumps is bounded above by f(b) − f(a). Proof If the number of discontinuities is finite, the result follows from Proposition 5.2.8. Otherwise, let the set of points of discontinuity of f in (a, b) be arranged in a sequence {xj}j  1. Then 1 X

½f ðxj þ Þ  f ðxj Þ  f ðbÞ  f ða þ Þ:

ð5:5Þ

j¼1

Indeed, any partial sum of the series on the left-hand side here satisfies the same inequality by Proposition 5.2.8 and hence (5.5) follows by taking the limit. Also, f ða þ Þ  f ðaÞ þ

1 X

½f ðxj þ Þ  f ðxj Þ þ f ðbÞ  f ðbÞ  f ðbÞ  f ðaÞ:

j¼1

Hence the sum of jumps at the points of discontinuity of f in [a, b] does not exceed [f(b) − f(a)]. h Proposition 5.2.11 [Needed in Example 5.5.4] Suppose {fn}n  1 is a uniformly convergent sequence of increasing functions defined on an interval and c is a point in the domain such that the sequence formed by the jumps fn(c+) − fn(c−) does not tend to 0. Then the limit function f is increasing and has a positive jump at c. Proof By passing to a subsequence if necessary, we may assume that there exists an a > 0 such that fn(c+) − fn(c−) > a for every n. The given uniform convergence

5.2 Monotone Functions and Continuity

217

makes it possible to apply Proposition 1.5.7, which leads to f ðc þ Þ  f ðcÞ ¼ lim ðfn ðc þ Þ  fn ðcÞÞ  a [ 0 . Since f must be increasing by virtue of

n!1

Proposition 1.6.7, the foregoing inequality means that f has a positive jump at c. h

Remark The hypothesis that the sequence formed by the jumps f ðc þ Þ  f ðcÞ does not tend to 0 is essential, because each member of a sequence of increasing functions can have a nonzero jump at the same point and yet converge uniformly to 1 a continuous function. An instance of this is /n ðxÞ ¼ 1  1n for x  x0 and 1  2n for x > x0. The sequence {/n} converges uniformly to the constant function 1, although each /n has a nonzero jump at x0. Problem Set 5.2 5.2.P1. Give an example of a monotone function which is discontinuous at each rational number in [0, 1] (see Example 5.2.7). 5.2.P2. Let f be defined on an open interval (a, b) and assume that for each point x of the interval, there exists an open interval Nx of x in which f is increasing. Prove that f is increasing throughout (a, b). 5.2.P3. Let f be continuous on a compact interval [a, b] and assume that f does not have a local minimum or local maximum at any interior point. Prove that f must be monotonic on [a, b]. 5.2.P4. A function f increases on a closed interval [a, b] and it is true that, if f (a) < k < f(b), there exists an x 2 [a, b] such f(x) = k. Prove that f is continuous on [a, b]. Does the same conclusion hold if the hypothesis that f increases is dropped but it is still required that f(a) < f(b)? A function f increases on an open interval I and, for each a, b 2 I, it is true that, if f(a) < k < f(b), then there exists an x 2 I such f(x) = k. Prove that f is continuous on I. Does the same conclusion hold if the hypothesis that f increases is dropped?

5.3

Monotone Functions and Differentiability (A)

Let f be an increasing function defined on [a, b]. Since the set of points of discontinuity is at most countable, there exist countably many points xk 2 [a, b] that include all points of discontinuity (if there be any). Denote the jump of f at xk by j(xk). With the increasing function f on [a, b], we associate a function S defined throughout the interval [a, b] by means of the relations SðaÞ ¼ 0 and SðxÞ ¼

X

jðxk Þ þ f ðxÞ  f ðxÞ for x 2 ða; b;

ð5:6Þ

xk \x

where the summation extends over all indices k for which P xk < x. If the xk are infinite in number, then the terms of the convergent series jðxk Þ are nonnegative,

218

5 Differentiation

and therefore the order of summation on the right-hand side of (5.6) is immaterial. Since j(xk) = 0 when xk is not a point of discontinuity, the sum is independent of how the set {xk}  [a, b] is chosen so long as it includes all points of discontinuity. So the function S is well defined on [a, b] and is obviously increasing. By choosing {xk} so as to include a, we find from (5.6) that S(x) is the sum of all jumps on [a, x] when x 2 (a, b]. This leads to the first part of the following double inequality, while Proposition 5.2.10 justifies the second part: f ða þ Þ  f ðaÞ  SðxÞ  f ðxÞ  f ðaÞ for x 2 ða; b:

ð5:7Þ

The function S is called the saltus function or jump function associated with f. Theorem 5.3.1 Let f:[a, b]!R be increasing and S denote the saltus function associated with f. The difference function g = f − S is a continuous increasing function on [a, b]. Proof For a, b 2 (a, b] satisfying a < b, let {xk}  [a, b] be any countable set that contains all points of discontinuity and has infinitely many points in (a, a) as well as in (a, b). We have b > a as well as a > a, and hence the definition of S yields X X SðbÞ ¼ jðxk Þ þ jðaÞ þ jðxk Þ þ f ðbÞ  f ðbÞ; a  xk \a

SðaÞ ¼

X

a\xk \b

jðxk Þ þ ðf ðaÞ  f ðaÞÞ:

a  xk \a

Therefore, X

SðbÞ  SðaÞ ¼ jðaÞ þ

jðxk Þ þ f ðbÞ  f ðbÞ  ðf ðaÞ  f ðaÞÞ;

a\xk \b

¼ ðf ða þ Þ  f ðaÞÞ þ

X

jðxk Þ þ f ðbÞ  f ðbÞ  ðf ðaÞ  f ðaÞÞ

a\xk \b

¼ ðf ða þ Þ  f ðaÞÞ þ ðf ðaÞ  f ðaÞÞ X jðxk Þ þ f ðbÞ  f ðbÞ  ðf ðaÞ  f ðaÞÞ þ

ð5:8Þ

a\xk \b

¼

X

½f ðxk þ Þ  f ðxk Þ þ f ða þ Þ  f ðaÞ þ f ðbÞ  f ðbÞ:

a\xk \b

The right-hand side is the sum of all the jumps of the function f restricted to the closed interval [a, b]. It follows by Proposition 5.2.10 that SðbÞ  SðaÞ  f ðbÞ  f ðaÞ:

ð5:9Þ

Though (5.9) has been proved for a > a, it holds even for a = a in view of the second part of (5.7). So, g(b) = f(b) − S(b)  f(a) − S(a) = g(a) for a  a < b  b, i.e. g is increasing on [a, b].

5.3 Monotone Functions and Differentiability (A)

219

Let x0 2 [a, b) be arbitrary. We shall show that g is right continuous at x0. Let x > x0. From (5.9), it follows that SðxÞ  Sðx0 Þ  f ðxÞ  f ðx0 Þ: On letting x ! x0+, we obtain Sðx0 þ Þ  Sðx0 Þ  f ðx0 þ Þ  f ðx0 Þ:

ð5:10Þ

On the other hand, SðxÞ  Sðx0 Þ  f ðx0 þ Þ  f ðx0 Þ; using the first part of (5.7) if x0 = a and using (5.8) if x0 > a. On letting x ! x0+, we obtain Sðx0 þ Þ  Sðx0 Þ  f ðx0 þ Þ  f ðx0 Þ: Comparing this with (5.10), we get Sðx0 þ Þ  Sðx0 Þ ¼ f ðx0 þ Þ  f ðx0 Þ; whence gðx0 þ Þ ¼ f ðx0 þ Þ  Sðx0 þ Þ ¼ f ðx0 Þ  Sðx0 Þ ¼ gðx0 Þ: Thus g is right continuous at x0, where x0 2 [a, b) is arbitrary. The left continuity of g on (a, b] is proved analogously, except that the first part of (5.7) is not needed. This completes the proof. h In preparation for Lebesgue’s Theorem on differentiation of monotone functions, namely, that an increasing function on a closed bounded interval has a finite derivative almost everywhere, we discuss below the definitions of one-sided limits superior and inferior of a function. Let x be a point in the closure of an interval I other than its right endpoint (if any). In particular, for any d > 0, there exists a t in I such that 0 < t − x < d. Suppose g is an extended real-valued function defined either on I \{x} or on I. We define lim sup gðtÞ ¼ inf d [ 0 fsup gðtÞ : 0\t  x\dg t!x

¼ inffsupfgðtÞ : 0\t  x\dg : d [ 0g:

220

5 Differentiation

Since the sets {g(t): 0 < t – x < d} decrease with d, their sups also decrease with d and it follows that the inf in the above definition can be replaced by a limit: lim sup gðtÞ ¼ lim supfgðtÞ : 0\t  x\dg:

t!x þ

d!0

Similarly, if x is a point in the closure of I other than its left endpoint (if any) lim sup gðtÞ ¼ inf fsup gðtÞ : 0\x  t\dg d[0

t!x

¼ inffsupfgðtÞ : 0\x  t\dg : d [ 0g ¼ lim supfgðtÞ : 0\x  t\dg: d!0

We define lim inf gðtÞ and lim inf gðtÞ analogously: t!x þ

t!x

If x is a point in the closure of I other than its right endpoint (if any), lim inf gðtÞ ¼ sup finf gðtÞ : 0\t  x\dg t!x þ

d[0

¼ supfinffgðtÞ : 0\t  x\dg : d [ 0g ¼ lim inffgðtÞ : 0\t  x\dg: d!0

Similarly, if x is a point in the closure of I other than its left endpoint (if any), lim inf gðtÞ ¼ sup finf gðtÞ : 0\x  t\dg t!x

d[0

¼ supfinffgðtÞ : 0\x  t\dg : d [ 0g ¼ lim inffgðtÞ : 0\x  t\dg: d!0

Since sup{g(t): 0 < t − x < d}  inf{g(t) : 0 < t − x < d}, it follows upon taking limits that lim sup gðtÞ  lim inf gðtÞ provided that x is not the right endpoint

t!x þ

t!x þ

and similarly that lim sup gðtÞ  lim inf gðtÞ provided that x is not the left endpoint.

t!x

t!x

For any function g, the above limits exist at the points indicated but they may be infinite. We consider the following examples to illustrate the above definitions. (1) We have lim sup½t ¼ 0 ¼ lim inf ½t; lim sup½t ¼ 1 ¼ lim inf ½t: þ t!0 þ

t!0

t!0

t!0

5.3 Monotone Functions and Differentiability (A)

221

(2) Consider the function f(x) = sin(1/x), for x 6¼ 0, defined on R\{0}. In this case, lim sup f ðtÞ ¼ 1 ¼ lim sup f ðtÞ and lim inf f ðtÞ ¼ 1 ¼ lim inf f ðtÞ: t!0 þ

t!0 þ

t!0

t!0

(3) Let f be defined on R by means of the relations f(t) = 1 if t is rational and 0 if t is irrational. If a 2 R is arbitrary but fixed, lim sup f ðtÞ ¼ 1 ¼ lim sup f ðtÞ, t!a þ

t!a

whereas lim inf f ðtÞ = 0 = lim inf f ðtÞ . The reader will note that the one-sided t!a þ

t!a

limits f(a+) and f(a−) do not exist. Remark The reader is invited to show that lim sup gðtÞ ¼ inffsupfgðtÞ : 0\t  x\dg : d1 [ d [ 0g t!x þ

for an arbitrary d1 > 0. Thus the values of g in an arbitrarily small interval to the right of x are the only relevant ones for purposes of the limsup as t ! x+. It is easy to show that the right limit g(x+) exists (possibly ±∞) if and only if lim sup gðtÞ ¼ lim inf gðtÞ, in which case lim gðtÞ ¼ lim sup gðtÞ ¼ lim inf gðtÞ. t!x þ

t!x þ

t!x þ

t!x þ

t!x þ

Similarly for g(x −). In order to prove Lebesgue’s Theorem on differentiation of monotone functions, we shall need the concept of Dini derivatives, which involve replacing the limit in the definition of derivative by lim inf and lim sup from the left and from the right. We consider real-valued functions whose domains are intervals in R, containing more than one point. The four Dini derivatives that we are about to introduce have the advantage that they exist for functions that are not necessarily differentiable in the usual sense. They are defined as follows: For an arbitrary real-valued function f and any x which is not the right endpoint of the domain, D þ f ðxÞ ¼ lim sup h!0 þ

f ðx þ hÞ  f ðxÞ f ðx þ hÞ  f ðxÞ ; D þ f ðxÞ ¼ lim inf h!0 þ h h

and for any x which is not the left endpoint of the domain, D f ðxÞ ¼ lim sup h!0

f ðx þ hÞ  f ðxÞ f ðx þ hÞ  f ðxÞ ; D f ðxÞ ¼ lim inf : h!0 h h

The values of these derivatives are in the extended real number system R and they are known as the right upper and lower and the left upper and lower Dini derivatives respectively. Remarks 5.3.2 (a) The inequalities

222

5 Differentiation

D þ f ðxÞ  D þ f ðxÞ and D f ðxÞ  D f ðxÞ obviously hold wherever they make sense (i.e., except at endpoints). (b) It is also easy ton see that D+of(x) [resp. D+f(x)] is the largest [resp. smallest] limit of a sequence

f ðx þ hn Þf ðxÞ hn n1

where {hn}n  1 is a sequence of nonnegative

numbers with limit 0 and such that each x + hn is in the domain of f. Similar statements hold for D−f(x) and D−f(x). (c) If D+f(x) = D+f(x) [resp. D−f(x) = D−f(x)], then f is said to have a right derivative [resp. left derivative] at x and we write f þ0 ðxÞ [resp. f0 ðxÞ] for the common value of D+f(x) and D+f(x) [resp. D−f(x) and D−f(x)]. If x is an interior point of the domain of f and f þ0 ðxÞ ¼ f0 ðxÞ, then f is said to have a derivative at 0 ðxÞ and f0 ðxÞ, an obvious x and we write f 0 ðxÞ for the common value of f þ modification being needed if x is an endpoint. Note that our definition includes ±∞ as values of f 0 ðxÞ. If f 0 ðxÞ 6¼ ±∞, we say that f is differentiable at x. In this case, f 0 ðxÞ is the derivative in the usual sense, as reiterated in Sect. 1.6. Examples 5.3.3 (a) Suppose f is defined by f(x) = xsin 1x for x 6¼ 0 and f(0) = 0. Then D+f(0) = D−f(0) = 1 and D+f(0) = D−f(0) = −1. 1 Indeed, the ratio f ðhÞ h = sin h assumes all values in [−1, 1] as h varies in the 1 interval (ð2n þ1 2Þp, 2np ), n 2 N. Therefore D+f(0) = 1 and D+f(0) = −1. Again by a similar argument, we have D−f(0) = 1 and D−f(0) = −1. (b) Let f be defined by f(x) = 1 if x is rational and 0 otherwise. For rational x and ðxÞ is either 0 or a negative number with large absolute h > 0, the ratio f ðx þ hÞf h + value. So, D f(x) = 0 and D+f(x) = −∞. On the other hand, for h < 0, the same ratio is either 0 or a positive number with large absolute value. Hence D−f(x) = ∞ and D−f(x) = 0. If x is irrational, then similar considerations show that D+f(x) = ∞ and D+f(x) = 0, while D−f(x) = 0 and D−f(x) = −∞. Thus the function does not have even one-sided derivatives. (c) Let f(x) = |x|. Then D+f(0) = D+f(0) = 1 and D−f(0) = D−f(0) = −1. (d) Let f:R ! R be defined by f(0) = 0, f(x) = x(1 + sin(lnx)) if x > 0 and pffiffiffiffiffiffi x þ x sin2(ln(−x)) if x < 0. Then D+f(0) = 2, D+f(0) = 0, D−f(0) = 1, D−f(0) = −∞. Indeed, sin(lnh) assumes all values between −1 and 1 in any 1 , where n 2 N, and hence for positive values interval from expð2np1 þ 2pÞ to expð2npÞ of h arbitrarily close to 0. Therefore D+f(0) = 2 and D+f(0) = 0. Again, by a similar argument, D−f(0) = 1, D−f(0) = −∞. F. Riesz’ proof in the case of a continuous function, given in [21], is both elegant and simple. We shall borrow from Riesz’ proof and combine it with an innovative

5.3 Monotone Functions and Differentiability (A)

223

idea of Rubel [18] to give a reasonably elementary proof of the theorem of Lebesgue. We begin by proving Riesz’ Lemma. Riesz’ Lemma 5.3.4 Let g be a continuous real-valued function on the interval [a, b] and let E ¼ fx 2 ða; bÞ : gðxÞ\gðyÞ for some y [ xg

ð5:11Þ

E 0 ¼ fx 2 ða; bÞ : gðxÞ\gðyÞ for some y\xg:

ð5:12Þ

and

Then the set E [resp. E′] is either empty or an open set; if E [resp. E′] is nonempty and (ak, bk) is any component interval of it, then g(ak)  g(bk) [resp. g(ak)  g(bk)]. Proof (F. Riesz) We first show that E is open. Let x0 be in E. Then there exists y > x0 with g(y) > g(x0). Let e = g(y) − g(x0). By continuity of g, there exists a d > 0 such that |x − x0| < d implies |g(x)−g(x0)| < e. Let d1 be min{d,y − x0, x0 − a}. Now |x − x0| < d1 implies x < y as well as g(x) < g(x0) + e = g(y), which in turn implies x 2 E. Therefore E contains the open interval (x0 − d1, x0 + d1) and is thus an open set. The set E, if not empty, decomposes into an at most countable collection of disjoint nonempty open intervals (ak, bk), k = 1, 2,… (see Theorem 1.3.17); the points ak, bk do not belong to the set E. Let ak < x0 < bk. We shall prove that g(x0)  g(bk); the desired result will follow upon letting x0 tend to ak. Let x1 be a point in the interval [x0, b] at which g assumes a maximum value. Then gðx0 Þ  gðx1 Þ:

ð5:13Þ

gðbk Þ  gðx1 Þ:

ð5:14Þ

Also, since bk 2 [x0, b], we have

Furthermore, every point in [x1, b] satisfies g(y)  g(x1), and it therefore follows that x1 62 E. Since the part of [x0, b] to the left of bk lies entirely within E, we have bk  x1  b. Again, bk 62 E and therefore the inequality g(bk) < g(x1) cannot hold. By (5.14), it follows that g(bk) = g(x1), and hence by (5.13), g(x0)  g(bk). This completes the proof for the set E. The proof for E′ is similar. h Lemma 5.3.5 Let f be a continuous increasing function defined throughout [a, b]. For any positive number M, let EM = {x 2 [a, b) : D+f(x) > M}. Then the measure m(EM) of EM satisfies the inequality mðEM Þ 

1 ðf ðbÞ  f ðaÞÞ: M

224

5 Differentiation

Proof By definition of D+f(x), whenever D+f(x) > M, there exists y > x such that f ðyÞ  f ðxÞ [ M: yx So, f ðyÞ  f ðxÞ [ My  Mx;

i:e:;

f ðyÞ  My [ f ðxÞ  Mx:

ð5:15Þ

If we set g(t) = f(t) − Mt for t 2 [a, b], then g is continuous and (5.15) implies that x 2 E, where E is the set as in Riesz’ Lemma 5.3.4. Thus EM  E. If EM is empty, there is nothing to prove; so we may assume that EM is nonempty, so that E is also nonempty. By Riesz’ Lemma, E is open and is, therefore, representable as a countable disjoint union of component intervals: E¼

1 [

ðak ; bk Þ;

k¼1

where g(ak)  g(bk) for each k, i.e. Mðbk  ak Þ  f ðbk Þ  f ðak Þ; whence mðEM Þ  mðEÞ ¼

X k

ðbk  ak Þ 

1X 1 ðf ðbk Þ  f ðak ÞÞ  ðf ðbÞ  f ðaÞÞ: k M M

The last inequality follows in the following manner from the fact that f is increasing: Any n among the disjoint intervals (ak, bk) can be arranged so that bk  ak+1 for k = 1, 2, …, n − 1 and n X

ðf ðbk Þ  f ðak ÞÞ ¼ f ða1 Þ þ ðf ðb1 Þ  f ða2 ÞÞ þ    þ ðf ðbn1 Þ  f ðan ÞÞ þ f ðbn Þ

k¼1

  f ða1 Þ þ f ðbn Þ  f ðbÞ  f ðaÞ: Taking the limit as n ! ∞ if there are infinitely many k yields the desired inequality. h Corollary 5.3.6 Let f be a continuous increasing function defined throughout [a, b]. Then m(E∞) = 0, where E∞ = {x 2 (a, b) : D+f(x) = ∞}. Proof E∞ is the intersection of the sequence of sets {En}, where

5.3 Monotone Functions and Differentiability (A)

225

En ¼ fx 2 ½a; b : D þ f ðxÞ [ ng: Since E1 E2 ⋯ and m(E1) < ∞, we have m(E∞) = lim m(En) by continuity n!1

of measure (see Proposition 2.3.21). But this limit is 0, in view of the inequality m (En)  1n ðf ðbÞ  f ðaÞÞ assured by the foregoing lemma. h Lemma 5.3.7 Let f be a continuous increasing function defined throughout [a, b]. Then D þ f ðxÞ  D f ðxÞ a.e. on (a, b). Proof Step I. Let N and M be positive numbers such that N < M and ENM ¼ fx 2 ða; bÞ : D f ðxÞ\N\M\D þ f ðxÞg: We shall show that m(ENM) = 0. We may assume that ENM is nonempty, because otherwise there is nothing to prove. Let x 2 ENM. Since D−f(x) < N, there exists a y < x such that f ðyÞ  f ðxÞ \N; yx whence we get (using the fact that y − x < 0) f ðyÞ  f ðxÞ [ Ny  Nx;

i:e:;

f ðyÞ  Ny [ f ðxÞ  Nx:

ð5:16Þ

If we set g(t) = f(t) − Nt, t 2 [a, b], then g is continuous and (5.16) implies that x 2 E′, where E0 ¼ ft 2 ða; bÞ : gðtÞ\gðyÞ for some y 2 ða; tÞg as in (5.12) of Riesz’ Lemma. Thus ENM is covered by an at most countable number of disjoint intervals (ak, bk) such that g(bk)  g(ak) and therefore f ðbk Þ  f ðak Þ  Nðbk  ak Þ:

ð5:17Þ

Next we consider inside each of the intervals (ak, bk) points x where D+f(x) > M. That includes all the points of the nonempty set ENM. Using arguments similar to the ones in the paragraph above, and applying Riesz’ Lemma to the function g1(t) = f(t) − Mt, t 2 [ak, bk], we see that these points x form an open nonempty set. There exists an at most countable number of disjoint intervals fðak‘ ; bk‘ Þg‘  1 each contained in (ak, bk) and such that

226

5 Differentiation

Mðbk‘  ak‘ Þ  f ðbk‘ Þ  f ðak‘ Þ and hence M

X

ðbk‘  ak‘ Þ  ‘

X ‘

ðf ðbk‘ Þ  f ðak‘ ÞÞ  f ðbk Þ  f ðak Þ:

On using (5.17), it follows that X X X M ðbk‘  ak‘ Þ  ðf ðbk Þ  f ðak ÞÞ  N ðbk  ak Þ: k;‘ k k If jS 0 j, jS 1 j and jS 2 j denote the total lengths of the systems S 0 = {(a, b)}, S 1 = {(ak, bk)} and S 2 = fðak‘ ; bk‘ Þg respectively, it follows that jS 2 j 

N N jS 1 j  jS 0 j: M M

Repeating the two arguments above alternately, we obtain a sequence S 0 ; S 1 ; S 2 ; . . . of systems of disjoint open intervals, each having union contained in that of the preceding one, and satisfying the inequalities jS 2n j 

N jS 2n2 j; M

n ¼ 1; 2; . . . ;

which implies  n N jS 2n j  jS 0 j ! 0 as n ! 1: M Thus the set ENM can be covered by a system of intervals of total length as small as we please; so mðENM Þ ¼ 0: Step II. We next show that mðfx 2 ða; bÞ : D þ f ðxÞ [ D f ðxÞgÞ ¼ 0: Observe that fx 2 ða; bÞ : D þ f ðxÞ [ D f ðxÞg ¼

[ r; s 2 Q r[s

where

Esr ;

5.3 Monotone Functions and Differentiability (A)

227

Esr ¼ fx 2 ða; bÞ : D þ f ðxÞ [ r [ s [ D f ðxÞg and r and s are rational numbers. Since by Step I, m(Esr) = 0 for every pair r and s of rational numbers (r > s), and since there are at most countably many such pairs, it follows on using Proposition 2.2.9 that mðfx 2 ða; bÞ : D þ f ðxÞ [ D f ðxÞgÞ 

X

mðEsr Þ ¼ 0:

r; s 2 Q r[s This completes the proof of Step II. It now follows immediately that D þ f ðxÞ  D ðxÞ a.e. on [a, b], which is what the lemma claimed. h We now present Lebesgue’s Theorem for continuous monotone functions, which is the case he treated. The generalisation to all monotone function s, originally due to Young, will follow in the next section. Theorem 5.3.8 (Lebesgue’s Theorem (continuous case)). Every continuous monotone function defined on [a, b] possesses a finite derivative almost everywhere. Proof We need consider only increasing functions. The above lemma, when applied to the function −f(−x), yields D f ðxÞ  D þ ðxÞ a.e. on (a, b). Since f is increasing on [a, b], each of the four Dini derivatives is nonnegative. Combining this observation with the above inequality, Lemma 5.3.7 and Corollary 5.3.6 gives 0  D þ f ðxÞ  D f ðxÞ  D f ðxÞ  D þ f ðxÞ  D þ f ðxÞ\1 a.e. on (a, b), i.e. f′(x) exists and is finite a.e. on [a, b].

h

Problem Set 5.3 5.3.P1. If the function f assumes its local maximum at an interior point c of its domain, then show that D+f(c)  0 and D−f(c)  0. 5.3.P2. Suppose f is continuous on [a, b] and D+f(x) > 0 for all x in [a, b). Show that f(b)  f(a). Also, give a counterexample to show that the continuity hypothesis cannot be dropped. 5.3.P3. Show that f may be discontinuous at x0 when all four Dini derivatives are equal to ∞. 5.3.P4. Let f:[0, ∞)!  R be differentiable and suppose that f(0) = 0 and that f′ is increasing. Prove that

228

5 Differentiation

gðxÞ ¼

f ðxÞ x 0

f ð0Þ

x[0 x¼0

defines an increasing function of x. 5.3.P5. (a) Show that if f′(x) exists, then D+(f + g)(x) = f′(x) + D+g(x) and similarly for other Dini derivatives. (b) Give an example when D+(f + g)(x) 6¼ D+f(x) + D+g(x). 5.3.P6. ðaÞ + (a) Let f be continuous on [a, b]. If f ðbÞf ba \C, then D f(x)  C for uncountably many x 2 [a, b]. ðaÞ (b) Let f be continuous on [a, b]. If f ðbÞf [ C, then D−f(x)  C for uncountably ba many x 2 [a, b].

5.3.P7. Let f be a continuous function defined on [a, b]. Suppose that there exist real constants a, b such that a  D+f(x)  b for all x 2 (a, b). Prove that ha  f(x + h) − f(x)  hb provided a  x < x + h  b. 5.3.P8. Let f be a continuous real-valued function defined on [a, b] and let x 2 (a, b) be such that D+f is finite in a neighbourhood of x and continuous at x. Prove that f′(x) exists. 5.3.P9. If one of the Dini derivatives of a continuous function is zero everywhere in an interval, the function is constant there. 5.3.P10. Construct monotonic jump functions f on [0, 1) whose discontinuities have 0 as a limit point and such that f+′(0) is (a) zero (b) ∞ (c) positive and finite. Also, compute the quantum of jump at each of the discontinuities. 5.3.P11. If all Dini derivatives of a function f satisfy |Df(x)|  K everywhere on an interval, then the function satisfies the condition |f(x) − f(y)|  K|x − y|.

5.4

Monotone Functions and Differentiability (B)

In this section, we shall present three proofs of Lebesgue’s Theorem on differentiation of arbitrary monotone functions. We begin by proving an extension of Riesz’ Lemma 5.3.4 that applies to the discontinuous case. It is our view that the extension is not so obvious and we spell out its proof in detail. The reader may note that the functions in this section too are real-valued. Suppose g is a function on a closed interval [n, η] having finite one-sided limits at each point. Let G be the function on the domain [n, η] defined by G(z) = max{g(z+), g(z−), g(z)} for z 2 (n, η) and G(n) = max{g(n+), g(n)}, G(η) = max{g(η−), g(η)}.

5.4 Monotone Functions and Differentiability (B)

229

It is a trivial consequence that g  G everywhere. Suppose z 2 [n,η] and a > b > G(z). By definition of G, it follows that not only g(z) < b but also g(z+) < b (for z 6¼ η) and g(z−) < b (for z 6¼ n), which further implies that g  b on a neighbourhood of z, and this in turn implies G < a on a neighbourhood of z. Thus, G(z) < a implies G < a on a neighbourhood of z. This property of G is called upper semicontinuity. It is obvious for any upper semicontinuous function / that a set of the form {z 2 (n,η): /(z) < a} is always open. Moreover, the same argument as for continuous functions in elementary analysis shows that an upper semicontinuous function on a closed bounded interval is bounded above and attains its supremum. This property of G makes it possible to adapt the reasoning of Riesz’ Lemma for the continuous case, where the fact that g had a maximum value on any closed subinterval of its domain played a crucial role. Proposition 5.4.1 (Extended Riesz Lemma) Let the notations be as in the paragraph above. Then the set E ¼ fx 2 ðn; gÞ : gðyÞ [ GðxÞ for some y [ xg is open; if E 6¼ ∅ and (a, b) is a component interval of E, then g(x)  G(b) for all x 2 (a, b); in particular, g(a+)  G(b). The set F ¼ fx 2 ðn; gÞ : gðyÞ [ GðxÞ for some y\xg is open, and if it is nonempty and (a′, b′) is a component interval of F, then we have G(a′)  g(x) for all x 2 (a′, b′); in particular, g(b′−)  G(a′). Proof Let x 2 E. By definition of E, there exists y > x such that g(y) > G(x). By upper semicontinuity of G, the inequality G(z) < g(y) holds for all z in some neighbourhood of x. The intersection of this neighbourhood with (n, y) then lies in E. This shows that E is open. By the defining condition of E, if x 62 E, then g is bounded above by G(x) on the subinterval [x,η], which implies that all one-sided limits on (x, η] are also bounded by G(x), which further implies that G is bounded by G(x) on [x, η]. This observation will be used below. The set E, if not empty, decomposes into an at most countable collection of disjoint open intervals (ak, bk), k = 1, 2, … (see Theorem 1.3.17); the points ak, bk do not belong to the set E. Let ak < x < bk. We shall prove that g(x)  G(bk); the desired result will follow upon letting x tend to ak. Since G is upper semicontinuous, there exists some x1 2 [x, η] such that G(x1) = sup{G(z): z 2 [x, η]}. Then gðxÞ  GðxÞ  Gðx1 Þ: Since x < bk, we have bk 2 [x, η], which implies

ð5:18Þ

230

5 Differentiation

Gðbk Þ  Gðx1 Þ:

ð5:19Þ

Furthermore, since [x1, η]  [x, η], every point y 2 [x1, η] satisfies g(y)  G(y)  G(x1), and it therefore follows that x1 62 E. Since the part of [x,η] to the left of bk lies entirely within E, we have bk  x1  η. Again, bk 62 E and the observation above permits the conclusion that G(x1)  G(bk). By (5.19), it follows that G(bk) = G(x1), and hence by (5.18), g(x)  G(bk). This completes the proof for the set E. The proof for the set F is similar. h Remark (a) The function G discussed above need not take the value c = sup{g(z): z 2 I} on I unless I = [n, η]. This is illustrated by the function g on [n, η] = [0, 2] defined to satisfy g(z) = z on [0, 1), g(z) = z + 2 on (1, 2] and g(1) = 2. Here g(1+) = 3 and therefore G(z) = z on [0, 1) and G(z) = z + 2 on [1, 2], which means that when I = [0, 1], the function G never takes the value 2 = sup{g(z): z 2 I}. However, G(1) = 3 > sup{g(z): z 2 I} and 1 2 I. (b) A function on a closed bounded interval having one-sided limits at each point can be discontinuous at every rational point (28, Problems 5.2.P5 and 6.2.P4). Nevertheless, such a function can be uniformly approximated by step functions (28, Corollary 6.2.7). (c) It is a matter of peripheral interest that one can tweak the proof of Proposition 5.4.1 slightly to obtain g(x) < G(bk). All it takes is to note that one of the inequalities in (5.18) is actually strict, because otherwise g takes its maximum value on [x, η] at x, making g(x) an upper bound for g on [x, η] and thus contradicting the fact that x 2 E. Proposition 5.4.2 Let f be an increasing function on [a, b] and let ER ¼ fx 2 ða; bÞ : f is continuous at x and D þ f ðxÞ [ Rg; where R is a fixed positive number. If ER is nonempty, it can be covered by an at most countable set of intervals {(ak, bk)}k  1 whose total length R(bk− ak) satisfies the inequality X

ðbk  ak Þ  R1 ½f ðbÞ  f ða þ Þ:

Proof If we alter the function at the single point b by resetting its value there to be f(b−), the altered function will be increasing on [a, b], have the same values of f(a+), f(b−) and D+f, have the same set ER, but will satisfy the extra condition that f(b−) = f(b). It is therefore sufficient to prove the required inequality with f(b−) replaced by f(b).

5.4 Monotone Functions and Differentiability (B)

231

Set g(z) = f(z) − Rz for z 2 [a, b]. Since f is increasing, not only does g have a one-sided limit at each point, but also, at any interior point, f(z+)  f(z)  f(z−) and hence f(z+) − Rz  f(z) − Rz  f(z−) − Rz, so that gðz þ Þ  gðzÞ  gðzÞ: Consequently, the function G defined on [a, b] as GðzÞ ¼ maxfgðz þ Þ; gðzÞ; gðzÞg

for every z 2 ða; bÞ;

GðaÞ ¼ maxfgða þ Þ; gðaÞg; GðbÞ ¼ maxfgðbÞ; gðbÞg actually satisfies GðzÞ ¼ gðz þ Þ for every z 2 ½a; bÞ;

GðbÞ ¼ gðbÞ:

ð5:20Þ

Now, consider any arbitrary x 2 ER. There exists y > x satisfying f(y) − f(x) > R(y − x), i.e., gðyÞ [ gðxÞ: From the continuity of f at x, which implies that of g, we obtain g(y) > g(x+), and hence by (5.20), also g(y) > G(x). Since we have shown such a y to exist for an arbitrary x 2 ER, on applying Proposition 5.4.1 to g, we find that ER is covered by an at most countable number of disjoint open intervals (ak, bk) such that Gðbk Þ  gðak þ Þ: By using (5.20), when bk < b, this inequality can be rewritten as g(bk+)  g(ak+) and when bk = b as g(b)  g(ak+). This can in turn be written as f(bk+) − Rbk  f(ak+) − Rak, or equivalently, R(bk − ak)  f(bk+) − f(ak+) when bk < b and as R(b − ak)  f(b) − f(ak+) when bk = b. It follows that X X ðbk  ak Þ  R1 ½f ðbk þ Þ  f ðak þ Þ; subject to the proviso that bk+ is to be replaced by b if bk = b. The fact that f is increasing and the intervals (ak, bk) are disjoint now leads to the inequality we seek. h Corollary 5.4.3 Let f be an increasing function on [a, b]. Then the set {x: D+f (x) = ∞} has measure zero. Proof Let H be the set of all points in [a, b] at which f is discontinuous. We know that H is at most countable and hence has measure zero. Now, for any R > 0, we have

232

5 Differentiation

fx : D þ f ðxÞ ¼ 1gER [ H and so, the measure of the set {x : D+f(x) = ∞} does not exceed m(ER) + m(H) = m(ER). If ER = ∅, the proof is finished. If ER 6¼ ∅, we know by the preceding Proposition 5.4.2 that mðER Þ ¼

X

ðbk  ak Þ  R1 ½f ðbÞ  f ða þ Þ:

Since R can be chosen as large as we please, it follows that m({x: D+f(x) = ∞}) = 0. h Proposition 5.4.4 Let f be an increasing function on [a, b] and let Er ¼ fx 2 ða; bÞ : f is continuous at x and D f ðxÞ\rg; where r is a fixed positive number. If Er is nonempty, then Er can be covered by an at most countable set of intervals {(ak, bk)}k whose total length R(bk− ak) satisfies the inequality X X ½f ðbk Þ  f ðak þ Þ  r ðbk  ak Þ  rðb  aÞ: Proof If x 2 Er, there exists a y < x such that f ðyÞ  f ðxÞ \r; yx that is, f ðyÞ  f ðxÞ [ rðy  xÞ; since y < x. This implies f ðyÞ  ry [ f ðxÞ  rx:

ð5:21Þ

Set g(x) = f(x) − rx. Since g is continuous at x, the function G used in Proposition 5.4.1 satisfies G(x) = g(x). From (5.21), it follows that g(y) > G(x). This means Er is a subset of what has been called F in that proposition. Therefore we conclude that Er can be covered by an at most countable set of disjoint intervals {(ak, bk)}k such that g(bk−)  G(ak). Hence f(bk−) − rbk  G(ak). [To compute G(ak), note that, as in the proof of Proposition 5.4.2, we have GðzÞ ¼ gðz þ Þ for every z 2 ½a; bÞ;

GðbÞ ¼ gðbÞ;

so that G(ak) = f(ak+) − rak.] So, f(bk−) − rbk  f(ak+) − rak and this implies that f(bk−) − f(ak+)  r(bk − ak). Now the first part of the desired inequality follows

5.4 Monotone Functions and Differentiability (B)

233

upon summation. The second part follows from the fact that the intervals are disjoint and contained in [a, b]. h Lemma 5.4.5 Suppose that f:[a, b]!R is an increasing function and 0 < a < b. If Ea,b = {x 2 (a, b) : D_f(x) < a < b < D+f(x)}, then m(Ea,b) = 0. Proof Let H be the at most countable set of points of discontinuity in [a, b]. It will be sufficient to show that m(Ea,b\H) = 0. If Ea denotes the set {x 2 (a, b) : f is continuous at x and D−f(x) < a} and Eb the set {x 2 (a, b) : f is continuous at x and D+f(x) > b}, then Ea,b\H = Ea \ Eb . It is therefore sufficient to show that m(Ea \ Eb ) = 0. Assume that Ea \ Eb is nonempty, because otherwise there is nothing to prove. By Proposition 5.4.4, Ea can be covered by an at most countable family {(ak, bk)}k of open intervals such that X

½f ðbk Þ  f ðak þ Þ  aðb  aÞ:

ð5:22Þ

Next, we consider inside each of the intervals (ak, bk) the points of Eb, i.e. where b < D+f(x). The set Fk formed by such points satisfies Fk ¼ ðak ; bk Þ \ Eb ðak ; bk Þ \ Ea \ Eb and their union taken over all k contains Ea \ Eb , considering that the family {(ak, bk)}k covers Ea. In view of the assumption that Ea \ Eb is nonempty, Fk must be nonempty for some k. By Proposition 5.4.2, the set Fk, if nonempty, can be covered by disjoint open intervals fðak‘ ; bk‘ Þg‘ such that b

X ‘

ðbk‘  ak‘ Þ  ½f ðbk Þ  f ðak þ Þ:

ð5:23Þ

From (5.22) and (5.23), it follows that b

X k;‘

ðbk‘  ak‘ Þ  aðb  aÞ;

where the summation is taken over only those k for which n Fk is nonempty. Thus, o if

jS 1 j denotes the total length of the system S 1 ¼ ðak‘ ; bk‘ Þk;‘ : Fk 6¼ £ , it follows that jS 1 j 

a ðb  aÞ; b

and that S 1 covers the union of the sets Fk, which (as already noted) contains Ea \ Eb , and hence covers Ea \ Eb . Applying the same procedure to each of the intervals in S 1 , we get an at most countable family S 2 of open intervals that covers Ea \ Eb and satisfies jS 2 j  ab jS 1 j . Continuing indefinitely in this manner, we obtain a sequence S 1 ; S 2 ; . . . of systems of intervals, each “imbedded” in the preceding one and such that

234

5 Differentiation

jS n þ 1 j 

a jS n j; n ¼ 1; 2; . . . b

and each S n covers Ea \ Eb . It follows that  n a jS n j  ðb  aÞ: b Thus the set Ea \ Eb can be covered by an at most countable family of intervals whose total length is as small as we please; so, m(Ea \ Eb ) = 0, which is all that was needed to be shown. h We now restate Lebesgue’s Theorem 5.3.8, this time without the assumption of continuity, and prove it. Theorem 5.4.6 (Lebesgue’s Theorem) Let f:[a, b]!R be an increasing function. Then f is differentiable almost everywhere. Proof Since f is increasing, each of the four Dini derivatives is nonnegative at each point of [a, b]. We assert that mðfx 2 ða; bÞ : D f ðxÞ\D þ f ðxÞgÞ ¼ 0: Observe that fx 2 ða; bÞ : D þ f ðxÞ [ D f ðxÞg ¼

[

Er;s ;

r; s 2 Q r[s where Er;s ¼ fx 2 ða; bÞ : D þ f ðxÞ [ r [ s [ D f ðxÞg and r and s are rational numbers. Consequently, by Lemma 5.4.5, X mðfx 2 ða; bÞ : D þ f ðxÞ [ D f ðxÞgÞ  mðEr;s Þ ¼ 0: r; s 2 Q r[s Hence D þ f ðxÞ  D f ðxÞ

ð5:24Þ

for almost all x 2 [a, b]. The inequality (5.24), when applied to −f(−x) yields

5.4 Monotone Functions and Differentiability (B)

235

D f ðxÞ  D þ f ðxÞ

ð5:25Þ

for almost all x 2 [a, b]. Using Remark 5.3.2(a) with (5.24) and (5.25), we obtain 0  D þ f ðxÞ  D f ðxÞ  D f ðxÞ  D þ f ðxÞ  D þ f ðxÞ for almost all x 2 [a, b]. Hence the equality sign must hold a.e. on [a, b]. h The next result is about the saltus function S of an increasing function f on an interval [a, b]. Recall that the definition of S (see beginning of Sect. 5.3) is that SðaÞ ¼ 0 and SðxÞ ¼

X

jðxk Þ þ f ðxÞ  f ðxÞ for x 2 ða; b;

xk \x

where x1,x2,… form any countable set containing all the points of discontinuity of f, whether finite or infinite in number, and the summation extends over all indices k for which xk < x. Here, j(x), called the “jump of f at x”, is understood in the sense that j(x) = |f(x+) − f(x−)| if x 2 (a, b), whereas j(a) = |f(a+) − f(a)| and j(b) = |f(b) − f(b−)| (see Definition 5.2.4). The saltus function, as noted in Sect. 5.3, is increasing and bounded above by f(b) − f(a). The next proof follows the ideas of Boas [5, pp. 81–82]. In preparation for it, we note the following P about saltus functions. Suppose a > a is a point of continuity. Then SðaÞ ¼ xk \a jðxk Þ . Hence if b > a is also a point of continuity, P SðbÞ  SðaÞ ¼ a\xk \b jðxk Þ, which is the sum of the jumps at those points of discontinuity that lie in the open interval (a,P b). The equality can be expressed as S(b) − S(a) = J((a, b)), where J(G) means xk 2G jðxk Þ for any G  [a, b]. This notation will continue to be used in the next proof. The obvious fact that J is a measure on [a, b] with every subset taken as measurable will be used. Theorem 5.4.7 Let f be an increasing function defined on [a, b] and S be its saltus function. Then S′(x) = 0 almost everywhere on [a, b]. Proof If there are only finitely many points of discontinuity, then S is constant between the points and there is nothing to prove. So, assume there are infinitely many, in which case, they must be countable. Let us enumerate them as {xk}k  1. The function S, being increasing, is measurable because {x 2 [a, b]: S(x) < c} is always an interval. Hence D+S(x), being the limit superior from the right of a sequence of quotients of measurable functions [see Remark 5.3.2(b)], is also measurable. It follows that any set 2 En ¼ fx : D þ SðxÞ [ g; n

n ¼ 1; 2; . . .;

being the inverse image of the open interval (2n, ∞] under the measurable function D+S, is measurable. We shall show that m(En) = 0 for n = 1, 2, …. Since the set {x: D+S(x) > 0} = [ n  1En, it will follow that m({x: D+S(x) > 0}) is zero. It is sufficient to show that m(En) = 0 with the assumption that a, b 62 En.

236

5 Differentiation

Suppose for some n, En is not of measure zero. Let d = m(En) > 0 and consider a positive number η such that d 0\g\ : 2 Since

1 P

jðxk Þ  SðbÞ\1, there exists a k0 such that

k¼1

X k [ k0

jðxk Þ\

d : 16n

ð5:26Þ

For each k  k0, choose a closed interval Ik of length 2−kη and containing xk, 0 and denote the complement [a, b]\ [ kk¼1 Ik by A. This choice of intervals Ik has two consequences. First, xk 2 A implies k > k0 and hence by (5.26), GA implies JðGÞ ¼

X

jðxk Þ\

xk 2G

d : 16n

ð5:27Þ

Second, mð

k0 [ k¼1

Ik Þ 

k0 X

‘ðIk Þ ¼

k¼1

k0 X

2k g\g

k¼1

and hence mðEn n

k0 [

Ik Þ  mðEn Þ  mð

k¼1

k0 [

Ik Þ [ d  g [ d 

k¼1

d d ¼ : 2 2

0 By Proposition 2.8.18, we can find a closed set F  En\ [ kk¼1 Ik such that

mððEn n

k0 [

d Ik ÞnFÞ\ ; 4 k¼1

so, mðFÞ ¼ mðEn n

k0 [

Ik Þ  mððEn n

k¼1

d d d [  ¼ : 2 4 4

k0 [ k¼1

Ik ÞnFÞ

ð5:28Þ

5.4 Monotone Functions and Differentiability (B)

237

0 Since the bounded set F is closed and [ kk¼1 Ik , being a finite union of closed 0 intervals, is also closed, it follows that F is at a positive distance from [ kk¼1 Ik . Keeping in mind that a, b 62 En F, to each x belonging to F, we assign a positive dx such that

• • • •

(x − dx, x + dx)  (a, b), neither x − dx nor x + dx is a point of discontinuity, S(x + dx) − S(x) > dnx and 0 Ik . (x − dx, x + dx)  A = [a, b]\ [ kk¼1

As neither x − dx nor x + dx is a point of discontinuity, as noted before the statement of the theorem, J((x − dx, x + dx)) = S(x + dx) − S(x − dx). Moreover, since f is increasing, we have Jððx  dx ; x þ dx ÞÞ Sðx þ dx Þ  Sðx  dx Þ Sðx þ dx Þ  SðxÞ 1 ¼ [ [ : dx dx dx n

ð5:29Þ

By the Heine–Borel Theorem, a finite collection {Gp} of the intervals (x − dx, x + dx) covers F. By Problem 5.4.P3, we can prune the collection {Gp} of open intervals in such a manner that no three have a nonempty intersection but the union [ pGp remains the same. Then by Problem 5.4.P4(b), X

JðGp Þ  2Jð p

[ p

Gp Þ;

ð5:30Þ

considering that J is a measure on the r-algebra of all subsets of [a, b]. Since F  [ pGp, we have X mðFÞ  ‘ðGp Þ: By (5.29), J(Gp) 

‘ðGp Þ 2n

and so, from the above inequality, we get mðFÞ  2n

X p

JðGp Þ:

ð5:31Þ

But [ pGp  A and therefore by (5.27), Jð

[ p

Gp Þ 

d : 16n

d Combining this with (5.30), we find that RpJ(Gp)  2(16n ) and hence by (5.31), d d d m(F)  4n(16n) = 4 . But m(F) > 4 by (5.28) and we have obtained a contradiction, thereby establishing that m(En) = 0. We have thus proved that D+S(x) = 0 almost everywhere. Since D+S(x)  D+S(x)  0, this shows that S has zero right-hand derivative almost everywhere. By considering the function −f(−x), we see that S has zero left-hand derivative everywhere. This completes the proof. h

238

5 Differentiation

Remark Every increasing function f can be decomposed into the sum of its continuous part g and saltus part S (Theorem 5.3.1). By Theorem 5.3.8, the function g has finite derivative a.e. and, by Theorem 5.4.7, the function S has zero derivative almost everywhere. It therefore follows that the increasing function f has a finite derivative almost everywhere. We thus have a second proof of Lebesgue’s Theorem 5.4.6. We now present a third proof due to Rubel [18]. But first, a lemma. Lemma 5.4.8 Let f be a strictly increasing function defined throughout [a, b]. Then f has a continuous increasing “left inverse” with domain [f(a), f(b)], that is, there exists a continuous increasing function F:[f(a), f(b)]!R such that F(f(x)) = x for each x 2 [a, b]. Proof Define F:[f(a), f(b)]!R by setting FðyÞ ¼ supft : f ðtÞ  yg:

ð5:32Þ

It is immediate that the range of F is contained in [a, b]. First we show that F(f(x)) = x for each x 2 [a, b]. To this end, consider any x 2 [a, b]. Since f is strictly increasing, we have f(t)  f(x) , t  x, so that {t: f(t)  f(x)} = {t: t  x}; hence sup{t: f(t)  f(x)} = x. By (5.32), this is the same as saying that F(f(x)) = x. Next, we show that F is increasing. If y1  y2, then f(t)  y1 ) f(t)  y2, and therefore {t: f(t)  y1}  {t: f(t)  y2}, so that sup{t: f(t)  y1}  sup{t: f(t)  y2}, i.e. F(y1)  F(y2). Thus F is an increasing function. It remains to prove continuity. Since we have shown that the function is increasing, its continuity will follow by Remark 5.2.3 if we can show that its range is the entire interval [a, b]. But this is a trivial consequence of the fact that F(f(x)) = x for each x 2 [a, b]. h Example 5.4.9 Suppose f has domain [0, 2] and f(x) = x for x 2 [0, 1), f(x) = x + 2 for x 2 (1, 2] and f(1) = a, where 1  a  3, then f is strictly increasing and f (0) = 0, f(2) = 4. A continuous left inverse F on [0, 4] is given by F(y) = y for y 2 [0,1), F(y) = 1 for y 2 [1, 3] and F(y) = y − 2 for y 2 (3, 4], regardless of the choice of a 2 [1, 3]. If we alter the values of F at some points in [1, 3] other than a, the resulting function is not monotone but is still a left inverse for f. We can alter the values in such a manner as to make it discontinuous as well, for example, by altering values at only a finite number of points. Remark 5.4.10 The left inverse F of Lemma 5.4.8 is constant in a neighbourhood of every point that lies in the interior of the complement of the range of f. To see why, let y be such a point. Then y belongs to an open interval lying within the complement of the range and therefore the set {t: f(t)  z} is the same for all z in the interval. It follows from the definition of the left inverse F in the statement of the lemma that the left inverse is constant on the interval. Now we are ready to present Rubel’s proof of the Lebesgue’s Theorem 5.4.6 on differentiation of arbitrary monotone functions.

5.4 Monotone Functions and Differentiability (B)

239

Without loss of generality, we may assume that f is strictly increasing, and also that it satisfies the condition f ðyÞ  f ðxÞ  y  x whenever y  x:

ð5:33Þ

For, otherwise we could consider the function g(x) = f(x) + x. This function satisfies the assumed conditions, namely, g is strictly increasing and gðyÞ  gðxÞ ¼ f ðyÞ  f ðxÞ þ y  x  y  x

whenever y  x:

Also, g′(x) < ∞ a.e. will imply f′(x) < ∞ a.e. If F is the continuous left inverse of f as in Lemma 5.4.8, then F′(y) < ∞ a.e. by Theorem 5.3.8. Thus the set EF where F′ fails to exist has measure zero. We write f ðyÞ  f ðxÞ f ðyÞ  f ðxÞ ¼ ¼ yx Fðf ðyÞÞ  Fðf ðxÞÞ



 Fðf ðyÞÞ  Fðf ðxÞÞ 1 : f ðyÞ  f ðxÞ

Thus, for every point x such that f is continuous at x and f(x) does not belong to the exceptional set EF, we see as in the elementary proof concerning the derivative of a two-sided inverse that f ðyÞ  f ðxÞ ¼ ½F 0 ðf ðxÞÞ1 ; y!x yx

lim

i.e. f′(x)  ∞ exists. But the set of points of discontinuity of f is at most countable. Furthermore, f−1(EF) has measure zero, because for any interval I, the set f−1(I) is an interval of length not exceeding that of I, using (5.33). Finally, in view of Corollary 5.4.3, the set of points x where f′(x) = ∞ has measure zero. Hence f′(x) < ∞ exists almost everywhere. Lebesgue’s Theorem may be proved in yet another way, using what is known as the Vitali Covering Theorem (see Problems 5.4.P1 and 5.4.P2). It does not use the continuous case, which two of the three previous proofs do. Problem Set 5.4 Definition 5.4.P1 Let E be a subset of R of finite outer measure. A family I of closed intervals, each of positive length is called a Vitali cover of E if for each x 2 E and every e > 0, there exists an I 2 I such that x 2 I and ‘ðIÞ < e, i.e., each point of E belongs to an arbitrarily short interval of I . Example Let {rn}n  1 be an enumeration of the rationals in [a, b]. Then the collection {In,i}, where In,i = [rn − 1i , rn+ 1i ], n, i 2 N, forms a Vitali cover of [a, b]. (Vitali Covering Theorem) Let E be a subset of R of finite outer measure and I be a Vitali cover of E. Then given any e > 0, there is a finite disjoint collection {I1,I2,…,IN} of intervals in I such that

240

5 Differentiation

m ðEn

N [

Ii Þ\e:

i¼1

5.4.P2. (Lebesgue’s Theorem) Let [a, b] be a closed interval in R and let f be a real-valued monotone function on [a, b]. Using the Vitali Covering Theorem, prove that f is differentiable almost everywhere. 5.4.P3. If three open intervals have a nonempty intersection, then at least one among them is contained in the union of the other two, or equivalently, the union of some two among them is the same as the union of all three. 5.4.P4. (a) Let l be a finite measure on a set X and A1,…,An be finitely many distinct measurable subsets of X, where n  3. If no three sets among the Ai have a nonempty intersection and each l(Ai) is finite, show that lð

n [

Ai Þ ¼

n X

i¼1

X

lðAi Þ 

i¼1

lðAi \ Aj Þ:

i\j  n

(b) Let l be a measure on a set X and A1,…,An be finitely many distinct measurable subsets of X, where n  3. If no three sets among the Ai have a nonempty intersection, show that n X

lðAi Þ  2lð

i¼1

5.5

n [

Ai Þ:

i¼1

Integral of the Derivative

Let f be Riemann integrable on [a, b] and suppose that there exists a function F on the same interval such that F′(x) = f(x) everywhere. Then Z

b

f ðxÞdx ¼ FðbÞ  FðaÞ:

ð5:34Þ

a

This is the Second Fundamental Theorem of Calculus [see Theorem 1.7.2(b)]. It is the purpose of this section to investigate the relationship (5.34) in the case when f is monotone on [a, b]. It is proposed to replicate (5.34) for Lebesgue measurable functions under appropriate conditions. A weaker form of (5.34) is the content of the next theorem.

5.5 Integral of the Derivative

241

Theorem 5.5.1 Let f be an increasing function defined on [a, b] and E be the set of points in [a, b] at which f′ exists. Then the derivative f′:E!R is integrable in the sense of Lebesgue and Z Z f 0 dm ¼ f 0 dm  f ðbÞ  f ðaÞ: ð5:35Þ ½a; b

E

Proof Since the function f is increasing, it is continuous a.e. and is, therefore, measurable because {x 2 [a, b]: S(x) < c} is always an interval. Let us extend it to the interval [a, b + 1] by putting f ðxÞ ¼ f ðbÞ for b\x  b þ 1: We then define un : [a, b]!R by 1 un ðxÞ ¼ n½f ðx þ Þ  f ðxÞ; n

a  x  b;

n ¼ 1; 2; . . .:

Since f is increasing, the functions un are nonnegative everywhere. It is clear that un (x) ! f′(x) if x 2 E. Also, the functions un are measurable and therefore f′ is measurable by Proposition 2.6.1. Since [a, b]\E is of measure zero by Lebesgue’s Theorem 5.4.6, we have Z

Z

b

un ðxÞdx ¼ E

Z

a

Z

¼ n½

b

un ðxÞdx ¼ n½ a

b þ 1n

a þ 1n

Z ¼ n½

Z f ðxÞdx 

1 f ðx þ Þdx  n

b

Z

b

f ðxÞdx

a

f ðxÞdx

a

b þ 1n

Z f ðxÞdx 

b

a þ 1n

ð5:36Þ f ðxÞdx

a

Z

a þ 1n

¼ f ðbÞ  n

f ðxÞdx:

a

By monotonicity of f, we have f(x)  f(a) and hence Z n

a þ 1n

f ðxÞdx  f ðaÞ:

ð5:37Þ

a

(5.36) and (5.37) together imply Z un ðxÞdx  f ðbÞ  f ðaÞ: E

By Fatou’s Lemma (Theorem 3.2.8) and (5.38), it follows that

ð5:38Þ

242

5 Differentiation

Z

f 0 dm  f ðbÞ  f ðaÞ:

ð5:39Þ

E

Once again using the fact that m([a, b]\E) = 0, we obtain from (5.39) Z

f 0 dm ¼

E

Z ½a; b

f 0 dm  f ðbÞ  f ðaÞ:

Remark The inequality (5.35) obtained in the above theorem seems somewhat unsatisfactory when compared with (5.34). However, equality need not always hold in (5.34), e.g. the function given by f ðxÞ ¼

0 1

0  x\1 x¼1

R satisfies f′(x) = 0 a.e. but f(1) − f(0) = 1 − 0 6¼ ½0; 1 f 0 . We next construct a continuous monotone function called the Cantor function or Lebesgue’s singular function which has the same feature. Example 5.5.2 Recall that we defined a function g from the Cantor set C to [0, 1] in Proposition 1.8.12. Recall also the terminology of removed and component intervals of Cn and that they are denoted by In,k and Jn,k respectively. It was proved that g is onto. We shall now show that the function assumes the same values at the endpoints of any interval that has been removed. Indeed, for the pair 13 = 0.0222 … and 13 = 0.2000 …, the endpoints of the   interval I1,1, we get g 13 = 0.0111 … and g 23 = 0.1000 …, which represent the same number in base 2; for (a, b) = In,k for some n and some k, 1  k  2n−1, n1 1 n1 P aj P P aj 2 2 where a = þ and b = j j 3 3 3 j þ 3n (see Remark 1.8.8, second paraj¼1

j¼n þ 1

graph), we find that gðaÞ ¼

nP 1 j¼1

bj 2j

þ

j¼1 1 P

j¼n þ 1

1 2j

and gðbÞ ¼

nP 1 j¼1

bj 2j

þ

1 2n

and they rep-

resent the same number in base 2. Here each bj is even and bj ¼ 12 aj . We extend the definition of g to the whole of [0, 1] as follows. On any interval that has been removed, we define g by giving it the same value that g has at the endpoints of that interval. The function g has now been extended to all of [0, 1] and the range of g is also [0, 1]. We next show that g is monotone and continuous. Let x and y be two points of C, where x < y, and suppose their (unique) ternary expansions with even digits (see Theorem 1.8.9) are x ¼ :a1 a2    an an þ 1    ; y ¼ :b1 b2    bn bn þ 1    :

5.5 Integral of the Derivative

243

There exists a smallest integer i such that ai 6¼ bi; call it n. Then bn > an (see the last paragraph of Remark 1.8.1). In the binary expansions of g(x) and g(y), the nth digit of g(y) will be 1 larger than the nth digit of g(x), while the preceding digits (if any) will be the same. This implies g(y)  g(x) (again by Remark 1.8.1). Thus the restriction of g to C is an increasing function. Also, on each of the removed intervals and their endpoints, the function is constant. Hence g is an increasing function on [0, 1]. We have already noted (two paragraphs ago) that the range of g is an interval, namely, [0, 1]. Since the function is increasing, it follows immediately from Remark 5.2.3 that it is continuous. It was noted in Remark 1.8.13 that the unextended function g has the property that there exist arbitrarily small numbers at which it takes positive values. Since its extension is increasing, we know that it too has the property that x > 0 implies g(x) > 0. Moreover, the equality g′(x) = 0 can be shown to hold for almost all x 2 [0, 1]. In fact, g′(x) = 0 holds on [0, 1]\C and C is a set of measure zero [see Remark 2.2.11(c)]. Therefore Z ½0; 1

g0 ¼ 0\1 ¼ gð1Þ  gð0Þ:

The continuous monotone function g is called the Cantor function or Lebesgue’s singular function. It will be seen later in Theorem 5.8.2 that equality holds in Theorem 5.5.1 under suitable conditions. We give below an alternative description of the Cantor function (notations as in Definition 1.8.7). For any x 2 [0, 1], put gn ¼

 n 2 v Cn 3

Z

x

and fn ðxÞ ¼

gn ðtÞdt:

0

 1 The reader may check that f1(x) = 23 x if 0  x  13 ; 12 if 15 \x\ 25 and 21   1 x  23 if 23  x  1 It will be found instructive to compute f2 and f3. 2 þ 3 The functions fn, n = 1, 2, … are such that fn(0) = 0 and since each Jn,k is of length 31n , we have fn ð1Þ ¼

2n Z X k¼1

1 0

 n  n   2 2 1 vJn;k ðtÞdt ¼ 2n n ¼ 1: 3 3 3

Since gn  0, it follows that fn is increasing. Moreover, fn is continuous [gn is a step function and therefore its Lebesgue integral is the same as its Riemann integral]. If J is one of the component intervals of Cn, then

244

5 Differentiation

Z J

Z  n 2 gn ðtÞdt ¼ dt ¼ 2n 3 J

and, by the same computation with n + 1 in place of n, Z Z Z gn þ 1 ðtÞdt ¼ gn þ 1 ðtÞdt þ gn þ 1 ðtÞdt ¼ 2n1 þ 2n1 ¼ 2n ; J

J1

ð5:40Þ

ð5:41Þ

J2

where J1 and J2 are the parts of the closed interval J after removing its middle third. Suppose x 62 Cn. Then x is in none of the component intervals of Cn and therefore in none of the component intervals of any Cm with m > n. So, the defining integrals for fn and fn+1 are to be taken over only the intervals lying to the left of x. Hence, it follows from (5.40) and (5.41) that fn þ 1 ðxÞ ¼ fn ðxÞ

for x 62 Cn :

Since Cn+1  Cn, it further follows that fm ðxÞ ¼ fn ðxÞ for

m [ n and x 62 Cn :

ð5:42Þ

Now suppose x 2 Cn and K be the component interval of Cn that x belongs to. Then fn(x) − fn+1(x) is an integral over all the component intervals lying to the left of K and over the part of K that lies to the left of x. It follows from (5.40) and (5.41) that the integral over the component intervals lying to the left of K is 0. Consequently, Z jfn ðxÞ  fn þ 1 ðxÞj  jgn ðtÞ  gn þ 1 ðtÞjdt K Z Z  jgn ðtÞjdt þ jgn þ 1 ðtÞjdt: K

K

Hence it follows from (5.40) and (5.41) that jfn ðxÞ  fn þ 1 ðxÞj  2n þ 2n ¼ 2n þ 1 for x 2 Cn :

ð5:43Þ

From (5.42) and (5.43), we see that {fn} converges uniformly on [0,1]. The uniform limit must then be a continuous increasing function f with f(0) = 0, f(1) = 1 and f′(x) = 0 for all x 2 Cc. Indeed, the functions fn being constant on each of the intervals removed up to and including the nth stage, the limit function f must be constant on each of the removed intervals and hence have derivative 0 at each point of it. In other words, f′(x) = 0 on Cc. Therefore

5.5 Integral of the Derivative

Z

245 1

f 0 ðxÞdx ¼ 0\1 ¼ f ð1Þ  f ð0Þ:

0

It remains to show that the functions f and g are the same. Consider any removed interval In,k of some Cn and let x 2 In,k. Then x 62 Cn, and as before, the defining integral for fn is to be taken only over the component intervals lying to the left of x, which means component intervals lying to the left of the interval In,k. Hence, it follows from (5.40) that fn(x) equals 2−n times the number of component intervals to the left of In,k. In fact, on the entire interval In,k, the function fn takes the constant value equal to 2−n times the number of component intervals to the left of In,k. On the basis of (5.42), we infer that the limit function f also takes that same value on the entire interval. Hence it follows from the continuity of the function (noted above), that f takes this value at the right endpoint of the removed interval under consideration. It is now a consequence of Remark 1.8.14 (f) that f and g agree there. But f and g are both constant on any removed interval and therefore they agree everywhere on every removed interval of Cn. Thus they agree on the complement in [0, 1] of the Cantor set C. But the complement is dense because C cannot contain an interval of positive length, as it has measure 0 [see Example 2.2.11(c)]. Since both f and g are continuous, it follows by Proposition 1.3.25 that they agree everywhere on [0, 1]. Lebesgue’s Theorem 5.4.6 asserts that every monotone function is differentiable almost everywhere. Perhaps of no less importance is the following theorem of Fubini on the termwise differentiability of series with monotone terms, which Theorem 5.5.1 enables us to prove. Fubini’s Theorem. 5.5.3 Let fn:[a, b]!R be increasing for n = 1, 2, … and assume that the series f ðxÞ ¼

1 X

fk ðxÞ

ð5:44Þ

k¼1

converges pointwise on [a, b]. Then f 0 ðxÞ ¼

1 X

fk0 ðxÞ

k¼1

for almost all x 2 [a, b]. Proof Denote the nth partial sum of the series (5.44) by sn(x). We may then write f ðxÞ ¼ sn ðxÞ þ rn ðxÞ;

246

5 Differentiation

where 1 X

rn ðxÞ ¼

fk ðxÞ:

k¼n þ 1

Let E denote the set of points in [a, b] where each of the functions fk, k = 1, 2, … and the function f are differentiable. By Lebesgue’s Theorem 5.4.6, these functions are differentiable almost everywhere. So the set [a, b]\E has measure zero. If x 2 E, then sn is differentiable at x and therefore so is rn = f − sn. Also, rn′ = f′− sn′. Since rn is increasing, its derivative rn′ is nonnegative, and it follows that f 0 ðxÞ ¼ s0n ðxÞ þ rn0 ðxÞ  s0n ðxÞ ¼

n X

fk0 ðxÞ

k¼1

for all x 2 E. On letting n ! ∞, we obtain f 0 ðxÞ 

1 X

fk0 ðxÞ:

ð5:45Þ

k¼1

An application of Theorem 5.5.1 to the increasing function rn yields Z

b

0 a

rn0 ðxÞdx  rn ðbÞ  rn ðaÞ ¼

1 P

½fk ðbÞ  fk ðaÞ:

k¼n þ 1

By hypothesis, the series series

1 X

1 P

fk ðbÞ and

r¼1

1 P

fk ðaÞ converge and hence so does the

k¼1

½fk ðbÞ  fk ðaÞ Consequently,

k¼1 1 X

lim

n!1

½fk ðbÞ  fk ðaÞ ¼ 0:

k¼n þ 1

This implies

Z lim

n!1

Now,

Z

b a

0

Z

f ðxÞdx ¼ a



a

b

rn0 ðxÞdx ¼ 0:

b

n X

Z

k¼1 1 bX

a

fk0 ðxÞdx þ

k¼1

Z a

fk0 ðxÞdx þ

ð5:46Þ

b

rn0 ðxÞdx

Z a

b

rn0 ðxÞdx:

5.5 Integral of the Derivative

247

Letting n ! ∞ and using (5.46), we get Z

b

Z

0

f ðxÞdx ¼

a

a

1 bX

fk0 ðxÞdx:

ð5:47Þ

k¼1

From (5.45), we obtain Z

b

f 0 ðxÞdx 

a

Z

1 bX

a

fk0 ðxÞdx;

k¼1

which, together with (5.47), implies Z

b

f 0 ðxÞdx ¼

a

Z a

1 bX

fk0 ðxÞdx;

k¼1

i.e., Z a

b

½f 0 ðxÞ 

1 X

fk0 ðxÞdx ¼ 0:

k¼1

As the integrand here is nonnegative by (5.45), it follows that f 0 ðxÞ ¼

1 P k¼1

fk0 ðxÞ

a.e. on [a, b]. h As an application of the theorem we give an alternative proof of Theorem 5.4.7, according to which the saltus function S of an increasing function f on [a, b] has derivative zero a.e. on [a, b]. Let us enumerate all the points of discontinuity of f in a sequence {xk}k  1. (As before, we may assume there are infinitely many.) To each k  1 such that a < xk < b, we assign a function fk defined on [a, b] as follows: fk ðxÞ ¼

8
xk. 1 P It is plain from the definition that fk(a) = 0 for all k. Therefore fk ðaÞ = 0 = k¼1

S(a) by definition of the latter. Now let x 2 (a, b) and suppose also that x 6¼ xk for all k. Then f(x) − f(x−) = 0. When p 6¼ k 6¼ q, by (5.48) and (5.50), we have fk ðxÞ ¼ 0 for x\xk and fk ðxÞ ¼ jðxk Þ for xk \x: When k = p, by (5.51), we have xk ¼ xp ¼ a\x and fk ðxÞ ¼ fp ðxÞ ¼ f ða þ Þ  f ðaÞ ¼ jðaÞ ¼ jðxp Þ: When k = q, by (5.52), we have xk ¼ xq ¼ b [ x and fk ðxÞ ¼ fq ðxÞ ¼ 0: It follows that 1 X k¼1

fk ðxÞ ¼ fp ðxÞ þ fq ðxÞ þ

X

fk ðxÞ ¼ jðxp Þ þ 0 þ

p6¼k6¼q

X X ¼ jðxp Þ þ jðxk Þ ¼ jðxk Þ xk \x;k6¼p xk \x X ¼ jðxk Þ þ f ðxÞ  f ðxÞ x \x k

¼ SðxÞ

by definition of S:

X p6¼k6¼q

fk ðxÞ

5.5 Integral of the Derivative

249

Continuing with x 2 (a, b), we drop the supposition that x 6¼ xk for all k. Let x = xr. We know that p 6¼ r 6¼ q because a 6¼ x 6¼ b. When p 6¼ k 6¼ q, by (5.48) and (5.50), we have fk ðxÞ ¼ 0 for x\xk and fk ðxÞ ¼ jðxk Þ for xk \x; provided k 6¼ r and, by (5.49), fr ðxÞ ¼ fr ðxr Þ ¼ f ðxr Þ  f ðxr Þ ¼ f ðxÞ  f ðxÞ: When k = p, we have (exactly as before) xk ¼ xp ¼ a\x and fk ðxÞ ¼ fp ðxÞ ¼ f ða þ Þ  f ðaÞ ¼ jðaÞ ¼ jðxp Þ: When k = q, we have (exactly as before) xk ¼ xq ¼ b [ x and fk ðxÞ ¼ fq ðxÞ ¼ 0: It follows that 1 X

X

fk ðxÞ ¼ fp ðxÞ þ fq ðxÞ þ fr ðxÞ þ

fk ðxÞ

p6¼k6¼q;k6¼r

k¼1

X

¼ jðxp Þ þ 0 þ f ðxÞ  f ðxÞ þ ¼

jðxk Þ

xk \x;k6¼p

X xk \x

jðxk Þ þ f ðxÞ  f ðxÞ

¼ SðxÞ by definition of S: So far, we have proved that S(x) =

1 P

fk ðxÞ if x 2 [a, b).

k¼1

Consider x = b. There are two cases to consider, b > xk for all k and b = xq. In the first case, f(b) − f(b−) = 0 and, by ( and (5.51), fk(5.50b) = j(xk) for all k (including p). It follows that 1 X

fk ðbÞ ¼

X xk \b

k¼1

¼

X

fk ðbÞ ¼

X

jðxk Þ

xk \b

jðxk Þ þ f ðbÞ  f ðbÞ

xk \b

¼ SðbÞ by definition of S: In the second case, b > xk for all k 6¼ q, and hence by (5.50), (5.51) and (5.52), fq(b) = f(b) − f(b −) and fk(b) = j(xk) for all other k 6¼ q (including p). It follows that

250

5 Differentiation 1 X

fk ðbÞ ¼ fq ðbÞ þ

X

fk ðbÞ ¼ f ðbÞ  f ðbÞ þ

k6¼q

k¼1

¼ f ðbÞ  f ðbÞ þ

X

X

fk ðbÞ

k6¼q

jðxk Þ

k6¼q

¼ f ðbÞ  f ðbÞ þ

X

jðxk Þ

xk \b

¼ SðbÞ by definition of S: This completes the proof of (5.54). By Fubini’s Theorem 5.5.3, the equality (5.54) leads to S0 ðxÞ ¼

1 X

fk0 ðxÞ almost everywhere:

k¼1

In view of (5.53), this completes the alternative proof of Theorem 5.4.7 that the saltus function S of an increasing function f on [a, b] has derivative zero a.e. on [a, b]. 1 P A slight variation in the above proof would be to avoid proving S(b) = fk ðbÞ k¼1

and to use Problem 5.5.P3 instead, after applying Fubini’s Theorem 5.5.3 on intervals [a, b] with a < b < b. The following examples illustrate the use of Fubini’s Theorem for producing nonconstant monotone functions, one with a dense set of discontinuities and the other without any discontinuities, and yet each having a zero derivative a.e. Examples 5.5.4 (a) Let H be the function given by HðxÞ ¼

0 1

x0 x [ 0:

That is, H is the characteristic function of the set (0, ∞). Clearly, H is increasing and H′(x) = 0 for all x 6¼ 0. Let {rk}k  1 be an enumeration of the rational numbers and set f ðxÞ ¼

1 X 1 Hðx  rk Þ: k 2 k¼1

ð5:55Þ

P1 Since 21k H ðx  rk Þ  21k and is convergent, the series on the right of (5.55) is 2k uniformly convergent. Observe that the function f is the limit of the sequence of partial sums of (5.55) and the nth partial sum has a nonzero jump at rk with jump 21k , k = 1, 2, …, n. The function f, being the sum of increasing functions, is itself an increasing function. Since

5.5 Integral of the Derivative

251

(a) the convergence of the partial sums is uniform and (b) each partial sum from the kth one onward have the same jump at rk, namely 21k ; it follows from Proposition 5.2.11 that f has a positive jump at each rk, k = 1, 2, …. Applying Fubini’s Theorem 5.5.3 to an arbitrary interval [a, b] and then letting a ! −∞ and b ! ∞, we obtain f 0 ðxÞ ¼

1 X 1 0 H ðx  rk Þ ¼ 0 2k k¼1

for almost all x 2 R. This function is strictly increasing, thus not constant on any interval. We next give an example of a function defined on R which is continuous and strictly increasing, and yet has a zero derivative a.e. (b) Let g be the Cantor function defined on [0, 1] (see Example 5.5.2). The function g is continuous and increasing on [0, 1] with g(0) = 0 and g(1) = 1 and is constant on each of the removed intervals which form the complement of the Cantor set. Also, g′(x) = 0 for all x 2 C. Let 8 x\0 < 0 GðxÞ ¼ gðxÞ 0  x  1 : 1 1\x: Then G has domain R, is continuous and increasing and has G′(x) = 0 almost everywhere. Let {rk}k  1 be an enumeration of the rational numbers in [0, 1] and set FðxÞ ¼

1 X 1 Gðx  rk Þ: 2k k¼1

Observe that F is continuous and increasing (being a limit of increasing functions), and using Fubini’s Theorem 5.5.3, it follows that F′(x) = 0 a.e. We shall argue that the function F is strictly increasing on [0, 1]. Suppose x, y 2 [0, 1] and x < y. Then x − rk < y − rk, so that Gðx  rk Þ  Gðy  rk Þ

for every k:

As the sequence {rk}k  1 goes through all the rational numbers in [0, 1], we must have x < rj < y for some j. Then x − rj < 0 < y − rj < 1. It follows from the definition of G that G(x − rj) = 0 and G(y − rj) = g(y − rj)  0. But g(y − rj) > 0 on account of the property of g noted in Example 5.5.2 that x > 0 implies g(x) > 0, and hence G(x − rj) < G(y − rj). On the other hand, G(x − rk)  G(y − rk) for every k. This implies F(x) < F(y).

252

5 Differentiation

Problem Set 5.5 5.5.P1. Using the functions constructed in Examples 5.5.4 (a) and (b), show that strict inequality can hold in Theorem 5.5.1. 5.5.P2. Let f be Lebesgue’s singular function. Compute all the four Dini derivatives of f at each point x 2 [0, 1]. [Cf. Example 5.5.2.] 5.5.P3. Suppose f:[a, b]!R has derivative 0 a.e. on [a, b] whenever b < b. Show that f has derivative 0 a.e. on [a, b].

5.6

Total Variation

The sum of two (or any finite number) of increasing functions is obviously an increasing function. However, their difference is, in general, not so. Indeed, the functions f1(x) = x and f2(x) = x2 defined on [0, 1] are increasing but the function f (x) = f1(x) − f2(x) = x(1 − x), 0  x  1, is such that f(0) = 0 = f(1) and assumes its maximum value 14 at x = 12 . The difference f of two increasing functions f1 and f2 defined on [a, b] satisfies the following property: For every partition a = x0 < x1 < ⋯ < xn = b of [a, b], the sum n X

jf ðxk Þ  f ðxk1 Þj 

k¼1

¼

n X

jf1 ðxk Þ  f1 ðxk1 Þj þ

k¼1

n X

½f1 ðxk Þ  f1 ðxk1 Þ þ

n X

jf2 ðxk Þ  f2 ðxk1 Þj

k¼1 n X

k¼1

½f2 ðxk Þ  f2 ðxk1 Þ

k¼1

since fi ðxk Þ  fi ðxk1 Þ  0 for i ¼ 1; 2 and k ¼ 1; 2; . . .; n ¼ f1 ðbÞ  f1 ðaÞ þ f2 ðbÞ  f2 ðaÞ ¼

2 X

ðfi ðbÞ  fi ðaÞÞ:

i¼1

A function for which the sum n X

jf ðxk Þ  f ðxk1 Þj

k¼1

does not exceed some finite bound, independent of the partition, is called a function of bounded variation, or finite variation. The class of such functions plays a fundamental role in several branches of mathematical analysis, including the theory of Fourier series and the theory of integration. We shall study below the regularity (continuity and differentiability properties) of these functions.

5.6 Total Variation

253

Definition 5.6.1 Let P: a = x0 < x1 < ⋯ < xn = b be a partition of [a, b] and f be a function defined on [a, b]. We introduce the notation TðP; f Þ ¼

n X

jf ðxk Þ  f ðxk1 Þj:

k¼1

The total variation of f on [a, b] is defined to be Vð½a; b; f Þ ¼ sup TðP; f Þ; P

the supremum being taken over all partitions P of [a, b]. If V([a, b],f) < ∞, we say that f is of bounded (or finite) variation on [a, b]. Remarks 5.6.2 (a) A monotone function on a closed and bounded interval is always of bounded variation. For details, see the introductory paragraph of this section. In fact, T(P, f) = jf ðbÞ  f ðaÞj for every monotone function f and every partition P of [a, b]. (b) Every function of bounded variation on [a, b] is bounded on the interval. In fact, for a  x  b, jf ðxÞ  f ðaÞj þ jf ðbÞ  f ðxÞj  Vð½a; b; f Þ: Therefore jf ðxÞj  jf ðaÞj þ jf ðxÞ  f ðaÞj  Vð½a; b; f Þ þ jf ðaÞj: (c) The sum, difference and product of functions of bounded variation are again of the same kind. To see why, let f and g be such functions on [a, b] and s = f + g. Then for any partition P: a = x0 < x1 < ⋯ < xn = b, we have n X

jsðxk Þ  sðxk1 Þj 

k¼1

n X k¼1

jf ðxk Þ  f ðxk1 Þj þ

n X

jgðxk Þ  gðxk1 Þj:

k¼1

It follows that Vð½a; b; sÞ  Vð½a; b; f Þ þ Vð½a; b; gÞ: The proof that f − g is of bounded variation is similar and is, therefore, not included. Now suppose p = fg. Let

254

5 Differentiation

M ¼ supfjf ðxÞj : a  x  bg and N ¼ supfjgðxÞj : a  x  bg: Then for any partition P: a = x0 < x1 < ⋯ < xn = b, we have jpðxk Þ  pðxk1 Þj  jf ðxk Þgðxk Þ  f ðxk1 Þgðxk Þj þ jf ðxk1 Þgðxk Þ  f ðxk1 Þgðxk1 Þj  N jf ðxk Þ  f ðxk1 Þj þ M jgðxk Þ  gðxk1 Þj: This implies Vð½a; b; pÞ  N  Vð½a; b; f Þ þ M  Vð½a; b; gÞ: Thus p = fg is of bounded variation whenever f and g are. (d) If f and g are of bounded variation and there exists an a > 0 such that g(x) > a for a  x  b, then gf is also of bounded variation. It is enough to show that 1g is of bounded variation. The result will then follow from (c) above. For any partition P: a = x0 < x1 < ⋯ < xn = b,

n

n X X

1  1  1 jgðxk Þ  gðxk1 Þj

gðx Þ gðx Þ a2 k k1 k¼1 k¼1 and this yields 1 1 Vð½a; b; Þ  2 Vð½a; b; gÞ: g a (e) Recall that if f:[a, b]!R is such that |f(x) − f(y)|  M|x − y| for every x and y in [a, b], where M is a fixed positive number, then f is said to satisfy a Lipschitz condition. Such a function is always of bounded variation, because for any partition P: a = x0 < x1 <  < xn = b, n X

jf ðxk Þ  f ðxk1 Þj  M

k¼1

n X

ðxk  xk1 Þ ¼ Mðb  aÞ

k¼1

and so, V([a, b],f)  M(b − a). (f) If f:[a, b]!R is continuous on [a, b] and has a derivative f′(x) at every point of (a, b) and if f′ is bounded, then from the Mean Value Theorem, we have

5.6 Total Variation

255

jf ðxÞ  f ðyÞj  jf 0 ðzÞjjx  yj; where z lies between x and y. The function therefore satisfies a Lipschitz condition. (g) If f1 and f2 are two continuous increasing functions, then f = f1 − f2 is a continuous function of bounded variation. This is an immediate consequence of parts (a) and (c). We list below some examples of functions that are of bounded variation and some that are not. Examples 5.6.3 (a) The function f(x) = sin x is of bounded variation on any compact interval [a, b]. This follows from Remark 5.6.2(f) above. In the paragraph just above Example 5.6.12, the notion of bounded variation will be extended to R. We shall see that f is not of bounded variation on R [see Example 5.6.12 (b)]. (b) The function f(x) = x − x2, 0  x  1, being the difference of two increasing functions, is of bounded variation on [0, 1] [see Remark 5.6.2(c).] p (c) The function f:[0, 1]!R defined by f(0) = 0 and f(x) = x sin 2x ; x 6¼ 0, is not of bounded variation. However, it is continuous. Indeed, for the partition 1 1 1 1 \ \    \ \ \1; 2n þ 1 2n 3 2 2n þ1 X 1 1 1 TðP; f Þ ¼ þ þ    þ þ 1: jf ðxk Þ  f ðxk1 Þj ¼ 2n þ 1 2n  1 3 k¼1

P : 0\

These sums become infinite as n ! ∞. That the function is continuous on its domain is well known. (d) The function f(x) = xpsin 1x, 0 < x  1, and f(0) = 0 can be shown to be of bounded variation on [0, 1] for p  2, as follows: The derivative f′(x) = pxp−1sin 1x − xp−2cos 1x is such that |f′(x)|  p + 1, 0 < x < 1, and f is continuous on [0, 1]. The assertion therefore follows by Remark 5.6.2(e) and (f). Theorem 5.6.4 Let f be any real-valued function on [a, b] (not necessarily of bounded variation) and let a < c < b. Then Vð½a; b; f Þ ¼ Vð½a; c; f Þ þ Vð½c; b; f Þ: Proof If either V([a, c],f) or V([c, b],f) is infinite, then the equality is trivial. So, we may assume that both are finite. If c is a point of the partition P: a = x0 < x1 < ⋯ < xn = b of [a, b], then P determines a partition P1 of [a, c] and a partition P2 of [c, b]. It is clear that

256

5 Differentiation

TðP; f Þ ¼ TðP1 ; f Þ þ TðP2 ; f Þ and therefore TðP; f Þ  Vð½a; c; f Þ þ Vð½c; b; f Þ:

ð5:56Þ

If c is not a point of the partition P, then the partition P′ obtained by adding the point c is a refinement of P. Moreover, (5.56) holds with P′ in place of P. Now, xi−1 < c < xi for some i. So, jf ðxi Þ  f ðxi1 Þj  jf ðxi Þ  f ðcÞj þ jf ðcÞ  f ðxi1 Þj; which implies TðP; f Þ  TðP0 ; f Þ: Since (5.56) holds with P′ in place of P, the preceding inequality shows that (5.56) holds for P in this case as well (i.e. when c is not a point of P). Taking the supremum over all P, we obtain Vð½a; b; f Þ  Vð½a; c; f Þ þ Vð½c; b; f Þ: It remains to prove the reverse inequality. Let e > 0 be arbitrary. There exist partitions P1 and P2 of [a, c] and [c, b] respectively such that TðP1 ; f Þ [ Vð½a; c; f Þ 

e 2

and e TðP2 ; f Þ [ Vð½c; b; f Þ  : 2 Let P = P1 [ P2. Then P is a partition of [a, b] and TðP; f Þ ¼ TðP1 ; f Þ þ TðP2 ; f Þ [ Vð½a; c; f Þ þ Vð½c; b; f Þ  e; which implies Vð½a; b; f Þ  Vð½a; c; f Þ þ Vð½c; b; f Þ  e: Since e > 0 is arbitrary, the required reverse inequality follows.

h

Example Suppose that f:[a, b]!R is an arbitrary step function. Then there exists a partition P: a = x0 < x1 < ⋯ < xn = b of [a, b] and numbers a1, a2,…,an such that f(x) = ai, xk−1 < x < xk, k = 1, 2,…, n. We shall show that f is of bounded variation.

5.6 Total Variation

257

By Theorem 5.6.4, V([a, b],f) is the sum of the variations over the n intervals [xk−1, xk]. If a partition Pk of [xk−1, xk] consists of only the two endpoints, then T(Pk,f) = |f(xk−1) − f(xk)|, but if it consists of more than just the two endpoints, then T(Pk, f) = |ak− f(xk−1)| + |ak− f(xk)|, because the function has the constant value ak on the interval (xk−1, xk). In view of the inequality |ak− f(xk−1)| + |ak− f(xk)|  |f(xk−1) − f(xk)|, it follows that V([xk−1, xk], f) = |ak− f(xk−1)| + |ak− f(xk)|. Therefore the total variation of f on [a, b] is Vð½a; b; f Þ ¼

n X

ðjak  f ðxk1 Þj þ jak  f ðxk ÞjÞ\1:

k¼1

Corollary 5.6.5 If it is possible to subdivide [a, b] into a finite number of parts, on each of which f is monotone, then f has finite variation on the interval. Proof The assertion is immediate from Theorem 5.6.4. h The following Theorem, due to C. Jordan, describes completely the class of functions of bounded variation. Theorem 5.6.6 A real-valued function on a compact interval is of bounded variation if and only if it can be expressed as the difference of two increasing functions. Proof The sufficiency of the condition follows from Remarks 5.6.2(a) and (c). To prove the necessity, we consider a function f:[a, b] ! R of bounded variation and set vðxÞ ¼ Vð½a; x; f Þ;

a  x  b;

where V([a, x], f) denotes the total variation of the function calculated on the interval [a, x]. For n > x, we have v(n) − v(x) = V([x, n], f) by Theorem 5.6.4 and Vð½x; n; f Þ  jf ðnÞ  f ðxÞj;

ð5:57Þ

since |f(n) − f(x)| is the sum T(P, f) for the particular partition P consisting only of x and n. So, vðnÞ  vðxÞ

ð5:58Þ

and vðnÞ  vðxÞ  f ðnÞ  f ðxÞ; which implies vðnÞ  f ðnÞ  vðxÞ  f ðxÞ:

ð5:59Þ

From (5.58) and (5.59), we conclude that m and m − f are increasing functions and

258

5 Differentiation

f ðxÞ ¼ vðxÞ  ðvðxÞ  f ðxÞÞ is a decomposition of f as the difference of increasing functions defined on [a, b]. From (5.57), we also have vðnÞ  vðxÞ   ðf ðnÞ  f ðxÞÞ and this implies vðnÞ þ f ðnÞ  vðxÞ þ f ðxÞ:

ð5:60Þ

It follows from (5.60) that v + f is also an increasing function on [a, b]. We thus obtain a symmetric decomposition of f as the difference of increasing functions: 1 1 f ðxÞ ¼ ðvðxÞ þ f ðxÞÞ  ðvðxÞ  f ðxÞÞ: 2 2

h

Remark 5.6.7 As a consequence of Jordan’s Theorem 5.6.6, a function f:[a, b]!R of bounded variation inherits many properties of monotone functions: (a) The limit from the right at x, denoted by f(x +), exists if a  x < b; the limit from the left at x, denoted by f(x−), exists if a < x  b. (b) The set of points of discontinuity of f is at most countable. (c) f′(x) exists a.e. on [a, b] and f′ 2 L1[a, b] by Theorem 5.5.1. (d) The representation of a function of bounded variation as a difference of two increasing functions is not unique, because one can add the same increasing function to both of the latter. We finally prove that v(x) = V([a, x], f) is increasing and continuous at each point of continuity of f. Theorem 5.6.8 Let f be a function of bounded variation on [a, b]. The function v defined by vðxÞ ¼ Vð½a; x; f Þ for x 2 ½a; b is increasing and is continuous at each point of continuity of f. Proof That the function v is increasing has been proved in Theorem 5.6.6 [statement (5.58) of its proof]. Suppose f is continuous at x, where a < x  b. We shall show that v(x−) = v(x). Let e > 0 be arbitrary. On using the definition of v(x), we can find a partition P of [a, x] such that e TðP; f Þ [ vðxÞ  : 2

ð5:61Þ

5.6 Total Variation

259

In view of the left continuity of f at x, we may choose a point y such that a < y < x and jf ðyÞ  f ðxÞj\

e 2

ð5:62Þ

and there is no point of subdivision between y and x. Let P′ be the refinement of P obtained by adjoining y to the points of P. Then TðP0 ; f Þ  TðP; f Þ

ð5:63Þ

(see the proof of Theorem 5.6.4). Let P″ be the partition of [a, y] formed by using all the points of P′. Then e e vð yÞ  TðP00 ; f Þ ¼ TðP0 ; f Þ  jf ðyÞ  f ðxÞj [ vðxÞ   ¼ vðxÞ  e; 2 2 using (5.61), (5.63) and (5.62). It may be noted that we have only used the left continuity of f at x. The use of right continuity will imply right continuity of v when a  x < b. h Remark We note in passing that when f is of bounded variation, we can express it as f = f1 − f2, where f1 and f2 are increasing nonnegative functions. This is because in the decomposition of f as a difference f = g1 − g2 of increasing functions in accordance with Theorem 5.6.6, we have 1 g1 ¼ ðv þ f Þ; 2

1 g2 ¼ ðv  f Þ: 2

By Remark 5.6.2 (b), both are bounded. Taking a common lower bound c for them, we find that f1 = g1 − c and f2 = g2 − c are increasing nonnegative functions such that f = f1 − f2. Moreover, by Theorem 5.6.8, both f1 and f2 are left and/or right continuous wherever f is. Theorem 5.6.6 of Jordan characterises a function of bounded variation as being decomposable into a difference of two increasing functions. In what follows, we shall obtain such a decomposition using positive and negative variations introduced below. Let f be a real-valued function of bounded variation defined on [a, b]. If P: a = x0 < x1 < ⋯ < xn = b is a partition of [a, b], let PðP; f Þ ¼

n X

½f ðxk Þ  f ðxk1 Þ þ

k¼1

and N ðP; f Þ ¼ 

n X k¼1

½f ðxk Þ  f ðxk1 Þ ;

260

5 Differentiation

where r+ means max{r, 0} and r_ means min{r, 0}; i.e. PðP; f Þ is the sum of all those differences f(xk) − f(xk−1) that are nonnegative, while N ðP; f Þ is the negative of the sum of those that are nonpositive. Note however that the definition makes N ðP; f Þ  0 . We also define Pð½a; b; f Þ ¼ sup PðP; f Þ P

N ð½a; b; f Þ ¼ sup N ðP; f Þ; P

where the supremum has been taken over all partitions P of [a, b]. Pð½a; b; f Þ and N ð½a; b; f Þ are called respectively the positive and negative variations of f over [a, b]. Proposition 5.6.9 If f is of bounded variation on [a, b], then its positive and negative variations are finite. Moreover, 2Pð½a; b; f Þ ¼ Vð½a; b; f Þ þ f ðbÞ  f ðaÞ;

ð5:64Þ

2N ð½a; b; f Þ ¼ Vð½a; b; f Þ  f ðbÞ þ f ðaÞ;

ð5:65Þ

Pð½a; b; f Þ þ N ð½a; b; f Þ ¼ Vð½a; b; f Þ

ð5:66Þ

Pð½a; b; f Þ  N ð½a; b; f Þ ¼ f ðbÞ  f ðaÞ:

ð5:67Þ

and

Proof The finiteness of the positive and negative variations follows from the fact that 0  PðP; f Þ  TðP; f Þ

and 0  N ðP; f Þ  TðP; f Þ:

The following relations are evident: PðP; f Þ  N ðP; f Þ ¼ f ðbÞ  f ðaÞ TðP; f Þ ¼ PðP; f Þ þ N ðP; f Þ: From these, we get 2PðP; f Þ ¼ TðP; f Þ þ f ðbÞ  f ðaÞ 2N ðP; f Þ ¼ TðP; f Þ  f ðbÞ þ f ðaÞ; from which we further get (5.64) and (5.65). These then lead to (5.66) and (5.67). h

5.6 Total Variation

261

Remark If f is of bounded variation on [a, b], it follows that f is of bounded variation on [a, x], a < x  b. The symbols Pð½a; x; f Þ, N ð½a; x; f Þ and V([a, x], f) are all meaningful. Moreover, relations (5.64)–(5.67) of the above proposition hold if we replace b by x in them. We define Pð½a; a ; f Þ ¼ N ð½a; a; f Þ ¼ Vð½a; a; f Þ ¼ 0: We know by Theorem 5.6.8 that v(x) = V([a, x], f) is an increasing function of x and is continuous wherever f is continuous. Proposition 5.6.10 The functions of x given by Pð½a; x; f Þ and N ð½a; x; f Þ are increasing. Proof Set g(x) = V([a, x], f) − f(x) and h(x) = V([a, x], f) + f(x). It has been shown in Theorem 5.6.6 that both are increasing. From Proposition 5.6.9 (5.64), (5.65), with b replaced by x, we have 1 Pð½a; x; f Þ ¼ ðhðxÞ  f ðaÞÞ 2 and 1 N ð½a; x; f Þ ¼ ðgðxÞ þ f ðaÞÞ: 2 So, Pð½a; x; f Þ and N ð½a; x; f Þ are increasing functions of x on [a, b].

h

Theorem 5.6.11 If f is a function of bounded variation on [a, b], there exist nonnegative increasing functions f1 and f2 on [a, b] such that f = f1 − f2. The functions Pð½a; x; f Þ, N ð½a; x; f Þ and V([a, x], f) are continuous at any x 2 [a, b] where f is continuous. The functions f1 and f2 can be so chosen that if f is continuous at x, then so are f1 and f2. Proof Although the existence of the functions f1 and f2 is a consequence of Theorem 5.6.8, we give a fresh proof here, again using Theorem 5.6.8. Set 1 f1 ðxÞ ¼ Pð½a; x; f Þ þ ðf ðaÞ þ jf ðaÞjÞ 2 and f2 ðxÞ ¼ N ð½a; x; f Þ þ

1 ðf ðaÞ þ jf ðaÞjÞ: 2

In view of Proposition 5.6.10, f1 and f2 are increasing functions defined on [a, b]. Moreover, they are nonnegative and

262

5 Differentiation

f1 ðxÞ  f2 ðxÞ ¼ Pð½a; x; f Þ  N ð½a; x; f Þ þ f ðaÞ ¼ f ðxÞ; where we have used (5.67) of Proposition 5.6.9 with x in place of b. Assume that f is continuous at x0. It follows from Theorem 5.6.8 that V([a, x], f) is continuous at x0. Consequently, the functions g and h of Proposition 5.6.10 are continuous at x0 and so are Pð½a; x; f Þ and N ð½a; x; f Þ . It follows that f1 and f2 are continuous at x0. h Remark Suppose g:[a, b]!R is summable in the Lebesgue sense. Define f:[a, b]!R by the formula Z f ðxÞ ¼

g; ½a; x

where it is understood that f(a) = 0. In view of Problem 3.2.P16, f is continuous on R [a, b]. Moreover, it is of bounded variation on [a, b] with V([a, b], f)  ½a; b jgj. (That equality holds here is left as Problem 5.6.P9). (i) Let x0 2 [a, b). Then for x > x0, we have Z Z g¼ f ðxÞ  f ðx0 Þ ¼ ½x0 ;x

½a; x

v½x0 ;x g:





Observe that v½x0 ;xgj  jgj and the latter is Lebesgue integrable. Moreover, v½x0 ;xg ! 0 as x ! x0. By the Dominated Convergence Theorem 3.2.16, we obtain Z jf ðxÞ  f ðx0 Þj 

½x0 ;x

jgj ! 0 as x ! x0 :

R Similarly, if x0 2 (a, b] and x < x0, then ½x;x0  jgj ! 0 as x ! x0. This shows that f is continuous at x0. Since x0 is arbitrary, it follows that f is continuous on [a, b]. (ii) Consider any partition P: a = x0 < x1 < ⋯ < xn = b of [a, b]. Then



Z

X n Z n Z X



g  jf ðxk Þ  f ðxk1 Þj ¼ j gj  jgj:

½xk1 ;xk  k¼1 ½xk1 ;xk  ½a; b k¼1 k¼1

n X

Since by hypothesis, g is integrableRon [a, b], it follows that f is of bounded variation on [a, b] with V([a, b], f)  ½a; b jgj. Obviously a function of bounded variation on R is of bounded variation on every compact subinterval of R, where variation on R means sup{V([a, b], f): a < b}.

5.6 Total Variation

263

Examples 5.6.12 (a) The function f(x) = x is of bounded variation on any compact interval. However it is not of bounded variation on R. In fact, V([a, b], f) = b − a, and hence sup {V([a, b], f): a < b} = ∞. (b) We have observed that the function f(x) = sin x, a  x  b, is of bounded variation there [Example 5.6.3(a)]. However, it is not of bounded variation on R. Indeed, for the partition p 3p ð2n  1Þp ð2n þ 1Þp P : \ \\ \ ; 2 2 2 2 the sum   

n

 X

f ð2k þ 1Þp  f ð2k  1Þp ¼ 2n



2 2 k¼1

exceeds every positive positive number as n ! ∞. R (c) Let g 2 L1(R). Then the function f(x) = ½1;x g is of bounded variation on R. Indeed, Z Vð½a; x; f Þ 

consequently, V(R, f) 

R ½1;1

Z ½a; x

jgj 

½1;1

jgj;

jgj .

Problem Set 5.6 5.6.P1. Show that Vð½a; b; f þ gÞ  Vð½a; b; f Þ þ Vð½a; b; gÞ and Vð½a; b; cf Þ ¼ jcjVð½a; b; f Þ: 5.6.P2. Let /:(a, b)!R have a (finite) left limit everywhere and y 2 [a, b) [resp. y 2 (a, b]] be a point where the right limit /(y+) [resp. left limit /(y−)] exists (and is finite). Then first, lim /ðxÞ ¼ /ðy þ Þ

and second,

x!y þ

lim /ðxÞ ¼ /ðyÞ:

x!y

264

5 Differentiation

In particular, limx!a þ /ðxÞ ¼ /ða þ Þ

and

limx!b /ðxÞ ¼ /ðbÞ.

5.6.P3. The function f defined by f(x) = sin px , 0 < x  1 and f(0) = 0, is not of bounded variation on [0, 1]. 5.6.P4. Define f:[0, 1]!R by f(0) = 0 and f(x) = x2 sin 1x, x 6¼ 0. Show that f is of bounded variation on [0, 1]. 5.6.P5. Show that the function defined on [0, 1] by 2 x sin x12 for x 6¼ 0 f ðxÞ ¼ 0 for x ¼ 0

is not of bounded variation. 5.6.P6. Suppose that f:[a, b] ! [c, d] is monotone and that g is of bounded variation on [c, d]. Prove that V([a, b],g f)  V([c, d], g). 5.6.P7. Let {fn}n  1 be a sequence of real-valued functions defined on [a, b] that converge pointwise to the function f. Prove that Vð½a; b; f Þ  lim inf Vð½a; b; fn Þ: n!1

of bounded variation on [a, b] such 5.6.P8. P P Let {fn}n  1 be a sequence of functions that n fn ðaÞ converges absolutely and n V ð½a; b; fn Þ\1 . Prove that P (i) nfn(x) converges absolutely for each x 2 [a, b], (ii) V([a, b],Rn fn Þ  Rn V ð½a; b; fn Þ . R 5.6.P9. Suppose g 2 L1[a, b] and set f(x) = ½a;x g, a  x  b. Show that V([a, b], R f) = ½a; b jgj. 5.6.P10. Deduce from Problem 5.6.P9 that sinx is of bounded variation on [0, 2p] and that its total variation is 4. 5.6.P11. Define f on [0, 1] by f ð0Þ ¼ 0

and

1 f ðxÞ ¼ x sin ; x 6¼ 0: x

Show that f is not an indefinite integral of a Lebesgue integrable function on [0, 1]. 5.6.P12. Define

f ðxÞ ¼

1 0

x2Q x 62 Q:

Show that f is not of bounded variation on any compact interval.

5.6 Total Variation

265

5.6.P13. Construct functions fn, n = 0, 1, 2, … on an interval [a, b] such that (i) fn! f uniformly, (ii) each fn is of bounded variation, (iii) f is not of bounded variation. 5.6.P14. Let f be defined on [0, 1] by the following formulae (see Fig. 5.1): 8 x > > < 0 f ðxÞ ¼ > linear > : 0

1 x ¼ 2n ;n 2 N x ¼ 2n 1þ 1 ; n 2 N 1 1 nþ1  x  n ; n 2 N x ¼ 0:

Show that f is not of bounded variation. R 5.6.P15. If f(x) ¼ ½a; x g, where g 2 L1[a, b], prove that the positive and negative R variations of f, namely, Pð½a; b; f Þ and N ð½a; b; f Þ, are given by ½a; b g þ and R  ½a; b g respectively. 5.6.P16. Let x1, x2, … be the points of discontinuity in (a, b) of the function f of bounded variation with domain [a, b]. We define j:[a, b]!R by the following formulae: jðaÞ ¼ f ðaÞ  f ða þ Þ and jðxÞ ¼ f ðxÞ  f ðxÞ þ

X

½f ðxi þ Þ  f ðxi Þ if a\x  b:

xi \x

ð5:68Þ By the “jump” of f at x, we mean |f(x) − f(x−)| + |f(x+)−f(x)| if a < x < b and |f(a+) − f(a)| if x = a and |f(b) − f(b−)| if x = b. [Note that this agrees with Definition 5.2.4 for a monotone function.]

Fig. 5.1 Function in Problem 5.6.P14

1/2

1/4

1/5 1/4 1/3

1/2

1

266

5 Differentiation

Show without using Theorem 5.3.1 that (a) The function on [a, b]. P j defined in (5.68) is of bounded variation P (b) jðyÞ ¼ ½f ðxi þ Þ  f ðxi Þ for y 2 (a, b], jðy þ Þ ¼ ½f ðxi þ Þ  f ðxi Þ xi \y

xi  y

for y 2 (a, b) and j(a+) = 0. (c) If F = f − j, then F is continuous and of bounded variation. (d) j′ = 0 a.e. (e) The representation of f as F + j, where F is continuous and j′ = 0 a.e. is not unique (not even up to a constant) (Cf. Problem 5.7.P4).

5.7

Absolute Continuity

In this section, we shall study a special class of functions of bounded variation, known as absolutely continuous functions. These functions are representable as the integral of a summable function with a variable limit of integration (see Sect. 5.8). These functions are important for a number of applications and are, in addition, interesting in their own right. We begin with a definition. Definition 5.7.1 A real-valued function f defined on an interval [a, b] or R or (−∞, b] or [a, ∞) is said to be absolutely continuous on its domain if, given any e > 0, there exists a d > 0 such that, for any finite system of disjoint intervals {(ak, bk)}1  k  n contained in the domain, the sum of whose lengths is less than d, n P ðbk  ak Þ\d, there holds the inequality i.e. k¼1 n X

jf ðbk Þ  f ðak Þj\e:

k¼1

Remarks 5.7.2 (a) We give an example of a function that is absolutely continuous on compact intervals but not on R, (−∞, b] or [a, ∞). The example is given by f(x) = x2. For any open subinterval (a,b) of [a, b], we have |f(b) − f(a)| = |b2 − a2|  2max {|a|,|b|}(b−a) and we may therefore choose d satisfying d(2max{|a|,|b|}) = e to establish absolute continuity on [a, b]. However, if we take R, (−∞, b] or [a,∞) as the domain, we find that, no matter what d > 0 is chosen, the single interval (a, a + c), where 0 < c < d and 2c|a| > e + c2, has length less than d and yet satisfies |f(a + c) − f(a)| = |2ac + c2|  |2ac| − c2 > e; note that such an a

5.7 Absolute Continuity

267

can always be found in an unbounded interval, so that the criterion for absolute continuity is violated. (b) It is readily seen that an absolutely continuous function is uniformly continuous (in the definition, take a system consisting of just one interval), and hence continuous. As the reader knows from elementary analysis, the function given by f(x) = x2 is not uniformly continuous on R, (−∞, b] or [a, ∞), which is another way to see what we have noted in part (a), namely, that this function is not absolutely continuous there. However, a uniformly continuous function need not be absolutely continuous, as we shall illustrate in Example 5.7.3(d). See also Problems 5.7.P2 and 5.7.P6. (c) The inequality in Definition 5.7.1 can be replaced by the weaker condition





X n



½f ðbk Þ  f ðak Þ \e:



k¼1 Indeed, given any e > 0, let d > 0 be such that this inequality is fulfilled for any finite system of disjoint intervals {(ak, bk)}1  k  n contained in [a, b] and having sum of lengths less than d. We divide the intervals into two disjoint classes A and B, where A consists of those for which [f(bk) − f(ak)]  0 and B consists of the rest. By hypothesis, the inequality in the weaker condition must be fulfilled by each of the classes A and B of intervals separately. Consequently, n X

jf ðbk Þ  f ðak Þj ¼

k¼1

X

jf ðbk Þ  f ðak Þj þ

k2A

X

jf ðbk Þ  f ðak Þj

k2B





X

X





¼

½f ðbk Þ  f ðak Þ þ

½f ðbk Þ  f ðak Þ \2e:

k2B

k2A

Since e > 0 is arbitrary, it follows that f is absolutely continuous on [a, b]. (d) Let f be a real-valued function defined on [a, b]. Then f is absolutely continuous if and only if for every e > 0, there exists a d > 0 such that for any finite system of disjoint intervals {(ak, bk)}1  k  n contained in [a, b] and having sum of lengths less than d, the inequality n X

xk \e

k¼1

holds, where xk ¼ supfjf ðpÞ  f ðqÞj : p; q 2 ½ak ; bk g: Let f be absolutely continuous on [a, b] and let e > 0 be given. Then there exists a d > 0 such that for any finite system {(ak, bk)}1  k  n of disjoint open

268

5 Differentiation

intervals contained in [a, b] and having sum of lengths less than d, the inequality n X

jf ðbk Þ  f ðak Þj\e

k¼1

holds. Let ak, bk in [ak, bk] be such that f(ak) = mk and f(bk) = Mk, where mk and Mk are respectively the least and greatest values of the function f in [ak, bk]. Then n X

jðbk  ak Þj 

k¼1

n X

ðbk  ak Þ\d

k¼1

and so, n X k¼1

xk ¼

n X

jf ðbk Þ  f ðak Þj\e:

k¼1

On the other hand, suppose the given condition holds. Consider e > 0 and let d > 0 be as in the condition. Then for any finite system {(ak, bk)}1  k  n of disjoint open intervals contained in [a, b] and having sum of lengths less than d, we have |f(bk) − f(ak)|  xk, k = 1, 2, …, n, and therefore n X

jf ðbk Þ  f ðak Þj 

k¼1

n X

xk \e:

k¼1

Thus the function is absolutely continuous. (e) A function f defined on [a, b], R, (−∞, b] or [a, ∞) and satisfying a Lipschitz condition (see Definition 4.2.12) jf ðxÞ  f ðyÞj  M jx  yj for every x and y in the domain, where M is a fixed positive number, is absolutely continuous. This is so because, for any finite system {(ak, bk)}1  k  n of disjoint open intervals contained in the domain, we have n X k¼1

jf ðbk Þ  f ðak Þj  M

n X

ðbk  ak Þ;

k¼1

from which it may be easily inferred that f is absolutely continuous. As a particular case, if f′ exists and is bounded in (a, b) or R and if f:[a, b]!R is continuous, then it is absolutely continuous.

5.7 Absolute Continuity

269

(f) If the functions f and g are absolutely continuous on [a, b], then their sum, difference and product are also absolutely continuous. If g vanishes nowhere, then the quotient gf is absolutely continuous. From the inequality jðf  gÞðxÞ  ðf  gÞðyÞj  jf ðxÞ  f ðyÞj þ jgðxÞ  gðyÞj; it may be easily inferred that f ± g are absolutely continuous provided f and g are absolutely continuous. We shall next show that fg is absolutely continuous. Let e > 0 be given. Then there exists a d > 0 such that, for any finite system {(ak, bk)}1  k  n of disjoint intervals contained in [a, b] and having sum of lengths less than d, we have n X

jf ðbk Þ  f ðak Þj\

k¼1

n X e e and ; jgðbk Þ  gðak Þj\ 2ðA þ BÞ 2ðA þ BÞ k¼1

where A  |f| and B  |g| on [a, b]. Then n X

jðfgÞðbk Þ  ðfgÞðak Þj ¼

k¼1



n X

jf ðbk Þgðbk Þ  f ðak Þgðak Þj

k¼1

n X

j½f ðbk Þ  f ðak Þgðbk Þj þ

k¼1 n X

B

n X

jf ðak Þ½gðbk Þ  gðak Þj

k¼1

jf ðbk Þ  f ðak Þj þ A

k¼1

n X

jgðbk Þ  gðak Þj

k¼1

\e: Finally, if g vanishes nowhere, then there exists an m > 0 such that |g(x)| > m > 0 for all x 2 [a, b]. So,



1 1

jgðxÞ  gðyÞj

:

gðxÞ  gðyÞ  m2 Let e > 0 be given. Then there exists a d > 0 such that, for any finite system {(ak,bk)}1  k  n of disjoint intervals contained in [a, b] and having sum of lengths less than d, we have n X k¼1

jgðbk Þ  gðak Þj\m2 e:

270

5 Differentiation

Together with the inequality further above, this yields

n

X

1  1 \e:

gðb Þ gða Þ

k k k¼1 Thus 1g is absolutely continuous. Combining it with the result proved in the preceding paragraph for products, we get the desired conclusion. If the domain of definition of the functions happens to be unbounded, then the sum and difference of absolutely continuous functions are absolutely continuous; however, the product and quotient can be shown to be absolutely continuous provided the functions involved are assumed to be bounded. Examples 5.7.3 (a) If f is nonnegative and absolutely continuous on [a, b], then so is f p , p  1. To prove this, consider any y > x  0. Then first of all, yp  xp ¼ pðy  xÞðx þ hðy  xÞÞp1 for some h such that 0\h\1: So,



jf ðyÞp  f ðxÞp j ¼ pjf ðyÞ  f ðxÞj f ðxÞ þ hðf ðyÞ  f ðxÞÞp1

 pðM þ ð2MÞp1 Þjf ðyÞ  f ðxÞj;

ð5:69Þ

where M  |f| on [a, b]. Now let e > 0 be given. Since f is absolutely continuous, there exists a d > 0 such that, for any finite system {(ak, bk)}1  k  n of disjoint intervals contained in [a, b] and having sum of lengths less than d, we have n X

jf ðbk Þ  f ðak Þj\

k¼1

e pðM þ ð2MÞp1 Þ

;

which, in view of (5.69), implies n X k¼1

jf ðbk Þp  f ðak Þp j  pðM þ ð2MÞp1 Þ

n X

jf ðbk Þ  f ðak Þj\e:

k¼1

Boundedness of the domain is essential here, because x is absolutely continuous on every interval, bounded or not, but x2 is not [see Remark 5.7.2(a)]. (b) Since the function f(x) = x, 0  x  1, is trivially absolutely continuous, it follows in view of (a) above that g(x) = xp, p  1, 0  x  1, is also absolutely continuous. Now, g′(x) is bounded for 0  x  A (any A > 0) and therefore Remark 5.7.2(e) also shows that g is absolutely continuous. We shall take up the case when 0  p < 1 in Problem 5.7.P6.

5.7 Absolute Continuity

271

(c) The function f(x) = sin x is absolutely continuous on R, and so is the function g (x) = sinpx, p  1. The function sin x satisfies a Lipschitz condition with M = 1 and is, therefore, absolutely continuous by Remark 5.7.2(e). The function g can also be seen to be absolutely continuous by using Remark 5.7.2(e). Indeed, it satisfies a Lipschitz condition with M = p. 1 (d) Absolute continuity of f on [a, b] does not imply that of f 2 . For example, p consider the function f:[0, 1]!R defined by f(0) = 0 and f(x)= x2sin2 2x , x 6¼ 0. 2 p p p Observe that f′(x)=sin 2x − psin 2x cos 2x for 0 < x < 1, so that |f′(x)|  2 + p. Thus, f is absolutely continuous on [0, 1] by Remark 5.7.2(e). However,



1 1 f ðxÞ2 ¼ x sin p when x > 0 and f ð0Þ2 ¼ 0 . For the partition 2x

0\ TðP; f Þ ¼ 1 2

2n þ1

X

1 1 1 1 \ \    \ \ \1; 2n þ 1 2n 3 2



f ðxk Þ  f ðxk1 Þ

1 2

1 2

k¼1



1

p

1 1 p

j sinð2n  1Þ j

þ   

sinð2n þ 1Þ þ

j sin npj  2n þ 1 2 2n 2n  1 2



p



1

1 3p 1



þ

jsin j  jsin pj

þ

jsin pj  sin

3 2 2 2 2 1 1 1 ¼ þ þ    þ þ 1: 2n þ 1 2n  1 3

¼

1

These sums become infinite as n ! ∞. Consequently, f 2 is not of bounded variation. As we shall see later (see Proposition 5.7.6), this implies that it cannot be absolutely continuous (despite being continuous on the closed and bounded interval [0, 1] and hence uniformly continuous). (e) A function / defined on (a, b) is said to be convex if for every x, y 2 (a, b) and every t 2 [0, 1], the following inequality holds: /ðð1  tÞx þ tyÞ  ð1  tÞ/ðxÞ þ t/ðyÞ: Graphically, the condition is that if x < z < y, then the point (z, /(z)) in the plane should lie below or on the line segment joining the points (x, /(x)) and (y, /(y)). Consider any x, y, z 2 (a, b) such that x < z < y. Then z¼

yz zx xþ y ¼ ð1  tÞx þ ty; yx yx

Therefore, convexity of / implies

where t ¼

zx : yx

272

5 Differentiation

/ðzÞ 

yz zx /ðxÞ þ /ðyÞ; yx yx

which leads to (y − x)/(z)  (y − z)/(x) + (z − x)/(y) and hence to ðy  z þ z  xÞ/ðzÞ  ðy  zÞ/ðxÞ þ ðz  xÞ/ðyÞ: It follows from here that ðz  xÞð/ðzÞ  /ðyÞÞ  ðy  zÞð/ðxÞ  /ðzÞÞ; from which we obtain /ðzÞ  /ðxÞ /ðyÞ  /ðzÞ  : zx yz The fact that a convex function / on (a, b) satisfies the above inequality for arbitrary x, y, z 2 (a, b) such that x < z < y is about to be used. A convex function / defined on (a, b) is absolutely continuous on every closed subinterval [c, d] (a, b), as we show below: Let a < c′ < c and d < d′ < b. For x, y 2 [c, d], x < y, we have /ðc0 Þ  /ðcÞ /ðxÞ  /ðcÞ /ðyÞ  /ðxÞ /ðdÞ  /ðyÞ /ðd 0 Þ  /ðdÞ     ; c0  c xc yx dy d0  d using the fact that / is convex on (a, b). The above inequalities imply



0



0

/ðyÞ  /ðxÞ







 max /ðd Þ  /ðdÞ ; /ðc Þ  /ðcÞ ¼ H; say:

yx

d 0  d c0  c

So, / satisfies a Lipschitz condition on [c, d] with Lipschitz constant H and is therefore absolutely continuous by Remark 5.7.2(e). (f) The composition of absolutely continuous functions need not be absolutely continuous. (However, see Propositions 5.7.4 and 5.7.5.) It is a consequence of Problem 5.7.P6 that the square root function is absolutely continuous on [0, 1]. Accordingly, it follows from the same example as in (d) above that the composition of absolutely continuous functions need not be absolutely continuous. Proposition 5.7.4 Let f:I ! [c, d] be an absolutely continuous function, where I is any interval. If F is a function on [c, d] which satisfies a Lipschitz condition, then the composite function F f is absolutely continuous. Proof If |F(y) − F(x)|  M|y − x|, then for an arbitrary finite system of disjoint open intervals {(ak, bk)}1  k  n contained in I, we have

5.7 Absolute Continuity

n X

273

jFðf ðbk ÞÞ  Fðf ðak ÞÞj  M

k¼1

n X

jf ðbk Þ  f ðak Þj:

k¼1

Since f is absolutely continuous on I, it follows that the right-hand side of the above n P ðbk  ak Þ is small. h inequality is small whenever k¼1

Proposition 5.7.5 Let f be an absolutely continuous function defined on any interval I and suppose that it is increasing. If F is absolutely continuous on f(I), then the composition F f is absolutely continuous on I. Proof Let e > 0 be given. Since F is absolutely continuous on f(I), there exists a d > 0 such that, for any finite system of disjoint open intervals {(Ak, Bk)}1  k  n contained in f(I) and having total length less than d, we have n X

jFðBk Þ  FðAk Þj\e:

k¼1

For this d > 0, there exists an η > 0 such that m P

m P

ðbk  ak Þ\g implies

k¼1

jf ðbk Þ  f ðak Þj\d whenever the intervals {(ak, bk)}1  k  m are disjoint. Choose

k¼1

any finite system fðck ; dk Þg1  k  ‘ of disjoint open intervals with total length ‘ P ðdk  ck Þ\g. Then the intervals fðf ðck Þ; f ðdk ÞÞg1  k  ‘ are disjoint, since f is

k¼1

increasing. Moreover the sum of their lengths,

‘ P

jf ðdk Þ  f ðck Þj\d . Hence

k¼1 ‘ X

jFðf ðdk ÞÞ  Fðf ðck ÞÞj\e:

h

k¼1

Our next result shows that the class of absolutely continuous functions is contained in the class of functions of bounded variation. Proposition 5.7.6 Let f be an absolutely continuous function on [a, b]. Then f is of bounded variation. Proof Since f is absolutely continuous, choose d > 0 corresponding to e = 1 such that, for every finite system {(ak, bk)}1  k  n of disjoint open intervals contained in [a, b] and having sum of lengths less than d, we have

274

5 Differentiation n X

jf ðbk Þ  f ðak Þj\1:

k¼1

Let N > ba d be any integer, and let P: a = x0< x1< ⋯ < xN = b be the partition ba of [a, b] such that xk − xk − 1 = ba N , i.e., xk = a + k N for 1  k  N. Then 0 < xk − xk−1 < d for 1  k  N. Therefore, for every subdivision of the subinterval [xk−1, xk], the sum of the absolute increments of f on these intervals corresponding to the subdivision of [xk−1, xk] is less than 1. Therefore V([xk−1, xk], f )  1. In view of Theorem 5.6.4, it follows that V([a, b], f)  N. h Remarks (a) The converse of Proposition 5.7.6 is false. The Cantor ternary function is continuous and increasing, and hence is of bounded variation. However, it fails to be absolutely continuous by Problem 5.7.P2. Moreover, in contrast to the example in Remark 5.7.2(a), it is a continuous function on a bounded interval that is not absolutely continuous. (b) The implication in Proposition 5.7.6 fails to be true if a or b is allowed to be infinite. In fact, the functions x and sin x are absolutely continuous but are not of bounded variation on (−∞, b] or on [a, ∞). By the Mean Value Theorem, |sinx − siny|  |x − y| for all x, y 2 R. Consequently, for any finite system {(ak, bk)}1  k  n of disjoint open intervals, we have n X

jsin bk  sin ak j 

k¼1

n X

ðbk  ak Þ;

k¼1

from which follows the absolute continuity of sin x. If we take the points p 3p 5p ð2n  1Þp ð2n þ 1Þp \ \ \\ \ 2 2 2 2 2 for points of division, then the sum

n

X

sin ð2k  1Þp  sin ð2k þ 1Þp ¼ 2n



2 2 k¼1

can be made large for large values of n. Hence the function sin x, x 2 R, is not of bounded variation even on [p2, ∞) and a fortiori not on R. Corollary 5.7.7 An absolutely continuous function on [a, b] has a finite derivative a.e. on the domain and the derivative is Lebesgue integrable.

5.7 Absolute Continuity

275

Proof By Proposition 5.7.6, the function is of bounded variation on [a, b]. The Corollary now follows by using Theorem 5.5.1. h Corollary 5.7.8 There exist continuous functions on bounded intervals that are not absolutely continuous. [The case of unbounded intervals has been taken care of in Remark 5.7.2(a).] p Proof The function f(x) = x sin 2x ; x 6¼ 0, and f(0) = 0 is continuous on [0, 1] but is not of bounded variation [see Example 5.6.3(c)], and hence by Proposition 5.7.6, cannot be absolutely continuous. h

Proposition 5.7.9 Let f:[a, b]!R be absolutely continuous and let vðaÞ ¼ 0; vðxÞ ¼ Vð½a; x; f Þ for x 2 ½a; b (notations as in Theorem 5.6.8). Then v is absolutely continuous on [a, b]; f can therefore be expressed as the difference of increasing absolutely continuous functions. Proof By Theorem 5.6.4, we see that v(b) − v(a) = V([a, b], f) if a  a  b  b. Let e > 0 be given and d > 0 correspond to e as in the definition of absolute continuity. Suppose {(ak,bk)}1  k  n is a finite system of disjoint open intervals contained in [a, b] and having sum of lengths less than d, and Pk : ak ¼ ck;1 \ck;2 \    \ck;mk ¼ bk is any partition of [ak, bk]. Then {(ck,j, ck,j+1): 1  j  mk, 1  k  n} is a finite system of disjoint open intervals contained in [a, b] with sum of lengths less than d. By absolute continuity of f, we have mk

n X X

f ðck;j þ 1 Þ  f ðck;j Þ \e: k¼1 j¼1

Since the partitions Pk are arbitrary, the above implies n X k¼1

jvðbk Þ  vðak Þj ¼

n X

Vð½ak ; bk ; f Þ  e:

k¼1

Thus v is absolutely continuous. Since the sum and difference of absolutely continuous functions are absolutely continuous, it follows that 12 (v(x) ± f(x), which are increasing functions (see proof of Theorem 5.6.6) and the difference of which is f, are also absolutely continuous. Alternatively, one can argue that since the sum and difference of absolutely continuous functions are absolutely continuous, it follows that P ([a, x], f), N ([a, x], f) are also absolutely continuous and so are the functions f1 and f2 (notations as in Theorem 5.6.11). h

276

5 Differentiation

Problem Set 5.7 5.7.P1. If f:[a, b]!R is absolutely continuous on [a, c] as well as on [c, b], then it is absolutely continuous on [a, b]. 5.7.P2. Let f:[a, b]!R be an absolutely continuous function and let E [a, b] be a set of measure zero. Then show that m(f(E)) = 0. Use this to show that the Cantor function is not absolutely continuous. Show that on a closed bounded interval, the class of absolutely continuous functions is a proper subclass of continuous functions of bounded variation. [This is trivial if the domain is unbounded, because sinx will do the trick.] 5.7.P3. Let f:[a, b]!R be an absolutely continuous function. Then f maps measurable sets into measurable sets. 5.7.P4. Let f be defined and measurable on the closed interval [a, b] and let E be any measurable subset on which f′ exists. Then prove that



Z

m ðf ðEÞÞ 

jf 0 j:

E

So far, we have proved (see Remark 5.7.2(b), Proposition 5.7.6 and Problem 5.7. P2) that an absolutely continuous function is continuous, of bounded variation and maps a set of measure zero into a set of measure zero. The following result shows that these three properties characterise an absolutely continuous function [Banach– Zarecki Theorem]. 5.7.P5. If f is continuous and of bounded variation on [a, b], and if f maps a set of measure zero into a set of measure zero, then f is absolutely continuous on [a, b]. 5.7.P6. Give an example to show that continuity on [0, 1], even when combined with absolute continuity on [e, 1] for each positive e < 1, does not imply absolute continuity on [0, 1]. What if the function is also of bounded variation on [0, 1]? Show that xp is absolutely continuous on [0, 1] when 0  p < 1. 1 5.7.P7. Show that the function f given on [0, ∞) by f(x) = x2 for 0  x < 1 and satisfying f(x + k) = f(x) + k for 0  x < 1 and k 2 N is absolutely continuous on any bounded subinterval of its domain but not on the entire domain. (Note that the example of such a function given in Remark 5.7.2(a) satisfies a Lipschitz condition on every bounded subinterval of [0, ∞) while the present one does not.) 5.7.P8. Let f be an increasing function defined on [a, b]. Show that f can be decomposed into a sum of increasing functions f = g + h, where g is absolutely continuous and h is increasing with h′ = 0 a.e.

5.8 Differentiation and the Integral

5.8

277

Differentiation and the Integral

In this section, we clarify the relation between differentiation and integration in the Lebesgue sense. In part (ii) of the remark preceding Example 5.6.12, we have seen R that f ðxÞ ¼ ½a;x g, where g is a Lebesgue integrable function on [a, b], is a function of bounded variation. It turns out that f indeed is absolutely continuous. It will be R shown in Theorem 5.8.2 and Corollary 5.8.3 that f(x) − f(a) = ½ax f 0 if and only if f is absolutely continuous. It generalises the Second Fundamental Theorem of Calculus [see Theorem 1.7.2 (b)]. Let g be an integrable function defined on [a, b]. The function Z f ðxÞ ¼ c þ

g; ½a;x

where c is an arbitrary constant is called an indefinite integral of g. We shall investigate whether the equality f′ = g holds if f is an indefinite integral of g. It will be shown in Theorem 5.8.5 that the equality f′(x) = g(x) holds for almost all x for an arbitrary integrable function g. This generalises the First Fundamental Theorem of Calculus [see Theorem 1.7.2 (a)]. We begin by proving that an indefinite integral is absolutely continuous. Proposition 5.8.1 Let g be a Lebesgue integrable function defined on [a, b]. Then the indefinite integral Z f ðxÞ ¼ c þ

g ½a;x

is absolutely continuous. Proof In view of Problem 3.2.P15, for e > 0, there is a d > 0 such that for every measurable set E with m(E) < d, we have Z jgj\e: E

Now, suppose that {(ak, bk)}1  k  n is a finite system of disjoint open intervals contained in [a, b] and having sum of lengths less than d. Then



Z

X n n Z n Z X X



g  jf ðbk Þ  f ðak Þj ¼ j gj ¼ jgj\e

½ak ;bk  k¼1 ½ak ;bk  [ k ½ak ;bk  k¼1 k¼1 because m( [ 1  k  n(ak, bk)) =

n P t¼1

uous.

ðbk  ak Þ\d. Therefore f is absolutely continh

278

5 Differentiation

Since the function given by f(x) = xp (0 < p < 1) on a bounded interval [0, A] is the indefinite integral of the function g(t) = ptp−1, the above proposition shows that it is absolutely continuous. An independent derivation of this fact has already been seen in Problem 5.7.P6. If f is absolutely continuous on [a, b], then by Theorem 5.5.1 and Proposition 5.7.9, f is differentiable a.e. and the derivative f′ is integrable in the Lebesgue sense. We shall show that its integral agrees with f(b) − f(a). Theorem 5.8.2 Suppose f:[a, b]!R is absolutely continuous. Let E be the set of points x 2 (a, b) such that f is differentiable at x. Then Z Z 0 f ¼ f 0 ¼ f ðbÞ  f ðaÞ: ½a; b

E

Proof The function f′ is summable over [a, b] by Theorem 5.5.1 and Proposition 5.7.9. In view of the observation above, m((a, b)\E) = 0. Consequently, Z ½a; b

0

Z

f ¼

f 0:

E

It remains to prove that Z

f 0 ¼ f ðbÞ  f ðaÞ:

ð5:70Þ

E

By Proposition 5.7.9, f = f1 − f2, where f1 and f2 are increasing and absolutely continuous. In view of this and the linearity of the integral, it suffices to prove (5.70) under the additional assumption that f is increasing on [a, b]. In the rest of the proof we make this assumption. Since f is continuous on [a, b], its Lebesgue integral over any interval [a, a + 1n], 1 n > ba , is the same as its Riemann integral. Extend the definition of f by setting f(x) = f(b) for x > b. Then it is continuous on [a, ∞) and therefore its Lebesgue integral over any bounded interval [a, K], where K > a, is the same as its Riemann integral. Define /n:[a, b]!R by the formula 1 /n ðxÞ ¼ n½f ðx þ Þ  f ðxÞ: n Observe that, since f is increasing on [a, b], we have /n ð xÞ  0 everywhere and f 0 ð xÞ  0 wherever this derivative exists:

5.8 Differentiation and the Integral

279

Clearly, /n is continuous and at each x 2 E, /n(x) ! f′(x). Now, Z Z Z Z 1 /n ¼ /n ¼ n f ðt þ Þdt  n f ðtÞdt n E ½a; b ½a; b ½a; b Z Z ¼n f ðtÞdt  n f ðtÞdt ½a þ 1n;b þ 1n ½a; b Z f ðtÞdt; ¼ f ðbÞ  n ½a;a þ 1n

where the integral on the right agrees with the corresponding Riemann integral for sufficiently large n. Since f is continuous, the First Fundamental Theorem of Calculus [see Theorem 1.7.2 (a)] yields Z /n ¼ f ðbÞ  f ðaÞ:

lim

n!1

E

Therefore (5.70) will follow if we prove that Z

Z /n ¼

lim

n!1

E

f 0:

E

Let e > 0 be given. Since the function f is constant on [b, b + 1] and is absolutely continuous on [a, b], it is absolutely continuous on [a, b + 1]. Hence there exists a d1 > 0 such that, for any finite system {(aj, bj)}1  j  p of disjoint open intervals contained in [a, b + 1] and having sum of lengths less than d, we have p

X

f ðbj Þ  f ðaj Þ \ e : 3 j¼1

ð5:71Þ

The function f′ is summable over [a, b], and so by Problem 3.2.P15, there exists a d2 > 0 (which we choose less than d1) such that, for every measurable set F  E with m(F) < d2, we have Z e ð5:72Þ jf 0 j\ : 3 F Now by Egorov’s Theorem 2.6.2, there is a measurable subset F  E with m(F) < d2 such that /n ! f 0 uniformly on E\F. Without loss of generality, we may assume that F  (a, b). With F fixed, there exists an N such that n  N implies

280

5 Differentiation

Z

e j/n  f 0 j\ : 3 EnF

ð5:73Þ

It then follows that

Z Z

Z Z





0 0 0

/n  f ¼

ð/  f Þ þ ð/  f Þ

n n



E E EnF F Z Z Z  /n þ f 0; j/n  f 0 j þ EnF

F

F

where we do not need to take the absolute value of the last two terms in view of the observation recorded above after defining /n. Upon using (5.72) and (5.73), it further follows that n  N implies

Z Z Z



2

/n  f 0  /n þ e



3 E E F

for

n  N:

ð5:74Þ

Since m(F) < d2 < d1, there exists an open set V such that F  V and m(V) < d1. [This is because, by Proposition 2.3.24, there exists an open set V F such that m(V\F) < d1 − d2, which implies m(V) = m(V\F) + m(F) < d1 − d2 + d2 = d1]. As F  (a, b), we may assume that V  (a, b) by replacing it with V \ (a, b) if necessary. We can express V as a countable union of disjoint open intervals (ak, bk)  (a, b)  [a, b], k = 1, 2, … . In what follows, the obvious simplification for the case of finitely many disjoint intervals is left to the reader. For any x such that 0  x  1 and for any positive integer m, the intervals (ak + x,bk + x)  [a, b + 1], k = 1, …, m, are disjoint and have sum of lengths equal to mð

m [

ðak þ x; bk þ xÞÞ  mðVÞ\d1

k¼1

and hence by (5.71), m X

e ½f ðbk þ xÞ  f ðak þ xÞ\ : 3 k¼1

ð5:75Þ

5.8 Differentiation and the Integral

281

Now, Z 1 f ðt þ Þdt  f ðtÞdt n ½ak ;bk  ½ak ;bk  Z Z ¼ n½ f ðtÞdt  f ðtÞdt ½ak þ 1n;bk þ 1n ½ak ;bk  Z Z f ðtÞdt  f ðtÞdt ¼ n½ ½bk ;bk þ 1n ½ak ;ak þ 1n Z Z f ðt þ bk Þdt  f ðt þ ak Þdt: ¼ n½

Z

Z

½ak ;bk 

/n ¼ n½

½0;1n

½0;1n

Therefore m Z X k¼1

½ak ;bk 

Z /n ¼ n

m X

½0;  1 n

! ½f ðt þ bk Þ  f ðt þ ak Þ dt 

k¼1

e ; 3

using (5.75) above. Consequently, Z

Z /n 

F

/n ¼ V

1 Z X k¼1

½ak ;bk 

/n 

e : 3

Therefore by (5.74),

Z Z



/n  f 0 \e for n  N:



E

h

E

Corollary 5.8.3 A function f:[a, b]!R is absolutely continuous if and only if it can be expressed in the form Z f ðxÞ ¼ c þ

g; ½a; x

where c is a constant and g:[a, b]!R is integrable. Proof Define g(x) = f′(x) at those points x in (a, b) where f is differentiable and assign arbitrary values at the other points of [a, b]. The R above theorem completes the argument in one direction. By Proposition 5.8.1, ½a; x g is absolutely continuous. Since a constant function is always absolutely continuous, it follows that f as defined above is absolutely continuous. h Corollary 5.8.4 If f:[a, b]!R is absolutely continuous and f′(x) = 0 almost everywhere, then f is a constant function.

282

5 Differentiation

R

Proof In fact, by Theorem 5.8.2, we have 0 ¼ [a, b], so that f(x) = f(a) everywhere on [a, b].

½a; x

f 0 ¼ f ðxÞ  f ðaÞ for any x 2 h

Theorem 5.8.5 If g:[a, b]!R is integrable and if f:[a, b]!R is defined by Z g; f ðxÞ ¼ ½a;x

then f′(x) = g(x) a.e. in [a, b]. Proof The function f is absolutely continuous by Proposition 5.8.1. Set /(x) = f′(x) if x 2 (a, b) and f is differentiable at x; let /(x) be arbitrary for other values of x. Then Z / ¼ f ðxÞ  f ðaÞ ¼ f ðxÞ ½a;x

by Theorem 5.8.2. Hence by the given definition of f, we get Z

Z 0¼

g

½a;x

Z ½a;x

/¼

½a;x

ðg  /Þ;

which implies g = / a.e. [see Problem 3-2.P14(b)]. So, g(x) = f′(x) for almost all x 2 [a, b]. h Remark 5.8.6 If g is integrable, the above theorem may be formulated by stating that, for almost every x 2 (a, b), the expressions R

½gðx þ tÞ  gðxÞdt and R 1 ½gðx þ tÞ  gðxÞdt h ½h;0 1 h ½0;h

both tend to 0 as h ! 0. In what follows, we shall prove a stronger result (Theorem 5.8.9), namely, for almost every x 2 (a, b), the expressions R

R jgðx þ tÞ  gðxÞjdt ¼ 1h ½x;x þ h jgðtÞ  gðxÞjdt and R R 1 1 h ½h;0 jgðx þ tÞ  gðxÞjdt ¼ h ½xh;x jgðtÞ  gðxÞjdt 1 h ½0;h

both tend to 0 as h ! 0. We begin with a definition. Definition 5.8.7 Let g 2 L1[a, b]. If lim 1h

R ½x;x þ h

h!0

lim 1h

h!0

R

jgðtÞ  gðxÞjdt ¼ 0

and gðtÞ  gðxÞjdt ¼ 0; j ½xh;x

the point x 2 (a, b) is said to be a Lebesgue point of the function g.

5.8 Differentiation and the Integral

283

Remark 5.8.8 (a) Every interior point of continuity of an integrable function g is a Lebesgue point of the function. Indeed, let x 2 (a, b) be a point of continuity of g. Then for e > 0, there exists a d > 0 such that jgðtÞ  gðxÞj\e whenever

jt  xj\d:

For 0 < h < d, we have R

jgðtÞ  gðxÞjdt\e and R 1 j gðtÞ  gðxÞjdt\e: h ½xh;x 1 h ½x;x þ h

This completes the proof. (b) The function g 2 L1[a, b] need not be continuous anywhere and yet almost every point in the domain of g may be a Lebesgue point. To see how, consider the function g defined on [0, 1] which is 1 at every rational point and 0 at every irrational Rpoint. It is discontinuous everywhere and Lebesgue integrable on [0, 1] with ½0;1 g ¼ 0 and, for every irrational number x, 1 h

Z ½x;x þ h

jgðtÞ  gðxÞjdt ¼

1 h

Z ½x;x þ h

jgðtÞjdt ¼ 0;

since the set {t : g(t) 6¼ 0} has measure 0. (c) We have proved that almost every point of the domain of the function g considered above is a Lebesgue point. In fact, the following theorem holds: Theorem 5.8.9 If g 2 L1[a, b], then almost every point of [a, b] is a Lebesgue point of g. Proof The function |g(t) − r|, where r is a rational number, is integrable on [a, b] and hence, for almost all x 2 [a, b], Z 1 lim ð5:76Þ jgðtÞ  r jdt ¼ jgðxÞ  r j h!0 h ½x;x þ h (see Theorem 5.8.5). Let E(r) be the set of points of [a, b] at which (5.76) does not hold. Let E¼

[ r2Q

ðEðrÞ [ ft : jgðtÞj ¼ 1gÞ:

284

5 Differentiation

Then mðEÞ 

X

mðEðrÞÞ þ mðft : jgðtÞj ¼ 1gÞ ¼ 0:

r2Q

We shall show that each point x0 2 (a, b)\E is a Lebesgue point of g. Let e > 0 be arbitrary and r be a rational number such that e jgðx0 Þ  r j\ : 3

ð5:77Þ

Then we have jjgðtÞ  r j  jgðtÞ  gðx0 Þjj\

e 3

for all t 2 [a, b]. It follows for h > 0 that

Z Z

1Z

1 1 e e



dt ¼ : jgðtÞ  r jdt  jgðtÞ  gðx0 Þjdt 

h ½x0 ;x0 þ h 3

h ½x0 ;x0 þ h h ½x0 ;x0 þ h 3 Since x0 62 E, the equality (5.76) yields a d > 0 such that 0 < h < d implies

Z

e

1



jgðtÞ  r jdt  jgðx0 Þ  r j \

3

h ½x0 ;x0 þ h and hence by (5.77), further implies 1 h

Z ½x0 ;x0 þ h

jgðtÞ  r jdt\

2e : 3

ð5:78Þ

It is a consequence of (5.77) and (5.78) that for any x0 62 E, there exists a d > 0 such that 0 < h < d implies Z Z Z 1 1 1 jgðtÞ  gðx0 Þjdt  jgðtÞ  r jdt þ jgðx0 Þ  r jdt\e: h ½x0 ;x0 þ h h ½x0 ;x0 þ h h ½x0 ;x0 þ h In other words, lim 1h h!0

R ½x0 ;x0 þ h

1 lim h!0 h

jgðtÞ  gðx0 Þjdt = 0. Similarly, it can be shown that

Z ½x0 h;x0 

Thus x0 is a Lebesgue point of g.

jgðtÞ  gðx0 Þjdt ¼ 0: h

5.8 Differentiation and the Integral

285

Corollary 5.8.10 Let g 2 L1[a, b]. Then we have Z 1 lim jgðx þ tÞ þ gðx  tÞ  2gðxÞjdt ¼ 0 h!0 h ½0;h at every Lebesgue point x of g and hence for almost all x 2 (a, b). Proof For fixed x 2 (a, b), write Z 1 jgðx þ tÞ þ gðx  tÞ  2gðxÞjdt h ½0;h Z Z 1 1  jgðx þ tÞ  gðxÞjdt þ jgðx  tÞ  gðxÞjdt h ½0;h h ½0;h Z Z 1 1 ¼ jgðtÞ  gðxÞjdt þ jgðtÞ  gðxÞjdt: h ½x;x þ h h ½xh;x Since 1 lim h!0 h

Z

1 jgðtÞ  gðxÞjdt ¼ 0 ¼ lim h!0 h ½x;x þ h

Z ½xh;x

jgðtÞ  gðxÞjdt

at each Lebesgue point of g, the “almost all x” part of the result now follows by Theorem 5.8.9 above. h Proposition 5.8.11 Let x be a LebesgueR point of an integrable function g defined on [a, b]. The indefinite integral f(x) = ½a; x g is differentiable at the point x and f′(x) = g(x). Proof It is easy to check that f ðx þ hÞ  f ðxÞ 1  gðxÞ ¼ h h

Z ½x;x þ h

½gðtÞ  gðxÞdt;

so,



Z

f ðx þ hÞ  f ðxÞ

1

  gðxÞ jgðtÞ  gðxÞjdt ! 0 as h ! 0 þ ;

h h ½x;x þ h since x is a Lebesgue point of g. Thus f+′(x) = g(x). Similarly it can be shown that f_′(x) = g(x). h Theorem 5.8.12 (Integration by Parts). Let f and g be functions in L1[a, b] and

286

5 Differentiation

Z FðxÞ ¼

½a; x

f þ a;

Z GðxÞ ¼

½a;x

g þ b;

where a and b are constants. Then Z Z Fg þ Gf ¼ GðbÞFðbÞ  FðaÞGðaÞ: ½a; b

½a; b

Proof By Proposition 5.8.1, F and G are absolutely continuous functions on [a, b] and hence so is FG by Remark 5.7.2(f), and ðFGÞ0 ¼ FG0 þ F 0 G a:e: on ½a; b as elementary calculations show. By Theorem 5.8.5, we have F′ = f and G′ = g a.e. Therefore the required equality follows upon applying Theorem 5.8.2. h We turn our attention to change of variables in a Lebesgue integral. The discussionf below is based on Serrin and Varberg [24]. Theorem 5.8.13 Let the function u:[a, b]!R be differentiable at each point of a subset E  [a, b] and u(E) have Lebesgue measure 0. Then {x 2 E: u′(x) 6¼ 0} also has measure 0. Proof Denote the set {x 2 E : u′(x) 6¼ 0} by E0. We have to show that E0 has measure 0. For each n 2 N, let 1 1 En fx 2 E0 : juðxÞ  uðyÞj  jx  yj for all y 2 ½a; b such that 0\jx  yj\ g: n n ð5:79Þ We shall prove that [

E0 

En :

ð5:80Þ

n1

So, consider any x 2 E0. By definition of E0, we have ju0 ðxÞj [ 0. By definition of derivative, there exists a d > 0 for which y 2 ½a; b;

0\jx  yj\d

)

juðxÞ  uðyÞj [

1 0 ju ðxÞjjx  yj: 2



Now choose n 2 N such that 1n < min d; 12 ju0 ðxÞj . For any such n 2 N, we obviously have x 2 En. This completes the proof that (5.80) must hold. The desired conclusion that m(E0) = 0 will follow from (5.80) if for each n 2 N we demonstrate that m(En) = 0.

5.8 Differentiation and the Integral

287

For this purpose, fix some n 2 N. In order to demonstrate that m(En) = 0, we need only argue that, for any interval I of length less than 1n the intersection Gn ¼ I \ En

ð5:81Þ

has measure 0. So, let e > 0 be given. Since Gn  En  E0  E, we have u(Gn)  u(E), which has measure 0 by hypothesis. Therefore u(Gn) has measure 0. By definition of measure being 0, there must exist a sequence {Ik}k  1 of intervals 1 P such that u(Gn)  [ k  1Ik and ‘ðIk Þ\e. By breaking up each Ik further if k¼1

necessary, we may assume that ‘ðIk Þ\ 1n for each k. Define An;k ¼ Gn \ u1 ðIk Þ:

ð5:82Þ

Considering that u(Gn)  [ k  1Ik, we must have Gn 

[

An; k :

ð5:83Þ

k1

With a view to estimating the outer measure m*(An,k), take any two distinct x, y 2 An,k. By (5.81) and (5.82), we have first, x, y 2 I and second, x, y 2 En. The first implies |x − y| < 1n because I has length less than 1n. As a consequence of this inequality, the second implies |u(x) − u(y)|  1n |x − y| by virtue of (5.79). Since this has been proved for any two distinct x, y 2 An,k, we arrive at supfjx  yj : x; y 2 An;k g  n  supfjuðxÞ  uðyÞj : x; y 2 An;k g  n  supfja  bj : a; b 2 Ik g

by (5:82Þ

 n  ‘ðIk Þ: By Problem 2.2.P13, the left-hand side here is greater than or equal to the outer measure m*(An,k). Therefore we arrive at the estimate m*(An,k)  n  ‘ðIk Þ. It now 1   P m An;k  ne. Hence follows upon using (5.83) that m*(Gn)  k¼1

m*(Gn) = 0 = m(Gn). This completes the argument that for any interval I of length less than 1n, the intersection I \ En has measure 0. As already noted, this is all that we needed to establish. h Theorem 5.8.14 (Chain Rule a.e.) Let F:[c, d]!R and f:[c, d]!R be functions such that F maps sets of measure 0 into sets of measure 0 and F′= f a.e. Suppose also that u:[a, b] ! [c, d] is differentiable a.e. Then the equality ðF uÞ0 ¼ ðf uÞu0 holds a.e. on [a, b].

ð5:84Þ

288

5 Differentiation

Proof According to the well-known Chain Rule (Proposition 1.6.1), if t 2 [a, b] has the property that u is differentiable at t and also F is differentiable at u(t), then the composition F u is differentiable at t, and (F u)′(t) = (F 0 u)(t)u′(t). Let U ¼ ft 2 ½a; b : u0 ðtÞ 6¼ 0g;

U0 ¼ ft 2 ½a; b : u0 ðtÞ ¼ 0g:

Since u is differentiable a.e., the complement (U [ U0)c has measure 0. Since F′= f a.e., there exists a set Z  [c, d] of measure 0 such that x 62 Z implies F′(x) = f(x). In view of the hypothesis that F maps sets of measure 0 into sets of measure 0, the set F(Z) has measure 0. Now, U = A [ B, where A ¼ U \ ft 2 ½a; b : uðtÞ 62 Zg; B ¼ U \ ft 2 ½a; b : uðtÞ 2 Zg: Also, U0 = A0 [ B0 [ C0, where A0 ¼ U0 \ ft 2 ½a; b : uðtÞ 62 Zg; B0 ¼ U0 \ ft 2 ½a; b : uðtÞ 62 Zg \ ft 2 ½a; b : ðF uÞ0 ðtÞ 6¼ 0g; C0 ¼ U0 \ ft 2 ½a; b : uðtÞ 62 Zg \ ft 2 ½a; b : ðF uÞ0 ðtÞ ¼ 0g: On the sets A as well as A0, the equality (5.84) is seen to hold everywhere by virtue of the Chain Rule (Proposition 1.6.1). It holds on C0 as well, because both sides of the equality are 0. Therefore the subset of U [ U0 on which it fails to hold is contained in B [ B0. We shall demonstrate that both B and B0 have measure 0. The definition of B implies that B  U and u(B)  Z. Together with the fact that Z has measure 0, this immediately implies by Theorem 5.8.13 that B has measure 0. As regards B0, its definition implies that ðF uÞ(B0)  F(Z), which has measure 0; since (F u)′ 6¼ 0 on B0, Theorem 5.8.13 implies that B0 has measure 0. Thus the equality (5.84) can fail to hold only on a subset of the union of the sets ðU [ U0 Þc ; B

B0 ;

and

each of which has measure 0. h For the purpose of the Change of Variables Formula for Lebesgue integrals in one dimension, it will be convenient to introduce Lebesgue integrals from right to left, exactly as for Riemann integrals: Z

a b

Z f dm ¼ 

Z ½a; b

f dm ¼ 

This enables us to restate Theorem 5.8.2 as

b

f dm when a\b: a

5.8 Differentiation and the Integral

289

Theorem 5.8.15 Suppose f:[a, b] ! R is absolutely continuous. Let E be the set of points x 2 (a, b) such that f is differentiable at x. Then Z

b

0

Z

f 0 dm ¼ f ðbÞ  f ðaÞ

f dm ¼

a

E

and Z

a

f 0 dm ¼ 

Z

f 0 dm ¼ f ðaÞ  f ðbÞ: E

b

Theorem 5.8.16 (Change of Variables Formula 1). Let f be an integrable function on [c, d] and u be an absolutely continuous function on [a, b] such that u[a, b]  [c ,d]. Suppose F u is absolutely continuous on [a, b], where Z

x

FðxÞ ¼

f dm: c

Then (f u)u′ is integrable on [a, b] and, for any a, b 2 [a, b], Z

uðbÞ

uðaÞ

Z f dm ¼

b

ðf uÞu0 dm:

a

Proof By Proposition 5.8.1, the function F is absolutely continuous, and hence by Problem 5.7.P2, maps sets of measure 0 into sets of measure 0. Moreover, F′ = f a.e. on [c, d] by Theorem 5.8.5. On the basis of the Chain Rule a.e. (Theorem 5.8.14), we therefore obtain ðF uÞ0 ¼ ðf uÞu0 a:e: on ½a; b: Since F u is given to be absolutely continuous, we now deduce from Corollary 5.7.7 that (f u)u′ is integrable. The absolute continuity of F u leads via Theorem 5.8.15 to Z

b

FðuðbÞÞ  FðuðaÞÞ ¼

ðF uÞ0 dm:

a

By definition of F, we have Z FðuðbÞÞ  FðuðaÞÞ ¼

uðbÞ

f dm: uðaÞ

The above three equalities together yield the required conclusion.

h

290

5 Differentiation

Theorem 5.8.17 (Change of Variables Formula 2) Let f be an integrable function on [c, d] and u be an absolutely continuous function on [a, b] such that u[a, b] [c, d]. Suppose (f u)u′ is integrable on [a, b]. Then for any a, b 2 [a, b], Z

uðbÞ

Z

b

f dm ¼

ðf uÞu0 dm;

a

uðaÞ

and F u is absolutely continuous on [a, b], where Z

x

FðxÞ ¼

f dm: c

Proof If f is bounded, then F satisfies a Lipschitz condition and F u is absolutely continuous in view of Proposition 5.7.4. It follows from Theorem 5.8.16 that the desired equality of integrals holds in the case of bounded f. Now suppose f is not bounded. Consider the sequence {fn}n  1 of functions on [c, d] given by fn(x) = f(x) or −n or n according as |f(x)| < n or f(x)  − n or f(x)  n. Then each fn is bounded and measurable. By what has been observed in the paragraph above, we have Z

uðbÞ

uðaÞ

Z

b

fn dm ¼

ðfn uÞu0 dm

for each n:

a

Also, fn! f everywhere and (fn u )u′ ! (f u)u′ a.e. Besides, jfn j  j f j; jðfn uÞu0 j  jðf uÞu0 j

and both f and ðf uÞu0 are integrable:

Therefore, upon taking limits in the above equality of integrals and appealing to Problem 3-2.P20(a), we arrive at the required equality. From the definition of F and the equality that has just been proved, it follows for any t 2 [a, b] that Z

uðtÞ

ðF uÞðtÞ  ðF uÞðaÞ ¼

Z

t

f dm ¼

ðf uÞu0 dm:

a

uðaÞ

Since (f u)u′ has been assumed integrable, Proposition 5.8.1 shows that the function of t defined by the left-hand side must be absolutely continuous, and hence so is F u. h Corollary 5.8.18 Let f be an integrable function on [c, d] and u be an absolutely continuous function on [a, b] such that u[a, b]  [c, d]. Then the function F u, where Z FðxÞ ¼

x

f dm; c

5.8 Differentiation and the Integral

291

is absolutely continuous if and only if (f u)u′ is integrable. When this is so, Z

uðbÞ

Z f dm ¼

b

ðf uÞu0 dm

a

uðaÞ

for any a, b 2 [a, b]. h

Proof This is trivial from the previous two theorems.

Remark 5.8.19 The following example, based on E. J. McShane [25, p. 214], shows that the hypothesis in Theorem 5.8.16 that F u be absolutely continuous is not redundant. Let [a, b] = [c, d] = [0, 1] and f ðxÞ ¼ x3 ; f ð0Þ ¼ 0; 2

uðtÞ ¼ t3 cos6

and

p ; uð0Þ ¼ 0: t

Then F(x) = 3x1/3 and (F u)(t) = 3t cos2(pt ) = 32 t(1 + cos(2pt )). The function u is absolutely continuous because it has a bounded derivative; the function F u can be seen to be not absolutely continuous by modifying the argument of Example 5.6.3(c). Moreover, it follows from Theorem 5.8.17 (or from Corollary 5.8.18) that (f u)u′ is not integrable on [u(0), u(1)] = [0, 1], although f is integrable on [0, 1]. Corollary 5.8.20 Let c 6¼ 0 and let f be an integrable function on [a, b]. Then the function given by U(x) = f(x/c) on [ca, cb] or [cb, ca] (according as c > 0 or < 0) is integrable and Z c

b

Z f ðxÞdx ¼

cb

Z UðtÞdt ¼

ca

a

cb

f ðt=cÞdt:

ca

Proof Consider the absolutely continuous function u:[ca, cb] ! [a, b] or u:[cb, ca] ! [a, R x b], as the case may be, for which u(x) = x/c. The function F such that FðxÞ ¼ a f dm is absolutely continuous by Corollary 5.8.3 and hence one can deduce directly from Definition 5.7.1 that F u is also absolutely continuous. It follows from Theorem 5.8.16 that (f u)u′ is integrable on [ca, cb] or [cb, ca], as the case may be, and Z

b a

Z f dm ¼

cb ca

ðf uÞu0 dm:

Since f u = U and u′ = 1/c everywhere, this is precisely what was to be proved. h

292

5 Differentiation

Remark 5.8.21 When c > 0 and [a, b] = [− l, l], where l > 0, the equality asserted by the above corollary takes the form Z c

l l

Z f ðxÞdx ¼

cl

cl

Z UðtÞdt ¼

cl

cl

f ðt=cÞdt:

When c = − 1 and [a, b] = [− l, 0], where l > 0, it takes the form Z

0

l

Z f ðxÞdx ¼

l

Z UðxÞdx ¼

0

l

f ðxÞdx:

0

Problem Set 5.8 5.8.P1. Suppose f is absolutely continuous on [a, b] and f′(x) = 0 a.e. on [a, b]. Then f is a constant function. Give a proof of this that is different from the one in Corollary 5.8.4. 5.8.P2. If f and g are absolutely continuous on [a, b], f′(x) = g′(x) a.e. on [a, b] and f(x0) = g(x0) for some x0 2 [a, b], then show that f(x) = g(x) for every x 2 [a, b]. 5.8.P3. Let f:[a, b]!R be absolutely continuous and let vðaÞ ¼ 0; vðxÞ ¼ Vð½a; x; f Þ for x 2 ½a; b: Then v is absolutely continuous on [a, b] by Proposition 5.7.9. Show that v′(x) = |f′(x)| for almost all x 2 [a, b]. 5.8.P4. Show that, if g is an integrable function defined on [a, b] with indefinite integral f, then f′(x) R = g(x) whenever x is a point of continuity of g. 5.8.P5. If f(x) = ½0; x g, where g is an integrable function defined on [0, 1], then f′ = g need not hold even when f′ exists. R 5.8.P6. (Cf. Theorem 5.5.1) Show that the inequality ½a; x f′  f(b) − f(a) need not hold when f is not monotone increasing. 5.8.P7. (Lebesgue Decomposition Theorem for Functions) Suppose f is of bounded variation on [a, b]. Then there exists an absolutely continuous function g and a function h such that f(x) = g(x) + h(x), x 2 [a, b], where h′(x) = 0 a.e. Up to constants, the decomposition is unique. [For the relation between this and Theorem 5.10.13, see Problems 5.11.P5 and 5.11.P8.] 5.8.P8. [Application to Fourier Series; due to Lebesgue] Let f be a function in L1([−p, p]). Then show that lim rn ðxÞ ¼ f ðxÞ n

at every Lebesgue point x of f (and hence a.e. on [−p, p] by Theorem 5.8.9). Here the notation is as in Fejér’s Theorem 4.3.4.

5.8 Differentiation and the Integral

293

Definition Let f(x) = g(x) + ih(x) be a complex function on the closed bounded interval [a, b], where g(x) and h(x) are real. For each partition P: a = x0 < x1 < ⋯ < xn = b of [a, b], we set TðP; f Þ ¼

n X

kf ðxk Þ  f ðxk1 Þk;

k¼1

where ||f(x) − f(y)||2 = (g(x) − g(y))2 + (h(x) − h(y))2. The number Vð½a; b; f Þ ¼ sup TðP; f Þ P

is called the total variation of f over [a, b]. Additivity over adjacent intervals as in Theorem 5.6.4 can be established by making obvious modifications in the proof of that theorem. It will be used in Problems 5.8.P9 and 5.8.P10. 5.8.P9. Show that the complex function f = g + ih satisfies Vð½a; b; f Þ  Vð½a; b; gÞ þ Vð½a; b; hÞ; is of finite variation if and only if g and h are so, and is absolutely continuous if and only if g and h are so. Definition Let g and h be continuous real-valued functions on the closed bounded interval [a, b]. The set of points (g(x), h(x)), a  x  b, in R2 is called a continuous curve and the total variation V([a, b], f) of f(x) = g(x) + ih(x) is called the length of the curve. Similarly, if vðxÞ ¼ Vð½a; x; f Þ;

a  x  b;

then v(x) is called the length of the arc of the curve between the points (g(a), h(a)) and (g(x), h(x)). If the curve has finite length, it is called a rectifiable curve. 5.8.P10. Show that the curve given by g and h is rectifiable if and only if the functions are of bounded variation on [a, b]. Show that in this case, Z vðxÞ 

½a; x

h

ð g0 Þ þ ð h0 Þ 2

2

i12

for all x

and equality holds for all x if and only if g and h are absolutely continuous. In particular, if g is a function of bounded variation and v(x) is its total variation R 0 over [a, x], then v(x)  ½a;x jg j for all x. Equality holds for all x if and only if g is absolutely continuous.

294

5 Differentiation

5.8.P11. Let g be a real-valued function on [0, 1] and c(t) = t + ig(t). The length of the graph of g is, by definition, the total variation of c on [0, 1]. Show that the length is finite if and only if g is of bounded variation. 5.8.P12. Assume that g is continuous and increasing on [0, 1], g(0) = 0, g(1) = a. By Lg we denote the length of the graph of g. Show that Lg = 1 + a if and only if g′(x) = 0 a.e. In particular, the length of the graph of the Cantor function constructed in Example 5.5.2 is 2. Rp Rp 5.8.P13. Let f 2 L1[−p, p] and p f ðxÞ dx ¼ 0, p f ðxÞ cos nx dx ¼ 0, Rp p f ðxÞ sin nx dx ¼ 0, n = 1, 2,…. Then f(x) = 0 a.e. on [−p, p].

5.9

Signed Measures

In this section, we consider a simple but rather useful generalisation of the concept of measure. It arises naturally if the measure is allowed to assume both positive and negative values. Note that if l1 and l2 are measures defined on the same measurable space ðX; F Þ, then a1l1 + a2l2, where a1 and a2 are nonnegative constants, is again a measure on ðX; F Þ. However, the situation is different if we try to define a set function l by lðEÞ ¼ l1 ðEÞ  l2 ðEÞ for E 2 F : It is obvious that l is not always nonnegative. This is an interesting phenomenon which leads to the consideration of signed measures studied below. The difficulty, namely, that l may not be defined (l1(E) = l2(E) = ∞ for some E 2 F Þ is overcome by assuming that either l1 or l2 is a finite measure. In what follows, we shall study signed measures and their properties. The Hahn and Jordan Decomposition Theorems are discussed as well. The second of these says that a signed measure can only arise as a difference in the manner described above. We begin with the following. Definition 5.9.1 Let ðX; F Þ be a measurable space. By a signed measure we mean an extended real-valued set function l defined on the class F of all measurable sets (l:F !R ) of the measurable space ðX; F Þ which satisfies the following: (a) l assumes at most one of the values ∞ and −∞, (b) l(∅) = 0, (c) l is countably additive (as in Definition 2.2.7), that is, for every disjoint sequence of sets {Aj}j  1 such that each Aj 2 F , we have

5.9 Signed Measures

295

lð

1 [

Aj Þ ¼

j¼1

1 X

lðAj Þ:

j¼1

A signed measure l on (X, F ) is said to be finite if |l(X)| < ∞ and it is said to be r-finite if there exists a sequence of sets {Aj}j  1 such that each Aj 2 F , X = [ j  1Aj and |l(Aj)| < ∞ for every j. A signed measure which only takes nonnegative values is a measure in the sense of Chap. 3 and will sometimes be referred to in the context of signed measures as a nonnegative measure. In the rest of this section, l will be understood to be a signed measure on (X, F ) unless the context demands otherwise. Remarks 5.9.2 (a) If the series in part (c) of Definition 5.9.1 were to be conditionally convergent, then it would be possible to rearrange it to converge to an arbitrary desired sum; however, the assertion of (c) implies that rearranging it does not alter the sum. Thus, it follows from (c) that the series cannot be conditionally convergent, which is to say, it must be either absolutely convergent or properly divergent to ±∞. (b) A nonnegative measure is a special case of a signed measure but a signed measure is in general not a nonnegative measure. (c) It follows from the definition that a signed measure, like any nonnegative measure, is finitely additive. Let E = [ 1  j  nEj, Ej 2 F for every j and Ej \ Ei = ∅ whenever i 6¼ j. Then 1   n   P P E = [ j  1Ej with Ej = ∅ for i > n and lðEÞ ¼ l Ej ¼ l Ej . j¼1

j¼1

(d) If |l(E)| < ∞ and F  E, then |l(F)| < ∞, |l(E\F)| < ∞ and l(E \F) = l(E) − l(F). Indeed, the set-theoretic identities E = (E\F) [ F and (E \F) \ F = ∅, in conjunction with finite additivity, lead to l(E) = l(F) + l(E \F). Therefore, when |l(E)| < ∞, we must have |l(F)| < ∞ and |l(E\F)| < ∞. Moreover, it follows that l(E\F) = l(E) − l(F). (e) l is finite if and only if |l(E)| < ∞ for every E 2 F . Since X 2 F , it follows from this condition that |l(X)| < ∞. Conversely, if |l(X)| < ∞, it follows from (c) that |l(E)| < ∞ for every E 2 F . (f) If {Ej}j  1 is a sequence of sets in F satisfying E1  E2  …, then l( [ j  1Ej) = lim lðEn Þ . This follows by the same argument as Proposition n!1

2.3.21 (a). The reader is reminded that lim En = [ j  1Ej here. (g) If {Ej}j  1 is a sequence! of sets in F satisfying E1 E2 …, and T |l(E1)| < ∞, then l Ej ¼ lim lðEn Þ . This follows by the same arguj1

n!1

ment as Proposition 2.3.21 (b). The reader is reminded that lim En = \ j  1Ej here.

296

5 Differentiation

(h) If {Ej}j  1 is a sequence of disjoint sets in F such that |l( [ j  1Ej)| < ∞, then 1   P the series l Ej is absolutely convergent. This follows from (a) above. i¼1

(i) Suppose l is finite and {Ei}i2I is a family of disjoint sets in F . Then {i 2 I: l(Ei) 6¼ 0} is at most countable, as we show below. For each positive integer m, let Im ¼ fi 2 I : lðEi Þ [

1 g m

and

1 Im0 ¼ fi 2 I : lðEi Þ\  g: m

We assert that Im is finite. If not, then choose an infinite sequence i1, i2,… of 1   P distinct elements of Im. Since Eij are disjoint, l Eij converges. But this is a j¼1

contradiction because l(Eij ) > m1 for all j. By a similar argument, Im′ is also finite. It follows that [ m  1Im and [ m  1Im′ are at most countable and hence so is their union. However, their union is the set we wanted to prove to be at most countable. In particular, when l(Ei) 6¼ 0 for every i 2 I, we can conclude that I is at most countable. (j) If l(E) = ∞ [resp. −∞] for some set E 2 F , then l(X) = ∞ [resp. −∞]. This follows from finite additivity: l(X) = l(E) + l(X\E). (k) When l1 and l2 are signed measures, the sum l1(E) + l2(E) fails to be defined if and only if l1(E) = −l 2(E) = ± ∞. By (j) above, this happens if and only if l1(X) = −l 2(X) = ±∞. When this situation does not occur, we have a well-defined set function l1 + l2 on F . We claim that l1 + l2 will be a signed measure. That (l1 + l2)(∅) = 0 is trivial. If l1 is finite, then l1 + l2 fails to assume whichever of the values ±∞ that l2 fails to assume. Suppose l1 is not finite. If it fails to assume the value ∞, then it must take the value −∞ and therefore l1(X) = −∞ by (j) above. It follows that l2(X) 6¼ ∞ and hence, by (j) again, l2 also fails to assume the value ∞. It follows that l1 + l2 fails to assume the value ∞. An analogous argument shows that if l1 fails to assume the value −∞, then l1 + l2 does the same. Thus, whether l1 is finite or not, l1 + l2 fails to assume one among the values ± ∞. Lastly, countable additivity of l1 + l2 follows from the fact that two series can be added term by term except when one diverges to ∞ and the other one to −∞. The condition that l1 + l2 is well defined will be described by saying that l1 + l2 is a signed measure. Examples 5.9.3 (a) Let l1 and l2 be signed measures on F such that at least one of them is finite. Then the set function l defined by l0 ðEÞ ¼ l1 ðEÞ  l2 ðEÞ is a signed measure on F .

5.9 Signed Measures

297

Clearly, l0(∅) = l1(∅) − l2(∅) = 0. Let E = [ j  1Ej, where the Ej are dis1 1     P P joint elements of F . Then l1(E) = l1 Ej and l2(E) = l2 Ej Suppose j¼1

j¼1

l1 is finite, so that l1(A) < ∞ for every A 2 F . If l2(E) < ∞ as well, then the 1      P series l1 Ej  l2 Ej is absolutely convergent to l1(E) − l2(E). Hence j¼1

l0 ðEÞ ¼ l1 ðEÞ  l2 ðEÞ ¼

1 X

ðl1 ðEj Þ  l2 ðEj ÞÞ ¼

j¼1

If l2(E) = ± ∞, then

1 X

l0 ðEj Þ:

j¼1

1      P l1 Ej  l2 Ej is divergent to −l2(E) = j¼1

l1(E) − l2(E). Thus l0 is a signed measure on F when l1 is finite. The case when l2 is finite is argued analogously. If l1 an l2 are both finite signed measures, then |l1(X)| < ∞ and|l2(X)| < ∞ and hence l0 is a finite signed measure. Similarly, if one among l1 and l2 is finite and the other is r-finite, then l0 is r-finite. (b) Let m be a nonnegative measure on F and fRbe an extended meaR real-valued þ  surable function on X such that one among X f dm and X f dm is finite, so R that X fdm is defined as an extended real number in accordance with Definition 3.2.9. Set Z f dm for E 2 F : lðEÞ ¼ E

Then R l is a signed measureR on F , as we now show: R If X f þ dm is finite, then E f þ dm is finite for every E 2 F , so that E f dm R  R  cannot be ∞ for any E 2 F . Similarly, if E f dm is finite, then E f dm is finite for R every E 2 F , so that E f dm cannot be −∞ for any E 2 F . Thus, (a) of Definition 5.9.1 holds. It is trivial that (b) also holds. Let {Ej}j  1 be a sequence of disjoint sets in F . For E 2 F , write Z Z þ þ  l ðEÞ ¼ f dm and l ðEÞ ¼ f  dm: E

E

It follows by using Corollary 3.2.6 that l+ and l− are measures on F . Moreover, l = l+ − l−. Therefore lð

1 [ j¼1

Ej Þ ¼ l þ ð

1 [ j¼1

Ej Þ  l ð

1 [

Ej Þ ¼

j¼1

as at no stage will “∞ − ∞” occur.

1 X j¼1

l þ ðEj Þ 

1 X j¼1

l ðEj Þ ¼

1 X j¼1

lðEj Þ;

298

5 Differentiation

Definition 5.9.4 A set A  X is a positive set (for l) if A is measurable and every measurable subset E  A satisfies l(E)  0. Similarly, a set B  X is a negative set (for l) if B is measurable and every measurable subset E  B satisfies l(E)  0. Note that every measurable subset of a positive set is a positive set. Let lA be defined on F by lA(E) = l(A \ E) for all E 2 F . Then lA is a nonnegative measure. Indeed, lA(∅) = l(∅ \ A) = l(∅) = 0, and if {Ej}j  1 is a sequence of disjoint sets in F , then lA ð

1 [

Ej Þ ¼ lðð

j¼1

1 [

Ej Þ \ AÞ ¼ lð

j¼1

1 [

ðEj \ AÞÞ ¼

j¼1

1 X

lðEj \ AÞ ¼

j¼1

1 X

lA ðEj Þ:

j¼1

A set B is a negative set for l if it is a positive set for −l. Every measurable subset of a negative set is again a negative set. Finally, A is a null set for l (or a l-null set) if it is a positive as well as a negative set for l. Equivalently, A is a null set if A 2 F and l(E) = 0 for every E 2 F such that E  A. Remarks 5.9.5 (a) The empty set is a null set for every signed measure. (b) A null set has measure zero. However, a set of measure zero may well be a union of two sets whose measures are not zero but are negatives of each other. (c) The condition l(A)  0 [resp. l(A)  0] is necessary but not in general sufficient for A to be a positive [resp. negative] set. The main result is the Hahn Decomposition Theorem 5.9.10, which states that X may be split into two disjoint measurable sets such that the respective restrictions of l and −l are nonnegative measures. The following results will be used to prove the main result. Lemma 5.9.6 The union of a countable or finite collection of positive sets is a positive set. Proof Let A be the union of a sequence {Aj}j  1 of positive sets. If E 2 F and E  A, set E1 ¼ E \ A1

and

Ej ¼ E \ Aj \ Acj1 \    \ Ac1 for j [ 1:

Then Ej is a measurable subset of Aj and so l(Ej)  0. Since the sets Ej are disjoint and E = [ j  1Ej, we have lðEÞ ¼

1 X

lðEj Þ  0:

j¼1

Thus A is a positive set. A corresponding argument applies to a finite collection of sets.

h

5.9 Signed Measures

299

Corollary 5.9.7 A countable union of negative [resp. null] sets is a negative [resp. null] set. Proposition 5.9.8 Suppose that E 2 F and 0 < l(E) < ∞. Then there exists an A  E such that A is a positive set and l(A) > 0. Proof If E contains no set of negative l-measure, then E is a positive set and A = E gives the result. Suppose no set A of the required sort exists. In particular, E is not a positive set. Let n1 be the smallest positive integer for which there exists a measurable set E1  E with l(E1) <  n11 . Then we have lðE \ E1c Þ ¼ lðEÞ  lðE1 Þ [ lðEÞ [ 0 and so by our assumption, E \ Ec1 is not a positive set for l. As before, let n2 be the smallest positive integer for which there exists a measurable set E2  E \ Ec1 with l(E2) <  n12 . Then lðE \ ðE1 [ E2 Þc Þ ¼ lðEÞ  lðE1 Þ  lðE2 Þ [ 0: Continuing this process, we obtain a sequence {nk}k  1 of minimal positive integers and a corresponding sequence {Ek}k  1 of sets in F such that E1  E and k1 S Ek  E \ ( Ej)c for k > 1and l(Ek) <  n1k for each k 2 N. Note that the property j¼1 k1 S

Ek  E \ (

Ej)c for k > 1 makes the Ek disjoint.

j¼1

If we set A = E\ [ j  1Ej, then E = A [ ( [ j  1Ej). Observe that 1 1 1 [ X X 1 1 [ lðAÞ ¼ lðEÞ  lð Ej Þ ¼ lðEÞ  lðEj Þ [ lðEÞ þ [ 0: n j¼1 j¼1 j¼1 j

In particular,

1 P 1 i¼1

nj \1 jand

hence nj! ∞ as j ! ∞; furthermore, A is not a

positive set. Choose F  A such that l(F) < 0 and choose k > 1 so large that l(F) <  n1k and also nk > 2. Then, k1 [

F [ Ek E \ ð

j¼1

and hence

Ej Þc

300

5 Differentiation

lðF [ Ek Þ ¼ lðFÞ þ lðEk Þ\  \

1 1  nk nk

1 ; as nk [ 2: nk  1

But this contradicts the minimality of nk. Consequently, a set A of the required sort exists. h Definition 5.9.9 A Hahn decomposition for a signed measure l on (X, F ) is an ordered pair ⟨A, B⟩ 2 F F such that A [ B = X, A \ B = ∅, A is a positive set for l and B is a negative set for l. It is not required in the foregoing definition that A or B should be nonempty. In fact, if l is a nonnegative measure, then ⟨X, ∅⟩ is a Hahn decomposition. Theorem 5.9.10 (Hahn Decomposition Theorem) Let l be a signed measure on (X, F ). Then l has a Hahn decomposition ⟨A, B⟩. It is unique in the sense that if ⟨A1, B1⟩ and ⟨A2, B2⟩ are both Hahn decompositions for l, then l(A1 \ E) = l(A2 \ E) and l(B1 \ E) = l(B2 \ E) for any E 2 F . Proof Without loss of generality, we may assume that l never takes the value ∞, for otherwise, we may switch to −l and interchange the rôles of positive and negative sets. Let k = sup{l(C) : C is a positive set for l}. Since the empty set is positive, k  0. Choose a sequence {An}n  1 of positive sets such that lim lðAn Þ ¼ k and n!1

define A = [ n  1An. By Lemma 5.9.6, A is itself a positive set and so k  l(A). But A\An  A and therefore l(A\An)  0 for each n. Thus, l(A) = l(An) + l(A \An)  l(An) for each n, so that lðAÞ  lim An ¼ k . It follows that k < ∞. n!1

Let B = Ac. Consider any positive subset E of B. Then E \ A = ∅ and E [ A is a positive set. Hence k  lðE [ AÞ ¼ lðEÞ þ lðAÞ ¼ lðEÞ þ k; which implies l(E)  0, because 0  k < ∞. Thus B contains no positive set of positive measure and hence contains no subset of positive measure by Proposition 5.9.8. Consequently, B is a negative set, so that ⟨A, B⟩ is a Hahn decomposition for l. Now suppose ⟨A1, B1⟩ and ⟨A2, B2⟩ are both Hahn decompositions for l. Let E 2 F be arbitrary. Since E \ A1 \ B2 is a subset of both A1 and B2, which are positive and negative sets for l, we have l(E \ A1 \ B2) = 0. Similarly, l(E \ A2 \ B1) = 0. Observe that E \ A1 ¼ E \ A1 \ X ¼ E \ A1 \ ðA2 [ B2 Þ ¼ ðE \ A1 \ A2 Þ [ ðE \ A1 \ B2 Þ and

5.9 Signed Measures

301

ðE \ A1 \ A2 Þ \ ðE \ A1 \ B2 ÞA2 \ B2 ¼ £: Since l(E \ A1 \ B2) = 0, this implies lðE \ A1 Þ ¼ lðE \ A1 \ A2 Þ: Similarly, lðE \ A2 Þ ¼ lðE \ A1 \ A2 Þ: Consequently, l(E \ A1) = l(E \ A2) and a similar argument shows that l(E \ B1) = l(E \ B2). This completes the proof. h Remark 5.9.11 A Hahn decomposition need not be unique in the strict sense and there may actually be uncountably many Hahn decompositions. Indeed, any null set can be taken either as a part of A or of B. The following example illustrates this. R Let X be the interval [0, 3] with Lebesgue measure m and lðEÞ ¼ E f dm, where the function f:[0, 3]!R is −v[0,1] + v[2,3]. Then f vanishes on (1, 2), so that l(E) = 0 for every measurable E  (1, 2), which is to say, (1, 2) is a null set. In particular, for an arbitrary a 2 (1, 2) we have f  0 on [a, 3] and f  0 on [0, a], whence it follows that ⟨[a, 3], [0, a]⟩ is a Hahn decomposition for l whenever a 2 (1, 2). This exhibits uncountably many Hahn decompositions. R 2 Example 5.9.12 Let l(E) ¼ E xex dx, E 2 M. We shall determine the positive, negative and null sets for l. Z lðEÞ ¼

xex dx þ 2

E \ ð0;1Þ

Z

xex dx: 2

E \ ð1;0Þ

Observe that E is a positive [resp. negative] set for l if and only if m(E \ (−∞, 0)) = 0 [resp. m(E \ (0, ∞)) = 0] and E is a null set if and only if m(E) = 0. The sets [0, ∞) and (−∞, 0) form one of the uncountably many possible Hahn decompositions for l. A Hahn decomposition for a signed measure l helps to obtain a decomposition of l as a difference of measures. The idea of decomposing a signed measure into a difference of nonnegative measures is attributed to H. Lebesgue. However, earlier C. Jordan obtained a decomposition of a real function of bounded variation as a difference of increasing functions, which is a precursor to the decomposition of signed measures. Hence the name Jordan Decomposition Theorem (see Theorem 5.9.16). Definition 5.9.13 Two nonnegative measures l1 and l2 on (X, F ) are said to be mutually singular [in symbols: l1 ⊥ l2] if there are disjoint measurable sets A, B with X = A [ B such that l2(A) = 0 = l1(B).

302

5 Differentiation

Remark 5.9.14 (a) In the above definition, one need not take A and B to be disjoint. If they were not disjoint, then A\B and B would constitute a pair of disjoint sets with the same properties. (b) If l1, l2 and m are nonnegative measures on (X, F ) and li ⊥ m for i = 1, 2, then l1 + l2 ⊥ m. Indeed, there exists sets Ai, Bi 2 F satisfying Ai [ Bi = X such that m(Ai) = 0 = li(Bi) for i = 1, 2. Then m(A1 [ A2)  m(A1) + m(A2) = 0 and ðl1 þ l2 ÞðB1 \ B2 Þ ¼ l1 ðB1 \ B2 Þ þ l2 ðB1 \ B2 Þ  l1 ðB1 Þ þ l2 ðB2 Þ ¼ 0: Moreover, ðA1 [ A2 Þ [ ðB1 \ B2 Þ ¼ ððA1 [ A2 Þ [ B1 Þ \ ððA1 [ A2 Þ [ B2 Þ ¼ X: Example 5.9.15 Let m be a nonnegative measure on (X, F ) and A, B be measurable sets such that m(A \ B) = 0. Define m1(E) = m(A \ E) and m2(E) = m(B \ E) for every E 2 F . Then m1 ⊥ m2. Indeed, m1(B) = m(A \ B) = 0 and m2(Bc) = m(B \ Bc) = m(∅) = 0. Theorem 5.9.16 (Jordan Decomposition Theorem) Let l be a signed measure on (X, F ). Then there exist nonnegative measures l+ and l− on (X, F ) such that l = l+ − l− and l+ ⊥ l−. Moreover, the decomposition is unique. Proof Let ⟨A, B⟩ be a Hahn decomposition for l. Define l+ and l− by l þ ðEÞ ¼ lðE \ AÞ

and l ðEÞ ¼ lðE \ BÞ

for E 2 F :

It follows from the paragraph after Definition 5.9.4 that l+ and l− are nonnegative measures on (X, F ). Moreover l+(B) = l(B \ A) = 0 and l−(A) = −l(B \ A) = 0. So, l+ ⊥ l−. Also, for E 2 F , lðEÞ ¼ lðE \ XÞ ¼ lðE \ AÞ þ lðE \ BÞ ¼ l þ ðEÞ  l ðEÞ: Hence l = l+ − l−. It remains to show that the decomposition is unique. Let l = l1 − l2 with l1 ⊥ l2 be another decomposition of l. Let A′, B′ 2 F be such that l1(B′) = 0 = l2(A′) with A′ \ B′ = ∅ and A′ [ B′ = X. If D 2 F and D  A′, then lðDÞ ¼ l1 ðDÞ  l2 ðDÞ ¼ l1 ðDÞ:

ð5:85Þ

Also, if D 2 F and D  B′, then l1 ðDÞ ¼ 0:

ð5:86Þ

5.9 Signed Measures

303

Since l1(D)  0, it follows from (5.85) that A′ is a positive set for l. Similarly, the analogues of (5.85) and (5.86) hold for l2. Since l2(D)  0, it follows from the analogue of (5.85) that B′ is a negative set for l. Thus ⟨A′, B′⟩ is a Hahn decomposition for l. By Theorem 5.9.10, it follows that for E 2 F , we have l þ ðEÞ ¼ lðE \ AÞ ¼ lðE \ A0 Þ ¼ l1 ðE \ A0 Þ by (5:85Þ ¼ l1 ðE \ A0 Þ þ l1 ðE \ B0 Þ by (5:86Þ ¼ l1 ðEÞ: A similar argument using the analogues of (5.85) and (5.86) leads to l−(E) = l2(E). h Remark Since l+ and l− are nonnegative measures, their sum |l| = l+ + l− is also a nonnegative measure. It satisfies |l|(E) = l+(E) + l−(E)  |l+(E) − l−(E)| = |l(E)| for every E 2 F . Definition 5.9.17 The decomposition of l given by l = l+ − l− as in Theorem 5.9.16 is called the Jordan decomposition of l. The measures l+ and l− are called the positive variation and negative variation respectively of l. The measure |l| = l+ + l− is called the total variation of l. Example 5.9.18 Let (X, F , m) be a measure space, with m a nonnegative measure let f be anRextended real-valued measurable function on X such that one among Rand þ  X f dm and X f dm is finite. Define a set function l on F by Z lðEÞ ¼

f dm for every E 2 F : E

As seen in Example 5.9.3 (b), l is a signed measure. Let A = {x 2 X: f(x)  0} and B = {x 2 X: f(x) < 0}. Then ⟨A, B⟩ can be shown to be a Hahn decomposition of l. Indeed, A [ B = X, A \ B = ∅ and for E 2 F such that E  A, we have l(E)  0, i.e., A is a positive set for l. Similarly, B is a negative set for l. We next show that l+ and l− given by Z Z f þ dm and l ðEÞ ¼ f  dm for E 2 F l þ ðEÞ ¼ E

E

form the Jordan decomposition of l. This is because clearly, for any E 2 F , Z lðEÞ ¼

Z f dm ¼

E

þ

Z

f  dm ¼ l þ ðEÞ  l ðEÞ

f dm  E

E

304

5 Differentiation

and l þ ðBÞ ¼

Z

f þ dm ¼ 0;

l ðAÞ ¼

B

Z

f  dm ¼ 0:

A

Furthermore, |l| is given by þ

Z



jljðEÞ ¼ l ðEÞ þ l ðEÞ ¼

Z

þ

f dm þ E

Z



f dm ¼ E

j f jdm

for E 2 F :

E

We note from the above definition of l+ and l− that l þ ðEÞ ¼

Z

f þ dm ¼ E Z Z   f dm ¼ l ðEÞ ¼ E

Z E\A

E\B

f dm ¼ lðE \ AÞ;

f dm ¼ lðE \ BÞ;

which is an alternative way to see that l+ and l− form the Jordan decomposition of l, keeping in view how the decomposition was obtained in the proof of the Jordan Decomposition Theorem 5.9.16. Remark Define integration with respect to a signed measure l by Z Z Z þ f dl ¼ f dl  f dl ; X

X

X

where l+ and l− form the Jordan decomposition of l. Then for E 2 F , we have

Z

Z

Z







f dl  f dl þ þ f dl :







E

E

E

If |f|  M on X, then

Z

Z Z



þ

f dl  j f jdl þ j f jdl



E E E Z Z  Mð dl þ þ dl Þ E

E

 Mðl þ ðEÞ þ l ðEÞÞ ¼ M jljðEÞ: Problem Set 5.9 5.9.P1. Let l be a signed measure on (X, F ). Show that n

X

lðEj Þ ; jljðEÞ ¼ sup j¼1

5.9 Signed Measures

305

where the sets Ej 2 F are disjoint and E = [ 1  j  nEj. 5.9.P2. If l = l1 − l2, where l1 and l2 are nonnegative measures and either l1 or l2 is finite, then l1  l+ and l2  l−, where l = l+ − l− is the Jordan decomposition of l. 5.9.P3. Let (X, F , l) be a finite signed measure space. Show that there is an M > 0 such that |l(E)| < M for every E 2 F . 5.9.P4. Show that a signed measure l on (X,F ) is finite [resp. r-finite] if and only if |l| is finite [resp. r-finite] if and only if l+ and l− are both finite [resp. r-finite]. 5.9.P5. Let l be a signed measure l on (X,F ). Then, for every E 2 F , the following hold: (a) l+(E) = sup{l(F): F  E, F 2 F }; (b) l_(E) = sup{−l(F): F  E, F 2 F }. 5.9.P6. Let l1 and l2 be nonnegative measures on (X, F ) such that for all a > 0 and b > 0, there exist sets Aa,b, Ba,b 2 F satisfying Aa,b [ Ba,b = X, l1(Aa,b) < a, l2(Ba,b) < b. Show that l1 ⊥ l2.

5.10

The Radon–Nikodým Theorem

In this section, a measure will be understood to be nonnegative unless specified as being a signed measure. Let (X, F , l) be a measure space and f be an extended real-valued measurable function on X. For E 2 F , let Z mðEÞ ¼

f dl

ð5:87Þ

E

R R provided one among X f þ dm and X f  dm is finite, so that the integral on the right-hand side is meaningful. Then m is a signed measure on X [see Example 5.9.3 (b)]. This seems to be a natural generalisation of the indefinite integral, as it leads to such theorems as the Radon–Nikodým Theorem 5.10.8 and the Lebesgue Decomposition Theorem 5.10.13. It is desired to determine when a signed measure is expressible as an integral, as is the case with m in (5.87). Here the essential concept is absolute continuity and the main result is the Radon–Nikodým Theorem. We begin with the definition of absolute continuity. Definition 5.10.1 Let (X, F ) be a measurable space and l,m be signed measures on X. We say that m is absolutely continuous with respect to l, in symbols, m  l if m(E) = 0 for every E 2 F such that |l|(E) = 0.

306

5 Differentiation

Remark 5.10.2 In view of the obvious fact that the total variation of |l| is nothing but |l|, absolute continuity with respect to |l| is the same thing as absolute continuity with respect to l. Examples 5.10.3 (a) Let (X, F ,l) be a measure space and f be an extended nonnegative real-valued measurable function on X. Define Z mðEÞ ¼ f dl; E 2 F: E

It is an immediate consequence of Corollary 3.2.6 that m is a countably additive measure on (X, F ). Moreover, for all E 2 F , l(E) = 0 implies m(E) = 0, i.e., a l-null set is also a m-null set. Consequently, m  l. (b) Let (R, M, m) be the Lebesgue measure space and l be the counting measure on M. If l(E) = 0, then E = ∅ and hence m(E) = 0. This shows that m  l. (c) Let (R, M, m) be the Lebesgue measure space and m:M![0, ∞] be defined by m(∅) = 0 and m(E) = ∞ for E 2 M and E 6¼ ∅. Clearly, m  m. P 2n and (d) Let X = N and F ¼ P ðNÞ. For E 2 F , define lðEÞ ¼ n2E P n 2 . Obviously, l and m are measures on P ðNÞ. Observe that mðEÞ ¼ n2E

m(E) = 0 if and only if l(E) = 0 if and only if E = ∅. This shows that m  l and also l  m. Proposition 5.10.4 Let l and m be signed measures on a measurable space (X,F ). Then the following conditions are equivalent: ðaÞ m  l

ðbÞ

m þ  l; m  l

ðcÞ jmj  jlj

ðdÞjmj  l:

Proof From Remark 5.10.2, we know that m  l if and only if m  |l|. So, without loss of generality, we may assume that l  0. Let (a) hold and X = A [ B be a Hahn decomposition for m. Then, whenever l(E) = 0, we have 0  lðE \ AÞ  lðEÞ ¼ 0 and 0  lðE \ BÞ  lðEÞ ¼ 0; and therefore m þ ðEÞ ¼ mðE \ AÞ ¼ 0;

m ðEÞ ¼ mðE \ BÞ ¼ 0:

This proves that (a) ) (b). The implications (b) ) (c) and (c) ) (a) follow from the relations

5.10

The Radon–Nikodým Theorem

jmjðEÞ ¼ m þ ðEÞ þ m ðEÞ

307

and

0  jmðEÞj  jmjðEÞ

(for the latter, see the Remark preceding Definition 5.9.17). Finally (c) , (d) by Remark 5.10.2. h The assertions in the proposition below follow from the definitions of absolute continuity and mutual singularity. (Two signed measures k1 and k2 on F are said to be mutually singular, and we write k1 ⊥ k2, if there exists disjoint measurable sets A and B such that A [ B = X and |k1|(A) = 0 = |k2|(B). Since |(|l|)| = |l| always, the relation k1 ⊥ k2 is equivalent to |k1| ⊥ k2, k1 ⊥ |k2| and to |k1| ⊥ |k2|. It is not required that A or B must be nonempty; consequently, the zero measure and any signed measure are always mutually singular.) Proposition 5.10.5 Let l1, l2- and l3 be signed measures on (X, F ). Then (a) If l1  l2 and l2  l3, then l1  l3. (b) If l1  l3 and l2  l3 and if l1 + l2 is a signed measure, then l1 + l2  l3. (c) If l1  l3 and l2⊥l3, then l1⊥l2. (d) If l1  l2 and l1⊥l2, then l1 = 0.

Proof (a) Since l2  l3, we know from Proposition 5.10.4 [the implication (a) ) (d) therein] that |l2|  l3. Let E 2 F be such that |l3|(E) = 0. Since |l2|  l3, we have |l2|(E) = 0. The hypothesis that l1  l2 now yields l1(E) = 0. (b) Obvious. (c) Since l2 ⊥ l3, there are disjoint sets A and B in F such that A [ B = X and |l3|(A) = 0 = |l2|(B). Since l1  l3, we have |l1|  l3 by Proposition 5.10.4 [the implication (a) ) (d) therein] and hence |l1|(E) = 0 for every E 2 F for which |l3|(E) = 0. In particular, |l1|(A) = 0. We have thus proved that there are disjoint sets A and B in F satisfying A [ B = X and |l1|(A) = 0 = |l2|(B), i.e., l1 ⊥ l2. (d) By (c), the hypothesis of (d) implies l1 ⊥ l1 and this clearly implies l1 = 0. h The next proposition essentially says, if m is a finite signed measure, then “m l” is equivalent to “m is small when l is small”. Proposition 5.10.6 Let m be a finite signed measure on (X, F ). Then the following conditions are equivalent: (a) m  l; (b) to every e > 0, there corresponds a d > 0 such that |m(E)| < e for every measurable set E for which |l|(E) < d. Proof Assume (b). Consider any F 2 F such that |l|(F) = 0. To show that m(F) = 0, take an arbitrary e > 0. Then there exists a d > 0 such that |l|(E) < d

308

5 Differentiation

implies |m(E)| < e for any E 2 F . Since |l|(F) = 0, we have |l|(F) < d. It follows that |m(F)| < e. Since e is arbitrary, we conclude that m(F) = 0. Thus, (b) implies (a). Now suppose (b) is false. Then there exists an e > 0 and sets En 2 F , n = 1, 2, … such that |l|(En) < 2−n but |m(En)|  e. Hence |m|(En)  e. Put An ¼

1 [

Ei ;

A¼

i¼n

1 \

An :

n¼1

Then jljðAn Þ 

1 X

jljðEi Þ\2n þ 1 ;

An An þ 1

i¼n

and so, Proposition 3.1.8 shows that |l|(A) = 0 and that, in view of the finiteness of m, jmjðAÞ ¼ lim jmjðAn Þ  e [ 0 n!1

since |m|(An)  |m|(En). It follows that we do not have, |m|  l and hence by Proposition 5.10.4 that we do not have m  l, which means (a) is false. h As noted in Example 5.9.3 (b), if (X, F , l) is a measure Rspace and f anR extended real-valued measurable function on X such that one among X f þ dl and X f  dl is finite, then the set function m defined on F by Z mðEÞ ¼

f dl;

E2F

ð5:88Þ

E

is a signed measure on F . It is clear that m  l. The Radon–Nikodým Theorem asserts that if m  l and both are r-finite, then m is of the form (5.88), where f is real-valued. The following lemma will be needed in the proof of the Radon–Nikodým Theorem. Lemma 5.10.7 Let l and m be finite measures on (X, F ) such that m  l and m is not identically zero. Then there exists a set E 2 F such that l(E) > 0 and an e > 0 such that m(F) − el(F)  0 for every F  E, F 2 F . Proof For every n 2 N, consider the signed measure m  1n l. Let X = An [ Bn be a Hahn decomposition for the measure m  1n l, n = 1, 2, … . Then for every n 2 N, (m  1n l)(Bn)  0 and (m  1n)(An)  0. Let

5.10

The Radon–Nikodým Theorem

A0 ¼

309

1 [

and B0 ¼

An

n¼1

1 \

Bn :

n¼1

Then x 62 A0 implies x 62 An for every n, which further implies x 2 Bn for every n, i.e., x 2 B0. Thus, X = A0 [ B0. Moreover, since B0  Bn, we have 1 1 0  mðB0 Þ  mðBn Þ  lðBn Þ  lðXÞ: n n Hence m(B0) = 0. Since X = A0 [ B0, we get m(A0) = m(X) > 0. Thus, m (An0 ) > 0 for some n0 2 N. Since m  l, we also have l (An0 ) > 0. Also, An0 is a positive set for v  n10 l . On choosing e ¼ n10 and E = An0 , the required conclusion is seen to hold. h Theorem 5.10.8 (Radon–Nikodým Theorem). Let (X,F ,l) be a r-finite measure space and m be a r-finite signed measure on F such that m  l. Then there exists a real-valued measurable function f on X such that Z mðEÞ ¼

f dl;

E 2 F:

E

The function f is unique in the sense R that if g is any extended real-valued measurable function on X with mðEÞ ¼ E g dl for all E 2 F , then f = g a.e.[l]. Proof The uniqueness is easily seen from Problem 3.2.P14(a). We proceed to prove the existence in three steps. Step 1. Assume that both l and m are finite measures. Let K be the class of nonnegative measurable functions f integrable with respect to l such that R f dl  m(E) for every E 2 F . Observe that K is nonempty, as 0 2 K. Let E Z a ¼ supf

f dl : f 2 Kg:

ð5:89Þ

X

Since m(X) < ∞, it follows that 0  a < ∞. Let {fn}n  1 be a sequence in K such that Z fn dl ¼ a:

lim

n!1

X

Let gn = max{f1, f2, … ,fn}. Clearly, gn  0 and gn is integrable with respect to l. We shall show that gn 2 K. Let E 2 F be fixed and let

310

5 Differentiation

E1 ¼ fx : gn ðxÞ ¼ f1 ðxÞg \ E; E2 ¼ fx : gn ðxÞ ¼ f2 ðxÞg \ ðEnE1 Þ; ... n1 S En ¼ fx : gn ðxÞ ¼ fn ðxÞg \ ðEn Ek Þ: k¼1

Then E = [ 1  k  nEk, the sets Ek are disjoint and belong to F . Besides, gn(x) = fj(x) on Ej, j = 1, 2,…,n. Consequently, we have Z gn dl ¼

n Z X

E

fj dl 

Ej

j¼1

n X

mðEj Þ ¼ mðEÞ:

ð5:90Þ

j¼1

RTherefore gn 2 K. Now, let g = supnfn. Clearly, gn  gn+1 and gn! g. Since g is integrable. By the Monotone Convergence E gn dl  mðXÞ,R we see that R Theorem 3.2.4, E gn dl ! E g dl and, by (5.90), it follows that g 2 K, i.e., Z g dl  mðEÞ;

E 2 F:

E

Since fn  gn, we see that a ¼

R

g dl . We define

X

f ðxÞ ¼

gðxÞ 0

if gðxÞ\1 if gðxÞ ¼ 1:

R R Then f(x) = g(x) a.e.[l], so that f is integrable and E f dl ¼ E g dl for all E 2 F . R Note that it follows that f 2 K. We shall prove that if m0(E) = m(E) − E f dl, then the measure m0 is identically zero. This will complete Step 1. Suppose, if possible, that m0 is not identically zero. Observe that m0 is a finite measure and m0  l. Using Lemma 5.10.7, we obtain an e > 0 and an F 2 F such that l(F) > 0, and for every set E  F, E 2 F , we have m0(E) − el(E)  0, i.e., Z mðEÞ 

f dl  elðEÞ  0: E

In particular, Z elðE \ FÞ  mðE \ FÞ  If h = f + evF, then

f dl; E\F

E 2 F:

5.10

The Radon–Nikodým Theorem

Z

311

Z h dl ¼ ZE

E

f dl þ elðE \ FÞ f dl þ mðE \ FÞ  mðEnFÞ þ mðE \ FÞ;

 EnF

using the fact that f 2 K. Consequently, Z h dl  mðEnFÞ þ mðE \ FÞ ¼ mðEÞ E

for every measurable set E, so that h 2 K. However, since Z Z h dl ¼ f dl þ elðFÞ [ a; X

X

equality (5.89) is violated, showing that m0 is identically zero, and Step 1 is therefore complete. Step 2. Assume that l and m are r-finite measures on (X, F ). With the assumption of r-finiteness, we can write X = [ j  1Aj, l(Aj) < ∞ and X = [ k  1Bk, m(Bk) < ∞ and {Aj}, {Bk} may be supposed to be sequences of disjoint sets. So, setting X = [ k  1,j  1(Aj \ Bk), we obtain X as the union of disjoint sets on which l and m are finite, say X = [ n  1Xn. Let F n = {E \ Xn : E 2 F }, a r-algebra in Xn. Considering R l and m restricted to F n , we obtain fn such that every E 2 F n satisfies mðEÞ ¼ E fn dl. So, if E 2 F , we have E = [ n  1En, where En = E \ Xn 2 F n , and therefore upon defining f = fn on Xn, we obtain a measurable function on X and 1 R R P mðEÞ ¼ En fn dl ¼ E f dl . This completes Step 2. n¼1

Step 3. Now suppose that l is a r-finite measure on (X, F ) and m a r-finite signed measure. Let m = m+ − m− be the Jordan decomposition of m. Since |m|  l by Proposition 5.10.4, the nonnegative measures m+ and m−, which are obviously both r-finite, satisfy m+l and m−  l. By the result of Step 2, there exist nonnegative real-valued measurable f1 and f2 such that Z Z f1 dl; v ðEÞ ¼ f2 dl v þ ðEÞ ¼ E

E

for every E 2 F . Since f1 and f2 are real-valued, their difference f = f1 − f2 is also real-valued and we claim that the inequalities f+  f1 and f−  f2 hold. Consider any x 2 X. If f(x)  0 then f1(x) = f(x) + f2(x)  f(x) = f+(x), while if f(x)  0 then f1(x)  0 = f+(x). Hence f+  f1 for all x 2 X. We deduce from here that f2(x) = f1(x) − f(x)  f+(x) − f(x) = f−(x) for all x 2 X. This justifies our claim.

312

5 Differentiation

R R Now, either m+(X) < ∞ or m−(X) < ∞, so that either X f1 dl\1 or X f2 dl\1. In conjunction with the inequalities proved in the preceding paragraph, R þ this implies that the function f = f1 − f2 has the property that either X f dl\1 or R R  (though not necessarily finite). X f dl\1, and hence thatR E f dl isRwell defined R Consequently, the equality E f dl ¼ E f1 dl  E f2 dl holds for every E 2 F , which validates the computation Z Z þ  f1 dl  f2 dl mðEÞ ¼ m ðEÞ  m ðEÞ ¼ E E Z f dl: ¼ E

This completes Step 3 and thereby, also the proof of the theorem.

h

Remarks 5.10.9 (a) The function f given by the Radon–Nikodým Theorem is called the Radon– dm Nikodým derivative of m with respect to l and is sometimes denoted by [dl ]. (b) The condition that l is r-finite is essential in the Radon–Nikodým Theorem. Let (R, M, m) be the Lebesgue measure space and l be the counting measure on M. So, l is not r-finite. RAlso, m  l, as seen in Example 5.10.3 (b). Suppose f exists such that mðEÞ ¼ E f dl whenever E 2 F . Then for any x 2 R, we have Z 0 ¼ mfxg ¼

fxg

f dl ¼ f ðxÞlfxg ¼ f ðxÞ:

Hence f is identically 0. This means for every E 2 M, we have Z 0 dl ¼ 0:

mðEÞ ¼ E

This contradicts the fact that m is Lebesgue measure. (c) The condition that m is r-finite is also essential in the Radon–Nikodým Theorem. Consider the set X consisting of a single point x0. Let l be the counting measure on X and m be the measure that m(X) = ∞. R Clearly, l is r-finite and m  l. If f is any real-valued function on X, then X f dl is the real R number f(x0) but m(X) = ∞, so that m(X) 6¼ X f dl. Remark 5.10.10 Suppose E 2 F . Then {F \ E: F 2 F } = {F 2 F : F  E} and is a r-algebra of subsets of E. For a signed measure l on (X, F ), the restriction to this r-algebra is called the restriction of the measure to E and we shall denote it by l↿E. The following facts about restrictions of measures are easy to justify in the order in which they are stated.

5.10

The Radon–Nikodým Theorem

313

(a) If ⟨A, B⟩ is a Hahn decomposition of X for l, then ⟨A \ E, B \ E⟩ is a Hahn decomposition of E for l↿E. (b) (l↿E)+ = l+↿E, (l↿E)− = l−↿E, |(l↿E)| = (|l|)↿E. (c) m  l ) m↿E l↿E. (d) m⊥l ) m↿E⊥l↿E. Let (X, F , l) be a r-finite measure space. In what follows, we shall show that all r-finite signed measures on (X, F ) can be analysed by considering only those that are absolutely continuous or singular with respect to l. More specifically, we have the following: Definition 5.10.11 Given a measure l on (X, F ) and a signed measure m on (X, F ), a Lebesgue decomposition of m with respect to l is a pair of measures ma and ms on (X, F ) with the property that m ¼ ma þ ms ;

where

ma  l and ms ?l:

Remark 5.10.12 Suppose E 2 F , l is a measure on (X, F ) and m a signed measure on (X, F ). If m = ma + ms is a Lebesgue decomposition of m, then by parts (c) and (d) of Remark 5.10.10, m↿E = va↿E + ms↿E is a Lebesgue decomposition of m↿E with respect to l↿E. Theorem 5.10.13 (Lebesgue Decomposition Theorem for Measures) Let (X, F , l) be a r-finite measure space and m be a r-finite signed measure on (X, F ). Then m has a unique Lebesgue decomposition with respect to l consisting of a pair of r-finite measures; i.e., there exists a unique pair of r-finite measures ma and ms on (X, F ) with the property that m ¼ ma þ m s ;

where

ma  l

and ms ?l:

Proof We shall prove the existence in two steps. Step 1. Let m be nonnegative and set k = l + m. Since both l and m are absolutely continuous with respect to k, by the Radon–Nikodým Theorem 5.10.8, there exist nonnegative measurable functions f and g such that, for E 2 F , Z Z f dk and mðEÞ ¼ g dk: lðEÞ ¼ E

E

Let A = {x: f(x) > 0} and B = {x: f(x) = 0}. Then X = A [ B, where A \ B = ∅ and l(B) = 0. If we define ms on F by ms ðEÞ ¼ mðE \ BÞ; we have ms(A) = m(A \ B) = 0 and so, ms ⊥ l. Define ma on F by

314

5 Differentiation

Z ma ðEÞ ¼ mðE \ AÞ ¼

g dk: E\A

Then, for any E 2 F , we have mðEÞ ¼ mðE \ AÞ þ mðE \ BÞ ¼ ma ðEÞ þ ms ðEÞ; i.e., m = ma + ms. It remains to show that ma  l. Let E 2 F be such that l(E) = 0. Then Z 0 ¼ lðEÞ ¼

f dk; E

which implies f = 0 a.e.[k] on E. Since f > 0 on A \ E, we must have k(A \ E) = 0. Hence m(A \ E) = 0 (recalling that m  k) and so, ma(E) = m(E \ A) = 0. Observe that if m does not take the value ∞, the same is true of ma and ms. Step 2. Suppose m is a r-finite signed measure. By the Jordan Decomposition Theorem 5.9.16, m = m+ − m−. We treat m+ and m− as in Step 1. We consider the case when m does not take the value ∞; then m+ does not take the value ∞. Also, the measures m+ and m_ are r-finite by Problem 5.9.P4. So, let + −  v þ ¼ vaþ þ vsþ and m− = v a þ vs be any Lebesgue decompositions of m and m . From the observation at the end of Step 1, it follows that both vaþ and vsþ do not þ  take the value ∞. The set functions vaþ  v a and vs  vs are thus well defined and therefore must be signed measures by Remark 5.9.2(k). This permits us to write þ  m ¼ m þ  m ¼ ðmaþ  m a Þ þ ðms  ms Þ:

In view of Proposition 5.10.5(b), we have (vaþ  v a )  l. The r-finiteness of þ  þ  va  va and vs  vs follows from Problem 5.9.P4. It remains only to demonstrate þ that (vsþ  v s ) ⊥ l. Since vs does not take the value ∞, we have þ þ ðmsþ  m s Þ  ms

  and ðmsþ  m s Þ  ms

ð5:91Þ

by Problem 5.9.P2. It is trivial to see that for (nonnegative) measures l1,l2,l3, l1  l2 and l2 ?l3

)

l1 ?l3 :

þ  + Therefore, (5.91) and fact that vsþ ⊥ l and v s ⊥ l together yield (vs  vs ) ⊥ l þ  − and (vs  vs ) ⊥ l. By Remark 5.9.14 (b), it follows that + − þ þ  + þ  − þ  ((vsþ  v ) + (v  v s s s ) ) ⊥ l. However (vs  vs ) + (vs  vs ) = |vs  vs |. þ  þ Thus we have demonstrated that |vs  vs | ⊥ l, which is equivalent to (vs  v s ) ⊥ l. The case when m does not take the value −∞ is analogous. This completes the proof of existence. We next prove the uniqueness of the decomposition.

5.10

The Radon–Nikodým Theorem

315

First suppose that m is finite. Then in any Lebesgue decomposition of it, both measures must be finite, as is evident from the requirement in Definition 5.10.11 that their sum is m. If m ¼ ma þ ms and m =  ma þ  ms are two Lebesgue decompositions of m, then the finiteness of all measures concerned permits the assertion that ms   ms ¼  ma  va . Since the left-hand side, namely, ms  ms can be shown to be singular in the same manner as in Step 2, and since the right-hand side, namely, ma  ma is absolutely continuous by Prop. 5.10.5(b), it follows on the basis of Proposition 5.10.5 (d) that ms   ms ¼  ma  ma ¼ 0. This proves uniqueness in the finite case. For a general r-finite m, we can express X as a countable disjoint union of measurable sets Xn with each m(Xn) < ∞. By Remark 5.10.12, for any n, the restrictions to Xn of any two Lebesgue decompositions provide Lebesgue decompositions of the restriction of m to Xn. It follows from the finite case that the restrictions of the Lebesgue compositions agree and it is then easy to deduce by “piecing together” the restrictions that the Lebesgue decompositions (on X) agree. h For the relation between the above theorem and Problem 5.8.P7, see Problems 5.11.P5 and 5.11.P8. The following example shows that the condition of r-finiteness cannot be dropped in the Lebesgue Decomposition Theorem. Example 5.10.14 Let l be the Lebesgue measure m on M, the r-algebra of Lebesgue measurable subsets of [0, 1] and suppose m is the counting measure on M. We argue that m cannot be written as m = ma + ms, where ma l and ms ⊥ l. Suppose the contrary. Then for each x 2 [0, 1], we have 1 ¼ mfxg ¼ ma fxg þ ms fxg ¼ 0 þ ms fxg: Since ms ⊥ l, there exists disjoint measurable sets A, B with A [ B = [0, 1] such that l(A) = 0 = ms(B). Now if B 6¼ ∅, take x 2 B. Then 0 ¼ ms ðBÞ  ms fxg ¼ 1; which is a contradiction. Consequently, B = ∅. Since A \ B = ∅ and A [ B = [0, 1], it follows that A = [0, 1], which implies l(A) = m(A) = m[0, 1] = 1. This contradicts the stipulation that l(A) = 0. Problem Set 5.10 5.10.P1. Let l = m + d, where m is Lebesgue measure on [0, 1] and d is the “Dirac” measure, which is defined by setting d(E) = 1 if 0 2 E and 0 if 0 62 E. Determine [dm dl ].

316

5 Differentiation

5.10.P2. Let pffiffiffiffiffiffiffiffiffiffiffi 1x f ðxÞ ¼ R 0 mðEÞ ¼ E f dx

2 x if x  0 if x  1 ; gðxÞ ¼ 0 if x\0 if x [ 1 R and lðEÞ ¼ E g dx; E 2 M:

Find the Lebesgue decomposition of m with respect to l. 5.10.P3. Let l = m + d, where m is the usual Lebesgue measure on R and d is the “Dirac” measure defined by dðEÞ ¼

1 0

if 0 2 E if 0 62 E;

E 2 M:

Determine the Lebesgue decomposition of l with respect to m. 5.10.P4. Let l1,l2 and m be r-finite measures on (X, F ). Prove the following: (a) If li  m, i = 1, 2, then l1 + l2  m and ½

dðl1 þ l2 Þ  dm

¼½

dl1 dl  þ ½ 2 dm dm

a:e:½m:

dl2 1 (b) If l1  l2 and l2  l1, then [dl dl2 ] [dl1 ] = 1 a.e.[l1] and a.e.[l2].

5.10.P5. (X, F , m) be a r-finite measure space and l1,l2 be r-finite signed measures such that l1 + l2 is also a signed measure on (X, F ). Prove the following: (a) If li  m, i = 1, 2, then l1 + l2  m and dðl1 þ l2 Þ  dm

½

dl1 dl  þ ½ 2 dm dm

¼½

a:e:½m:

(b) If l is a r-finite signed measure on (X, F ) such that l  m, then



dl

d jl j ½  ¼ ½  a:e:½m: dm

dm

5.10.P6. Let (X, F , l) be a r-finite measure space. Let m be a measure on (X, F ) for which the conclusion of the Radon–Nikodým Theorem holds. Prove that m is r-finite.

5.11

The Lebesgue–Stieljes Measure

In what follows, we shall study the connection between bounded increasing functions and finite measures—the functions and measures will both be on R, the domain of the measure being the r-algebra of Borel subsets of R (see Definition 2.3.13).

5.11

The Lebesgue–Stieljes Measure

317

Let B denote the r-algebra of Borel subsets of R and l be a finite (nonnegative) measure on B. Define a function fl on R by the rule fl ðxÞ ¼ lðð1; xÞÞ ¼ lðfy 2 R : y\xgÞ: Theorem 5.11.1 (a) The function fl is a bounded increasing function on R which is left continuous. Moreover lim fl ðxÞ ¼ 0. x!1

(b) The function fl is continuous at x if and only if l({x}) = 0. Proof (a) Observe that for a < b, lð½a; bÞÞ ¼ lðð1; bÞnð1; aÞÞ ¼ fl ðbÞ  fl ðaÞ: Since l is nonnegative, fl(b)  fl(a), i.e., fl is an increasing function on R. It follows from the fact that l is finite that fl is also bounded. Also, [a, b) being the intersection of the sets [a − 1n, b), outer continuity of measure (see Proposition 3.1.8) implies 1 lð½a; bÞÞ ¼ lim lð½a  ; bÞÞ n!1 n and so, 1 fl ðbÞ  fl ðaÞ ¼ fl ðbÞ  lim fl ða  Þ n!1 n ¼ fl ðbÞ  fl ðaÞ by monotonicity of fl : Consequently, fl(a) = fl(a−), i.e., fl is left continuous. Since \ n  1(−∞, −n) = ∅, by outer continuity of measure again, we have fl(−n) = 0 and hence lim fl ðxÞ ¼ 0 by monotonicity of fl. x!1

  (b) Since l({x}) = l( \ n  1[x, x + 1n)) = lim fl x þ 1n  fl ðxÞ = fl(x +)−fl(x), the n!1

function fl is continuous at x if and only if the set {x} consisting of x alone has measure 0. h The finite measure l on B and the monotone increasing function fl are closely related. Indeed, we have the following theorem. Theorem 5.11.2 If l is a finite measure on the r-algebra B of Borel subsets of R, then the monotone function fl is absolutely continuous if and only if l  m. Proof Suppose l  m. By Proposition 5.10.6, to every e > 0, there corresponds a d > 0 such that l(E) < e for every Borel set E with m(E) < d. Let {(ai, bi): i = 1,…, n} be finite disjoint collection of bounded open intervals for which

318

5 Differentiation

mð

n [

½ai ; bi ÞÞ ¼

n X

i¼1

ðbi  ai Þ\d:

i¼1

Then n

n n X [

X

fl ðbi Þ  fl ðai Þ ¼ lð½ai ; bi ÞÞ ¼ lð ½ai ; bi ÞÞ\e; i¼1

i¼1

i¼1

this proves that fl is absolutely continuous. Suppose conversely that fl is absolutely continuous. Let e > 0 be arbitrary and d > 0 be such that n X

ðbi  ai Þ\d

implies

i¼1

n

X

fl ðbi Þ  fl ðai Þ \e; i¼1

where {(ai, bi) : i = 1,…,n} is any finite disjoint collection of open intervals. Let E be a Borel set of Lebesgue measure 0. By using Proposition 2.3.24, we can find a pairwise disjoint sequence {[ai, bi)}i  1 of bounded intervals such that E

1 [

½ai ; bi Þ and

i¼1

1 X

ðbi  ai Þ\d:

i¼1

Since it follows that n

X

fl ðbi Þ  fl ðai Þ \e i¼1

for every positive integer n, we have lðEÞ 

1 X i¼1

lð½ai ; bi ÞÞ ¼

1

X

fl ðbi Þ  fl ðai Þ  e: i¼1

Since e > 0 is arbitrary, we have l(E) = 0. h The intervals of the form [a, b), where a  b, will be referred to in the present section as left closed intervals. Conversely, let f be a bounded increasing function on R which is left continuous. We shall show that there is a unique measure lf on B such that lf ð½a; bÞÞ ¼ f ðbÞ  f ðaÞ for all left closed intervals [a, b). See Theorem 5.11.22. Indeed, it will be shown in Remark 5.11.24 (b) that, if l = lf, then fl will be equal to f + constant. We will call

5.11

The Lebesgue–Stieljes Measure

319

lf as the Lebesgue–Stieltjes measure induced by f. To begin with, let l be defined on the family of all left closed intervals [a, b) by lð½a; bÞÞ ¼ f ðbÞ  f ðaÞ: Observe that l assumes only nonnegative values and l(∅) = 0. Indeed, ∅ = [a, b) if and only if a = b. We begin with the following lemmas. Lemma 5.11.3 Let {Ei}1  i  n be disjoint left closed intervals (Ei = [ai, bi), 1  i  n) such that [ 1  i  nEi  I, where I is a left closed interval. Then n P lðEi Þ  lðIÞ. i¼1

Proof Let I = [a, b). Without loss of generality, we may assume that a1 < a2 <    < an. Since [ 1  i  n Ei  I, we know that a  a1 and bn  b. The disjointness of the intervals then tells us that we also have bi  ai+1 for 1  i  n − 1. Since f is increasing, these inequalities imply that n X i¼1

ðf ðbi Þ  f ðai ÞÞ 

n X

ðf ðbi Þ  f ðai ÞÞ þ

i¼1

n1 X

ðf ðai þ 1 Þ  f ðbi ÞÞ ¼ f ðbn Þ  f ða1 Þ

i¼1

 f ðbÞ  f ðaÞ: Up to this stage, the argument is valid for any kind of bounded intervals I and Ei with [ 1  i  n Ei  I without even a continuity hypothesis about the increasing function f. But when I and Ei are left closed, the inequality above asserts precisely what the lemma claims. This completes the proof. h Lemma 5.11.4 If I = [a0, b0] is contained in the union of a finite number of bounded open intervals Ui = (ai, bi), 1  i  n, then f ðb0 Þ  f ða0 Þ 

n X

ðf ðbi Þ  f ðai ÞÞ:

i¼1

Proof Let k1 be such that a0 2 Uk1 . Then a0 b0. Without loss of generality, we assume that m = n and that Uki = Ui = (ai, bi) for 1  i  n. This is easily achieved by omitting superfluous Ui’s and changing the notation if necessary. In other words, we may assume that

and, if n > 1, then ai+1

a1 \a0 \b1 ; an \b0 \bn < bi < bi+1 for 1  i  n − 1. It follows that

320

5 Differentiation

f ðb0 Þ  f ða0 Þ  f ðbn Þ  f ða1 Þ ¼ f ðb1 Þ  f ða1 Þ þ

n1 X

ðf ðbi þ 1 Þ  f ðbi ÞÞ

i¼1

 f ðb1 Þ  f ða1 Þ þ

n1 X

ðf ðbi þ 1 Þ  f ðai þ 1 ÞÞ

i¼1

¼

n X

ðf ðbi Þ  f ðai ÞÞ:

i¼1

h

This completes the proof. Lemma 5.11.5

(a) If the left closed interval [a0, b0) is contained in the union of a sequence of left closed intervals [ai, bi), i = 1, 2, …, then lð½a0 ; b0 ÞÞ 

1 X

lð½ai ; bi ÞÞ:

i¼1

(b) If the intervals in the sequence [ai, bi), i = 1, 2, …, are disjoint and [a0, b0) = [ i  1[ai, bi), then lð½a0 ; b0 ÞÞ ¼

1 X

lð½ai ; bi ÞÞ:

i¼1

Proof (a) If a0 = b0, then the result is trivial. Let e be a positive number such that e < b0 − a0. Since f is left continuous at ai, to every number d and positive integer i, there corresponds a positive number ei such that f ðai Þ  f ðai  ei Þ\

d ; 2i

i ¼ 1; 2; . . .:

ð5:92Þ

If F0 = [a0, b0 − e] and Ui = (ai − ei, bi), i = 1, 2, …, then F0  [ i  1Ui and therefore by the Heine–Borel Theorem, there is an integer n such that F0 is contained in the union of n among the intervals Ui. By renumbering the Ui if necessary, we may assume that F0 

[ 1in

It follows from Lemma 5.11.4 that

Ui :

5.11

The Lebesgue–Stieljes Measure

f ðb0  eÞ  f ða0 Þ 

321 n X

ðf ðbi Þ  f ðai  eÞÞ

i¼1

¼ 

n X i¼1 n X

ðf ðbi Þ  f ðai ÞÞ þ

n X

ðf ðai Þ  f ðai  eÞÞ

i¼1

ðf ðbi Þ  f ðai ÞÞ þ d

by (5:92Þ:

i¼1

Since e and d are arbitrary, the proof of (a) is complete. (b) It follows from Lemma 5.11.3 that So,

1 P

n P

lð½ai ; bi ÞÞ  l([a0, b0)) for n = 1, 2, ….

i¼1

lð½ai ; bi ÞÞ  l([a0, b0)). The proof is completed by using (a) above. h

i¼1

Remark 5.11.6 (a) Let [a, b) and [a, b) be left closed intervals. Then x belongs to their intersection if and only if max{a, a}  x < min{b, b}. Thus the intersection is empty if max{a, a}  min{b, b} and is the left closed interval ½maxfa; ag; minfb; bgÞ otherwise. Therefore the intersection is always a left closed interval. (b) The set-theoretic difference [a, b)\[a, b) is a union of at most two disjoint left closed intervals. This is trivial if a = b. Suppose a < b. Clearly, the complement of [a, b) is the union (−∞, a) [ [b, ∞) and therefore [a, b)\[a, b) is the union of the intersections ½a; bÞ \ ð1; aÞ and ½a; bÞ \ ½b; 1Þ: The first of these is empty if a  a and is the left closed interval [a, min{b, a}) otherwise; the second is empty if b  b and is the left closed interval [max{a, b}, b) otherwise. (c) The union of two left closed intervals [a, b) and [a, b) that have a nonempty intersection can be shown to be a left closed interval by reasoning as follows. First of all, the union of any family of intervals with a nonempty intersection is an interval. This is easy to prove by means of the characterisation of an interval as being a set that contains any number that lies between two of its numbers. It follows that the union of [a, b) and [a, b) is an interval. It obviously contains min{a, a} and nothing smaller; it also contains numbers arbitrarily close to max{b, b} and nothing bigger or equal. It follows that the union is precisely the left closed interval ½minfa; ag; maxfb; bgÞ:

322

5 Differentiation

(d) A union of finitely many left closed intervals is also a union of finitely many disjoint left closed intervals, as we now prove by induction. If there is only one interval in the union, there is nothing to prove. Suppose the claim is true when there are k intervals in the union and let E = [ 1  i  k+1Ei, where Ei, 1  i  k + 1 are left closed intervals. If all the Ei are disjoint, we are done. Suppose instead that there are distinct p and q such that Ep and Eq have a nonempty intersection. By (c) their union is a left closed interval F. Then F and the remaining Ei (p 6¼ i 6¼ q) constitute k left closed intervals whose union is E. Therefore by the induction hypothesis, E is a union of finitely many disjoint left closed intervals. Definition 5.11.7 A collection < of subsets of a nonempty set X is called a ring if whenever E, F 2 0 is arbitrary, it follows that l*(E) 

1 P i¼1

l ðEi Þ .

An outer measure is not necessarily countably or even finitely additive. We single out a family of subsets of R on which l* is countably, and hence also finitely, additive. The precise formulation uses Carathéodory’s condition of measurability along the lines of Definition 2.3.1.

5.11

The Lebesgue–Stieljes Measure

327

Definition 5.11.17 Let l* be the outer measure induced by l (see Definition 5.11.14). A set E  R is called l*-measurable if l ðAÞ ¼ l ðA \ EÞ þ l ðA \ E c Þ for all AR: Remark 5.11.18 Since the outer measure is subadditive [see Remark 5.11.16 (c)], we have l*(A)  l*(A \ E) + l*(A \ Ec), and it follows that E is l*-measurable if l ðAÞ  l ðA \ EÞ þ l ðA \ Ec Þ for all A R: This observation is frequently used in the discussion that follows. Remark 5.11.19 By Remark 5.11.16 (a), we know that l*(∅) = 0. Therefore it is immediate from Definition 5.11.17 that ∅ and R are l*-measurable. Next, we have the following theorem. Theorem 5.11.20 Let l* be the outer measure induced by l and let S  denote the class of l*-measurable sets. Then S  is a r-algebra and l* restricted to S  is a complete measure. Proof The argument of Theorems 2.3.12 and 2.3.13 carries over upon replacing m* there by what we have called l* here. The argument of Remark 2.3.2 (d) carries over to show that l* restricted to S  is complete. h Theorem 5.11.21 Every set in B is l*-measurable. In particular, the restriction of l* to B is a measure. Moreover, l* coincides with l on ~I and it is therefore an extension of l. Proof First consider E 2 ~I and let A  R be arbitrary. By definition of l*, for any e > 0, there exists a sequence {Ei}i  1 of sets in ~I such that A  [ i  1Ei and l ðAÞ þ e 

1 X

lðEi Þ:

i¼1

Since l is a measure on ~I (see Remark 5.11.13 and Theorem 5.11.12), 1 X i¼1

lðEi Þ ¼

1 X

ðlðEi \ EÞ þ lðEi \ E c Þ

i¼1

 l ðA \ EÞ þ l  ðA \ Ec Þ

by definition of l :

Combining this with the previous inequality, we get l ðAÞ þ e  l ðA \ EÞ þ l ðA \ Ec Þ: Since this holds for any e > 0, the sufficient condition for l*-measurability noted in Remark 5.11.18 is found to hold. We thus conclude that E is l*-measurable. Since E is an arbitrary element of ~I, this proves ~I  S  . Now, S  is a r-algebra by Theorem 5.11.20. Therefore S(~I)  S  . However, S(~I) = B by Proposition 5.11.11 and hence B  S  . This means every set in B is l*-measurable.

328

5 Differentiation

By Theorem 5.11.20, l* is a measure on S  and is now seen to be a measure on B because B  S  . To see why l* coincides with l on ~I, consider an arbitrary E 2 ~I. Then E can be written as E = [ 1  i  nEi, where Ei, 1  i  n, are disjoint left closed intervals. Recall from Remark 5.11.13 and the proof of Theorem 5.11.12 that n n P P l(E) = lðEi Þ . By Remark 5.11.16 (a), this leads to lðEÞ ¼ l ðEi Þ . Now that i¼1

i¼1

l* has been shown to be a measure on B, which contains I by Prop. 5.11.11, we n P also have l ðEi Þ ¼ l ðEÞ . Thus l(E) = l*(E). h i¼1

Theorem 5.11.22 Let f be a bounded increasing function on R which is left continuous. Then there exists a unique measure lf on B such that lf ð½a; bÞÞ ¼ f ðbÞ  f ðaÞ for all left closed intervals [a, b). Proof Denote the restriction of l* to the Borel r-algebra B by lf. By Theorem 5.11.21, lf coincides with l on ~I and it therefore satisfies lf ð½a; bÞÞ ¼ lð½a; bÞÞ ¼ f ðbÞ  f ðaÞ; for all left closed intervals [a, b). This proves the existence of lf. Suppose m is a measure on B such that m = l on all left closed intervals. Since every E 2 ~I can be written as a finite union of disjoint left closed intervals [Remark 5.11.6 (d)], it follows from the additivity of m and the definition of l on ~I that m = l on ~I. It follows that whenever A = [ i  1Ei and Ei are disjoint sets in ~I, the definition of lf and Theorem 5.11.21 lead to lf ðAÞ ¼

1 X i¼1

lf ðEi Þ ¼

1 X

l ðEi Þ ¼

1 X

i¼1

lðEi Þ ¼

i¼1

1 X

mðEi Þ ¼ mðAÞ:

ð5:98Þ

i¼1

We wish to show that m = lf on B. If E 2 B and e > 0, Proposition 5.11.15 shows that there exists a sequence {Ei}i  1 of disjoint left closed intervals such that E  [ i  1Ei and lf ðEÞ þ e 

1 X

lðEi Þ ¼

i¼1

1 X

mðEi Þ:

i¼1

Since the Ei are disjoint, we further get lf ðEÞ þ e  mð

1 [ i¼1

Ei Þ  mðEÞ;

5.11

The Lebesgue–Stieljes Measure

329

which implies lf ðEÞ  mðEÞ; because e > 0 is arbitrary. Conversely, suppose that E 2 B. Then lf(E) < ∞. Let e > 0 be given. Then there exists an A E such that lf(A) < lf(E) + e, where A = [ i  1Ei and Ei are disjoint sets in ~I. Observe that lf(A) = m(A) on the basis of (5.98). So, lf ðEÞ  lf ðAÞ ¼ mðAÞ ¼ mðEÞ þ mðAnEÞ: But by what has been proved in the preceding paragraph, m(A\E)  lf(A\E). Also, since lf(E) < ∞, we have lf(A\E) = lf(A) − lf(E) < e. This implies lf ðEÞ  mðEÞ þ e: Since e > 0 is arbitrary, we have lf(E)  m(E).

h

Definition 5.11.23 If f is a bounded increasing function on R which is left continuous, the unique measure lf on B such that lf ð½a; bÞÞ ¼ f ðbÞ  f ðaÞ for all left closed intervals [a, b) is called the Lebesgue–Stieltjes measure induced by f. Remark 5.11.24 (a) Since two bounded increasing left continuous functions on R that differ by a constant obviously lead to the same l and l*, they induce the same Lebesgue– Stieltjes measure. (b) Let f be a bounded increasing function on R which is left continuous and lf be the Lebesgue–Stieltjes measure on B induced by f, i.e., lf([a, b)) = f(b) − f(a) for all left closed intervals [a, b). Then lf is bounded. Conversely, starting with a bounded measure l on B, we can generate a bounded increasing function fl on R which is left continuous. Although there is no one-to-one correspondence here, the following is true: Starting with the bounded measure lf on B induced by a bounded increasing function f on R, if we generate the bounded increasing function flf on R, then flf differs from f by a constant. It is sufficient to show that when F is also a bounded increasing function and lF = lf, the functions F and f differ by a constant. Since f and F are both bounded and increasing, the limits lim f ðxÞ and lim FðxÞ must x!1

both exist. It follows for any real b that

x!1

330

5 Differentiation

FðbÞ  lim FðxÞ ¼ lim ðFðbÞ  FðxÞÞ ¼ lim lF ð½x; bÞÞ x!1

x!1

x!1

¼ lim lf ð½x; bÞÞ ¼ lim ðf ðbÞ  f ðxÞÞ ¼ f ðbÞ  lim f ðxÞ: x!1

x!1

x!1

Hence, FðbÞ  f ðbÞ ¼ lim FðxÞ  lim f ðxÞ: x!1

x!1

Since this holds for any real b, the functions F and f differ by the constant lim FðxÞ  lim f ðxÞ .

x!1

x!1

(c) Observe the consequence of part (b) that Theorem 5.11.2 has the companion result that if f is a bounded increasing function on R then lf  m if and only if f is absolutely continuous. (d) For any x 2 R, we have lf({x}) = f(x+) − f(x). This is because {x} = \ n  1[x, x + 1n ), which leads via outer continuity of measure (see Proposition 3.1.8) to     lf({x}) = lim lf x; x þ 1n ¼ lim f x þ 1n  f ðxÞ = f(x+) − f(x). n!1

n!1

Remark 5.11.25 The relation between the Riemann–Stieltjes integral and the integral with respect to Lebesgue–Stieltjes measure is as described below. If u is a Borel measurable function and f is a monotone increasing function which is left continuous, we define the Lebesgue–Stieltjes integral of u as Z Z u df ¼ u dl; where l = lf is the Lebesgue–Stieltjes measure induced by f as in this section. It is R clear what is meant by Lebesgue–Stieltjes integral ½a; b u df , where a  b. We shall show for a certain class of functions u on [a, b], which includes continuous functions, that Z

Z ½a; bÞ

udf ¼

b

uðxÞdf ðxÞ;

a

where the integral on the right is the Riemann–Stieltjes integral. If u = v[a,b), where [a, b)  [a, b), then u is right continuous and hence by Theorem 11.2.10 of [28], Z a

b

uðxÞ df ðxÞ ¼ f ðbÞ  f ðaÞ:

5.11

Also,

The Lebesgue–Stieljes Measure

331

Z ½a; bÞ

u df ¼ lf ½a; bÞ ¼ f ðbÞ  f ðaÞ:

The equality in question holds therefore for linear combinations of characteristic functions of left closed intervals. Let u be any continuous function on [a, b]. It is uniformly continuous and therefore there exists a sequence of functions un , each of which is a linear combination of characteristic functions of left closed intervals, such that un converges uniformly to u. By Theorem 11.3.1 of [28], the uniform convergence yields Z

b

Z un ðxÞ df ðxÞ !

a

b

uðxÞ df ðxÞ:

a

Since lf is a finite measure, the uniform convergence also yields Z Z un df ! u df : ½a; bÞ

½a; bÞ

Since the equality in question has been shown to hold for every un , it follows from the above limits that the equality holds for u. Let l be a finite nonnegative measure on (R,B ) and fl be the related function as defined at the beginning of this section: fl ðxÞ ¼ lðð1; xÞÞ ¼ lðfy 2 R : y\xgÞ: Recall that by Theorem 5.11.1, the function fl is bounded, increasing and left continuous. Moreover, lim fl ðxÞ ¼ 0 Being an increasing function, it has a x!1

derivative a.e. (see Lebesgue’s Theorem 5.4.6). In the result that follows, we shall compute the derivative fl′ wherever it exists and motivate a definition of the derivative of a measure on R, independently of the Radon–Nikodým derivative. Its relation to the latter will be described later in Proposition 5.11.28. Theorem 5.11.26 Let l and f = fl be as in the paragraph above and let x0 2 R. Then the following statements are equivalent: (a) f is differentiable at x0 with f′(x0) = k; (b) for every e > 0, there exists a d > 0 such that



lðIÞ





mðIÞ  k \e for every open interval I with x0 2 I and ‘ðIÞ ¼ mðIÞ < d. Proof (a) ) (b). Suppose that (a) holds. Let e > 0 be given. There exists a d > 0 such that

332

5 Differentiation



f ðtÞ  f ðx0 Þ



 k

 e

tx 0 provided |t − x0| < d. Suppose x0 2 I = (s, t) and t − s < d. Choose {sn}n  1 such that x0 > s1 > s2 >   >sn! s. Then







lð½sn ; tÞÞ

f ðtÞ  f ðsn Þ

f ðtÞ  f ðx0 Þ f ðx0 Þ  f ðsn Þ



¼

¼

 k  k þ  k

ts

ts

ts

t  sn n n n



t  x0 f ðtÞ  f ðx0 Þ x0  sn f ðx0 Þ  f ðsn Þ t  x0 x0  sn

¼

þ  k k x0  s n t  s n t  x0 t  sn t  sn t  sn







x0  sn f ðx0 Þ  f ðsn Þ

t  x0

f ðtÞ  f ðx0 Þ

þ



  k  k





t  x0 x0  s n t  sn t  sn t  x0 x0  s n  eþ e ¼ e: t  sn t  sn Since I = [ n  1[sn,t), it follows by using inner continuity of measure (see Proposition 3.1.8) that lðIÞ ¼ lim lð½sn ; tÞÞ . But m(I) = t − s = lim ðt  sn Þ. n!1

n!1

Therefore



lðIÞ



 e:  k

mðIÞ

An alternative presentation of the above argument that makes the use of convexity more explicit is as follows. Let I be an open interval (a, b) with x0 2 I. i.e. a < x0 < b. The simple equality f ðbÞ  f ðaÞ x0  a f ðx0 Þ  f ðaÞ b  x0 f ðbÞ  f ðx0 Þ ¼ þ ba ba x0  a b  x0 ba ðaÞ ðaÞ ðx0 Þ shows that f ðbÞf is a convex combination of f ðxx00Þf and f ðbÞf a ba bx0 . Therefore, if the latter two lie within some positive distance η of f′(x0), so does the first. For any e > 0, there exists a d > 0 such that 0 < x0 − a < d and 0 < b − x0 < d imply that the latter two numbers mentioned in the preceding paragraph both lie within 12 e of f′(x0). Now, consider x0 2 I, i.e. a < x0 < b and suppose m(I) < d, i.e., b − a < d. Let a1 such that a < a1 < x0 be arbitrarily chosen. Then b − a1 < d and the inequalities 0 < x0 − a < d, 0 < x0 − a1 < d and 0 < b − x0 < d all hold, and hence by the convexity noted above,

f ðbÞ  f ðaÞ ba

and

both lie within a distance e of f′(x0). Now,

f ðbÞ  f ða1 Þ a1 !a b  a1 lim

5.11

The Lebesgue–Stieljes Measure

333

lðIÞ lðða; bÞÞ lð½a; bÞÞ f ðbÞ  f ðaÞ ¼  ¼ mðIÞ ba ba ba and lðIÞ lðða; bÞÞ ¼  mðIÞ ba It follows that

lðIÞ mðIÞ

lim lð½a1 ; bÞÞ

a1 !a

ba

¼ lim

a1 !a

lð½a1 ; bÞÞ f ðbÞ  f ða1 Þ ¼ lim : a1 !a b  a1 b  a1

lies within a distance e of f′(x0).

lðIÞ (b) ) (a). On the other hand, suppose (b) holds. Since mðIÞ  0 always, we must have k  0. First we prove that f is continuous at x0. By taking e = 1 in (b), we obtain an increasing sequence {sn} converging to x0 and a decreasing sequence {tn} converging to x0 such that l((sn, tn))  (1 + k)(tn− sn). Since {x0} = \ n  1(sn, tn), we use outer continuity of measure (see Proposition 3.1.8) to arrive at the equality l({x0}) = 0. We deduce from this and Theorem 5.11.1 that f is continuous at x0. The same theorem also tells us that it is an increasing function. Next, for e > 0, choose d as in (b). If s < x0 < t and t − s < d, then



lððs  1n; tÞÞ





t  s þ 1  k \e n

for all sufficiently large n. Since [s, t) = \ n  1(s − 1n, t) and f(t) − f(s) = l([s, t)), it follows that



f ðtÞ  f ðsÞ



ð5:99Þ

t  s  k  e for s\x0 \t\s þ d; using outer continuity of measure once again. Now we would like to take limits as s ! x0 and as t ! x0 separately in (5.99); however, the range of values of s or that of t, as described in (5.99), is not explicitly an interval. Therefore we resort to an artifice. The inequalities 1 x0  d\s\x0 2

and

x0 \t\x0 þ

together imply s\x0 \t\s þ d: In conjunction with (5.99), this leads to

1 d 2

334

5 Differentiation



f ðtÞ  f ðsÞ





t  s  k  e

1 for x0  d\s\x0 2

and x0 \t\x0 þ

1 d: 2

This permits us to take limits as s ! x0 and as t ! x0 separately. Taking the limits and keeping in mind the continuity of f at x0, we obtain respectively



f ðtÞ  f ðx0 Þ



 e for x0 \t\x0 þ 1 d ð5:100Þ  k

tx

2 0 and





f ðx0 Þ  f ðsÞ

e

 k

x s 0

1 for x0  d\s\x0 : 2

ð5:101Þ

Since we have shown that, for every e > 0, there exists a d > 0 such that (5.100) and (5.101) hold, it follows that the right and left derivatives of f at x0 both exist and are both equal to k. h Definition 5.11.27 Let l be a finite nonnegative measure on (R, B). We say that l is differentiable at x 2 R if the limit DlðxÞ ¼ lim

lðIÞ

mðIÞ!0 mðIÞ

;

x2R

exists in the sense of part (b) of Theorem 5.11.26. The limit Dl(x) is called the derivative of l at x. Proposition 5.11.28 Let lf be the Lebesgue–Stieltjes measure induced by an absolutely continuous bounded increasing function f on R. Then h

dlf dm

i

¼ f 0 ¼ Dlf

a:e:

Proof Since f is absolutely continuous, Remark 5.11.24(c) allows us to infer that lf  m. Therefore the Radon–Nikodým derivative in question exists and is characterised by the equality lf ðEÞ ¼

Z h

dlf E dm

i dm

for all Borel sets E:

Since lf is finite, the Radon–Nikodým derivative is integrable. It follows by Theorem 5.5.1 that f′ is integrable and by Theorem 5.8.2 that whenever a < b, we have

5.11

The Lebesgue–Stieljes Measure

Z ½a; b

335

Z

0

f dm ¼ f ðbÞ  f ðaÞ ¼ lf ð½a; bÞÞ ¼

½a; b

hdl i f dm: dm

The first of the desired equalities is now a consequence of Problem 3.2.P14(b) and the integrability of both the Radon–Nikodým derivative and f′. The second follows upon applying Theorem 5.11.26 and Definition 5.11.27. h Remark 5.11.29 Suppose the bounded increasing left continuous function f on R is constant on some interval (a,∞). Then it is trivial for every left closed interval E  (a, ∞) that lf(E) = l(E) = 0. It follows in two obvious steps that l*(E) = 0 for an arbitrary set E  (a, ∞) and hence l(E) = 0 for every Borel set E  (a, ∞). Similarly, if f is constant on some interval (−∞, a), then l(E) = 0 for every Borel set E  (−∞, a). Problem Set 5.11 5.11.P1. Suppose l is a measure on I and l* is the outer measure induced by l. Let Ir denote the family of countable unions of sets of I. Given any set E and any e > 0, show that there is a set A 2 Ir such that E  A and l ðAÞ  l ðEÞ þ e: 5.11.P2. Suppose l is a measure on I and l* is the outer measure induced by l. Let Ir denote the family of countable unions of sets of I and Ird denote the family of countable intersections of sets of Ir. Given any set E, show that there is a set A 2 Ird such that E  A and l*(E) = l*(A). 5.11.P3. Let F be a real-valued bounded increasing right continuous function on R and for any right closed interval (a, b], define l((a, b]) = F(b) − F(a). Show that: (i) Let {Ei}1  i  n be disjoint right closed intervals (Ei = (ai, bi], 1  i  n) such that [ 1  i  nEi  I, where I is a right closed interval. Then n P lðEi Þ  lðIÞ . i¼1

(ii) If the right closed interval (a0, b0] is contained in the union of a sequence of right closed intervals (ai, bi], i = 1, 2, …, then lðða0 ; b0 Þ 

1 X

lððai ; bi Þ;

i¼1

moreover, if the intervals in the sequence (ai, bi], i = 1, 2, …, are disjoint and (a0, b0] = [ i  1(ai, bi], then lðða0 ; b0 Þ ¼

1 X i¼1

lððai ; bi Þ:

336

5 Differentiation

5.11.P4. Let h be a real-valued bounded increasing left continuous function on R such that h′ = 0 a.e. and l = lh be the Lebesgue–Stieltjes measure induced by h. Suppose also that there is a closed bounded interval I such that h is constant on each of the unbounded open intervals that constitute the complement Ic. Show that l and Lebesgue measure m (on Borel sets) are mutually singular (see Definition 5.9.13). 5.11.P5. Suppose f is an increasing function on [a, b] that is left continuous and let f = g + h be a Lebesgue decomposition in accordance with Problem 5.8.P7. Extend each of the three functions to R so as to take the same value on (−∞, a) as at a and to take any constant value on (b, ∞) not less than the value at b, but so chosen that the equality f = g + h holds on (b, ∞). Prove the following: R (a) There is some constant c such that gðxÞ ¼ c þ ½a; x f 0 for all x 2 R. (b) The functions f, g and h on R are also bounded, increasing and left continuous; moreover, f = g + h on R. (c) If lf,lg and lh are the Lebesgue–Stieltjes measures induced by f, g and h respectively, then lf = lg + lh and this equality constitutes the unique Lebesgue decomposition of lf with respect to Lebesgue measure m. 5.11.P6. Let l be a finite nonnegative measure on (R,B ) and f = fl. Suppose [a, b] is an arbitrary interval and / is the function on R obtained by extending the restriction of f to [a, b] so as to be equal to f(a) on (−∞, a) and equal to f(b+) on (b, ∞). Then obviously, / is bounded, increasing and left continuous. Show that the Lebesgue–Stieltjes measure l/ induced by / is given by l/(E) = l(E \ [a, b]) for every Borel set E. 5.11.P7. Let l be a finite nonnegative measure on (R, B) and f = fl. If l and Lebesgue measure m (on B) are mutually singular, show that f′ = 0 a.e. 5.11.P8. Let l be a finite nonnegative measure on (R, B) and l = m + k be its Lebesgue decomposition with respect to Lebesgue measure m. Then fm is absolutely continuous on any interval [a, b] and fk′ = 0 a.e. Also, fl = fm + fk. [Remark: This means the equality fl = fm + fk is a Lebesgue decomposition of fl on any interval [a, b] in accordance with Problem 5.8.P7.]

Chapter 6

Lebesgue Spaces and Modes of Convergence

6.1

The Spaces Lp as Normed Linear Spaces

The theory of integration developed in Chap. 3 enables us to define certain spaces of functions, namely, Lp spaces, 1
p
1, whose importance in analysis is hard to overemphasise. We considered the Lp norm to define the distance between two points of the space. With this notion of distance, a satisfying and useful property of the space, namely, completeness, was proved (Theorems 3.3.10 and 3.3.12). Certain dense subsets of the Lp spaces, 1
p\1, were also listed. In what follows, we shall be concerned with a characterisation of what are known as bounded linear functionals on LP spaces, 1
p\1. A real linear (or vector) space is a set V, whose elements are called vectors and, in which two operations called addition and scalar multiplication are defined with the following familiar algebraic properties: (a) To each pair of vectors x and y, there corresponds a vector x + y satisfying the following: x þ y ¼ y þ x; x þ ðy þ zÞ ¼ ðx þ yÞ þ z; V contains a unique vector 0 (the zero vector or the origin of V) such that xþ0 ¼ x

for x 2 V

and to each x 2 V, there corresponds a unique vector −x such that x þ ðxÞ ¼ 0:

© Springer Nature Switzerland AG 2019 S. Shirali and H. L. Vasudeva, Measure and Integration, Springer Undergraduate Mathematics Series, https://doi.org/10.1007/978-3-030-18747-7_6

337

338

6 Lebesgue Spaces and Modes of Convergence

(b) To each pair (a, x), where x 2 V and a 2 R (called a scalar in this context), there corresponds a vector ax 2 V in such a way that 1x ¼ x; aðbxÞ ¼ ðabÞx; aðx þ yÞ ¼ ax þ ay; ða þ bÞx ¼ ax þ bx for x, y 2 V and scalars a, b. The concept defined above makes sense when the scalars are allowed to be rational or complex numbers, but we shall be concerned only with real numbers as scalars. So, we shall drop the word “real” and speak of simply a “linear space”. Examples (i) Let V ¼ Rn . For x ¼ ðx1 ; x2 ; . . .; xn Þ; y ¼ ðy1 ; y2 ; . . .; yn Þ and a 2 R, define x þ y ¼ ðx1 þ y1 ; x2 þ y2 ; . . .; xn þ yn Þ and ax ¼ ðax1 ; ax2 ; . . .; axn Þ. Then V is a linear space. In particular, R is seen to be a linear space when n = 1. (ii) Let V be the collection of all functions f : X ! R on a nonempty set X. Define (f + g)(x) = f(x) + g(x) and (af)(x) = af(x) for all x 2 X and a 2 R. The preceding example is just a special case with X chosen to be the finite set {1, 2, …, n}. (iii) Let V be the collection of all bounded functions f : X ! R on a nonempty set X. Define (f + g)(x) = f(x) + g(x) and (af)(x) = af(x) for all x 2 X and a 2 R. Then V is a linear space, as can be easily verified. The first step is to observe that sums and constant multiples of bounded functions are again bounded. (iv) Let X = [a, b], a closed bounded interval, and C[a, b] be the set of all continuous real-valued functions on [a, b]. A sum of continuous functions and a constant multiple of a continuous function are always continuous and hence the operations f + g and af can be set up in C[a, b] in the same manner as in (ii) above. It is then a trivial verification that C[a, b] becomes a linear space. (v) By virtue of Theorem 3.3.8, operations f + g and af can be set up in the space Lp of functions, where 1
p
∞. It is as easy as in the case of C[a, b] to verify that the space Lp of functions thereby becomes a linear space. The Lp spaces studied so far are examples of a complete normed linear space, also known as a Banach space. This concept, and some related ones that we shall work with, are defined below: Definition 6.1.1 Let V be a linear space. A function kk : V ! R, whose value at x 2 V is written as kxk, is said to be a norm on V if it satisfies the following conditions for all x, y 2 V and all a 2 R : k xk  0 with equality if and only if x ¼ 0; kaxk ¼ jajk xk; and kx þ yk
k xk þ k yk. If kk is a norm on V, then V is called a normed linear space. When V is a normed linear space, d ðx; yÞ ¼ kx  yk defines a metric on V, called the metric induced by the norm. If V is a complete metric space with this metric, then V is called a complete normed linear space or a Banach space.

6.1 The Spaces Lp as Normed Linear Spaces

339

Examples (a) Rn is a normed linear space. The following are examples of norms on Rn : k x kp ¼ ð

n P

1

jxi jp Þp ; where x ¼ ðx1 ; x2 ; . . .; xn Þ 2 Rn and 1
p\1;

i¼1

kxk1 ¼ supfjxi j : 1
i
ng; where x ¼ ðx1 ; x2 ; . . .; xn Þ 2 Rn : In the particular case when n = 1, the norm kxkp agrees with the absolute value j xj for 1
p
∞. (b) The space C(I) of continuous real-valued functions defined on a closed bounded interval IR may be normed by k f k ¼ k f k1 ¼ supx2I jf ðxÞj. It is understood that I contains more than one point. It can be verified that kk1 is a norm on C(I) and C(I) is a complete normed linear space, the latter being essentially a reformulation of the well-known fact that a uniformly Cauchy sequence of continuous functions on I isR uniformly convergent with a continuous limit. However, C(I) with k f k1 ¼ I j f j dl is a normed space that is not complete. Lack of completeness is a consequence of Proposition 3.3.14. (c) The list of properties of k f kp in Remark 3.3.9 says that Lp(X), 1
p
∞, is a normed linear space. By Theorems 3.3.10 and 3.3.12, it is complete and therefore a Banach space. Definition A linear transformation of a linear space V into a linear space V1 is a mapping F of V into V1 such that Fðax þ byÞ ¼ aFðxÞ þ bFðyÞ for x, y 2 V and scalars a, b. It is trivial to verify that F(0) = 0. In the particular case when V = {0}, there is only one linear transformation possible. In the special case in which V1 is R, the field of scalars, F is called a linear functional. Examples (i) Let X be an arbitrary nonempty set and V be the linear space of all real-valued functions on X. Given any fixed x0 2 X, the map F0 of V into R defined by F0(f) = f(x0) is a linear functional on V. More generally, given finitely many x1, x2, …, xn 2 X and corresponding real numbers k1, k2, …, n   P kn, the mapping F of V into R defined by Fðf Þ ¼ kj f xj is a linear j¼1

functional on V. (ii) Let V ¼ Rn . Recall that this is just a special case of (i) above, wherein X has been chosen to be the finite set {1, 2, …, n}. Thus, given real numbers

340

6 Lebesgue Spaces and Modes of Convergence

k1, k2, …, kn, the mapping F such that F ðx1 ; x2 ; . . .; xn Þ ¼

n P

kj xj is a linear

j¼1

functional. In fact, every linear functional F is of this kind: Given F, put kj = F(0, 0, …, 0, 1, 0, …, 0), where the only “1” is in the jth position, and one can then check by computation that the equality F ðx1 ; x2 ; . . .; xn Þ ¼ n P

kj xj holds. Thus all linear functionals look alike and we know exactly how

j¼1

they look. R (iii) The map F : C½a; b ! R defined by Fðf Þ ¼ ½a;b f dm is a linear functional on the linear space C[a, b]. As before, given finitely many x1, x2, …, xn 2 [a, b] and corresponding real numbers k1, k2, …, kn, the mapping F of n   P C[a, b] into R defined by Fðf Þ ¼ kj f xj is a linear functional on C[a, b]. j¼1

The norm of any linear transformation T from a normed linear space V 6¼ {0} into a normed linear space V1, denoted by kT k, is defined to be  kT k ¼ sup

 kTxk : x 6¼ 0 : k xk

In the event that V = {0}, there is only one linear transformation T possible and its norm is 0. A linear transformation is said to be bounded if the norm is finite; when V 6¼ {0}, this is equivalent to boundedness of the set mentioned above. When we speak of the norm of a linear transformation or linear functional, it is often presumed to be finite unless the context demands otherwise. Examples In the course of the preceding two examples, we described infinitely many norms on Rn and all linear functionals on it. The linear functionals can be shown to be bounded no matter which norm one takes on Rn . The norm of a linear functional naturally depends on what norm one takes on Rn and the focus here is on the different norms of the same linear functional when one takes the different norms kk1 ; kk2 and kk1 on Rn . A generalisation will be established in Propositions 6.1.2(v) and 6.1.3. Recall that a linear functional F on Rn is given by n P k1, k2, …, kn via the equality F ðx1 ; x2 ; . . .; xn Þ ¼ kj xj . i¼1

(i) First consider R with the norm kk1 . We have n

   X X n n     x   maxfjki j : 1
i
ng; kj xj 
jFðx1 ; x2 ; . . .; xn Þj ¼   j¼1 j  j¼1

6.1 The Spaces Lp as Normed Linear Spaces

341

which means the value M ¼ maxfjki j : 1
i
ng fulfils the inequality jFðxÞj
M k x k1

whenever 0 6¼ x 2 Rn :

Furthermore, M has to be |kk| for some k. Consider the vector z ¼ ðz1 ; z2 ; . . .; zn Þ 2 Rn such that zk = 1 but all other zj are 0. This z has the property that kzk1 ¼ 1 and    X n   kj zj  ¼ jkk j ¼ jkk j  1 ¼ M kzk1 : jFðzÞj ¼    j¼1 Since kzk1 [ 0, this shows that the value of M that we have given is the largest possible value of jFðxÞj=kxk1 ; kxk1 6¼ 0. It is therefore the supremum involved in the definition of kF k. Thus max fjki j : 1
i
ng is the norm of the linear functional given by k1, k2, …, kn when Rn is taken with the norm kk1 . In other words, if we think of the linear functional F as given by ðk1 ; k2 ; . . .; kn Þ 2 Rn , then its norm as a linear functional on Rn taken with the norm kk1 is kðk1 ; k2 ; . . .; kn Þk1 . This is generalised in Proposition 6.1.2(v). (ii) Now consider Rn with the norm kk2 . We have    X X n n n X 1 1   kj xj 
ð k2j Þ2 ð x2j Þ2 jFðx1 ; x2 ; . . .; xn Þj ¼    j¼1 j¼1 j¼1 which means the value M¼ð

n X

1

k2j Þ2

j¼1

fulfils the inequality jFðxÞj
M whenever 0 6¼ x 2 Rn : k x k2

342

6 Lebesgue Spaces and Modes of Convergence

In the case when kj = 0 for every j, it is trivial to see that F(x) = 0 for every x and n P hence that kF k ¼ 0 ¼ ð k2j Þ1=2 . Suppose kj 6¼ 0 for some j. Consider the j¼1

vector z ¼ ðk1 ; k2 ; . . .; kn Þ 2 Rn . This z has the property that kzk2 ¼ ð

n P j¼1

k2j Þ1=2

and    X X n n n n X X 1 1   kj zj  ¼ k2j ¼ ð k2j Þ2 ð k2j Þ2 ¼ M kzk2 : jFðzÞj ¼    j¼1 j¼1 j¼1 j¼1 Since kzk2 [ 0, this shows that the value of M that we have given is the largest possible value of jFðxÞj=kxk2 ; kxk2 6¼ 0. It is therefore the supremum involved n P in the definition of kF k: Thus ð k2j Þ1=2 is the norm of the linear functional j¼1

given by k1, k2, …, kn when Rn is taken with the norm kk2 . In other words, if we think of the linear functional F as given by ðk1 ; k2 ; . . .; kn Þ 2 Rn , then its norm as a linear functional on Rn taken with the norm kk2 is kðk1 ; k2 ; . . .; kn Þk2 . This is generalised in Proposition 6.1.3. (iii) It is left to the reader to check that the norm of the linear functional given by n   P kj  when Rn is taken with the norm kk . That is to say, k1 ; k2 ; . . .; kn is 1 j¼1

if we think of the linear functional F as given by ðk1 ; k2 ; . . .; kn Þ 2 Rn , then its norm as a linear functional on Rn taken with the norm kk1 is kðk1 ; k2 ; . . .; kn Þk1 . This too is generalised in Proposition 6.1.3 below. Remark The above ideas extend to the case when the linear space admits complex scalars. However, we shall not be concerned with the extension. Hölder’s Inequality suggests a way of defining bounded linear functionals on Lp spaces. We begin by doing this in the case p = 1. Proposition 6.1.2 Let g 2 L∞(X) and let Fg be the mapping on L1(X) defined by Z fg dl; f 2L1 ðXÞ: ð6:1Þ Fg ðf Þ ¼ X

Then Fg has the following properties: (i) Fg is well defined; (ii) Fg is linear, which is to say, for a; b 2 R and f, h 2 L1(X), we have Fg(af + bh) = aFg(f) + bFg(h); (iii) Fg ðf Þ
kgk1 k f k1 ; f 2 L1 ðXÞ; (iv) Fg is bounded and hence continuous everywhere on L1(X): If kfn  f k1 ! 0 as n ! ∞, then Fg(fn) ! Fg(f) as n ! ∞;

6.1 The Spaces Lp as Normed Linear Spaces



jFg ðf Þj

343



¼ kgk1 provided that every set of positive measure   contains a subset of finite positive measure; in particular, Fg  ¼ kgk1 under the proviso.

(v) sup

k f k1

: k f k1 6¼ 0

Proof (i) The R mapping Fg is well defined because f = f1 a.e. and g = g1 a.e. implies R fg dl ¼ X X f1 g1 dl, so that the value of Fg given by (6.1) depends only on the equivalence classes that f and g represent. Since Z  Z    fg dl
jfgjdl
kgk1 k f k1 ;   X

X

Fg(f) is a real number. (ii) and (iii) are trivial. (iv) follows from (ii) and (iii). (v) It follows from (iii) that the supremum here does not exceed the right-hand side. To obtain the reverse inequality, we proceed as follows: Since the matter is trivial when kgk1 ¼ 0, we may assume kgk1 [ 0. Given any positive e\kgk1 , let E  X be a measurable subset of positive measure on which jgj [ kgk1  e [ 0. Since every set of positive measure contains a subset of finite positive measure, we may take E to have finite positive measure, so that vE 2 L1 ðXÞ. Consequently, the function f = (sgn g)vE 2 L1(X) and is nonzero. Besides, Z Z ððsgn gÞgÞdl ¼ Fg ðf Þ ¼ jgjdl  ðkgk1  eÞk f k1 : E

E

This shows that Fg ðf Þ  kgk1  e; k f k1 and since the positive number e\kgk1 is arbitrary, the desired inequality is obtained. h In (v) of the above proposition, the proviso cannot be omitted. Let X consist of two points called a and b. Consider the measure which is 1 on {a} and ∞ on {b}. Then L1 consists of functions that are 0 at b. The function g defined to be 0 at a and 1 at b belongs to L∞ with kgk1 ¼ 1. However, Fg = 0. Fg is the prototype of a bounded linear functional defined on Lp(X). Remarks (a) If Lp(X) contains only the zero element, then any real-valued function on it is trivially continuous. Suppose Lp(X) contains a nonzero element. Then a linear functional F that is defined on Lp(X) is continuous if and only if it is bounded. Suppose that F is

344

6 Lebesgue Spaces and Modes of Convergence

continuous but unbounded. Then for each n, there is an fn 2 Lp ðXÞ such that jF ðfn Þj  nkfn kp 6¼ 0, or  !  fn jFðfn Þj   ¼ F   1: nkfn kp  nkfn kp  n Now, the sequence

fn nkfn kp

o

    is such that nkffnn k ¼ 1n ! 0 as n ! ∞, contra-

dicting the continuity of F at 0 in view of the above inequality. On the other hand, if fn ; f 2 Lp ðXÞ; n ¼ 1; 2; . . ., then by linearity and boundedness of F, we have jFðfn Þ  Fðf Þj ¼ jFðfn  f Þj
M kfn  f kp : If the right-hand side of the foregoing inequality tends to zero as n tends to infinity, so does the left-hand side. Thus the continuity of F is proved. Note that this argument also works in a general normed linear space that contains a nonzero element. 1 P (b) Let ‘p, 0 < p < ∞, be the set of all sequences {xn}n  1 such that jxn jp \1. n¼1

When p  1, it is easy to see by using Problem 3.3.P21 that ‘p is a linear space 1   P and fxn gn  1 p ¼ ð jxn jp Þ1=p defines a norm on it. In Problem 3.3.P22, it has n¼1

been proved that it is a complete metric space. When p < 1, it can be proved that ‘p is a linear space on the basis of the inequality (a + b)p
ap + bp for a, b  0. To see why the inequality (a + b)p
ap + bp holds for a, b  0 when 0 < p
1, take f(t) = (b + t)p − bp − tp, t  0, and note that f 0 ðtÞ ¼ pðb þ tÞp1  ptp1
0 for t  0. Consequently, f(t)
0 for t  0 and hence for t = a. 1 P The reader may note that, when p < 1, ð jxn jp Þ1=p does not provide a norm on n¼1

‘p . Indeed, take x = (1, 0, 0, …) and y = (0, 1, 0, 0, …); p = 12. Then k xkp ¼ kykp ¼ 1. So, kxkp þ kykp ¼ 2. But kx þ ykp ¼ 4. Let ‘1 be the set of all bounded sequences {rn}n  1 of real numbers, i.e. all 1 sequences {rn}n  1 such that sup  fjrn j : n  1g\1. The proof that ‘ is a linear space with norm given by frn gn  1 1 ¼ supfjrn j : n  1g is as below. Suppose q = {qn}n  1 and r = {rn}n  1 are sequences in ‘1 . Then kq þ rk1 ¼ supfjqn þ rn j : n  1g
supfjqn j : n  1g + supfjrn j : n  1g ¼ kqk1 + krk1 :

6.1 The Spaces Lp as Normed Linear Spaces

345

Also, kqk1 ¼ supn jqn j ¼ 0 if and only if qn = 0, n = 1, 2, …, and kaqk1 ¼ supn jaqn j ¼ jajkqk1 , where a is any real number. The above results about ‘p ð1
p\1Þ, including those of Problems 3.3.P21 and 3.3.P22, can also be obtained from the corresponding results about Lp by using 1 P Problem 3.2.P31, according to which the sum jxn jp is the integral with respect to n¼1

the counting measure of the pth power of the absolute value of the function on N that is given by {xn}n  1. The result about ‘1 can also be obtained from the corresponding result about L∞ because the supremum is the same as the essential supremum for the counting measure. For r ¼ frn gn  1 2 ‘1 , define a linear mapping F on ‘p , 0 < p
1, by the following rule: FðcÞ ¼

1 X

r n cn ;

c ¼ fcn gn  1 2 ‘p :

n¼1

It is obvious that, if F is real-valued (i.e. is a functional), then it is linear. Now, jFðcÞj
krk1

1 X

jcn j:

n¼1

Since ð

k X n¼1

jcn jÞp


k X n¼1

jcn jp


1 X

jcn jp for 0\p
1;

n¼1

using (a + b)p
ap + bp, for a, b  0, it follows that jFðcÞj
krk1 kckp . Thus F is real-valued and also a bounded linear mapping on ‘p . Observe that F is not a bounded linear functional in the sense defined earlier, because ‘p need not be a normed linear space when p < 1. It is easy to check that the linear space Lp(X), where 1
p
∞, contains a nonzero element if and only if there exists a measurable set E  X such that 0 < l(E) < ∞. In what follows in this section, whatever we have to say about an Lp space is trivial if the space consists of only the zero element; so, we shall consider only Lp spaces containing a nonzero element. (c) There are plenty of bounded linear functionals on Lp(X), 1
p < ∞, and it is possible to characterise them as we do below. Let p and q be a pair of conjugate exponents. If 1
q < ∞ (in which case it is also true that 1 < p
∞) and g 2 Lq(X), we define a linear functional Fg on Lp(X) by

346

6 Lebesgue Spaces and Modes of Convergence

Z Fg ðf Þ ¼

f 2 Lp ðXÞ:

fg dl; X

Then Fg is bounded. In fact, we have the following: Proposition 6.1.3 Let p and q be a pair of conjugate exponents, 1
q < ∞, 1 < p
∞, and g 2 Lq(X). Then the linear functional Fg defined on Lp(X) by Z Fg ðf Þ ¼

f 2 Lp ðXÞ:

fg dl;

ð6:2Þ

X

  is bounded and Fg  ¼ kgkq . Proof The Rmapping Fg is well defined because f = f1 a.e. and g = g1 a.e. implies R fg dl ¼ X f1 g1 dl, so that the value of Fg given by (6.2) above depends only on X the equivalence classes that f and g represent. If kgkq ¼ 0, then g = 0 a.e. and there is nothing to prove. So, assume kgkq 6¼ 0. Observe that by Hölder’s Inequality, we have   Fg ðf Þ
kgk k f k q p p and  indeed Fg is a bounded, hence continuous, linear functional on L (X), satisfying F g 
k g k . q   To show that Fg 
kgkq , we first suppose that p < ∞, so that q > 1. Select an f defined by q

f ¼ ðsgn gÞjgjp and note that Z k f kpp ¼

X

jgjq dl ¼ kgkqq \1;

i.e. f 2 Lp ðXÞ

fg ¼ ðsgn gÞjgjp g ¼ jgjp þ 1 ¼ jgjq : q

and

q

Thus for this particular f 2 Lp(X), we have Z Fg ðf Þ ¼

Z fg dl ¼

X

X

jgjq dl ¼ kgkqq :

6.1 The Spaces Lp as Normed Linear Spaces

347

Consequently,     Fg ðf Þ kgkqq Fg   ¼ q ¼ kgk : q k f kp kgkpq Now consider the case when p = ∞ and q = 1. Select f = sgn g. Since kgkq 6¼ 0, the function cannot be 0 a.e. and hence sgn g 6¼ 0 on a set of positive measure, which implies k f kp ¼ k f k1 ¼ 1. Also, fg = |g|. Proceeding as before, we obtain   F g   k g k ¼ k g k . h 1 q q The next theorem shows that L (X) is, in some sense, the set of all bounded linear functionals on Lp(X) when X  R is of finite Lebesgue measure and p < ∞. We begin with the following lemma, which will prove useful in the sequel. Lemma 6.1.4 Let ðX; F ; lÞ be a measure space with l(X) < ∞ and g an integrable function on X. Suppose M is a nonnegative real number such that Z     fg dl
M k f k p   X

for all bounded measurable functions f. Then g 2 Lq(X) and kgkq
M. Proof Define a sequence of bounded measurable functions as follows:  gn ðxÞ ¼

if jgðxÞj
n if jgðxÞj [ n:

gðxÞ 0

Since gn is bounded and l(X) < ∞, we have kgn kq \1. First we assume p > 1, so that q < ∞. Set q

fn ¼ ðsgn gn Þjgn jp ;

n ¼ 1; 2; . . .

Now, Z

Z p

jfn j dl ¼ X

X

jgn jq dl ¼ kgn kqq

and q

jgn jq ¼ jgn jp jgn j ¼ fn gn ¼ fn g: Hence Z kgn kqq ¼

X

q

fn g dl
M kfn kp ¼ Mðkgn kq Þp ;

348

6 Lebesgue Spaces and Modes of Convergence

using the hypothesis. Since q − q/p = 1 and kgn kq \1, we have kgn kq
M and R q q q q X jgn j dl
M . Since jgn j ! jgj as n ! 1, it follows by Fatou’s Lemma 3.2.8 that Z Z q jgj dl
lim inf jgn jq dl
M q : X

X

Thus g 2 Lq(X) and kgkq
M, proving the result for p > 1. Now, let p = 1, so that q = ∞. Let the real number M0 > M be arbitrary and A ¼ Xðjgj [ M0 Þ. If we can show that l(A) = 0, it will follow that kgk1
M0 , with the consequence that not only g 2 L∞(X) but also kgk1
M, because M0 > M is arbitrary. Suppose otherwise: l(A) > 0. Write 1 En ¼ Xðjgj [ ð1 þ ÞM0 Þ: n   Then A = [ n  1En and therefore l(En) > 0 for some n. Since jgðxÞj [ 1 þ 1n M0 for all x 2 En, we have Z

1 jgjdl  ð1 þ ÞM0 lðEn Þ: n En

ð6:3Þ

Set f = (sgn g)vEn . Then f is a bounded measurable function such that k f k1 ¼ lðEn Þ
lðXÞ\1, so that 0 < k f k1 \1. From (6.3), we obtain Z

Z fg dl ¼

X

1 1 jgjdl  ð1 þ ÞM0 lðEn Þ ¼ ð1 þ ÞM0 k f k1 n n En

[ M0 k f k1 ; considering that 0\M0 \1 and 0\k f k1 \1  M k f k1 ; in violation of the hypothesis about M. This contradiction shows that l(A) = 0, which is what we needed to prove. h It is convenient in certain parts of the next proof to assume that X = [0, 1], and we make this assumption. The necessary changes when X  R is an arbitrary measurable subset are cumbersome but minor. The theorem provides an instance when all bounded linear functionals look alike and we know how they look. Theorem 6.1.5 (Riesz Representation) Let F be a bounded linear functional on Lp(X), 1
p < ∞. Then there exists a unique element g 2 Lq(X), where 1p þ 1q ¼ 1, such that Z Fðf Þ ¼ fg dm; f 2Lp ðXÞ: ð6:4Þ X

Moreover,

6.1 The Spaces Lp as Normed Linear Spaces

349

kF k ¼ kgkq :

ð6:5Þ

Proof The uniqueness of g is clear, for if g and g1 both satisfy (6.4), then Z f ðg  g1 Þdm ¼ 0; f 2 Lp ðXÞ: X

particular, for f = vE, where E  X is measurable, we have v X E ðg  g1 Þdm ¼ 0. So, g = g1 a.e. by Problem 3.2.P14(a). Next, if (6.4) holds, then (6.5) follows in the case p > 1 by Proposition 6.1.3 and in the case p = 1 by Proposition 6.1.2(v). So, it remains to prove that g exists such that equality (6.4) holds. If kF k ¼ 0, then (6.4) and (6.5) hold with g = 0. Therefore we may assume that kF k [ 0. Let f = v[0,s], where v[0,s] denotes the characteristic function of the interval [0, s]. Let /(s) = F(v[0,s]). Clearly, / defines a real-valued function on [0, 1] and /(0) = 0. We first show that / is absolutely continuous. Suppose e > 0 is given. Let {(xi, x0i ): 1
i
n} be a finite collection of disjoint subintervals of [0, 1] n   P x0i  xi \d, where d will be appropriately chosen later. Then, for such that RIn

i¼1

f ¼

n X

ðv½0;x0i   v½0;xi  Þsgnð/ðx0i Þ  /ðxi ÞÞ;

i¼1

we have n  X  /ðx0 Þ  /ðxi Þ ¼ Fðf Þ: i

i¼1

Since k f kpp \d, we have n  X  /ðx0 Þ  /ðxi Þ ¼ Fðf Þ
kF kk f k \d1p kF k: p i i¼1

If we choose d ¼ ep =kF kp , then it follows that

n     P / x0  /ðxi Þ \ e for any finite i

i¼1

collection of disjoint intervals of total length less than d ¼ ep =kF kp, that is, / is absolutely continuous on [0, 1]. By Corollary 5.8.3, there exists an integrable function g defined on X such that Z g for all s 2 ½0; 1; /ðsÞ ¼ ½0;s

that is,

350

6 Lebesgue Spaces and Modes of Convergence

Z Fðv½0;s Þ ¼

gv½0;s dm: X

Since every step function on [0, 1] is equal (except at finitely many points) to a n P linear combination / ¼ cj v½0;sj  , we have j¼1

Z Fð/Þ ¼

g/ dm X

for every step function U by linearity of F and that of the integral. Let f be any nonnegative bounded measurable function on [0, 1] with bound M. By Proposition 3.4.3, there exists a sequence {Un}n  1 of nonnegative step functions such that limn kf  /n kp ¼ 0. By passing to a subsequence if necessary, we may assume that Un ! f a.e. (see Corollary 3.3.11). Now, the functions Wn = min{Un, M} are also nonnegative step functions and satisfy Wn ! f a.e. as well as limn kf  Wn kp ¼ 0 in view of the simple fact about real numbers that a
M ) jminfr; Mg  aj
jr  aj: Moreover, they R have a Rcommon bound M. Then F(Wn) ! F(f) because F is continuous and X Wn g ! X fg by the Dominated Convergence Theorem, the applicability of which follows from the existence of a common bound for the Wn. So, the equality (6.4) holds for an arbitrary nonnegative bounded measurable function f and hence for an arbitrary bounded measurable R  function f. Together with the fact that jFðf Þj
kF kk f kp , this implies that  X fg
kF kk f kp for an arbitrary bounded measurable function f. By appealing to Lemma 6.1.4, we can conclude that g 2 Lq(X). We proceed to prove that g satisfies (6.4). Let e > 0 be given and f 2 Lp(X) be arbitrary. By Proposition 3.4.3, there is a step function U such that kf  /kp \e. Since U is a step function, we have Z Fð/Þ ¼ /g dm: X

Hence     Z Z Z     Fðf Þ  fg dm ¼ Fðf Þ  Fð/Þ þ /g dm  fg dm    X X ZX    
jFðf Þ  Fð/Þj þ  ð/  f Þg dm X


kF kkf  /kp þ kgkq kf  /kp
ðkF k þ kgkq Þ  e:

6.1 The Spaces Lp as Normed Linear Spaces

351

Since e > 0 is arbitrary, it follows that (6.4) holds. The equality (6.5) now follows upon applying Proposition 6.1.3 if p > 1 and Proposition 6.1.2(v) if p = 1. h The next result shows that the translates of a function f in Lp ðRÞ are uniformly continuous in the norm. More precisely, we have the following: Proposition 6.1.6 (Cf. Problem 3.3.P4) If f 2 Lp ðRÞ; 1
p\1 and h 2 R, then the function fh : R ! R such that fh(x) = f(x + h) for all x 2 R belongs to Lp ðRÞ. Moreover, the function from R to Lp ðRÞ that maps h 2 R into fh 2 Lp ðRÞ is uniformly continuous. Proof That fh belongs to Lp ðRÞ is a consequence of Problem 6.1.P4. We shall prove the uniform continuity of the map h ! fh. Fix e > 0. Since f 2 Lp ðRÞ, there exists some positive number A and a continuous function g : R ! R whose support lies in [−A, A] such that kf  gkp \e (see Proposition 3.4.6). It follows from the uniform continuity of g that there exists a positive d < A such that js  tj\d implies jgðsÞ  gðtÞj\ð3AÞp  e: 1

Note that the provision that d < A ensures that (3A)−1(2A + d) < 1. Consider any s and t such that js  tj\d. Then Z jgðx þ sÞ  gðx þ tÞjp dm\ð3AÞ1  ep  ð2A þ dÞ; R

from which we obtain kgs  gt kp \e; bearing in mind that (3A)−1(2A + d) < 1. Observe that k f kp ¼ kfs kp for all s 2 R (see Problem 6.1.P4). Consequently, kfs  ft kp
kfs  gs kp þ kgs  gt kp þ kgt  ft kp    
ðf  gÞ  þ ðg  f Þ  þ e s p

\3e:

t p

h

Proposition 6.1.7 Let p and q be conjugate exponents, where 1
p < ∞, and let R f 2 Lp ðRÞ, g 2 Lq ðRÞ. Then FðtÞ ¼ R f ðx þ tÞgðxÞdm is continuous in t.

352

6 Lebesgue Spaces and Modes of Convergence

Proof Indeed, given t, h 2 R, note that Z    jFðt þ hÞ  FðtÞj ¼  ½f ðx þ t þ hÞgðxÞ  f ðx þ tÞgðxÞdm ZR
jf ðx þ t þ hÞ  f ðx þ tÞjjgðxÞjdm R 
ðft Þh  ft p kgkq ; which tends to zero as h ! 0 in view of the proposition above. h As an application of the Radon–Nikodým Theorem 5.10.8, we discuss the Riesz Representation Theorem 6.1.5 on a general finite measure space, not just a measurable subset of R. We present only that portion of the proof which differs significantly from the proof of the case above. As the rest of the arguments are no different from those in Theorem 6.1.5 above, they are not included. Theorem 6.1.8 Let (X, F , l) be a measure space with l(X) < ∞ and F be a bounded linear functional on Lp ðXÞ ¼ Lp ðX; F ; lÞ; 1
p\1. Then there exists a unique element g 2 Lq ðXÞ ¼ Lq ðX; F ; lÞ, where 1p þ 1q ¼ 1, such that Z Fðf Þ ¼

fg dl;

f 2 Lp ðXÞ:

ð6:6Þ

X

Moreover, kF k ¼ kgkq :

ð6:7Þ

Proof If kF k ¼ 0, then (6.6) and (6.7) hold with g = 0. Therefore we may assume that kF k [ 0. For E 2 F , define mðEÞ ¼ FðvE Þ: Observe that m is a signed  measure: Clearly, mð£Þ ¼ F v£ ¼ Fð0Þ ¼ 0. Since vA [ B = vA + vB for disjoint A and B, we have m(A [ B) = F(vA [ B) = F(vA + vB) = F(vA) + F(vB) = m(A) + m(B). It follows that m is finitely additive. Let E ¼ [ i  1 Ei , where the measurable sets Ei are disjoint, and   Fn ¼ [ 1
k
n Ek . We have vFn  vE p ¼ ðlðEnFn ÞÞ1=p ! 0 as n ! 1. Since F is n P continuous, we have m(Fn) ! m(E). Observe that mðFn Þ ¼ mðEk Þ because the k¼1

sets Ei are disjoint and we have proved that m is finitely additive. So, the convergence just proved says that m is countably additive.

6.1 The Spaces Lp as Normed Linear Spaces

353

Also, if l(E) = 0 then kvE kp ¼ 0 and so, m(E) = 0, which shows that m  l. By the Radon–Nikodým Theorem 5.10.8, there exists a g 2 L1(l) such that, for every E 2 F, Z

Z

FðvE Þ ¼ mðEÞ ¼

g dl ¼ E

vE g dl: X

It follows from the linearity of F and of the integral that, for every simple function s, Z FðsÞ ¼

sg dl: X

Let f be a nonnegative bounded measurable function. By Theorem 2.5.9, there is a sequence {sn}n  1 of simple functions, which converges uniformly to f, and satisfies 0
sn
f. Since l(X) < ∞, we have kf  sn kp ! 0 as n ! ∞ and hence Fðf Þ ¼ lim Fðsn Þ: n!1

Moreover, {sng}n  1 converges pointwise to fg and satisfies jsn gj
jfgj. Since f is bounded and g 2 L1(l), the function jfgj is integrable. Hence by the Dominated Convergence Theorem, Z

Z sn g dl ¼

lim

n!1

X

fg dl: X

R However, each n satisfies Fðsn Þ ¼ X sn g dl because each sn is a simple function. Combining this with the previous two equalities established, we obtain Z Fðf Þ ¼

fg dl X

on the assumption that f be a nonnegative bounded measurable function. By taking differences, we can deduce the same for any bounded measurable function f. It now follows exactly as in the proof of Theorem 6.1.5 that g 2 Lq(X). Moreover, R p X jfgj dl\1 for every f 2 L (X), a fact that will be used below. We shall now prove that (6.6) holds. For every n, let 8 < n f ðxÞ /n ðxÞ ¼ : n

if f ðxÞ [ n if jf ðxÞj
n if f ðxÞ\  n:

Then each Un is a bounded measurable function and hence

354

6 Lebesgue Spaces and Modes of Convergence

Z Fð/n Þ ¼

/n g dl: X

Also, j/n  f jp
j f jp and tends to 0 pointwise. Since j f jp is integrable, it follows by the Dominated Convergence Theorem that {Un}n  1 converges to f in the Lp norm, so that Fðf Þ ¼ lim Fð/n Þ: n!1

Moreover, /n g ! fg; fj/n gjgn  1 is dominated by |fg| and the Dominated Convergence Theorem, Z

R X

jfgjdl\1, so that by

Z fg dl ¼ lim

/n g dl:

n!1

X

X

From the three preceding equalities, we conclude that Fðf Þ ¼

R X

fg dl.

h

Problem Set 6.1

R 6.1.P1. If f 2 Lp(X), ∞ > p > 1, then k f kp ¼ supf X jfgj dl : g 2 Lq ðXÞ, kgkq ¼ 1g, where q is given by 1p þ 1q ¼ 1. R If f 2 L1(X), then k f k1 ¼ sup X jfgj dl : g 2 L1 ðXÞ; kgk1 ¼ 1 . [Note that this differs from Proposition 6.1.3 in that the absolute value is inside the integral rather than outside it.] R 6.1.P2. If f 2 L∞(X), then k f k1 ¼ supf X jfgj dl : g 2 L1 ðXÞ; kgk1 ¼ 1g, provided that every set of positive measure contains a subset of finite positive measure. Show also that this is false if the proviso about the measure is omitted. [Note that this differs from Proposition 6.1.2(v) in that the absolute value is inside the integral rather than outside it.] 6.1.P3. Let {fn}n  1 be a sequence of functions in L∞(X). Prove that {fn}n  1 converges to f 2 L∞(X) if and only if there is a set E of measure zero such that fn ! f uniformly on Ec. 6.1.P4. Suppose f is integrable on R and for fixed h 2 R, let fh(x) = f(x + h) be a translate of f. Show that fh is also integrable and that Z Z fh dm ¼ f dm: R

R

6.1.P5. Show that if jfn j
M a.e. and fn ! f in Lp(X), where l(X) < ∞ and p  1, then fn ! f in Lp′(X) for 1
p′ < ∞.

6.1 The Spaces Lp as Normed Linear Spaces

355

6.1.P6. Let I ¼ ½a; b R and 1 < p < ∞. If f 2 Lp(I) and F ð xÞ ¼ C þ then show that sup

Xn1 jFðxk þ 1 Þ  Fðxk Þjp ðxk þ 1  xk Þp1

k¼0

P

R ½a;x

f dm,

Z j f jp dm;




ð6:8Þ

I

where P : a ¼ x0 \x1 \    \xn ¼ b is a partition of [a, b]. Conversely, if Xn1 jFðxk þ 1 Þ  Fðxk Þjp

K ¼ sup

k¼0

P

ðxk þ 1  xk Þp1

\1;

where P isR a partition of I, thenR there exists a / 2 Lp(I) such that FðxÞ ¼ C þ ½a;x / dm; show also that I j/jp dm
K. Use this to show that equality holds in (6.8). 6.1.P7. (Chebychev’s Inequality) Let f 2 Lp(X), where 1
p < ∞. Show that, for every k > 0, Z j f jp dl:

k lðfx : jf ðxÞj [ kgÞ
p

X

Moreover, lim kp lðfx : jf ðxÞj [ kgÞ ¼ 0:

k!1

6.1.P8. Let {fn}n  1 be a sequence in Lp(X), 1
p
∞, which converges to a function f in Lp(X). Then show that, for each g in Lq(X), where 1p þ 1q ¼ 1, we have Z

Z fg dm ¼ lim

X

6.2

n!1

fn g dm: X

Modes of Convergence

Up to this point we have encountered various types of convergence of a sequence of measurable functions, such as pointwise, pointwise a.e., uniform and Lp convergence. There are other modes of convergence that are of importance in dealing with measurable functions, namely, almost uniform convergence and convergence in measure.

356

6 Lebesgue Spaces and Modes of Convergence

The definitions of various modes of convergence that we have already encountered are restated for the sake of convenience and ready reference. Throughout this section, we shall consider only real-valued functions defined on a set X with a r-algebra F of subsets and a measure l on it. We shall soon restrict the functions to be measurable as well. Suppose that we have a sequence of functions fk on X. We may fix our attention either on the sequence {fk}k  1 whose terms are the functions themselves or on the sequence {fk(x)}k  1 whose terms are the values of the functions at the individual points x of the domain X. Let f be a real-valued function defined on X. (i) The sequence {fk}k  1 converges uniformly to f on X [in symbols, unif

limkfk = f (unif) or fk ! f ] if for every e > 0, there exists a natural number N such that jfk ðxÞ  f ðxÞj\e

whenever

k  N and x 2 X:

(ii) The sequence {fk}k  1 converges pointwise to f on X [in symbols, limkfk(x) = f(x), x 2 X or fk ! f] if the sequence {fk(x)}k  1 converges to f(x) for each x 2 X, that is, if for every e > 0 and every x 2 X, there exists a natural number N such that jfk ðxÞ  f ðxÞj\e

whenever

k  N:

Uniform and pointwise convergence make sense even when there is no r-algebra. However, the other modes of convergence to be discussed only make sense if (X, F , l) is a measure space. (iii) The sequence {fk}k  1 converges almost everywhere to f on X [in symbols, ae limkfk(x) = f(x) for almost all x or fk !f ] if there exists a set S 2 F with l(S) = 0 such that the sequence {fk(x)}k  1 converges to f(x) for each x 2 X\S, that is, for every e > 0 and every x 2 Sc, there exists a natural number N such that jfk ðxÞ  f ðxÞj\e

whenever

k  N:

(iv) The sequence {fk}k  1 in Lp(X), 1
p < ∞, converges in the Lp norm to f 2 Lp

Lp(X) [in symbols, limkfk = f in Lp(X) or fk !f ] if for every e > 0, there exists a natural number N such that Z kf k  f kp ¼ ð

jfk  f jp dlÞp \e whenever 1

X

k  N:

6.2 Modes of Convergence

357

Lp norm convergence is also called convergence in the mean of order p or simply L convergence. This kind of convergence applies only to a sequence of Lp functions while the other three modes of convergence just described apply to any kinds of real-valued functions, whether measurable or not. Besides the above four modes of convergence, we discuss two other modes of convergence—alluded to in the opening paragraph—making six in all. The first of the remaining two modes of convergence applies to functions that need not be measurable. p

Definition 6.2.1 Let {fk}k  1 be a sequence of functions on X and f be a function on X. Then we say that fk converges to f almost uniformly [in symbols, limkfk = f a.u. unif

au

or fk ! f ] if for every η > 0, there exists a set E 2 F with l(E) < η such that fk ! f on X\E. Definition 6.2.2 Let {fk}k  1 be a sequence of measurable functions on X and f be a measurable function on X. Then we say that fk converges to f in measure [in meas symbols, limkfk = f in measure or fk ! f ] if lim lðfx 2 X : jfk ðxÞ  f ðxÞj [ agÞ ¼ 0

k!1

for every a > 0. Using the notation explained at the beginning of Sect. 2.4, the condition can be abbreviated as lim lðXðjfk  f j [ aÞÞ ¼ 0:

k!1

We shall continue to use this shorter notation in this section whenever possible. Definition 6.2.3 The sequence {fk}k  1 of measurable functions on X is said to be Cauchy in measure if limp,kl(X(|fk − fp| > a)) = 0 for every a > 0. Definition 6.2.4 The sequence {fk}k  1 of real-valued functions on X is said to be almost uniformly Cauchy if, for every η > 0, there exists a set of measure less than η, on the complement of which, the sequence is uniformly Cauchy. Remarks 6.2.5 (a) If a sequence of functions converges a.e. or almost uniformly, then the limit function is unique a.e. For almost everywhere convergence, this is no different from Problem 3.2.P30 ae ae (b). Nonetheless, we discuss the matter here ab initio. Let fk !f and fk !g. Then there exist sets S 2 F , T 2 F with l(S) = l(T) = 0 such that the sequence {fk(x)}k  1 converges to f(x) for each x 2 X\S and to g(x) for each x 2 X\T. It follows that it converges to f(x) as well as g(x) for each x 2 XnðS [ TÞ. Therefore f(x) = g(x) for each x 2 XnðS [ TÞ. Since l(S [ T) = 0, we conclude ae that f = g a.e. The following converse is also true: If fk !f and f = g a.e., then

358

6 Lebesgue Spaces and Modes of Convergence ae

fk !g. This is easy to see because there exist two sets of measure zero, outside one of which fk ! f pointwise and outside the other, f = g everywhere; their union provides a set of measure zero, outside which fk ! f pointwise and ae f = g everywhere, so that fk ! g pointwise. Thus, fk !g. au au Next, suppose fk !f and fk !g. Then, for each p 2 N, there exist sets Ep, Fp unif

unif

with l(Ep) < 1/2p, l(Fp) < 1/2p and fk ! f on the complement Ecp and fk ! g on  c the complement Fcp; in particular, f = g on Epc \ Fpc ¼ Ep [ Fp . Thus Gp ¼ Ep [ Fp is a measurable set with l(Gp) < 1/p such that f = g on its complement Gcp . The set S ¼ \ p  1 Gp has measure 0 and x 2 Sc ) x 2 Gcp for some p ) au

f(x) = g(x). Thus f = g a.e. Here again, a converse is true: If fk !f and f = g a.e., au then fk !g. The proof proceeds along the same lines as in the preceding paragraph, using the fact the union of a set of measure zero and a set of measure less than e is a set of measure less than e. However, it may happen that f is measurable and g is not. Example: X ¼ fa; b; cg, F ¼ f£; fag; fb; cg; Xg, lð£Þ ¼ 0, lðfagÞ ¼ 1, lðfb; cgÞ ¼ 0; lðXÞ ¼ 1; fk = f = 1 everywhere, g(a) = 1, g(b) = 2, g(c) = 3. ae ae au au Then fk !f and fk !g as well as fk !f and fk !g but only f is measurable. This cannot happen when the measure is what will later be called “complete” (see Definition 7.2.1), because in that event, a function that agrees with a measurable function a.e. can be shown to be measurable (see Problem 3.2.P30(e)). (b) If a sequence of measurable functions converges in measure, then the limit function (measurable by the very definition of convergence in measure) is unique a.e. meas meas Let fk ! f and fk ! g. Since jf  gj
jf  fk j þ jfk  gj, it follows that, for any a > 0, a a Xðjf  gj [ aÞXðjfk  f j [ Þ [ Xðjfk  gj [ Þ: 2 2 So, a a lðXðjf  gj [ aÞÞ
lðXðjfk  f j [ ÞÞ þ lðXðjfk  gj [ ÞÞ: 2 2 Since the right-hand side of the above inequality tends to 0 as k ! ∞, it meas follows that f = g a.e. It is left to the reader to show that, if fk ! f and f = g a.e., meas then fk ! g. meas meas (c) If fk ! f and gk ! g, then meas

(i) fk þ gk ! f þ g; meas (ii) afk ! af for any real a; meas (iii) jfk j ! j f j;

6.2 Modes of Convergence

359

meas

(iv) maxffk ; gk g ! maxff ; gg, meas meas (v) fkþ ! f þ ; fk ! f  ;

meas

minffk ; gk g ! minff ; gg;

meas

meas

(vi) If l(X) < ∞, then fk2 ! f 2 and fk gk ! fg. Part (i) follows from the relation a a Xðjðfk þ gk Þ  ðf þ gÞj [ aÞXðjfk  f j [ Þ [ Xðjgk  gj [ Þ 2 2 and the arguments in (b) above. Part (ii) follows from the relation Xðjafk  af j [ aÞ ¼ Xðjfk  f j [

a Þ: jaj

Part (iii) is a consequence of the relation kfk j  jf k
jfk  f j. One now deduces part (iv) from the fact that 1 maxfb; cg ¼ ½ðb þ cÞ þ jb  cj and 2 1 minfb; cg ¼ ½ðb þ cÞ  jb  cj 2 for any real numbers b, c. For part (v), one just notes that f þ ¼ maxff ; 0g and f  ¼ maxff ; 0g. We shall establish (vi) in several steps. meas

Step 1. If f = 0, then fk2 ! 0. This is because X(|f2k − 0| > a) = X(|fk − 0| > a1/2). Step 2. If / is any measurable real-valued function, then for any d > 0, there exists a measurable set E  X and a constant M > 0 such that lðEÞ\d and |/|
M on the complement XnE ¼ Ec . To see why, let En ¼ Xðj/j [ nÞ, n = 1, 2, …. The sets En form a descending sequence with empty intersection because / is real-valued. By Proposition 3.1.8, and the hypothesis that l(X) < ∞, we have lim lðEn Þ ¼ 0. n!1

Choose n large enough to make l(En) < d. Then E = En and M = n have the required properties. meas Step 3. If / is any measurable real-valued function, then fk / ! f /. To prove this, consider any a > 0. We have to prove that lim lðX ðjfk /  f /j [ aÞÞ ¼ 0. So, let k!1

e > 0 be arbitrary and we need to produce N such that k  N ) lðX ðjfk /  f /j [ aÞÞ\e. As warranted by Step 2, there exists some measurable E and a constant M > 0 such that lðEÞ\ 2e and j/j
M on the complement Ec. Observe that this complement must then have the property that Xðjfk /  f /j [ aÞ \ E c Xðjfk  f j [

a Þ: M

360

6 Lebesgue Spaces and Modes of Convergence

Now choose the integer N to be so large that k  N ) lðXðjfk  f j [

a e ÞÞ\ : M 2

We claim that this N has the requisite property. Indeed, Xðjfk /  f /j [ aÞ ¼ Xðjfk /  f /j [ aÞ \ ðE c [ EÞðXðjfk /  f /j [ aÞ \ E c Þ [ E a Xðjfk  f j [ Þ [ E by the observation above: M

In conjunction with the fact that lðEÞ\ 2e and the manner in which N was chosen, this inclusion shows that k  N ) lðX ðjfk /  f /j [ aÞÞ\e, as claimed. meas

Step 4. The convergence fk2 ! f 2 holds. This follows from the identity fk2 ¼ ðfk  f Þ2 þ 2fk f  f 2 upon applying Step 1, Step 3, (i) and (ii). meas Step 5. The convergence fk gk ! fg holds. The identity 1 fk gk ¼ ½ðfk þ gk Þ2  ðfk  gk Þ2  4 yields the convergence when taken in conjunction with Step 4, (i) and (ii). The analogues of (i)–(v) hold for almost uniform convergence, but obviously not (vi), which is false even for uniform convergence. (d) In part (vi) of (c) above, the condition that l(X) < ∞ cannot be dispensed with, as the following example shows: Let X = (0, ∞) and fk(x) = x for each k 2 N and each x 2 X. Then f(x) = x. Let gk(x) = ak, where {ak}k  1 is a sequence of positive real numbers converging to 0. Here l(X) = ∞. Also, lðXðjfk gk  fgj [ aÞÞ ¼ lðXðak x [ aÞÞ ¼ 1 for each k and each a > 0. Consequently, fkgk does not converge to fg in measure. The space of measurable functions is “complete” with respect to convergence in measure:

6.2 Modes of Convergence

361

Theorem 6.2.6 A sequence of measurable functions is convergent in measure if and only if it is Cauchy in measure. Proof Suppose {fk}k  1 converges in measure to some (measurable) function f. It follows from the relation     a a Xðfk  fj  [ aÞXðjfk  f j [ Þ [ Xðfj  f  [ Þ 2 2 that     a a lðXðfk  fj  [ aÞÞ
lðXðjfk  f j [ ÞÞ þ lðXðfj  f  [ ÞÞ: 2 2 This implies that {fk}k  1 is Cauchy in measure. For the converse, suppose {fk}k  1 is Cauchy in measure. We can, for each natural number j, choose a natural number kj such that lðXðjfr  fs j [

1 1 ÞÞ\ j whenever r; s  kj : 2j 2

Furthermore, the integers kj can be selected to be strictly increasing. Now, let   1 Ej ¼ Xðfkj  fkj þ 1  [ j Þ: 2   Observe that l Ej \ 21j . Let FJ ¼ [ j  J Ej , so that FJ 2 F and l(FJ) < 2−(J−1). If i > j > J and x 62 FJ, then     fk ðxÞ  fk ðxÞ
jfk ðxÞ  fk ðxÞj þ    þ fk ðxÞ  fk ðxÞ i j i i1 jþ1 j ð6:9Þ 1 1 1
i1 þ    þ j \ j1 : 2 2 2 1 for all J, so that l(F) = 0. Let F ¼ \ J  1 FJ . Then F 2 F and lðFÞ
lðFJ Þ\ 2 J1 By (6.9), ffkj gj  1 converges on X\F. If we define

( f ðxÞ ¼

lim fkj ðxÞ

if x 62 F

0

if x 2 F;

j!1

then ffkj gj  1 converges a.e. to the measurable function f. Letting i ! ∞ in (6.9), we infer that, if j  J and x 62 FJ, then   f ðxÞ  fk ðxÞ
1
1 : j 2j1 2J1

ð6:10Þ

362

6 Lebesgue Spaces and Modes of Convergence

We shall show by using (6.10) that ffkj gj  1 converges uniformly to f on the complement of each set FJ and hence, for every positive a,   a lðXðfkj  f  [ ÞÞ ! 0 as j ! 1: 2

ð6:11Þ

   Indeed, given a > 0, choose J such that 21J \a2. Then X fkj  f  [ a2     X fkj  f  [ 21J for all j, which implies by (6.10) that X(|fkj − f| > a2)  FJ+1 for j  J + 1, thus ensuring that l(X(|fkj − f| > a2))
l(FJ+1) < 2−J for j  J + 1. Now,     a a Xðjfk  f j [ aÞXðfk  fkj  [ Þ [ Xðfkj  f  [ Þ: 2 2 If k and kj are sufficiently large, the measure of the first set on the right-hand side is arbitrarily small because {fk}k  1 is Cauchy in measure, and the measure of the second set is arbitrarily small because of (6.11). h Corollary 6.2.7 If a sequence of measurable functions converges in measure to some limit, then it has a subsequence which converges a.e. to the same limit. Proof Follows from the argument of Theorem 6.2.6. h If a sequence converges in measure to some limit, the sequence itself need not converge to it a.e. or almost uniformly. This is illustrated in Example 6.2.20 below. Corollary 6.2.8 If a sequence of measurable functions converges in measure to some limit, then it has a subsequence which converges almost uniformly to the same limit. Proof Follows from the argument of Theorem 6.2.6. h We note in passing that Corollary 6.2.7 is also a consequence of Corollary 6.2.8 in conjunction with Theorem 6.2.11. In what follows, we study the interrelationships between the six types of convergence described in this section up to Definition 6.2.2. We begin with some that the reader is perhaps familiar with. In order to ensure transparency in our discussions, we shall consider the interrelationship under: (A) no restriction on the total measure l(X), (B) l(X) < ∞, (C) the sequence {fk}k  1 dominated by a suitable function. We begin by considering (A). It is obvious that uniform convergence implies pointwise convergence, which in turn, implies almost everywhere convergence.

6.2 Modes of Convergence

363

⇒

Uniform convergence

Almost uniform convergence

Trivial

⇓ Trivial

Theorem 6-2.11

⇒

Pointwise convergence

Trivial

⇓

Almost everywhere convergence

Also, uniform convergence obviously implies almost uniform convergence. The reverse implications all fail, as Examples 6.2.9(a)–(c) and 6.2.12 show. Examples 6.2.9 (a) On [0, 1], the sequence given by fk(x) = xk converges pointwise to the limit function f that vanishes on [0, 1) but has value 1 when x = 1. However, the convergence is not uniform, because otherwise the limit would have to be continuous. (b) The sequence {fk}k  1 defined on [0, 1] by 

0 if x 62 Q ð1Þk if x 2 Q converges to the identically zero function a.e. but has no pointwise limit at all. If the only set of measure zero is the empty set (as is the case with the counting measure), almost everywhere convergence is the same as pointwise convergence. fk ðxÞ ¼

(c) Let X = [0, 2], F denote the r-algebra of Lebesgue measurable subsets of X and m be Lebesgue measure on F . Define fk to be v½1;2 for k = 1, 2, … and k k set f(x) = 0 for x 2 X. Then fk ! f uniformly on the complement of [0, d], where d is any positive number less than 2. In fact, if k0 > 2d, then fk vanishes on au

fk does not converge uniformly to (d, 2] for k  k0. Thus fk !f . However,

 f because fk takes the value 1 on 1k ; 2k . (d) The sequence in part (a) above converges almost uniformly to the function which is 0 everywhere on [0, 1], but does not converge to it pointwise. Thus almost uniform convergence does not imply pointwise convergence. However, almost uniform convergence does imply almost everywhere convergence (Theorem 6.2.11). It will be seen later (Theorem 6.2.14) that it implies convergence in measure if all functions in the sequence are measurable. That almost everywhere convergence does not imply almost uniform convergence is demonstrated in Example 6.2.12. The notion of almost uniformly Cauchy is equivalent to that of almost uniform convergence.

364

6 Lebesgue Spaces and Modes of Convergence

Theorem 6.2.10 Let {fk}k  1 be an almost uniformly Cauchy sequence. Then there au exists a real-valued function f such that fk !f . Proof For each n, there exists an En such that l(En)
1n and {fk}k  1 is uniformly Cauchy on Ecn. Then the set E ¼ \ n  1 En is measurable and lðEÞ
lðEn Þ
1n for all n, so that l(E) = 0. Moreover, for each x 2 E c ¼ [ n  1 Enc , the sequence {fk(x)}k  1 is Cauchy. For each x 2 Ec, define f ðxÞ ¼ lim fk ðxÞ 2 R and define k!1

f(x) = 0 for x 2 E. Then for every n, we have fk(x) ! f(x) on Ecn and {fk}k  1 is uniformly Cauchy on Ecn. We shall show that {fk}k  1 converges almost uniformly to f. With this in view, consider an arbitrary η > 0. On account of {fk}k  1 being almost uniformly Cauchy, there exists a set F of measure less than η such that {fk}k  1 is uniformly Cauchy on Fc. Considering that l(E) = 0, the set G = F [ E has the property that not only is l(G) < η but also fk(x) ! f(x) at each x 2 Gc. But {fk}k  1 is uniformly Cauchy on Fc and a fortiori on its subset Gc. It follows that {fk}k  1 converges uniformly on Gc to some limit function g. However, Gc  Ec and we already know that {fk}k  1 converges to f on Ec. Hence g = f on Gc. Thus, {fk}k  1 converges uniformly on Gc to f, while at the same time, l(G) < η. h We go on to consider the relation between convergence a.e. and almost uniform convergence. The result may be summarised as: Almost uniform convergence implies convergence a.e. Theorem 6.2.11 Let {fk}k  1 be a sequence of functions converging to f almost uniformly. Then it converges to f almost everywhere. Proof For each k, let Ek 2 F be such that lðEk Þ\ 1k and {fk}k  1 converges uniformly to f on Eck. Then, for x 2 [ k  1 Ekc , we have x 2 Ecj for some j, and so, h fk(x) ! f(x). But ð [ k  1 Ekc Þc ¼ \ k  1 Ek is in F and lð \ k  1 Ek Þ ¼ 0. Example 6.2.12 The converse of the theorem is false (see the example after Egorov’s Theorem 2.6.2): The sequence {fk}k  1, where fk ðxÞ ¼ v½k;k þ 1 on ½0; 1Þ converges pointwise to the function f = 0. To see this, fix x 2 [0, ∞) and choose k0 > x. Then jf ðxÞ  fk ðxÞj ¼ 0 for k  k0. However, the sequence does not converge almost uniformly. If it did, it would have to converge to f (and also to functions equal a.e. to it). But for e ¼ 12, any measurable set Ee with m(Ee) < e   satisfies m ½k; k þ 1 \ Eec [ 12, which has the consequence that ½k; k þ 1 \ Eec is nonempty, and any point x in it satisfies jf ðxÞ  fk ðxÞj ¼ 1. This is true for all k and thus fk cannot converge uniformly to f on Ece. It should be noted that fk does not converge to f either in measure or in the Lp norm. This is because Xðjf ðxÞ  fk ðxÞj [ 12Þ ¼ ½k; k þ 1 has measure 1 and

6.2 Modes of Convergence

365

!1p

Z ½0;1Þ

jfk  f j

p

¼ 1;

neither of which tends to 0. This shows that pointwise convergence does not imply convergence in measure or in Lp norm, and hence neither does convergence a.e. However, it will be seen in Theorem 6.2.22 that when l(X) < ∞, convergence a.e. implies convergence in measure. On the other hand, pointwise convergence does not imply convergence in the Lp norm even when l(X) < ∞. This will be shown in Example 6.2.16. Recall from the last paragraph in Remark 6.2.5(a) that a sequence {fk}k  1 of measurable functions can converge almost uniformly to a nonmeasurable limit f. However, as with convergence a.e., there always exists a measurable limit function to which the sequence converges. Proposition 6.2.13 Suppose a sequence {fk}k  1 of measurable functions converges almost uniformly to f. Then there exists a measurable function g such that au g = f a.e. and any such g has the property that fk !g. ae

Proof By Theorem 6.2.11, fk !f . Using the results of part (c) and then part (b) of ae Problem 3.2.P30, there exists some measurable g such that fk !g and g = f a.e. It au now follows from Remark 6.2.5(a) that fk !g. h It is trivial to check that uniform convergence of a sequence of measurable functions implies convergence in measure. We shall prove the corresponding result for almost uniform convergence, which may be summarised as: Almost uniform convergence implies convergence in measure. Theorem 6.2.14 If a sequence {fk}k  1 of measurable functions converges almost uniformly to f, then some measurable function g is equal to f a.e. and the sequence converges in measure to any such function g. Proof By Proposition 6.2.13, some measurable function g is equal to f a.e. and au meas au fk !g for any such g. In order to prove that fk ! g, consider any a > 0. Since fk !g, unif

for any e > 0, there exists a set Xe 2 F such that l(X\Xe) < e and fk ! g on Xe. So, for sufficiently large k, the set Xðjfk  gj [ aÞ must be contained in X\Xe. Since fk and g are both measurable functions, the set is measurable and satisfies lðXðjfk  gj [ aÞÞ
lðXnXe Þ\e: meas

Thus fk ! f .

h

366

6 Lebesgue Spaces and Modes of Convergence

Corollary 6.2.8 is a partial converse of the preceding theorem. Lp norm convergence implies convergence in measure. Theorem 6.2.15 Let {fk}k  1 be a sequence of functions in Lp(X) such that Lp

meas

fk !f 2 Lp ðXÞ, i.e., kfk  f kp ! 0 as k ! ∞. Then fk ! f . Proof First suppose p < ∞. For a > 0, set Ek ¼ Xðjfk  f j [ aÞ; then Z Z 1 1 1 ð jfk  f jp dlÞp  ð jfk  f jp dlÞp [ alðEk Þp : X

Ek

Since kfk  f kp ! 0 as k ! ∞, it follows that l(Ek) ! 0 as k ! ∞. Now suppose p = ∞. Consider any a > 0 and any e > 0. Since lim kfk  f k1 ¼ 0, there exists an N such that k  N implies kfk  f k1 \a.

k!1

On the basis of the definition of the norm kk1 , this means k  N implies l(X(|fk − f| > a)) = 0 < e. So, lim lðX ðjfk  f j [ aÞÞ ¼ 0. h k!1

Example 6.2.16 The converse of the above theorem fails. In fact, none among almost uniform convergence, convergence in measure, pointwise convergence and almost everywhere convergence implies convergence in the Lp norm. To see why, let X = [0, 2], F be the r-algebra of Lebesgue measurable subsets of X and l be the Lebesgue measure m on F . The sequence {fk}k  1, where fk ¼ kv½1k;2k ;

k ¼ 1; 2; . . .;

converges to the zero function f pointwise as well as almost uniformly and in measure. However, Z  p 1   k f k  f kp ¼ ð kv½1;2  dmÞp ½0;2

¼k

11p

k k

 1 for all k and all p  1;

that is, fk does not converge to f in the Lp norm, p < ∞. Since kfk k1 ¼ k, neither does the sequence converge to f in the L∞ norm. We shall show in Theorem 6.2.23 that uniform convergence implies convergence in the Lp norm, provided that l(X) < ∞. Here we record an example to show that this is not so when l(X) = ∞. Example 6.2.17 Let X ¼ R, F ¼ M and l be the Lebesgue measure m. The sequence {fk}k  1, where fk = k−1/pv[0,k], k = 1, 2, … converges to the zero function f uniformly. Indeed, |fk − f| equals k−1/p on [0, k] and 0 elsewhere. So, |fk(x) − f(x)| is arbitrarily small for all large k and all x 2 R. But

6.2 Modes of Convergence

367

Z kf k  f kp ¼ ð

1

R

jfk  f jp dmÞp ¼ 1 for every k;

that is, fk does not converge to f in the Lp norm. In the reverse direction, we have the following two theorems: Theorem 6.2.18 Let {fk}k  1 be a sequence of functions in Lp(X) such that Lp

au

fk !f 2 Lp ðXÞ. Then it has a subsequence ffkj gj  1 such that fkj !f . h

Proof An immediate consequence of Theorem 6.2.15 and Corollary 6.2.8. Lp

Theorem 6.2.19 Let {fk}k  1 be a sequence of functions in Lp(X) such that fk !f 2 ae Lp(X). Then it has a subsequence ffkj gj  1 such that fkj !f . Proof See Corollary 3.3.11. Also an immediate consequence of Theorem 6.2.15 and Corollary 6.2.7 as well as of Theorems 6.2.11 and 6.2.18. h We shall now present an example to show that neither Lp norm convergence nor convergence in measure implies convergence a.e. It would be pertinent to note first that convergence in measure implies the existence of a subsequence that converges to the same limit function almost uniformly (Corollary 6.2.8) and hence by Theorem 6.2.11 also a.e. Moreover, by Theorem 6.2.19, Lp norm convergence also implies the same. Example 6.2.20 Let X = [0, 1] with Lebesgue measure on the family of all

 j Lebesgue measurable subsets of it. Set Ek;j ¼ j1 k ; k , j = 1, 2, …,k and enumerate

 all these intervals as follows: E1,1 = [0, 1], E2;1 ¼ 0; 12 , E2;2 ¼ 12 ; 1 ,





 E3;1 ¼ 0; 13 , E3;2 ¼ 13 ; 23 , E3;3 ¼ 23 ; 1 , …. Let {Fk}k  1 denote the above sequence of intervals. Define fk ¼ vFk , k = 1, 2, …. Since l(Fk) ! 0 as k ! ∞, Lp

meas

we have fk !0 and also fk ! 0 [the latter convergence can be seen directly but also as a consequence of the former in view of Theorem 6.2.15]. On the other hand, we observe that X(|fk| > 0) = Fk. So, for each x 2 X = [0, 1], fk(x) = 1 for infinitely many values of k. Hence fk(x) does not converge to 0 for any x 2 X. Thus fk does not converge a.e. to 0 and hence also not almost uniformly (Theorem 6.2.11). However, the subsequence fvEn;1 gk  1 does converge to 0 almost uniformly and (hence) a.e., which confirms Corollaries 6.2.7 and 6.2.8. Example 6.2.20 leads to the (converse) question: Is convergence in (a) Lp norm or (b) measure (c) pointwise (d) almost uniformly implied by convergence a.e.? The answer to (a) is in the negative even if l(X) < ∞ [see Example 6.2.16; this example also shows that pointwise ; Lp and a.u. ; Lp ]. However, the answer to (b), though negative in general (Example 6.2.12), is in the affirmative when l(X) < ∞, as we demonstrate in the theorem below. The answers to (c) and (d) are both in the negative, as demonstrated in Examples 6.2.9(b) and 6.2.12 respectively.

6 Lebesgue Spaces and Modes of Convergence

a.e.

a.u.

368

Meas

6.2.18

6.2.9(b)

Trivial

Unif

6.2.16

Trivial

6.2.7

6.2.11

L

p

Ptwise

6.2.16

Fig. 6.1 Relations between various modes of convergence with no restriction on l(X)

The relations between various modes of convergence with no restriction on l(X) have been indicated in Fig. 6.1 on the next page. The dotted arrows mean that convergence of a suitable subsequence is implied. In order to avoid cluttering, some non-implications that follow from those that have been shown have been left out. For instance it follows from what has been shown that convergence a.e. does not imply convergence in measure (not shown) and hence that it also does not imply almost uniform convergence (also not shown).

6.2 Modes of Convergence

369

We go on to consider the situation in (B), namely, l(X) < ∞. The finiteness hypothesis plays a significant role in determining the relation between modes of convergence. For instance, we saw in an example above that uniform convergence does not imply Lp norm convergence; in that example l(X) was not finite. We shall see below (Theorem 6.2.23) that when l(X) < ∞, uniform convergence indeed implies Lp norm convergence. There are other instances of a similar nature. Almost uniform convergence does not follow from pointwise convergence (see Example 6.2.12), but if l(X) < ∞, the situation can be retrieved. Recall that almost everywhere convergence (hence pointwise convergence) implies almost uniform convergence when the total measure is finite (see Egorov’s Theorem 2.6.2). Although the theorem was proved only for Lebesgue measure m, the argument carries over to a general measure l. It may be noted that the argument need not use any analogue of Proposition 2.6.1, because Proposition 6.2.13 allows us to replace the limit by a measurable function and then Remark 6.2.5(a) allows us to switch back to the given limit. Moreover, Proposition 2.3.21 and Problem 2.3.P11, which were used in the proof of Egorov’s Theorem 2.6.2, are valid for a general measure l (see Proposition 3.1.8). We restate the theorem below for the sake of completeness. Theorem 6.2.21 (Egorov) Suppose that a sequence {fk}k  1 of measurable functions converges a.e. to f and l(X) < ∞. Then for any e > 0, there is a measurable subset Xe such that l(X\Xe) < e and the sequence {fk}k  1 converges to f uniformly on Xe. In other words, when the measure of the space is finite, convergence a.e. implies almost uniform convergence. Remark The requirement that l(X) < ∞ cannot be dropped: see the example provided soon after Theorem 2.6.2 and repeated in Example 6.2.12. Theorem 6.2.22 Let {fk}k  1 be a sequence of measurable functions such that ae fk !f . Assume that l(X) < ∞. Then there exists a measurable function g such that meas g = f a.e. and any such g has the property that fk ! g. Proof Immediate from Theorems 6.2.14 and 6.2.21. h The converse of Theorem 6.2.22 is false in view of the Example 6.2.20, in which the measure was finite. We have already seen in Example 6.2.17 that, in general, uniform convergence does not imply Lp norm convergence. We shall now show that the situation is different when l(X) < ∞. Theorem 6.2.23 Suppose that l(X) < ∞ and that the sequence {fk}k  1 in Lp Lp

converges uniformly to f on X. Then f 2 Lp and fk !f . Proof The case p = ∞ is trivial; so we consider only 1
p < ∞. Let e > 0 be unif

given. Since fk ! f , there exists a natural number k0(e) such that

370

6 Lebesgue Spaces and Modes of Convergence

jfk ðxÞ  f ðxÞj\e for k  k0 ðeÞ and x 2 X: For k  k0(e), Z kf k  f kp ¼ ð

Z p

1 p

1

jfk  f j dlÞ
ð X

1

ep dlÞp ¼ elðXÞp ; X

so that lim kfk  f kp ¼ 0:

h

k!1

Remarks 6.2.24 (a) The converse of the above theorem does not hold even if the measure is finite. The sequence with fk ¼ v½1;2 on [0, 1] illustrates this, but Example 6.2.20 k k shows this and more. (b) In Example 6.2.12, we showed that pointwise convergence does not imply convergence in measure when the measure is infinite. When the measure is finite however, convergence a.e. implies almost uniform convergence as already noted in Theorem 6.2.21, which then implies convergence in measure by Theorem 6.2.14. The relations between various modes of convergence with l(X) < ∞ have been indicated in Fig. 6.2 on the next page. As before, the dotted arrows mean that convergence of a suitable subsequence is implied, and some implications that follow from those that have been shown are left out. For instance it follows from what has been shown that pointwise convergence implies convergence in measure (not shown) but not vice versa (also not shown). Lastly, we take up case (C), when there is a dominating function. The next result, also known as the Dominated Convergence Theorem, is restated below without proof, as it follows easily from Theorem 3.2.16 and Problem 3.3. P16(b). ae

Theorem 6.2.25 Let {fk}k  1 be a sequence of functions in Lp such that fk !f . Lp

Assume that |fk|
g a.e. for every k 2 N, where g 2 Lp. Then f 2 Lp and fk !f . That the condition jfk j
g cannot be dropped follows from the remark following the proof of Theorem 3.2.16. When l(X) is not necessarily finite, almost uniform convergence, and hence also convergence in measure, does not imply Lp norm convergence (Example 6.2.16). However, this implication does hold when the convergence is dominated. meas

Theorem 6.2.26 Let {fk}k  1 be a sequence of functions in Lp such that fk ! f . Lp

Assume that jfk j
g a.e. for every k 2 N, where g 2 Lp. Then f 2 Lp and fk !f .

a.u.

371

a.e.

6.2 Modes of Convergence

6.2.7

6.2.21

Meas

6.2.18

Unif

Trivial

6.2.9(b)

6.2.16

Trivial

6.2.11

L

p

Ptwise

6.2.16

Fig. 6.2 Relations between various modes of convergence with l(X) < ∞

Proof Since the hypothesis implies that jf j
g a.e. and it is given that g 2 Lp, it follows that f 2 Lp. Suppose lim kfk  f kp 6¼ 0. Then there exists e > 0 and a subsequence {gj}j  1 k!1

of {fk}k  1 such that   gj  f   e for every j 2 N: p

ð6:12Þ

372

6 Lebesgue Spaces and Modes of Convergence meas

meas

Since {gj}j  1 is a subsequence of {fk}k  1 and fk ! f , it follows that gj ! f . By ae

Corollary 6.2.7, there is a subsequence {hi}i  1 of {gj}j  1 such that hi !f . Since ae hi !f and |hi|
g, it follows that limi khi  f kp ¼ 0 by Theorem 6.2.25. This contradicts (6.12). h Almost uniform convergence does not imply Lp norm convergence in general, even if the measure is finite (Example 6.2.16), although it does if the convergence is dominated by a function in Lp. au

Theorem 6.2.27 Let {fk}k  1 be a sequence of functions in Lp such that fk ! f 2 Lp. Lp

Assume that |fk|
g a.e. for every k 2 N, where g 2 Lp. Then fk ! f . Proof The result follows from Theorems 6.2.11 and 6.2.25.

h

Problem Set 6.2 6.2.P1. For the sequence {fk}k  1 of Example 6.2.17, determine whether it converges (a) almost uniformly (b) in measure. 6.2.P2. Suppose that {fk}k  1 is a sequence of nonnegative measurable functions defined on X that converges in measure to f. Prove that Z

Z f dl
lim inf X

fk dl: X

6.2.P3. If a sequence {fk}k  1 of measurable functions is Cauchy in measure and there exists a measurable function f to which a subsequence {fj}j  1 converges in meas measure, then fk ! f . 6.2.P4. Let l(X) < ∞ and {fk}k  1 be a sequence of measurable functions such that, 1 P meas for every e > 0, lðX ðjfk  f j  eÞÞ\1. Show that fk ! f . n¼1 R 6.2.P5. Let l(X) < ∞. Define qðf ; gÞ ¼ X 1 þjf gj jf gj dl for every pair of measurable functions f and g. Show that (a) 1 [ qðf ; gÞ  0; qðf ; gÞ ¼ qðg; f Þ and q(f, g) = 0 if and only if f = g a.e.; (b) q(f, h)
q(f, g) + q(g, h) [triangle inequality]; Remark This means q is a pseudometric in the sense of Definition 1.3.3. meas

(c) fk ! f if and only if q(fk, f) ! 0 as k ! ∞; (d) A sequence {fk}k  1 of measurable functions is Cauchy in measure if and only if it is Cauchy in the sense of the pseudometric q; (e) setting k f k ¼ qðf ; 0Þ does not provide a norm on the space of equivalence classes of measurable functions that agree a.e., except in the trivial case when l(X) = 0.

6.2 Modes of Convergence

373

6.2.P6. For any two measurable functions f and g on X, define qðf ; gÞ ¼ inffc þ lðXðjf  gj [ cÞÞ : c [ 0g: Show that, if either l(X) < ∞ or f, g, fk 2 L∞, then (a) ∞ > q(f, g)  0, and q(f, g) = 0 if and only if f = g a.e.; also, q(f, g) = q(g, f); (b) q(f, h)
q(f, g) + q(g, h) [triangle inequality]. Remark This means q is a pseudometric in the sense of Definition 1.3.3. meas

(c) fk ! f if and only if q(fk, f) ! 0 as k ! ∞; (d) a sequence {fk}k  1 of measurable functions is Cauchy in measure if and only if it is Cauchy in the sense of the pseudometric q; (e) if A  X is measurable and a > 0, then q(avA,0) = min{a, l(A)}; (f) setting k f k ¼ qðf ; 0Þ does not provide a norm on the space of equivalence classes of measurable functions that agree a.e., except in the trivial case that no subset A  X satisfies 0 < l(A) < ∞. meas

meas

6.2.P7. If fk ! f and g 2 L∞(X), then show that fk g ! fg. 6.2.P8. If f and {fk}k  1 are measurable functions defined on X with l(X) < ∞, show that the following are equivalent: meas

(i) fk ! f , (ii) every subsequence of {fk}k  1 has a subsequence converging to f a.e. 6.2.P9. For Lebesgue measure on [0, 1], show that there cannot exist a q satisfying (a) and (b) of Problems 6.2.P5 and 6.2.P6, but satisfying ae

fk ! f if and only if qðfk ; f Þ ! 0 as k ! 1 instead of (c). 6.2.P10. meas

ae

(a) If fk ! f , and fk !g, then g = f a.e. meas ae (b) If fk ! f , and fk
fk+1 a.e. for each k, then fk !f . 6.2.P11. Show that if X ¼ Z with the counting measure, then convergence in measure is equivalent to uniform convergence. Does this equivalence hold for the counting measure on an arbitrary set?

374

6 Lebesgue Spaces and Modes of Convergence

6.2.P12. Let (X, F , l) be a measure space with l(X) < ∞ and suppose f and {fk}k  1 are measurable functions on X. For e > 0 and integer k  1, put Eke ¼ Xðjfk  f j  eÞ: Prove that fk ! f (pointwise) if and only if, for all e > 0, to ∅.



[ k  n Eke

n1

decreases

6.2.P13. Let (X, F , l) be a measure space with l(X) < ∞ and suppose f and {fk}k  1 are measurable functions on X. For e > 0 and integer k  1, put Eke ¼ Xðjfk  f j  eÞ:   ae Prove that fk !f if and only if lim l [ k  n Eke ¼ 0 for all e > 0. n!1

6.2.P14. Let (X, F , l) be a measure space with l(X) < ∞ and suppose f and {fk}k  1 are measurable functions on X. For e > 0 and integer m  1, let Eme ¼ Xðjfm  f j  eÞ:   au Prove that fk ! f if and only if lim l [ m  n Eme ¼ 0 for all e > 0. n!1

Chapter 7

Product Measure and Completion

7.1

Product Measure

In Chap. 2, the intuitive notion of volume defined for cuboid s in Rn was extended to the class of Lebesgue measurable sets. The class of Lebesgue measurable sets includes, amongst others, open and closed subsets of Rn , their countable unions as well as countable intersections. In fact, the r-algebra of Lebesgue measurable subsets of Rn contains the r-algebra of Borel subsets of Rn . Since Rn þ p ¼ Rn  Rp , it is natural to ask (i) whether the r-algebra Mn þ p of Lebesgue measurable subsets of Rn þ p is the “Cartesian product” Mn  Mp of the r-algebras Mn and Mp of Lebesgue measurable subsets of Rn and Rp respectively; (ii) the Lebesgue measure mn+p in Rn þ p is the “Cartesian product” of the Lebesgue measure mn in Rn with Lebesgue measure mp in Rp . In general, we shall be concerned with the following problem. Given two measure spaces (X; F ; l) and (Y; G; m), how to define an appropriate measure space (X  Y; H; k), where H and k are related to F , G and l, m in some reasonable way, so as to include the following: A 2 F; B 2 G

)

A  B 2 H and kðA  BÞ ¼ lðAÞmðBÞ:

ð7:1Þ

We shall address these problems in reverse order, starting with the problem raised in the general context. In order to consider multiple integrals, we need to deal with measure and integration on the Cartesian product of measure spaces. To help motivate such a definition of measure, we consider the following concrete examples: (i) Let I and J be intervals in R. Then the Cartesian product I  J is a rectangle in R2 , whose area can be expressed as area(I  JÞ ¼ ‘ðIÞ‘ðJÞ ¼ mðIÞmðJÞ;

© Springer Nature Switzerland AG 2019 S. Shirali and H. L. Vasudeva, Measure and Integration, Springer Undergraduate Mathematics Series, https://doi.org/10.1007/978-3-030-18747-7_7

375

376

7 Product Measure and Completion

where m denotes the Lebesgue or the Borel measure on R. This is a restatement of (7.1) in the present context and the product measure k will be a generalisation of area to M  M-measurable sets. (ii) Let c be the counting measure on PðNÞ. If A; B 2 PðNÞ, then the number of elements in A  B can be expressed as jA  Bj ¼ j AjjBj ¼ cðAÞcðBÞ; or c0 ðA  BÞ ¼ cðAÞcðBÞ; where c0 is the counting measure on PðN  NÞ. In what follows, (X; F ; l) and (Y; G; m) are understood to be given measure spaces. Definition 7.1.1 A rectangle is a subset A  B  X  Y, where A 2 F , B 2 G. An elementary set is a finite union of disjoint rectangles. The family of elementary sets will be denoted by E. Some authors prefer the name “measurable rectangle”, but we avoid it, because it suggests that we already have a r-algebra of subsets of X  Y. A part of what (7.1) says is that the r-algebra H, which we are yet to construct, must contain all rectangles, and therefore, by virtue of being a r-algebra, must contain all elementary sets as well. This can be arranged for by using Proposition 2.3.17 [see the terminology “generated by” mentioned just after that Proposition]. The representation of a rectangle in the form A  B need not be unique. Indeed, £ ¼ X  £ ¼ £  Y: However, if P  Q = R  S is nonempty, then P = R and Q = S. If y 2 Q, then for x 2 P, (x, y) 2 P  Q = R  S, so that x 2 R and y 2 S, which implies P  R and Q  S. Similarly, one may show that R  P and S  Q. Let A be a measurable subset of R. Then A  £ ¼ £  £ ¼ £: Definition 7.1.2 The product r-algebra of F and G is the r-algebra generated by elementary sets and is denoted by the same symbol F  G as the Cartesian product (there will be no confusion on this account). Next, we wish to extend the set function defined on the family of all rectangles A  B as l(A)m(B) to the product r-algebra F  G so as to be a measure. This will be facilitated if we establish that the collection of elementary sets is an algebra. To this end, we begin with the following: Lemma 7.1.3 A finite disjoint union of elementary sets is an elementary set.

7.1 Product Measure

377

Proof Let E1, E2, …, En be disjoint elementary sets. Then their union F is a union of rectangles. The rectangles occurring in the union forming Ei are disjoint not only from each other but also from the rectangles occurring in the union forming Ej, j 6¼ i. This means all the rectangles occurring in all the n unions forming the respective elementary sets Ei are disjoint from each other. Hence F is an elementary set. h Lemma 7.1.4 If P1, P2, …, Pm are disjoint rectangles and Q is a rectangle, then ðP1 nQÞ [ ðP2 nQÞ [    [ ðPm nQÞ is an elementary set. Proof It is sufficient to prove that if P and Q are rectangles, then P\Q is an elementary set. The rest will follow by Lemma 7.1.3. Let P = A  B and Q = C  D, where A; C 2 F and B; D 2 G. Then PnQ ¼ ððAnCÞ  BÞ [ ðA  ðBnDÞÞ ¼ ððAnCÞ  BÞ [ ððA \ CÞ  ðBnDÞÞ; h

which is an elementary set.

Proposition 7.1.5 E is closed under taking finite intersections and differences. In symbols, P 2 E; Q 2 E

)

P \ Q 2 E; PnQ 2 E:

ð7:2Þ

Proof Let P ¼ P1 [ P2 [    [ Pp

and

Q ¼ Q1 [ Q2 [    [ Qq ;

where P1, P2, …, Pp are disjoint rectangles and so are Q1, Q2, …, Qq. Then P\Q ¼

p [ q [

ðPi \ Qj Þ

i¼1 j¼1

and Pi \ Qj (1  i  p, 1  j  q) are disjoint rectangles. So, P \ Q 2 E. To complete the proof of (7.2), we must show that PnQ 2 E. We accomplish this by induction on n. For n = 1, it is a consequence of Lemma 7.1.4 that PnQ 2 E. Now suppose PnQ 2 E whenever Q is a finite disjoint union of k rectangles. For a finite disjoint union Q of k + 1 rectangles Qj (1  j  k + 1), we have PnQ ¼ PnðQ1 [ Q2 [    [ Qk þ 1 Þ ¼ P \ ðQc1 \ Qc2 \    \ Qck \ Qck þ 1 Þ ¼ ðP \ ðQc1 \ Qc2 \    \ Qck ÞÞ \ Qck þ 1 ¼ ðPnðQ1 [ Q2 [    [ Qk ÞÞ \ Qck þ 1 : By the induction hypothesis, the set PnðQ1 [ Q2 [ . . . [ Qk Þ on the right-hand side here is an elementary set:

378

7 Product Measure and Completion

PnðQ1 [ Q2 [    [ Qk Þ ¼ R1 [ R2 [    [ Rr ; where R1, …, Rr are disjoint rectangles. Therefore we can further recast P\Q as PnQ ¼ ðR1 [ R2 [    [ Rr Þ \ Qck þ 1 ¼ ðR1 nQk þ 1 Þ [ ðR2 nQk þ 1 Þ [    [ ðRr nQk þ 1 Þ: By Lemma 7.1.4, P\Q is an elementary set. Induction is complete and (7.2) is established. Corollary 7.1.6 E is closed under taking finite unions and under complementation. In symbols, P 2 E; Q 2 E

)

P [ Q 2 E; Pc 2 E:

Thus E is an algebra. Proof Since P [ Q ¼ P [ ðQnPÞ, the corollary follows immediately from Lemma 7.1.3 and Proposition 7.1.5. Corollary 7.1.7 F  G ¼ M0 ðEÞ, where M0 ðEÞ denotes the monotone class generated by E. Proof The result follows from Corollary 7.1.6 and Problem 2.3.P19(b). Definition 7.1.8 The x-section of a subset E  X  Y, where x 2 X, is the subset Ex = {y 2 Y: (x, y) 2 E} of Y. Similarly for the y-section Ey, where y 2 Y. Examples 7.1.9 (a) Let A = S  T, where S and T are subsets of X and Y respectively. Then  Ax ¼

T £

x2S x 62 S



 and Ay ¼

S £

y2T y 62 T



(b) Suppose A = {(x, y) : x2 + 4y2  4}. Then  Ax ¼

1

1

½ 12 ð4  x2 Þ2 ; 12 ð4  x2 Þ2  j xj  2 £ otherwise

and  A ¼ y

1

1

½2ð1  y2 Þ2 ; 2ð1  y2 Þ2  j yj  1 £ otherwise:

(c) Let A = {(x, y) : 0  y  x2 and x  0}. Then

:

7.1 Product Measure

379

 Ax ¼

½0; x2  £

x0 otherwise

and Ay ¼

 pffiffiffi ½ y; 1Þ y0 £ otherwise:

Lemma 7.1.10 (Ex)c = (Ec)x and (Ey)c = (Ec)y. Moreover, for any sequence of sets E1, E2, …, we have [ j  1 ðEj Þx ¼ ð [ j  1 Ej Þx and \ j  1 ðEj Þy ¼ ð \ j  1 Ej Þy . Similarly for \ j  1 ðEj Þx and [ j  1 ðEj Þy . Proof y 2 (Ex)c , y 62 Ex , (x, y) 62 E , (x, y) 2 Ec , y 2 (Ec)x. So, (Ex)c = (Ec)x. An analogous argument shows that (Ey)c = (Ec)y. y2

1 [

ðEj Þx , y 2 ðEj Þx

for some j , ðx; yÞ 2 Ej

for some j

j¼1

, ðx; yÞ 2 ð

1 [

Ej Þ , y 2 ð

j¼1

This means [ j  1 ðEj Þx ¼ ð [ j  1 Ej Þx . An analogous argument shows that \ j  1 ðEj Þy ¼ ð \ j  1 Ej Þy .

1 [

Ej Þx :

j¼1

h

Lemma 7.1.11 If E is a rectangle A  B, then Ex = B for x 2 A and Ex ¼ £ for x 62 A; also, Ey = A for y 2 B and E y ¼ £ for y 62 B. In particular, if A 2 F and B 2 G, then m(Ex) = m(B)vA(x) and l(Ey) = l(A)vB(y). Proof By definition of an x-section of a set, we have y 2 Ex , (x, y) 2 E. Let x 2 A. Since E = A  B, we have (x, y) 2 E , y 2 B. Thus y 2 Ex , y 2 B. This means Ex = B. Now let x 62 A. Then (x, y) 62 E = A  B for every y 2 Y, which means Ex ¼ £: The proof for Ey is similar. The last part now follows trivially. h Proposition 7.1.12 Let E 2 F  G. Then Ex 2 G for every x 2 X, and E y 2 F for every y 2 Y. Proof Let H be the family of all E 2 F  G which satisfy the condition that Ex 2 G for every x 2 X, and E y 2 F for every y 2 Y. It is an immediate consequence of Lemma 7.1.11 that every rectangle belongs to H. Also, it is an immediate consequence of Lemma 7.1.10 that E 2 H ) E c 2 H and that for any sequence of sets E1, E2, … in H we have [ j  1 Ej 2 H. Thus H is a r-algebra containing all rectangles. By definition of the r-algebra F  G, it follows that F  G  H. On the other hand, it follows by the definition of H that H  F  G. Hence H ¼ F  G. h

380

7 Product Measure and Completion

Definition 7.1.13 Given x 2 X, the x-section of an extended real-valued function f on X  Y is the extended real-valued function fx on the domain Y given by fx(y) = f(x, y) for all y 2 Y. Similarly for the y-section fy, where y 2 Y. Note: The reader is cautioned that the y-section and the yth power are denoted by the same symbol and judgement has to be exercised to determine which is intended in any particular instance. R When fx is integrable or has integral ± ∞, we shall denote Y fx by the more R elaborate symbol Y f ðx; yÞdmðyÞ, as it will maintain greater clarity. Similarly, the R R y integral X f , when meaningful, will be denoted by X f ðx; yÞdlðxÞ. This notation obviates the need to introduce the symbols for the sections fx and fy. 2

Example 7.1.14 Let f : ð0; 1Þ  ð0; 1Þ ! R be defined by f ðx; yÞ ¼ xy þ xy . Then 1 1 and þ y 2 4y x2 3x2 : f 2 : ð0; 1Þ ! R is given by x2 þ ¼ 2 2 f12 : ð0; 1Þ ! R is given by

Proposition 7.1.15 If E  X  Y, then vEx ðyÞ ¼ ðvE Þx ðyÞ ¼ ðvE Þy ðxÞ ¼ vEy ðxÞ ¼ vðx; yÞ

for all x 2 X and y 2 Y:

If moreover E 2 F  G, then Z Z Z vE ðx; yÞdmðyÞ ¼ ðvE Þx ¼ vEx ¼ mðEx Þ Y

Y

for all x 2 X

Y

and Z

Z

X

Z ðvE Þy ¼

vE ðx; yÞdlðxÞ ¼ X

vEy ¼ lðEy Þ

for all y 2 Y:

X

Proof By definition of section, we have ðvE Þx ðyÞ ¼ ðvE Þy ðxÞ ¼ vE ðx; yÞ ¼ 1 or 0 according as ðx; yÞ 2 E or ðx; yÞ 62 E and vEx ðyÞ ¼ 1 or 0 according as y 2 Ex or y 62 Ex ¼ 1 or 0 according as ðx; yÞ 2 E or ðx; yÞ 62 E:

7.1 Product Measure

381

Similarly, vEy ðxÞ ¼ 1 or 0 according as ðx; yÞ 2 E or ðx; yÞ 62 E: This proves the string of equalities regarding characteristic functions. Since the integral of the characteristicR function of a measurable set is always R equal to measure of that set, we have Y vEx ¼ mðEx Þ and X vEy ¼ lðE y Þ. The equalities concerning the remaining integrals now follow from the string of equalities regarding characteristic functions. h Proposition 7.1.16 If E is a rectangle A  B, then ðvE Þx ¼ vEx ¼ vB

and

ðvE Þy ¼ vEy ¼ vA :

Proof Simple consequence of Proposition 7.1.15 and Lemma 7.1.11.

h

Proposition 7.1.17 Let the extended real-valued function f on X  Y be measurable. Then its sections fx and fy are also measurable. Proof Let E = {(x, y) 2 X  Y : f(x, y) > a}, where a is a real number. Then for fixed x 2 X, the set Ex = {y 2 Y : fx(y) > a} belongs to G by Proposition 7.1.12. That is, fx is a measurable function. Similarly, fy is also a measurable function. h Recall the concept of r-finiteness from Definition 5.9.1. We restate it formally here for ready reference. Definition 7.1.18 A measure space (X; F ; l) is called a r-finite measure space if there is a sequence {Xn}n  1 of F -measurable subsets such that [ j  1 Xn ¼ X and l(Xn) < ∞ for each n. (R; M; m) is r-finite. Indeed, the sets Xn = [−n, n], n 2 N, satisfy [ j  1 Xn ¼ R and m(Xn) < ∞ for each n. Let c be the counting measure on PðNÞ. Then the sets Xn = {n}, consisting of a single point each, satisfy the requirements: [ j  1 Xn ¼ N and c(Xn) = 1 for each n. Thus, (N; PðNÞ; c) is a r-finite measure space. Lemma 7.1.19 Suppose that (X; F ; l) and (Y; G; m) are r-finite measure spaces. Then for each E 2 F  G, (a) the function x ! m(Ex) defined on X is measurable, (b) Rthe function y ! Rl(Ey) defined on Y is measurable, (c) X mðEx ÞdlðxÞ ¼ Y lðE y ÞdmðyÞ: Proof To begin with, we suppose that l and m are finite measures. Let X be the family of all subsets E 2 F  G such that the assertions of the lemma hold. In what follows, we shall show that X ¼ F  G. This will be accomplished by showing that X contains every measurable rectangle and every elementary set, that is, X contains the algebra E, and that it is a monotone class contained in F  G. The desired equality X ¼ F  G will then follow by Corollary 7.1.7. If E = A  B, where A 2 F and B 2 G, then by Lemma 7.1.11, m(Ex) = m(B)vA(x) and therefore m(Ex) is a measurable function on X satisfying

382

7 Product Measure and Completion

Z mðEx ÞdlðxÞ ¼ mðBÞlðAÞ:

ð7:3Þ

X

Similarly, l(Ey) is a measurable function on Y and its integral over Y equals the right-hand side of (7.3). Therefore the assertions of the lemma are true for a measurable rectangle. If E is a finite union of disjoint rectangles, that is, if E ¼ [ 1  n  N ðAn  Bn Þ is an elementary set, then vEx ðyÞ ¼

N X

vAn ðxÞvBn ðyÞ;

n¼1

from which it readily follows that  X Z N  N X vAn ðxÞ vBn ðyÞdmðyÞ ¼ mðBn ÞvAn ðxÞ: mðEx Þ ¼ Y

n¼1

n¼1

Thus m(Ex), being a sum of measurable functions, is itself measurable. Moreover, Z mðEx ÞdlðxÞ ¼ X

N X

mðBn ÞlðAn Þ:

n¼1

Similarly, l(Ey) is a measurable function on Y and Z lðEy ÞdmðyÞ ¼ Y

N X

mðBn ÞlðAn Þ:

n¼1

Thus the assertions of the lemma hold for all elementary sets. It remains to show that X is a monotone class. Let En 2 X, n  1, be such that En  En+1 for every n and E ¼ [ n  1 En . Then (En)x  (En+1)x and (En)y  (En+1)y for every x 2 X and y 2 Y. Thus {m((En)x)}n  1 and {l((En)y)}n  1 are increasing sequences of nonnegative measurable functions, and lim mððEn Þx Þ ¼ mðEx Þ; lim lððEn Þy Þ ¼ lðE y Þ. Consequently, m(Ex) and l(Ey), being limits of measurable functions, are themselves measurable. Moreover, the functions m((En)x) and l((En)y) are nonnegative. By the Monotone Convergence Theorem 3.2.4, we have Z Z mðEx ÞdlðxÞ ¼ lim mððEn Þx ÞdlðxÞ X

n!1

X

7.1 Product Measure

383

and Z

Z lððEn Þy ÞdmðyÞ:

lðE y ÞdmðyÞ ¼ lim

n!1

Y

Y

Since En 2 X for each n  1, we have Z

Z X

lððEn Þy ÞdmðyÞ:

mððEn Þx ÞdlðxÞ ¼

Y

Therefore, it follows from the preceding two equalities that Z

Z mðEx ÞdlðxÞ ¼

X

lðEy ÞdmðyÞ; Y

that is, E ¼ [ n  1 En 2 X. Similarly, if En 2 X and En En þ 1 , n  1, we can conclude that E ¼ \ n  1 En 2 X on using the Dominated Convergence Theorem 3.2.16 and the hypothesis that l and m are both finite measures. To handle the general case, we may write X ¼ [ n  1 Xn , Y ¼ [ k  1 Yk , where {Xn} and {Yn} are disjoint sequences of sets of finite measure. In the paragraph above, we have proved that the assertions of the lemma hold for each rectangle Xn Yk when we are considering l and m restricted to measurable subsets of Xn and Yk respectively. Let E 2 F  G and write En;k ¼ E \ ðXn  Yk Þ. Then for each x, Ex ¼ [ n;k  1 ðEn;k Þx . By the finite case proved above, m((En,k)x) is a measurable 1 P function of x defined on Xn for each k; so, mððEn;k Þx Þ is a measurable function k¼1

defined on Xn. Hence mðEx Þ ¼

1 X

mððEn;k Þx Þ

n;k¼1

is F -measurable. Similarly, l(Ey) is G-measurable. An application of the Monotone Convergence Theorem shows that Z mðEx ÞdlðxÞ ¼ X

¼

1 Z X 1 X n¼1 Xn k¼1 1 X 1 Z X n¼1 k¼1

This completes the proof.

Yk

mððEn;k Þx ÞdlðxÞ ¼

1 X 1 Z X n¼1 k¼1

Z lððEn;k Þy ÞdmðyÞ ¼

Xn

mððEn;k Þx ÞdlðxÞ

lðE y ÞdmðyÞ: Y

h

Remark Problem 7.1.P8 shows that the r-finiteness condition in the above lemma is essential. Definition 7.1.20 Let (X; F ; l) and (Y; G; m) be r-finite measure spaces. Then the set function l  m defined on F  G by

384

7 Product Measure and Completion

Z

Z

ðl  mÞðEÞ ¼

mðEx ÞdlðxÞ ¼

lðE y ÞdmðyÞ

X

Y

is called the product of l and m. The equality of the integrals is assured by Lemma 7.1.19. It is an immediate consequence of this definition and (7.3) of Lemma 7.1.19 that (l  m)(A  B) = l(A)m(B), where A 2 F and B 2 G. The next result shows that the set function l  m defined on F  G is indeed a measure which is r-finite. Theorem 7.1.21 Let (X; F ; l) and (Y; G; m) be r-finite measure spaces. Then the set function l  m defined on the r-algebra F  G as Z ðl  mÞðEÞ ¼

Z mðEx ÞdlðxÞ ¼

lðE y ÞdmðyÞ

X

Y

is a r-finite measure. Proof Since the function m(Ex) defined on X is nonnegative and measurable, it has an integral, the value of which may be infinite. We shall only show that l  m is a measure on F  G, because the r-finiteness will then follow easily. Clearly, (l  m)(E)  0 for all E 2 F  G and (l  m)(£) = 0. Assume that {En}n  1 is a sequence of disjoint members of F  G. Then {(En)x}n  1 is a sequence of disjoint members of G. Consequently, ðl  mÞð

1 [

Z En Þ ¼

mðð X

n¼1

¼

1 [

Z En Þx ÞdlðxÞ ¼

mðð X

n¼1

Z X 1 X n¼1

mððEn Þx ÞdlðxÞ ¼

1 Z X

1 [

X

n¼1

ðEn Þx ÞÞdlðxÞ

n¼1

mððEn Þx ÞdlðxÞ ¼

1 X

ðl  mÞðEn Þ;

n¼1

using the Monotone Convergence Theorem. Hence l  m is a measure. h The measure l  m of the above theorem is called the product measure of l and m. R R Remark 7.1.22 In view of Proposition 7.1.15, Y vE ðx; yÞdmðyÞ ¼ Y ðvE Þx ¼ mðEx Þ R R while X vE ðx; yÞdlðxÞ ¼ X ðvE Þy ¼ lðE y Þ. The conclusions (a) and (b) of Lemma 7.1.19 can be expressed by saying that and these two functions of x and y respectively are measurable; conclusion (c) can be written as Z

Z ð

X

or in alternative notation as

Y

Z ðvE Þx Þ ¼

Z ðvE Þy Þ

ð Y

X

7.1 Product Measure

385

Z

Z

Z

Z

vE ðx; yÞdmðyÞ ¼

dmðyÞ vE ðx; yÞdlðxÞ: R X Also, one can write (l  m)(E) as the integral XY vE dðl  mÞ or alternatively, R as XY vE ðx; yÞdðl  mÞðx; yÞ. Therefore by Definition 7.1.20, we further have dlðxÞ

X

Y

Z

Y

Z ð

X

Y

Z ðvE Þx Þ ¼

Z

Z

vE dðl  mÞ ¼

ðvE Þy Þ

ð

XY

Y

X

or in alternative notation, Z Z Z dlðxÞ vE ðx; yÞdmðyÞ ¼ vE ðx; yÞdðl  mÞðx; yÞ X Y Z ZXY dmðyÞ vE ðx; yÞdlðxÞ: ¼ Y

X

Another way to say this is that the “double” integral of the characteristic function of a measurable subset of the product space equals both “iterated” integrals of that function. It is understood here that integrability of the functions involved is part of the assertion. We go on to prove it for a larger class of measurable functions on the product space. Theorem 7.1.23 (Tonelli) Let (X; F ; l) and (Y; G; m) be r-finite measure spaces and let (X  Y, F  G, l  m) be the product measure space. If f is a nonnegative extended real-valued F  G-measurable function on X  Y, then (a) (b) (c) (d)

the the the the

function function function function

fy is F -measurable, fx is G-measurable, R x ! Y f ðx; yÞdmðyÞ is F -measurable, R y ! X f ðx; yÞdlðxÞ is G-measurable, and the equation Z

Z

ðeÞ

X

Z f ðx; yÞdmðyÞ ¼

dlðxÞ Y

f ðx; yÞdðl  mÞðx; yÞ Z Z ¼ dmðyÞ f ðx; yÞdlðxÞ XY

Y

X

holds. Proof It follows from Proposition 7.1.17 that (a) and (b) hold. We next deal with (c), (d) and (e). Because of the symmetry between x and y, it is enough to prove (c) and the first half of (e). As recorded in Remark 7.1.22, the result is valid when f is the characteristic function of an (F  G)-measurable set. Therefore it is also valid for any simple function [the fact that (f + g)x = fx + gx is relevant here]. Now consider an arbitrary extended real-valued nonnegative (F  G)-measurable function f on X  Y. There exists an increasing sequence of nonnegative simple functions sk, k 2 N,

386

7 Product Measure and Completion

converging to f [see Theorem 2.5.9].R Since the required conclusion is valid for simple functions, we know that each Y sk ðx; yÞdmðyÞ is an F -measurable function on X and Z

Z

Z sk ðx; yÞdmðyÞ ¼

dlðxÞ X

sk dðl  mÞðx; yÞ:

Y

ð7:4Þ

XY

Now, the Monotone Convergence Theorem 3.2.4 leads to the equality Z k!1

Z sk ðx; yÞdmðyÞ ¼

lim

f ðx; yÞdmðyÞ;

Y

Y

R showing that Y f ðx; yÞdmðyÞ is an F -measurable function on X, thereby establishing R (c). Moreover, the F -measurable functions Y sk ðx; yÞdmðyÞ form an increasing sequence, and hence another application of the Monotone Convergence Theorem 3.2.4 leads to the equality Z Z Z Z dlðxÞ sk ðx; yÞdmðyÞ ¼ dlðxÞ f ðx; yÞdmðyÞ; lim k!1

X

Y

which is the same as Z lim k!1

XY

X

Z sk dðl  mÞðx; yÞ ¼

Y

Z f ðx; yÞdmðyÞ;

dlðxÞ X

Y

in view of (7.4). Finally, yet another application of the Monotone Convergence Theorem 3.2.4 leads to the equality Z Z lim sk ðx; yÞdðl  mÞðx; yÞ ¼ f ðx; yÞdðl  mÞðx; yÞ: k!1

XY

XY

The required equality, that is the first half of (e), is now immediate from the preceding two. h Remark 7.1.24 (a) Suppose F is a function defined almost everywhere and there exists an integrable function U defined everywhere that agrees with F a.e. Then U agrees a.e. with any integrable function that agrees with F a.e. and therefore both have the same integral. We speak of F as being integrable and the integral of U as being the integral of F. The statements (c) and (d) in the next theorem are to be understood in the light of this observation. (b) Let f be a complex-valued F  G-measurable function. Then it follows from R Tonelli’s Theorem that the three integrals XY jf ðx; yÞjdðl  mÞðx; yÞ; R R R R X dlðxÞ Y jf ðx; yÞjdmðyÞ; Y dmðyÞ X jf ðx; yÞjdlðxÞare equal.

7.1 Product Measure

387

Theorem 7.1.25 (Fubini) Let (X; F ; l) and (Y; G; m) be r-finite measure spaces and let (X  Y, F  G, l  m) be the product measure space. Suppose f is a R complex-valued F  G-measurable function on X  Y such that XY jf ðx; yÞj dðl  mÞðx; yÞ\1. Then (a) (b) (c) (d)

the function fy 2 L1(X; F ; l) for almost all y 2 Y, the function fx 2 L1(Y; G; m) for R almost all x 2 X, the a.e. defined function x ! Y f ðx; yÞdmðyÞ belongs to L1(X; F ; l), R the a.e. defined function y ! X f ðx; yÞdlðxÞ belongs to L1(Y; G; m), and the equation Z Z Z f ðx; yÞdmðyÞ ¼

f ðx; yÞdðl  mÞðx; yÞ Z ZXY dmðyÞ f ðx; yÞdlðxÞ ¼

dlðxÞ (e)

X

Y

Y

X

holds. Proof It follows from Proposition 7.1.17 (the proposition being evidently true for complex-valued functions) that the functions in (a) and (b) are measurable. From the finiteness in the hypothesis and Remark 7.1.24(b), we have Z Z dmðyÞ jf ðx; yÞjdlðxÞ\1 Y

X

and hence Z jf ðx; yÞjdlðxÞ\1

for almost all y 2 Y:

X

The latter inequality means the same thing as (a). The assertion (b) is proved in like manner. We shall first prove the result for real-valued f. f+ is measurable and is nonnegative, Tonelli’s Theorem 7.1.23 shows that R Since þ Y f ðx; yÞdmðyÞ is a measurable function on X satisfying Z

Z

f þ ðx; yÞdmðyÞ ¼

dlðxÞ X

Y

Z

f þ dðl  mÞðx; yÞ: XY

Now, f is integrable and therefore Z Z dlðxÞ f þ ðx; yÞdmðyÞ\1: X

ð7:5Þ

Y

R It follows that the measurable function on X given by Y f þ ðx; yÞdmðyÞ is finite almost everywhere, which is the same as (f+)x being integrable for almost all x. Similarly,

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7 Product Measure and Completion

Z

Z dlðxÞ X

f  ðx; yÞdmðyÞ\1;

ð7:6Þ

Y

R and the measurable function on X given by Y f  ðx; yÞdmðyÞ is finite almost everywhere, which is the same as (f–)x being integrable for almost all x. By virtue of the obvious equality fx ¼ ðf þ  f  Þx ¼ ðf þ Þx  ðf  Þx ;

ð7:7Þ

we can assert that fx is integrable for almost all x, thereby justifying (b). Part (a) can be justified on the basis of an analogous argument. R By (7.5) and (7.6), the functions defined on the set X by Y f þ ðx; yÞdmðyÞ and R  Y f ðx; yÞdmðyÞ are integrable. Therefore they may have ∞ as a value only on a set of measure 0, so that the function U given by Z

Z

f þ ðx; yÞdmðyÞ 

UðxÞ ¼ Y

f  ðx; yÞdmðyÞ

ð7:8Þ

Y

is defined almost everywhere. Upon extending it to be 0 on the set where it may fail to be defined, we obtain a measurable function U such that (7.8) holds for almost all x. The functions U+ and U_ then agree a.e. with the respective functions occurring on the right-hand side of (7.8), both of which are already known to be integrable, and accordingly, U is integrable with Z Z Z Z Z þ U¼ dlðxÞ f ðx; yÞdmðyÞ  dlðxÞ f  ðx; yÞdmðyÞ: ð7:9Þ X

X

Y

X

Y

As a consequence of (7.7), Z

Z

ðf þ Þx 

fx ¼ Y

Y

Z Y

ðf  Þx

for almost all x:

In other words, Z

Z f ðx; yÞdmðyÞ ¼

Y

Z

þ

f ðx; yÞdmðyÞ  Y

f  ðx; yÞdmðyÞ

for almost all x:

Y

Since (7.8) holds for almost all x, this leads to Z UðxÞ ¼

f ðx; yÞdmðyÞ

for almost all x:

ð7:10Þ

Y

R Thus there exists an integrable function on X that agrees with Y f ðx; yÞdmðyÞ almost everywhere. This proves (c). A similar argument proves (d).

7.1 Product Measure

389

We obtain from (7.9) and (7.10) that Z Z Z Z dlðxÞ f ðx; yÞdmðyÞ ¼ dlðxÞ f þ ðx; yÞdmðyÞ X Y XZ YZ  dlðxÞ f  ðx; yÞdmðyÞ: X

ð7:11Þ

Y

Now, by definition of integral, Z

Z f ðx; yÞdðl  mÞðx; yÞ ¼

f þ ðx; yÞdðl  mÞðx; yÞ

Z

XY

XY

f  ðx; yÞdðl  mÞðx; yÞ:

 XY

By Tonelli’s Theorem 7.1.23, the right-hand side here agrees with the one in (7.11). Therefore so does the left-hand side, which is what the first equality of part (e) says. The second equality is also valid for analogous reasons. This completes the proof for real-valued f. The case when f is complex-valued now follows by splitting into real and imaginary parts. h Problem Set 7.1 7.1.P1. Let X = Y = [0, 1], F ¼ G be the r-algebra of measurable subsets of [0, 1] and l = m be Lebesgue measure on [0, 1]. Let

f ðx; yÞ ¼

x2  y2 ðx2 þ y2 Þ2

;

ðx; yÞ 2 ð0; 1Þ  ð0; 1Þ:

Prove that each of the iterated integrals of f exists. The function f is, however, not in L1([0, 1]  [0, 1]). 7.1.P2. Let X = Y = [–1, 1] and F ¼ G be the r-algebra of measurable subsets of [–1,1] and l = m be Lebesgue measure on [–1, 1]. Let  f ðx; yÞ ¼

xy ðx2 þ y2 Þ2

0

if ðx; yÞ 6¼ ð0; 0Þ otherwise:

Show that Z ½1;1

Z ð

Z ½1;1

f ðx; yÞdmðyÞÞdlðxÞ ¼ 0 ¼

½1;1

Z ð

½1;1

f ðx; yÞdlðxÞÞdmðyÞ;

390

7 Product Measure and Completion

but the function is not Lebesgue integrable over [–1, 1]  [–1, 1]. 7.1.P3. Let f 2 L1(X; F ; l) and g 2 L1(Y; G; m), and suppose /(x, y) = f(x)g(y), x 2 X, y 2 Y. Show that / 2 L1(X  Y, F  G, l  m) and Z

Z

Z

/ðx; yÞdðl  mÞ ¼ ð XY

f ðxÞdlðxÞÞð

gðyÞdmðyÞÞ:

X

Y

7.1.P4. Let X = Y = [0, 1], F ¼ G = the r-algebra of Lebesgue measurable subsets of [0, 1] and l = m = Lebesgue measure. Suppose that either f 2 L1([0, 1]  [0, 1]) or f is a measurable nonnegative function on [0, 1]  [0, 1]. Prove that Z

!

Z

½0;1

½0;x

f ðx; yÞdy dx ¼

!

Z

Z ½0;1

½y;1

f ðx; yÞdx dy:

R 1 7.1.P5. Let f 2 L1((0, a)) and gðxÞ ¼ ½x;a f ðtÞ t dt, 0 < x  a. Show that g 2 L ((0, R R a)) and that ½0;a gðxÞdx ¼ ½0;a f ðtÞdt. 7.1.P6. Let X = Y = [0, 1], F ¼ G = the r-algebra of Lebesgue measurable subsets of [0, 1] and l = m = Lebesgue measure. Define for x, y 2 [0, 1],  f ðx; yÞ ¼

1 x2Q 2y x 62 Q:

Compute Z ½0;1

Z ½0;1

!

Z

f ðx; yÞdmðyÞ dlðxÞ

Z

and ½0;1

½0;1

! f ðx; yÞdlðxÞ dmðyÞ:

Does f 2 L1([0, 1]  [0, 1])? 7.1.P7. Suppose that {am,n}m,n  1 is a double sequence of nonnegative real numbers. Then show that 1 X 1 X m¼1 n¼1

am;n ¼

1 X 1 X

am;n :

n¼1 m¼1

7.1.P8. Let X = Y = [0, 1], F ¼ G = the r-algebra of Borel subsets of [0, 1]; suppose l is Lebesgue measure on Borel subsets of X and let c be the counting measure on Y. Show that V = {(x, y) 2 [0, 1]  [0, 1] : x = y} is F  G-measurable and

7.1 Product Measure

391

Z

Z

vV dl ¼ 0

dc Y

Z

Z

but

X

vV dc ¼ 1:

dl Y

X

7.1.P9. Let X ¼ Y ¼ N and l = m = counting measure on N. Let 8 < 2  2x f ðx; yÞ ¼ 2 þ 2x : 0

if x ¼ y if x ¼ y þ 1 otherwise:

Show that the iterated integrals of f are not equal and f 62 L1(l  m). [This shows that neither the integrability condition in Fubini’s Theorem nor the nonnegativity condition in Tonelli’s Theorem can be omitted.] 7.1.P10. Let (X; F ; l) and (Y; G; m) be measure spaces with l(X) = 1 and m r-finite. If E is a measurable subset of X  Y and if mðEx Þ  12 for almost all x 2 X, then 1 mðfy 2 X : lðEy Þ ¼ 1gÞ  : 2 7.1.P11. Let X = Y = [0, 1], equipped with the r-algebra of Lebesgue measurable subsets and Lebesgue measure. Consider a real sequence {an}n  1, 0 < a1 < a2 < …, satisfying lim an ¼ 1. For each n, choose a continuous function gn such that n!1 R {t : gn(t) 6¼ 0}  (an, an+1) and also ½0;1 gn ðtÞdt ¼ 1. Define f ðx; yÞ ¼

1 X

½gn ðxÞ  gn þ 1 ðxÞgn ðyÞ:

n¼1

Note that for each (x, y), at most two terms in the sum can be nonzero. Thus no convergence problem arises in the definition of f. Show that f is not integrable and that its repeated integrals do not agree. 7.1.P12. This problem deals with a measurable function on ½0; 1  ½0; 1  R2 for which the iterated integrals exist and are equal though the function fails to be integrable. equal squares. Let   Let  I = [0, 1]  [0, 1]; divide  I into  four  I1 ¼ 0; 12  0; 12 . Next, divide the square 12 ; 1  12 ; 1 into four equal squares     and set I2 ¼ 12 ; 232  12 ; 232 , and so on. On each square Ik, k = 1, 2, …, define a function uk as follows; divide Ik into four equal squares and let uk be −1 on the interiors of the left bottom and right upper squares, 1 on the interiors of the right bottom and left upper squares and zero elsewhere. Set

392

7 Product Measure and Completion

f ðx; yÞ ¼

1 X 1 u ðx; yÞ; jI j k k¼1 k

where |Ik| is the area of the square Ik. The function f(x, y) is well defined since each point (x, y) 2 I belongs to the interior of at most one of the Ik. Show that the repeated integrals exist and are equal but the function is not integrable, though it is measurable. 7.1.P13. This is another example (besides 7.1.P2 and 7.1.P12) which shows that the integrability condition cannot be omitted from Fubini’s Theorem. Let X ¼ Y ¼ Z, the set of integers, PðZÞ the r-algebra of all subsets of Z and c be the counting measure on PðZÞ. Let 8 < x f ðx; yÞ ¼ x : 0

if y¼x if y ¼ xþ1 otherwise:

Show that the repeated integrals exist and are unequal. Also, show by a direct computation that the function is not integrable. 7.1.P14. Let (X; F ; l) and (Y; G; m) be measure spaces, A  X and B  Y. If A  B 2 F  G show that, if A 6¼ ∅, then B is measurable and if B 6¼ ∅, then A is measurable. Hence show that A  B 2 F  G if and only if A 6¼ ∅ 6¼ B and either A or B is nonmeasurable. 7.1.P15. [Cf. Problem 3.2.P13(d)] Let (X; F ; l) be a measure space. Verify for every real-valued measurable function f that Z

Z p

j f j dl ¼ X

7.2

ð0;1Þ

ptp1 lðfj f j [ tgÞdt;

0\p\1:

The Completion of a Measure

Let (X; F ; l) be a measure space. A measurable set of measure zero is called a null set. If N is a null set and A  N, then by monotonicity of l, it follows that l(A) = 0 provided that A 2 F . But it may happen that some null set may have a subset that is not measurable. Consider for instance the following simple example: let X = {a, b, c}, F ¼ f£; fbg; fa; cg; Xg, and l(X) = l({b}) = 1, l(£) = l({a, c}) = 0. Then l is a measure on F . The set N = {a, c} is a null set, but its subsets {a} and {c} are not F -measurable. The reader will recall that there is a subset of a Borel set of Lebesgue measure zero which is not a Borel set [Problem 2.3.P20].

7.2 The Completion of a Measure

393

R If N is a null set and f is a nonnegative measurable function defined on X, then N f ¼ 0, i.e., null sets are negligible in integration. It is desirable that a subset of a negligible set is also negligible, i.e., a subset of a null set be measurable—in which case, its measure must necessarily be 0. The product m  m of Lebesgue measure m on R with itself is a measure on R2 , which turns out to be different from Lebesgue measure on R2 but closely related to it. To clarify the relation between m  m on R2 and Lebesgue measure on R2 , we shall need the following concept, which has already occurred in Problem 3.2.P30(e): Definition 7.2.1 A measure space (X; F ; l) is said to be complete if whenever N 2 F is a null set and A  N, A is measurable (in which case it has to be a null set). It means the same thing to say that the measure l is complete. It is trivial that a measure space (X; PðXÞ; l) is complete no matter what l is. The Lebesgue measure space (Rn ; Mn ; mn ) is a complete measure space. Indeed, if N 2 Mn is a null set and A  N, then A is a Lebesgue measurable null set [see Remark 2.8.2(d)]. Since a subset of a Borel set of measure zero need not be a Borel set [see 2.3.P20], the Borel measure space (R; B; m) is not complete. In particular, B M. It is worth noting that when (X; F ; l) is complete, a function on X that agrees with a measurable function almost everywhere is itself measurable. This is essentially Problem 3.2.P30(e). By extending a given measure on a r-algebra, we mean obtaining a measure on a bigger r-algebra of subsets of the same space, which agrees with the given measure when restricted to the given r-algebra. One also speaks of extending a measure space. We next consider whether an incomplete measure can be extended to a complete measure. An affirmative answer is provided in the theorem below. We begin with some simple examples of the relation between the r-algebras on which the incomplete measure and its complete extension are defined. The incomplete measure space described in the opening paragraph of this section can be extended to a complete measure space by enlarging F to PðXÞ, the  ¼ l on those sets that are in F and r-algebra of all subsets of X, and setting l ðfagÞ ¼ l ðfcgÞ ¼ 0, l ðfa; bgÞ ¼ 1 ¼ l ðfc; bgÞ. Note that, like any measure on l  is complete. Observe that every set in PðXÞ is the union of a set belonging PðXÞ, l to F and a subset of a null set belonging to F . The other incomplete space we have encountered is the Borel measure space (R; B; m). In contrast, (R; M; m) is a complete measure space. Whereas B M, there is a relation between B and M that is analogous to the one between F and PðXÞ in the previous example, as we now show. A subset E of R is Lebesgue measurable, i.e., E 2 M, if and only if there exists an F r set F and a Gd -set G such that F  E  G with m(G\F) = 0 [Remark 2.3.25]. Now E\F  G\F, a Borel set of measure zero and E ¼ F [ ðEnF Þ. So, E is the union of the Borel set F and a subset E\F of a Borel set of measure zero. Thus every set of M is the union of a set belonging to B and a subset of a null set belonging to B.

394

7 Product Measure and Completion

Theorem 7.2.2 Let (X; F ; l) be a measure space and let N ¼ fN 2 F : lðNÞ ¼ 0g. The collection F of all sets of the form E [ F, where E 2 F and ðE [ FÞ ¼ lðEÞ. F  N 2 N, is a r-algebra containing F . For such sets, define l  is a measure on F , ðX; F ; l Þ is a complete measure space and the Then l  to F agrees with l, i.e. l jF ¼ l. Moreover, l  is the unique extension restriction of l of l to F .  ¼ l.) (Note: If (X; F ; l) is complete, then obviously F ¼ F and l Proof The collection F ¼ fE [ F : E 2 F and F  N 2 Ng has X as an element. We shall show that it is a r-algebra. (i) F is closed under complementation, i.e., A 2 F ) Ac 2 F . Let A ¼ E [ F, where E 2 F and F  N 2 N. Then Ac ¼ ðAc \ N c Þ [ ðAc \ N Þ ¼ ððE [ FÞc \ N c Þ [ ðAc \ N Þ ¼ ððE c \ F c Þ \ N c Þ [ ðAc \ N Þ ¼ ðEc \ ðF c \ N c ÞÞ [ ðAc \ N Þ ¼ ðE c \ N c Þ [ ðAc \ N Þbecause F  N: Now, ðE c \ N c Þ 2 F and Ac \ N  N. Therefore Ac 2 F . (ii) F is closed under countable unions, i.e., if {Ak}k  1 is a sequence of subsets belonging to F and A ¼ [ k  1 Ak , then A 2 F . By definition of F , for each k, we have Ak ¼ Ek [ Fk , where Ek 2 F and Fk  Nk 2 N. Therefore A ¼ [ k  1 Ak ¼ ð [ k  1 Ek Þ [ ð [ k  1 Fk Þ ¼ E [ F, say. Clearly, E ¼ [ k  1 Ek 2 F and F  [ k  1 Nk is a subset of a set of measure zero. Thus A 2 F . Since F has X as an element (as already noted), it follows from (i) and (ii) that F is a r-algebra.  is a measure on F and ðX; F ; l Þ is a complete Next, we shall prove that l measure space.  ¼ l; so there is nothing to prove. If (X; F ; l) is complete, then F ¼ F and l  is well defined. Suppose for Suppose (X; F ; l) is not complete. We first show that l i = 1 and 2, we have A ¼ Ei [ Fi , where Ei 2 F and Fi  Ni 2 N. Then l(E1) = l(E2), because the fact that F2  N2 implies E1  E1 [ F1 ¼ E2 [ F2  E2 [ N2 2 F

7.2 The Completion of a Measure

395

and the fact that N2 2 N further implies lðE1 Þ  lðE2 [ N2 Þ  lðE2 Þ þ lðN2 Þ ¼ lðE2 Þ;  is well reversing the roles of E1 and E2, we get l(E2)  l(E1). Consequently, l defined.  is nonnegative-valued and l ð£Þ ¼ 0, it remains only Since it is obvious that l to check its countable additivity. To this end, let {Ak}k  1 be a sequence of disjoint sets in F , so that Ak ¼ Ek [ Fk , where Ek 2 F and Fk  Nk 2 N, k = 1, 2, …. Then F ¼ [ k  1 Fk  [ k  1 Nk 2 N. Since the sets Ek must also be disjoint, we have lð

1 [ k¼1

Ak Þ ¼ lðð

1 [ k¼1

Ek Þ [ FÞ ¼ lð

1 [ k¼1

Ek Þ ¼

1 X k¼1

lðEk Þ ¼

1 X

lðAk Þ:

k¼1

This proves the countable additivity that was claimed. jF ¼ l and that is complete. Suppose E 2 F . Then we have Next, we show that l ðEÞ ¼ lðEÞ. Thus l  extends l. E ¼ E [ £ and £ 2 N, and it is clear that l ðNÞ ¼ 0. Then N ¼ E0 [ F0 ,  is complete, let A  N 2 F and l To show that l  (N) = l(E0) by definition of l  and where F0  N0 and E0 2 F , l(N0) = 0. Since l ðNÞ ¼ 0, it follows that l(E0) = 0. It follows that A  E0 [ N0 and lðE0 [ N0 Þ ¼ 0. l Therefore A ¼ £ [ A, where £ 2 F and A  E0 [ N0 2 N, so that A 2 F . 1 is  is the only measure on F that extends l. Suppose l Finally, we show that l any extension of l to F , which may or may not be complete. For F  N 2 N, we 1 ðFÞ  l 1 ðNÞ ¼ lðNÞ ¼ 0. Thus if have F ¼ £ [ F 2 F , N ¼ £ [ N 2 F and l A ¼ E [ F 2 F , it readily follows that 1 ðAÞ  l 1 ðEÞ þ l 1 ðFÞ ¼ l 1 ðEÞ ¼ lðEÞ: 1 ðEÞ  l lðEÞ ¼ l 1 ðAÞ ¼ lðEÞ ¼ l ðAÞ. This completes the proof of uniqueness. Therefore l h As the following example shows, not every complete extension is of the kind described in the theorem. The space (X; F ; l), where X = {a, b}, F ¼ f£; Xg, l(£) = 0, l(X) = 1, is complete. Nevertheless it has several extensions which are again complete: Let G ¼ PðXÞ and set m(£) = 0, m(X) = 1, m({a}) = a, m({b}) = 1 − a, where 0  a  1. Note that in this instance, F ¼ F 6¼ G. Definition 7.2.3 The measure l of the above theorem is called the completion of l ) and F is called the completion of F with respect to l. The measure space (X; F ; l is called the completion of (X; F ; l). ) be its completion. The folLet (X; F ; l) be a measure space and let (X; F ; l lowing theorem describes in another manner the relationship between the members of F and those of F . ) be its comTheorem 7.2.4 Let (X; F ; l) be a measure space and let (X; F ; l pletion. Then A 2 F if and only if there exist sets F and G in F such that

396

7 Product Measure and Completion

F  A  G and lðGnFÞ ¼ 0: Proof Suppose A satisfies the condition stated in the theorem. That is, there exist sets F and G in F such that F  A  G and l(G\F) = 0. We may write A ¼ F [ ðAnFÞ. Then F 2 F , A\F  G\F and l(G\F) = 0. On the other hand let A 2 F . Then A ¼ E [ F, where E 2 F and F  N with l(N) = 0, i.e., F is a subset of a null set. Observe that E  A  E [ N; where E 2 F , E [ N 2 F (as F is a r-algebra) and lððE [ NÞnEÞ  lðNÞ ¼ 0. h The example discussed just before Theorem 7.2.2 showed that the Lebesgue measure space (R; M; m) is the completion of the Borel measure space (R; B;mjB ), although the term “completion” had not been introduced then. The same is true in n dimensions and we record it as a theorem, without prejudice to whether the ndimensional Borel measure space (Rn ; Bn ; mn jBn ) is complete or not. In fact, we shall use it to demonstrate in Theorem 7.2.17 later that (Rn ; Bn ; mn jBn ) is indeed not complete. A somewhat different demonstration without using the next theorem directly will be called for in a problem. Theorem 7.2.5 The Lebesgue measure space (Rn ; Mn ; mn ) is the completion of the Borel measure space (Rn ; Bn ; mn jBn ). Proof The argument is exactly as in the one-dimensional case above but using Remark 2.8.19 instead of Remark 2.3.25. h Remark 7.2.6 Let F be a r-algebra generated by a class of sets S. Then the r-algebra generated by any class of sets T such that S  T  F is obviously the same as F . In what follows, this will be of significance when T is the class of countable (or finite) unions of countable (or finite) intesections of sets belonging to S and, in particular, finite unions of disjoint sets belonging to S. A rephrasing of Definition 2.7.1 is that a cuboid in Rq ðq 2 NÞ is either the empty set or a Cartesian product of q bounded intervals, each of positive length. (Recall the requirement “ai ; bi 2 R and ai < bi” stated there.) For the purposes of the present section, it will be convenient to extend the notion to include Cartesian products of intervals of any kind, including those having length ∞ as well as those having length 0, such as single point intervals and the empty set. From the elementary fact that such intervals are countable unions of countable intersections of intervals of finite positive length, it is straightforward to deduce that a Cartesian product of intervals of any nonnegative length is a countable union of countable intersections of cuboids. We shall call them “extended cuboids” or “e-cuboids” for short. Note that Rq is an e-cuboid, though not a cuboid in the sense of Definition 2.7.1.

7.2 The Completion of a Measure

397

Definition 7.2.7 An e-cuboid in Rq is a Cartesian product of q intervals, called its edges. A nonempty e-cuboid has uniquely determined edges and is a cuboid if and only if all its edges are of finite positive length. Proposition 7.2.8 The r-algebra generated by finite unions of disjoint e-cuboids is the same as the one generated by cuboids. Proof In the light of the observations preceding the above definition, the r-algebra generated by cuboids contains all e-cuboids and hence also all finite unions of disjoint e-cuboids. The result now follows from Remark 7.2.6. h Proposition 7.2.9 The Borel r-algebra Bq is generated by the class of all cuboids in Rq and also by the class of all finite unions of disjoint e-cuboids in Rq . In particular, every e-cuboid is a Borel set. Proof Denote by C the r-algebra generated by open cuboids in Rq . Now, every open set is a countable union of open cuboids; indeed, for each x in an open set O  Rq , there exists an open cuboid with rational endpoints and containing x and contained in O. It therefore follows by Remark 7.2.6 that the r-algebra generated by open sets is also C. Thus Bq ¼ C. As every cuboid is a countable intersection of open cuboids, it follows by Remark 7.2.6 that the r-algebra generated by all cuboids is also C. Since we know that Bq ¼ C, we conclude that the r-algebra generated by all cuboids is Bq . By Proposition 7.2.8, Bq is also the r-algebra generated by finite unions of disjoint e-cuboids. It is now trivial that every e-cuboid is a Borel set. h Remark 7.2.10 In the context of Cartesian products in general, it is obvious that • (E1 [ E2)  F = (E1  F) [ (E2  F), (E1 \ E2)  F = (E1  F) \ (E2  F), • (E1\E2)  F = (E1  F)\(E2  F), • if E1 and E2 are disjoint, then E1  F and E2  F are disjoint and that the assertions in the first line hold also for infinite unions and intersections. Proposition 7.2.11 For any q 2 N, finite unions of disjoint e-cuboids in Rq form an algebra of sets. Proof Denote the family of finite unions of disjoint e-cuboids in Rq by E q . Step 1. A finite union of disjoint sets belonging to E q belongs to E q . This is proved exactly like Lemma 7.1.3 with sets of E q playing the role of “elementary” sets of Sect. 7.1 and e-cuboids playing the role of “rectangles”. Step 2. (Analogue of Lemma 7.1.4) If P1, P2, …, Pk are disjoint e-cuboids and Q is an e-cuboid, then ðP1 nQÞ [ ðP2 nQÞ [ . . . [ ðPk nQÞ 2 E q . As in the proof of Lemma 7.1.4, it is sufficient to prove that if P and Q are e-cuboids, then PnQ 2 E q . The rest will follow by Step 1. We proceed by induction on q.

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7 Product Measure and Completion

Suppose q = 1. Then a cuboid is nothing but an interval. It is an elementary case by case argument that a difference P\Q of intervals P and Q is a finite union of disjoint intervals. Now assume as induction hypothesis that the assertion in question holds for e-cuboids in Rq and consider e-cuboids P and Q in Rq þ 1 . We can represent P and Q as P = A  B and Q = C  D, where A, C are intervals and B,D are e-cuboids in Rq . We have (as in Proposition 7.1.4). PnQ ¼ ððAnCÞ  BÞ [ ðA  ðBnDÞÞ ¼ ððAnCÞ  BÞ [ ððA \ CÞ  ðBnDÞÞ: ð7:12Þ Since A\C is a finite union of disjoint intervals and B is an e-cuboid in Rq , it follows from Remark 7.2.10 that ðAnCÞ  B 2 E q þ 1 . Since A \ C is an interval and BnD 2 E q by the induction hypothesis, it follows from Remark 7.2.10 that ðA \ CÞ  ðBnDÞ 2 E q þ 1 . Thus we find that both sets in the union on the right-hand side of (7.12) belong to E q þ 1 . Since the two sets are disjoint, Step 1 leads to the required conclusion that PnQ 2 E q þ 1 . This completes Step 2. Step 3. (Analogue of Proposition 7.1.5) E q is closed under taking finite intersections and differences. In symbols, P 2 Eq; Q 2 Eq )

P \ Q 2 E q ; PnQ 2 E q :

This is proved exactly like Proposition 7.1.5 with sets of E q playing the role of “elementary” sets of Sect. 7.1 and e-cuboids playing the role of “rectangles”. Step 4. (Analogue of Corollary 7.1.6) E q is closed under taking finite unions and under complementation. In symbols, P 2 Eq; Q 2 Eq

)

P [ Q 2 E q ; Pc 2 E q :

This is proved exactly like Corollary 7.1.6 with sets of E q playing the role of “elementary” sets of Sect. 7.1 and e-cuboids playing the role of “rectangles”. h By Proposition 7.2.9, an e-cuboid I1  ⋯  Iq is a Borel set and therefore has q-dimensional measure mq(I1  ⋯  Iq). If it is a nonempty cuboid, i.e., each edge Ik is a bounded interval of positive length, we know from Proposition 2.7.8 that mq(I1  ⋯  Iq) equals the volume, which is defined as the product of the lengths of the edges, P1  k  q ‘ðIk Þ. Proposition 7.2.12 For any e-cuboid with edges I1, …, Iq, the q-dimensional measure is the product of the lengths of the edges: q

Y mq I1      Iq ¼ ‘ðIk Þ: k¼1

Proof For q = 1, this is trivial because the Lebesgue measure of an interval, whether bounded or not, is its length. We need argue the matter only for q > 1.

7.2 The Completion of a Measure

399

If the e-cuboid is a nonempty cuboid, then the required equality is essentially Proposition 2.7.8. We may assume that none of the edges is ∅ because otherwise the e-cuboid reduces to ∅ and there is nothing to argue. So, suppose I1  ⋯  Iq is not a cuboid. Then one of the edges has length 0 or ∞. We consider two cases. Case 1. One of the edges has length 0. The product of the lengths of the edges is then 0 and we therefore have to show that mq(I1  ⋯  Iq) = 0. We break up the matter into two subcases. Subcase 1(a). All the other edges also have finite length. Now, any edge of length 0 consists of a single point. By replacing every such edge by an interval having arbitrarily small positive length and containing the single point, we get a cuboid containing the given e-cuboid I1  ⋯  Iq and also having arbitrarily small positive volume, which is also its mq-measure. By monotonicity of Lebesgue measure, mq(I1  ⋯  Iq) = 0. Subcase 1(b). One of the other edges has infinite length. Any edge of infinite length, which means an unbounded interval, is a countable union of disjoint intervals of finite length. Consequently, Remark 7.2.10 permits us to represent the entire e-cuboid I1  ⋯  Iq as a countable union of disjoint e-cuboids of the kind discussed in subcase 1a. Each e-cuboid in the countable union must then have mqmeasure 0 and hence so must the entire e-cuboid. This completes the discussion of Case 1. Case 2. None of the edges has length 0. Since I1  ⋯  Iq is not a cuboid, one of the edges must have infinite length. The product of the lengths of the edges is then ∞ and we therefore have to show that mq(I1  ⋯  Iq) = ∞. If there are any edges of finite length, let a be the product of those lengths. If there are no edges of finite length, then let a = 1. In either event, a > 0. Now, any unbounded interval is a countable union of disjoint intervals of length 1. Consequently, Remark 7.2.10 permits us to represent the entire e-cuboid I1  ⋯  Iq as a countable union of disjoint cuboids of volume a each. Since a > 0, it follows that mq(I1  ⋯  Iq) = ∞. This completes the discussion of Case 2. h Theorem 7.2.13 Let (X; F ; l) be a r-finite measure space. Suppose that A is an algebra of subsets of X such that the r-algebra generated by A is F . If m is a measure on F such that m(A) = l(A) for all A 2 A, then m(A) = l(A) for all A 2 F . Proof Case (i): l(X) < ∞. Let D ¼ fE 2 F : vðEÞ ¼ lðEÞg Observe that A  D  F . We shall show that D is a monotone class. Let  Ej j  1 ; Ej 2 D for each j, be such that Ej  Ej+1. Since l and m are measures, using Proposition 3.1.8(a), we get

400

7 Product Measure and Completion

m

1 [

! Ej

! 1 [



Ej : ¼ lim m Ej ¼ lim l Ej ¼ l j!1

j¼1

j!1

j¼1

Thus [ j  1 Ej 2 D.  Similarly, if Ej j  1 ; Ej 2 D for each j, are such that Ej Ej+1, then we have \ j  1 Ej 2 D, using Proposition 3.1.8 (b) and the fact that m(E1) = l(E1) < ∞. Thus D is a monotone class. By 2.3.P19 (b), it follows that F  D. This completes the proof in this case. Case (ii): l(X) = ∞. Since the measure space is r-finite, there must exist a sequence {Xk} k  1 of measurable subsets Xk such that each l(Xk) is finite and X ¼ [ k  1 Xk . By replacing each Xk by Xk n [ k0 \k Xk0 for k > 1, we may assume that the sets Xk are disjoint. Now, A ¼ A \ X ¼ [ k  1 ðA \ Xk Þ. Because l(Xk) < ∞, it follows that Case (i) is applicable to Xk, keeping in view Problem 2-3.P24. Hence m(A \ Xk) = l(A \ Xk) for each n, and consequently, mðAÞ ¼

1 X k¼1

mðA \ Xk Þ ¼

1 X

lðA \ Xk Þ ¼ lðAÞ:

h

k¼1

In what follows, it is understood that r; s 2 N and we shall be concerned with the Euclidean spaces Rr , Rs and Rr þ s ¼ Rr  Rs . Theorem 7.2.14 Br  Bs ¼ Br þ s : Proof For any q 2 N, let I q denote the class of all cuboids in Rq . It follows from the associativity of the Cartesian product that a cuboid in Rr þ s is exactly the same as a Cartesian product I  J : I 2 I r , J 2 I s of a cuboid in Rr and a cuboid in Rs . Thus, K 2 I r þ s ) K ¼ I  J for some I 2 I r ; J 2 I s ) K ¼ I  J for some I 2 Br ; J 2 Bs ) K 2 Br  Bs ; i:e:; I r þ s  Br  Bs ;

ð7:13Þ

K ¼ I  J for some I 2 I r ; J 2 I s ) K 2 I r þ s :

ð7:14Þ

and also

By Proposition 7.2.9, Br þ s is the r-algebra generated by I r þ s . Therefore it follows from (7.13) that

7.2 The Completion of a Measure

401

Br þ s Br  Bs : For the reverse inclusion, fix any I 2 I r and consider D ¼ fT  Rs : I  T 2 Br þ s g: Since Ið

1 [

Ti Þ ¼

i¼1

1 [

ðI  Ti Þ

i¼1

and I  ðRs nTÞ ¼ ðI  Rs ÞnðI  TÞ; it follows that D is a r-algebra. By Proposition 7.2.9, we have Ir þ s  Br þ s and hence it follows from (7.14) that the r-algebra D contains I s ; we therefore conclude, again on the strength of Proposition 7.2.9, that D contains Bs . So, if I 2 I r and J 2 Bs , then I  J 2 Br þ s . Similarly, fix J 2 Bs and consider the collection of all S  Rr such that S  J 2 Br þ s ; this collection is a r-algebra and it contains I r because of what was proved in the preceding paragraph. Hence it contains Br . So, if I 2 Br and J 2 Bs , then I  J 2 Br þ s . Thus Br þ s contains every Cartesian product of Borel sets (“rectangles” in the terminology of Sect. 7.1) and hence also a union of finitely many disjoint Cartesian products of Borel sets (“elementary sets” in the terminology of Sect. 7.1). By Definition 7.1.2 of a product r-algebra, it follows that Br  Bs  Br þ s ; which is the required reverse inclusion.

h

Theorem 7.2.15 The measures mr+s and mr  ms agree on Br þ s . Proof It follows from the associativity of the Cartesian product that an e-cuboid in Rr þ s exactly the same as a Cartesian product I  J of an e-cuboid I in Rr and an ecuboid J in Rs . Moreover, it follows from Proposition 7.2.12 that mr þ s ðI  JÞ ¼ mr ðIÞms ðJÞ: However, mr(I)ms(J) = (mr  ms)(I  J) by definition of the product measure. Therefore mr+s and mr  ms agree on all e-cuboids and hence also on all finite unions of disjoint e-cuboids. Now, finite unions of disjoint e-cuboids form an algebra according to Proposition 7.2.11. Moreover, the r-algebra generated by the algebra is seen to be Br þ s on the basis of Proposition 7.2.9. As mr+s is r-finite, Theorem 7.2.13 yields the desired conclusion. h Theorem 7.2.16 Let n > 1 and 1  r,s  n – 1, where r + s = n. Then

402

7 Product Measure and Completion

B n  M r  Ms  Mn : Proof Since Br  Mr and Bs  Ms , we have Br  Bs  Mr  Ms . But Br  Bs ¼ Bn by Theorem 7.2.14. Next, suppose that A 2 Mr and B 2 Ms . Then both A  Rs and Rr  B belong to Mn , as we now show: Since A 2 Mr , there exists an F r-set E  Rr and a Gd -set F  Rr such that E  A  F and mr(F\E) = 0 by Remark 2.8.19; so E  Rs  A  Rs  F  Rs , and ðmr  ms ÞððF  Rs ÞnðE  Rs ÞÞ ¼ mr ðFnEÞms ðRs Þ ¼ 0: However, ðF  Rs ÞnðE  Rs Þ 2 Br  Bs ¼ Br þ s and hence by Theorem 7.2.15, we have mn ððF  Rs ÞnðE  Rs ÞÞ ¼ 0: Since E  Rs is an F r -set in Rr þ s ¼ Rn and F  Rs a Gd -set in Rr þ s ¼ Rn , it follows on using Remark 2.8.19 that A  Rs 2 Mn . By a similar argument, we can show that Rr  B 2 Mn . The same is therefore true of their intersection A  B. It follows that Mr  Ms  Mn . h Theorem 7.2.17 The Borel measure space (Rn ; Bn ; mn jBn ) is not complete. Proof If (Rn ; Bn ; mn jBn ) were to be complete, it would be its own completion. Now, we know from Theorem 7.2.5 that its completion is (Rn ; Mn ; mn ). Therefore it suffices to show that Bn Mn . We do so by induction. For n = 1, we already know the result to be true. Assume it to be true for some n 2 N. Then there exists an E 2 Mn such that E 62 Bn . Consider the set F ¼ E  R  Rn þ 1 . Trivially, F 2 Mn  M and hence F 2 Mn þ 1 by Theorem 7.2.16. Now, fix any y 2 R. Since {x 2 Rn : (x, y) 2 F} = E and E 62 Bn , we have F 62 Bn  B by Proposition 7.1.12. But Bn  B ¼ Bn þ 1 by Theorem 7.2.14. Therefore it follows that F 62 Bn þ 1 , although F 2 Mn þ 1 . The existence of such a set F shows that Bn þ 1 Mn þ 1 . h Theorem 7.2.18 Let n > 1 and 1  r,s  n – 1, where r + s = n. Then mn is the completion of the product measure mr  ms. Proof Let A 2 Mr  Ms . By Theorem 7.2.16, A 2 Mn . So, by Remark 2.8.19, there exists an F r -set E  Rn and a Gd -set F  Rn such that E  A  F and mn(F \E) = 0. Since E; F 2 Bn and Br  Bs ¼ Bn by Theorem 7.2.14, it follows that E; F 2 Mr  Ms and ðmr  ms ÞðAnEÞ  ðmr  ms ÞðFnEÞ

ð7:15Þ

because E  A  F. But the measures mn and mr  ms agree on Bn by Theorem 7.2.15. Therefore

7.2 The Completion of a Measure

403

ðmr  ms ÞðEÞ ¼ mn ðEÞ

ð7:16Þ

and also (mr  ms)(F\E) = mn(F\E) = 0. The latter equality and (7.15) together lead to ðmr  ms ÞðAnEÞ ¼ 0: Consequently, ðmr  ms ÞðAÞ ¼ ðmr  ms ÞðE [ ðAnEÞÞ ¼ ðmr  ms ÞðEÞ ¼ mn ðEÞ by (7:16Þ ¼ mn ðAÞ: So, mr  ms agrees with mn on Mr  Ms , which is to say mr  ms is the restriction of mn to Mr  Ms . On the other hand, Bn  Mr  Ms  Mn by Theorem 7.2.16 while Mn is the completion of Bn [see Theorem 7.2.5]. Therefore Problem 7.2.P3 leads to the required conclusion. h Remark 7.2.19 From the argument above, we know independently of Theorem 7.2.5 that mr  ms agrees with mn on Mr  Ms . A somewhat different proof of incompleteness of (Rn ; Bn ; mn jBn ) that uses Remark 7.2.19 but not Theorem 7.2.5 is as follows. By Remark 2.8.19, a set of Lebesgue measure 0 is always contained in a Borel set of measure 0. Therefore incompleteness of (Rn ; Bn ; mn jBn ) will follow from the existence of a set of Lebesgue measure 0 that is not a Borel set. As before, we proceed by induction on n. For n = 1, we already know that there exists such a set. Assume the same for some n 2 N. That is to say, there exists an E 2 Mn such that mn(E) = 0 and E 62 Bn . Consider the set F ¼ E  R  Rn þ 1 . Trivially, F 2 Mn  M and hence F 2 Mn þ 1 by Theorem 7.2.16. Moreover, ðmn  mÞðFÞ ¼ mn ðEÞmðRÞ ¼ 0, which implies mn+1(F) = 0 in view of Remark 7.2.19. Now, fix any y 2 R. Since fx 2 Rn : ðx; yÞ 2 Fg ¼ E and E 62 Bn , we have F 62 Bn  B by Proposition 7.1.12. But Bn  B ¼ Bn þ 1 by Theorem 7.2.14. Therefore it follows that F 62 Bn þ 1 . Thus F  Rn þ 1 is a set of Lebesgue measure 0 that is not a Borel set. Problem Set 7.2 7.2.P1. Show that if l is not complete, then f measurable and f = g a.e. do not imply that g is measurable. 7.2.P2. Suppose F and G are r-algebras of subsets of X with measures l and m respectively. If F  G and l ¼ mjF , show that F  G.

404

7 Product Measure and Completion

7.2.P3. Suppose F and G are r-algebras of subsets of X with measures l and m jG , show that G ¼ F and m ¼ l . respectively. If F  G  F and m ¼ l 7.2.P4. [Application to Fourier Series] Prove the following: Let f be a function in Lp[−p, p], where 1  p < ∞. Then the Cesàro means of the Fourier series for f converge to f in the Lp norm.

Chapter 8

Hints

Problem Set 2.1 2.1.P1. From properties (i) and (iii) of m, show that m(£) = 0, provided that at least one set C in the domain of m satisfies m(C) < ∞. Hint: Let E1 = C and Ej = £ for each j > 1. Then {Ej}j
1 is a sequence of disjoint subsets in the domain of m. Also, [ j
1Ej = C and property (iii) implies mðCÞ ¼ mðCÞ þ

1 P

mð£Þ: Since 0  m(C) < ∞, this leads to m(£:) = 0.

i¼2

2.1.P2. From properties (i) and (iii) of m, show that, if A \ B = £, where A, B, A [ B are in the domain of m, then m(A [ B) = m(A) + m(B). Hint: If m(C) = ∞ for every C in the domain of m, then there is nothing to prove. So, assume m(C) < ∞ for some C. Let E1 = A, E2 = B and Ej = £ for j > 2. Then {Ej}j
1 is a sequence of disjoint subsets in the domain of m. Also, [ j
1Ej = A [ B is in the domain of m. The result now follows from property (iii) and 2.1.P1. 2.1.P3. From properties (i) and (iii) of m, show that, if A\B, B\A, A, B, A [ B and A \ B are in the domain of m, then m(A [ B) + m(A \ B) = m(A) + m(B). Hint: A is the disjoint union of A\B and A \ B. Therefore by 2.1.P2, m(A) = m(A\B) + m(A \ B). Similarly, m(B) = m(B\A) + m(A \ B). Now, A [ B is the disjoint union of A\B, B\A and A \ B. Therefore m(A [ B) = m(A\B) + m(B\A) + m(A \ B). The three equalities proved, when taken together, imply the required equality. Problem Set 2.2 2.2.P1. Use the results of this section to show that [0, 1] is uncountable. Hint: If it were countable or finite, its outer measure would be 0 by Example 2.2.5, but the outer measure must be 1 by Proposition 2.2.6. © Springer Nature Switzerland AG 2019 S. Shirali and H. L. Vasudeva, Measure and Integration, Springer Undergraduate Mathematics Series, https://doi.org/10.1007/978-3-030-18747-7_8

405

406

8 Hints

2.2.P2. If the domain of a countably subadditive set function m includes £ and if m(£) = 0, show that m is finitely subadditive. Hint: Imitate the argument of Corollary 2.2.10. 2.2.P3. Show that for any two sets A and B with union [0, 1], the outer measure satisfies m ðAÞ
1  m ðBÞ: Hint: m ðA [ BÞ ¼ m ð½0; 1Þ ¼ ‘ð½0; 1Þ ¼ 1. Also, m ðA [ BÞ  m ðAÞ þ m ðBÞ by Corollary 2.2.10. 2.2.P4. Let {Ij}1  j  n be a finite sequence of open intervals covering the rationals n   P in [0, 1]. Show that ‘ Ij
1: [Note: With a little extra effort, it can be shown j¼1

that the sum of lengths is strictly greater than 1; however, we shall not need this fact.] Hint: First assume that each Ij (1  j  n) has rational endpoints. Since 0 belongs to [ 1  j  nIj, it must be in one of the Ij. Let this interval be (a1, b1). Then we have a1 < 0 < b1. If b1  1, then b1 2 [0, 1] and since b1 62 (a1, b1), there must be an interval (a2, b2) in the collection {Ij: 1  j  n} such that b1 2 (a2, b2), i.e. a2 < b1 < b2. Continuing in this manner, we obtain a sequence of intervals (ai, bi) belonging to the collection {Ij: 1  j  n} such that ai < bi−1 < bi. The second part of this double inequality ensures that the intervals in our sequence are distinct. Since {Ij: 1  j  n} is a finite collection, our sequence is also finite and therefore terminates with some (ak, bk). However the sequence terminates with (ak, bk) only if bk > 1. Thus a1 < 0 < 1 < bk and ai < bi-1 for 1  i  k. Now n X

‘ðIj Þ


j¼1

k X

‘ðai ; bi Þ ¼ ðbk  ak Þ þ ðbk1  ak1 Þ þ    þ ðb1  a1 Þ:

i¼1

Therefore, in view of the inequalities noted just earlier, we have n X

‘ðIj Þ [ bk  ðak  bk1 Þ  ðak1  bk2 Þ     ða2  b1 Þ  a1

j¼1

[ bk  a1 [ 1  0 ¼ 1: If some of the Ij do not have rational endpoints, replace each Ij by Ij0 , where Ij0 has rational endpoints, Ij0  Ij and ‘ðIj0 Þ\‘ðIj Þ þ ne : Then, from what has already been proved, we have

8 Hints

407 n X

‘ðIj Þ [

j¼1

n X

‘ðIj0 Þ  e [ 1  e:

j¼1

Since e > 0 is arbitrary, it follows that

n   P ‘ Ij
1: j¼1

2.2.P5. Show that if we were to define outer measure as approximation by finitely many open intervals, i.e. if m (A) were to be

inff

n X i¼1

‘ðIi Þ: A

n [

Ii ; each Ii an open intervalg;

i¼1

then it would not be countably subadditive. Hint: The total length of finitely many open intervals covering the rationals in [0, 1] is at least 1 by 2.2.P4. Thus the outer measure of the rationals in [0, 1] would turn out to be at least 1. On the other hand, the outer measure of a set consisting of a single point would still be zero. Therefore we would not have countable subadditivity. 2.2.P6. Show that if we were to define outer measure as approximation from within, i.e. if m (A) were to be

supf

1 X i¼1

‘ðIi Þ: A 

1 [

Ii ; each Ii an open interval and i 6¼ j ) Ii \ Ij ¼ £g;

i¼1

then it would not be finitely subadditive. Hint: This procedure would assign outer measure 0 to the set of rationals in [0, 1] as well as to the set of irrationals in [0, 1], because neither of them contains a nonempty open interval [see Remark 2.2.2(a)]. But the union of the two sets, which is [0, 1] would be assigned outer measure at least 1 [it contains the open interval (0, 1)]. So, finite subadditivity would not hold. 2.2.P7. Prove that, if the open set A is the union of a sequence fIn gn
1 of disjoint open intervals, then m ðAÞ ¼

1 X

‘ðIn Þ:

n¼1

P  Hint: Since A ¼ [ n
1 In , we have m ðAÞ ¼ m ð [ n
1 In Þ  1 n¼1 m ðIn Þ P1 ¼ n¼1 ‘ðIn Þ. To prove the reverse inequality, consider an arbitrary e > 0. There   exists a sequence In0 n [ 1 of open intervals such that [ n
1 In0  A and

408

8 Hints

1   P ‘ In0  m ðAÞ þ e. As [ n
1 In0  A ¼ [ n
1 In and i 6¼ j ) Ii \ Ij ¼ £, it

n¼1

follows that ‘ðIn Þ ¼ m ðIn Þ ¼ m ð

1 [

ðIn \ Ik0 ÞÞ 

k¼1

1 X

m ðIn \ Ik0 Þ ¼

k¼1

1 X

‘ðIn \ Ik0 Þ:

k¼1

Therefore 1 X n¼1

‘ðIn Þ 

1 X 1 X

‘ðIn \ Ik0 Þ 

n¼1 k¼1

1 X 1 X

‘ðIn \ Ik0 Þ 

k¼1 n¼1

1 X

‘ðIk0 Þ  m ðAÞ þ e:

k¼1

The last but one inequality is a consequence of the fact that if a countable number of disjoint open intervals are contained in a single open interval, then the sum of their lengths is less than or equal to the length of the single interval. This follows from the corresponding fact for finitely many intervals, for which the reader can supply an argument suggested by Fig. 8.1 for three intervals. Since e > 0 is arbitrary, the required reverse inequality now follows. Remark: The above result is an easy consequence of Proposition 2.2.6, 2.3.13, 2.3.16 of the next section. 2.2.P8. Show that using closed intervals instead of open intervals in the definition of outer measure does not change the evaluation of m (A). Hint: Denote the outer measure obtained by using closed intervals by mC . From the definition, mC ðAÞ  m ðAÞ: Let e [ 0 be given. There exists a sequence {In}n
1 of 1 P closed intervals such that [ n
1 In  A and ‘ðIn Þ  mC ðAÞ þ e. For each closed n¼1   interval In, let In0 be an open interval containing In with ‘ In0 ¼ ð1 þ eÞ‘(In). Then 1 1   1     P P P mC ðAÞ þ e
‘ðIn Þ ¼ ð1 þ eÞ1 ‘ In0 : So, ‘ In0  ð1 þ eÞ mC ðAÞ þ e : n¼1

n¼1

n¼1

Clearly, [ n
1In0  A, and therefore the preceding m (A)  mC (A), considering that e > 0 is arbitrary.

inequality

implies

2.2.P9. Show that using intervals closed only on the left [or on the right, or a mixture of various types of intervals] rather than open intervals does not change the evaluation of m (A). Hint: Let {In}n
1 be a sequence of intervals which cover A. Then {Inº}n
1 is the sequence of corresponding open intervals and let {ri}i
1 be an enumeration of the points of [ n
1(In\Inº). Enclose these points in open intervals of lengths e/2i+1.

Fig. 8.1 For Problem 2.2.P7

8 Hints

409

These intervals together with Inº form an open cover of A of total length less than 1 P ‘ðIn Þ þ e . n¼1

2.2.P10. (a) For k > 0 and A R, let kA denote {x: k−1x 2 A}. Show that m (kA) = k⋅m (A). (b) For A R, let −A denote {x: −x 2 A}. Show that m (−A) = m (A). Hint: (a) For e > 0, there exists a sequence {In}n
1 of open intervals such that 1 P kA [ n
1 In and m ðkAÞ þ e
‘ðIn Þ: But then A [ n
1 k 1 In . So, m (A) 

1 P



1

m ðk In Þ ¼

n¼1

1 P

n¼1 1

‘ðk In Þ ¼ k 1

n¼1

P1 n¼1

‘ðIn Þ  k 1 ðm ðkAÞ þ eÞ:

Hence

m ðkAÞ
k  m ðAÞ. By a similar argument, m ðAÞ ¼ m ðk1 ðkAÞÞ
k1 ðm ðkAÞÞ: (b) For e > 0, there exists a sequence {In}n
1 of open intervals such that 1 P A [ n
1 In and m ðAÞ þ e
‘ðIn Þ: But then A [ n
1 ðIn Þ. So, n¼1

m ðAÞ 

1 X n¼1

m ðIn Þ ¼

1 X n¼1

‘ðIn Þ ¼

1 X

‘ðIn Þ  m ðAÞ þ e:

n¼1

Since e > 0 is arbitrary, we have m (A)  m (−A). Replace A by −A to obtain the reverse inequality. 2.2.P11. Let A = {x 2 [0, 1]: x has a decimal expansion not containing the digit 5}. Note that 0.5 2 A, because 0.5 = 0.4999 …, but 0.51 62 A. Then show that m (A) = 0. 5 6 ] [ [10 , 1], which is Hint: In fact, A = \ n
1An, where A0 = [0, 1], A1 = [0, 10 5 6 obtained from A0 on removing the open interval 10 ; 10 ; A2 is obtained from A1 on removing the open intervals 5 6 15 16 25 26 35 36 45 46 65 66 75 76 ; Þ; ð 2 ; 2 Þ; ð 2 ; 2 Þ; ð 2 ; 2 Þ; ð 2 ; 2 Þ; ð 2 ; 2 Þ; ð 2 ; 2 Þ; 102 102 10 10 10 10 10 10 10 10 10 10 10 10 85 86 95 96 ð 2 ; 2 Þ; ð 2 ; 2 Þ: 10 10 10 10

ð

 9 2 Thus A2 is a union of 10 closed intervals of total length 10 . In general, An is the  9 n n union of 10 closed intervals of total length 10 . Since A An, it follows that  9 n ! 0 as n ! ∞. m (A)  10 2.2.P12. Suppose m ðA \ IÞ  12 m ðIÞ for every interval I. Prove that m (A) = 0. Hint: Suppose 0 < m (A) < ∞. Then we may take e ¼ 12 m ðAÞ. From the definition of outer measure, there exists a cover {In}n
1 consisting of open intervals such that

410

8 Hints 1 1 [ 3 3 3X ‘ðIn Þ\ m ðAÞ ¼ m ðA \ In Þ  m ðA \ In Þ: 2 2 2 n¼1 n¼1 n¼1

1 X

For at least one n, we must have 0\‘ðIn Þ  32 m ðA \ In Þ; which is a contradiction. Now drop the hypothesis that m*(A) < ∞. Let J be an arbitrary bounded interval and put A′ = A \ J. Then for any interval I, the intersection J \ I is an interval and hence we have m ðA0 \ I Þ ¼ m ðA \ ðJ \ IÞÞ  12 m ðJ \ IÞ  12 m ðIÞ. Since m*(A′)  m*(J) < ∞, it follows from what has already been proved that m*(A′) = 0. This means m*(A \ J) = 0 for every bounded interval J. In particular, m*(A \ [n, n + 1]) = 0 for every integer n. This implies m*(A) = 0. 2.2.P13. For nonempty E R, the diameter is diam E = sup{|x−y|: x, y 2 E}. Show that m*(E)  diamE. Hint: Let d denote diamE. If d = 0, E consists of a single point and there is nothing to prove. So, suppose d > 0 and consider an arbitrary positive e < 2d. By definition, there exist x, y 2 E such that jx  yj [ d  2e [ 0. Without loss of generality, we   may assume that x < y. Note that E x  2e ; y þ 2e . For, if z 2 E but   z 62 x  2e ; y þ 2e , then y þ 2e  z [ y  x þ e if z lies to the left of x  2e and z  x þ 2e [ y  x þ e if z lies to the right of y þ 2e : In the first case, y  z [ y  x þ 2e [ d and in the second case, z − x > d, a contradiction in either case. So, m*(E)  y − x + e  d + e. Since this is true for each e > 0, it follows that m*(E)  d. Problem Set 2.3 2.3.P1. Let A, B be subsets of R and A B. Show that (a) If m*(B\A) = 0, then m*(B) = m*(A). (b) If m*(B) = m*(A) < ∞ and A 2 M, then m*(B\A) = 0. Hint: (a) By the subadditivity of outer measure, m*(B) = m*(A [ (B\A))  m*(A) + m*(B\A) = m*(A) since m*(B\A) = 0. The reverse inequality follows from Proposition 2.2.3(c). (b) Since A is measurable, m*(B) = m*(B \ A) + m*(B \ Ac) = m*(A) + m*(B\A). 2.3.P2. Suppose that A is a subset of R with the property that, for e > 0, there exist measurable sets B and C such that B A C and m(C \ Bc) < e. Show that A is measurable. Hint:  For n 2 ℕ, there exist measurable sets Bn and Cn such that Bn A Cn and m Cn \ Bcn \ 1n. Set B = [ n
1Bn and C = \ n
1Cn. Then B and C are measurable (the collection M is a r-algebra) and m(C \ Bc)  m(Cn \ Bcn)  1n for each n 2 ℕ; so, m(C \ Bc) = 0. Now, A \ Bc C \ Bc and hence m*(A \ Bc)  m*(C \ Bc) = 0. This implies A \ Bc 2 M [Remark 2.3.2(d)] and hence A = (A \ Bc) [ B 2 M:

8 Hints

411

2.3.P3. Let {En}n
1 be a sequence of sets such that E1 E2 ⋯. Show that m ðlim En Þ ¼ lim m ðEn Þ. n!1

Hint: Since the sequence {m*(En)}n
1 is monotone, limh!1 m*(En) exists, although it may be infinite. Since m*(En)  m*( [ n
1En), we have limn!1 m*(En)  m*( [ n
1En). Choose measurable Fn  En such that m(Fn) = m*(En) (see Proposition 2.2.13). Writing Bn = \ j
nFj, we have B1 B2 ⋯, with each Bn measurable and En Bn Fn. So, m*(En)  m(Bn)  m(Fn), which implies m(Bn) = m*(En). Then by Proposition 2.3.21(a), m*( [ n
1En)  m( [ n
1Bn) = limn!1 m(Bn) = limn!1 m*(En). 2.3.P4. Given a subset A of R, let An = A \ [−n, n] for n 2 ℕ. Show that m*(A) = limn!1 m*(An). Hint: Note that A1 A2 ⋯, and A = [ n
1An = limAn. Apply Problem 2.3.P3. 2.3.P5. The symmetric difference of sets A, B is defined to be ADB = (A\B) [ (B\A), consisting of points belonging to one of the two sets but not to both. Show that if A 2 M and m*(ADB) = 0, then B 2 M: Hint: From Remark 2.3.2(d), we have ADB 2 M and so also the subsets A\B and B\A. Since M is a r-algebra, it follows that A \ B = A\(A\B) 2 M and hence also that B = (A \ B) [ (B\A) 2 M: 2.3.P6. Show that every nonempty open subset O has positive measure. Hint: Since O is nonempty, there is some a 2 O. Since O is open, it is measurable, so that m(O) = m*(O), and furthermore, (a − d, a + d) O for some d > 0. Hence, m(O) = m*(O)
m*(a − d, a + d) = ‘(a − d, a + d) by Proposition 2.2.6. But ‘(a − d, a + d) = 2d > 0. 2.3.P7. Let O = [ n
1(xn − n12 ; xn + n12 Þ, where x1, x2, … is an enumeration of all the rationals. Prove that m(ODF) > 0 for any closed set F. Hint: If m(O\F) > 0, then m(ODF) = m(O\F) + m(F\O)
m(O\F) > 0. Suppose m(O\F) = 0. Since O\F is open, it follows from Problem 2.3.P6 that it is empty and hence O F. Now, O contains Q, which has closure R. So, F = R and m(F) = ∞. P 1 But mðOÞ  2 1 n¼1 n2 \1. So, m(F\O) = ∞ and hence m(ODF) = m(O\F) + m(F\O) = ∞. 2.3.P8. The number of elements in a r-algebra generated by n given sets cannot n exceed 22 . Hint: Let A be a family of n distinct subsets A1, A2, …, An of a nonempty set X. Set up the family B of all possible intersections A*1 \ A*2 \ ⋯ \ A*n, where each A*k is either Ak or its complement Ack. Such intersections are at most 2n in number, so that the cardinality N of B is at most 2n. Now let C be the family of all possible finite unions of sets in B; including £. Trivially, C  B: Also, any such union is a ! N union of at most N sets. Since there are ways to choose m sets to include in a m ! PN N union of m sets (m
1), the cardinality of C is at most 1 þ m¼1 ¼ 2N . m

412

8 Hints

We record the observation here that C is the smallest family of sets that (i) contains B (ii) is closed under unions and (iii) has £ as an element. We shall argue that C is an algebra containing A: Since any algebra containing A must necessarily contain C; it will follow immediately that C is the algebra generated by A: Being finite, it must then be also the r-algebra generated by A: It is obvious that C is closed under (finite) unions and has £ as an element. It is sufficient therefore to show that C contains A; and is closed under intersections and complements. In order to show that C  A; consider an arbitrary set in A: For ease of notation, we shall take it to be A1. We have to produce sets in B whose union is precisely A1. Consider the family B1 of intersections A1 \ A*2 \ ⋯ \ A*n, where each A*k (k > 1) is either Ak or its complement Ack. Any such intersection belongs to B by definition of the latter. We shall demonstrate that the union of all the sets belonging to B1 is precisely A1. Since each of the sets in question is a subset of A1, their union is a subset of A1. In order to show that that A1 is a subset of their union, consider any x 2 A1. For each k > 1, select A*k to be Ak if x 2 Ak and Ack if x 62 Ak. This selection ensures x 2 A*k for all k > 1, so that x 2 A1 \ A*2 \ ⋯ \ A*n. Thus each element of A1 is seen to belong to at least one of the sets belonging to B1, which means that A1 is a subset of their union. This completes the demonstration that the union of all the sets belonging to B1 is precisely A1, thereby showing that C  A: We proceed to argue that C is closed under intersections. To start with, we observe that the intersection of any two distinct sets in B is empty. This is because in the respective representations of the distinct sets as intersections A*1 \ A*2 \ ⋯ \ A*n, the kth sets must differ for some k, which means the kth sets must be complements of each other, thereby rendering the intersections disjoint. In particular, the intersection of any two sets of B (whether distinct or not) is in C: Since C is closed under unions, we further conclude that the intersection of a set of B with any set of C belongs to C: Let D be the family of all sets in C whose intersection with any set of C belongs to C: It is immediate that £ 2 D and that D is closed under unions (because C is). Also, C  D: In view of the conclusion of the preceding paragraph, D  B: We have thus shown that D (i) contains B (ii) is closed under unions and (iii) has £ as an element. In conjunction with the observation recorded earlier, this implies that C D: Thus C = D and the definition of D now instantly yields the consequence that C is closed under intersections. It remains to prove that C is closed under complements. To begin with, observe that the argument given above for establishing that A C can be easily modified to establish that the complement of any set in A is also in C : all one has to do is to replace A1 \ A*2 \ ⋯ \ A*n by Ac1 \ A*2 \ ⋯ \ A*n. Since C is closed under unions, the foregoing observation implies that any union A*1 [ A*2 [ ⋯ [ A*n belongs to C: But the complement of any set in B is a union of this form and therefore belongs to C: Now, the complement of any set in C is a finite intersection of complements of sets in B: Since C has been shown to be closed under intersections, it is now plain that C is closed under complements.

8 Hints

413

2.3.P9. Show that a r-algebra consisting of infinitely many distinct sets contains an uncountable number of sets. Hint: Let F be an infinite r-algebra and A1, A2, ⋯ be an enumeration of a countable subfamily A F : Set up the family B of all possible countable intersections A*1 \ A*2 \ ⋯, where each A*k is either Ak or its complement Ack. Clearly, F  B: We shall demonstrate that an arbitrary set of A is an intersection of sets belonging to B: Consider an arbitrary set in A: For ease of notation, we shall take it to be A1. We shall produce a subfamily of B whose union is precisely A1. Consider the family B 1 of intersections A1 \ A*2 \ A*3 \ ⋯, where each A*k (k > 1) is either Ak or its complement Ack. Any such intersection belongs to B by definition of the latter, and so B1 B: We shall demonstrate that the union of all the sets belonging to B1 is precisely A1. Since each of the sets in question is a subset of A1, their union is a subset of A1. In order to show that that A1 is a subset of their union, consider any x 2 A1. For each k > 1, select A*k to be Ak if x 2 Ak and Ack if x 62 Ak. This selection ensures x 2 A*k for all k > 1, so that x 2 A1 \ A*2 \ A*3 \ ⋯. Thus each element of A1 is seen to belong to at least one of the sets belonging to B1, which means that A1 is a subset of their union. This completes the demonstration that an arbitrary set of A is an intersection of sets belonging to B: It is a consequence of this that if B were to be finite, then the same would be true of A: Therefore B is seen to be infinite. Next, we observe that the intersection of any two distinct sets in B is empty. This is because in the respective representations of the distinct sets as intersections A*1 \ A*2 \ ⋯, the kth sets must differ for some k, which means the kth sets must be complements of each other, thereby rendering the intersections disjoint. Since B is infinite, there exists a sequence {Bn}n
1 of distinct nonempty sets belonging to B: The set of all sequences in the two-element set {0, 1} has cardinality c of R (see Sect. 1.2). Therefore, if we can exhibit an injective map from the set into F ; it will follow that F has cadinality at least c. We claim that the map H from this set into F given by HðaÞ ¼

1 [

Bn ; where Bn means Bn if aðnÞ ¼ 1 and Bcn otherwise;

n¼1

is injective. Let a 6¼ b be sequences in the two-element set {0, 1}. Then a(k) 6¼ b(k) for some k and hence one among H(a) and H(b) contains the nonempty set Bk and the other is a union of Bck with sets of B that are distinct from Bk. However, distinct sets of B have been shown to be disjoint, and therefore sets of B that are distinct from Bk are disjoint from Bk and hence so is their union with Bck. Thus, one among H(a) and H(b) contains the nonempty set Bk and the other is disjoint from it. Now, a set that is disjoint from a nonempty set must be distinct from any set containing the latter. Therefore H(a) 6¼ H(b). This justifies the claim that H is injective, thereby completing the proof that F has cadinality at least c.

414

8 Hints

2.3.P10. Show that, if E1 and E2 are measurable, then mðE1 [ E2 Þ þ mðE1 \ E2 Þ ¼ mðE1 Þ þ mðE2 Þ: Hint: Observe that E1 [ E2 as well as E1 \ E2 are measurable. Now, m(E1 [ E2) = m((E1 [ E2) \ E1) + m((E1 [ E2) \ Ec1) = m(E1) + m((E1 [ E2) \ Ec1), since E1 is measurable and (E1 [ E2) \ E1 = E1. Adding m(E1 \ E2) to both sides, we get mðE1 [ E2 Þ þ mðE1 \ E2 Þ ¼ mðE1 Þ þ mðE2 \ E1c Þ þ mðE1 \ E2 Þ ¼ mðE1 Þ þ mðE2 Þ; since E1 is measurable. Note that the measurability of E2 is required only for obtaining the measurability of E1 [ E2 and E1 \ E2. Alternatively (using finite additivity; Proposition 2.3.8): The sets E1 \ Ec2, Ec1 \ E2, E1 \ E2 are disjoint. Moreover, E1 ¼ ðE1 \ E2c Þ [ ðE1 \ E2 Þ; and

E2 ¼ ðE1c \ E2 Þ [ ðE1 \ E2 Þ

    E1 [ E2 ¼ E1 \ E2c [ ðE1 \ E2 Þ [ E1c \ E2 :

From finite additivity of m, we get the three equalities   mðE1 Þ ¼ m E1 \ E2c þ mðE1 \ E2 Þ; and

  mðE2 Þ ¼ m E1c \ E2 þ mðE1 \ E2 Þ

    mðE1 [ E2 Þ ¼ m E1 \ E2c þ mðE1 \ E2 Þ þ m E1c \ E2 :

Upon adding m(E1 \ E2) to both sides of the third equality and using the first two, we get the desired result. 2.3.P11. If E1 2 B and E2 2 B are such that E1  E2 and m(E2) < ∞, then m(E1\E2) = m(E1) _ m(E2). [Since B M; it follows trivially that this is true also when E1 2 B and E2 2 B:] Hint: In fact, E1 = (E1\E2) [ E2. The result follows on using the hypothesis m(E2) < ∞ and the fact that m is finitely additive. To do this without finite additivity, use Problem 2.3.P10 in the following manner:

8 Hints

415

mðE1 nE2 Þ þ mðE2 Þ ¼ mððE1 nE2 Þ [ E2 Þ þ mððE1 nE2 Þ \ E2 Þ ¼ mðE1 [ E2 Þ ¼ mð£Þ becauseðE1 nE2 Þ [ E2 ¼ E1 [ E2 and ðE1 nE2 Þ [ E2 ¼ £ ¼ mðE1 Þ because E1  E2 : 2.3.P12. Let 0 < a < 1. Construct a measurable set E [0, 1] of measure 1 − a and containing no interval of positive length. 1 P Hint: Let {an}n
1 be a sequence of positive real numbers such that an = a n¼1

(an = a/2n is one such sequence). Remove from the middle of the interval [0, 1] the open interval of length a1, leaving behind two closed intervals of length (1−a1)/2 each. Note that (1−a1)/2 > (a−a1)/2 > a2/2. From the middle of each of these intervals, remove an open interval of length a2/2, leaving behind 22 closed intervals of length (1−a1−a2)/22 each. Note that (1−a1−a2)/22 > (a−a1−a2)/22 > a3/22. Repeating this process, at the (n + 1)st stage, from the middle of each of the 2n closed intervals, remove an open interval of length an+1/2n, leaving behind 2n+1 closed intervals of length (1−a1−⋯−an+1)/2n+1 each. Note that (1−a1−⋯−an+1)/ 2n+1 > (a−a1−⋯−an+1)/2n+1 > an+2/2n+1. The total length of the open intervals so removed is 1 X   a1 þ 2ða2 =2Þ þ 22 a3 =22 þ    þ 2n ðan þ 1 =2n Þ þ    ¼ an ¼ a: n¼1

Thus the set E which is left behind after successively removing open intervals as described, being the difference of the measurable sets [0,1] and the countable union of intervals removed, is measurable, and its measure is m(E) = m([0, 1])−a = 1−a. Observe that, at the nth stage, each remaining interval has length less than 21n , which tends to 0. Consequently, E contains no interval of positive length. Remark The set constructed above is called the generalised Cantor set. 2.3.P13. Show that, if E is such that 0 < m*(E) < ∞ and 0 < a < 1, then there exists an open interval I such that m*(I \ E) > a‘(I). Hint: By Proposition 2.2.12 (second part), there exists an open set O  E such that m(O) < a−1m*(E). But O = [ i
1Ii, where Ii are disjoint open intervals. So, 1 1 P P m*(E \ Ii). But then there must exist some n such that a ‘(Ii) < m*(E)  i¼1

i¼1

a‘(In) < m*(E \ In). 2.3.P14. (a) Suppose m*(E) < ∞ and, for every e > 0, there exists an open set U such that m*(EDU) < e. Show that E 2 M: (b) If E is measurable and m(E) < ∞, then show that, for every e > 0, there exists a finite union U of disjoint open intervals such that m*(EDU) < e. Hint: (a) First assume that m*(E) < ∞.

416

8 Hints

Let e > 0. By definition of outer measure, there exists an open set V  E such that m ðVÞ\m ðEÞ þ e:

ð8:1Þ

Set S = U \ V. Then S U, S V and hence SDE ¼ ðSnEÞ [ ðEnSÞ ðUnEÞ [ ðEnSÞ because S U ðUnEÞ [ ðE \ ðU \ VÞc Þ because S ¼ U \ V ¼ ðUnEÞ [ ðE \ ðU c [ V c ÞÞ ¼ ðUnEÞ [ ððE \ U c Þ [ ðE \ V c ÞÞ ¼ ðUnEÞ [ ðE \ U c Þ because E V ¼ ðUnEÞ [ ðEnUÞ ¼ EDU: Consequently, m ðSDEÞ  m ðEDUÞ\e:

ð8:2Þ

We claim that E S [ (SDE). S [ ðSDEÞ ¼ S [ ðSnEÞ [ ðEnSÞ  ðS \ EÞ [ ðSnEÞ [ ðEnSÞ because S  S \ E ¼ ðS \ EÞ [ ðS \ Ec Þ [ ðE \ Sc Þ  ðS \ ðE [ Ec ÞÞ [ ðE \ Sc Þ  S [ ðE \ Sc Þ ¼ ðS [ EÞ \ ðS [ Sc Þ ¼ S [ E  E: It follows that

m ðEÞ  m ðSÞ þ e:

Now, VnE ¼ ððVnSÞ [ SÞnE because S V ¼ ððV \ Sc Þ [ SÞ \ Ec ¼ ððV \ Sc Þ \ Ec Þ [ ðS \ E c Þ ððV \ Sc Þ [ ðS \ E c ÞÞ [ ðEnSÞ ¼ ðVnSÞ [ ðSDEÞ: Therefore by (8.2), m ðVnEÞ  m ðVnSÞ þ e By (8.1) and (8.3), we further obtain

ð8:3Þ

8 Hints

417

m ðVÞ\m ðSÞ þ 2e: Since S V and m*(V) < ∞, so that m*(S) < ∞, it follows that m ðVnSÞ ¼ m ðVÞ  m ðSÞ\2e and hence that

m ðVnEÞ\3e

By Proposition 2.3.24, we conclude that E is measurable. (b) By Proposition 2.3.24, for every e > 0, there exists an open set O  E such that m*(O\E) < e. Let O = [ i
1Ii, where Ii are disjoint open intervals. Then m* 1 P (O) = ‘(Ii) < ∞. Since this series is convergent, there exists an N 2 ℕ such that

i¼1 1 P i¼N þ 1

m*(O\U) 

‘ðIi Þ < e. Let U = [ 1  i  NIi. Then O\U [ i
N+1Ii, so that 1 P i¼N þ 1

‘ðIi Þ < e.

Now,

EDU = (E\U) [ (U\E) (O\U) [ (O\E).

Therefore m*(EDU)  m*(O\U) + m*(O\E) < 2e. Remark Part (b) is false without the condition m(E) < ∞. To see why, let 1 P E = [ n
2En, where En = [n, n + 1n), n
2. Then mðEÞ ¼ m ðEÞ ¼ ‘ðEn Þ ¼

1 P 1 n¼2

n

n¼2

¼ 1. There can be no finite union U of disjoint open intervals for which

m*(EDU) < e for arbitrary e. Indeed, if U contains an infinite interval, then m*(U \E) = ∞, and if each of the intervals comprising U is bounded, then m*(E\U) = ∞. 2.3.P15. Give an example of a nonmeasurable set. Hint: Define an equivalence relation in R by setting x y , x  y 2 Q: Clearly, the relation ‘*’ partitions R into disjoint equivalence classes Ea. Any two elements of the same class differ by a rational number while those of two different classes differ by an irrational number. Moreover [ aEa = R. For any real number x, let [x] be its integer part. Then x ðx  ½ xÞ

and

ðx  ½ xÞ 2 ð0; 1:

418

8 Hints

We construct a set A using the Axiom of Choice 1.1.1 by choosing one element from each set Ea \ [0, 1). Observe that [ aEa \ [0, 1) = [0, 1). Let x1, x2, … be an enumeration of the rationals in (−1, 1) and set An ¼ A þ xn : The sets {An}n
1 are disjoint. In fact, if there is a point x 2 Am \ An, then x = x′ + xm = x″ + xn, where x′ and x″ are points in A. This implies x′ − x″ = xn − xm is a rational number, which forces x′ * x″ and hence xm = xn; this implies m = n. Moreover, the following inclusions hold: ð0; 1Þ

1 [

An ð1; 2Þ

n¼1

To justify the first inclusion, consider any x 2 (0, 1). Then x 2 [0, 1) = [ aEa \ [0, 1), as observed above. Therefore x − x′ is a rational number for a unique x′ 2 A [0, 1); so x − x′ = xn for some n and hence x 2 An. The second inclusion follows from the fact that each An is obtained by translating the elements of A [0, 1) by xn 2 (−1, 1). If A were to be measurable, then it would follow from the preceding paragraph that 1 ¼ mðð0; 1ÞÞ  m

1 [

! An

n¼1

¼

1 X

mðAn Þ  mðð1; 2ÞÞ ¼ 3

n¼1

and also that m

1 [ n¼1

! An

¼

1 X n¼1

mðAn Þ ¼

1 X

mðAÞ ¼ 0 or 1;

n¼1

a contradiction. 2.3.P16. Show that the outer measure m* is not finitely additive. Hint: Let E R be a nonmeasurable set. Then there exists an A R for which m*(A) < m*(A \ E) + m*(A \ Ec). Here A \ E and A \ Ec are disjoint and have union A. Thus m* is not finitely additive. 2.3.P17. Let f be defined on [0, 1] by f(0) = 0 and f(x) = xsin 1x for x > 0. Show that the measure of the set {x: f(x) > 0} is 1−(ln2)/p.     1 Hint: The function f is positive on the set S ¼ p1 ; 1 [ [ n
1 ð2n þ1 1Þp ; 2np [0, 1]. This is a union of countably many disjoint intervals and so,

8 Hints

419

 1 1 1 X 1 1 1 1 1X 1 þ  ¼1 þ p p 2n 2n þ 1 p p n¼1 ð2nÞð2n þ 1Þ n¼1 1 ¼ 1  ln 2: p

mðSÞ ¼ 1 

2.3.P18. Let E be Lebesgue measurable with 0 < m(E) < ∞ and let e > 0 be given. Then there exists a compact set K E such that m(E\K) = m(E) − m(K) < e. Hint: Let En = E \ [−n, n] for n
1. Then {En}n
1 is an increasing sequence of Lebesgue measurable sets. Therefore m(En)  m(E) < ∞. Also, [ n
1En is E and hence the continuity property (Proposition 2.3.21) yields m(E) = limnm(En). It follows that there exists an N 2 ℕ such that m(E) − m(EN) < min{ 2e, m(E)}. Then m(EN) > 0. By Proposition 2.3.24, there exists a closed set K EN [−N, N] such that m(EN\K) = m(EN) − m(K) < min{ 2e, m(EN)}. Note that m(K) > 0, so that K 6¼ £ and also, K is compact, K E, and m(E) < ∞, so that mðEnKÞ ¼ mðEÞ  mðKÞ ¼ mðEÞ  mðEN Þ þ mðEN Þ  mðKÞ\e: Remark 1 Let (R; B, m) be the Borel measure space. It has been proved that m is an extended real-valued nonnegative function defined on B which is translation invariant and is both inner and outer regular, i.e. (i) for E 2 B with 0 < m(E) < ∞ and for e > 0, there exists a compact set K E with m(E\K) < e (2.3.P18) and (ii) for E 2 B and e > 0, there exists an open set O  E such that m(O\E) < e (Proposition 2.3.24). Therefore m is often referred to as a regular Borel measure. Moreover, it is translation invariant (Proposition 2.3.23). We next show that m is determined uniquely up to a constant multiple. Indeed, we have the following result. Let l be a r-finite measure on the Borel r-algebra B of subsets of R, satisfying the following properties: (i) l(U) > 0 for an arbitrary nonempty open subset U R; (ii) l(K) < ∞ for an arbitrary compact K R; (iii) l(E + x) = l(E) for an arbitrary E 2 B and arbitrary x 2 R. Then there exists a positive c 2 R such that l(E) = cm(E) for all E 2 B: Step 1. Let l be a measure on the Borel sets of R which is translation invariant and let l(0, 1) < ∞. Then l{x} = 0 for any real number x, and l(0, 1) = l[0, 1) = l(0, 1] = l[0, 1]. Proof: Suppose if possible that there is some real number x0 such that l{x0} 6¼ 0. Then l{x0} > 0. It follows by translation invariance that l{x} = l{x0} > 0 for all x. Now the interval (0, 1) contains infinitely many real numbers and it follows by countable additivity that any l(0, 1) = ∞, contrary to hypothesis. It now follows that l(0, 1) = l[0, 1) = l(0, 1] = l[0, 1].

420

8 Hints

Step 2. Let l be a measure on the Borel sets of R which is translation invariant and let l(0, 1) < ∞. Then l(I) = l(0, 1)m(I) for any interval I and l(O) = l(0, 1) m(O) for any open set O. Proof: By Step 1, l(0, 1) = l[0, 1) = l(0, 1] = l[0, 1]. Call this nonnegative real number c. We have to prove that l(I) = c•m(I) for any interval I and l(O) = c•m (O) for any open set O. Since an open set is a countable union of disjoint open intervals, we need prove the equality only for intervals. Since l{x} = 0 = m{x} for any real number x, we need consider only open intervals. By (Proposition 2.3.23), any two intervals of the same length have the same l-measure as well as m-measure. It follows that the l-measure of an interval of rational length is c times the length, i.e. c times the m-measure. For an arbitrary bounded interval (a, b), we choose a rational strictly increasing sequence {an} with limit a and a rational strictly decreasing sequence {bn} with limit b. Then l[a, b] = limnl(an, bn) = c(bn − an) = c(b − a). It follows from the fact that the measure of a singleton set is 0 that l(a, b) = l[a, b) = l(a, b] = l[a, b] = c(b − a). Since an unbounded interval is a countable disjoint union of bounded intervals, the same equality must hold for unbounded intervals as well. Step 3. According to Theorems 5.11.12 and 5.11.21, it can be concluded that for every Borel set E, the equality l(E) = cm(E) holds. In what follows, we provide a characterisation of the r-algebra generated by an algebra A in terms of what are called “monotone classes”. A family M of subsets of a nonempty set X is called a monotone class if it satisfies the following two conditions: (i) if A1 A2 ⋯ and each Aj 2 M; then [ j
1Aj 2 M; (ii) if B1  B2  ⋯ and each Bj 2 M; then \ j
1Bj 2 M: Any r-algebra is a monotone class. Let A X, where X is any nonempty set and M = {A}. Then M is a monotone class which is not a r-algebra. 2.3.P19. (a) If Y is any class of subsets of a nonempty set X, then show that there exists a smallest monotone class containing Y. We shall denote it by M0 (Y). (b) If A is an algebra, show that S ðAÞ = M0 (A); that is, the r-algebra generated by an algebra is also the smallest monotone class containing the algebra. Hint: (a) Let {Ma } be the collection of all those monotone classes of subsets of X that contain Y. Since the collection of all subsets of X is a monotone class containing Y, the collection { Ma} is nonempty. Also, any intersection of monotone classes is a monotone class. So, \ aMa is the required monotone class M 0(Y). (b) Let us write M0 in place of M0(A). Since every r-algebra is a monotone class, it is sufficient to show that M0 is a r-algebra. We first show that M0 is closed under complementation. To that end, let M′0 = {A 2 M0: Ac 2 M0}. Because A is an algebra and M0  A, it follows that M′0  A. Also, because M0 is a monotone class, it is easy to see that M00 is a

8 Hints

421

monotone class. Therefore M′0  M0, which implies that M0 is closed under complementation. We next show that M0 is closed under finite unions. Suppose A 2 A and let K(A) = {F 2 M0: A [ F 2 M0}. We shall show that K(A) is a monotone class containing A. Since A is an algebra and M0  A, it follows that K(A)  A. Now suppose that E1 E2 ⋯ and each Ej 2 K(A). Since K(A) M0 by definition and the latter is a monotone class, we have [ j
1Ej 2 M0. Again by definition of K(A), we have A [ Ej 2 M0 for each j; also, A [ E1 A [ E2 ⋯. Since M0 is a monotone class, we have [ j
1(A [ Ej) 2 M0, i.e. A [ ( [ j
1Ej) 2 M0. Therefore [ j
1Ej 2 K(A). Similarly, K(A) is closed under intersections of nonincreasing sequences. Thus K(A) is a monotone class containing A and, consequently, K(A)  M0. But, K(A) M0 by definition. Thus K(A) = M0. In other words, A [ F 2 M0 for all A 2 A and F 2 M0. Now suppose B 2 M0, and let K(B) = {F 2 M0: B [ F 2 M0}. We shall show that K(B) is a monotone class containing A. From the previous paragraph, we know that K(B)  A and using the same argument as in that paragraph, we find that K(B) is a monotone class. This implies that K(B) = M0. In other words, F [ G 2 M0 for all F, G 2 M0. Thus M0 is closed under finite unions. It remains to prove that M0 is a r-algebra. Let Fn 2 M0 for every n. For each n, define En = [ 1  k  nFk. Then E1 E2 ⋯ and [ n
1En = [ n
1Fn. Since M0 is an algebra, each En belongs to it and, therefore, since M0 is a monotone class, [ n
1En = [ n
1Fn 2 M0. 2.3.P20. Show that (a) the cardinality of M is greater than c and (b) the Cantor set has a subset which is not a Borel set. Hint: As usual, denote the Cantor set by C. Since the cardinality of C is c (see Proposition 1.8.12), the power set P (C) has cardinality 2c, which is greater than c. Now, we know from Remark 2.2.11(c) that m(C) = 0 and hence by Remark 2.3.2 (d) that P(C) M: Therefore the cardinality of M is also greater than c, which proves (a). On the other hand, the cardinality of the family of all Borel subsets of C is at most c. Since the cardinality of P(C) is greater than c, it follows that there exists a subset of C which is not a Borel set, which proves (b). 2.3.P21. Let F be an algebra of subsets of a set X. If A and B are subsets of X such that B, B \ Ac, A \ Bc all belong to F , show that A also belongs to F . Hint: The set A \ B ¼ ðB \ Bc Þ [ ðA \ BÞ ¼ B \ ðBc [ AÞ ¼ B \ ðB \ Ac Þc belongs to F , because both B and B \ Ac do. Hence A ¼ A \ ðB [ Bc Þ ¼ ðA \ BÞ [ ðA \ Bc Þ belongs to F , because both A \ B and A \ Bc do.

422

8 Hints

2.3.P22. Let f be a real-valued function on [a, b]. Suppose E [a, b], f′ exists and satisfies | f′(x)|  M for all x 2 [a, b]. Prove that m*(f(E))  Mm*(E). Hint: Assume that E does not contain the points a and b, so that E (a, b). There exists an open set G  E such that m(G) < m*(E) + Me . Without loss of generality, suppose that G (a, b). G = [ n
1In, where In are disjoint open intervals. Let In′ denote In with its endpoints adjoined to it. Observe that In′ [a, b]. Since f is continuous on In′, there exist a and b such that f(a) = min{f(x): x 2 In′} and f(b) = max{f(x): x 2 In′}. By the Mean Value Theorem, | f(b) − f(a)|  Mm(In′) = M‘(In′). This implies ‘(f(In))  ‘(f(In′))  M(‘(In)). So, m*(f(E))  m*(f(G)) = m*(f( [ n
1In)) = m*( [ n
1f(In))  Rnm*(f(In))  MRn ‘(In) = Mm(G)  M(m*(E) + Me ) = Mm*(E) + e. Note that the removal of points a and b from the set E implies the removal at most two points f(a) and f(b) from f(E). 2.3.P23. (a) Let F and Y be subsets of a set X. Then show that f c \ Y is the complement in Y of F \ Y. (b) Let Y X and G be a r-algebra of subsets of Y. Show that the family F 0 ¼ fF X : F \ Y 2 Gg of subsets of X is a r-algebra. Hint: (a) f c \ Y = (f c \ Y) [ (Yc \ Y) = (f c [ Yc) \ Y = (F \ Y)c \ Y. (b) It is trivial that X 2 F 0. If {Fn}n
1 is any sequence of sets belonging to F 0, then the equality [ n
1(Fn \ Y) = ( [ n
1Fn) \ Y shows that [ n
1Fn 2 F 0. As for complements, suppose F 2 F 0. Then F \ Y 2 G and f c \ Y is the complement in Y of F \ Y, as assured by part (a). It follows that f c \ Y 2 G and thus f c 2 F 0. 2.3.P24. Let A be a family of subsets of a set X and F the r-algebra generated by it. Suppose Y X and AY ¼ fA \ Y : A 2 Ag;

F Y ¼ fF \ Y : F 2 F g:

Show that F Y is the r-algebra of subsets of Y generated by AY. Hint: It is plain that F Y contains AY. That it is a r-algebra follows (as in the preceding proof) by using the equality [ n
1(Fn \ Y) = ( [ n
1Fn) \ Y and Problem 2.3.P23(a). Suppose G is a r-algebra of subsets of Y that contains AY. All we need to show is that F Y G. Let F 0 = {F X: F \ Y 2 G}. By Problem 2.3.23(b), F 0 is a r-algebra. Any set A 2 A satisfies A \ Y 2 AY G, so that it also satisfies A 2 F 0. Hence the r-algebra F 0 contains A and consequently, F F 0. So, any F 2 F satisfies F \ Y 2 G. This means F Y G.

8 Hints

423

Problem Set 2.4 2.4.P1. Let the function f: [0, 1]!R be defined by f ðxÞ ¼ 1x if 0\x  1; f ð0Þ ¼ 0. Show that f is measurable. Hint: For any real number a, 8 < ½0; 1 fx 2 ½0; 1 : f ðxÞ [ ag ¼ ð0; 1  : 0; a1

a\0 0  a\1 a
1:

2.4.P2. Show that, if f is a real-valued function on a measurable subset X R such that X(f
r) is measurable for every rational number r, then f is measurable. Hint: Let a be any real number. There is a decreasing sequence {rn}n
1 of rational numbers such that lim rn = a and rn > a for each n. Then Xðf [ aÞ ¼

1 [

X ðf
rn Þ:

n¼1

Since each of the sets on the right is measurable, their countable union is measurable too. So, f is measurable. 2.4.P3. Let X R be measurable. Without using Theorem 2.4.6, show that if f and g are measurable functions defined on X, then the set {x 2 X: f(x) > g(x)} is measurable. Hint: For each x 2 X such that f(x) > g(x), we can find a rational number r such that f(x) > r > g(x). Since fx 2 X : f ðxÞ [ r [ gðxÞg ¼ fx 2 X : f ðxÞ [ rg \ fx 2 X : gðxÞ\rg; it follows that the set on the left is measurable, being the intersection of two measurable sets. Now, fx 2 X : f ðxÞ [ gðxÞg ¼

[

fx 2 X : f ðxÞ [ r [ gðxÞg:

r2Q

Since Q is countable and a countable union of measurable sets is measurable, it follows that the set {x 2 X: f(x) > g(x)} is measurable. 2.4.P4. Let f be a real-valued function defined on a measurable subset X ℝ. Then f is measurable if and only if for every open set V R,

424

8 Hints

f 1 ðVÞ ¼ fx 2 X : f ðxÞ 2 Vg is measurable. Hint: Suppose that f is measurable. Being an open subset of R, V can be written as V = [ n
1In, where In = (an, bn), n = 1, 2, …, are disjoint open intervals (see Theorem 1.3.17). Then f 1 ðVÞ ¼ ¼

1 [ n¼1 1 [

f 1 ðIn Þ ¼

1 [

fx 2 X : an \f ðxÞ\bn g

n¼1

½fx 2 X : f ðxÞ [ an g \ fx 2 X : f ðxÞ\bn g:

n¼1

It follows from the measurability of f that f −1(V) is measurable. The reverse conclusion follows on taking V = (a, ∞), where a is an arbitrary real number. 2.4.P5. Let f be a measurable function defined on a measurable subset X R and / be defined and continuous on the range of f. Then /  f is a measurable function on X. Hint: Let a be an arbitrary real number. Xð/  f [ aÞ ¼ fx 2 X : f ðxÞ 2 Ug; where U = {u: /(u) > a}. But U is open, since / is continuous. Hence in view of 2.4.P4, the set X(/  f > a) is measurable. This proves that /  f is measurable. 2.4.P6. Show that any function f defined on a set X of Lebesgue measure zero is Lebesgue measurable. Hint: For an arbitrary real number a, {x 2 X: f(x) > a} X and so m*({x 2 X: f (x) > a}) = 0. Consequently, {x 2 X: f(x) > a} is Lebesgue measurable [see Remark 2.3.2(d)]. 2.4.P7. Let X R and f: X!R be any function. Show that the family of subsets of X given by F = {f −1(V): V 2 B} is a r-algebra. (The same is true if B is replaced by M:) Hint: It is easy to verify the set-theoretic identities

f

1

ðRÞ ¼ X; f

1

ðV Þ ¼ f c

1

c

ðVÞ and f

1

1 [ n¼1

! Vn

¼

1 [

f 1 ðVn Þ:

n¼1

This immediately implies that F = {f −1(V): V 2 B} is a r-algebra. 2.4.P8. Let X R and f: X!R be any function. If F is a r-algebra of subsets of X, show that the family of subsets of R given by G = {V R: f −1(V) 2 F } is a r-algebra. Hence show that, if f is Borel measurable, then V 2 B ) f −1(V) 2

8 Hints

425

{X \ U: U 2 B}. Is it true that, if f is Lebesgue measurable, then V 2 B ) f −1(V) 2 {X \ U: U 2 M}? Hint: The first part follows from the same set-theoretic identities as 2.4.P7. Now, B X = {X \ U: U 2 B} is a r-algebra of subsets of X (easy verification), and hence by the first part, G = {V R: f −1(V) 2 BX } is also a r-algebra. By Corollary 2.4.3, all open sets of the form (a, ∞) as well as (−∞, a), with a 2 R are in G. Since the latter is a r-algebra, all countable unions of finite intersections of such sets are in it. This means all open sets are in it. By definition of B and the fact that G a r-algebra, it follows that B G. Thus V 2 B ) V 2 G. By definition of G, this means that V 2 B ) f −1(V) 2 BX = {X \ U: U 2 B}. The last part is true, because the same argument works with MX in place of BX but not M in place of B; the details being as below. Now, MX = {X \ U: U 2 M} is a r-algebra of subsets of X (easy verification), and hence by the first part, G = {V R: f −1(V) 2 MX } is also a r-algebra. By Corollary 2.4.3, all open sets of the form (a, ∞) as well as (−∞, a), with a 2 R are in G. Since the latter is a r-algebra, all countable unions of finite intersections of such sets are in it. This means all open sets are in it. By definition of B and the fact that G a r-algebra, it follows that B G. Thus V 2 B ) V 2 G. By definition of G, this means that V 2 B ) f −1(V) 2 MX = {X \ U: U 2 M}. 2.4.P9. Let f be a real-valued measurable function defined on a measurable set X ℝ. Prove that (a) if a > 0, then the function | f |a is measurable; (b) if f(x) 6¼ 0 on X and a < 0, then | f |a is measurable. Hint: (a) For a > 0, the function /(t) = |t|a 8 t 2 R is continuous on R  range (f) and hence, | f |a = /  f is measurable by Problem 2.4.P5. (b) If a < 0, the function /(t) = |t|a 8 real t 6¼ 0 is continuous on R\{0}  range (f) and hence, | f |a = /  f is measurable by Problem 2.4.P5. 2.4.P10. A complex-valued function f with domain a measurable set X R is said to be measurable if its real and imaginary parts, which are real-valued functions on X, are measurable. Prove that a complex-valued function is measurable if and only if f −1(V) is measurable for every open set V C (the complex plane). Hint: Suppose f = g + ih is measurable. Any open V C is a union of a countable family {In}n
1 of Cartesian products of open intervals (see Theorem 1.3.18). Let In = (an, bn) (cn, dn). Then f 1 ðVÞ ¼

1 [ n¼1

f 1 ðIn Þ ¼

1  [

g1 ððan ; bn ÞÞ \ h1 ððcn ; dn ÞÞ



n¼1

is measurable. Conversely, suppose f −1(V) is measurable for every open set V C. Define a: C!R by a(z) = ℜ(z), the real part of z. Then a is a real-valued continuous function defined on C. Also, g = a  f and

426

8 Hints

  g1 ðða; 1ÞÞ ¼ ða  f Þ1 ðða; 1ÞÞ ¼ f 1 a1 ðða; 1ÞÞ ¼ f 1 ðða; 1Þ RÞ is a measurable set. So g is measurable. Similarly, one may show that h is measurable. Problem Set 2.5 2.5.P1. Let g(x) = f′(x) exist for every x 2 [a, b]. Prove that g is Lebesgue measurable. Hint: Define f(x) = f(b) for x > b. Let



  1 gn ðxÞ ¼ n f x þ  f ðxÞ n

for a  x  b and n 2 N:

The functions gn (n 2 ℕ), being continuous on [a, b], are measurable. Moreover, limngn(x) = f′(x) = g(x) for x 2 [a, b) and 0 when x = b. On using Corollary 2.5.5 and Proposition 2.5.13, it follows that g is measurable. 2.5.P2. Prove that the set of points at which a sequence of measurable functions converges or diverges to ±∞ is a measurable set. Hint: The set E = {x 2 R: limsupfn − lim inf fn = 0} is clearly the set where the sequence converges. Since lim sup fn and lim inf fn are measurable by Theorem 2.5.4, it follows upon using the (easily proven) analogue of Proposition 2.4.5 for extended real-valued functions that the set E is measurable. The set on which the sequence diverges is the complement of E and therefore measurable. 2.5.P3. Show that sup{fa: a 2 K} is not necessarily measurable even if each fa is. Hint: Let E be a nonmeasurable set, K = E and let fa = v{a} for each a 2 K = E. Each fa, being the characteristic function of a singleton set, is measurable. However, sup{fa: a 2 K} = vE is nonmeasurable. 2.5.P4. If f is a real-valued Lebesgue measurable function defined on a Borel set X R, then there exists a Borel measurable function g on X such that f(x) = g(x) a.e. Hint: First suppose X = ℝ. Let f = vE, where E is Lebesgue measurable. By Proposition 2.3.24, there exists an F r-set F E such that m(E\F) = 0 and hence also a Gd-set G  E\F such that m(G) = 0. Since F is Borel measurable, vF is a Borel measurable function such that {x 2 R: vE 6¼ vF} = {x 2 R: vE\F 6¼ 0} = E \F G, which is a Borel set of measure zero. So, the characteristic function of a Lebesgue measurable set E has the property that it agrees with the characteristic function of a certain Borel set F on the complement of some Borel set of measure zero (which we called G above). It follows that a Lebesgue measurable simple function has the same property, because a finite union of Borel measurable sets of measure zero is again a set of the same kind. If f is Lebesgue measurable and nonnegative, there exists (see Theorem 2.5.9) a sequence {sn}n
1 of Lebesgue measurable simple functions converging to f. Therefore there exists a sequence {tn}n
1 of Borel measurable simple functions and a sequence {Gn}n
1 of Borel

8 Hints

427

sets of measure zero such that tn = sn on the complement of Gn. Now G = [ n
1Gn is a Borel set of measure zero and tn(x) = sn(x) 8 n 2 ℕ 8 x 2 Gc. Reset each tn to be 0 on Gc. Then g(x) = limn tn(x) exists for every x 2 R, and g is Borel measurable (because each tn is still Borel measurable). Moreover, x 2 Gc ) g(x) = limn sn(x) = f(x). Since G has measure zero, this means g = f a.e. To remove the assumption that X = ℝ, now consider any Borel set X R. Extend f to R by setting it equal to 0 outside X. Then the extended function is also Lebesgue measurable and, by what has already been proved, there exists a Borel measurable g on R which agrees with f a.e. The restriction of this g to X serves our purpose. 2.5.P5. Let f and g be extended real-valued measurable functions defined on a measurable subset X ℝ. Then their product fg is also measurable. Hint: Let n 2 ℕ and let the “truncated” function fn be defined by 8 < f ðxÞ if j f ðxÞj  n if f ðxÞ [ n fn ðxÞ ¼ n : n if f ðxÞ\  n: Let gn be defined similarly. It follows from Remark 2.5.3(a) that fn and gn are measurable. Since they are real-valued (do not take ±∞ as values), it follows by Theorem 2.4.6 that each fngn is measurable. Since f(x)gm(x) = limnfn(x)gm(x), it follows from Corollary 2.5.5 that fgm is measurable. The result is now a consequence of the identity ðfgÞðxÞ ¼ f ðxÞgðxÞ ¼ lim f ðxÞgm ðxÞ for x 2 X m!1

and another application of Corollary 2.5.5. 2.5.P6. Let s be a simple function on a measurable set X such that s = R1  i  paivAi , where i 6¼ i0 ) Ai \ Ai0 ¼ £ and

p [

Ai ¼ X:

i¼1

(a) If c1, c2, …, cn are the distinct elements of the range of s, show that 

 cj : 1  j  n ¼ fai : 1  i  p; Ai 6¼ £g:

(b) Show that, for any j, 1  j  n, there must exist some index i (1  i  p) for which the coefficient ai equals cj and, at the same time, Ai 6¼ £. If for each j (1  j  n), Nj is the set of all such indices i corresponding to a given j, namely,

428

8 Hints

 Nj ¼ i : 1  i  p; ai ¼ cj

and

 Ai 6¼ £ ;

show that [ i2Nj Ai = X(s = cj). (c) Finally, show that the canonical representation of s is R1  j  n cjvBj , where Bj = [ i2Nj Ai . Hint: (a) Observe that x 2 Ai ) s(x) = ai, because the Ai are disjoint. It follows from this observation that Ai 6¼ £ ) ai 2 s(X), i.e. fai : 1  i  p; Ai 6¼ £g sðXÞ ¼ fc1 ; c2 ; . . .; cn g: Since [ Ai = X it follows from the same observation that the reverse inclusion also holds. (b) Consider any j, 1  j  n. By (a), there must exist some index i (1  i  p) for which the coefficient ai equals cj and Ai 6¼ £. Therefore the set of all such indices i corresponding to a given j, namely,  Nj ¼ i : 1  i  p; ai ¼ cj

and Ai 6¼ £



is nonempty for 1  j  n. Moreover, s(x) = cj for precisely those points x that belong to some Ai with i 2 Nj, which is precisely the assertion in question. (c) Immediate from (b) and Remark 2.5.8(c). 2.5.P7. Let R1  i  paivAi be the canonical representation of a simple function s. If 0 < a < ∞, show that R1  i  p(aai)vAi is the canonical representation of as. Hint: Clearly, R1  i  p(aai)vAi = as. We need show only that it is a canonical representation. Since R1  i  paivAi is a canonical representation, the sets Aj are nonempty, disjoint with [ 1  j  nAj = X and are measurable; moreover the real numbers a1, a2, …, an are distinct and nonnegative and hence so are the products aai (keeping in view that 0 < a < ∞). This means R1  i  p(aai)vAi is a canonical representation. 2.5.P8. Give an example when (f + g)+ is not the same as f + + g+. Hint: Take f to be a nonnegative function with at least one positive value, and g = −f. 2.5.P9. If f and g are extended real-valued functions and g
0, show that (fg)+ = f +g and (fg)− = f −g. Hint: Consider any x in the domain and let f(x)
0. Then f +(x) = f(x), and therefore f(x)g(x) = f +(x)g(x) = (f +g)(x). Also, f(x)g(x)
0 and so ((fg)+)(x) = f(x)g(x). Thus (fg)+(x) = (f +g)(x) when f(x)
0. The second equality holds at x because both sides are 0. A similar argument works when f(x)  0.

8 Hints

429

2.5.P10. Let f be a bounded measurable function defined on a bounded closed interval [a, b] and let e > 0 be arbitrary. Show that there exists a step function g on [a, b] such that mðfx 2 ½a; b : j f ðxÞ  gðxÞj
egÞ\e: Hint: Step I. Let f = vE, where E [a, b] is measurable. By Problem 2.3.P14(b), there exist disjoint open intervals I1, …, Ik such that1 m(ED [ 1  r  kIr) < η. Let g be the restriction to [a, b] of the characteristic function of [ 1  r  kIr. Then g is a step function on [a, b]. Furthermore, {x 2 [a, b]: | f(x)−g(x)|
e} = £ if e > 1 and for 0 < e  1, {x 2 [a, b]: | f(x)−g(x)|
e} ED [ 1  r  kIr. Hence mðfx 2 ½a; b : j f ðxÞ  gðxÞj
egÞ\g: Step II. Let f be any simple function on [a, b] with representation f ¼

n P i¼1

ai vE i ,

where E1, …, En are disjoint measurable sets with [ 1  i  nEi = [a, b]. For each i = 1, …, n, let gi be the step function on [a, b] (Step I) such that m({x 2 [a, b]: n P gi . Then |aivEi (x) − gi(x)|
1n e}) < 1nη. Let g ¼ i¼1

mðfx 2 ½a; b : j f ðxÞ  gðxÞj
egÞ\g: n n n P P P ai vE ðxÞ  gi ðxÞ ; so that This is because jf ðxÞ  gðxÞj ¼ ai vEi  gi  i i¼1 i¼1 i¼1 | f(x) − g(x)|
e implies ai vEi ðxÞ  gi ðxÞ
1n e for some i, and hence

fx 2 ½a; b : j f ðxÞ  gðxÞj
eg

n [ i¼1

 1 x 2 ½a; b : ai vEi ðxÞ  gi ðxÞ
e ; n

which implies mðfx 2 ½a; b : j f ðxÞ  gðxÞj
egÞ 

1

 n X 1 m x 2 ½a; b : ai vEi ðxÞ  gi ðxÞ
e : n i¼1

The fact that the intervals can be chosen to be disjoint is of no consequence here.

430

8 Hints

Step III. Let f be a bounded measurable function defined on [a, b]. By Theorem 2.5.9, there exists a simple function s such that | f − s| < e/2. By Step II, corresponding to the simple function s, there is a step function g such that n e o m x 2 ½a; b : jsðxÞ  gðxÞj
\g; 2 Since | f(x) − s(x)| < e/2, the inequality |s(x) − g(x)| < e/2 implies | f(x) − g(x)| < e, which means {x 2 [a, b]: | f(x)−g(x)|
e} {x 2 [a, b]: |s(x)−g(x)|
e/2}. It follows that n e o mðfx 2 ½a; b : jf ðxÞ  gðxÞj
egÞ  m x 2 ½a; b : jsðxÞ  gðxÞj
\g: 2 This completes the argument. Problem Set 2.7 2.7.P1. Prove that |mn (A) − mn (B)|  max{mn (A\B), mn (B\A)}  mn (ADB) provided mn (A) and mn (B) are finite. Hint: Since A = (A\B) [ (A \ B), we have mn (A)  mn (A\B) + mn (A \ B)  mn (A\B) + mn (B). So, mn (A) − mn (B)  mn (A\B). Similarly, mn (B) − mn (A)  mn (B\A). 2.7.P2. An outer measure on Rn is an extended real-valued, nonnegative, monotone and countably subadditive set function l* defined on all subsets of Rn and satisfying l*(£) = 0. If {ln }k
1 is a sequence of outer measures and {ak} k
1 a sequence of positive real numbers, then show that the set function defined by 1 P ak lk ðEÞ is an outer measure. l ðEÞ ¼ k¼1

Hint: l ð£Þ ¼

1 P k¼1

ak lk ð£Þ ¼ 0: E F ) lk ðEÞ  lk ðFÞ. Monotonicity now

follows. lk 1 P k¼1

ak lk

1 S j¼1

! Ej



1 P k¼1

ak

1 S j¼1 1 P j¼1

! Ej



1 P j¼1

  lk Ej

1 P 1 1   P   P   lk Ej ¼ ak lk Ej ¼ l Ej : j¼1 k¼1

j¼1

8 Hints

431

Problem Set 2.8 2.8.P1 Let {Ei}i
1 be a sequence of measurable sets. Then (a) if Ei Ei+1, then mn(lim En) = lim mn ðEk Þ; k

(b) if Ei  Ei+1 and mn(E1) < ∞, then mn(lim En) = lim mn ðEk Þ. k

Hint: Same as Proposition 2.3.21. 2.8.P2. Prove that for subsets A, B, C of any nonempty set X, ADC ðADBÞ [ ðBDCÞ: Hint: ADC = (A\C) [ (C\A). Consider any x 2 ADC. Suppose x 2 A\C; then x 2 A and x 62 C. If x 2 B, then x 2 B\C and if x 62 B, then x 2 A\B. Thus A\C (ADB) [ (BDC). Upon interchanging A and C, we get C\A (CDB) [ (BDA). 2.8.P3. Let A, B be subsets of Rn. Show that mn ðADBÞ  mn ðAÞ þ mn ðBÞ: Hint: Since ADB = (A\B) [ (B\A), it follows that mn ðADBÞ  mn ðAnBÞ þ mn ðBnAÞ  mn ðAÞ þ mn ðBÞ

because A\B A and B\A B. 2.8.P4. If l* is an outer measure on Rn and if A and B are subsets of Rn, of which at least one is l*-measurable, then show that l ðAÞ þ l ðBÞ ¼ l ðA [ BÞ þ l ðA \ BÞ: Remark Under the additional hypothesis that at least one among A and B has finite outer measure, the above equality can be written as l ðA [ BÞ ¼ l ðAÞ þ l ðBÞ  l ðA \ BÞ: Hint: We may assume that A is l*-measurable. Then

432

8 Hints

l ðBÞ ¼ l ðA \ BÞ þ l ðAc \ BÞ and l ðA [ BÞ ¼ l ðA \ ðA [ BÞÞ þ l ðAc \ ðA [ BÞÞ ¼ l ðAÞ þ l ðAc \ BÞ: Hence l ðA [ BÞ þ l ðA \ BÞ ¼ l ðAÞ þ l ðAc \ BÞ þ l ðA \ BÞ ¼ l ðAÞ þ l ðBÞ: 2.8.P5. Let A E, where E is measurable and mn(E) < ∞. Show that A is measurable provided mn ðEÞ ¼ mn ðAÞ þ mn ðEnAÞ: Hint: By Propositions 2.8.18 and 2.8.14, we have A A′ and E\A B′, where A′ and B′ are measurable and mn*(A) = mn*(A′) and mn*(E\A) = mn*(B′). Replacing A′ by A′ \ E and B′ by B′ \ E, we may assume that A′ E and B′ E. Since A′ and B′ are measurable and A′ [ B′ = E, we have mn ðA0 DB0 Þ þ mn ðA0 \ B0 Þ ¼ mn ðA0 [ B0 Þ ¼ mn ðEÞ:

ð8:4Þ

mn ðA0 nB0 Þ þ mn ðA0 \ B0 Þ ¼ mn ðA0 Þ;

ð8:5Þ

mn ðB0 nA0 Þ þ mn ðA0 \ B0 Þ ¼ mn ðB0 Þ:

ð8:6Þ

Also,

On adding (8.5) and (8.6), we get mn ðA0 nB0 Þ þ mn ðB0 nA0 Þ þ 2mn ðA0 \ B0 Þ ¼ mn ðA0 Þ þ mn ðB0 Þ ¼ mn ðAÞ þ mn ðEnAÞ ¼ mn ðEÞ: From (8.4) and (8.7), it follows that

ð8:7Þ

8 Hints

433

mn ðA0 nB0 Þ ¼ 0: Since A′ E and A E, we have A′\A = A′ \ (E\A) A′ \ B′. So, mn ðA0 nAÞ  mn ðA0 \ B0 Þ ¼ mn ðA0 \ B0 Þ ¼ 0; which implies A′\A is measurable. Consequently, A = A′\(A′\A) is measurable. 2.8.P6. If F 2 Mn and mn*(FDG) = 0, then show that G is measurable. Hint: By Remark 2.8.2(d), FDG is measurable and so are the subsets F\G and G\F. Now, F = (F\G) [ (F \ G). Therefore F \ G = F\(F\G) is measurable. Consequently, G = (G\F) [ (F \ G) is measurable. 2.8.P7. Show that every nonempty open set has positive measure. Hint: A nonempty open set must contain a nonempty cuboid, the volume of which is always positive. 2.8.P8. Let q1, q2, … be an enumeration of points in Rn with rational coordinates and let G = [ k
1Ik, where Ik is an open cuboid centred at qk with volume 1/k2. Prove that for any closed set F, mn(GDF) > 0. Hint: If mn(G\F) > 0, there is nothing to prove. Suppose mn(G\F) = 0 and since G\F is open, we must have G F by Problem 2.8.P7. But G contains the set of all points with rational coordinates, whose closure is Rn. So, F = Rn and mn(F) = ∞. But mn(G)  R(1/k2) < ∞. Therefore mn(F\G) = ∞ and hence mn ðGDFÞ ¼ mn ðGnFÞ þ mn ðFnGÞ ¼ 1 [ 0: Remark In fact, we have proved more than what is demanded in the problem, namely, either mn(G\F) > 0 or mn(F\G) = ∞. 2.8.P9. Let E be Lebesgue measurable with 0 < mn(E) < ∞ and let e > 0 be given. Then there exists a compact set K E such that mn(E\K) = mn(E) − mn(K) < e. Hint: Follow the ideas of Problem 2.3.P18. Problem Set 3.1

R 3.1.P1. Let X = [0, 4] and s = 2v[0,2] + 3v(2,4] and t = 6v(1,3]. Find X s dm and R R X t dm: Also, find the canonical form of s + t and use it to compute X ðs þ tÞdm from the R R definition. Hint: X s dm = 2(2 − 0) + 3(4 − 2) = 10, X t dm = 6(3 − 1) = 12. In canonical form, s þ t ¼ 2v½0; 1 þ 8vð1; 2 þ 9vð2; 3 þ 3vð3; 4 : So,

R X

ðs þ tÞ dm ¼ 2ð1  0Þ þ 8ð2  1Þ þ 9ð3  2Þ þ 3ð4  3Þ ¼ 22:

434

8 Hints

3.1.P2. Let X = (0, 4] and An = [1n, 4], so that A1 A2 ⋯ and [ n
1An = X. For   R  R  R s = v(0,2], find X svAn dm; X s dm and limn!1 X svAn dm.     R  R  Hint: svAn ¼ v½1;2 . Therefore X svAn dm ¼ 1  2  1n ; hence limn!1 X svAn Rn dm ¼ 2. Also, X s dm ¼ 1  ð2  0Þ ¼ 2. 3.1.P3. Let X be a measurable subset of R and {An}n
1 be a sequence of measurable subsets such that A1 A2 ⋯. If X\ [ n
1An Rhas positive measure, show  R that there exists a simple function s such that limn!1 X svAn dl 6¼ X s dl.  R R  Hint: Take s = vE, where E = X\ [ n
1An. Then X s dl [ 0 but X svAn dl ¼ 08 n 2 N. 3.1.P4. Let X be a measurable set, Ai (1  i  p) disjoint measurable subsets of it and ai (1  i  p) nonnegative real numbers. (a) If the sets Ai do not form a partition of X, show that there are infinitely many simple functions taking the respective values ai on Ai. (b) If the sets Ai form a partition of X, show that there is a unique simple function s taking the respective values ai on Ai and that it is given by s = R1  i  paivAi . Hint: (a) Since the Ai are disjoint, their not forming a partition means that their union has a nonempty complement. The specification that a function takes the respective values ai on Ai allows any nonnegative real number to be taken as a value on the nonempty complement. (b) Here the union is the whole of X and the specification that a function takes the respective values ai on Ai determines the function completely. Therefore such a function must be unique. Since s = R1  i  paivAi fulfils the specification (in view of the given disjointness), the unique function must be none other than s. 3.1.P5. Is Proposition 3.1.5 valid without the hypothesis that the Ai and the Bi form partitions of X? Hint: Yes, by Proposition 3.1.7. 3.1.P6. [Needed in 3.2.P13] Let s be a simple function on X. Define /s: [0, ∞) ! [0, ∞] as /s(u) = l(X(s > u)). If l(X) < ∞, then /s takes values in [0, ∞). Show that (a) /s(u) = 0 if u
M = max{s(x): x 2 X}. (b) /s is a step function. RM (c) The Riemann integral 0 /s ðuÞdu, which is equal to the improper integral R1 R 0 /s ðuÞdu in view of (a), is also equal to the measure space integral X sdl. (d) What happens if l(X) = ∞? Hint: (a) If u
M, then X(s > u) = £ and hence /s(u) = l(£) = 0. (b) Let the values of s be ai, 1  i  p, in increasing order, i.e. 0  a1 < a2 < ⋯ < ap = M. Then s = R1  i  paivAi, where Ai is the set on which s takes the value ai. It will be convenient to introduce the notation Si = Ri  j  p l(Aj). Suppose 0 < a1 and consider any u such that 0  u < a1. Then s(x) > u for all x. Therefore X(s > u) = X and hence /s(u) = l(X) = R1  i  p l(Ai) by the additivity of measure. This shows that, if 0  u < a1, then /s(u) = S1. Therefore, unless the interval [0, a1) is empty, /s has the constant value S1 on it. Now suppose p > 1 and consider any u such that ai−1  u < ai, where i > 1. Then s(x) > u ,

8 Hints

435

s(x)
ai , s(x) = aj for some j
i. This means X(s > u) = [ i  j  pAj and hence /s(u) = Si by the additivity of measure. This shows that /s has the constant value Si on [ai−1,ai). Finally, consider any u such that M = ap  u. Since s(x)  M everywhere, X(s > u) is now £ and therefore /s(u) = 0. This shows that /s has the constant value 0 on [ap, ∞). Thus /s is the following step function: 8 < S1 /s ðuÞ ¼ Si : 0

if0  u\a1 ifai1  u\ai ifap  u:

The only difference if a1 = 0 is that the case 0  u < a1 is to be omitted. If p = 1, then the case ai−1  u < ai has to be omitted. If p = 1 and also a1 = 0, then both the aforementioned cases are vacuous and s as well as /s are zero everywhere. (c) If p = 1, there are two subcases, namely, a1 = 0 and a1 > 0. Both are trivial and we go on to consider the case p
2. Since /s is a step function as described above, its Riemann integral over [0, M] = [0, ap] is (regardless of whether a1 = 0 or not) Z

M

/s ðuÞdu ¼ a1 S1 þ

0

p X

Si ðai  ai1 Þ

i¼2

¼

p1 X i¼1

ai ðSi  Si þ 1 Þ þ ap Sp ¼

p X

ai lðAi Þ:

i¼1

(d) If l(X) = ∞, then /s may have ∞ as a value on some interval of positive length. It is nonetheless a nonnegative extended real-valued step function and has Riemann integral ∞. But in that event, the measure space integral of s is also ∞ and the equality proved in (c) is still valid. 3.1.P7. Let {sn}n
1 be the sequence of simple functions given by sn(x) R = 1/n for |x|  n and 0 for |x| > n. Show that sn! 0 uniformly on X = R, but X sn dm ¼ 2 for every n 2 N. [Uniform convergence on a set X of finite measure to a bounded limit function does imply convergence of integrals; this will be seen to be true in the next section even for more general functions in the light of the Dominated Convergence Theorem 3.2.16] Hint: Let e > 0. Choose n0 2 N such that n0 > 1/e. For n
n0, we have sn(x) = 1/n < 1/n0 < e provided |x|  n. Since |x| > n ) sn(x) = 0, it follows that |sn(x)| < e for n
n0 and all x 2 X = R. Also, sn is a simple with canonical R function 1 1 representation sn ¼ n  v½n;n þ 0  vð1;nÞ [ ðn;1Þ so that X sn ¼ n 2n ¼ 2. 3.1.P8. Optional. [The principle that permits grouping of terms in a sum according to any scheme whatsoever can be expressed in this manner: Suppose we have a sum R1  i  pbi to evaluate and that the set {i: 1  i  p} of indices has been partitioned into n subsets Nj, 1  j  n, meaning thereby that these subsets are

436

8 Hints

nonempty and disjoint with union equal to {i: 1  i  p}. Then the required sum can be evaluated as the grand total of the n subtotals Ri2Nj bi . In other words, p X

bi ¼

i¼1

provided

n S

n X X

! bi

i2Nj

j¼1

Nj ¼ f1; 2; . . .; pg, every Nj 6¼ £ and j 6¼ j′ ) Nj \ Nj′ = £.

j¼1

This is valid even if one or more of the bi are ∞ as long as each bi is nonnegative. We shall refer to it as the grouping principle.] Let s be a simple function on a measurable set X such that

s¼

p X

ai vAi ;

i¼1

where i 6¼ i0 ) Ai \ Ai0 ¼ £ and

each Ai is measurable

and p [

Ai ¼ X:

i¼1

Using Problem 2.5.P6 and the grouping principle, but not Proposition 3.1.5, show that p X i¼1

Z ai lðAi Þ ¼

s dl:

ð8:8Þ

X

Hint: On the left-hand side of (8.8), the terms with Ai = £ can be omitted: (*) R1  i  p ail(Ai) = RAi 6¼£ ail(Ai). Now let c1, c2, …, cn be the distinct elements of s(X) and Nj = {i: 1  i  p, ai = cj and Ai 6¼ £}. Then the Nj are disjoint because the cj are distinct. Moreover, by Problem 2.5.P6, we have {cj: 1  j  n} = {ai: 1  i  p, Ai 6¼ £} and [ i2Nj Ai = X(s = cj). The first of these two equalities shows that [ 1  j  nNj = {i: 1  i  p, Ai 6¼ £} and therefore the sets Nj provide the kind of partition of {i: 1  i  p, Ai 6¼ £} needed for applying the grouping principle. The second of the equalities leads to: (**) Ri2Nj l(Ai) = l(X(s = cj)). Therefore

8 Hints

437

X Ai 6¼£

ai lðAi Þ ¼ ¼ ¼

" n X X j¼1

i2Nj

j¼1

i2Nj

" n X X n X j¼1

¼

n  X

cj

# ai lðAi Þ by the grouping principle # cj lðAi Þ by definition of Nj

X

! lðAi Þ

i2Nj

   by ðÞ cj l X s ¼ cj

j¼1

Z ¼

s dl by definitionof the integral: X

In conjunction with (*), this leads immediately to the required equality (8.8). 3.1.P9. If in a measure space, {An}n
1 is a sequence of sets of measure 0 and {Bn}n
1 is a descending sequence of sets of finite measure such that their measure tends to 0, show that \ n
1(An [ Bn) has measure 0. Hint: \ n
1(An [ Bn) ( [ n
1An) [ ( \ n
1Bn) and both [ n
1An and \ n
1Bn have measure 0 (the latter by Proposition 3.1.8). 3.1.P10. Let F be a r-algebra of subsets of a set X and f: X!R be any function whatsoever. Show that the family G = {A R: f −1(A) 2 F } of subsets of R is a r-algebra. Hence show that if f is F -measurable, then f −1(A) 2 F whenever A R is any Borel subset of R. Hint: Clearly, R 2 G. Since f −1(A)c = f −1(Ac), we have A 2 G ) Ac 2 G. If {Aj}j
1 is a sequence of sets in G, then f −1(Aj) 2 F for each j and hence f −1( [ j
1Aj) = [ j
1f −1(Aj) 2 F . Thus, G is a r-algebra. Now suppose f is F -measurable. The result of Problem 2.4.P4, which is easily seen to carry over to any set X with a r-algebra of subsets, shows that every open subset of R is in G. From what has been proved in the paragraph above, it therefore follows that the entire r-algebra of Borel subsets of R is contained in G, which is what we needed to prove. 3.1.P11. Suppose f:R!R and g:R R!R are both Borel measurable. Show that the composition f  g:R R!R is Borel measurable. Hint: Let V = {x 2 R: x > a}, where a 2 R is arbitrary. Then f −1(V) is a Borel subset of R because f is given to be Borel measurable. Since g is also given to be Borel measurable, it follows by Problem 3.1.P10 that g−1(f −1(V)) is a Borel subset of R R. Since (f  g)−1(V) = g−1(f −1(V)), and a 2 R is arbitrary, we have shown that the composition f  g:R R!R is Borel measurable.

438

8 Hints

3.1.P12. Let l be a measure on the Borel r-algebra B of subsets of R, satisfying the following properties: (i) (ii) (iii) (iv)

l(0, 1) > 0; l(0, 1) < ∞; for E 2 B and e > 0, there exists an open set O  E such that l(O\E) < e; l(E + x) = l(E) for an arbitrary E 2 B and arbitrary x 2 R (translation invariance).

Show that there exists a positive a 2 R such that l(E) = a⋅ m(E) for all E 2 B. Hint: This can be accomplished in three steps. Step 1. Let l be a measure on the Borel sets of R which is translation invariant and let l(0, 1) < ∞. Then l{x} = 0 for any real number x, and l(0, 1) = l[0, 1) = l(0, 1] = l[0, 1]. Proof: Suppose if possible that there is some real number x0 such that l{x0} 6¼ 0. Then l{x0} > 0. It follows by translation invariance that l{x} = l{x0} > 0 for all x. Now the interval (0, 1) contains infinitely many real numbers and it follows by countable additivity that any l(0, 1) = ∞, contrary to hypothesis. It now follows that l(0, 1) = l[0, 1) = l(0, 1] = l[0, 1], whereby the proof of Step 1 is complete. Step 2. Let l be a measure on the Borel sets of R which is translation invariant and let l(0, 1) < ∞. Then l(I) = l(0, 1)m(I) for any interval I and l(O) = l(0, 1)⋅m(O) for any open set O. Proof: By Step 1, l(0, 1) = l[0, 1) = l(0, 1] = l[0, 1]. Call this nonnegative real number a. Then, for the particular interval I = [0, 1] of length 1, the equality l(I) = a⋅⋅m(I) holds because m(I) = 1. We have to prove that l(I) = a•m(I) for any interval I and l(O) = a⋅⋅m(O) for any open set O. Since an open set is a countable union of disjoint open intervals, we need prove the equality only for intervals. Since l{x} = 0 = m{x} for any real number x, we need consider only open intervals. By translation invariance of m (Proposition 2.3.23) and of l, any two intervals of the same length have the same m-measure as well as l-measure. Upon combining this with fact that for the particular interval I = [0, 1] of length 1, the equality l(I) = a⋅⋅m(I) holds, it follows that the l-measure of an interval of rational length is a times the length, i.e. a times its m-measure For an arbitrary bounded interval (a, b), we choose a rational strictly increasing sequence {an} with limit a and a rational strictly decreasing sequence {bn} with limit b. Upon appealing to the continuity property of an arbitrary measure [noted just before Proposition 3.1.8], we arrive at l½a; b ¼ lim lðan ; bn Þ ¼ n!1

lim aðbn  an Þ ¼ aðb  aÞ: It follows from the fact that the measure of a singleton

n!1

set is 0 that l(a, b) = l[a, b) = l(a, b] = l[a, b] = a(b − a). Since an unbounded interval is a countable disjoint union of bounded intervals, the same equality must hold for unbounded intervals as well.

8 Hints

439

Since a is defined to mean l(0, 1), the proof of Step 2 is now complete. Step 3. Let l be a measure on the Borel sets of R which is translation invariant and let 0 < l(0, 1) < ∞. Suppose also that, for E 2 B and e > 0, there exists an open set O  E such that l(O\E) < e. Then there exists a positive a 2 R such that l(E) = a•m(E) for all E 2 B. Proof: By Step 2, the number a = l(0, 1) satisfies l(O) = a•m(O) for any open set O. Since it is assumed in this Step that 0 < l(0, 1), we have a > 0. Consider an arbitrary E 2 B and an arbitrary e > 0. By hypothesis, there exists an open set Ol  E such that l(Ol\E) < e. By Proposition 2.3.24, there exists an open set Om  E such that m(Om\E) < e. It follows that the open set O = Ol \ Om satisfies O  E;

mðOnEÞ\e and

lðOnEÞ\e:

Hence mðOÞ
mðEÞ ¼ mðOÞ  mðOnEÞ
mðOÞ  e and lðOÞ
lðEÞ ¼ lðOÞ  lðOnEÞ
lðOÞ  e: The former leads to a  mðOÞ
a  mðEÞ
a  mðOÞ  a  e and the latter leads to a  mðOÞ
lðEÞ
a  mðOÞ  e: If m(O) = ∞, then we have l(E) = ∞ = m(E), so that l(E) = a•m(E). So, suppose m(O) < ∞. Then it follows from the preceding two inequalities that e
a mðEÞ  lðEÞ
 a  e: Since e > 0 is arbitrary, it follows that l(E) = a•m(E). Since this has been established for arbitrary E 2 B, the proof of Step 3 is complete. Remark The result holds with M in place of B.

440

8 Hints

Problem Set 3.2 R 3.2.P1. (a) If X f dl [ 0, where f is a measurable extended real-valued function on X, show that X(f > 0) has positive measure. (b) [Needed in Theorem 3.2.20, Problems 3.2.P6 and 3.2.P14 and Theorem 3.3.5] R When f is nonnegative, show that if X(f > 0) has positive measure, then X f dl [ 0. R R (c) When f and g are measurable, f
g and X f dl ¼ X g dl, does it follow that f = g a.e.? R Hint: (a) Some simple function s such that 0  s  f + must satisfy X s dl [ 0. R P P Let s ¼ 1  i  p ai vAi canonically, so that s ¼ 1  i  p ai lðAi Þ. Then some i must satisfy ai > 0 as well as l(Ai) > 0. For any x 2 Ai, we have f +(x)
s(x) = ai > 0. Thus X(f > 0) = X(f + > 0)  Ai and hence l(X (f > 0))
l(Ai) > 0. (b) f ðxÞ [ 0 , f ðxÞ [ 1n for some n 2 N. This means X(f > 0) = [ n2N En, where   En ¼ X f [ 1n . Since En En+1, continuity of measure (see Proposition 3.1.8) implies that l(En) > 0 for some n. When f is nonnegative, we have f
fvA for any A X and by definition of En, we have f vEn
1n vEn . So, f
1n vEn , which leads to R f
1n lðEn Þ [ 0. R R (c) No. But if X f dl ¼ X g dl\1, then it does follow, because we can infer that R X ðf  gÞdl ¼ 0. 3.2.P2. Let {fn}n
1 be the sequence of functions on X = [0, 2] defined R R by fn = v[0,1] if n is even and v(1,2] if n is odd. Find X ðlim inf fn Þdl and lim inf X fn dl. Hint: For each x 2 X, the sequence {fn(x)}n
1 alternates between 0 and 1, starting with 0 or 1 according as Rx 2 [0, 1] or x 2 (1, 2]. Therefore lim inf fn is 0 everywhere and has integral 0. But X fn dl ¼ 1 for every n, with the consequence that lim inf R X fn dl ¼ 1. 3.2.P3. Let R L be the class of all Lebesgue integrable functions on X. Define q(f, g) to be X j f  gjdl. Show that q(f, g) can be 0 when f 6¼ g but that q has all the other properties of a metric (i.e. q is a “pseudometric”). In what way is this different Rb if X = [a, b] and we set q(f, g) = a j f ðxÞ  gðxÞjdx in the class of Riemann integrable functions? Hint: If f and R g differ at onlyR finitely many points, then f 6¼ g. However, q(f, g) = 0. Also, 0  X j f  gjdl ¼ X jg  f jdl; i.e. 0  q(f, g) = q(g, f). For Lebesgue integrable functions f, g, h, we have | f − h|  | f − g| R + |g − h| almost R everywhere on X and hence, by monotonicity and linearity, X jf  hjdl  X jf  gjdl þ R X jg  hjdl; i.e. q(f, h)  q(f, g) + q(g, h). Not different. 3.2.P4. Give an example of a nonnegative real-valued function on [0, 1] which is unbounded on every subinterval of positive length (so that it cannot have an improper Riemann integral) but has Lebesgue integral 0.

8 Hints

441

Hint: Let r1, r2, … be an enumeration of the rationals in [0, 1] and define f(x) to be 0 if x is irrational and f(rk) = k. Since every interval of positive length contains infinitely many rationals, it must contain rk with arbitrarily large k; this means that f is unbounded on it. Now consider the sequence of simple functions sn defined as sn(rk) = k if k  n and sn(x) = 0 for all other x. This is an increasing sequence converging to f and each function having integral 0. It follows by the Monotone Convergence Theorem 3.2.4 that f has integral 0. 3.2.P5. (a) [Needed in 3.2.P8(a) & 3.2.P9] Let Y be a measurable subset of X. If f is a measurable function on X, then its restriction to Y, denoted by f|Y, is a measurable function on Y. Show that the product fvY (defined on X) hasR an integralR if and only if the restriction f|Y has an integral, and that when this is so, X (fvY) = Y (f|Y). [Note: R R It is customary to denote this common value by Y f or Y fdl.] (b) Suppose f is a measurable extended nonnegative real-valued function on R. Let R E 2 M or B, and let E f ¼ 0. Show that f vanishes a.e. on E, i.e. {x 2 E: f(x) > 0} has measure zero. Hint: (a) A nonnegative measurable function always has an integral, though it may be ∞. First consider aR characteristic function f = vA, where A X is measurable. Then fvY = vA \ Y and X (fvY) = l(A \ Y). On the other hand, f|Y is the function on R Y that is 1 on A \ Y and 0 elsewhere, so that X (f|Y) = l(A \ Y). This proves the equality in question for characteristic functions. Since (af + bg)vY = a(fvY) + b(gvY) and (af + bg)|Y = a(f|Y) + b(g|Y), it follows by the linearity of the Lebesgue integral that the equality holds for all simple functions. Next, let f be a measurable nonnegative extended real-valued function and consider any increasing sequence {sn} of simple functions converging to it pointwise. Then {sn|Y} is an increasing sequence of simple functions on Y converging to f|Y pointwise, while {snvY} is an increasing sequence of simple functions on X converging to fvY pointwise. Since the equality in question holds for simple functions, it follows by the Monotone Convergence Theorem 3.2.4 that it holds for f. Finally, consider any measurable f. Observe that    þ     f þ vY ¼ ðf vY Þ þ ; f  vY ¼ ðf vY Þ and f þ jY ¼ f jY ; f  jY ¼ f jY : R The equality just established for the nonnegative case shows that X (f + vY) = R R R + |Y) and X (f − vY) = Y (f − |Y). In view of the above observation, we therefore X (f have Z Z Z Z  þ   ð f vY Þ þ ¼ f jY and ðf vY Þ ¼ f jY : X

Y

X

Y

The rest is now clear. R R (b) By definition, E f = R (fvE). Now, R(fvE > 0) = {x 2 E: f(x) > 0}. Apply Problem 3.2.P1(b).

442

8 Hints

3.2.P6. [Theorem due to Lebesgue] Show that a bounded function f:[a, b]!R is Riemann integrable if and only if it is continuous a.e. Hint: Let /, w be as in Proposition 3.2.19. If f is continuous a.e. then by (iv) we have / = w a.e. and hence by (iii), f is Riemann integrable. Conversely, assume f is R Riemann integrable. Then by (iii), we have ½a; b ðw  /Þdx ¼ 0, and by (i) and (ii), we also have w − /
0. By Problem 3.2.P1(b), it follows that / = w a.e. By (v) of (a), f is continuous at all points except perhaps those where / 6¼ w or which belong to one of the partitions Pn. But this exceptional set has measure zero. 3.2.P7. Show that there does not exist a nonnegative Lebesgue integrable function g on [0, 1] such that g(x)
n2xn(1 − x) everywhere. Hint: lim n2 xn ð1  xÞ ¼ 0 everywhere on [0, 1]. By Theorem 3.2.20, n!1

Z

Z

½0; 1

1

n2 !1 n2 xn ð1  xÞdx ¼ ðn þ 1Þðn þ 2Þ 0 Z 6 0¼ ¼ lim n2 xn ð1  xÞdmðxÞ:

n x ð1  xÞdmðxÞ ¼ 2 n

½0;1 n!1

By the Dominated Convergence Theorem 3.2.16, h i no such g can exist. 1 1 3.2.P8. (a) Let I denote the interval ð2k þ 1Þp ; 2kp ; where k 2 N. Using Problem 3.2. P5 and Theorem 3.2.20, show that the Lebesgue integral Z

is no less than

 1 1 sin vI ðxÞdmðxÞ x ½0;1 x

2 ð2k þ 1Þp :

R (b) Hence show that, for f ðxÞ ¼ 1x sin 1x, we have ½0;1 f þ dm ¼ 1. R (c) Does ½0; 1 f dm exist? Hint: (a)RBy Problem 3.2.P5(a), the given Lebesgue integral equals the Lebesgue integral I 1x sin 1x dmðxÞ, and by Theorem 3.2.20, this in turn equals the Riemann integral Z

1=2kp

1 1 sin dx ¼ x 1=ð2k þ 1Þp x

 1 1 x 2 sin dx: x x 1=ð2k þ 1Þp

Z

1=2kp

h i 1 Now, for x 2 ð2k þ1 1Þp ; 2kp ; we have 2kp  1  1  1 1 that x x2 sin x
ð2k þ 1Þp x2 sin 1x : Hence

1 x



 ð2k þ 1Þp and hence sin 1x
0, so

8 Hints

443

Z



1=2kp

  Z 1=2kp

1 1 1 1 1 sin sin dx
dx: x2 x ð2k þ 1Þp 1=ð2k þ 1Þp x2 x

x 1=ð2k þ 1Þp

But 1 ð2k þ 1Þp

(b) Let Ik ¼

Z



1=2kp 1=ð2k þ 1Þp

h

 1 1 1 ðcosð2kpÞ  cosðð2k þ 1ÞpÞÞ sin dx ¼ x2 x ð2k þ 1Þp 2 ¼ : ð2k þ 1Þp

i

1 1 ð2k þ 1Þp ; 2kp for each k 2 N. Then f ðxÞ
0 , þ ¼ f þ vK and f þ vIk ¼ f vIk 8 k 2 N. Set Jn ¼

x 2 [ k2N Ik ¼ K, say.

Therefore f [ 1  k  n Ik . Since the intervals Ik are disjoint, we have vJn ¼ R1  k  n vIk . Since Jn Jn+1, we also have f þ vJn  f þ vJn þ 1 . Also, [ n2N Jn ¼ K; so, lim vJn ¼ vK and hence n!1

limn!1 f þ vJn ¼ f þ vK ¼ f þ . By the Monotone Convergence Theorem 3.2.4, it  R R  follows that ð0;1 f þ dm ¼ lim ð0;1 f þ vJn dm. Now, x!1

Z f ð0;1

þ

n X

! vIk dm ¼

k¼1

¼


n Z X



k¼1 ð0;1 n Z X  k¼1 n X

ð0;1

 f þ vIk dm  f vIk dm

2 by partðaÞ: ð2k þ 1Þp k¼1

But this is the partial sum of a divergent series. R (c) No, because similar reasoning shows that (0,1] f − dm = ∞. 3.2.P9. Suppose that the function f:(0, 1]!R is Riemann integrable on [c,1] R1 R1 whenever 0 < c  1, and that limc!0 c j f ðxÞjdx and limr!0 c f ðxÞdx exist. Then R1 R1 by definition, these limits are respectively equal to 0 | f(x)|dx and 0 f(x)dx. Show R R1 that f is measurable and that its Lebesgue integral (0,1] f is the same as 0 f(x)dx. [Remark The corresponding assertion is true when f:[A, ∞)!R is RRiemann inte1 grable over [A, B] for any finite B > A and the improper integral A f(x)dx converges absolutely.] Hint: The restriction of fv[1/n,1] to [1/n, 1] is Riemann integrable and hence measurable on [1/n, 1]; since fv[1/n,1] vanishes outside [1/n, 1], it is measurable on (0, 1].

444

8 Hints

Now, f ¼ lim f v½1=n;1 and is therefore measurable. Using Problem 3.2.P5(a) and n!1 R R 1 R Theorem 3.2.20, we get (0,1](| f |v[1/n,1]) = [1/n,1]| f | = 1/n| f(x)|dx. Since  j f j ¼ lim j f jv½1=n;1 , the Monotone Convergence Theorem 3.2.4 now yields h!1 R1 R1 R j ð0;1 f j ¼ limn!1 1=n j f ðxÞjdx ¼ 0 j f ðxÞjdx: Thus | f | is Lebesgue integrable. A similar argument using the Dominated Convergence Theorem 3.2.16 shows that R1 R R1 ð0;1 f ¼ limn!1 1=n f ðxÞdx ¼ 0 f ðxÞdx. 3.2.P10. Suppose f: [0, 1]!R is Lebesgue integrable and is also Riemann inteR1 grable on [e, 1] for every positive e < 1. Show that lim e f ðxÞdx exists and equals e!n R1 R [0,1] f dm. In other words, the Riemann integral 0 f(x)dx exists, possibly as an improper integral, and whether improper or not, equals the Lebesgue integral of f over [0, 1]. [Remark The corresponding assertion is true when f: [A, ∞)!R is Lebesgue integrable over [A, ∞).] Hint: Let {an}n
1 be a sequence of positive real numbers less than 1 such that lim an ¼ 0. Denote f v½an ;1 by fn. By Problem 3.2.P5 and Theorem 3.2.20, we get: n!1 R R R1 (*) [0,1] fn = ½an ;1 = an f(x)dx. Also, | fn|  | f | and f is given to be Lebesgue R integrable. By the Dominated Convergence Theorem 3.2.16, it follows that [0,1] R fn! [0,1] f as n ! ∞. In view of (*), this is precisely what was required to be proved. 3.2.P11. Let fn: [0, 1]!R be defined by fn ðxÞ ¼ nx=ð1 þ n10 x10 Þ. Find the pointwise limit f of {fn}n
1 and show that the convergence is neither uniform nor monotone. R1 R1 Also, establish the convergence of Riemann integrals 0 fn(x)dx! 0 f(x)dx.  Hint: It is elementary that f is 0 everywhere. Since f 1n ¼ 12 8n; convergence is not uniform. Also, fn(x)
f(x) on [0, 1] and therefore the convergence, if monotone, must be decreasing. But this is not so, because fn+1(x) > fn(x) when x = 1/n(n + 1). It is easy to verifyRthat 0  fn(x) < R1 everywhere. By the Dominated Convergence Theorem 3.2.16, [0,1] fn(x)dm(x)! [0,1] f(x)dm(x). But all functions involved are continuous and therefore, by Theorem 3.2.20, the Lebesgue integrals are the same as Riemann integrals. 3.2.P12. Let f: [0, 1] [0, 1]!R satisfy | f(x, y)|  1 everywhere. Suppose that for each fixed x 2 [0, 1], f(x, y) is a measurable function of y, and also that for each fixed yR 2 [0, 1], f(x, y) is a continuous function of x. If g: [0, 1]!R is defined as g (x) = [0,1] f(x, y)dm(y), show that g is continuous. Hint: Let {xn}n
1 be any sequence in [0, 1] converging to x0 2 [0, 1]. We have to show g(xn) ! g(x0). Consider the sequence of functions fn on [0, 1] defined by fn(y) = f(xn, y). According to the hypothesis, the sequence {fn} converges pointwise to f0, where f0(y) = f(x0, y), and | fn(y)|  1, | f0(y)|  1 everywhere. By the Dominated Convergence Theorem 3.2.16,

8 Hints

445

Z

Z ½0;1

fn ðyÞdmðyÞ !

½0;1

f0 ðyÞdmðyÞ:

This says precisely that g(xn) ! g(x0).R 1 Note The improper Riemann integral 0 g(u)du is the sum of limits of Riemann integrals Z c Z b gðuÞdu þ lim gðuÞdu; lim a!0

a

b!1

c

where c can be taken as any positive real number without affecting the value of the sum. The improper integral is finite if and only if each of the limits is finite. This will be needed in the next problem. 3.2.P13. Let f be a measurable extended nonnegative real-valued function on X. Define /f: [0, ∞) ! [0, ∞] as /f(u) = l(X(f > u)). Show that (a) /f is a decreasing function; (b) f  g ) /f  /g; (c) For an increasing sequence {hn} of measurable functions converging to h, we have /hn ! /h. R1 R (d) 0 /f(u)du is equal to the integral X f dl. R1 R (e) 0 l(X(f > u))pup−1du = X f pdl, where 0 < p < ∞. Hint: (a) Let 0  u < m. Then f(x) > m ) f(x) > u, which means X(f > m) X(f > u), which implies l(X(f > u))
l(X(f > m )), i.e. /f(u)
/f(m ). (b) f(x) > u ) g(x) > u. So, X(f > u) X(g > u), which implies l(X (f > u))  l(X(g > u)), i.e. /f(u)  /g(u). (c) By definition, /hn m ) = l(X(hn > mÞ) ! l(X(h > mÞ) = /h(mÞ. Indeed, from the hypothesis that {hn} is an increasing sequence of measurable functions converging to h, it follows that the sets X(hn > mÞ form an increasing sequence with union X(h > mÞ. (d) Let a sequence of simple functions increasing to f. By Problem 3.1. R {sn}n
1R be 1 P6, X sndl = 0 /sn (u)du and by the Monotone Convergence Theorem 3.2.4, R R R1 R1 X sndl! X f dl. So, it is enough to show that 0 /sn (u)du! 0 /f(u)du. In view of (a), (b) and (c), this is a consequence of the following result about improper Riemann type integrals: If {gn}n
1 is an increasing sequence of nonnegative extended real-valued decreasing functions on [0, ∞), so that g ¼ lim gn exists n!1 R1 R1 everywhere on the domain and is decreasing there, then 0 gn(u)du! 0 g(u)du. We give the proof of this assertion below. Suppose that g(a) = R∞ for some aR > 0. Since g is decreasing, we have g = ∞ 1 a on [0, a]. Therefore 0 g(u)du
0 g(u)du = ∞. So we need to show that R1 0 gn(u)du ! ∞. To this end, consider any K > 0. Since gn(a) increases to

446

8 Hints

g(a) = ∞, there exists a natural number N such that gn ðaÞ [ Ka for n
N. For such R1 Ra n, we have gn ðuÞ [ Ka for 0  u  a and hence 0 gn ðuÞdu
0 gn ðuÞdu [ R R 1 1 a Ka ¼ K: This proves that 0 gn(u)du ! ∞ = 0 g(u)du if g(a) = ∞ for some a > 0. In view of what has just been proved, we may assume that g(a) < ∞ for every a > 0. This permits us when 0 < a < b < ∞ to form differences such as gn(a) − Rb Rb gn(b) and g(a) − g(b), and we know that the integrals a gn(u)du and a g(u)du are finite. To begin with, we shall show for 0 < a < b < ∞ that Z

b

Z gn ðuÞdu !

b

gðuÞdu:

a

ð8:9Þ

a

Let e > 0 be arbitrary. Since gn(a) − gn(b) ! g(a) − g(b), there exists an M > 0 such that 0  g(a) − g(b) < M as well as 0  gn(a) − gn(b) < M for all n. Let P: a = u0 < u1 < ⋯ 0. In view

of the finiteness of the aforementioned limits, there exist a and b such that 0 < a < b < ∞ and Z 0

a

gðuÞdu\ 0

g ; 4

Z

1

0

gðuÞdu\ b

g : 4

Since 0  gn  g for all n, the above two inequalities are satisfied by every gn as well. Now an improper integral from 0 to ∞ is a sum of three integrals in the obvious way, of which two are governed by the inequalities just mentioned. It follows that Z

0

1

Z gn ðuÞdu 

1

0

Z gðuÞdu  g þ

b

Z gn ðuÞdu 

a

a

b

gðuÞdu :

By R 1 (8.9), the second term on the right-hand side approaches 0 as n ! ∞. Since 0 gn(u)du is an increasing sequence and must therefore have a limit, it follows from the above inequality that Z lim n!1

0

1

Z gn ðuÞdu  0

1

gðuÞdu  g:

R1 R1 This holds for all η > 0 and hence 0 gn(u)du! 0 g(u)du, on the assumption that R1 0 g(u)du < ∞. R1 It remains to prove that the same convergence obtains even if 0 g(u)du = ∞. In Rb this situation, for any K > 0, there exist a and b such that 0 < a < b < ∞ and a g Rb Rb (u)du > K. Since a gn(u)du! a g(u)du by (8.9), there exists an N 2 N such that R1 Rb Rb n
N. Now, 0 gn(u)du
a gn(u)du because each gn a gn(u)du > K whenever R1 R1 is nonnegative. So, 0 gn(u)du > K whenever n
N. Thus 0 gn(u)du ! ∞ = g(u)du as n ! ∞. (e) If l(X(f p > u)) = ∞ for some finite positive u, then both sides of the equality are ∞. Suppose this does not happen. Then from part (d), we have

448

8 Hints

R1 Rb f pdl = 0 l(X(f p > u))du. Now, for 0 < a < b < ∞, we have a l(X(f p > u)) Rb du = a l(X(f > u1/p))du. Since this is a Riemann integral, we may use the substiRb R b1=p tution t = u1/p (see Proposition 1.7.3) to obtain a l(X(f > u1/p))du = a1=p l(X (f > t))ptp−1dt. We arrive at the desired result by taking limits as a ! 0 and b ! ∞. 3.2.P14. (a) Let f be an Rintegrable function on a set R X such that, for every measurable E X, we have E f dl = 0 (which means X fvE dl = 0, as explained in 3.2.P5). Show that f = 0 a.e. R (b) Let f be a Lebesgue integrable function on [a, b]. If [a,x] f dm = 0 for all x 2 [a, b], show that f = 0 a.e. on [a, b]. + RHint:+(a) Let E = X(f
0). Then f = fvE and it follows+from the hypothesis that X f dl = 0. From Problem 3.2.P1(b), it follows that f = 0 a.e. A similar argument shows that f − = 0 a.e. (b) Suppose not. Then there exists a measurable set E [a, b] with m(E) > 0 such that f(t) 6¼ 0 for t 2 E. Obviously, we may suppose f > 0 on E. By Proposition 2.3.24 [the part that says (a) ) (d)], there exists a closed set F E with m(F) > 0. Set A = [a, b]\F. Then A is open and A = [ iIi, a countable union of disjoint intervals Ii. Now, R

X

Z 0¼

½a;b

Z f dm ¼

A[F

Z f dm ¼

Z f dm þ

A

f dm; F

R R which implies A f dm ¼  F f dm 6¼ 0, since f > 0 on E  F. Therefore R R 6¼ 0. Consequently, there exists an i such that Ii f dm 6¼ 0. This implies that A f dm R R either [a,a] f dm 6¼ 0 or [a,b] f dm 6¼ 0, where a and b are the endpoints of Ii. This contradicts the hypothesis. 3.2.P15. Let f be an integrable function on a set E. Show that for every eR> 0, there exists a d > 0 such that whenever A E with l(A) < d, we have | Afdl| < e R (which means | E(fvA)dl| < e, as explained in Problem 3.2.P5). R R Hint: Since | Af|  A| f |, it is sufficient to consider f
0. Suppose, if possible, that there exists Rsome η > 0 such R that we can find subsets An E for which l(An) < 2−n and An f
η, i.e. E(f vAn )
η. Let gn = f vAn . Then gn! 0 except on the set \ n
1( [ i
nAi). Since l( [ i
nAi) < 2−n+1, we have gn! 0 a.e. Let fn = f − gn = f(1 − vAn ). Then {fn}n
1 is a sequence of measurable nonnegative R functions and fn! f a.e. Hence by Fatou′s Lemma 3.2.8, Ef  lim inf R R R R R Efn = Ef − limsup Egn  Ef − η. This inequality cannot hold unless Ef = ∞, contrary to the hypothesis. R 3.2.P16. Let f be an integrable function on [a, b]. If F(x) = [a,x] f dm, prove that F is continuous on [a, b]. If f is an integrable function on [a, ∞), is it true that F is uniformly continuous on [a, ∞)?

8 Hints

449

Hint: Z

½a;b



Z f v½a;x þ h 

 f v½a;x ½a;b ½a;b Z  Z  Z f  v½a;x þ h  v½a;x ¼ f vðx;x þ h ¼ ¼

Fðx þ hÞ  FðxÞ ¼

½a;b

½x;x þ h

f;

assuming hR > 0 and x < b. Now, let e > 0 be given. By 3.2.P15, there exists a d > 0 such that | [x,x+h] f | < e provided h = m((x, x + h)) < d, i.e. such that 0 < h < d ) |F(x + h) − F(x)|. This proves right continuity at any x 2 [a, b). Left continuity at an arbitrary x 2 (a, b] is proved analogously. Since d does not depend on x, the preceding argument proves uniform continuity and is valid even if [a, b] is replaced by [a, ∞). 3.2.P17. Let {fn}n
1 be a sequence nonnegative functions on R such R of measurable R that fn! f a.e. and suppose that R fndm! R f dm < ∞. Show that for each meaR R surable set E R, Efndm! Ef dm. [The result also holds on a general measure space.] Hint: Suppose there is a measurable subset A for which the convergence fails. Then R there exists a r > 0 for which there are infinitely many n such that Afn lies outside R R R R the interval ( Af − 2r, Af + 2r). If Afn  Af − 2r for infinitely many n, then there is a subsequence for which the inequality holds. However, this contradicts Fatou’s Lemma 3.2.8 and therefore Rthe inequality can hold only for finitely many R n. It must therefore be the case that Afn
Af + 2r for infinitely many n. Again there must the inequality holds. But by hypothesis, R R be a subsequence R for which R | R fn − R f | < r, and hence R fn < R f + r, for all sufficiently large n, in particular for all terms of the last mentioned In terms of B = R\A, this means R subsequence. R that the subsequence satisfies B fn < B f − r. This leads to a contradiction to Fatou’s Lemma 3.2.8 again. 3.2.P18. Let f be a measurable nonnegative function on R and E be a measurable subset of finite measure. Show that R R (a) E f = limN!∞ E fN, where fN(x) = min{f(x), N}; R R (b) R f = limN!∞ [−N,N] f. Hint: (a) {fN}N
1 is an increasing sequence of measurable nonnegative functions converging to f pointwise. The required conclusion follows by the Monotone Convergence Theorem 3.2.4. (b) Set fN(x) = f(x) if |x|  N and 0 otherwise. Then {fN}N
1 is an increasing sequence of measurable nonnegative functions converging to f pointwise. Again, the required conclusion follows by the Monotone Convergence Theorem 3.2.4. 3.2.P19. Let E R be measurable and g an integrable function on E. Suppose {fn}n
1 is a sequence of measurable functions on E such that | fn|  g a.e. Show that

450

8 Hints

Z

Z

Z

lim inf fn  lim inf E

Z

fn  lim sup

fn 

E

E

lim sup fn : E

Hint: lim inf fn = limn(infk
nfk). Define the sequence {pn}n
1 by the following rule: pn ðxÞ ¼ inf ffn ðxÞ; fn þ 1 ðxÞ; . . .g: Observe that {pn}n
1 is an increasing sequence of measurable functions such that |pn|  g a.e. and that limn!∞pn = lim inf fn a.e. The sequence g + pn is therefore an increasing sequence of measurable nonnegative R functions. RBy the Monotone Convergence Theorem 3.2.4, we have limn!∞ E(g + pn) = Elimn!∞(g + pn). R R R Since g is integrable, this leads to limn!∞ Epn = E(limn!∞pn) = E(lim inf fn). R R R Since fn
pn everywhere, we have lim inf E fn
limn!∞ Epn = E(lim inffn). This proves the first of the three inequalities in question. The third follows from the first by considering −fn and the second is true for any sequence. 3.2.P20(a) Let {fn}n
1 and {gn}n
1 be sequences of measurable functions on X such that | fn|  gn for every n. Let f and R g beR measurable functions such that limnfn = f a.e. and limngn = g a.e. If limn Xgn = Xg < ∞, show that Z

Z fn ¼

lim

n!1

X

f: X

(b) Let {fn}n
1 be a sequence of integrable functions on X such that fn! Rf a.e., R where f is also integrable. Show that X| fn − f|dl ! 0 if and only if X| fn| R dl! X| f |dl. Hint: (a) Since | fn|  gn, it follows first, that gn± fn are both nonnegative and second, that | f |  g, R so that, by Fatou’s Lemma and the Rhypothesis, R X(g ± f)  lim inf X(gn± fn). On using the two facts that (i) limn Xgn exists R and (ii) Xg < ∞ in that order, we obtain Z

Z

Z

f  lim inf X

fn  lim sup X

Z fn 

X

(b) The “only if” part follows from the simple inequality kfn jj f k  jfn  f j:

f: X

8 Hints

451

For the “if” part, observe that jfn  f j  jfn j þ j f j: R Since | fn − f| ! 0, | fn| + | f | ! 2| f | as n ! ∞, and X| f | < ∞, the conclusion follows from the result of part (a). 3.2.P21. Let the function f:I1 I2!R, where I1 and I2 are intervals, satisfy the following conditions: (i) The function f(x, y) is measurable for each fixed y 2 I2 and f(x, y0) is Lebesgue integrable on I1 for some y0 2 I2; @ (ii) @y f ðx; yÞ exists for each interior point (x, y) 2 I1 I2; (iii) There exists a nonnegative integrable function g on I1 such that @ @y f ðx; yÞ  gðxÞ for each interior point (x, y) 2 I1 I2. R Show that the Lebesgue integral I1 f(x, y)dx exists for every y 2 I2 and that the function F on I2 given by Z f ðx; yÞdx FðyÞ ¼ I1

is differentiable at each interior point of I2, the derivative being 0

F ðyÞ ¼

Z

@ f ðx; yÞdx: @y I1

Hint: Fix y 2 I2. By the Mean Value Theorem, we have f(x, y) − f(x, y0) = (y − y0) @ 0 @v f ðx; y Þ; where y′ lies between y0 and y. By (iii), it follows that | f(x, y)|  | f(x, y0)| + |y − R y0|g(x). Hence by (i) and (iii), f(x, y) is integrable on I1, i.e. the Lebesgue integral I1 f(x, y)dx exists for every y 2 I2. For differentiability, consider any sequence {yn}n
1 in I2 such that yn 6¼ y (8 n) and yn! y. The functions hn(x) = @ (f(x, yn) − f(x, y))/(yn − y) are integrable on I1 and hn ðxÞ ! @v f ðx; yÞ pointwise. @ 00 Moreover, jhn ðxÞj ¼ @v f ðx; y Þ  gðxÞ by the Mean Value Theorem and hypothesis (iii). Now the Dominated Convergence Theorem 3.2.16 implies that R R R lim I1 hn ¼ I1 limhn ¼ I1 @@v f ðx; yÞdx. n

n

But Z

Z hn ¼

I1

f ðx; yn Þ  f ðx; yÞ F ðyn Þ  FðyÞ dx ¼ ! F 0 ðyÞ: y y  y  y n n I1

R1 3.2.P22. Use the result of Problem 3.2.P21 to prove that 0 xnexp(−x)dx = n! R1 n R Hint: By the remark in Problem 3.2.P9, 0 x exp(−x)dx = [0,∞)xnexp(−x)dx and exp(−x/2) is Lebesgue integrable on [0, ∞). Consider the function f(x, y) = exp(−xy)

452

8 Hints

on I1 I2, where I1 = [0, ∞) and I2 = (0, a), a > 1. The function f is continuous in x for each fixed as noted  y 2 I2 and is therefore a measurable function of x. Moreover, @n above, f x; 12 ¼ expðx=2Þ is Lebesgue integrable on I1. For each n, @y f ðx; yÞ ¼ n n n ð1Þ x expðxyÞ exists at each interior point (x,y) 2 I1 I2. Observe that, firstly, 0  xnexp(−xy)  xnexp(−xy0) < exp  12 xy0 , where y
y0 > 0, for all suffi  ciently large x, and secondly, exp  12 xy0 is integrable. [The last inequality follows     from the fact that xnexp ðxy0 Þ exp 12 xy0 = xnexp  12 xy0 < 1, as the exponential tends to ∞ faster than any power of x.] So, we have Z

ðnÞ

F ðyÞ ¼ ð1Þ where F(y) = therefore

R

[0,∞]exp(−xy)dx,

n

xn expðxyÞdx; ½0;1

ð8:11Þ

using Problem 3.2.P21. Also, F(y) = 1/y and

F ðnÞ ðyÞ ¼ ð1Þn n!=yn þ 1 :

ð8:12Þ

R From (8.11) and (8.12), we obtain [0,∞]xnexp(−xy)dm(x) = n!/yn+1. Substitute y = 1. [Remark This can be more easily worked out using induction and the Riemann theory.] R1 3.2.P23. Prove that the improper integral 0 exy sinx x dx; where it is understood that sin x x is to be replaced by 1 when x = 0, converges absolutely for y > 0, and evaluate it. Hint: sinx x is bounded above on [0, ∞); so let K be an upper bound for it. Then xy sin x xy e ; now, for each y > 0, e−xy is Lebesgue integrable (as a function of x  Ke x) on [0, ∞), and therefore so is exy sinx x. By the remark in Problem 3.2.P10, it follows that the given improper converges absolutely for y > 0 and agrees R integral xy sin x with the Lebesgue integral ½0;1Þ e x dmðxÞ ¼ FðyÞ; say. Let I1 = [0, ∞) and I2 = (a, ∞), where a is any positive real number. Then f(x, y) = exy sinx x satisfies a the hypotheses of Problem 3.2.P21 with g(x) = ex . Therefore R F 0 ðyÞ ¼  ½0;1 exy sin x dmðxÞ. By Theorem 3.2.20, the integrals involved are Riemann integrals, permitting integration by parts. Doing this twice, for any y 2 I2 and b > 0, we have Z ½0;b

exy sin x dmðxÞ ¼

eby ðy sin b  cos bÞ 1 þ : 1 þ y2 1 þ y2

8 Hints

453

R1 Letting b ! ∞, we find that 0 exy sin xdx ¼ 1 þ1 y2 ; keeping in mind that y > 0. By the remark in Problem 3.2.P9, this integral equals the corresponding Lebesgue integral. Hence F 0 ðyÞ ¼  1 þ1 y2 for y 2 I2 , i.e. for y > a. Since a > 0 is arbitrary, the equality holds for all y > 0. Upon integrating both sides, we find that F(y) − F(b) = arctan b − arctan y. Therefore, for any sequence bn
1 tending to ∞, it follows that FðyÞ  lim F ðbn Þ ¼ n!1

p  arctan y: 2

ð8:13Þ

But limnF(bn) = 0, because on (0, ∞), exbn sinx x ! 0 pointwise, exbn sinx x  Kex , where K is an upper bound of sinx x , and Ke−x is integrable, so that the Dominated Convergence Theorem 3.2.16 is applicable. Therefore (8.13) shows that F(y) = (p/2) − arctan y. It may be noted that the argument here does not work when y = 0, which is the case in the next R 1 problem. R1 3.2.P24. Assuming that 0 exy sinx x dx converges for y = 0, i.e. 0 sinx x dx converges, find its value by using the Dominated Convergence Theorem 3.2.16. It is understood that sinx x is to be replaced by 1 when x = 0. [In Problem 4.2.P10, we shall obtain the value without employing Lebesgue integration.] Show that sinx x is nevertheless not Lebesgue integrable over [1, ∞). Rn Hint: Since the integral is assumed to exist, it is enough to compute lim 0 sinx x dx. n Rn Consider the functions fn on [0, ∞) given by fn ðyÞ ¼ 0 exy sinx x dx. Then Z 0

1

sin x dx ¼ lim fn ð0Þ: n!1 x

ð8:14Þ

By Theorem 3.2.20, the Riemann integral that defines fn(y) equals the corre R xy sin x @ sponding Lebesgue integral Since @y exy sinx x ¼ ½0;n e x dmðxÞ. exy sin x and exy is Lebesgue integrable over R [0, n], it follows by using Problem 3.2.P21 with g(x) = 1 on [0, n] that fn0 ðyÞ ¼  ½0;n exy sin x dmðxÞ. By making use of the Dominated Convergence Theorem 3.2.16, we can show that fn0 is continuous:   R  indeed,fn0 ðy þ hÞ þ fn0 ðyÞ ¼ ½0;n exy sin x exh  1 dmðxÞ ! 0 as h ! 0 because the integrand is bounded in absolute value by en for |h|  1 and tends to 0  pointwise as h ! 0 Reason: x 2 ½0; n; jhj  1 ) jxhj  n ) 0\exh  en . Being continuous, fn′ has a Riemann integral, and by Theorem 3.2.20, Z ½0;n

fn0 ðyÞdmðyÞ ¼

Z

n 0

fn0 ðyÞdy ¼ fn ðnÞ  fn ð0Þ:

ð8:15Þ

454

8 Hints

Rn Since jfn ðnÞj  0 Kenx dx  K=n, where K is an upper bound of sinx x on [0, ∞), we have limnfn(n) = 0. Using this limit in (8.15) leads to Z  lim

n!1

n

fn0 ðyÞdy ¼ lim fn ð0Þ: n!1

0

ð8:16Þ

Using Theorem 3.2.20 again and then integrating by parts twice, we have

fn0 ðyÞ ¼

Z ½0;n

exy sin x dmðxÞ ¼

Z

n

exy sin xdx ¼

0

1  eny ðy sin n þ cos nÞ ; 1 þ y2

so that v½0;n ðyÞfn0 ðyÞ ! 1 þ1 y2 as n ! 1 . Now, 1 þ ey ð1 þ yÞ 1 1þy 1 3  þ ey  þ ey ; v½0;n ðyÞfn0 ðyÞ  1 þ y2 1 þ y2 1 þ y2 1 þ y2 2 which is Lebesgue integrable over [0, ∞) in view of Theorem 3.2.20 and the Remark in Problem 3.2.P9. On using the Dominated Convergence Theorem 3.2.16, we get Z



lim

n!1

½0;n

fn0 ðyÞ



Z dmðyÞ ¼

1 dmðyÞ: 2 ½0;1Þ 1 þ y

Together with Theorem 3.2.20 and the Remark in Problem 3.2.P10, the above equality yields Z lim

n!1

0

n



fn0 ðyÞ



Z

1

dy ¼ 0

1 p dy ¼ : 1 þ y2 2

ð8:17Þ

R1 By (8.14), (8.16) and (8.17), we have 0 sinx x dx ¼ p=2: If sinx x were Lebesgue integrable over [1, ∞), then by the remark in Problem 3.2. P10, we would have Z

Z 1 Z B sin x sin x sin x dmðxÞ ¼ dx ¼ lim dx: x B!1 0 x 0 ½1;1Þ x

8 Hints

455

  R B sin x R R R dx ¼ 1=B x sin 1  12 dx ¼  1=B 1 sin 1 dx. Hence lim 1 Now, 1 1 1 c x x x x x c!0 1 1 R 1 1 sin dx would exist, in which case, by Problem 3.2.P9, sin dmðxÞ would x

ð0;1 x

x

x

also exist, contradicting Problem 3.2.P8(c). 3.2.P25. Let f be a nonnegative integrable function on [a, b]. For each n 2 N, let 1 P ðk  1Þ  mðEk Þ\1: En = X(n − 1  f < n). Prove that k¼1

Hint: Let vk,v be the characteristic functions of Ek and [ k
1Ek respectively. Since 1 P f
0, we have f = fv; since the Ek are disjoint, we further have f ¼ f v ¼ ðf vk Þ, k¼1

where the partial sums are increasing and pointwise convergent. By the Monotone Convergence Theorem 3.2.4, it follows that Z 1[

½a;b

f ¼

1 Z X k¼1

½a;b

ð f vk Þ


1 X

ðk  1Þ  mðEk Þ:

k¼1

3.2.P26. Let xcf(x) be integrable over (0, ∞) for R c = a and c = b, where 0 < a < b. Show that for each c 2 (a, b), the integral (0,∞)xcf(x)dm(x) exists and is a continuous function of c. Hint: Consider any c 2 (a, b). Then xc  xa for 0 < x  1, while xc  xb for x > 1. So, |xcf(x)|  (xa + xb)| f(x)| for all x 2 (0,∞). Since the right-hand side is integrable, so is the left-hand side. For continuity, choose h so small that c + h 2 (a, b). Now, |(xc+h − xc)f(x)|  2(xa + xb)| f(x)|. The right-hand side being integrable, the Dominated Convergence Theorem 3.2.16 implies that Z ð0;1Þ



 xc þ h  xc f ðxÞdmðxÞ ! 0 as h ! 0:

R 3.2.P27. Suppose f is an integrable real-valued function on R. Define lf(E) = E fdl for every measurable E R. Show that: lf(E) = 0 8 E , f = 0 a.e. R on R.R Hint: ) . Put f = f + − f −. Let E+ = {x 2 R: f(x)
0}. Then 0 = E þ f ¼ E þ f þ : By 3.2.P1(b), f + vanishes a.e. on E+. It follows that f vanishes a.e. on E+. Similarly, f − vanishes a.e. on E− = {x 2 R: f(x)  0} and hence so does f. But E+ [ E− = R. (. Suppose f vanishes a.e. on R. Then so does | f |. Moreover, R R | f | = (| f |v ) E E = 0 by 3.2.P1(b), considering that | f |vE vanishes a.e. and is R R R nonnegative on R. Hence | E f |  E| f | = 0. 3.2.P27. [Complement to Fatou’s Lemma] Let {fn}n
1 be a sequence of measurable Rfunctions on X such that |fnR|  g, where g is an integrable function on X. Prove that X limsupfndl
limsup X fndl and give an example to show that the condition about the function g cannot be dropped. measurable Hint: Set gn = g − fn. Then {gn}n
R1 is a sequence of nonnegative R functions. By Fatou’s Lemma 3.2.8, X lim infgn  lim inf X gn, which implies

456

8 Hints

R R by Theorem 3.2.15 and the assumed finiteness of X g that X limsupfn
limsup R R take X = R and fn = n2v[−1/n,1/n]. Then X fn = n2(2/ X fn. For the example, R n) = 2n and so limsup Xfn = ∞. On the other hand, limsupfn(x) = 0 if x 6¼ 0 and R ∞ if x = 0, so that Xlimsupfn = 0. Alternatively, take fn = v[n,n+1] instead. Then the fn are also bounded. n R1  sin nx dx has limit 0 3.2.P28. Show that the improper Riemann integral 0 1 þ nx as n ! ∞.  n x2 x2 Hint: For n > 1 and x > 0, we have 1 þ nx ¼ 1 þ x þ nðn1Þ 2 þ    [ 4 : We n2 define g(x) to be x42 for x
1 and x1=2 for 0 < x < 1. Then by an elementary R1 computation, the improper Riemann integral 0 ex=2 dx is absolutely convergent and hence by the remark in Problem 3.2.P9, g is Lebesgue integrable over  n ½0; 1Þ: Also, 1 þ nx sin nx  gðxÞ for n [ 1 for n > 1 and all x > 0. [Indeed,  n x2 þ    ¼ 1 þ nx . for 0 < x < 1 and for all n, we have x1=2 \1\1 þ x þ nðn1Þ 2 2 n  n Therefore 1 þ nx sin nx is also Lebesgue integrable on [0, ∞). Now,    n n 0  1 þ nx sin nx  x42 sin nx ; so that 1 þ nx sin nx ! 0 pointwise as n ! ∞. By the Dominated Convergence Theorem 3.2.16, it follows that n R1  1 þ nx sin nx dmðxÞ ! 0 as n ! 1. However, the Lebesgue integral here is 0 the same as the improper Riemann integral in question in virtue of the remark in Problem 3.2.P10. n Rn  3.2.P29. Evaluate the limit as n ! ∞ of the Riemann integral 0 1  nx ex=2 dx.  n Hint: Set fn ðxÞ ¼ v½0;n 1  nx ex=2 . Observe that fn ðxÞ  ex=2 . Indeed,  n 0\x\n ) 0\ 1  nx \1. By an elementary computation, the improper R 1 x=2 Riemann integral 0 e dx is absolutely convergent with value 2 and hence by the remark in Problem 3.2.P9, g(x) = ex=2 has Lebesgue integral 2 over [0, ∞). x=2 . Therefore by the Dominated Moreover, fn(x)!e R R R RConvergence Theorem 3.2.16, it follows that [0,∞)fn! [0,∞)g = 2. But [0,∞)fn = [0,n)fn by Problem 3.2.P5 and this equals the Riemann integral in question by virtue of Theorem 3.2.20. 3.2.P30. Let (X,F , l) be a measure space and {fk}k
1 be a sequence of functions on X which converges almost everywhere to a limit function f. In other words, there exists some N 2 F such that l(N) = 0 and fk(x) ! f(x) for every x 62 N. In symbols, ae fk ! f . (a) Give an example to show that f need not be measurable even if each fk is. ae (b) Show that fk ! g if and only if f = g a.e. (c) If each fk is measurable, show that there always exists a measurable g such that ae fk ! g: (d) State the Monotone and Dominated Convergence Theorems for the case when the inequalities and the convergence in the hypotheses are assumed to hold only almost everywhere.

8 Hints

457 ae

(e) If fk ! f and each fk is measurable, then under the additional hypothesis that X; F ; l is complete, which is to say, F E; lðEÞ ¼ 0 implies

F 2 F;

show that f has to be measurable. Hint: (a) Let X = {a, b, c}, F = {£, {a}, {b, c}, X}, l(£) = 0, l({a}) = 1, ae l({b, c}) = 0, l(X) = 1; fk = 1 everywhere, f(a) = 1, f(b) = 2, f(c) = 3. Then fk ! f but f is not measurable. ae ae (b) Suppose fk ! f and fk ! g: There exists sets N1 2 F and N2 2 F such that l(N1) = 0 = l(N2), fk(x) ! f(x) for every x 2 Nc1 and fk(x) ! g(x) for every x 2 Nc2. It follows that fk(x) ! f(x) as well as fk(x) ! g(x) for every x 2 Nc1 \ Nc2. Thus f(x) = g(x) for every x 2 Nc1 \ Nc2. Since Nc1 \ Nc2 = (N1 [ N2)c and ae l(N1 [ N2) = 0, this means that f = g a.e. Conversely, suppose fk ! f and f = g a.e. Then there exists sets N1 2 F and N2 2 F such that l(N1) = 0 = l(N2), fk(x) ! f (x) for every x 2 Nc1 and f(x) = g(x) for every x 2 Nc2. Therefore fk(x) ! g(x) for every x 2 Nc1 \ Nc2. Once again, since Nc1 \ Nc2 = (N1 [ N2)c and l(N1 [ N2) = 0, ae this means that fk ! g: (c) Let N 2 F , l(N) = 0 and fk(x) ! f(x) for every x 62 N. Then ðfk vN c ÞðxÞ ! ðf vN c ÞðxÞfor every x 2 X: Since each function fx vN c is measurable, it follows that g = f vN c is measurable. Now, (fkvN c )(x) = fk(x) and g(x) = (f vN c )(x) = f(x) for every x 2 Nc. Therefore fk(x) ! g(x) for every x 2 Nc. Since g is measurable and l(N) = 0, the function g has the required property (d) Monotone Convergence Theorem: Let {fn}n
1 be a sequence of measurable nonnegative extended real-valued functions on X such that 0  f1  f2    

a:e:

and fn ! f a:e: as n ! 1: Then some measurable function equals f a.e. and for any such function F,

458

8 Hints

Z

Z fn dl ¼

lim

n!1

X

F dl: X

Dominated Convergence Theorem: Let {fn}n
1 be a sequence of measurable extended real-valued functions on X, converging almost everywhere to a function f and let g be an integrable function on X such that jfn ðxÞj  gðxÞ

almost everywhere:

Then some measurable function equals f a.e. and for any such function F, Z Z Z fn dl ¼ F dl: lim jfn  F jdl ¼ 0 and lim n!1

n!1

X

X

X

ae

(e) By part (c) above, fk ! g for some measurable g. By part (b), f = g a.e. So, we need only show that, if f = g a.e. and g is measurable, then f is measurable. Consider a real number a and let A = X(f > a), B = X(g > a). Then B is measurable and we must argue that A has to be measurable as well. Since f = g a.e., there exists some N with l(N) = 0 such that {x 2 X: f(x) 6¼ g(x)} N. Since A \ Bc and B \ Ac are both subsets of {x 2 X: f(x) 6¼ g(x)}, they are measurable by virtue of the completeness hypothesis. It follows by the argument of Problem 2.3.P5 that A is measurable. 3.2.P31. Let c be the counting measure [see Example 3.1.2(a)] on either N or {1, 2, …, n}. Show for any nonnegative extended real-valued function f that the integral is given by the sum Rkf(k). Hint: The conclusion holds trivially if f is permitted to take the value ∞. So we consider only real-valued f. Let c be the counting measure on N. Then any function is measurable. It is given that f:N!R is nonnegative. Define

f ðkÞ 1  k  n fn ðkÞ ¼ 0 otherwise: Clearly, for each k 2 N, we have fn(k) ! f(k) and 0  f1(k)  f2(k)  ⋯, the latter because f is nonnegative. By the Monotone Convergence Theorem 3.2.4, Z Z fn dc ¼ f dc: lim n!1

X

X

Now, fn = f(1)v{1} + f(2)v{2} + ⋯ + f(n)v{n} + 0⋅v{n,n+1,⋯} and therefore by n R P Pn Proposition 3.1.5, N fn dc ¼ f ðjÞcðfjgÞ; which is the same as j¼1 f ðjÞ by j¼1

definition of the counting measure. Hence

8 Hints

459

Z f dc ¼ lim

n!1

X

n X

f ðjÞ ¼

1 X

j¼1

f ðkÞ:

k¼1

If instead of N, the domain is {1, 2, …, n}, then all nonnegative real-valued functions are simple and a similar conclusion with the obvious modification that the summation is finite can be derived; the argument then does not need the Monotone Convergence Theorem and we merely need to represent f as f(1)v{1} + f(2)v{2} + ⋯+ f(n)v{n} before applying Proposition 3.1.5. Problem Set 3.3 3.3.P1. Give an example of a nonnegative extended real-valued function on [0, 1] which has Lebesgue integral 1 and is (a) the limit of an increasing sequence of continuous functions (b) unbounded on every subinterval of positive length (so that it cannot have an improper Riemann integral). Hint: The limit of the sequence of functions in the proof of Proposition 3.3.14 has the required features. R 3.3.P2. Give an example when X j fn  f jdl ! 0 but fn does not converge pointwise to f anywhere.   Hint: Let X = [0, 1] and consider the intervals j1 ; j ; 1  j  k; k 2 N; arranged in  1k 1k   1 1 2 this order. The first five intervals are [0, 1], 0; 2 ; 2 ; 1 ; 0; 3 ; 3 ; 3 . Let fn be the characteristic function of the nth interval and let f be zero everywhere. If n
k(k + 1)/2, thenR fn is the characteristic function of an interval of measure at most 1/k. Therefore X j fn  f j  1/k ! 0. For any given x 2 [0, 1], any subdivision of [0, 1] into 3 or more subintervals must include a subinterval containing x and also one not containing x, which means {fn(x)}n
1 has a subsequence with each term 1 and also a subsequence with each term 0. So the sequence {fn}n
1 converges nowhere on [0, 1]. 3.3.P3. Let {fk}k
1 be a sequence of integrable functions on a measurable set 1 R 1 P P X such that fk converges a.e. to an integrable X jfk j\1. Show that the series k¼1

sum f and that Hint: Let g ¼

R X

f ¼

1 P

k¼1

1 R P k¼1

X fk .

jfk j. By the Monotone Convergence Theorem 3.2.4 and the given R finiteness of the sum of integrals, it follows that Xg < ∞ and so g is finite a.e. (see Remark 3.2.2). Thus there exists a measurable set H X such that g is finite on 1 P H and l(Hc) = 0. Then the series fk is convergent on H, and denoting its sum on k¼1

k¼1

H by f0, we have | f0|  g on H. Since H is measurable and restrictions of the 1 P fk to H are also measurable, it follows that f0 is a measurable partial sums of k¼1

460

8 Hints

function on H. Therefore its extension f to X obtained by setting it equal to 0 on Hc is measurable (the reader is invited to supply the simple justification). Since the n P extension satisfies | f |  g on X, it is integrable. Write gn ðxÞ ¼ fk . Then k¼1

|gn(x)|  g(x) and lim gn ðxÞ ¼ f ðxÞ8x 2 H. By the Dominated Convergence n!1 n R P Theorem 3.2.16 and the fact that l(Hc) = 0, we obtain lim X fk ¼ n R P

fk lim n!1 k¼1 H

¼ lim

n R P

fk n!1 H k¼1

¼ lim

R

n!1 H

gn ¼

R H

f ¼

R X

n!1 k¼1

f

3.3.P4. For f 2 L1(R), show that Z lim

h!0

R

jf ðx þ hÞ  f ðxÞjdx ¼ 0;

ð8:18Þ

i.e., the mapping φ:R!L1(R) defined by φ(h) = fh, where fh(x) = f(x + h), x 2 R, is continuous. Hint: Let g 2 Cc(R), the space of continuous functions on R with compact support, be such that kf  gk1 \ 3e . Observe that jf ðx þ hÞ  f ðxÞj  jf ðx þ hÞ  gðx þ hÞj þ jgðx þ hÞ  gðxÞj þ jgðxÞ  f ðxÞj: On integrating both sides over R, we find that the integral of the left-hand side of (8.18) does not exceed Z

Z R

jf ðx þ hÞ  gðx þ hÞjdx þ

Z R

jgðx þ hÞ  gðxÞjdx þ

R

jgðxÞ  f ðxÞjdx

= A + B + C, say. Observe that A = C for all h (translation invariance of measure) and by our choice of g, we have A; C  3e. Note that, g, being a continuous function with compact support, is actually uniformly continuous. Let I be any interval containing the support of g. Now, for any η > 0, there exists a d > 0 such that jgðx þ hÞ  gðxÞj\g for all jhj\d and x 2 R: R Then if |h| < d, it follows that R jgðx þ hÞ  gðxÞjdx < 2η‘(I) < 2η‘(I) < 3e, by choosing η sufficiently small. This shows that the left-hand side of (8.18) does not exceed e when |h| < d.

8 Hints

461

3.3.P5. Suppose that f 2 L1(R). Show that Z

Z

lim

jhj!1

R

jf ðx þ hÞ þ f ðxÞjdx ¼ 2

R

jf ðxÞjdx:

Hint: Assume that suppf I R, where I is a bounded interval. In this case, the sum f ðxÞ þ f ðx þ hÞ equals either f(x), f(x + h) or zero, provided |h| > ‘(I). Therefore, for these values of h, we have Z R

Z

Z

jf ðx þ hÞ þ f ðxÞjdx ¼

jf ðx þ hÞjdx þ jf ðxÞjdx ZIh Z I Z ¼ jf ðx þ hÞjdx þ jf ðxÞjdx ¼ 2 jf ðxÞjdx: R

R

R

Consequently, Z

Z

lim

jhj!1

R

jf ðx þ hÞ þ f ðxÞjdx ¼ 2

R

jf ðxÞjdx:

So far the result has been shown to hold for an integrable function whose support is bounded. Now, let f be an arbitrary integrable function. Define

fn ðxÞ ¼

f ðxÞ n  x  n 0 jxj [ n:

R Clearly, | f − fn|  | f | and | f − fn| ! 0 pointwise. Therefore, R | f(x) − fn(x) |dx ! 0 by the Dominated Convergence Theorem 3.2.16. Let e > 0 be given. There R exists an n0 such that n
n0 implies R | f(x) − fn(x)|dx <  4e. It follows that R R n
n0 also implies R | fn(x)|dx > R | f(x)|dx  4e The inequality jf ðx þ hÞ þ f ðxÞj
jfn0 ðx þ hÞ þ fn0 ðxÞj  jfn0 ðx þ hÞ þ fn0 ðxÞ  f ðx þ hÞ  f ðxÞj; on integration, yields Z R

Z jf ðx þ hÞ þ f ðxÞjdx


R

Z

jfn0 ðx þ hÞ þ fn0 ðxÞjdx  2 

e 4

e ¼ 2 jfn0 ðxÞjdx  2  forjhj [ 2n0 4 ZR e
2 jf ðxÞjdx  4  forjhj [ 2n0 : 4 R Thus we have shown that for any e > 0, there exists an M > 0, namely, M = 2n0, such that

462

8 Hints

Z

Z R

jf ðx þ hÞ þ f ðxÞjdx
2

R

jf ðxÞjdx  e for jhj [ M:

On the other hand, Z R

Z jf ðx þ hÞ þ f ðxÞjjdx 

R

Z jf ðx þ hÞjdx þ

R

Z jf ðxÞjdx ¼ 2

R

jf ðxÞjdx:

This proves the required result. 3.3.P6. If f 2 L2[0, 1], show that f 2 L1[0, 1]. However, if f 2 L2[0, ∞) then f need not belong to L1[0, ∞). R Hint: Let g = 1 everywhere on [0, 1]. Then [0,1]g2 = 1 and by the Cauchy– R R R R Schwarz Inequality, we have ( [0,1]| f |)2 = ( [0,1](| f |⋅g))2  ( [0,1]| f |2)( [0,1]g2) = R ( [0,1]| f |2) < ∞ when f 2 L2[0, 1]. However, f ðxÞ ¼ 1 þ1 x defines a function in L2[0,  2 R ∞) that does not belong to L1[0, ∞). Indeed, ½0;1 1 þ1 x dmðxÞ ¼ 1; whereas R 1 ½0;1 1 þ xdmðxÞ ¼ 1: 3.3.P7. The functions in L2[−p, p] given by fn(x) = sinnx, n 2 N, form a bounded and closed subset which is not compact. Hint: By an elementary computation, for n 6¼ k, we have kfn  fk k2 = R R 2 2 2 [−p,p](sinnx − sinkx) dm(x) = [−p,p](sin nx + sin kx − 2sinnxsinkx)dm(x) = 2p. pffiffiffiffiffiffi Thus kfn  fk k ¼ 2p when n 6¼ k. So, the sequence {fn}n
1 cannot have a convergent subsequence and the set {fn: n 2 N} therefore cannot be compact. As the pffiffiffiffiffiffi distance between any two distinct points of the set has been shown to be 2p, it has no limit points, which means that it is vacuously true that it contains all its limit points; consequently, the set is closed. It is also bounded, because all its points lie in pffiffiffiffiffiffi a closed ball of radius 2p around any one of its points. 3.3.P8. Let f and g be positive measurable functions on [0, 1] such that fg
1. Prove that ! Z ! Z f dm g dm
1: ½0;1

½0;1

Hint: If one of the integrals on the left is infinite, there is nothing to prove. Assume both to be finite. Then f 1=2 2 L2 ½0; 1 and g1=2 2 L2 ½0; 1. Also, f 1=2 g1=2
1, so  R  that ½0;1 f 1=2 g1=2
1. But by the Cauchy–Schwarz Inequality, we have R 1=2 R 1=2 R   f g
½0;1 f 1=2 g1=2 . ½0;1 ½0;1

8 Hints

463

3.3.P9. Show that the following inequalities are inconsistent for a function f 2 L2[0, p]: 2 kf  cosk  ; 3

kf  sink 

1 : 3

Hint: The given inequalities imply that kcos  sink  kf  cosk þ kf  sink R  23 þ 13 ¼ 1: However, this is false because kcos  sink2 = [0,p](cosx − sinx)2 Rp dm(x) ¼ 0 ð1  sin 2xÞdx ¼ p by an elementary computation. 3.3.P10. Let fn ðxÞ ¼ 1 þnnpffix for x 2 [0, 1] and n 2 N. (a) Show that the sequence {fn}n
1 has a pointwise limit a.e.; (b) Does fn belong to L2[0, 1] for each (or some) n 2 N? (c) Does the pointwise limit (a.e.) of fn belong to L2[0, 1]? Hint: Proceeding as in Example 3.3.13, we find that (a) the pointwise limit is p1ffix when x 6¼ 0 (b) fn belongs to L2[0, 1] for every n 2 N. For (c), we note that Z ½0;1



1 pffiffiffi x

2

Z dmðxÞ ¼ 0

1

1 dx ¼ 1: x

3.3.P11. [Cf. Problem 3.3.P14] Let (X, F ,l) be a measure space with l(X) < ∞. Show that L∞(X) Lp(X) for all p, 0 < p < ∞. Also, show that, if f is measurable, then lim k f kp ¼ k f k1 :

p!1

Hint: Let f 2 L∞(X). Then | f(x)|p 2 Lp(X) because

Z

Z X

j f jp dl  k f kp1 lðXÞ

or X

1p 1 jf jp dl  k f k1 ðlðXÞÞp :

Since (l(X))1/p! 1 as p ! ∞, we have lim supk f kp  k f k1 :

ð8:19Þ

On the other hand, suppose that | f(x)|
B on a set F of positive measure. Then 1

BðlðFÞÞp  k f kp ;

464

8 Hints

which implies B  lim inf k f kp : Let e > 0 be arbitrary. The set {x 2 X: | f(x)| > k f k1 e} has positive measure by definition of k f k1 . Therefore the preceding inequality yields k f k1 e  lim inf k f kp : Since e > 0 is arbitrary, we get k f k1  lim inf k f kp . Combining this with (8.19), we get the desired inequality. Next, suppose that f 62 L∞(X). We have to show that lim k f kp ¼ 1 . Consider an p!1

arbitrary B > 0. Since k f k1 ¼ 1, the inequality | f(x)|
2B must hold on a set F of (finite) positive measure l(F). The inequality (l(F))1/p > 12 holds for all sufficiently large p. It follows (without any presumption of finiteness) that k f kp
2B (l(F))1/p
B. Since this has been shown to hold for all B and all sufficiently large p, we have lim k f kp ¼ 1, as desired. p!1

3.3.P12. (a) Let 0 < p < 1 and let q be such that 1p þ 1q ¼ 1: Then q < 0. If f 2 R R Lp(X) and the function g on X is such that g 6¼ 0 a.e. and X|g|qdl < ∞, then X| g|qdl 6¼ 0 unless l(X) = 0, and

Z

Z

1p Z 1q jf jp dl jgjq dl :

jfgjdl
X

X

X

R (This is Hölder’s Inequality for 0 < p < 1.) If the hypothesis that X|g|qdl < ∞ is omitted, then the inequality is valid trivially, provided we interpret a negative power of ∞ to mean 0. (b) Let 0 < p < 1 and f, g be measurable functions on X such that f
0, g
0. Then kf þ gkp
k f kp þ kgkp : (This is Minkowski’s Inequality for 0 < p < 1.)

R R (c) For a complex-valued measurable function / on X, show that | X/dl|  X|/| dl. Hint: (a) First note that q < 0 because p < 1 and 1p þ 1q ¼ 1: Set

8 Hints

465

P¼ Then P > 1 and Q > 1 and

1 P

þ

1 p

q Q¼ : p

and

 ¼ p  pq ¼ p 1  1q ¼ 1: Set

1 Q

j f j ¼ ðuvÞp ; jgj ¼ vp : 1

1

These equations define nonnegative measurable functions u and v a.e., considering that g 6¼ 0 a.e., and

Z

Z uvdl 

P1 Z

P

u dl

X

v dl

X

p Z

p

jf j dl  X

;

X

Z

Z

Q1

Q

q

jfgjdl

jgj dl

X

Q1

Z

p Z

¼

q

jfgjdl

X

jgj dl

pq

X

:

X

R Taking the pth root of both sides and multiplying by ( X |g|qdl)1/q, which is permissible because it has been assumed finite, we get

Z

Z

jf jp dl

jfgjdl
X

q1 Z

X

jgjq dl

1q

:

X

(b) Since f
0, g
0, we have f + g
f
0 and f + g
g
0, so that kf þ gkp
k f kp as well as kf þ gkp
kgkp . Therefore we may assume f + g, f, g 2 Lp(X). We may also assume that kf þ gkp [ 0; because otherwise we would also have k f kp ¼ 0 ¼ kgkp , so that the inequality in question would be trivially true. Observe that Z

Z

Z

p

ðf þ gÞ dl ¼ X

f ðf þ gÞ X

p1

gðf þ gÞp1 dl:

dl þ X

We intend to apply (a) to each Rterm on the right-hand side here.R In order to do so, we need to know not only that X(f + g)(p−1)qdl < ∞, i.e., that X(f + g)pdl < ∞, but also that f + g 6¼ 0 a.e. It is legitimate to assume that f + g > 0 everywhere because, on the set where f + g is 0, the functions f and g must also be 0 and integrals need be taken only over the complement of that set; in effect, we may take X to be the complement and f + g > 0 everywhere. Now, applying (a) and following the proof of Minkowski’s Inequality 3.3.6, we arrive at the required conclusion.

466

8 Hints 1 2

(c) The case p =

of (b) is that

Z

1 2

" Z

2

ðf þ gÞ dl

2 1 2




f dl

X

2 # g dl ;

Z

1 2

þ

X

X

which is equivalent to Z



f2 þg

X

1 2 2

" Z dl


2 Z 2 #12 jf jdl þ jgjdl

X

X

with real-valued measurable R f and g, not R presumed nonnegative. This in turn is equivalent to the inequality X |/|dl
| X /dl| for complex-valued /. In fact, for a measurable / = fR+ ig, where f and g are the real and imaginary parts respectively of /, the integral X |/|dl is the left-hand side of the inequality displayed above and R the absolute value | X /dl| is the right-hand side. R However, here is a quicker proof: Note that for complex /, the integral X /dl is R R defined only when X |/|dl < ∞. Since X /dl is a complex number, there exists a R R complex number a such that |a| = 1 and a X /dl = | X /dl|. Then ℜ(a/)  |a/| = |/|. Therefore Z

Z 

Z 

Z  / dl ¼ < j / dlj ¼ < a / dl ¼ < a/ dl X X X X Z Z ¼ 1. Let s be such that 1r þ 1s ¼ 1: Then

8 Hints

467

Z

Z

Z p

p

j f j dl ¼

pr

j f j  1dl 

X

j f j dl

X

X

Z ¼

1r Z

1s 1 dl s

X

pq 1 q jf j dl ðlðXÞÞ s \1:

X

For q = ∞, | f |p  k f kp1 a.e. and so, | f |p is integrable. However, if l(X) = ∞, the inclusion Lq(X) Lp(X) need not hold because the function f: [1, ∞] ! ℝ given by f(x) = 1/x belongs to Lq[1, ∞] for every q > 1 but not to L1[1, ∞]. 3.3.P15. Let X be a measurable subset of ℝ and 0 < p < q < ∞. If f 2 Lp(X) \ Lq(X), then f 2 Lr(X) for all r such that p < r < q. Hint: Since p < r < q, there is a number t, 0 < t < 1, such that r = tp + (1 − t)q. Observe that | f |tp 2 L1/t(X) and | f |(1−t)q 2 L1/(1−t)(X). Hence by Hölder’s Inequality, we have j f jr ¼ j f jtp j f jð1tÞq 2 L1 ðXÞ: 3.3.P16. (a) Let p
1 and let kfn  f kp !0 as n ! ∞. Show that kfn kp ! k f kp as n ! ∞. (b) Suppose {fn}n
1 is a sequence in Lp(X), f 2 Lp(X), fn! f a.e. and kfn kp ! k f kp as n ! ∞. Prove that kfn  f kp !0 as n ! ∞. Hint: (a) By Minkowski’s Inequality, we have kfn kp k f kp  kfn  f kp : The result follows on using the convergence kfn  f kp ! 0: (b) Analogous to Problem 3.2.P20(b). R −1/4 3.3.P17. Show that [0,p]x sinx dx  p3/4. Hint: Apply the Cauchy–Schwarz Inequality to obtain Z x ½0;p

14

Z sin x dx 

x ½0;p

12

!12 Z

!12 2

dx

sin x dx ½0;p

12 1 12 3 p ¼ p4 : ¼ 2p 2 

1 2

3.3.P18. For each n, consider the functions fn: ℝ ! ℝ given by fn = v[n,n+1], n = 1, 2, …. Show that fn(x) ! 0 as n ! ∞ for each x 2 ℝ. Also, show that for all p such that 1  p  ∞, we have fn 2 Lp(ℝ), kfn kp ¼ 1 for all n, so that || fn||p 9 0 as n ! ∞. (This example shows that pointwise convergence does not imply convergence in any Lp norm, 1  p  ∞.)

468

8 Hints

Hint: Fix x 2 ℝ. There exists an integer n0 such Rthat n0 > x. RSo, fn(x) = 0 for n > n0 + 1 and hence fn(x) ! 0 as n ! ∞. Also, R jfn jp dx ¼ R vp½n;n þ 1 dx ¼ 1; which implies kfn kp ¼ 1 for all n and for all p such that 1  p < ∞. Being the characteristic function of a set of positive measure, fn has essential supremum 1, i.e., kfn k1 ¼ 1 for all n. Consequently, || fn||p 9 0 as n ! ∞ for 1  p  ∞. 3.3.P19. Give an example to show that convergence in the pth norm does not imply convergence a.e. Hint: Consider the interval [0, 1]. For each n, divide [0, 1] into n subintervals:       1 1 2 n1 ; ; . . .; ;1 : 0; ; n n n n Write En;k ¼

  k1 k ; ; n n

k ¼ 1; 2; . . .; n:

Enumerate all these intervals as follows: E11 ; E21 ; E22 ; E31 ; E32 ; E33 ; . . .: Let {En}n
1 denote the above sequence of intervals. Define fn ¼ vEn : R Since m(En) ! 0 as n ! ∞, [0,1] f pn = m(En) ! 0 as n ! ∞, i.e. kfn  0kp ! 0 as n ! ∞. However, fn(x) converges for no x 2 [0, 1]. Indeed,

x 2 ½0; 1 : jfn ðxÞj


1 2

 ¼ En

for all n;

i.e., for each x 2 [0, 1], fn(x) = 1 for infinitely many values of n. Hence fn(x) 9 0 as n ! ∞. 3.3.P20. Let p and q be conjugate indices and lim kfn  f kp ¼ 0 ¼ lim kgn  gkq , n!1

n!1

where fn, f 2 Lp(X) and gn, g 2 Lq(X), n = 1, 2, …. Show that lim kfn gn  fgk1 ¼ 0: n!1

8 Hints

469

Hint: We have kfn gn  fgk1  kfg  fn gk1 þ kfn g  fn gn k1  kf  fn kp jjgjjq þ kfn kp kg  gn kq ; using Hölder’s Inequality. Since kfn kp k f kp  kfn  f kp , using the Minkowski Inequality, for all large n, we have kfn kp  k f kp þ 1: Hence, for large n,  k fn gn  fgk1  k f  fn kp kgkq þ k f kp þ 1 kg  gn kq : The result now follows. 3.3.P21. For 1  p < ∞, we denote by ‘p the space of all sequences {xm}m
1 such 1 P that jxm jp \1. m¼1

(a) Without interpreting sums as integrals with respect to the counting measure, show that, if fxm gm
1 2 ‘p and fym gm
1 2 ‘q , where 1 < p, q < ∞ and 1p þ 1q ¼ 1; then 1 X

j xm ym j 

m¼1

1 X

!1p jxm j

p

m¼1

1 X

!1q jym j

q

:

m¼1

For p = 1, q = ∞, show that 1 X

1 X

j xm ym j 

m¼1

!  jxm j fym g

m¼1

 

m
1 1:

(b) Using the inequality of part (a), show also that, if {xm}m
12 ‘p and {ym}m
12 ‘p, then 1 X m¼1

!1p p

j xm þ ym j



1 X m¼1

!1p j xm j

p

þ

1 X

!1p j ym j

p

:

m¼1

[The inequalities in (a) and (b) are known as the Hölder and Minkowski Inequalities respectively for sequences.]

470

8 Hints

Hint: (a) We need only consider the case when

1 P m¼1

jxm jp 6¼ 0 6¼

1 P m¼1

jym jq . To begin

with, we assume that 1 X

jxm jp ¼ 1 ¼

m¼1

1 X

j ym j q :

m¼1

In this case, the inequality in question reduces to 1 X

jxm ym j  1:

m¼1

To obtain this, we put successively a = |xm|p and b = |ym|q for m = 1, 2, …, n in Lemma 3.3.3 and then add up the inequalities so obtained to arrive at n X

j xm ym j 

m¼1

n n 1 1 1X 1X 1X 1X 1 1 j xm j p þ j ym j q  j xm j p þ jym jq ¼ þ ¼ 1; p m¼1 q m¼1 p m¼1 q m¼1 p q

provided that p > 1 (so that q 6¼ ∞). This is true for each n 2 N: Letting n ! ∞, we obtain the inequality in question for the special case when 1 1 P P jxm jp ¼ 1 ¼ jym jq . The general case can be reduced to the foregoing special

m¼1

m¼1

case if we take in place of xm and ym the numbers x0m ¼ xm =

1 X

!1p j xm j p

;

1 X

y0m ¼ ym =

m¼1

!1q j ym j q

m¼1

for which 1 1 X X x0 p ¼ 1 ¼ y0 q : m m m¼1

m¼1

It follows by what has been proved in the paragraph above that 1 X m¼1

j xm ym j 

1 X m¼1

!1p jxm j

p

1 X

!1q jym j

m¼1

provided that p > 1.The case p = 1, q = ∞ is trivial.

q

;

8 Hints

471

(b) For p = 1, observe that jxm þ ym j  jxm j þ jym j;

m ¼ 1; 2; . . .:

Summing the inequalities for m = 1, 2, …,n, we obtain n X

j xm þ ym j 

m¼1

n X

j xm j þ

n X

m¼1

jym j 

m¼1

1 X

jxm j þ

m¼1

1 X

jym j;

m¼1

which implies the desired inequality on letting n ! ∞. Now suppose p > 1. It is obvious that "

n X

#1p p

j xm þ ym j

" 

m¼1

n X

#1p ðjxm j þ jym jÞ

p

:

ð8:20Þ

m¼1

Now, ðjxm j þ jym jÞp ¼ ðjxm j þ jym jÞp1 jxm j þ ðjxm j þ jym jÞp1 jym j: Summing these identities for m = 1, 2, …, n, we obtain n X

ð j xm j þ j ym j Þ p ¼

m¼1

n X

ðjxm j þ jym jÞp1 jxm j þ

m¼1

n X

ðjxm j þ jym jÞp1 jym j:

m¼1

The first term on the right may be estimated by the Hölder Inequality of part (a) (similarly for the second term), leading to n X

ð j xm j þ j ym j Þ

p1

jxm j 

m¼1

n X

!1p j xm j

p

m¼1

¼

n X

!1p j xm j p

m¼1

n X

ð j xm j þ j ym j Þ

ðp1Þq

m¼1 n X

!1q

!1q ð j xm j þ j ym j Þ p

:

m¼1

Thus n X

0 ðjxm j þ jym jÞp  @

m¼1 n P

m¼1

obtain

!1p jxm jp

m¼1

Dividing by

n X

þ

n X m¼1

!1p 1 !1q n X ðjxm j þ jym jÞp : j ym j p A m¼1

1=q ð j xm j þ j ym j Þ

p

, which we may suppose to be nonzero, we

472 n X ð j xm j þ j ym j Þ p

8 Hints

!1p 

m¼1

n X

!1p j xm j

p

n X

þ

m¼1

!1p jym j

p

1 X



m¼1

8 !1=p 9 < P = n ð j xm j þ j ym j Þ p Thus : y¼1 ;

!1p jxm j

m¼1

p

þ

1 X

!1p jym j

p

:

m¼1

is an increasing sequence of nonnegative n
1

numbers bounded above by the sum 1 X

!1p p

j xm j

1 X

þ

m¼1

!1p p

jym j

:

m¼1

The desired inequality now follows by using (8.20). 3.3.P22. For the space ‘p (1  p < ∞) of Problem 3.3.P21, show that

dp ðx; yÞ ¼

1 X

!1p jxm  ym jp

m¼1

defines a complete metric, without using Theorem 3.3.10 Hint: It is a consequence of Minkowski’s Inequality [Problem 3.3.P21(b)] that dp(x, y) 2 ℝ. Evidently, dp(x, y) = 0 if and only if x = y. The triangle inequality for dp also follows from Minkowski’s Inequality.   ðmÞ  ðmÞ ðmÞ ðmÞ ; x ¼ x ; x ; . . . denote a Cauchy sequence in ‘p. Then for Let x 1 2 m
1 given e > 0, there exists a positive integer n0(e) such that 



dp xðnÞ ; xðmÞ ¼

1 X xðnÞ  xðmÞ p m

m

!1p \e

ð8:21Þ

m¼1

ðnÞ ðmÞ for all n, m
n0(e). This implies xm  ym \e for n; m
n0 ðeÞ, that is, for each n o ðmÞ is a Cauchy sequence of numbers. So by Cauchy’s m, the sequence xm m
1

ðmÞ

principle of convergence, limm xm ¼ xm , say. Let x be the sequence (x1, x2, …). It will be shown that x 2 ‘p and limmx(m) = x. From (8.21), we have N X xðnÞ  xðmÞ p \ep m

m¼1

m

ð8:22Þ

8 Hints

473

for any positive integer N, provided n, m
n0(e). Letting m ! ∞ in (8.22), we obtain N X xðnÞ  xðmÞ p \ep m

m

m¼1

p  N P ðnÞ for any positive integer N and all n
n0(e). The sequence xm  xm m¼1 N
1 p 1 P ðnÞ is increasing and bounded above, and therefore has a finite limit xm  xm , m¼1

which is less than or equal to ep. Hence

1 X xðnÞ  xm p m

!1p  e for all n
n0 ðeÞ:

ð8:23Þ

m¼1

Observe that 1 X

!1p jxm j

m¼1

p

1 X xðnÞ  xm p  m m¼1

!1p

1 X xðnÞ p þ m

!1p

m¼1

by Minkowski’s inequality, and consequently, x 2 ‘p and x(m) ! x in view of (8.23). n P 3.3.P23. Show that, if k1 ; k2 ; . . .; kn [ 1; k1i ¼ 1 and fi 2 Lki ðXÞ for each i, then i¼1

k1 Z Y n n Z Y i ki jfi j dl : fi dl  X i¼1 X i¼1

ð8:24Þ

Equality occurs if and only if either fi = 0 a.e. for some i or there exist positive constants ci, 1  i  n, such that k ci jfi jki ¼ cj fj j a:e:1  i; j  n:

ð8:25Þ

Hint: We omit writing “dl” in order to reduce clutter. For n = 1, the inequality (8.24) is trivial. For n = 2, it is just Hölder’s Inequality of −1 Theorem 3.3.5. Suppose that it holds for some n > 1. Then if a = (1 − k−1 n+1) , we n n   P P 1 aki ¼ a ki1 ¼ a 1  kn1þ 1 ¼ 1. So, have i¼1

i¼1

474

8 Hints

Z X

n Y

! a

jfi j

i¼1



n Z Y

jfi jki

a=ki

:

X

i¼1

But from Hölder’s Inequality, Z Z nY þ1 fi  X i¼1 X

a !!a1 Z 1=kn þ 1 nY 1=ki Y n þ 1 Z kn þ 1 ki  : jfn þ 1 j jfi j fi i¼1 X X i¼1

So, (8.24) holds for n + 1. Thus by induction, the desired inequality follows for arbitrary n 2 N. If fi = 0 a.e. for some i, then equality clearly holds in (8.24). If on the other hand (8.25) holds, then a little computation shows that both sides of (8.24) are equal to

Z

c1 1=k1 1=k2 1=k c2 . . .cn n

c1

 jf1 j

k1

:

X

This shows that for equality to hold in (8.24), it is sufficient that either fi = 0 a.e. for some i or (8.25) holds for some positive c1, …, cn. We go on to consider necessity. When n = 2, the case of equality in (8.24) has been discussed in Theorem 3.3.5. Assume that the necessity in question is valid for n functions, where n > 1 (induction hypothesis). Consider the case of n + 1 functions, none of them being zero a.e., such that 1=ki Z nY þ1 n þ 1 Z Y ki fi ¼ : ð8:26Þ jfi j X i¼1 X i¼1 It is to be shown that there exist positive constants ci, 1  i  n + 1, such that k ci jfi jki ¼ cj fj j

a:e: 1  i; j  n þ 1:

Let a be as above. From (8.26) and Hölder’s Inequality, we have nY þ 1 Z i¼1

1=ki jfi j X

ki

a !a1 Z 1=kn þ 1 Z Y Z nY þ1 n ¼ fi  ; jfn þ 1 jkn þ 1 fi X i¼1 X i¼1 X

8 Hints

475

which, upon cancelling the common last factor on each side and raising to power a, implies a ! a=ki Z Y n Z n Y ki  jfi j fi : X X i¼1 i¼1 Recalling from above that a

n P l¼1

ki1 ¼ 1 and applying (8.24) to the right-hand side

here, we obtain a ! a=ki Z Y n n Z Y ki : jfi j fi  X i¼1 X i¼1 The preceding two inequalities lead to a ! a=ki Z Y n n Z Y : jfi jki fi ¼ X i¼1 X i¼1

ð8:27Þ

n Q In particular, fi cannot be 0 a.e. Besides, it follows by the induction hypothesis i1

that there exist positive constants ci, 1  i  n, such that k ci jfi jki ¼ cj fj j

a:e: 1  i; j  n:

ð8:28Þ

It remains to show that there exists a positive cn+1 such that cn þ 1 jfn þ 1 jkn þ 1 ¼ c1 jf1 jk1 . To this end, we substitute (8.27) into (8.26), thereby obtaining Z nY þ1 fi ¼ X i¼1

a !a1 Z 1=kn þ 1 Z Y n kn þ 1 jfn þ 1 j : fi X i¼1 X

−1 −1 Since 1/a + 1/kn+1 = 1 (by definition of a as being (1 − kn+1 ) ) and neither n Q fi nor | fn+1| is 0 a.e., the second assertion of Theorem 3.3.5 yields a positive i¼1

n  n a Q Q a=ki number c such that cjfn þ 1 jkn þ 1 ¼ fi . We claim that cn þ 1 ¼ ci c serves

i¼1

i¼1

our purpose. Indeed, on the basis of (8.28), almost everywhere we have

476

8 Hints

cn þ 1 jfn þ 1 j

kn þ 1

¼

n Y

a=k ci i

! 

cjfn þ 1 j

i¼1

¼

n  Y

1=ki

ci

jfi j



kn þ 1



n Y

¼

a=k ci i

i¼1

!a ¼

i¼1

n  Y

1=k1

c1

a ! Y n fi i¼1 ! a

jf1 jk1 =ki

i¼1

 a 1=k þ  þ 1=kn ¼ c1 1 jf1 jk1 ð1=k1 þ  þ 1=kn Þ ¼ c1 jf1 jk1 ; once again using the equality a

n X

ki1 ¼ 1:

i¼1

3.3.P24. (a) Let ∞
p
1 and fi 2 Lp(X) for i = 1, 2, …, n. Show that    X n n X   fi   kf i kp :   i¼1  i¼1 p

(b) Let ∞ > p > 1. If equality holds in (a), show that there exist nonnegative constants ci, 1  i  n, not all 0, such that cifi = cjfj a.e. for 1  i, j  n. (c) If either p = 1 or p = ∞, show that the analogue of (b) does not hold. Hint: (a) For n = 2, this is Theorem 3.3.8. Suppose the result holds for n = j, where j > 1. Then, using Theorem 3.3.8 once again,       jþ1  j j   X  j X X      X     fi  ¼  fi þ fj þ 1    fi  þ fj þ 1 p  kfi kp þ fj þ 1 p    i¼1   i¼1  i¼1  i¼1 p

¼

jþ1 X

p

p

kf i kp :

i¼1

So the result holds for n = j + 1. By induction, the inequality now follows. (b) If fi = 0 a.e. for some i, then we need only choose ci = 1 and cj = 0 for j 6¼ i. So, we consider only the case when none of the fi is 0 a.e. When n = 2, what is to be proved is immediate from the second part of Minkowski’s Inequality in Theorem 3.3.6. Assume the assertion true for some n þ 1  n þ 1 P  P n > 1 and consider n + 1 functions fi such that  fi kfi kp , none of them  ¼  l¼1

p

l¼1

being 0 a.e. Then by Minkowski’s Inequality,     nX X  nX þ1 þ1  n     fi    fi  þ kfn þ 1 kp ; kf i kp ¼      i¼1 i¼1 i¼1 p

p

8 Hints

477

  n  n P P  so that  fi  kfi kp : But the reverse inequality must hold by (a) and   l¼1

p

i¼1

therefore n X i¼1

   X n   fi  : kf i kp ¼   i¼1  p

Now the induction hypothesis yields nonnegative constants ci, 1  i  n, not all 0, such that cifi = cjfj a.e. for 1  i, j  n. It remains to show that there exists a nonnegative cn+1 such that cn+1fn+1 = c1f1 a.e. Since none of the fi is 0 a.e., we know that none of the ci is 0. It follows that fi = (c1/ci)f1 for every i and hence that n n P P fi ¼ ðc1 =ci Þf1 . Now, it follows from the statements displayed above that l¼1

i¼1

    nX  n  þ1   X   f ¼ f  þ kf n þ 1 kp :   i¼1 i   i¼1 i  p

p

Applying the second part of Minkowski’s Inequality in Theorem 3.3.6, we find that n P there exist nonnegative constants a and b, not both zero, such that a fi ¼ bfn þ 1 i¼1

a.e. Since fn+1 is not 0 a.e., we know that a 6¼ 0. Combining this with the equality n n P P fi ¼ ðc1 =ci Þf1 proved earlier, we get i¼1

i¼1 n X

ðb=aÞ

! ð1=ci Þ fn þ 1 ¼ c1 f1 ;

i¼1

which means cn þ 1 ¼ ðb=aÞ

n P

 ð1=ci Þ , obviously nonnegative, serves the

i¼1

purpose. (c) Let X consist of two points called a and b, with the counting measure. Consider the functions f1 and f2 on X given by f1 ðaÞ ¼ f1 ðbÞ ¼ f2 ðaÞ ¼ 1 and f2 ðbÞ ¼ 2: Then there can be no numbers c1 and c2 such that c1f1 = c2f2. But kf1 k1 ¼ 2; kf2 k1 ¼ 3; kf1 þ f2 k1 ¼ 5 ¼ kf1 k1 þ kf2 k1 ; kf1 k1 ¼ 1; kf2 k1 ¼ 2; kf1 þ f2 k1 ¼ 3 ¼ kf1 k1 þ kf2 k1 :

478

8 Hints

3.3.P25. Show that, if for some p, where 0 < p < ∞, f 2 Lp(X) \ L∞(X), then for all q such that p < q < ∞, we have f 2 Lq(X) and p ð1pÞ k f kq  k f kqp k f k1 q :

Hint: Observe that p p ð1pÞ p j f j ¼ j f jq j f jð1qÞ  k f k1 q j f jq ;

which implies Z

ð1pÞq j f jq dl  k f k1 q

Z j f p jdl;

X

X

i.e. ð1pÞq k f kqq  k f k1 q k f kpp and hence p ð1pÞ k f kq  k f k1 q k f kqp :

Problem Set 4.1 4.1.P1. (a) [Needed in (b)] Prove Abel’s Lemma: If b1
b2
⋯
bn
0 and p P k ur  K for 1  p  n, where u1, u2, …,un are any n real numbers, then r¼1

b1 k 

n X

br ur  b1 K:

r¼1

The next two parts together constitute what is called Bonnet’s form of the Second Mean Value Theorem for Integrals and will be needed in Theorem 4.2.8 below. Rb Rb (b) If a f and a g both exist and f is decreasing on [a, b] with f(b)
0, then there exists a n 2 [a, b] such that Z a

b

Z fg ¼ f ðaÞ

n

g: a

8 Hints

479

Rb Rb (c) If a f and a g both exist and f is increasing on [a, b] with f(a)
0, then there exists a n 2 [a, b] such that Z

b

Z

g: n

a

Hint: (a) Let Sp ¼

Pp r¼1

b

fg ¼ f ðbÞ

ur ; 1  p  n. Then

n P

br ur ¼ b1 S1 þ b2 ð S2  S1 Þ þ    þ

r¼1

bn ðSn  Sn1 Þ ¼ ðb1  b2 ÞS1 þ ðb2  b3 ÞS2 þ P   þ ðbn1  bn ÞSn1 þ bn Sn . [This n is Abel’s summation formula.] Therefore b u
ðb1  b2 Þk þ ðb2  b3 Þ Pnr¼1 r r k þ    þ ðbn1  bn Þk þ bn k ¼ b1 k and also r¼1 br ur  ðb1  b2 ÞK þ ðb2  b3 Þ K þ    þ ðbn1  bn ÞK þ bn K ¼ b1 K: (b) Let P: a = x0 < x1 < ⋯ < xn = b be a partition of [a, b], and mr and Mr be such that mr  g(x)  Mr for all x 2 [xr−1, xr]. Let tr be any point in [xr−1, xr]. We then have Z

xr

mr ðxr  xr1 Þ 

g  Mr ðxr  xr1 Þ

xr1

and mr ðxr  xr1 Þ  gðtr Þðxr  xr1 Þ  Mr ðxr  xr1 Þ: Summing these over 1  r  p, we get p X

Z

xp

mr ðxr  xr1 Þ  a

r¼1

g

p X

Mr ðxr  xr1 Þ

r¼1

and p X

mr ðxr  xr1 Þ 

r¼1

p X

gðtr Þðxr  xr1 Þ 

r¼1

p X

Mr ðxr  xr1 Þ;

r¼1

p p R x P P which give a p g  gðtr Þðxr  xr1 Þ  ðMr  mr Þðxr  xr1 Þ  xðg; PÞ; r¼1

r¼1

where x(g, P) is the difference between the upper and lower sums of g over the partition P. It follows that Z a

xp

g  xðg; PÞ 

p X r¼1

Z

xp

gðtr Þðxr  xr1 Þ  a

g þ xðg; PÞ:

480

8 Hints

Rt Since a g is a continuous function of t, it has a minimum value A and a maximum value B. Using the preceding inequality, we get A  xðg; PÞ 

p X

gðtr Þðxr  xr1 Þ  B þ xðg; PÞ;

for 1  p  n:

r¼1

From this inequality, the hypothesis that f is decreasing with f(b)
0 and from Abel’s Lemma [see part (a)], it follows that f ðaÞðA  xðg; PÞÞ 

n X

f ðtr Þgðtr Þðxr  xr1 Þ  f ðaÞðB þ xðg; PÞÞ;

r¼1

Rb Rb which implies f ðaÞA  a fg  f ðaÞB: So, a fg ¼ f ðaÞl for some l between A and Rt B. But these are the minimum and maximum values of a g and the Intermediate Rn Value Theorem 1.3.26 yields a point n 2 [a, b] such that l = a g . Thus Rn Rb a fg ¼ f ðaÞ a g, as required. (c) If we reset the value f(a) to be 0, then f continues to be increasing and Riemann integrable over the interval. So we assume f(a) = 0. Since f is increasing, it follows that /(x) = f(b) − f(x) is decreasing; also /(b) = 0. Applying (b) with / in place of Rb Rb f and collecting terms suitably, we get a fg ¼ f ðbÞ n g. 4.1.P2. Find the Fourier coefficients of the periodic function f for which

f ðxÞ ¼

a a

p  x\0 0  x\p:

Hint: From Example 4.1.5(a), the Fourier series of the function g such that

gðxÞ ¼ is gðxÞ a þ

4a p

1 P sinð2n þ 1Þx n¼0

2n þ 1

0 2a

p  x\0 0  x\p

. The given function f here can be obtained from g as

f = g − a and therefore its Fourier series is

f ðxÞ

1 4a X sinð2n þ 1Þx : p n¼0 2n þ 1

8 Hints

481

4.1.P3. Using Bonnet’s form of the Second Mean Value Theorem for Integrals of 4.1.P1(c), prove that for a monotone function f on [−p, p], the Fourier coefficients an and bn (n 6¼ 0) satisfy the inequalities jan j 

1 j f ðpÞ  f ðpÞj; np

jbn j 

1 j f ðpÞ  f ðpÞj: np

Hint: When f is increasing, the function f(x) − f(−p) is increasing as well as Rp nonnegative, and since pan = p (f(x) − f(−p))cos nx dx, by 4.1.P1(c), we have Z pan ¼ ½f ðpÞ  f ðpÞ

p

n

sin np  sin nn n  sin nn : ¼ ½f ðpÞ  f ðpÞ n

cos nx dx ¼ ½f ðpÞ  f ðpÞ

4.1.P4. Suppose f is Lebesgue integrable with period 2p and “modulus of continuity” x, which means xðdÞ ¼ supfj f ðxÞ  f ðyÞj : jx  yj  dg: Show that its Fourier coefficients satisfy |an|  x(p/n), |bn|  x(p/n), (n 6¼ 0). Rp Hint: On substituting t ¼ x þ pn in an ¼ p1 p f ðxÞ cos nx dx; we get another integral for an over [−p, p]:

1 an ¼ p

Z

p 1 f x cosðnx  pÞdx ¼  n p p þ p=n p þ p=n



Z

 p f x cos nx dx: n p p

Adding these two integrals over [−p, p], we get

2an ¼

1 p

Z

h  p i f ðxÞ  f x  cos nx dx: n p p

Therefore

2j an j 

1 p

Similarly for bn.

Z

 p p 1 p  2p ¼ 2x : f ðxÞ  f x  j cos nxjdx  x n p n n p p

482

8 Hints 1 2 a0

4.1.P5. For the trigonometric series 1 P

þ

1 P

ðan cos nx þ bn sin nxÞ, suppose

n¼1

ðjak j þ jbk jÞ converges. Show that the trigonometric series converges uniformly

k¼1

and that it is the Fourier series of its own sum. p P Hint: Let sp ðxÞ ¼ 12 a0 þ ðak cos kx þ bk sin kxÞ. Then for n [ m; jsn ðxÞ  sm ðxÞj k¼1 Pn P ¼ k¼m þ 1 ðak cos kx þ bk sin kxÞ  nk¼m þ 1 ðjak j þ jbk jÞ. Thus the sequence of partial sums {sn(x)} is uniformly Cauchy and hence converges (uniformly) to a continuous limit f(x). Apply Proposition 4.1.3. 4.1.P6. A finite set of continuous functions g1, …, gn on [a, b] is said to be an orthonormal system with weight function w (Lebesgue integrable and nonnegative) if 

Z



b

gi ; g j ¼

gi ðxÞgj ðxÞwðxÞdx ¼

a

0 1

i 6¼ j : i¼j

(Example: By Lemma 4.1.2, any finite number of functions among 12 ; cos kx; sin kx form an orthonormal set with domain [−p, p] and weight wðxÞ ¼ p1 :Þ For the n P function ak gk ðxÞ; show that k¼1

n X

pffiffiffi j ak j  n 

Z

wðxÞdx a

k¼1

Hint:

n P

pffiffiffi jak j  n 



k¼1

n P

k¼1

n X k¼1

12

b

* a2k

¼

1=2 a2k

n X j¼1

( ) X n max a g ðxÞ : x 2 ½a; b : k¼1 k k

aj gj ;

by the Cauchy–Schwarz Inequality. Also,

n X ‘¼1

+ a‘ g‘

Z ¼ a

b

n X

!2 ak gk ðxÞ

wðxÞdx

k¼1

( )!2 Z n X b  max ak gk ðxÞ : x 2 ½a; b wðxÞdx: k¼1 a

4.1.P7. Suppose the sequence of partial sums sn of the trigonometric series a0 + (ancosnx + bnsinnx) has a subsequence sn(k) that converges uniformly to a function f. Show that the trigonometric series is the Fourier series corresponding to f. Rp Hint: Consider any p 2 N. We shall show that ap ¼ p1 p f ðxÞ cos px dx: To this end, let e be any positive number. Choose k 2 N such that |sn(k) − f | < e/2p

8 Hints

483

Rp everywhere on [−p, p] and also n(k) > p. Then | p1 p [sn(k)(x) − f(x)]cospx dx| < e. R p But p1 p sn(k)(x)cospx dx = ap in view of the choice n(k) > p. This shows that R p |ap − p1 p f(x)cospx dx| < e. Problem Set 4.2 4.2.P1. Suppose that f is a 2p-periodic function that is Lebesgue integrable on [−p, p] and is differentiable at a point x0. Prove that lim sn ðx0 Þ ¼ f ðx0 Þ;

n!1

where sn denotes the partial sum of the Fourier series of f. Hint: Using Remark 4.2.4 and Proposition 4.2.3(c), we may write

1 s n ð x0 Þ  f ð x0 Þ ¼ p 1 ¼ p

Z

p

Z

p p

½f ðx0  tÞ  f ðx0 ÞDn ðtÞdt

 f ðx0  tÞ  f ðx0 Þ 12 t 1 sin n þ t dt: t 2 sin 12 t p

ðx0 Þ Since lim f ðx0 tÞf ¼ f 0 ðx0 Þ; it follows that, for e > 0, there exists a d > 0 such t t!0

that f ð x0  t Þ  f ð x0 Þ 0  f ðx0 Þ \e for 0\t\d t that is, f 0 ðx0 Þ  e\

f ð x0  t Þ  f ð x0 Þ \f 0 ðx0 Þ þ e t

for 0\t\d:

ð x0 Þ Thus by the assumption on f, the difference quotient f ðx0 tÞf is bounded near t zero and is Lebesgue integrable on [−p, p]. The factor (t/2)/(sin(t/2)) is bounded on [−p, p]. Therefore

f ðx0  tÞ  f ðx0 Þ 12 t t sin 12 t is Lebesgue integrable over [−p, p]. The required conclusion now follows from the Riemann Lebesgue Theorem 4.2.5.

484

8 Hints

4.2.P2. (Bessel’s Inequality) If f is any Riemann integrable periodic function, its Fourier coefficients satisfy the inequality Z 1   1 p 1 2 X 2 2 a þ a þ bn  f ðxÞ2 dx: 2 0 n¼1 n p p Note that it is part of the assertion that the series on the left is convergent. [It follows from this result that the Fourier coefficients an and bn of a Riemann integrable function tend to zero as n ! ∞.] Hint: For n
1, let sn(x) be the partial sum of the Fourier series of f. By using Lemma 4.1.2 and Definition 4.1.4, it follows that

1 p

Z

Z n  X  1 p 1 sn ðxÞ2 dx ¼ a20 þ a2k þ b2k ¼ f ðxÞsn ðxÞdx: 2 p p p k¼1 p

Now,

1 p

Z

Z Z Z 1 p 2 p 1 p 2 ½f ðxÞ  sn ðxÞ dx ¼ f ðxÞ dx  f ðxÞsn ðxÞdx þ sn ðxÞ2 dx p p p p p p p Z n   1 p 1 2 X 2 ¼ f ðxÞ dx  a0  a2k þ b2k : p p 2 k¼1 p

2

Since [f(x) − sn(x)]2
0 and the integral of a nonnegative function is nonnegative, the required inequality follows. Moreover, the square of a Riemann integrable function must be Riemann integrable, and therefore it also follows that the series involved is convergent. 1 P cos pffiffinx is uniformly convergent on every closed subinterval [a, 4.2.P3. The series n n¼1

2p − a], 0 < a < p of (0, 2p). Show that it is not the Fourier series of a Riemann integrable function. Hint: The uniform convergence is proved as in Example 4.2.14(b) using Dirichlet’s Test [28, Theorem 4.4.2] and the summation formula      12 sin n  12 x  sin 2x x coskx ¼ for sin 6¼ 0: x 2 sin 2 k¼1

n1 X

Since the series

1 P 1 n¼1

n

is divergent, it follows from 4.2.P2 that the given series is not

a Fourier series of a Riemann integrable function.

8 Hints

485

4.2.P4. Is the series

1 P n¼1

sin nx lnðn þ 1Þ

the Fourier series of a continuous function?

Hint: No. If it were the Fourier series of any Riemann integrable function, then by 1 P 1 4.2.P2 the series would be convergent, which it is not, as can be seen ðlnðn þ 1ÞÞ2 n¼1 P1 . by comparison with n n 1   R P sin 2 p ðn þ 2 t Þ 1 4 1 4.2.P5. Show that p 0 2 sin1 t dt
p42 k
p2 n þ ln n : 2

k¼1

Hint: Since |sinx|  |x| for all x, we get

2 p

Z

p 0

  Z

Z 1 sin n þ 1 t 2 p 1 1 2 ðn þ 2Þp 1 2 dt
t dt ¼ sin n þ j sin tj dt 2 sin 12 t p 0 2 t p 0 t Z n n kp 2X 1 4 X1 :
j sin tjdt ¼ 2 p k¼1 kp ðk1Þp p k¼1 k

It may be noted that since

n P 1 k¼1

2 p

Z 0



4.2.P6. Show that

p

1 k
n

þ

Rn 1 1 1 x dx ¼ n þ ln n; it follows that



 sin n þ 1 t 4 1 2 dt
þ ln n : 2 sin 12 t p2 n

1 R 2 p sinðn þ 2 tÞ p 0 2 sin12 t dt 

ln n þ

1 np

þ ln p þ p2 :

Hint: Split the integral as I + J, where

2 I¼ p

 Z 1  n sin n þ 1 t 2 dt 1 0 2 sin 2 t

2 and J ¼ p

Z

p 1 n

 sin n þ 1 t 2 dt: 2 sin 12 t

n sinðn þ 12Þ t 1   P 1 Since 2 sin1 t ¼ 2 þ coskt  n þ 12 ; we have I  p2 1 þ 2n : Moreover, 2 k¼1 R p since sint
2t/p when 0  t  p/2, we also have J  p2 p2 1=n 1t dt ¼ ln n þ ln p: Now add these two inequalities. 4.2.P7. Let the Fourier series corresponding to a continuous periodic function 1 P f(x) be given by 12 a0 þ (ancosnx + bnsinnx). Show that the series obtained by n¼1 Rx integrating this Fourier series term by term converges to 0 f(t)dt. [The remarkable thing about this result is that the Fourier series of f is not assumed to converge to it. Note however that the integrated series is not a trigonometric series.]

486

8 Hints

 Rx  Hint: Consider gðxÞ ¼ 0 f ðtÞ  12 a0 dt: By the Fundamental Theorem of Calculus, the function g is differentiable at each x and therefore satisfies a Lipschitz condition of order 1. By Theorem 4.2.13, it has a Fourier series which converges to it uniformly. Denote its Fourier coefficients by ck, dk. We evaluate them by integrating by parts and using the obvious fact that g(0) = 0 = g(2p):   Z p 1 1 1 p 0 gðtÞ sin ktjp  ck ¼ g ðtÞ sin kt dt p k k p  Z p Z Z 1 1 1 p 1 1 p ¼ f ðtÞ  a0 sin kt dt ¼  f ðtÞ sin kt dt þ a0 sin kt dt kp p 2 kp p 2 kp p 1 ¼  bk ; k and   Z p 1 1 1 p 0 dk ¼ gðtÞ cos ktjp þ g ðtÞ cos kt dt p k k p  Z p Z Z 1 1 1 p 1 1 p ¼ f ðtÞ  a0 cos kt dt ¼ f ðtÞ cos kt dt  a0 cos kt dt kp p 2 kp p 2 kp p 1 ¼ ak ; k where we have used the fact that g(p) = g(−p). Using these values of the Fourier coefficients of g and the fact that the series converges uniformly to g, we get 1 X 1 ak sin kx  bk cos kx : gðxÞ ¼ c0 þ 2 k k¼1

Put x = 0 to obtain 0 = gð0Þ ¼ 12 c0 

gðxÞ ¼

n P 1 k¼1

k bk

or 12 c0 ¼

n P 1 k bk . From this we get

k¼1

1 X ak sin kx þ bk ð1  cos kxÞ k¼1

k

:

8 Hints

487

Together with the definition of g, this implies Z

x

0

1 X 1 ak sin kx þ bk ð1  cos kxÞ : f ðtÞdt ¼ a0 x þ 2 k k¼1

This is exactly what we would have obtained if we had integrated the Fourier series of f term by term. 4.2.P8. If f is continuously differentiable on [a, b], use integration by parts (but not the Riemann–Lebesgue Theorem 4.2.5) to show that Z lim

k!1

b

Z f ðtÞ sin kt dt ¼ 0 ¼ lim

k!1

a

b

f ðtÞ cos kt dt:

a

Hint: Integrating by parts, we get Z

b

lim

k!1

a

 b Z b cos k t cos k t 0 f ðtÞ dt: f ðtÞ sin kt dt ¼ f ðtÞ  þ k k a a

Now, f ðaÞ cos ka  f ðbÞ cos kb 2M1  ; k k

where M1
supfj f ðxÞj : x 2 ½a; bg

and Z

b a

M2 cos kt 0 f ðtÞdt  ; k k

Z where M2


b

jf 0 ðtÞj dt:

a

Rb

Hence | a f(t)sinkt dt |  (2M1 + M2)/k, which tends to 0 as k ! ∞. The argument for the other integral is similar. 4.2.P9. Let f be continuous with period 2p and Fourier coefficients ak, bk. Show that n X k¼0

1

ðjak j þ jbk jÞ  2ð2n þ 1Þ2 supfj f ðxÞj : x 2 ½p; pg:

488

8 Hints

Hint: By the Cauchy–Schwarz Inequality,

n X

"

#1 n  X  2 2 2 ðjak j þ jbk jÞ  ð2n þ 1Þ ak þ bk : 1 2

k¼0

k¼0

In view of Bessel’s Inequality [see Problem 4.2.P2] Z n  X  1 2 1 p 2 2 a k þ bk  a0 þ f ðxÞ2 dx 2 p p k¼0

Z 2 Z 1 1 p 1 p  f ðxÞdx þ f ðxÞ2 dx 2 p p p p

2 1 1 2pðsupfj f ðxÞj : x 2 ½p; pgÞ  2 p 1 þ 2pðsupfj f ðxÞj : x 2 ½p; pgÞ2 p  2ðsupfj f ðxÞj : x 2 ½p; pgÞ2 þ 2ðsupfj f ðxÞj : x 2 ½p; pgÞ2  4ðsupfj f ðxÞj : x 2 ½p; pgÞ2 : Therefore n X

1

ðjak j þ jbk jÞ  2ð2n þ 1Þ2 supfj f ðxÞj : x 2 ½p; pg:

k¼0

R1 4.2.P10. [Cf. Problem 3.2.P24] Show that the improper integral 0 sinx x dx converges and evaluate it by applying Dirichlet’s Theorem 4.2.10. It is understood that sin x x is to be replaced by 1 when x = 0. R1 RA Hint: Since sinx x is continuous on [0, 1], 0 sinx x dx exists. Now, 1 sinx x dx ¼ RA x R 1 cos x cos 1  cosA A  1 cos x2 dx and Rthe improper integral 1 x2 dx converges, as can be 1 1 seen by comparing it with 1 x2 dx: To evaluate, consider the periodic function f:R!R given on the interval [−p, p] by f ðxÞ ¼ sinðx=2Þ with the understanding that x f ð0Þ ¼ 12 : Then f is continuous everywhere and satisfies the hypotheses of Dirichlet’s Theorem 4.2.10. Therefore

8 Hints

489

  Z 1 p sin n þ 12 u sin 12 u du f ð0  uÞDn ðuÞdu ¼ lim  n!1 p p u 2 sin 12 u p   Z 1 p sin n þ 12 u ¼ lim du n!1 p p 2u Z Z 1 1 ðn þ 2pÞ sin x 1 1 sin x dx ¼ dx ¼ lim n!1 2p  n þ 1p p 0 x ð 2Þ x

1 1 ¼ lim 2 n!1 p

Z

p

since this integral converges. Problem Set 4.3 4.3.P1. If f: [a, b]!R is continuous and e > 0, then prove by using Fejér’s Theorem 4.3.4 that there exists a polynomial P(x) such that supfjf ðxÞ  PðxÞj : x 2 ½a; bg\e: [This is known as the Weierstrass Polynomial Approximation Theorem.] Hint: Step 1. Suppose f is a continuous even function on R which is of period 2p. By Fejér’s Theorem 4.3.4, for e > 0, there must exist an integer, n + 1 say, such that supfjf ðxÞ  rn þ 1 ðxÞj : x 2 ½p; pg\e: where rn þ 1 ðxÞ ¼

s0 ðxÞ þ s1 ðxÞ þ    þ sn ðxÞ ; nþ1

sk being the partial sums of the Fourier series of f. Observe that, in view of the evenness of f, the Fourier coefficients bn are all zero and therefore 1 sk ðxÞ ¼ a0 þ a1 cos x þ    þ ak cos kx: 2 Step 2. There exists a polynomial Tm such that cos mx ¼ Tm ðcos xÞ for x 2 ½p; p:

490

8 Hints

[This is of independent interest apart from being a step in the present proof.] The result is true for m = 0 (take T0(t) = 1) and for m = 1(take T1(t) = t). Suppose it is true for all n < m, where m
2. Then cos mx ¼ cos mx þ cosðm  2Þx  cosðm  2Þx ¼ cosððm  1Þx þ xÞ þ cosððm  1Þx  xÞ  cosðm  2Þx ¼ 2 cosðm  1Þx cos x  cosðm  2Þx ¼ Tm ðcos xÞ; where Tm(t) = 2t•Tm–1(t) − Tm−2(t). The results of these two steps may be summarised as below. Let f be a continuous even function defined on R, having period 2p. Then, for e > 0, there exists an integer n such that n X supf f ðxÞ  ck Tk ðcos xÞ : x 2 ½p; pg\e: k¼0 Next we prove the existence of the polynomial P when the domain of f is [0, 1]. Let f: [0, 1]!R be a continuous function and let e > 0 be given. Define g:R!R by gðxÞ ¼ f ðjcos xjÞ: Then g is continuous on R, has period 2p and g(−x) = g(x). It follows from what has been proved above that there exist polynomials T0, …, Tn and constants c0, …, cn such that n X supf gðxÞ  ck Tk ðcos xÞ : x 2 ½p; pg\e: k¼0 Hence n X supf f ðxÞ  ck Tk ðxÞ : x 2 ½0; 1g k¼0 n X  supf f ðjcos xjÞ  ck Tk ðjcos xjÞ : x 2 ½p; pg\e: k¼0 Writing PðxÞ ¼

n P k¼0

ck Tk ðxÞ, we obtain the required polynomial.

8 Hints

491

Finally we prove the existence of P when the domain is any closed bounded interval [a, b]. Consider f(a + y(b − a)). This function of y is defined and continuous on [0, 1]. Hence it follows from what has been proved so far that there exists a polynomial Q(y) such that supfjf ða þ yðb  aÞÞ  QðyÞj : y 2 ½0; 1g\e: xa For x 2 [a, b], we have y ¼ ba 2 [0, 1] and x = a + y(b − a). Thus

 x  a supf f ðxÞ  Q : x 2 ½a; bg\e: ba xa has the required property. Therefore PðxÞ ¼ Q ba 4.3.P2. Use the Weierstrass Polynomial Approximation Theorem to prove that if f and g are continuous functions on [a, b] such that Z

b

Z xn f ðxÞdx ¼

a

b

xn gðxÞdx

for all n
0;

a

then f = g. Rb Hint: We need only show that if a xnf(x)dx = 0 for all n
0, then f(x) = 0 Rb everywhere on [a, b]. If f satisfies this hypothesis, then a f(x)P(x)dx = 0 for any polynomial function P. Let M be an upper bound for | f | on [a, b]. For any e > 0, there exists [by the Weierstrass Polynomial Approximation Theorem] a polynomial function P such that | f − P| < e/M on [a, b]. Then Z 0 a

b

Z f ðxÞ dx  2

a

b

Z f ðxÞðf ðxÞ  PðxÞÞdx þ

a

b

f ðxÞPðxÞdx

 Mðe=MÞ þ 0 ¼ e: Since e > 0 is arbitrary, it follows that

Rb

2 a f(x) dx

= 0. It is now sufficient to show Rb that if f is a continuous nonnegative-valued function on [a, b] with a f = 0, then f(x) = 0 everywhere on [a, b]. If not, then f(x0) > 0, where x0 2 [a, b]. As f is continuous, there exists a d > 0 for which | f(x) − f(x0)| < 12 f(x0) whenever |x − x0| < d. Then f(x) > 12 f(x0) whenever |x − x0| < d. Consequently,

492

8 Hints

Z

b

Z

x0 d

f ðxÞdx ¼

a

a

Z




Z f ðxÞdx þ

x0 þ d x0 d

[

x0 þ d

x0 d

Z f ðxÞdx þ

b

x0 þ d

f ðxÞdx

f ðxÞdx by nonnegativity of f

1 f ðx0 Þ2d ¼ f ðx0 Þd [ 0; a contradiction: 2

4.3.P3. If an and bn are the Fourier coefficients of f, show that the Fejér sums rn are  n1

X 1 k 1  ðak cos kx þ bk sin kxÞ: rn ðxÞ ¼ a0 þ 2 n k¼1 Use this fact to show that 1 p

Z

 n1

X  1 k 2 2 rn ðxÞ2 dx ¼ a20 þ 1 ak þ b2k : 2 n p k¼1 p

Deduce Parseval’s Theorem that if f has period 2p and is continuous with Fourier coefficients an, bn, then 1 p

Z

1  X  1 f ðxÞ2 dx ¼ a20 þ a2k þ b2k : 2 p k¼1

Hint: Since sp ðxÞ ¼ 12 a0 þ rn ðxÞ ¼

p

p P

ðak cos kx þ bk sin kxÞ, we have

k¼1

n1 1X 1 n1 n2 sp ðxÞ ¼ a0 þ ða1 cos x þ b1 sin xÞ þ ða2 cos 2x þ b2 sin 2xÞ n p¼0 2 n n

nk 1 ðak cos kx þ bk sin kxÞ þ    þ ðan1 cosðn  1Þx þ bn1 sinðn  1ÞxÞ n n  n1

X 1 k 1  ðak cos kx þ bk sin kxÞ: ¼ a0 þ 2 n k¼1 þ  þ

8 Hints

Now nP 1 

1

k¼1

493

R 1 p p p



2

rn ðxÞ dx ¼

 k 2 2 n (ak

+

R 1 p 1 p p 2 a0

nP 1  k¼1

2  1  ðak cos kx þ bk sinÞkx dx ¼ 12 a20 þ k n

by Lemma 4.1.2. Since rn! f uniformly on [−p, p] by

b2k )

Fejér’s Theorem 4.3.4, it follows that 1 p

Z

p p

Z 1 p rn ðxÞ2 dx n!1 n!1 p p p " #  n1

 1 2 X k  2 2 ¼ lim a0 þ 1 ak þ bk n!1 2 n k¼1

f ðxÞ2 dx ¼



1 p

Z

p

lim rn ðxÞ2 dx ¼ lim

1   1 2 X a0 þ a2k þ b2k : 2 k¼1

Now apply Bessel’s Inequality (Problem 4.2.P2). 4.3.P4. Two different periodic functions can have the same Fourier series (e.g. if they differ only on a finite set of points). Can this happen if the functions are continuous? Hint: No, in view of Corollary 4.3.6. 4.3.P5. If f and g are continuous functions of period 2p with Fourier coefficients an, bn and an, bn respectively, show that 1 p

Z

1 X 1 f ðxÞgðxÞdx ¼ a0 a0 þ ðan an þ bn bn Þ: 2 p n¼1 p

Hint: Observe that the Fourier coefficients of the function f − g are  a0 Þ; an  an and bn − bn, n = 1, 2, 3, … . Applying Parseval’s Theorem (see Problem 4.3.P3) to the function f − g and also to f and g, we obtain

1 2 ð a0

1 p

Z

1 1 X X 1 ðf  gÞ2 ¼ ða0  a0 Þ2 þ ð a n  an Þ 2 þ ðbn  bn Þ2 ; 2 p n¼1 n¼1 p

from which the desired equality follows. 4.3.P6. Suppose 0 < d < p and f(x) = 1 if |x|  d, f(x) = 0 if d < |x| < p, and f(x + 2p) = f(x) for all x. (a) Compute the Fourier coefficients of f. 1 P sin nd pd (b) Conclude that n ¼ 2 when 0 < d < p. n¼1

494

8 Hints

(c) Deduce from Parseval’s Theorem (see Problem 4.3.P3) that

1 P sin2 nd n¼1

n2 d

¼ pd 2

when 0 < d < p. R 1  2 (d) Let d ! 0 and prove that 0 sinx x dx ¼ p2 : Rd R 1 d 2 Hint: (a) a0 ¼ p1 d dx ¼ 2d p ;an ¼ p d cosnx dx ¼ np sin nd and every bn is 0. (b) Since the function is continuous at 0 and is monotone on (−d, 0) and on (0, d), the Fourier series converges to f(0) = 1 at 0 by Dirichlet’s Theorem 4.2.10. So, 1 1 1 P P P 2 pd sin nd 1 ¼ 12 a0 þ an ¼ pd þ np sin nd and hence 2 ¼ n . n¼1

n¼1

(c) By Parseval’s Theorem, Therefore

d2 p2

þ

2 p2

1 P sin2 nd n¼1

n2

1 2 2 a0

þ

1 P n¼1

¼ pd ; so that

a2n

d p

¼

þ

2 p

1 4d2 2 p2

n¼1 1 P

þ

n¼1

1 P sin2 nd n¼1

n2 d

4 sin2 nd n 2 p2

¼ p1

Rd

d

dx ¼ 2d p :

¼ 1; which implies for

0 < d < p that 1 1 pd X sin2 nd X sin2 nd ¼ ¼ d: 2 2 2 n2 d n¼1 n¼1 n d

ð8:29Þ

R 1  2 (d) Now let e > 0 be given. Since the improper integral 0 sinx x dx is convergent (by comparison (see Proposition 1.7.6) with 1/x2) there exists an X such that Z

 Z X

1 sin x2 sin x 2 dx  dx \e 0 x x 0 and X[

1 : e

ð8:30Þ

ð8:31Þ

R X  2 Approximating the Riemann integral 0 sinx x dx by Riemann sums, we find that there exists a positive integer m′ such that d = X/m′ satisfies d\2e

ð8:32Þ

8 Hints

495

and Z

X sin x2 m0 X sin2 nd dx  d \e: 2 2 0 x n¼1 n d

ð8:33Þ

In view of (8.29), there exists an m > m′ such that p  d X m sin2 nd  d \e; 2 2 2 n¼1 n d i.e. p  d X m0 m X sin2 nd sin2 nd  d d \e: 2 2 2 2 2 n¼1 n d n¼m0 þ 1 n d Now,

m P n¼m0

Moreover,

sin2 nd d n2 d2

þ1 R 1 1 1 d m0 x2 dx

¼

1 P n¼m0

þ1 1 1 ¼ 0 md X

R 1 1 1 1 n2 d  d m0 x2 dx;

because

1 n2



1 x2

ð8:34Þ

on [n − 1, n].

because d = X/m′. It follows by (8.31) that 1 X sin2 nd d\e: 2 2 n¼m0 þ 1 n d

Taken with (8.34), (8.33) and (8.30), this implies p  d Z 1 sin x2  dx \4e: 2 x 0 R 1  2 Using (8.32), we further obtain p2  0 sinx x dx \5e: 4.3.P7. Show that

1 P cos nx n¼1

n3=2

is the Fourier series of a continuous function and con-

verges to it uniformly. 1 nx P 1 1  3=2 Hint: Since cos for all x and is convergent, it follows by the n3=2 n n3=2 n¼1

Weierstrass M-test that the given series is uniformly convergent. The conclusion now follows from Proposition 4.1.3 and Corollary 4.3.6. Problem Set 4.4 4.4.P1. Find the Fourier cosine series for the function given on [0, p] by 1 P 1 sinx. Deduce from the series that 12 ¼ 4m2 1. n¼1

496

8 Hints

Hint: The function f(x) = sinx may be extended to an even function on [−p, p] by setting f(x) = |sinx| and then extended to a periodic function on the whole of R. The (extended) function satisfies a Lipschitz condition on [−p, p] with M = 1 and a = 1. In fact by the Mean Value Theorem, jj sin xj  j sin yjj  j sin x  sin yj  j cos njjx  yj  jx  yj; where n lies between x and y. Thus its Fourier series converges uniformly on [−p, p] by Theorem 4.2.13. We proceed to obtain the series. 2 a0 ¼ p

Z

p

sinx dx ¼

0

4 p

and Z Z 2 p 1 p sinx cos nx dx ¼ ½sinðn þ 1Þx  sinðn  1Þxdx p 0 p 0   1 cosðn þ 1Þx cosðn  1Þx p 2  þ ¼ ½ð1Þn þ 1 for n 6¼ 1; ¼ 2 p nþ1 n1 p n ð  1 Þ 0

an ¼

While a1 ¼ p2

Rp 0

sinx cos x dx ¼ 0: So,

j sin xj ¼

1 2 X 4 cos 2nx  : p n¼1 pð4n2  1Þ

Set x = 0 to obtain the required equality. 4.4.P2. Find the Fourier series of the function 8 0 be given. Deduce from Bessel’s Inequality that the number of positive integers k such that sinkx > d for all x 2 E is finite. R Hint: For any k of the kind in question, ½p;p vE sin kx dmðxÞ
mðEÞd, i.e. the Fourier coefficient bk of vE exceeds m(E)d > 0. If this happens for infinitely many k, then the series in Bessel’s InequalityR diverges, contrary to the assertion of the inequality that its sum cannot exceed ½p;p v2E ¼ mðEÞ\1. Problem Set 5.2 5.2.P1. Give an example of a monotone function which is discontinuous at each rational number in [0, 1] (see Example 5.2.7). Hint: Let {rk}k
1 be an enumeration of the rational numbers in [0, 1]. Define f on [0, 1] by the rule X 2n f ðxÞ ¼ rn  x

where the summation is extended over all indices n for which rn  x. Clearly, the function f is increasing and, for each rational number ri, f(ri) − f ðri Þ = 2−i. It may be noted that f is right continuous at each rational number ri. Moreover, it is continuous at each irrational number. 5.2.P2. Let f be defined on an open interval (a, b) and assume that for each point x of the interval, there exists an open interval Nx of x in which f is increasing. Prove that f is increasing throughout (a, b). Hint: Let x < y. Consider the closed interval [x, y]. By hypothesis, there is an open interval about each of the points in [x, y] in which f is increasing. By the Heine– Borel Theorem, there exists a finite cover fNxi gm i¼1 of intervals which covers [x, y] and in each of which f is increasing. Without loss of generality, we may assume x 2 Nx1 . If y 2 Nx1 , then f(x)  f(y) because f is increasing on Nx1 . If y does not belong to Nx1 , then there exists an interval amongst fNxi gm i¼2 which has a nonempty intersection with Nx1 . Call it Nx2 and let z1 2 Nx1 \ Nx2 . Clearly, f is increasing in Nx1 [ Nx2 . If y 62 Nx2 , then continue the process till [ ki¼1 Nxi contains y, which must happen at some stage because there are only finitely many intervals and they cover [x, y]. Obviously, the function is increasing on [ ki¼1 Nxi . Consequently, f(x)  f(y).

500

8 Hints

5.2.P3. Let f be continuous on a compact interval [a, b] and assume that f does not have a local minimum or local maximum at any interior point. Prove that f must be monotonic on [a, b]. Hint: Since f is continuous on the compact interval [a, b], it assumes its minimum as well as maximum. It cannot have a minimum or maximum at an interior point, for that point will also be a local minimum or local maximum. If the function is constant, there is nothing to prove. Assume that the minimum is attained at a, in which case the maximum is attained at b. We shall show that f is increasing on [a, b]. Suppose not. Then there exist x, y 2 [a, b] such that x < y but f(x) > f(y). Set a ¼ f ðxÞ þ2 f ðyÞ : Clearly, f(a)  f(y) < a < f(x). By the Intermediate Value Theorem 1.3.26, the value a is attained between a and x and again between x and y, say, at u and v respectively. Between u and v, it assumes its maximum, which cannot be at u or v, because f(x) > a = f(u) = f(v). Thus f has a maximum between u and v (by the Extreme Value Theorem 1.3.20). Hence f has a local maximum, which contradicts the hypothesis. A similar argument shows that, when a is a maximum and b a minimum, the function is decreasing. 5.2.P4. A function f increases on a closed interval [a, b] and it is true that, if f(a) < k < f(b), there exists an x 2 [a, b] such f(x) = k. Prove that f is continuous on [a, b]. Does the same conclusion hold if the hypothesis that f increases is dropped but it is still required that f(a) < f(b)? A function f increases on an open interval I and, for each a, b 2 I, it is true that, if f(a) < k < f(b), then there exists an x 2 I such f(x) = k. Prove that f is continuous on I. Does the same conclusion hold if the hypothesis that f increases is dropped? Hint: In either case, the hypothesis implies that the range is an interval. Therefore continuity follows directly from Remark 5.2.3. In the first case, the same conclusion does not hold if the hypothesis that f increases is dropped. Consider the function (see Fig. 8.2)

f ðxÞ ¼

x x1

0  x\1 1  x  2:

In the second case, just drop the endpoints.

1 Fig. 8.2 Function in Problem 5.2.P4

2

8 Hints

501

Problem Set 5.3 5.3.P1. If the function f assumes its local maximum at an interior point c of its domain, then show that D+f(c)  0 and D−f(c)
0. Hint: If D+f(c) > 0, then there exist arbitrarily small positive values of h such that f ðc þ hÞf ðcÞ [ 0; which implies f(c + h) − f(c) > 0 for some c + h in every neighh bourhood of c, contradicting the fact that f assumes a local maximum at c. If D−f(c) < 0, then there exist arbitrarily small positive values of h such that f ðchÞf ðcÞ \0; which implies f(c − h) − f(c) > 0 for some c − h in every neighh bourhood of c, contradicting the fact that f assumes its local maximum at c. 5.3.P2. Suppose f is continuous on [a, b] and D+f(x) > 0 for all x in [a, b). Show that f(b)
f(a). Also, give a counterexample to show that the continuity hypothesis cannot be dropped. Hint: Let E = {x 2 [a, b]: f(x)
f(a)}. The set E is a closed subset of [a, b] because f is continuous on [a, b]. If c = supE, then c 2 E since E is closed. Therefore f(c)
ðcÞ f(a). We need only show that c = b. If c < b, then D+f(c) > 0 implies f ðxÞf [0 xc for some x arbitrarily close to c and on the right of it. Thus f(x) > f(c)
f(a) for some x > c. This is a contradiction since c = sup{x 2 [a, b]: f(x)
f(a)}. Hence c = b. The function f on [0, 2] given by f(x) = x on [0, 1) and f(x) = x − 3 on [1, 2] is discontinuous at 1 and satisfies D+f(c) = 1 everywhere, but f(2) = −1 < 0 = f(0). 5.3.P3. Show that f may be discontinuous at x0 when all four Dini derivatives are equal to ∞. Hint: Let 0 < k < 1 and consider the function 8 < 1 x [ x0 f ðxÞ ¼ 0 x\x0 : k x ¼ x0 : Then D þ f ðx0 Þ ¼ lim sup

f ð x0 þ h Þ  f ð x0 Þ 1k ¼ lim sup ¼ 1; h h h!0 þ

D þ f ðx0 Þ ¼ lim inf

f ð x0 þ h Þ  f ð x0 Þ 1k ¼ lim inf ¼ 1; h!0 þ h h

h!0 þ

h!0 þ

D f ðx0 Þ ¼ lim sup h!0

f ð x 0 þ hÞ  f ð x 0 Þ k ¼ lim sup ¼ 1; h h!0 h

502

8 Hints

D f ðx0 Þ ¼ lim inf h!0

f ð x 0 þ hÞ  f ð x 0 Þ k ¼ lim inf ¼ 1: h!0 h h

Thus all four Dini derivatives of f at x0 are equal to ∞ and the function is discontinuous at x0. 5.3.P4. Let f: [0, ∞)!R be differentiable and suppose that f(0) = 0 and that f′ is increasing. Prove that gðxÞ ¼

f ðxÞ x

f 0 ð0Þ

x[0 x¼0

defines an increasing function of x. Hint: Let 0  a < b. Applying the Lagrange Mean Value Theorem to g on [a, b], we obtain some c 2 (a, b) such that   gðbÞ  gðaÞ 1 0 f ðcÞ 0 ¼ g ðcÞ ¼ f ðcÞ  : ba c c Applying the Lagrange Mean Value Theorem to f on [0, c], we obtain some d 2 (0, c) such that f ðcÞ  f ð0Þ ¼ f 0 ðdÞ: c Therefore   gðbÞ  gðaÞ 1 0 f ðcÞ 1 ¼ f ðcÞ  ¼ ½f 0 ðcÞ  f 0 ðdÞ
0: ba c c c 5.3.P5. (a) Show that if f′(x) exists, then D+(f + g)(x) = f′(x) + D+g(x) and similarly for other Dini derivatives. (b) Give an example when D+(f + g)(x) 6¼ D+f(x) + D+g(x). Hint: (a) Let η > 0 be given. There exists a d′ such that for all h satisfying 0 < h < d′, we have

8 Hints

503

f ðx þ hÞ  f ðxÞ [ f 0 ðxÞ  g h and f ðx þ hÞ  f ðxÞ gðx þ hÞ  gðxÞ þ \D þ ðf þ gÞðxÞ þ g: h h Now, gðx þ hÞ  gðxÞ ðf þ gÞðx þ hÞ  ðf þ gÞðxÞ f ðx þ hÞ  f ðxÞ ¼  h h h \D þ ðf þ gÞðxÞ þ g  ðf 0 ðxÞ  gÞ ¼ D þ ðf þ gÞðxÞ  f 0 ðxÞ þ 2g: Therefore

D þ gðxÞ ¼ inf d [ 0 sup0\h\d

gðx þ hÞ  gðxÞ h

 D þ ðf þ gÞðxÞ

f 0 ðxÞ þ 2g. Since η is arbitrarily chosen, it follows that D+g(x)  D+(f + g)(x) − f′(x), which is equivalent to D+(f + g)(x)
D+g(x) + f′(x). The reverse inequality is an immediate consequence of the definition of the lim sup. (b) Let 8 < 0 f ðxÞ ¼ x : x

x¼0 x2Q x 62 Q

and let g(x) = −f(x). Then D+(f + g)(0) = 0. However, D+f(0) = D+g(0) = 1. In fact, for h > 0, f ðhÞ  f ð0Þ f ðhÞ ¼ ¼ h h

1 x2Q 1 x 62 Q:

ð0Þ Consequently, lim sup f ðhÞf ¼ 1: Similarly D+g(0) = 1. h h!0 þ

ðaÞ + 5.3.P6. (a) Let f be continuous on [a, b]. If f ðbÞf ba \C; then D f(x)  C for uncountably many x 2 [a, b]. ðaÞ (b) Let f be continuous on [a, b]. If f ðbÞf [ C; then D−f(x)
C for uncountably ba many x 2 [a, b]. Hint: (a) Let k = C(b − a). Then k > f(b) − f(a). Consider the function

504

8 Hints

gðxÞ ¼ f ðxÞ  f ðaÞ  k

xa : ba

Then g(a) = 0 and g(b) = f(b) − f(a) − k < 0. Let s be such that 0 = g(a) > s > g (b). Let E ¼ fx 2 ½a; b : gðxÞ ¼ sg; E is closed, being the inverse image of a closed subset of R under the continuous map g. Since E is also bounded, it has a largest point xs, say, with g(xs) = s. Since g(b) < s, we have xs < b. Since g(xs+ h) < s for all 0 < h < b − xs (Intermediate Value Theorem 1.3.26), we have D+g(xs)  0, whence by Problem 5.3.P5, D þ f ðxs Þ ¼ D þ gðxs Þ þ

k k  ¼ C: ba ba

Different s’s generate different xs’s and there are uncountably many s’s between 0 and g(b), i.e., there are uncountably many points x such that D+f(x)  C. (b) Similar. Remark It is a consequence that if f is continuous and one Dini derivative is nonnegative except perhaps at a countable or finite number of points, then f is increasing. In fact, if D+f(x)
0 for a  x  b except for a countable or finite number of points and f fails to be increasing, there must exist points x and y such ðxÞ + that y > x and f(y) < f(x), then f ðyÞf yx \C\0; which implies D f(z)  C < 0 for uncountably many points z between x and y, contradicting our hypothesis. 5.3.P7. Let f be a continuous function defined on [a, b]. Suppose that there exist real constants a, b such that a  D+f(x)  b for all x 2 (a, b). Prove that ha  f (x + h) − f(x)  hb provided a  x < x + h  b. Hint: Suppose D+f(x)
a. Let g(x) = f(x) − ax. Since D+g(x)
0 [see Problem 5.3.P5], the function g is increasing [see Remark above]. If h > 0, we therefore have g(x + h)
g(x), or in other words, f ðx þ hÞ  f ðxÞ  aðx þ hÞ þ ax
0; i.e. f ðx þ hÞ  f ðxÞ
a: h

ð8:35Þ

This proves the first of the two inequalities in question. A similar argument proves the other.

8 Hints

505

Remark From (8.35), it follows that D+ f(x)
a. Again, g is increasing. If h > 0, we therefore have g(x) − g(x − h)
0, f ðxÞ  ax  f ðx  hÞ þ aðx  hÞ
0;

f ðx  hÞ  f ðxÞ
a; h whence D f ðxÞ
D f ðxÞ
a: If f is continuous, all four Dini derivatives have the same upper and lower bounds in any interval. 5.3.P8. Let f be a continuous real-valued function defined on [a, b] and let x 2 (a, b) be such that D+f is finite in a neighbourhood of x and continuous at x. Prove that f′(x) exists. Hint: Since D+f is finite and continuous at x, its upper and lower bounds are arbitrarily close to D+f(x) in a sufficiently small neighbourhood of x. By the remark following Problem 5.3.P7, the same is true of the upper and lower bounds of the other three Dini derivatives. This means all the four Dini derivatives at x coincide with f′(x). 5.3.P9. If one of the Dini derivatives of a continuous function is zero everywhere in an interval, the function is constant there. Hint: D þ f ¼ 0 , D þ f
0 and D þ f  0: From the continuity of f and the fact that D+f
0, we conclude that f is increasing: Similarly from the continuity of f and the fact that D+f  0, we conclude that f is decreasing: Since f has been shown to be increasing as well as decreasing, it follows that f is constant. 5.3.P10. Construct monotonic jump functions f on [0, 1) whose discontinuities have 0 as a limit point and such that f þ0 (0) is (a) zero (b) ∞ (c) positive and finite. Also, compute the quantum of jump at each of the discontinuities. Hint: (a) Define a function on the domain [0,1) as

506

8 Hints

f ðxÞ ¼

3n 0

2n  x\2n þ 1 ; n ¼ 1; 2; . . . x ¼ 0:

Let x1 < x2. Then there exist m and n, m
n, such that x1 2 [2−m, 2−m+1) and x2 2 [2−n, 2−n+1). So, f(x1) = 3−m and f(x2) = 3−n, i.e. f is increasing. At each of the points 2−n, we have f(2−n+) − f(2−n) = 3−n − 3−n = 0 and f(2−n) − f(2−n−) = 3−n − 3−n−1 > 0, i.e., at the point x = 2−n, f is right continuous and has a jump of magnitude 3n  3n1 , which tends to 0 as n ! ∞. Moreover, f þ0 ð0Þ ¼ 0: In fact, if h > 0, we have lim

h!0

f ð0 þ hÞ  f ð0Þ f ðhÞ ¼ lim ¼ 0: h!0 h h

Indeed, for 2−m−1  h < 2−m



m þ 1 1 2 m f ðhÞ 2   \ : 3 3 h 3 (b) Define a function on the domain [0, 1) as

f ðxÞ ¼

2n 0

3n  x\3n þ 1 ; n ¼ 1; 2; . . . x ¼ 0:

As in (a) above, it is right continuous at 3−n, with a jump of magnitude 2−n − 2−n−1. 0 Moreover, f þ (0) = ∞. In fact, if h > 0, we have f ð0 þ hÞ  f ð0Þ f ðhÞ ¼ ; h h and if 3−m−1  h < 3−m, then

f ðhÞ h

[

3m 2m þ 1

; which tends to ∞ as m ! ∞.

(c) Define a function on the domain [0, 1) as f ðxÞ ¼ Then f has jumps of amount

1 n

1 n

0 1 nðn þ 1Þ

1  x\ n1 ; n ¼ 2; 3; . . . x¼0

at the points 1n : Also, for h > 0, we have

8 Hints

507

f ð0 þ hÞ  f ð0Þ f ðhÞ ¼ h h and if (m + 1)−1  h < m−1, then f ðhÞ ¼

1 ; mþ1

so that 1 f ðhÞ m\  1; mþ1 h which implies f ðhÞ ¼ 1: h!0 þ h lim

5.3.P11. If all Dini derivatives of a function f satisfy |Df(x)|  K everywhere on an interval, then the function satisfies the condition | f(x) − f(y)|  K|x − y|. Hint: Suppose not. Then we have | f(a)−f(b)| > K′|a − b| for some a and b in the interval and some K′ > K. We may suppose a < b. Then the inequality | f(a) − f(b)| > K′(b − a) holds over the interval (a, b). For any c 2 (a, b), the same inequality holds either over the interval (a, c) or over (c, b), as can be deduced by applying the triangle inequality. By the usual bisection argument, we get a nested sequence of intervals (an, bn), such that | f(an) − f(bn)| > K′(bn − an) for each n and 0 < (bn − an) ! 0. Let n be the unique common point of all these intervals. Then either an < n for all n or bn > n for all n. In the former case, |D − f(n)|
K′ > K and in the latter case, |D+f(n)|
K′ > K. Problem Set 5.4 5.4.P1. Definition. Let E be a subset of R of finite outer measure. A family I of closed intervals, each of positive length, is called a Vitali cover of E if for each x 2 E and every e > 0, there exists an I 2 I such that x 2 I and ‘(I) < e., i.e., each point of E belongs to an arbitrarily short interval of I : Example. Let {rn}n
1 be an enumeration  of the rationals in [a, b]. Then the collection {In,i}, where In;i ¼ rn  1i ; rn þ 1i ; n; i 2 N; forms a Vitali cover of [a, b]. (Vitali Covering Theorem) Let E be a subset of R of finite outer measure and I be a Vitali cover of E. Then given any e > 0, there is a finite disjoint collection {I1, I2, …, IN} of intervals in I such that

508

8 Hints

m



En

N [

! Ii \e:

i¼1

Hint: Let O be an open subset of R of finite measure containing E. Let I 0 = {I 2 I : I O}. Obviously, I 0 is still a Vitali cover of E. Let h1 = sup{ ‘(I): I 2 I 0}. Choose an interval I 2 I 0 whose length is greater than 12 h1, i.e., ‘ðIn Þ [ 12 h1 . If E I1, the construction is complete. If not, suppose I1, I2, …, In 2 I 0 have been selected such that they are disjoint. If E [ nk¼1 Ik, the construction is complete. Otherwise, write An ¼

n [

Ik ; Un ¼ O \ Acn :

k¼1

Clearly, An is closed, Un is open and Un \ E is nonempty. By definition of a Vitali covering, there must exist an I 2 I 0 such that I Un. Let hn ¼ supf‘ðIÞ : I 2 I 0 ; I Un g: Choose In+1 2 I 0 such that In+1 Un and ‘(In+1) > 12hn. Thus we have a sequence {In} of disjoint intervals of I 0. Let A = [ n
1In. Since [ n
1In O; we have Rn‘(In)  m(OÞ < ∞. Hence we can find an integer N such that 1 X

e ‘ðIn Þ\ : 3 n¼N þ 1

ð8:36Þ

Also, lim hk  2  lim ‘ðIk þ 1 Þ ¼ 0. Let k!1

k!1

R ¼ En

N [

Ii :

i¼1

We need to show that m*(R) < e. Let x 2 R be arbitrary. Since [ Ni¼1 Ii is a closed set not containing x, we can find an interval I 2 I 0 which contains x and whose length is so small that I does not meet any of the intervals I1, I2, …, IN. The interval I must have one point in common with some interval Ik, k > N. For, if I were disjoint from IN+1, IN+2, …, then it would follow on noting that ‘(I)  hk, k = N + 1, N + 2, … (by definition of hk) and limk!1 hk ¼ 0 (proved above) that ‘(I) = 0, so that I would not be an interval in a Vitali cover. Let m be the smallest integer such that I meets Im. Then we have m > N and ‘(I)  hm−1 < 2 ‘(Im).

8 Hints

509

Im x

I Fig. 8.3 For Problem 5.4.P1

Since x is in I and has a point in common with Im, it follows that the distance of the midpoint of Im from x is at most ‘ðIÞ þ 12 ‘ðIm Þ  52 ‘ðIm Þ (see Fig. 8.3 above and produce a formal proof!). Thus x belongs to the interval Jm having the same midpoint as Im and five times its length. Thus we have shown R

1 [

Jm :

m¼N þ 1

Hence m ðRÞ 

P1

m¼N þ 1

‘ðJm Þ ¼ 5

P1

m¼N þ 1

‘ðIm Þ\e by (8.36).

5.4.P2. (Lebesgue’s Theorem) Let [a, b] be a closed interval in R and let f be a real-valued monotone function on [a, b]. Using the Vitali Covering Theorem, prove that f is differentiable almost everywhere. Hint: Without loss of generality, we may assume that f is increasing (otherwise consider −f). Let us show that the set where some two Dini derivatives are unequal is of measure zero. Let E ¼ fx 2 ½a; b : D þ f ðxÞ\D þ f ðxÞg: For every pair of positive rational numbers u and v, such that u < v, let Eu;v ¼ fx 2 E : D þ f ðxÞ\u\v\D þ f ðxÞg: Clearly, E = [ Eu,v, where u and v are positive rational numbers such that u < v. Since E is a countable union of the sets [ Eu,v, it suffices to show that m(Eu, v) = 0 for all 0 < u < v in Q. Suppose not. Then there exist positive rational numbers u and v such that m(Eu,v) = a > 0. For a given e > 0, there exists an open set O  Eu,v such that m(O) < a + e. For each x 2 Eu,v, there exists an arbitrarily small positive h such that [x, x + h]  [a, b) \ O and

510

8 Hints

f ðx þ hÞ  f ðxÞ\uh;

ð8:37Þ

using the fact that D+ f(x) < u. This way, we obtain a Vitali covering I of Eu,v. By the Vitali Covering Theorem 5.4.P1, there is a finite subcollection {I1, I2, …, IN} of disjoint closed intervals in I such that

m Eu;v n

N [

! Ii \e:

i¼1

Write Ii = [xi,xi + hi], i = 1, 2, …, N. Then the above inequality leads to " m Eu;v \

N [

#c ! ð x i ; x i þ hi Þ

\e:

ð8:38Þ

i¼1

Then summing (8.37) over the n disjoint intervals, we have N X

½f ðxi þ hi Þ  f ðxi Þ\

i¼1

N X

uhi \u  mðOÞ\u  ða þ eÞ:

i¼1

Let y be an arbitrary point of the set " E0 ¼ Eu;v \

N [

# ð x i ; x i þ hi Þ :

i¼1

Since D+f(y) > v, there is a small interval of the form [y, y + k] Ii for some i = 1, 2, …, N such that f ðy þ kÞ  f ðyÞ [ vk:

ð8:39Þ

This way, we obtain a Vitali covering I * of E0. Using the Vitali Covering Theorem 5.4.P1, we obtain a subcollection {J1, J2, …, JM} of disjoint intervals in I * such that m E0 n

M [ j¼1

! Jj \e;

8 Hints

511

i.e., " m Eu;v \

N [

#

M [

ð x i ; x i þ hi Þ n

i¼1

! Jj \e:

ð8:40Þ

j¼1

If we write Jj = [yj, yj + kj], j = 1, 2, …, M, then N [

m

! ð x i ; x i þ hi Þ

¼m

i¼1

M [

! Jj



j¼1

M X

kj :

j¼1

Consequently, " #c ! " #! N N [ [   a ¼ m Eu;v  m Eu;v \ ð x i ; x i þ hi Þ ðxi ; xi þ hi Þ þ m Eu;v \ M [

\e þ e þ m

!! Jj

i¼1

i¼1

by ð8:38Þand ð8:40Þ

j¼1

\e þ e þ

M X

!

kj ;

j¼1

which implies

M P

kj [ a  2e.

j¼1

Summing (8.39) over the intervals J1, J2, …, JM, we obtain M   M X X    f yj þ kj  f yj [ v kj [ v  ða  2eÞ: j¼1

j¼1

Since each Jj is contained in some Ii, we sum over those j for which Jj  Ii to obtain X     f yj þ kj  f yj  f ðxi þ hi Þ  f ðxi Þ; using the fact that f is an increasing function. Therefore N X i¼1

½ f ð x i þ hi Þ  f ð x i Þ 


M   X    f yj þ kj  f yj ; j¼1

512

8 Hints

which implies u  ða þ eÞ
v  ða  2eÞ: Since this is true for each e > 0, we must have u
v. But u < v. This contradiction implies a = 0. Hence m(E) = 0. We next show that the set F ¼ fx 2 ða; bÞ : f 0 ðxÞ ¼ 1g has measure zero. Let b > 0 be arbitrary. For each x 2 F, there exists an arbitrarily small positive h such that [x, x + h]  (a, b) and f(x + h) − f(x) > bh. Repeated application of the Vitali Covering Theorem 5.4.P1 followed by an application of Proposition 2.3.21(b) yields a countable disjoint family {[xn, xn + hn]} of these intervals such that

m Fn

1 [

! ½ x n ; x n þ hn 

¼ 0:

n¼1

Now, bmðFÞ  b

X

hn 

X

½f ðxn þ hn Þ  f ðxn Þ  f ðbÞ  f ðaÞ;

which implies m(F) = 0, since b > 0 is arbitrary. 5.4.P3. If three open intervals have a nonempty intersection, then at least one among them is contained in the union of the other two, or equivalently, the union of some two among them is the same as the union of all three. Hint: The union of any family of intervals with a nonempty intersection is an interval. This is easy to prove by means of the characterisation of an interval as being a set that contains any number that lies between two of its numbers. So, the union of any two of the three intervals is an open interval and the union of all the three intervals is also an open interval, which we shall call (a, b). Then a must be the left endpoint of one of them, which we shall name as A. If the other endpoint of A is b, then both the other intervals are contained in A and the required conclusion follows. So, suppose b is the right endpoint of some other interval, which we shall call B. Now, A [ B is an interval. Its left endpoint must be a and its right endpoint must be b. Thus A [ B = (a, b), which is also the union A [ B [ C of all three intervals. Therefore C A [ B [ C = A [ B. This shows that C is contained in the union of the other two.

8 Hints

513

5.4.P4. (a) Let l be a finite measure on a set X and A1, …, An be finitely many distinct measurable subsets of X, where n
3. If no three sets among the Ai have a nonempty intersection and each l(Ai) is finite, show that ! n n [ X  X  l Ai ¼ lðAi Þ  l Ai \ Aj : i¼1

ijn

i¼1

(b) Let l be a measure on a set X and A1, …, An be finitely many distinct measurable subsets of X, where n
3. If no three sets among the Ai have a nonempty intersection, show that n X

lðAi Þ  2l

i¼1

n [

! Ai :

i¼1

Hint: (a) We use induction. First consider three sets, i.e. n = 3. Since their intersection is empty, the sets A1 \ A3 and A2 \ A3 are disjoint. Therefore lððA1 [ A2 Þ \ A3 Þ ¼ lððA1 \ A3 Þ [ ðA2 \ A3 ÞÞ ¼ lðA1 \ A3 Þ þ lðA2 \ A3 Þ: Using the equality l(A [ B) = l(A) + l(B) − l(A \ B) twice and the above equality once, we obtain lðA1 [ A2 [ A3 Þ ¼ lðA1 [ A2 Þ þ lðA3 Þ  lððA1 [ A2 Þ \ A3 Þ ¼ lðA1 Þ þ lðA2 Þ  lðA1 \ A2 Þ þ lðA3 Þ  lðA1 \ A3 Þ  lðA2 \ A3 Þ ¼ lðA1 Þ þ lðA2 Þ þ lðA3 Þ  lðA1 \ A2 Þ  lðA1 \ A3 Þ  lðA2 \ A3 Þ: Now assume the equality for n − 1 sets A1, …, An−1 and consider n distinct sets A1, …, An such that no three have a nonempty intersection. The same must be true of A1, …, An−1, and moreover, the n−1 sets A1 \ An, A2 \ An, …, An−1 \ An must be disjoint. So, l

n[ 1 i¼1

! Ai

! \ An

¼l

n[ 1 i¼1

! ðAi \ An Þ

¼

n1 X

lðAi \ An Þ:

i¼1

Using the equality l(A [ B) = l(A) + l(B) − l(A \ B) once, then the induction hypothesis and finally the above equality, we obtain

514

8 Hints

l

n [

! Ai

n[ 1

¼l

i¼1

! Ai

þ lðAn Þ  l

i¼1

¼

n1 X

n X

¼

X

lðAi Þ 

! \ An

Ai

  l Ai \ Aj þ lðAn Þ  l

i\j  n1

X

lðAi Þ  lðAi Þ 

i¼1

X

n[ 1

! Ai

! \ An

i¼1 n1   X l Ai \ Aj  lðAi \ An Þ

i\j  n1

i¼1 n X

!

i¼1

i¼1

¼

n[ 1

i¼1

  l Ai \ Aj :

i\j  n

(b) If one of the measures l(Ai) is ∞, the inequality holds trivially. So, assume each l(Ai) is finite. By part (a), we have the equality n X

lðAi Þ ¼ l

i¼1

n [ i¼1

! Ai

þ

X

  l Ai \ Aj :

i\j  n

However, the sets Ai \ Aj are disjoint and their union is contained in [ ni =1Ai; therefore X i\j  n

! n [   l Ai \ Aj  l Ai : i¼1

Problem Set 5.5 5.5.P1. Using the functions constructed in Examples 5.5.4(a) and (b), show that strict inequality can hold in Theorem 5.5.1. R Hint: Using Example 5.5.4(a): Since f′(x) = 0 a.e., we have 0;1 f 0 ¼ 0: However, P 1 f ð1Þ  f ð0Þ ¼ 2k [ 0: 0\rk  1 R (b) Using Example 5.5.4(b): Since F′(x) = 0 a.e., we have ½0;1 F 0 ¼ 0: However, F(1) − F(0) > 0 because F is strictly increasing. 5.5.P2. Let f be Lebesgue’s singular function. Compute all the four Dini derivatives of f at each point x 2 [0, 1]. (Cf. Example 5.5.2). Hint: Let x be a point in the complement of the Cantor set C. Since f is constant in a neighbourhood of x, the derivative f′(x) exists and is zero. We now investigate the differentiability at points of the Cantor set. At a left endpoint x of a complementary interval (the function being constant on the right of x),

8 Hints

515

f þ0 ðxÞ ¼ lim

h!0 þ

f ðx þ hÞ  f ðxÞ ¼ 0 ¼ D þ f ðxÞ ¼ D þ f ðxÞ; h

and at the right endpoint x of a complementary interval (the function being constant on the left of x), f ðx  hÞ  f ðxÞ ¼ 0 ¼ D f ðxÞ ¼ D f ðxÞ: h!0 þ h

f0 ðxÞ ¼ lim

Let x denote the right endpoint of a complementary interval. Then x written in the ternary representation without using 1’s has the form x = .a1a2 ⋯ an200⋯ and f(x) = . b1b2 ⋯ bn100⋯, where bi = ai/2, i = 1, 2, …, n, interpreted as a number in the binary representation. If h is between 3−m−1 and 3−m, and m > n, then x + h would lie between .a1a2 ⋯ an200⋯ + 3−m−1 and .a1a2⋯an200⋯ + 3−m. Using the definition of 1 1 P P 1 1 1 1 f, the difference f(x + h) − f(x) would lie between 2k ¼ 2m þ 1 and 2k ¼ 2m ; k¼m þ 2

k¼m þ 1

consequently, 2m1 f ðx þ hÞ  f ðxÞ 2m \ \ : h 3m 3m1 Hence, as h ! 0 (so that m ! ∞) this difference quotient becomes positively infinite. That is, at a right endpoint x, f þ0 (x) = ∞. Similarly, f0 (x) = ∞ at a left endpoint. Let x be a point of the Cantor set but not the endpoint of a removed interval. Then x = .a1a2⋯an⋯, where an = 2 for infinitely many n. Let xn = .a1a2⋯an00⋯. Then {xn}n
1 is a sequence of right endpoints and is an increasing sequence converging to x. Therefore f ðxÞ  f ðxn Þ :0    0bn þ 1 bn þ 2    ð binary Þ ¼ x  xn :0    0an þ 1 an þ 2    ð ternary Þ If bN is the first nonzero bk with k
n + 1, then the right-hand side is greater than or equal to 1 2N 1 3N1

¼

3N1 ! 1 as n ! 1: 2N

So, D−f(x) = ∞. Similarly, it may be shown that D+f(x) = ∞. The computation of D+ f(x) and D−f(x) is left to the reader.

516

8 Hints

5.5.P3. Suppose f: [a, b] ! R has derivative 0 a.e. on [a, b] whenever b < b. Show that f has derivative 0 a.e. on [a, b]. Hint: We may assume for convenience that b − a > 1. For every n, there exists an An a; b  1n with measure 0 such that f has derivative 0 everywhere on [a, b]\(An   [ Bn), where Bn ¼ b  1n ; b : Consider any x 2 [a, b] such that x 62 \ n
1(An [ Bn). Then x 62 (An [ Bn) for some n i.e., x 2 [a, b]\(An [ Bn) for some n and therefore f has derivative 0 at x. Therefore f has derivative 0 everywhere on the complement of \ n
1(An [ Bn). Since \ n
1(An [ Bn) has measure 0 by Problem 3.1.P9, this means f has derivative 0 a.e. on [a, b]. Problem Set 5.6 5.6.P1. Show that Vð½a; b; f þ gÞ  Vð½a; b; f Þ þ Vð½a; b; gÞ and Vð½a; b; cf Þ ¼ jcjVð½a; b; f Þ: Hint: The first inequality has already been proved [see Remark 5.6.2(c)]. For a partition P: a = x0 < x1 < ⋯ < xn = b of [a, b],

TðP; cf Þ ¼

n X

jcf ðxk Þ  cf ðxk1 Þj ¼

k¼1

n X

jcjjf ðxk Þ  f ðxk1 Þj  jcjVð½a; b; f Þ;

k¼1

which implies V([a, b], cf)  |c|V([a, b], f). Conversely, let e > 0 be given. There exists a partition P: a = x0 < x1 < ⋯ < xn = b of [a, b] such that n X

jf ðxk Þ  f ðxk1 Þj [ Vð½a; b; f Þ 

k¼1

e : jcj

Now, n X

jcf ðxk Þ  cf ðxk1 Þj ¼ jcj

k¼1

n X

jf ðxk Þ  f ðxk1 Þj [ jcjVð½a; b; f Þ  e;

k¼1

which implies Vð½a; b; cf Þ  jcjVð½a; b; f Þ  e:

8 Hints

517

5.6.P2. Let /: (a, b) ! ℝ have a (finite) left limit everywhere and y 2 [a, b) [resp. y 2 (a, b]] be a point where the right limit /(y+) [resp. left limit /(y−)] exists (and is finite). Then first, lim /ðxÞ ¼ /ðy þ Þ

x!y þ

and second, lim /ðxÞ ¼ /ðyÞ:

x!y

In particular, lim /ðxÞ ¼ /ða þ Þ x!a þ

and

lim /ðxÞ ¼ /ðbÞ.

x!b

Hint: Consider any e > 0. By definition of the right-hand limit, there exists a d > 0 such that /ðy þ Þ  e\/ðzÞ\/ðy þ Þ þ e

for all z 2 ðy; y þ dÞ:

ð8:41Þ

Now let x 2 (y, y + d) be arbitrary. Then (y, x)  (y, y + d), and therefore it follows in accordance with (8.41) that /ðy þ Þ  e\/ðuÞ\/ðy þ Þ þ e

for all u 2 ðy; xÞ:

Since this holds for all u 2 (y, x), we may take the limit as u ! x, which automatically means u ! x−. Upon taking the limit, we get /ðy þ Þ  e  /ðxÞ  /ðy þ Þ þ e: As this has been shown to hold for an arbitrary x 2 (y, y + d), we may take the limit as x ! y, which automatically means x ! y+. Upon taking the limit, we obtain /ðy þ Þ  e  lim /ðxÞ  /ðy þ Þ þ e: x!y þ

As e > 0 is arbitrary, we have lim /ðxÞ ¼ /ðy þ Þ:

x!y þ

For y = a, this means limx!a+/(x−) = /(a+).

518

8 Hints

Consider any e > 0. By definition of the left-hand limit, there exists a d > 0 such that /ðyÞ  e\/ðzÞ\/ðyÞ þ e for all z 2 ðy  d; yÞ:

ð8:42Þ

Now let x 2 (y − d, y) be arbitrary. Then (y − d, x)  (y − d, y), and therefore it follows in accordance with (8.42) that /ðyÞ  e\/ðuÞ\/ðyÞ þ e for all u 2 ðy  d; xÞ: Since this holds for all u 2 (y − d, x), letting u ! x (u ! x−), we get /ðyÞ  e  /ðxÞ  /ðyÞ þ e: As this has been shown to hold for an arbitrary x 2 (y − d, y), we may take the limit as x ! y (x ! y−), thereby obtaining /ðyÞ  e  lim /ðxÞ  /ðyÞ þ e: x!y

Therefore lim /ðxÞ ¼ /ðyÞ;

x!y

When y = b, the above equality becomes limx!b−/(x−) = /(b−). 5.6.P3. The function f defined by f ðxÞ ¼ sin px ; 0\x  1 and f(0) = 0, is not of bounded variation on [0, 1]. Hint: Choose the partition of [0, 1] consisting of the points 0; 1 and

2 ; k ¼ 1; 2;    ; n; 2k þ 1

i.e., the points of the partition (in increasing order) are 0\

2 2 2 2 \ \    \ \ \1: 2n þ 1 2n  1 5 3

8 Hints

519

Then 





 n

X 2 2 2 2 f 2k  1  f 2k þ 1 þ f ð1Þ  f 3 þ f 2n þ 1  f ð0Þ k¼2

¼ 2n þ 2 ! 1 as n ! 1: 5.6.P4. Define f: [0, 1]!R by f(0) = 0 and f ðxÞ ¼ x2 sin 1x ; x 6¼ 0: Show that f is of bounded variation on [0, 1]. Hint: f 0 ðxÞ ¼ 2x sin 1x  cos 1x and so, | f′(x)|  3. Using parts (e) and (f) of Remark 5.6.2, it follows that f is of bounded variation. 5.6.P5. Show that the function defined on [0, 1] by

f ðxÞ ¼

x2 sin x12 0

for x 6¼ 0 for x ¼ 0

is not of bounded variation. Hint: Modify the argument of Example 5.6.3(c). 5.6.P6. Suppose that f: [a, b] ! [c, d] is monotone and that g is of bounded variation on [c, d]. Prove that V([a, b],g  f)  V([c, d], g). Hint: Assume that f is increasing. Let P: a = x0 < x1 < ⋯ < xn = b be a partition of [a, b]. Then c  f ðx0 Þ  f ðx1 Þ      f ðxn Þ  d; though not necessarily a partition, gives rise to a partition Q of [c, d], and

TðP; g  f Þ ¼

n X

jgðf ðxk ÞÞ  gðf ðxk1 ÞÞ þ jgðf ðx0 ÞÞ  gðcÞj þ jgðdÞ  gðf ðx0 ÞÞj

k¼1

¼ TðQ; gÞ  Vð½c; d; gÞ: 5.6.P7. Let {fn}n
1 be a sequence of real-valued functions defined on [a, b] that converge pointwise to the function f. Prove that Vð½a; b; f Þ  lim inf V ð½a; b; fn Þ: n!1

Hint: Let P: a = x0 < x1 < ⋯ < xp = b be a partition of [a, b]. Since fn! f pointwise,

520

8 Hints

1. for x 2 [a, b], and any e > 0, there exists an mx such that n
mx ) | fn(x) − f(x)| < e ;   2p               2. f xj  f xj1  f xj  fn xj þ fn xj  fn xj1 þ fn xj1  f xj1       e  2 2p þ fn xj  fn xj1 for n
m0
max mx0 ; mx1 ; . . .; mxp . So, p p X     X     f xj  f xj1  fn xj  fn xj1 þ e for n
m0 : j¼1

j¼1

Thus, TðP; f Þ  V ð½a; b; fn Þ þ e for n
m0 : Consequently, TðP; f Þ  lim inf V ð½a; b; fn Þ þ e: n!1

Considering that P is an arbitrary partition of [a, b], we obtain V([a, b], f)  lim inf V ð½a; b; fn Þ þ e The rest is clear. n!1

5.6.P8. Let {fn}n
1 be a sequence of functions of bounded variation on [a, b] such that Rnfn(a) converges absolutely and RnV([a, b], fn) < ∞. Prove that (i) Rnfn(x) converges absolutely for each x 2 [a, b], (ii) V([a, b],Rnfn)  RnV([a, b], fn). Hint: (i) | fn(x)|  | fn(x) − fn(a)| + | fn(a)|  V([a, b], fn) + | fn(a)|. Therefore X X X V ð½a; b; fn Þ þ jfn ðxÞj  jfn ðaÞj\1: n

n

n

(ii) For n = 2, the result has already been established. Assume that it is true for n = m. Then

V ½a; b;

m þ1 X k¼1

! fk

¼ V ½a; b;

m X

! fk þ fm þ 1  V ½a; b;

k¼1

þ V(½a; b; fm þ 1 Þ 

m X k¼1

m þ1 X k¼1

V ð½a; b; fk Þ:

! fk

8 Hints

521 n P

This completes the induction proof that Vð½a; b; implies lim inf Vð½a; b; n!1

n P

fk Þ 

k¼1

1 P

fk Þ 

k¼1

n P

Vð½a; b; fk Þ. This

k¼1

Vð½a; b; fk Þ. The required conclusion (ii) now

k¼1

follows upon applying Problem 5.6.P7.

R 5.6.P9. Suppose g 2 L1[a, b] and set f ðxÞ ¼ ½a:x g; a  x  b: Show that R Vð½a; b; f Þ ¼ ½a; b jgj: Hint: We have already shown just before Example 5.6.12 that f is of bounded variation on [a, b] and Z jgj: ð8:43Þ Vð½a; b; f Þ  ½a; b

It is sufficient to prove the reverse inequality. Suppose g is continuous. Let e > 0 be given. By the uniform continuity of g, there is a d > 0 such that |g(x) − g(y)| < e/2(b − a) whenever |x − y| < d. Now let P: a = x0 < x1 < ⋯ < xn = b be a partition of [a, b] such that xk+1 − xk < d, k = 0,1, 2, …, n − 1. Then Z Z g ¼ jf ðxk þ 1 Þ  f ðxk Þj ¼ fgðxk Þ  ½gðxk Þ  gg ½xk ;xk þ 1  ½xk ;xk þ 1  Z Z Z e
gðxk Þ  ½g  gðxk Þ
gðxk Þ  ðxk þ 1  xk Þ ½xk ;xk þ 1  ½xk ;xk þ 1  ½xk ;xk þ 1  2ðb  aÞ Z e ¼ jgðxk Þj  ðxk þ 1  xk Þ 2ðb  aÞ ½x ;x  Z k kþ1 Z e
jgj  jg  gðxk Þj  ðxk þ 1  xk Þ 2ðb  aÞ ½xk ;xk þ 1  ½xk ;xk þ 1  Z e
: jgj  ðxk þ 1  xk Þ ðb  aÞ ½xk ;xk þ 1 

Summing up both sides, we have n X

Z jf ðxk þ 1 Þ  f ðxk Þj


k¼1

½a; b

jgj  e;

which implies Z Vð½a; b; f Þ


½a; b

jgj  e:

522

8 Hints

Since e > 0 is arbitrary, we arrive at Z Vð½a; b; f Þ


jgj;

½a; b

ð8:44Þ

R where f is the function related to g by f ðxÞ ¼ ½a; x g; a  x  b: This has been obtained by assuming that g is continuous. For g 2 L1[a, b] and arbitrary η > 0, there exists a continuous g0 on [a, b] such that Z ð8:45Þ jg0  gj\g; ½a;b

using Proposition 3.4.2. Now, Z

Z ½a; b

j g0 j ¼

Z ½a; b

j g þ ð g0  gÞ j


Z ½a; b

jgj 

Z ½a; b

j g0  gj


½a; b

jgj  g: ð8:46Þ

Since g0 is continuous, we may apply (8.44) to it, thereby obtaining Z V ð½a; b; f0 Þ
jg0 j; ½a; b

where f0 is the function related to g0 by f0 ðxÞ ¼ with (8.46), we obtain

R ½a; x

g0 ; a  x  b: Combining this

Z V ð½a; b; f0 Þ


½a; b

jgj  g:

ð8:47Þ

Now, applying (8.43) to g − g0 yields Z V ð½a; b; f0  f Þ 

½a; b

jg0  gj:

ð8:48Þ

However, by the first part of Problem 5.6.P1, we have Vð½a; b; f Þ
 V ð½a; b; f0  f Þ þ V ð½a; b; f0 Þ Z
 jg0  gj þ V ð½a; b; f0 Þ byð8:48Þ ½a; b

[ V ð½a; b; f0 Þ  g Z [ jgj  2g ½a; b

by ð8:45Þ by ð8:47Þ:

8 Hints

523

Since η > 0 is arbitrary, the result follows. 5.6.P10. Deduce from Problem 5.6.P9 that sinx is of bounded variation on [0, 2p] and that its total variation is 4. Hint: [We have already noted in Example 5.6.3(a) without using Problem 5.6.P9 that sinx is of bounded variation on any compact interval.] Since Z sin x ¼

cos t dt ½0;x

and cost 2 L1[0, 2p], it follows by Problem 5.6.P9 that sinx is of bounded variation on [0, 2p] and that Z Vð½0; 2p; sin xÞ ¼ Z ¼

½0;2p

½0;p2

j cos tjdt Z

cos t dt 

Z ½p2;3p2 

cos t dt þ

½3p2 ;2p

cos t dt ¼ 4:

5.6.P11. Define f on [0, 1] by

f ð0Þ ¼ 0

and

1 f ðxÞ ¼ x sin ; x 6¼ 0: x

Show that f is not an indefinite integral of a Lebesgue integrable function on [0, 1]. Hint: It is clear that f is continuous on [0, 1]. However, it is not of bounded variation on [0, 1]. The argument is essentially that of Example 5.6.3(c). Indeed, for the partition

0\

2 2 2 2 2 \ \ \    \ \ \1; ð4n þ 1Þp 4np ð4n  1Þp 2p p

we have the inequality



 

 2 2 f þ f 2  f þ    þ sin 1  2  f ð0Þ ð4n þ 1Þp 4np ð4n þ 1Þp p

 4 1 1
1þ þ  þ ; p 3 4n þ 1 the right-hand side of which tends to ∞ as n ! ∞.

524

8 Hints

1 Consequently, by Problem R 5.6.P9 above, it is impossible to find an L [0, 1] function 1 g such that x sin x ¼ ½0;x g; 0\x  1; i.e. f is not an indefinite integral of a Lebesgue integrable function on [0, 1]. 5.6.P12. Define

f ðxÞ ¼

1 0

x2Q x 62 Q:

Show that f is not of bounded variation on any compact interval. Hint: In Example 5.3.3(b), we have seen that for rational x, D þ f ðxÞ ¼ 0; D þ f ðxÞ ¼ 1; D f ðxÞ ¼ 1 and D f ðxÞ ¼ 0 and for irrational x, D þ f ðxÞ ¼ 1; D þ f ðxÞ ¼ 0; D f ðxÞ ¼ 0 and D f ðxÞ ¼ 1: Thus the function f does not possess even a one-sided derivative at any x in the compact interval. By Remark 5.6.7(c), it cannot be of bounded variation. Alternative argument: Call the compact interval [a, b] and let P: a = x0 < x1 < ⋯ < xn+1 = b be a partition of it such that the interior points x1 < ⋯ < xn are n   P f xj  f ðxi1 Þ ¼ n  1. alternately rational and irrational. Then TðP; f Þ
j¼2

Since n is arbitrary, it follows that f fails to be of bounded variation. 5.6.P13. Construct functions fn, n = 0, 1, 2, … on an interval [a, b] such that (i) fn! f uniformly, (ii) each fn is of bounded variation, (iii) f is not of bounded variation.

0 x¼0 0 0  x\ 1n ; where f ðxÞ ¼ Hint: Let fn ðxÞ ¼ p 0\x  1: x cos 2x f ðxÞ 1n  x  1 (i) Let e > 0 be given.

Choose n10 so large that 1/n0 < e. For n
n0, we have 1 0 n x1 p \ n \e whenever and x cos 2x jf ðxÞ  fn ðxÞj ¼ p x cos 2x 0\x\ 1n 0\x\ 1n : So; fn ! f uniformly on [a, b].  p  1    p  2x2 (ii) The restriction of fn to 1n ; 1 satisfies fn0 ðxÞ ¼ cos 2x  x sin 2x 0 p 1 p n ¼ cos 2x þ 2x sin 2x. Therefore fn ðxÞ  1 þ 2 : So, by Remark 5.6.2(f), the     restriction of fn to 1n ; 1 is of bounded variation; but the restriction to 0; 1n is a step function and is therefore also of bounded variation. It follows by Theorem 5.6.4 that fn is of bounded variation on [0, 1].

8 Hints

525

(iii) It is not difficult to check that f is not of bounded variation. Indeed, for the 1 1 1 \ 2n1 \ 2n3 \    \ 13 \ 12 \1; we have partition P : 0\ 2n TðP; f Þ ¼

1 1 1 1 1 1 þ þ þ þ...þ þ 2n 2n 2n  2 2n  2 2 2

¼ 1þ

1 1 1 þ  þ þ ; 2 n1 n

which tends to ∞ as n ! ∞. 5.6.P14. Let f be defined on [0, 1] by the following formulae [see Fig. 8.4]: 8 1 x x ¼ 2n ;n 2 N > > > < 0 1 x ¼ 2n þ 1 ; n 2 N f ðxÞ ¼ 1 1 > linear > nþ1  x  n ; n 2 N > : 0 x ¼ 0: Show that f is not of bounded variation. Hint: Clearly, the function is continuous. For the partition P : 0;

1 1 1 1 ; ; ;; ;1 2n þ 1 2n 2n  1 2

of [0,1], we have TðP; f Þ ¼

1 1 1 1 1 1 þ þ þ þ...þ þ 2n 2n 2n  2 2n  2 2 2

¼ 1þ

1 1 1 þ  þ þ ; 2 n1 n

which tends to ∞ as n ! ∞. R 5.6.P15. If f ðxÞ ¼ ½a;x g, where g 2 L1[a, b], prove that the positive and negative R variations of f, namely, P([a, b], f) and N ([a, b], f), are given by ½a;b g þ and R  ½a;b g respectively.

1/2

1/4

1/5 1/4 1/3

Fig. 8.4 Function in Problem 5.6.P14

1/2

1

526

8 Hints

Hint: By (5.64) of Proposition 5.6.9, 2Pð½a; b; f Þ ¼ Vð½a; b; f Þ þ f ðbÞ  f ðaÞ Z ¼ jgj þ f ðbÞ  f ðaÞby Problem 5:6:P9 ½a;b Z Z ¼ jgj þ g þ 0 by definition of f ½a;b ½a;b Z Z Z ¼ ð g þ þ g Þ þ ð g þ  g Þ ¼ 2 ½a;b

½a;b

½a;b

gþ :

The argument for N ([a, b], f) is analogous. 5.6.P16. Let x1, x2, … be the points of discontinuity in (a, b) of the function f of bounded variation with domain [a, b]. We define j: [a, b]!R by the following formulae: jðaÞ ¼ fP ðaÞ  f ða þ Þ and jðxÞ ¼ f ðxÞ  f ðxÞ þ ½f ðxi þ Þ  f ðxi Þ if a\x  b:

ð8:49Þ

xi \x

By the “jump” of f at x, we mean | f(x) − f(x−)| + | f(x+) − f(x)| if a < x < b and | f (a+) − f(a)| if x = a and | f(b) − f(b−)| if x = b. [Note that this agrees with Definition 5.2.4 for a monotone function.] Show without using Theorem 5.3.1 that (a) The function P j defined in (8.49) is of bounded variation on [a, b]. (b) jðyÞ ¼ ½f ðxi þ Þ  f ðxi Þ for y 2 ða; b, xi \y P ½f ðxi þ Þ  f ðxi Þ for y 2 ða; bÞ j ðy þ Þ ¼ xi  y

and j(a+) = 0. (c) If F = f − j, then F is continuous and of bounded variation. (d) Show that j′ = 0 a.e. (e) Show that the representation of f as F + j, where F is continuous and j′ = 0 a.e. is not unique (not even up to a constant) (cf. Problem 5.7.P4). Hint: (a) We shall show first that Vð½a; b; jÞ  J: Let P: a = y0 < y1 < ⋯ < yn = b be a partition of [a, b]. We shall estimate |j(yi) − j(yi−1)| for i = 1, …, n. For i = 1,

8 Hints

527

jjðy1 Þ  jðy0 Þj ¼ jjðy1 Þ  jðaÞj X ½f ðxk þ Þ  f ðxk Þ  f ðaÞ þ f ða þ Þ ¼ f ð y1 Þ  f ð y1  Þ þ x \y k

1

 jf ðy1 Þ  f ðy1 Þj þ j f ðaÞ  f ða þ Þj X ðjf ðxk Þ  f ðxk Þj þ jf ðxk þ Þ  f ðxk ÞjÞ: þ xk \y1

For i
2, jjðyi Þ  jðyi1 Þj ¼ jf ðyi Þ  f ðyi Þ þ

X

½f ðxk þ Þ  f ðxk Þ

xk \yi

X

 f ðyi1 Þ þ f ðyi1 Þ 

½f ðxk þ Þ  f ðxk Þj

xk \yi1

¼ f ðyi Þ  f ðyi Þ  f ðyi1 Þ þ f ðyi1 Þ þ

X yi1 \xk \yi

½f ðxk þ Þ  f ðxk Þ

¼ j f ðyi Þ  f ðyi Þ  f ðyi1 Þ þ f ðyi1 Þ þ f ðyi1 þ Þ  f ðyi1 Þ X ½ f ðx k þ Þ  f ðx k  Þ þ y \x \y i1

k

i

 jf ðyi Þ  f ðyi Þj þ jf ðyi1 þ Þ  f ðyi1 Þj X þ ðjf ðxk Þ  f ðxk Þj þ jf ðxk þ Þ  f ðxk ÞjÞ: yi1 \xk \yi

Upon writing these estimates consecutively for i = 2, …, n, we obtain

j ( y2 ) - j ( y1 ) £ f ( y2 ) - f ( y2 -) + f ( y1 +) - f ( y1 ) å

y1 < xk < y2

( f ( xk ) - f ( xk -) +

f ( xk +) - f ( xk ) )

j ( y3 ) - j ( y2 ) £ f ( y3 ) - f ( y3 -) + f ( y2 +) - f ( y2 ) å

y2 < xk < y3

( f ( xk ) - f ( xk -) +

f ( xk +) - f ( xk ) )

j ( y4 ) - j ( y3 ) £ f ( y4 ) - f ( y4 -) + f ( y3 +) - f ( y3 ) å

y3 < xk < y4

( f ( xk ) - f ( xk -) +

f ( xk +) - f ( xk ) )

j ( yn ) - j ( yn-1 ) £ f ( yn ) - f ( yn -) + f ( yn-1 +) - f ( yn-1 ) å

yn-1 < xk < yn

( f ( xk ) - f ( xk -) +

f ( xk +) - f ( xk ) ).

528

8 Hints

When we take the summation over i = 1, …, n, the first term on the right-hand side of each estimate except the last one can be combined with the second term on the right-hand side of the next estimate. Upon doing so, we obtain TðP; jÞ  J: Consequently, V([a, b], j)  J, as was to be shown. We proceed to prove that J  V([a, b], f), which will complete the demonstration that j is of bounded variation. First we shall prove that if y1, y2, …, yn are distinct points of (a, b), then V ðð½a; b; f ÞÞ
sðaÞ þ

n X

sðyi Þ þ sðbÞ;

i¼1

where s(x) means the jump at x. Without loss of generality, we may suppose that y1 < y2 < ⋯ < yn. There exists a d > 0 such that the n + 2 intervals (a, a + d), (yi − d, yi + d), (b − d, b) are all disjoint. Let η > 0 be arbitrary and put e ¼ 13 g. By definition of jump, for each i (1  i  n), we can select points ui 2 (yi − d, yi) and vi 2 (yi, yi + d) which satisfy | f(yi) − f(ui)| + | f(yi) − f(vi)| > sðyi Þ  ne . We can also select points v0 2 (a, a + d) and un+1 2 (b − d, b) which satisfy | f(a) − f(v0)| > s(a) − e and | f(b) − f(vn+1)| > s(b) − e. On the basis of the disjointness noted earlier, we know that a\v0 \u1 \y1 \v1 \    \un \yn \vn \un þ 1 \b: Consequently, these points form a partition of [a, b], which we shall denote by P. It follows from the manner in which the points ui and vi have been selected that TðP; f Þ
sðaÞ þ

n X

sðyi Þ þ sðbÞ  e  n

i¼1

e n

 e ¼ sðaÞ þ

n X

sðyi Þ þ sðbÞ  g:

i¼1

Since η > 0 is arbitrary and V([a, b], f)
T(P, f), it follows that V([a, b], n P f)
sðaÞ þ sðyi Þ þ sðbÞ . On the other hand, J is the supremum of such sums, i¼1

taken over all choices of finitely many distinct points y1, y2, …, yn of (a, b). So, the desired inequality is now immediate. (b) For y 2 (a, b],

jðyÞ ¼ f ðyÞ  f ðyÞ þ

X xk \y

½f ðxk þ Þ  f ðxk Þ;

8 Hints

529

therefore X

jðy  hÞ ¼ f ðy  hÞ  f ððy  hÞÞ þ

½f ðxk þ Þ  f ðxk Þ;

0\h\y  a;

xk \yh

jðyÞ  jðy  hÞ ¼ f ðyÞ  f ðyÞ  f ðy  hÞ þ f ððy  hÞÞ X ½f ðxk þ Þ  f ðxk Þ: þ y [ xk
yh

Hence jðyÞ ¼ jðyÞ  f ðyÞ þ f ðyÞ þ ¼

X

X

½f ðxk þ Þ  f ðxk Þ

xk \y

½f ðxk þ Þ  f ðxk Þ:

xk \y

For y 2 (a, b), jðyÞ ¼ f ðyÞ  f ðyÞ þ

X

½f ðxk þ Þ  f ðxk Þ;

xk \y

therefore jðy þ hÞ ¼ f ðy þ hÞ  f ððy þ hÞÞ þ

X

½f ðxk þ Þ  f ðxk Þ;

0\h\b  y;

xk \y þ h

jðy þ hÞ  jðyÞ ¼ f ðy þ hÞ  f ððy þ hÞÞ  f ðyÞ þ f ðyÞ X ½f ðxk þ Þ  f ðxk Þ: þ y  xk \y þ h

530

8 Hints

Hence jðy þ Þ ¼ jðyÞ þ ½f ðy þ Þ  f ðyÞ X ½f ðxk þ Þ  f ðxk Þ þ f ðy þ Þ  f ðyÞ ¼ f ðyÞ  f ðyÞ þ ¼

X

xk \y

½f ðxk þ Þ  f ðxk Þ:

xk  y

Finally, consider y(a+). We have jðaÞ ¼ f ðaÞ  f ða þ Þ and jða þ hÞ ¼ f ða þ hÞ  f ðða þ hÞÞ þ

X

½f ðxk þ Þ  f ðxk Þ;

0\h\b  a

xk \a þ h

jða þ hÞ  jðaÞ ¼ f ða þ hÞ  f ðða þ hÞÞ  f ðaÞ þ f ða þ Þ X þ ½f ðxk þ Þ  f ðxk Þ: xk \a þ h

(c) By (8.49), we have F(a) = f(a+). We shall first show that F is continuous and of bounded variation on [a, b]. First we prove continuity at any y 2 (a, b). By Problem 5.6.P2,

8 Hints

531

jðyÞ ¼

X

½f ðxi þ Þ  f ðxi Þ

ð8:50Þ

½f ðxi þ Þ  f ðxi Þ:

ð8:51Þ

xi \y

and jðy þ Þ ¼

X xi  y

From (8.49) and (8.50), jðyÞ ¼ jðyÞ  f ðyÞ þ f ðyÞ; whence FðyÞ ¼ f ðyÞ  jðyÞ ¼ f ðyÞ  jðyÞ þ f ðyÞ  f ðyÞ ¼ FðyÞ:

ð8:52Þ

From (8.50), (8.51) and (8.52), we obtain Fðy þ Þ ¼ f ðy þ Þ  jðy þ Þ ¼ f ðy þ Þ  ðf ðy þ Þ  f ðyÞÞ  ¼ f ðyÞ 

X

X

½f ðxi þ Þ  f ðxi Þ;

by ð8:51Þ

xi \y

½ f ð xi þ Þ  f ð xi  Þ 

xi \y

¼ f ðyÞ  jðyÞby ð8:50Þ ¼ FðyÞ

by ð8:52Þ:

Hence Fðy þ Þ ¼ FðyÞ: Thus F is continuous at every y 2 (a, b). By Problem 5.6.P2, we know that jða þ Þ ¼ 0

and

jðbÞ ¼

X

½f ðxi þ Þ  f ðxi Þ:

xi \b

Recall that F(a) = f(a+). Now, F(a+) = f(a+) − j(a+) = f(a+) = F(a). Therefore F is continuous at a. FðbÞ ¼ f ðbÞ  jðbÞ ¼ f ðbÞ  f ðbÞ þ f ðbÞ  ¼ f ðbÞ 

X

X xi \b

½f ðxi þ Þ  f ðxi Þ

xi \b

¼ f ðbÞ  jðbÞ ¼ FðbÞ:

½f ðxi þ Þ  f ðxi Þ

532

8 Hints

Thus the function F is continuous on [a, b]. F is of bounded variation because f and j are. (d) Since f is of bounded variation, we have f = f1 − f2, where f1 and f2 are nonnegative [nonnegativity will not be relevant in what follows] increasing functions on [a, b]. Let S1 and S2 be the saltus functions associated with f1 and f2 respectively. Now, jðxÞ ¼ f ðxÞ  f ðxÞ þ

X

½f ðxi þ Þ  f ðxi Þ for a\x  b

xi \x

¼ f1 ðxÞ  f2 ðxÞ  f1 ðxÞ þ f2 ðxÞ X ½ f 1 ð xi þ Þ  f 2 ð xi þ Þ  f 1 ð xi  Þ þ f 2 ð xi  Þ  þ xi \x

¼ f1 ðxÞ  f1 ðxÞ þ

X

½f1 ðxi þ Þ  f1 ðxi Þ

xi \x

 ½f2 ðxÞ  f2 ðxÞ þ

X

½f2 ðxi þ Þ  f2 ðxi Þ:

xi \x

Recall that in the summation occurring in the definition of the associated saltus functions [see the beginning of Sect. 5.3], the left endpoint a is not excluded from being among the points xk; but in the summation occurring in the definition of j, it is specifically excluded by taking x1, x2, … to be the points of discontinuity in (a, b). Taking this distinction into account, we find that f1 ðxÞ  f1 ðxÞ þ

X

½f1 ðxi þ Þ  f1 ðxi Þ ¼ S1 ðxÞ  ½f1 ða þ Þ  f1 ðaÞ

x1 \x

and f2 ðxÞ  f2 ðxÞ þ

X

½f2 ðxi þ Þ  f2 ðxi Þ ¼ S2 ðxÞ  ½f2 ða þ Þ  f2 ðaÞ:

x1 \x

Substituting these two equalities into the one derived for j(x) above, we arrive at jðxÞ ¼ S1 ðxÞ  S2 ðxÞ  ½f ða þ Þ  f ðaÞ for a\x  b: Since S1 and S2 are differentiable a.e. and S1′(x) = 0 = S2′(x) a.e. on [a, b] by Theorem 5.4.7, it now follows that j is differentiable a.e. and j′(x) = 0 a.e. on [a, b]. (e) Suppose the domain [a, b] is [0, 1]. Let g be the Cantor function as in Example 5.5.2. It was shown there that g is increasing (hence of bounded variation), continuous and has derivative 0 a.e. It follows on the one hand that F + g is of bounded variation and continuous, and on the other hand that j + g has derivative 0 a.e. However, their difference is f.

8 Hints

533

Problem Set 5.7 5.7.P1. If f: [a, b]!R is absolutely continuous on [a, c] as well as on [c, b], then it is absolutely continuous on [a, b]. Hint: For any interval I = [a, b] [a, b], denote by D(I) the nonnegative number | f(b) − f(a)|. P Let e > 0 be given. There exists a d > 0 such that DðIÞ\ 2e for an arbitrary finite I2K

system K of disjoint intervals contained in either [a, c] or [c, b] and having total length less than d. Now consider an arbitrary finite system J of disjoint intervals contained in [a, b] and having total length less than interval of the system Pd. If every lies in either [a, c] or [c, b], it is clear that DðIÞ\ 2e þ 2e = e. If not, then i2J

precisely one interval J fails to do so, in which case, c belongs to the interior of J. Moreover, all the other intervals are contained either in [a, c] or [c, b]. Denote by A the subsystem of the former and by B the subsystem of the latter. If either of these systems is empty, then the sum relating to it is empty and therefore taken to be 0. Then X i2J

DðIÞ ¼

X

DðIÞ þ

X

DðIÞ þ DðJÞ:

ð8:53Þ

i2B

i2A

Let JA be the part of J that lies to the left of c and JB be the part that lies to the right of c. Then DðJÞ  DðJA Þ þ DðJB Þ:

ð8:54Þ

Also, the intervals of A and the interval JA together of disjoint P form a finite system intervals contained in [a, c]. Therefore DðIÞ þ DðJA Þ\ 2e : Similarly, I2A P DðIÞ þ DðJB Þ\ 2e : It follows that

I2B

X i2A

DðIÞ þ DðJA Þ þ

X

DðIÞ þ DðJB Þ\

i2B

Upon combining this with (8.53) and (8.54), we get

e e þ ¼ e: 2 2

P

DðIÞ\e .

I2J

5.7.P2. Let f: [a, b]!R be an absolutely continuous function and let E  [a, b] be a set of measure zero. Then show that m(f(E)) = 0. Use this to show that the Cantor function is not absolutely continuous. Show that on a closed bounded interval, the class of absolutely continuous functions is a proper subclass of continuous functions of bounded variation. [This is trivial if the domain is unbounded, because sinx will do the trick.]

534

8 Hints

Hint: Let e > 0 be given. We shall show that there exists a d > 0 such that, for an arbitrary countable system of disjoint intervals {(ak, bk)}k
1, the sum of whose lengths is less than d, the inequality 1 X

ðMk  mk Þ\e

ð8:55Þ

k¼1

holds, where mk = min{f(x): ak  x  bk} and Mk = max{f(x): ak  x  bk}. By absolute continuity of f and Remark 5.7.2(d), there exists a d > 0 such that the inequality (8.55) holds with 2e in place of e for any finite system of disjoint intervals {(ak, bk)}n
k
1, the sum of whose lengths is less than d, i.e., n X

e ðMk  mk Þ\ : 2 k¼1

Letting n ! ∞, we obtain (8.55). Assume that the points a and b do not belong to E, so that E  (a, b). Since m(E) = 0, there exists a bounded open set G  E such that m(G) < d. Without loss of generality, we may assume that G (a, b), for otherwise we replace G by G \ (a, b) and call the intersection G. Now, G¼

1 [

ðak ; bk Þ and

k¼1

1 X

ðbk  ak Þ\d:

k¼1

This implies f(E) f(G) = [ k
1f((ak, bk)) [ k
1f([ak, bk]). Therefore m ðf ðEÞÞ 

1 X

m ðf ð½ak ; bk ÞÞ:

k¼1

It is also clear that f ð½ak ; bk Þ ¼ ½mk ; Mk  and consequently, m ðf ðEÞÞ 

1 X k¼1

ðMk  mk Þ\e;

8 Hints

535

which implies mðf ðEÞÞ ¼ 0; since e > 0 is arbitrary. Note that the removal of the points a and b from the set E implies removal of at most two points, namely, f(a) and f(b), from f(E). This has no effect on the measure of f(E). The Cantor function maps the Cantor set, which has measure 0, into the set [0, 1], which does not have measure 0. Therefore it is not absolutely continuous. Consequently, the Cantor function provides an example of a function that is of bounded variation on a closed bounded interval but is not absolutely continuous. 5.7.P3. Let f: [a, b]!R be an absolutely continuous function. Then f maps measurable sets into measurable sets. Hint: Let E be a measurable set in [a, b]. Then there exist an increasing sequence {Fn} of closed sets contained in E such that mðEÞ ¼ lim mðFn Þ n!1

by Proposition 2.3.24. Therefore we can write

E¼

1 [

! [ N;

Fn

n¼1

where N is a set of measure zero; hence

f ðEÞ ¼ f

1 [

! Fn [ f ðNÞ ¼

n¼1

1 [

f ðFn Þ [ f ðNÞ:

n¼1

By Problem 5.7.P2 above, f(N) is of measure zero. Each f(Fn), being the image of a compact subset Fn under the continuous map f, is compact and hence measurable. Consequently, f(E) is measurable. 5.7.P4. Let f be defined and measurable on the closed interval [a, b] and let E be any measurable subset on which f′ exists. Then prove that Z m ðf ðEÞÞ  jf 0 j: E

536

8 Hints

Hint: First we suppose that | f′(t)| < M on E, where M denotes a positive integer. Let

Ekn ¼

t2E:

 k1 k 0 n \ f ðtÞ ; k ¼ 1; 2; . . .; M2 ; n ¼ 1; 2; . . . : j j\ 2n 2n

Then for each n, 

m ðf ðEÞÞ ¼ m 



X

f

[

!! ¼m

Ekn



[   f Ekn

k

   m f Ekn ;

!

k

using Proposition 2:2:9

k

Xk  1   1 X   m Ekn þ n m Ekn : ¼ 2n 2 k k Therefore "

# Z Xk  1   1 X   n n m ðf ðEÞÞ  lim m Ek þ n m Ek ¼ jf 0 j: n n!1 2 2 E k k 

Now, assume that f′ is not bounded. Let Ak ¼ ft 2 E : ðk  1Þ  f 0 ðtÞ\kg; k ¼ 1; 2; . . .: Then 

m ðf ðEÞÞ ¼ m



f

[ k

Z

!! Ak



¼m

[ k

! f ðAk Þ 

X k



m ðf ðAk ÞÞ 

XZ k

jf 0 j

Ak

jf 0 j:

¼ E

So far, we have proved (see Remark 5.7.2(b), Proposition 5.7.2 and Problem 5.7. P2) that an absolutely continuous function is continuous, of bounded variation and maps a set of measure zero into a set of measure zero. The following result shows that these three properties characterise an absolutely continuous function [Banach– Zarecki Theorem]. 5.7.P5. If f is continuous and of bounded variation on [a, b], and if f maps a set of measure zero into a set of measure zero, then f is absolutely continuous on [a, b]. Hint: Let {(ak, bk)}k
1 be a finite system of disjoint open intervals contained in [a, b] and let Ek = {x 2 [ak, bk]: f′ exists}. Since m([ak, bk]\Ek) = 0 and f maps sets of measure zero into sets of measure zero,

8 Hints

537

mðf ð½ak ; bk ÞÞ ¼ mðf ðEk ÞÞ; and therefore n X

j f ð bk Þ  f ð ak Þ j 

k¼1

  

which tends to 0 as

n P

n X k¼1 n X

mðf ð½ak ; bk ÞÞ mðf ðEk ÞÞ

k¼1 n Z X

jf 0 j

k¼1

Ek

k¼1

½ak ;bk 

n Z X

jf 0 j;

ðbk  ak Þ ! 0, since f′ is integrable in view of

k¼1

Theorem 5.5.1. 5.7.P6. Let f be absolutely continuous on [e, 1] for each positive e < 1. Give an example to show that continuity of f on [0, 1] does not imply absolute continuity on [0, 1]. What if f is also of bounded variation on [0, 1]? Show that xp is absolutely continuous on [0, 1] when 0  p < 1. Hint: The function

f ðxÞ ¼

x sin 1x 0

x 6¼ 0 x¼0

is such that it is absolutely continuous on [e, 1] for positive e < 1, since 1 1 1 1 jf 0 ðxÞj ¼ sin  cos  1 þ : x x x e The function f is continuous on [0, 1]. However, it is not of bounded variation on [0, 1]. The answer is Yes if f is also of bounded variation on [0, 1]: The function v(x) = V([0, x], f), 0  x  1, is continuous, since f is continuous on [0, 1]. Let e > 0 be given. Then there exists a d1 > 0 such that v(x) < 2e provided x < d1. We assume, as we may, that d1 < 1. On [d1, 1], the function is absolutely continuous. There exists a d2 > 0 such that for any finite system {(ak, bk)}1  k  n of disjoint intervals contained in [d1, 1] and where the sum of lengths is less than d2, we have

538

8 Hints n X

e jf ðbk Þ  f ðak Þj\ : 2 k¼1

Let d = min{d1, d2} and let {(ai, bi)}1  i  m be a system of disjoint open intervals contained in [0,1] and having sum of lengths less than d. If an (ai, bi) is such that d1 2 (ai, bi), we may replace (ai, bi) by (ai, d1) [ (d1, bi). This new system of intervals {(ai, bi)}1  i  m′ are disjoint and the sum of their lengths is less than d. We then have m X

jf ðbi Þ  f ðai Þj 

m X

i¼1

jf ðbi Þ  f ðai Þj

i¼1

¼

X

jf ðbi Þ  f ðai Þj þ

i2A

X

jf ðbi Þ  f ðai Þj\

i2B

e e þ ¼ e; 2 2

where A consists of those indices i in {1, 2, …, m} for which (ai, bi) lie in [0, d1] and B consists of the remaining indices. For every positive e < 1, the function xp (0  p < 1) is continuous on [e, 1] with bounded derivative and hence absolutely continuous by Remark 5.7.2(e). Since it is increasing on [0, 1], it is also of bounded variation on [0, 1]; besides, it is continuous on [0, 1]. It follows from what has been proved above that the function is absolutely continuous on [0, 1]. 1 5.7.P7. Show that the function f given on [0, ∞) by f ðxÞ ¼ x2 for 0  x\1 and satisfying f(x + k) = f(x) + k for 0  x < 1 and k 2 N is absolutely continuous on any bounded subinterval of its domain but not on the entire domain. [Note that the example of such a function given in Remark 5.7.2(a) satisfies a Lipschitz condition on every bounded subinterval of [0, ∞) while the present one does not.] Hint: From the second part of the definition of f, it follows that 1 1 f ð1Þ ¼ f ð0 þ 1Þ ¼ f ð0Þ þ 1 ¼ 12 , so that the equality f ðxÞ ¼ x2 holds for x = 1 as well. Therefore the function is absolutely continuous on [0, 1] by Problem 5.7.P6. Now, for any k 2 N; we find that x 2 [k, k + 1] implies x − k 2 [0, 1], which further 1

implies f ðxÞ ¼ f ððx  kÞ þ kÞ ¼ f ðx  kÞ þ k ¼ ðx  kÞ2 þ k. Therefore, by Problem 5.7.P6 and the fact that k is a constant, the function is absolutely continuous on the interval [k, k + 1] for any k 2 N: It now follows by Problem 5.7.P1 that it is absolutely continuous on any bounded interval [0, k + 1] and hence on any bounded subinterval of [0, ∞). To see why it is not absolutely continuous on [0, ∞), consider some positive d < 1 1 P 1 d and A [ i2 :For xi ¼ i; yi ¼ i þ Ai2 ; i ¼ 1; . . .; n; i¼1

n X i¼1

j yi  xi j ¼

n n X X d 1 ¼ d \d: 2 Ai Ai2 i¼1 i¼1

8 Hints

539

However, n X

j f ð yi Þ  f ð xi Þ j ¼

i¼1

12 X n d 1 : A i¼1 i

As the right-hand side of the above equality tends to ∞ as n ! ∞, the function f cannot be absolutely continuous on [0,∞). 5.7.P8. Let f be an increasing function defined on [a, b]. Show that f can be decomposed into a sum of increasing functions f = g + h, where g is absolutely continuous and h is increasing with h′ = 0 a.e. Hint: For an increasing function f, the derivative f′ exists a.e. (see Theorem 5.4.6) and is integrable (see Theorem 5.5.1). Set Z gðxÞ ¼

½a;x

Then g is absolutely continuous and Z gðbÞ  gðaÞ ¼

½a;b

f 0 ðtÞdt:

f 0 ðtÞdt  f ðbÞ  f ðaÞ:

ð8:56Þ

If h = f − g, then hðbÞ  hðaÞ ¼ ½f ðbÞ  f ðaÞ  ½gðbÞ  gðaÞ
0; using (8.56). Consequently, h is increasing. However, h0 ðxÞ ¼ f 0 ðxÞ  g0 ðxÞ ¼ 0 a:e:

Problem Set 5.8 5.8.P1. Suppose f is absolutely continuous on [a, b] and f′(x) = 0 a.e. on [a, b]. Then f is a constant function. Give a proof of this that is different from the one in Corollary 5.8.4. Hint: Let E = {x 2 (a, b): f′(x) = 0}. Then Z mðf ðEÞÞ 

jf 0 j ¼ 0

ð8:57Þ

E

by Problem 5.7.P4. Also, f is absolutely continuous, and [a, b]\E is a set of measure 0; so it follows that

540

8 Hints

mðf ð½a; bnEÞÞ ¼ 0

ð8:58Þ

by Problem 5.7.P2. Now, mðf ð½a; bÞÞ ¼ mðf ð½a; bnEÞ [ EÞ ¼ mðf ðð½a; bnEÞÞ [ f ðEÞÞ

ð8:59Þ

 mðf ðð½a; bnEÞÞÞ þ mðf ðEÞÞ ¼ 0; by (8.57) and (8.58). Since f is continuous, unless f is constant, f([a, b]) contains an interval of positive length, contradicting (8.59). Therefore f must be constant. 5.8.P2. If f and g are absolutely continuous on [a, b], f′(x) = g′(x) a.e. on [a, b] and f (x0) = g(x0) for some x0 2 [a, b], then show that f(x) = g(x) for every x 2 [a, b]. Hint: If h(x) = f(x) − g(x) for x 2 [a, b], then h is absolutely continuous and h′(x) = 0 a.e. on [a, b]. By Problem 5.8.P1, h is a constant. Since it vanishes at x0, it follows that f(x) = g(x) for every x 2 [a, b]. 5.8.P3. Let f: [a, b]!R be absolutely continuous and let vðaÞ ¼ 0; vðxÞ ¼ Vð½a; x; f Þfor x 2 ½a; b: Then v is absolutely continuous on [a, b] by Proposition 5.7.9. Show that v′(x) = | f′(x)| for almost all x 2 [a, b]. Hint: Since v is absolutely continuous, Theorem 5.8.2 leads to Z vðxÞ ¼ v0 ; ½a; x

which, in turn, implies Z vðyÞ  vðxÞ ¼

½x;y

v0

for all y [ x:

Let P be an arbitrary partition of [a, b]: P : a ¼ x0 \x1 \    \xn ¼ b: Then another application of Theorem 5.8.2 leads to Z X n Z n Z X 0 0 TðP; f Þ ¼ f  jf j ¼ jf 0 j: jf ðxk Þ  f ðxk1 Þj ¼ ½xk1 ;xk  k¼1 ½xk1 ;xk  ½a;b k¼1 k¼1 n X

So,

8 Hints

541

Z vðbÞ  vðaÞ ¼ Vð½a; b; f Þ 

½a;b

jf 0 j:

Similarly, Z vðyÞ  vðxÞ 

½x;y

jf 0 j for all y [ x:

Hence v′(x)  | f′(x)| for almost all x 2 [a, b] by Theorem 5.8.5. Conversely, since vðyÞ  vðxÞ
j f ðyÞ  f ðxÞj

for all y [ x;

we obtain v0 ðxÞ
jf 0 ðxÞj

for all x 2 ½a; b:

5.8.P4. Show that, if g is an integrable function defined on [a, b] with indefinite integral f, then f′(x) = g(x) whenever x is a point of continuity of g. Hint: For any x, Z f ðxÞ ¼

g and ½a;x

f ðx þ hÞ  f ðxÞ 1 ¼ lim h!0 h!0 h h

f 0 ðxÞ ¼ lim

Z ½x;x þ h

g ¼ gðxÞ

since every point of continuity of g is a Lebesgue point of g. R 5.8.P5. If f ðxÞ ¼ ½0;x g, where g is an integrable function defined on [0, 1], then f′ = g need not hold even when f′ exists. R 1 if x is rational Hint: Let gðxÞ ¼ Then for every x, [0,x]g = 0 because 0 if x is irrational. the set over which g is not zero, being a subset of the set of all rationals, has measure zero. So, f′(x) = 0 for all x 2 [0, 1]. However, f′ = g at all irrational x and f′ 6¼ g on the set of rationals in [0, 1]. R 5.8.P6. (Cf. Theorem 5.5.1) Show that the inequality [a,x] f′  f(b) − f(a) need not hold when f is not monotone increasing. Hint: Take any monotone increasing function for which strict inequality holds and multiply by −1. In particular, for the Cantor R function (see Example 5.5.2), f′(x) = 0 a.e. on [0, 1] and f(0) = 0, f(1) = 1. So, [0,1] f′ = 0 < f(1) − f(0) = 1. On multiR plying f by −1, we obtain 0 = [0,1] –f′ > −1.

542

8 Hints

5.8.P7. (Lebesgue Decomposition Theorem for Functions) Suppose f is of bounded variation on [a, b]. Then there exists an absolutely continuous function g and a function h such that f(x) = g(x) + h(x), x 2 [a, b], where h′(x) = 0 a.e. Up to constants, the decomposition is unique. Hint: It is sufficient to prove this on the assumption that f is monotonic on [a, b]. Then the derivative f′ exists a.e. by Lebesgue’s Theorem 5.4.6 and is integrable by Theorem 5.5.1. Let Z gðxÞ ¼ f 0 for a  x  b: ½a;x

Then g is absolutely continuous by Proposition 5.8.1 and by Theorem 5.8.5, we have f′(x) − g′(x) = 0 a.e. on [a, b]. Now, all we need to do is to set hðxÞ ¼ f ðxÞ  gðxÞ;

x 2 ½a; b:

As for uniqueness, suppose f = g1 + h1, where g1 is absolutely continuous and h1′ = 0 a.e. Then g − g1 = h1 − h is absolutely continuous with derivative 0 a.e., which implies g = g1 + c by Corollary 5.8.4 and hence h = h1 − c. 5.8.P8. [Application to Fourier Series; due to Lebesgue] Let f be a function in L1([−p, p]). Then show that lim rn ðxÞ ¼ f ðxÞ

n!1

at every Lebesgue point x of f (and hence a.e. on [−p, p] by Theorem 5.8.9). Here the notation is as in Fejér’s Theorem 4.3.4. Hint: All integralsR in this solution are to be understood as Lebesgue integrals. Write Ux(t) = t0|/x(u)|du, where /x(u) = f(x + u) + f(x − u) − 2f(x). Then Ux(u) = |/x(u)| a.e. by Theorem 5.8.5. In view of Corollary 5.8.10, we have lim Ux ðtÞ=t ¼ 0 because x is a Lebesgue point of f. t!0

Consider any e > 0 and choose d > 0 such that 1 Ux ðtÞ  e t

0\t  d:

whenever

ð8:60Þ

We may suppose that d\p: If n [ 1d, we consider Z p½rn ðxÞ  f ðxÞ ¼ 0

p

/x ðuÞ

sin2 nu 2 du ¼ 2n sin2 u2

and estimate each of the quantities I1, I2, I3.

Z

1 n

0

Z þ 1 n

d

Z þ d

p

¼ I1 þ I2 þ I3 ;

8 Hints

543

Since sint < t for t > 0, and have

1 p sin t \ 2t

for 0\t\ p2 ; therefore for 0\t\ 1n ; we

nu2  sin2 nu p 2 p2 n 2 :  2 ¼ 2u 8 2n u 2n sin 2

Hence j I1 j 

p2 n 8

Z

1 n

0

using (8.60). We next estimate I2. For Z jI2 j 

p2 1  8 1n

j/x ðuÞjdu ¼

1 n

d

1 n

Z

1 n

j/x ðuÞjdu\

0

p2 e; 8

ð8:61Þ

 u  d; we have

j/x ðuÞj

sin2 nu p2 2 du  2n 2n sin2 u2

Z

d

1 n

j/x ðuÞ du; u2

since |sint|  1 for every t and sin1 t \ 2tp when 0\t\ p2 : Now, integration by parts for Lebesgue integrals can be justified by the usual argument, taking care to note that a product of absolutely continuous functions is absolutely continuous, whence the product rule for derivatives holds a.e. On integrating by parts, and using the fact that Ux(u) = |/x(u)| a.e., we obtain ) 2 þ Ux ðuÞdu 1 u3 1 n n ( )

 Z d p2 1 1 eu 2  Ux ðdÞ  2  Ux du using ð8:60Þ  n þ2 1 n u3 2n d n



 p2 1 1 1 2 Ux ðdÞ  2  Ux ¼  n þ 2e n  n d 2n d



 2 p e 1 1  Ux   n2 þ 2e n  using ð8:60Þ again n d 2n d p2 f2eng ¼ p2 e;  2n

p2 j I2 j  2n

(

1 Ux ðuÞ  2 u

d

Z

d

ð8:62Þ

neglecting the terms which are negative. Finally, we estimate I3. Z

p

j I3 j 

j/x ðuÞj

d

Z

p

 Mn ðdÞ d

sin2 nu 2 du 2n sin2 u2

j/x ðuÞjdu;

ð8:63Þ

544

8 Hints

 1 sin2 mu where Mn ðdÞ ¼ maxd  u  p 2n sin22 u ¼ 2n sin2 d2 , which tends to 0 as n ! ∞. 2

On using (8.61), (8.62) and (8.63), we obtain jrn ðxÞ  f ðxÞj ! 0 as n ! 1:

Definition Let f(x) = g(x) + ih(x) be a complex function on the closed bounded interval [a, b], where g(x) and h(x) are real. For each partition P: a = x0 < x1 < ⋯ < xn = b of [a, b], we set TðP; f Þ ¼

n X

kf ðxk Þ  f ðxk1 Þk;

k¼1

where || f(x) − f(y)||2 = (g(x) − g(y))2 + (h(x) − h(y))2. The number Vð½a; b; f Þ ¼ sup TðP; f Þ P

is called the total variation of f over [a, b]. Additivity over adjacent intervals as in Theorem 5.6.4 can be established by making obvious modifications in the proof of that theorem. It will be used in Problems 5.8.P9 and 5.8.P10. 5.8.P9. Show that the complex function f = g + ih satisfies Vð½a; b; f Þ  Vð½a; b; gÞ þ Vð½a; b; hÞ; is of finite variation if and only if g and h are so, and is absolutely continuous if and only if g and h are so. Hint: This is an immediate consequence of the inequalities maxfTðP; gÞ; TðP; hÞg  TðP; f Þ  TðP; gÞ þ TðP; hÞ and additivity over adjacent intervals. Definition Let g and h be continuous real-valued functions on the closed bounded interval [a, b]. The set of points (g(x), h(x)), a  x  b, in R2 is called a continuous curve and total variation V([a, b], f) of f(x) = g(x) + ih(x) is called the length of the curve. Similarly, if vðxÞ ¼ Vð½a; x; f Þ;

axb

then v(x) is called the length of the arc of the curve between the points (g(a), h(a)) and (g(x), h(x)). If the curve has finite length, it is called a rectifiable curve.

8 Hints

545

5.8.P10. Show that the curve given by g and h is rectifiable if and only if the functions are of bounded variation on [a, b]. Show that in this case, Z

½ðg0 Þ2 þ ðh0 Þ2 2 1

vðxÞ


½a;x

for all x

and equality holds for all x if and only if g and h are absolutely continuous. In particular, if g is a function of bounded variation and v(x) is its total variation R over [a, x], then v(x)
[a,x]|g′| for all x. Equality holds for all x if and only if g is absolutely continuous. Hint: The curve is rectifiable if and only if the complex function f = g + ih is of finite variation and, by Problem 5.8.P9, this is so if and only if g and h are of finite variation. When this so, each of v, g and h is differentiable a.e. and hence they are simultaneously differentiable a.e. Let x 2 (a, b) be an arbitrary point where all three are differentiable. Then for sufficiently small D > 0, by additivity of V over adjacent intervals, we have h i12 vðx þ DÞ  vðxÞ ¼ Vð½x; x þ D; f Þ
ðgðx þ DÞ  gðxÞÞ2 þ ðhðx þ DÞ  hðxÞÞ2 : h i12 Upon dividing by D and letting D ! 0, we obtain v0 ð xÞ
g0 ð xÞ2 þ h0 ð xÞ2 . This holds for almost all x 2 [a, b]. It follows by Theorem 5.5.1 that Z vðxÞ


½a;x

v0 ðuÞdu


Z ½a;x

h i1 2 2 2 ð g0 Þ þ ð h0 Þ :

Now suppose g and h are absolutely continuous. We wish to show that Z vðxÞ ¼

½a;x

h i1 2 2 2 ð g0 Þ þ ð h0 Þ

for almost all x:

Since g and h are absolutely continuous, we have Z gðbÞ  gðaÞ ¼

½a;b

g0

Z and hðbÞ  hðaÞ ¼

½a;b

h0 ;

where a  a < b  b. Hence n o12 kf ðbÞ  f ðaÞk ¼ ½gðbÞ  gðaÞ2 þ ½hðbÞ  hðaÞ2 8 2 Z 2 912 = < Z ¼ g0 þ h0 : ½a;b ; : ½a;b

ð8:64Þ

546

8 Hints

Let us write A = || f(b) − f(a)||. Assume that A > 0. Choose c1 and c2 defined by Z Z g0 ; c 2 A ¼ h0 : c1 A ¼ ½a;b

½a;b

It follows from (8.64) that c21 + c22 = 1. Moreover, Z ½c1 g0 þ c2 h0  ¼ c21 A þ c22 A ¼ A: ½a;b

But jc1 g0 þ c2 h0 j  ðc21 þ c22 Þ2 ððg0 Þ2 þ ðh0 Þ2 Þ2 1

1

¼ ððg0 Þ2 þ ðh0 Þ2 Þ2 : 1

So,

Z A¼

½a;b

Z

½c1 g0 þ c2 h0  

½a;b

Z

jc1 g0 þ c2 h0 j 

½ðg0 Þ2 þ ðh0 Þ2 2 : 1

½a;b

As A was defined to mean || f(b) − f(a)||, the above inequality says that Z 1 ½ðg0 Þ2 þ ðh0 Þ2 2 : kf ðbÞ  f ðaÞk  ½a;b

ð8:65Þ

Now, let a = x0 < x1 < ⋯ < xn = x be any partition of [a, x]. TðP; f Þ ¼ 

n X

kf ðxk Þ  f ðxk1 Þk

k¼1 n Z X k¼1

½xk1 ;xk 

½ðg0 Þ2 þ ðh0 Þ2 2 by (8:65Þ: 1

R So v(x) does not exceed [a,x][(g′)2 + (h′)2]1/2. R Conversely, suppose that v(x) = [a,x][(g′)2 + (h′)2]1/2. Appealing to Proposition 5.8.1, we obtain absolute continuity of v. The inequalities |g(x + D) − g(x)|  || f(x + D) − f(x)||  v(x + D) − v(x) and |h(x + D) − h(x)|  || f(x + D) − f(x)|| v(x + D) − v(x) now lead to the conclusion that g and h are absolutely continuous, as was to be proved. [By appealing to Theorem 5.8.5, we further obtain v′(x) = [g′ (x)2 + h′(x)2]1/2, which the problem does not ask for.] 5.8.P11. Let g be a real-valued function on [0, 1] and c(t) = t + ig(t). The length of the graph of g is, by definition, the total variation of c on [0, 1]. Show that the length is finite if and only if g is of bounded variation. Hint: Immediate from Problem 5.8.P9 above.

8 Hints

547

5.8.P12. Assume that g is continuous and increasing on [0, 1], g(0) = 0, g(1) = a. By Lg we denote the length of the graph of g. Show that Lg = 1 + a if and only if g′(x) = 0 a.e. In particular, the length of the graph of the Cantor function constructed in Example 5.5.2 is 2. Hint: If a = 0, then g, being increasing, is identically zero. Moreover, the length of the graph of g equals the length of the interval between 0 and 1, which is 1. Next, assume that a > 0. Let g′(x) = 0 almost everywhere and write f(x) = x + ig(x). Since Lg = V([0, 1], f) by definition, we have Lg ¼ Vð½0; 1; x þ igðxÞÞ  1 þ Vð½0; 1; gÞ by Problem 5:8:P9 ¼ 1 þ a: It remains to show that Lg
1 + a. Select a sequence {hn}n
1 of nonnegative real numbers satisfying hn! 0 and let e > 0. If A ¼ fx : g0 ðxÞ ¼ 0g and An ¼ fx : gðx þ hn Þ  gðxÞ  ehn g; then m(A) = 1. Moreover, x 2 A implies x 2 An for all n
nx, so A lim infAn. Then m(lim infAn) = 1 and since m(lim infAn)  lim infm(An) by Proposition 2.3.21(c), we have also lim infm(An) = 1. Hence limm(An) = 1. Now select the index p and the integer N such that m(Ap) > 1 − e and hpN > 1. Let a = infAp and b = supAp. Then b − a > 1 − e since [a, b]  Ap and m(Ap) > 1 − e. Take x1 2 Ap such that x1 − a < e1 = e/N and y1 = x1 + hp. Then y1 ¼ x1 þ hp \a þ

e þ hp : N

Claim. a + e/N + hp < b i.e. b − a > e/N + hp. It is enough to show that 1  e [ Ne þ hp , 1  hp [ eð1 þ N1 Þ , eð1 þ N1 Þ\1  hp \ð1  N1 Þ because hp [ , eðN þ 1Þ\N  1 , e\ NN1 þ 1 ; which is true: Let e1 = e/N. Take x1 2 Ap such that

1 N

548

8 Hints

x1  inffx : x 2 Ap g\e1 and let y1 = x1 + hp. Next, take x2 2 Ap such that 0  x2  inffx : x 2 Ap \ ½y1 ; 1g\e1 and let y2 = x2 + hp. Continuing with this procedure, we obtain at most N closed intervals [xi, yi] with disjoint interiors, and mð½0; 1n½xi ; yi Þ  mð½0; 1nAp Þ þ Ne1 \e þ e ¼ 2e: So, m( [ i[xi, yi]) > 1 − 2e. Now consider the partition P of [0, 1] having any partition points, besides the endpoints of [0, 1], all xi and all yi. The part X

jf ðyi Þ  f ðxi Þj

of T(P, f) is at least 1 − 2e and since X X ðyi  xi Þ  e; jgðyi Þ  gðxi Þj  e where g(1) − g(0) = a, the remaining terms of T(P, f) yield at least a − e. Hence T(P, f)
1 − 2e + a − e = 1 + a − 3e. Therefore Lg ¼ Vð½0; 1; f Þ
1 þ a: R Conversely, let Lg = 1 + a and let g = g1 + g2, where g1 ðxÞ ¼ ½0;x g0 and g2′(x) = 0 a.e,. be the Lebesgue decomposition of g. Since g1 and g2 are increasing, V([0, 1], g) = V([0, 1], g1) + V([0, 1], g2). Then 1 þ a ¼ Vð½0; 1; x þ ig1 þ ig2 Þ  Vð½0; 1; x þ ig1 Þ þ Vð½0; 1; g2 Þ  1 þ Vð½0; 1; g1 Þ þ Vð½0; 1; g2 Þ ¼ 1 þ Vð½0; 1; gÞ ¼ 1 þ a; so that V([0, 1], x + ig1) = 1 + V([0, 1], g1). It follows by Problem 5.8.P10 that Z ½0;1

n

1 þ ðg01 Þ2

o12

Z ¼

½0;1



 1 þ g01 :

But {1 + (g1′)2}1/2  1 + g1′ for almost all x. Hence the functions are equal a.e., and so, g1′ = 0 a.e. Since g′(x) = g1′(x) holds a.e., we arrive at the desired result. Rp Rp Rp 5.8.P13. Let f 2 L1[−p, p] and p f(x)dx = 0, p f(x)cosnx dx = 0, p f(x)sinnx dx = 0, n = 1, 2, … R y . Then f(x) = 0 a.e. on [−p, p]. R p Hint: Let F(y) = p f(x)dx. Observe that F is continuous and F(p) = p f(x) dx = 0 = F(−p). Also, F′(y) = f(y) by Theorem 5.8.5. Using integration by parts (see Theorem 5.8.12), we have

8 Hints

549

Z

p

p

Z FðyÞ cos ny dy ¼

p

Z

y

f ðxÞdxÞ cos ny dy  Z p y sin ny p sin ny dy ¼ 0: f ðxÞdxÞ  f ðyÞ n n p p p

 p Z ¼ ð

ð

p

Rp Similarly, p F(y)sinny dy = 0. By Corollary 4.3.6, it follows that F = 0 R y everywhere on [−p, p]. Now, p f(x)dx = 0 for every y 2 [−p, p], which implies f = 0 a.e. on [−p, p] by Problem 3.2.P14(b). Problem Set 5.9 5.9.P1. Let l be a signed measure on (X, F Þ. Show that jljðEÞ ¼ sup

n X lðEj Þ ; j¼1

where the sets Ej 2 F are disjoint and E = [ 1  j  nEj. Hint: Fix E 2 F and suppose that E = [ 1  j  nEj, where the sets Ej 2 F are disjoint. Let b denote the right-hand side of the equality to be proved. Then we have n n X X lðEj Þ ¼ l þ ðEj Þ  l ðEj Þ j¼1



j¼1 n X

n X l þ ðEj Þ þ l ðEj Þ

j¼1

¼

n X

j¼1

l þ ðEj Þ þ

j¼1

¼

n X

l ðEj Þ

j¼1

n  X

 l þ ðEj Þ þ l ðEj Þ

j¼1

¼

n X

jljðEj Þ ¼ jljðEÞ:

j¼1

Consequently, b  jljðEÞ: Consider {E \ A, E \ B}, where A, B is a Hahn decomposition of X for l. b
jlðE \ AÞj þ jlðE \ BÞj ¼ jl þ ðEÞj þ jl ðEÞj as seen in the proof of the Jordan Decomposition Theorem 5:9:16 ¼ l þ ðEÞ þ l ðEÞ ¼ jljðEÞ:

550

8 Hints

5.9.P2. If l = l1 − l2, where l1 and l2 are nonnegative measures and either l1 or l2 is finite, then l1
l+ and l2
l−, where l = l+ − l− is the Jordan decomposition of l. Hint: Let E be a measurable subset of X and E1 = E \ A, E2 = E \ B, where A, B is a Hahn decomposition of X for l. Then l−(E1) = 0 and therefore l2(E1)
l−(E1). But l1 − l+ = l2 − l− and hence l1(E1)
l+(E1). Now, l+(E) = l+(E1) + l+(E2)  l1(E1)  l1(E) since l+(E2) = 0. This shows that l1
l+, which immediately leads to l2
l− when combined with the fact that l1 − l+ = l2 − l−. 5.9.P3. Let (X, F , l) be a finite signed measure space. Show that there is an M > 0 such that |l(E)| < M for every E 2 F : Hint: Let l = l+ − l− be the Jordan decomposition of l. Since l(X) = l+(X) − l−(X) is given to be finite, both l+(X) and l−(X) must be finite. It follows that l+(X) + l−(X) is finite. Now, l+ and l− are nonnegative measures and therefore l+(E)  l+(X) and l−(E)  l−(X) for every E 2 F : Hence jlðEÞj ¼ jl þ ðEÞ  l ðEÞj  l þ ðEÞ þ l ðEÞ  l þ ðXÞ þ l ðXÞ for every E 2 F : Thus M = l+(X) + l−(X) has the required property. 5.9.P4. Show that a signed measure l on (X, F Þ is finite [resp. r-finite] if and only if |l| is finite [resp. r-finite] if and only if l+ and l− are both finite [resp. r-finite]. Hint: If l is finite, it follows as in Problem 5.9.P3 that l+ and l− are both finite and hence that |l| = l+ + l− is finite. Conversely, if |l| = l++l− is finite, it is immediate that l+ and l− are both finite and hence that l = l+ − l− is finite. The corresponding results on r-finiteness are straighforward consequences. 5.9.P5. Let l be a signed measure on (X, F Þ. Then, for every E 2 F ; the following hold: (a) l+(E) = sup{l(F): F E, F 2 F }; (b) l_(E) = sup{−l(F): F E, F 2 F }. Hint: (a) Let A, B be a Hahn decomposition of X for l. For E 2 F ; l þ ðEÞ ¼ lðE \ AÞ  supflðFÞ : F E; F 2 F g: Since B is a negative set for l, if F E with F 2 F ; then lðFÞ ¼ lðF \ AÞ þ lðF \ BÞ  lðF \ AÞ ¼ l þ ðFÞ  l þ ðEÞ:

8 Hints

551

Hence supflðFÞ : F E; E 2 F g ¼ l þ ðEÞ:

(b) Since l_ = (−l)+, assertion (b) follows from (a). 5.9.P6. Let l1 and l2 be nonnegative measures on (X, F Þ such that for all a > 0 and b > 0, there exist sets Aa,b, Ba,b 2 F satisfying Aa,b [ Ba,b = X, l1(Aa,b) < a, l2(Ba,b) < b. Show that l1 ⊥ l2. Hint: We prove this in two steps. Step 1: Given e > 0, there exist Ae, Be 2 F such that Ae [ Be = X, l1(Ae) = 0 and l2(Be) < e. Proof Consider an arbitrary p 2 N: Choose a ¼ 1p and b ¼ 2ep : In accordance with the hypothesis, we obtain sets Aa,b, Ba,b 2 F satisfying Aa;b [ Ba;b ¼ X; l1 ðAa;b Þ\a ¼ 1p ; l2 ðBa;b Þ\b ¼ 2ep : In order to emphasise the special choice of a and b, we shall denote the sets Aa,b, Ba,b respectively by Ap,e, Bp,e. Now let Ae = \ p
1Ap,e and Be = [ p
1Bp,e. Since Ap,e, Bp,e have union equal to X, the same is 1 P true of Ae, Be. Moreover, l1(Ae) = 0 and l2 ðBe Þ\ ð2ep Þ ¼ e. This completes the p¼1

proof of Step 1. Step 2: l1 ⊥ l2. Proof Consider an arbitrary p 2 N and choose e ¼ 1p. From Step 1, we obtain disjoint Ae, Be 2 F such that Ae [ Be = X, l1(Ae) = 0 and l2(Be) < e = 1p. In order to emphasise the special choice of e, we shall denote the sets Ae, Be respectively by Ap and Bp. Now let A = [ p
1Ap and B = \ p
1Bp. Since Ap, Bp have union equal to X, the same is true of A, B. Moreover, l1(A) = 0 and l2(B) = 0. Using Remark 5.9.14(a), we arrive at l1 ⊥ l2. Problem Set 5.10 5.10.P1. Let l = m + d, where m is Lebesgue measure on [0, 1] and d is the “Dirac” measure,   which is defined by setting d(E) = 1 if 0 2 E and 0 if 0 62 E. Determine dm du :   Hint: The Radon–Nikodým derivative dm du is defined [see Remark 5.10.9(a)] as being characterized a.e.[l] by the equality Z  mðEÞ ¼ E

Note that d satisfies

 dm dl; dl

for every measurable E:

ð8:66Þ

552

8 Hints

Z f dd ¼ f ð0Þfor any Lebesgue measurable f :

ð8:67Þ

X

Since v(0,1] = 1 a.e.[m], for arbitrary Lebesgue measurable E, we have Z

Z

Z

1dm ¼

mðEÞ ¼ ZE

Z

v½0;1 dm ¼ E

v½0;1 vE dm ¼ X

 v½0;1 vE dm þ v½0;1 vE ð0Þ Z ZX v½0;1 vE dm þ v½0;1 vE dd by ð8:67Þ ¼ X ZX v½0;1 vE dl because l ¼ m þ d ¼ ZX ¼ v½0;1 dl:

v½0;1 vE dm þ 0 X

¼

E

h i Taking into account (8.66), we conclude that

dm dl

¼ vð0; 1 a:e:½l:

5.10.P2. Let ( pffiffiffiffiffiffiffiffiffiffiffi

2 if x
0 x 1x if x  1 f ðxÞ ¼ gðxÞ ¼ 0 if x\0 0 if x [ 1; Z Z f dx and lðEÞ ¼ g dx; E 2 M. mðEÞ ¼ E

E

Find the Lebesgue decomposition of m with respect to l. Hint: Observe that l and m are nonnegative measures. Now, Z vðEÞ ¼ Claim: m0(E) = m1  l.

R

E \ (−∞,0)

E \ ð1;0Þ

Z f dx þ

f dx and m1(E) =

R

f dx: E \ ð0;1Þ

E \ (0, ∞)f

dx are such that m0 ⊥ l and

(i) Since m0[0, ∞) = 0 and l(−∞, 0) = 0, it follows that m0 ⊥ l. (ii) In order to show that m1  l, one needs to consider only those E 2 which satisfy E (0, ∞), because m1(−∞, 0] = 0. Let E be one such set with l(E) = 0. From the definition of l, it follows that m(E) = 0 since g > 0 on (0, ∞). Consequently, m1(E) = 0, which implies m1  l. This proves the claim.

8 Hints

553

It now follows from the obvious equality m = m0 + m1 and Definition 5.10.11 that m = m0 + m1 is a Lebesgue decomposition of m with respect to l. By Theorem 5.10.13, it is unique. Comment: The solution depends only on the fact that f is nonnegative while g is 0 on (−∞, 0) but > 0 on (0, ∞). 5.10.P3. Let l = m + d, where m is the usual Lebesgue measure on ℝ and d is the “Dirac” measure defined by

dðEÞ ¼

if 0 2 E if 0 2 6 E;

1 0

E 2 M.

Determine the Lebesgue decomposition of l with respect to m. Hint: For E 2 M, lðEÞ ¼ lðE \ Rnf0gÞ þ lðE \ f0gÞ ¼ m1 ðEÞ þ m0 ðEÞ; say, where m0 ⊥ m since m0 is concentrated at 0 and m{0} = 0. Also, m1  l. Indeed, if E 2 M and l(E) = 0, then l(E \ ℝ\{0})  l(E) = 0. Consequently, m1(E) = 0. 5.10.P4. Let l1,l2 and m be r-finite measures on (X,F Þ. Prove the following: (a) If li  m, i = 1, 2, then l1 + l2  m and



     d ðl1 þ l2 Þ dl1 dl2 ¼ þ dm dm dm

(b) If l1  l2 and l2  l1, then

h

dl1 dl2

ih

dl2 dl1

i

a:e:½m:

¼ 1 a.e.[l1] and a.e.[l2].

Hint: (a) Clearly, l1 + l2 is a r-finite measure and l1 + l2  m. For E 2 F ; Z   Z   dl1 dl2 ðl1 þ l2 ÞðEÞ ¼ l1 ðEÞ þ l2 ðEÞ ¼ dm þ dm: dm E E dm So, the huniqueness ofh thei Radon–Nikodým derivative gives the result. i R dl1 2 (b) Write dl ¼ f and dl dl ¼ g: Since l1 is nonnegative, E f dl2 = l1(E)
0 2

1

for every E 2 F : Consequently, f
0 a.e.[l2]. Let {fn}n
1 be an increasing sequence of simple nonnegative functions converging everywhere to f. It follows by using the Monotone Convergence Theorem 3.2.4 that

554

8 Hints

Z

Z fn dl2 ¼

lim

n!1

E

Z

Z f dl2

and

E

fn gdl1 ¼

lim

n!1

E

fgdl1

for every E 2 F

E

Let F 2 F : Then Z

Z vF dl2 ¼ l2 ðE \ FÞ ¼

E

E\F

Z g dl1 ¼

vF g dl1 : E

So, Z

Z fn dl2 ¼ E

fn g dl1 ;

n ¼ 1; 2; . . .

E

and therefore Z l1 ðEÞ ¼

Z f dl2 ¼

E

fg dl1 ; E

which implies    dl1 dl2 ¼1 dl2 dl1

a:e:½l1  and a:e:½l2 :

5.10.P5. (X, F ;m) be a r-finite measure space and l1, l2 be r-finite signed measures such that l1 + l2 is also a signed measure on (X, F Þ. Prove the following: (a) If li  m, i = 1, 2, then l1 + l2  m and 

     d ðl1 þ l2 Þ dl1 dl2 ¼ þ dm dm dm

a:e:½m:

(b) If l is a r-finite signed measure on (X, F Þ such that l  m, then 

   dl djlj ¼ dm dm

a:e:½m:

þ Hint: (a) Let li ¼ liþ þ l i be the Jordan decomposition of li. Since li þ  m þ þ  and li þ  m (see Proposition 5.10.4), it follows that ðl1 þ l2 Þ  m and  ðl 1 þ l2 Þ  m. Consequently, it follows from Problem 5.10.P4 that

8 Hints

555

  þ   þ    þ d l1 þ l2þ dl1 dl2 ¼ þ a:e: ½m dm dm dm and

        d l1 þ l dl dl2 2 1 þ a:e: ½m: ¼ dm dm dm

Since l1 + l2 is a signed measure, it follows first of all that ðl1þ þ l2þ Þ(E) and  ðl 1 þ l2 Þ(E) cannot both be ∞ for any E 2 F : For, such an E would satisfy either + l1 (E) = ∞ or l2+(E) = ∞ and we need handle only the possibility that l1+(E) = ∞. When this occurs, we have l1(E) = ∞ together with l1−(E) = 0, which leads to l2−(E) = ∞ and hence to l2(E) = −∞, contradicting the hypothesis that (l1 + l2)(E) is well defined.  Next, since (l1+ + l2+)(E) and ðl 1 þ l2 Þ(E) cannot both be ∞ for any E 2 F ; it follows from the equalities displayed above that, for every E 2 F with |(l1 + l2) (E)| < ∞, we have      ðl1 þ l2 ÞðEÞ ¼ l1þ þ l2þ ðEÞ  l 1 þ l2 ðEÞ   Z   þ Z    d l1 þ l2þ d l1 þ l 2 ¼ dm  dm dm dm E E Z  þ    Z  þ    dl1 dl1 dl2 dl2 ¼  dm þ  dm dm dm dm dm E E   Z  Z  dl1 dl2 dm þ dm: ¼ E dm E dm

ð8:68Þ

In the last step, we have used the easily proven equality [employing the Radon– Nikodým Theorem and Example 5.9.18] Z  E

   Z   dl þ dl dl dm;  dm ¼ dm dm E dm

where m is a r-finite measure and l is a r-finite signed measure, m  l. Now, from (8.68) and the uniqueness of the Radon–Nikodým derivative, 

     d ðl1 þ l2 Þ dl1 dl2 ¼ þ a:e: ½m dm dm dm

(b) l  m if and only if |l|  m, using Proposition 5.10.4. So, for E 2 F ; we have

556

8 Hints

Z  lðEÞ ¼ E

 dl dm dm

Z djlj jljðEÞ ¼ dm dm:

and

E

Let A, B be a Hahn decomposition of X for l. Then l+(E) = l(E) = |l|(E) for every E A and l−(E) = −l(E) = |l|(E) for every E B. Thus, 

   dl djlj ðxÞ ¼ ðxÞ a:e: ½m for x 2 A dm dm

and  

   dl djlj ðxÞ ¼ ðxÞ a:e: ½m for x 2 B: dm dm

Hence     dl ðxÞj ¼ djlj ðxÞ a:e: ½m: dm dm 5.10.P6. Let (X, F ;l) be a r-finite measure space. Let m be a measure on (X, F Þ for which the conclusion of the Radon–Nikodým Theorem holds. Prove that m is r-finite. Hint: It follows from the hypothesis that Z mðXÞ ¼

f dl; X

where f is a real-valued nonnegative measurable function on X. Since l is r-finite, we have X = [ n
1Xn, l(Xn) < ∞. Let Yk = {x 2 X: 0  f(x)  k}. Then X can be written as X = [ k
0Yk, considering that f is real-valued. Consequently, X = [ n
1,k
0(Xn \ Yk) and Z vðXn \ Yk Þ ¼

Xn \ Yk

f dl  klðXn Þ\1:

This implies that m is r-finite. Problem Set 5.11 5.11.P1. Suppose l is a measure on I and l* is the outer measure induced by l. Let Ir denote the family of countable unions of sets of I. Given any set E and any e > 0, show that there is a set A 2 Ir such that E A and

8 Hints

557

l ðAÞ  l ðEÞ þ e: Hint: By the definition of l*, there exists a sequence {Ei}i
1 of sets Ei 2 I such that 1 P E [ i
1Ei and lðEi Þ\l ðEÞ þ e. Set A = [ i
1Ei. Then A 2 Ir and by i¼1

countable subadditivity of l* [Remark 5.11.16(c)] and the fact that l* coincides with l on I [Theorem 5.11.21], we have l ðAÞ 

1 X

l ðEi Þ ¼

i¼1

1 X

lðEi Þ\l ðEÞ þ e:

i¼1

5.11.P2. Suppose l is a measure on I and l* is the outer measure induced by l. Let Ir denote the family of countable unions of sets of I and Ird denote the family of countable intersections of sets of Ir. Given any set E, show that there is a set A 2 Ird such that E A and l*(E) = l*(A). Hint: For every positive integer n, there exists a set An 2 Ir with E An and l*(An)  l*(E) + 1n. Let A = \ n
1An. Then A 2 Ird and E A. Since A An, we have l*(A)  l*(An)  l*(E) + 1n for each n. Therefore l*(A)  l*(E). But E A and so, l*(E)  l*(A) by monotonicity. Hence l*(E) = l*(A). 5.11.P3. Let F be a real-valued bounded increasing right continuous function on R and for any right closed interval (a, b], define l((a, b]) = F(b) − F(a). Show that: (i) Let {Ei}1  i  n be disjoint right closed intervals (Ei = (ai, bi], 1  i  n) such that [ 1  i  nEi I, where I is a right closed interval. Then n P lðEi Þ  lðIÞ. i¼1

(ii) If the right closed interval (a0, b0] is contained in the union of a sequence of right closed intervals (ai, bi], i = 1, 2, …, then lðða0 ; b0 Þ 

1 X

lððai ; bi Þ;

i¼1

moreover, if the intervals in the sequence (ai, bi], i = 1, 2, …, are disjoint and (a0, b0] = [ i
1(ai, bi], then lðða0 ; b0 Þ ¼

1 X

lððai ; bi Þ:

i¼1

Hint: Modify the proofs of Lemmas 5.11.3 and 5.11.5.

558

8 Hints

5.11.P4. Let h be a real-valued bounded increasing left continuous function on R such that h′ = 0 a.e. and l = lh be the Lebesgue–Stieltjes measure induced by h. Suppose also that there is a closed bounded interval I such that h is constant on each of the unbounded open intervals that constitute the complement Ic. Show that l and Lebesgue measure m (on Borel sets) are mutually singular (see Definition 5.9.13). Hint: Since h′ = 0 a.e., there exist disjoint Borel sets A0 and B0 with union R [i.e., mutually complementary Borel subsets of R such that mðA0 Þ ¼ 0 and h′(x) = 0 for all x 2 B0. We may assume that B0 contains the complement Ic = R\I of the interval I. Note that lðI c Þ ¼ 0; because h is constant on each of the unbounded open intervals that constitute the complement Ic. Let a > 0 and b > 0. Each x 2 B0 is contained in a closed bounded interval [s, t] of arbitrarily small length such that h(t) − h(s)  (b/m(I))•(t − s). If [s, t] lies outside I, then h(t) − h(s) = 0. Such intervals provide a Vitali cover of B0 \ I in the sense defined in Problem 5.4.P1 and therefore by the Vitali Covering Theorem of that problem, there exists a finite disjoint collection of intervals in the Vitali cover, whose union B1 satisfies mððB0 \ I ÞnB1 Þ\a: If we replace each closed interval [a,b] by the corresponding left closed interval [a, b), their union satisfies the same inequality. We may thus assume that B1 is the union of the left closed intervals. Then we can assert on the basis of the disjointness assured by the Vitali Covering Theorem that

 b lðB1 Þ   mðB1 Þ\b: mðIÞ Since l(Ic) = 0, it follows that l(B1 [ Ic) < b. Now take A ¼ A0 [ ððB0 \ I ÞnBÞ

and B ¼ B1 [ I c :

Then we have mðAÞ  mðA0 Þ þ mððB0 \ I ÞnB1 Þ ¼ 0 þ mððB0 \ I ÞnB1 Þ\a

8 Hints

559

and lðBÞ  lðB1 Þ þ lðI c Þ ¼ lðB1 Þ þ 0\b: In view of Problem 5.9.P6, all we need prove is that A [ B = R; which we do in the next paragraph. Consider any x 2 R such that x 62 A; we must show that x 2 B. Since x 62 A, we must have x 62 A0, which implies x 2 B0, considering that A0,B0 are mutually complementary. By the definition of A, we must also have x 62 (B0 \ I) \ Bc = B0 \ (I \ Bc). However, as already noted, x 2 B0. It follows that x 62 (I \ Bc). But the definition of B implies I \ Bc = I \ (B1 [ Ic)c = I \ Bc1 = Bc. Since we have shown that x 62 (I \ Bc), it is now immediate that x 2 B, as desired. This completes the argument that A [ B = R; which is all that we needed to prove. 5.11.P5. Suppose f is an increasing function on [a, b] that is left continuous and let f = g + h be a Lebesgue decomposition in accordance with Problem 5.8.P7. Extend each of the three functions to R so as to take the same value on (−∞, a) as at a and to take any constant value on (b, ∞) not less than the value at b, but so chosen that the equality f = g + h holds on (b, ∞). Prove the following: R (a) There is some constant c such that g(x) = c + [a,x] f 0 for all x 2 R: (b) The functions f, g and h on R are also bounded, increasing and left continuous; moreover, f = g + h on R: (c) If lf, lg and lh are the Lebesgue–Stieltjes measures induced by f, g and h respectively, then lf = lg+ lh and this equality constitutes the unique Lebesgue decomposition of lf with respect to Lebesgue measure m. Hint: (a) As in the solution of Problem 5.8.P7, g is given on [a, b] by g(x) = c + R 0 f ½a;x , where c is some constant. The manner of extending the functions to all of R makes f 0 vanish outside [a, b] and therefore g is given by the same equality on all of R: (b) Since f is increasing on [a, b], the derivative f 0 must be nonnegative wherever it exists, and therefore g is increasing on [a, b]. By the inequality (5.35) of Theorem 5.5.1, the same is true of h. Moreover h is left continuous on (a, b] because f and g both are. The manner of extending the functions to all of R ensures that the functions f, g and h, when extended to all of R; continue to be increasing and left continuous, and to satisfy f = g + h. (c) The equality f = g + h shows by an easy computation that when E is a left closed interval, lf(E) = lg(E) + lh(E). By the uniqueness part of Theorem 5.11.22, we obtain lf = lg+ lh. Since g is absolutely continuous, it follows from Remark 5.11.24(c) that lg m. By Problem 5.8.P7, in accordance with which the functions g and h are formed, we have h′ = 0 a.e. on [a, b] and hence also on R: It follows by Problem 5.11.P4 that lh and m are mutually singular. By the uniqueness part of Theorem 5.10.13, lf = lg+ lh is the unique Lebesgue decomposition of lf with respect to Lebesgue measure.

560

8 Hints

5.11.P6. Let l be a finite nonnegative measure on ðR; BÞ and f = fl. Suppose [a, b] is an arbitrary interval and / is the function on R obtained by extending the restriction of f to [a, b] so as to be equal to f(a) on (−∞,a) and equal to f(b+) on (b,∞). Then obviously, / is bounded, increasing and left continuous. Show that the Lebesgue–Stieltjes measure l/ induced by / is given by l/(E) = l(E \ [a, b]) for every Borel set E. Hint: Since l(E \ [a, b]) defines a finite measure on Borel sets E, it is sufficient in view of the uniqueness part of Theorem 5.11.22 to show that l/(E) = l(E \ [a, b]) for every left closed interval E = [a, b). That is, l([a, b) \ [a, b]) = /(b) − /(a) whenever a  b. Since this is trivial when a = b, we may assume a < b. Case 1: a < a. Here, f(a) = /(a). Subcase b  a. This makes ½a; bÞ \ ½a; b ¼ £

and

/ðbÞ ¼ f ðaÞ:

So, l([a, b) \ [a, b]) = 0 = f(a) − f(a) = /(b) − /(a), as desired. Subcase a < b  b. Here, ½a; bÞ \ ½a; b ¼ ½a; bÞ and f ðbÞ ¼ /ðbÞ: So, l([a, b) \ [a, b]) = f(b) − f(a) = /(b) − /(a). Subcase b < b. In this subcase, ½a; bÞ \ ½a; b ¼ ½a; b and

f ðb þ Þ ¼ /ðbÞ

So, on the basis of Remark 5.11.24(d), lð½a; bÞ \ ½a; bÞ ¼ lð½a; bÞ ¼ lð½a; bÞÞ þ lðfbgÞ ¼ f ðbÞ  f ðaÞ þ lðfbgÞ ¼ f ðbÞ  f ðaÞ þ f ðb þ Þ  f ðbÞ ¼ f ðb þ Þ  f ðaÞ ¼ /ðbÞ  /ðaÞ: Case 2: a  a  b. Here, f(a) = /(a). Subcase b  b. This makes ½a; bÞ \ ½a; b ¼ ½a; bÞ

and

f ðbÞ ¼ /ðbÞ:

So, l([a, b) \ [a, b]) = l([a, b)) = f(b) − f(a) = /(b) − /(a). Subcase b < b. Here, ½a; bÞ \ ½a; b ¼ ½a; b and So, on the basis of Remark 5.11.24(d),

f ðb þ Þ ¼ /ðbÞ:

8 Hints

561

lð½a; bÞ \ ½a; bÞ ¼ lð½a; bÞ ¼ lð½a; bÞÞ þ lðfbgÞ ¼ f ðbÞ  f ðaÞ þ lðfbgÞ ¼ f ðbÞ  f ðaÞ þ f ðb þ Þ  f ðbÞ ¼ f ðb þ Þ  f ðaÞ ¼ /ðbÞ  /ðaÞ: Case 3: b < a. Here, /(a) = /(b) = f(b+) and [a, b) \ [a, b] = £. So, l([a, b) \ [a, b]) = 0 = f(b+) − f(b+) = /(b) − /(a). 5.11.P7. Let l be a finite nonnegative measure on ðR; BÞ and f = fl. If l and Lebesgue measure m (on B) are mutually singular, show that f 0 = 0 a.e. Hint: Let [a, b] be an arbitrary interval and / be the function on R obtained as in Problem 5.11.P6 by extending the restriction of f to [a, b] so as to be equal to f(a) on (−∞, a) and equal to f(b+) on (b, ∞); it is sufficient to show that /′ = 0 a.e. Note that f(b+)
f(b), so that the constant value assigned to / on (b,∞) is not less than /(b) and thus / is an extension that fulfils the requirement of Problem 5.11.P5 as well. Let f = g + h be a Lebesgue decomposition of the restriction of f to [a, b] in accordance with Problem 5.8.P7. Then h′ = a.e. on [a, b]. The function f already has an extension / to R of the kind required in Problem 5.11.P5. Now extend g and h also to R in a corresponding manner while ensuring the equality / = g + h. Then h′ = 0 a.e. on R: Moreover, the Lebesgue–Stieltjes measures l/,lg and lh induced by /, g and h respectively satisfy l/ = lg+ lh and this equality constitutes the unique Lebesgue decomposition of l/ with respect to Lebesgue measure m. Consider any Borel set E. By Problem 5.11.P6, l/ ðEÞ ¼ lðE \ ½a; bÞ: Since l(E \ [a, b])  l(E) for every Borel set E, the measures l/ and m must be mutually singular. Since l/ = 0 + l/, this equality constitutes the unique Lebesgue decomposition of l/, and hence lg = 0. By Remark 5.11.24(b), the function g must be a constant. It is now immediate that f′ = g′ + h′ = 0 a.e. 5.11.P8. Let l be a finite nonnegative measure on ðR; BÞ and l = m + k be its Lebesgue decomposition with respect to Lebesgue measure m. Then fm is absolutely continuous on any interval [a, b] and fk′ = 0 a.e. Also, fl = fm + fk. [Remark This means the equality fl = fm + fk is a Lebesgue decomposition of fl on any interval [a, b] in accordance with Problem 5.8.P7.] Hint: The equality fl = fm + fk is trivial from the definition of the function fl associated with a measure l. Since m  m, the function fm is absolutely continuous by Remark 5.11.24(c). And since k and m are mutually singular, fk′ = 0 a.e. by Problem 5.11.P7. Problem Set 6.1 6.1.P1. If f 2 Lp(X), ∞ > p > 1, then k f kp = sup{ where q is given by 1p þ 1q ¼ 1:

R X

| fg|dl: g2Lq(X), kgkq = 1},

562

8 Hints

R If f 2 L1(X), then k f k1 = sup{ X| fg|dl: g 2 L∞(X), kgk1 = 1}. Note that this differs from Proposition 6.1.3 in that the absolute value is inside the integral rather than outside it. Hint: We assume k f kp 6¼ 0; because otherwise there is nothing to prove. By Hölder’s Inequality, Z X

jfgjdl  k f kp kgkq  k f kp if kgkq ¼ 1:

Hence

Z kf kp
sup

 jfgj dl : g 2 L ðXÞ; kgkq ¼ 1 : q

X

Taking into account that kf kp 6¼ 0; let p

g ¼ j f jp1 kf kp q : Then Z

Z jgjq dl ¼

X

X

j f jðp1Þq k f kp p dl ¼ 1;

so that kgkq ¼ 1: But Z

Z jfgjdl ¼

X

X

p

pp

jf jp dlk f kp q ¼ k f kp q ¼ k f kp :

For the assertion about f 2 L1(X), proceed as above but take g = 1 everywhere. R 6.1.P2. If f 2 L∞(X), then k f k1 = sup{ X| fg|dl: g 2 L1(X), kgk1 ¼ 1}, provided that every set of positive measure contains a subset of finite positive measure. Show also that this is false if the proviso about the measure is omitted. Note that this differs from Proposition 6.1.2(v) in that the absolute value is inside the integral rather than outside it. Hint: The case when k f k1 ¼ 0 is trivial. Given any positive e < k f k1 , let E X be a measurable subset of positive measure on which | f | > k f k1  e [ 0: Since every set of positive measure contains a subset of finite positive measure, we may take E to have finite positive measure, so that vE 2 L1(X). Consequently, the function g = (sgnf)vE 2 L1(X) and is nonzero. Besides,

8 Hints

563

Z

Z

Z

jgjdl ¼ X

jðsgnf Þf jdl ¼ E

E

  jf jdl
k f k1  e kgk1 :

R

Replacing g by g/kgk1 , we get X| fg|dl
k f k1  e and kgk1 ¼ 1: Since the positive number e\k f k1 is arbitrary, we arrive at

Z sup X

 jfgjdl : g 2 L1 ðXÞ; kgk1 ¼ 1
k f k1 :

The reverse inequality is trivial. For the second part, let X consist of two points called a and b. Consider the measure which is 1 on {a} and ∞ on {b}. Then L1 consists of functions that are 0 at b. The function f defined to be 0 at a and 1 at b belongs to L∞ with k f k1 ¼ 1: However, R 1 X| fg|dl = 0 for every g 2 L (X). 6.1.P3. Let {fn}n
1 be a sequence of functions in L∞(X). Prove that {fn}n
1 converges to f 2 L∞(X) if and only if there is a set E of measure zero such that fn! f uniformly on Ec. Hint: Let En = {x 2 X: | fn(x) − f(x)| > kfn  f k1 } and E = [ n
1En. Since 1 P m(En) = 0, it follows that mðE Þ  mðEn Þ ¼ 0. Then | fn(x) − f(x)|  kfn  f k1 n¼1

for all n and all x 2 Ec. The hypothesis that kfn  f k1 \ e for all n
n0 implies | fn(x) − f(x)| < e for all n
n0 and all x 2 Ec, that is, {fn}n
1 converges to f uniformly on Ec. On the other hand, suppose {fn}n
1 converges to f uniformly a.e., that is, for all e > 0, there exists an n0 such that n
n0 implies jfn ðxÞ  f ðxÞj\e for all x 2 Ec ; where m(E) = 0; that is, n
n0 implies mðfx 2 X : jfn ðxÞ  f ðxÞj
egÞ ¼ 0: Then, for n
n0, kfn  f k1  e: 6.1.P4. Suppose f is integrable on R and for fixed h 2 R; let fh(x) = f(x + h) be a translate of f. Show that fh is also integrable and that

564

8 Hints

Z

Z

R

fh dm ¼

f dm: R

Hint: Clearly, (fh)+ = (f +)h and (fh)− = (f −)h; so it is sufficient to prove the result for f
0. m P Let s ¼ ck vEk , where ck > 0 and Ek is a measurable set of finite measure, k¼1

1  k  m. Then sh ¼

m X

ck ðvEk Þh ¼

k¼1

m X

ck vEk h

k¼1

and Z R

sh dm ¼

m X

ck mðEk  hÞ ¼

k¼1

m X

Z ck mðEk Þ ¼

s dm; R

k¼1

since Lebesgue measure is translation invariant (Proposition 2.3.23). By Theorem 2.5.9, there exists an increasing sequence {sn} of measurable simple functions such that sn! f. But then the sequence {(sn)h} is also increasing and (sn)h! fh. So, by the Monotone Convergence Theorem, and the result of the paragraph above, we have Z

Z R

fh dm ¼ lim

n!1

R

Z

Z ðsn Þh dm ¼ lim

n!1

R

sn dm ¼

f dm: R

6.1.P5. Show that if | fn|  M a.e. and fn! f in Lp(X), where l(X) < ∞ and p
1, then fn! f in Lp′(X) for 1  p′ < ∞. Hint: Suppose p′ < p and p1 = p/p′; let q1 be such that p11 þ q11 ¼ 1: Then Z X

  0 0 0 1 1  jfn  f jp dl  jfn  f jp  ðlðXÞÞq1 ¼ ðkfn  f kp Þp ðlðXÞÞq1 ! 0 p1

as n ! ∞, that is, fn! f in Lp′(X). Assume p′ > p. Since | fn|  M a.e., we have | fn − f|  2M a.e. for all n. Now, 0

0

ðkfn  f kp0 Þp  ð2MÞp p kfn  f kpp ! 0 as n ! ∞.

8 Hints

565

Remark The reader will note that the hypothesis | fn|  M a.e. is used only for p′ > p and the hypothesis m(X) < ∞ is used only for p′ < p. R 6.1.P6. Let I = [a, b]  R and 1 < p < ∞. If f 2 Lp(I) and F(x) = C + [a,x] f dm, then show that sup P

n1 X jFðxk þ 1 Þ  Fðxk Þjp

ðxk þ 1  xk Þp1

k¼0

Z j f jp dm;



ð8:69Þ

I

where P : a = x0 < x1 < ⋯ < xn = b is a partition of [a, b]. Conversely, if K ¼ sup

n1 X jFðxk þ 1 Þ  Fðxk Þjp

P

k¼0

ðxk þ 1  xk Þp1

\1;

ð8:70Þ

p where R P is a partition of I, Rthen pthere exists a / 2 L (I) such that F(x) = C + [a,x]/dm; show also that I |/| dm  K. Use this to show that equality holds in (8.69). Hint: By Hölder’s Inequality,

Z Z 1 1 q jFðxk þ 1 Þ  Fðxk Þj ¼ f dm  ðxk þ 1  xk Þ ð j f jp dmÞp ; ½xk ;xk þ 1  ½xk ;xk þ 1  where

1 p

þ 1q ¼ 1: Hence jFðxk þ 1 Þ  Fðxk Þjp ðxk þ 1  xk Þp1

Z 

½xk ;xk þ 1 

j f jp dm;

which is to say, inequality (8.69) holds. For the converse, note that the given condition (8.70) is strengthened if some of the terms in it are omitted. So, for arbitrary disjoint open intervals (ak, bk), k = 1, 2, 3, …, n, contained in [a, b], we have n X jFðbk Þ  Fðak Þjp k¼1

By Hölder’s Inequality,

ðbk  ak Þp1

 K:

566

8 Hints n X

jFðbk Þ  Fðak Þj ¼

k¼1

n X jFðbk Þ  Fðak Þj 11p

ðbk  ak Þ

k¼1

( 

1

ðbk  ak Þq

n X jFðbk Þ  Fðak Þjp

)1p (

ðbk  ak Þp1

k¼1

n X

)1q ðbk  ak Þ

:

k¼1

Therefore n X

( jFðbk Þ  Fðak Þj  K

1 p

n X

k¼1

)1q ðbk  ak Þ

;

k¼1

which implies F is absolutely continuous and is, therefore, representable in the form Z FðxÞ ¼ C þ

½a;x

/ dm;

where /2 L1(I) by Corollary 5.8.3. The reader will note that / = F′ a.e. by Theorem 5.8.5 (although this will play no role in the argument). It remains to show that / 2 Lp(I). For this purpose, we divide [a, b] into n equal parts by the points ðnÞ

xk ¼ a þ

k ðb  aÞ; n

k ¼ 0; 1; 2; . . .; n

and introduce the function fn(t) by setting fn ðtÞ ¼

8 ðnÞ ðnÞ < Fðxk þ 1 ÞFðxk Þ ðnÞ

:

ðnÞ

xk þ 1 xk

ðnÞ

ðnÞ

xk \t\xk þ 1 ðnÞ

t ¼ xk :

0

We shall show that lim fn ¼ / almost everywhere. Let x be a point which is not a point n!1

of subdivision and for which F′(x) exists and is finite. Thus x lies in some open interval ðnÞ ðnÞ ðnÞ ðxkn ; xkn þ 1 Þ for all natural numbers n. Since xkn þ 1  xðnÞ kn ¼ ba n ! 0 as n ! 1, it follows that each of the expressions ðnÞ

Fðxkn þ 1 Þ  FðxÞ ðnÞ

xkn þ 1  x tends to F′(x) as n ! ∞. However,

ðnÞ

and

Fðxkn Þ  FðxÞ ðnÞ

xkn  x

ð8:71Þ

8 Hints

567 ðnÞ

ðnÞ

Fðxk þ 1 Þ  Fðxk Þ

fn ðxÞ ¼

ðnÞ

ðnÞ

xk þ 1  xk

;

and accordingly, the number fn(x) lies between the numbers (8.71). Therefore lim fn ðxÞ ¼ F 0 ðxÞ. By Fatou’s Lemma,

n!1

Z

Z j/jp dm  sup

I

I

p f dm n

and since Z jfn jp dm ¼ I

n1 Z X k¼0

p ðnÞ ðnÞ n1 Fðx Þ  Fðx Þ X kþ1 k  K; jfn jp dm ¼  ðnÞ ðnÞ ðnÞ ðnÞ p1 ½xk ;xk þ 1  k¼0 xk þ 1  xk

it follows that Z j/jp dm  K; I

completing the proof of the converse. We now proceed to prove that equality holds in (8.69). Since f 2 Lp(I), inequality (8.69) implies that condition R (8.70) holds. R It follows that there exists a / 2 Lp(I) such that F(x) = C + [a,x]/dm and I |/|pdm  K. However, it is given that R R R F(x) = C + [a,x] f dm and therefore / = f a.e. Hence I | f |pdm = I |/|pdm  K, which is precisely the reverse of inequality (8.69). Thus, equality must hold in (8.69). 6.1.P7. (Chebychev’s Inequality) Let f 2 Lp(X), where 1  p < ∞. Show that, for every k > 0, Z j f jp dl:

kp lðfx : jf ðxÞj [ kgÞ  X

Moreover, lim kp lðfx : jf ðxÞj [ kgÞ ¼ 0:

k!1

Hint: Let Ak = {x: | f(x)| > k}. Then Ak is measurable and kp vAk  | f |p a.e. So,

568

8 Hints

Z

Z k vAk dl ¼ k lðfx : jf ðxÞj [ kgÞ  p

X

j f jp dl:

p

X

Also, Z j f jp dl ! 0 as k ! 1

kp lðfx : jf ðxÞj [ kgÞ  Ak

since f 2 Lp(X). 6.1.P8. Let {fn}n
1 be a sequence in Lp(X), 1  p  ∞, which converges to a function f in Lp(X). Then show that, for each g in Lq(X), where 1p þ 1q ¼ 1; we have Z

Z fg dm ¼ lim

X

n!1

fn g dm: X

R Hint: The mapping that carries f 2 Lp(X) into X fg dm is continuous by Proposition 6.1.3 if 1  p < ∞ and by Proposition 6.1.2(iv) if p = ∞. Problem Set 6.2 6.2.P1. For the sequence {fk}k
1of Example 6.2.17, determine whether it converges (a) almost uniformly (b) in measure. Hint: (a) Yes (b) Yes. 6.2.P2. Suppose that {fk}k
1is a sequence of nonnegative measurable functions defined on X that converges in measure to f. Prove that Z Z f dl  lim inf fk dl: X

R

X

R

Hint: Suppose that lim inf X fk dl < X f dl < ∞. Then there exists a suitable R R d > 0 and a subsequence {fi}i
1of {fk}k
1such that X fi dl < X f dl − d for all ae

i. By Corollary 6.2.7, we can find a further subsequence {fj}j
1such that fj ! f . But then by Fatou’s Lemma 3.2.8, we have Z Z Z f dl  lim inf fj dl  f dl  d: X

X

X

This is a contradiction. R R On the other hand, suppose that lim inf X fk dl < X f dl = ∞. Then there exists R a constant M > 0 and a subsequence {fi}i
1of {fk}k
1such that X fi dl < M for all i. By Corollary 6.2.7, we can find a further subsequence {fj}j
1 such that ae fj ! f . But then by Fatou’s Lemma 3.2.8, we have

8 Hints

569

Z

Z f dl  lim inf X

fj dl  M: X

This is again a contradiction. 6.2.P3. If a sequence {fk}k
1 of measurable functions is Cauchy in measure and there exists a measurable function f to which a subsequence {fj}j
1 converges in meas measure, then fk !f . Hint: The following is the key to the result: a a Xðjfk  f j [ aÞ Xð fk  fj [ Þ [ Xð fj  f [ Þ: 2 2 6.2.P4. Let l(X) < ∞ and {fk}k
1 be a sequence of measurable functions such that, 1 P meas for every e [ 0; lðXðj fk  f j
eÞÞ\1. Show that fk !f . n¼1

Hint: Since the series of nonnegative terms is convergent, it follows that, for large k, meas lðXðj fk  f j
eÞÞ\e; i:e: fk !f . R 6.2.P5. Let l(X) < ∞. Define qðf ; gÞ ¼ X 1 þj f jgj f gj dl for every pair of measurable functions f and g. Show that (a) ∞ > q(f, g)
0, q(f, g) = q(g, f) and q(f, g) = 0 if and only if f = g a.e.; (b) q(f, h)  q(f, g) + q(g, h) [triangle inequality]; Remark This means q is a peusdometric in the sense of Definition 1.3.3. meas (c) fk ! f if and only if q(fk, f) ! 0 as k ! ∞; (d) a sequence {fk}k
1 of measurable functions is Cauchy in measure if and only if it is Cauchy in the sense of the pseudometric q; (e) setting k f k ¼ q (f, 0) does not provide a norm on the space of equivalence classes of measurable functions that agree a.e., except in the trivial case when l(X) = 0. Hint: (a) Clearly, q(f, g)
0 and q(f, g) = q(g, f). Since l(X) < ∞, we have q(f, g) < ∞. Also, q(f, g) = 0 if and only if 1 þj f jgj f gj ¼ 0 a.e., which is equivalent to f = g a.e. (b) For all real a and b, we have 1 þjajaþþbjbj  1 þjajjaj þ 1 þjbjjbj : [See Theorem 1.1.2 of [26, p. 24]]. Put a = f − g and b = h − g and integrate to obtain the required triangle inequality. (c) Let fk and f all be measurable functions and set Xk = X(| fk − f| > a), where a > 0. On using the fact that the function given by 1 þt t ; t [ 0; is increasing, we obtain

570

8 Hints

Z jfk  f j jfk  f j dl
dl qðfk ; f Þ ¼ 1 þ f 1 þ  f j j jfk  f j k X Xk Z a a dl ¼ lðXk Þ: [ 1 þ a 1 þ a Xk Z

meas

So, if q(fk, f) ! 0, then l(Xk) ! 0, that is fk ! f : meas

On the other hand, suppose fk ! f : Now we have Z qðfk ; f Þ ¼

jfk  f j dl
1 þ jfk  f j Xk

Z

jfk  f j dl 1 þ jfk  f j XnXk

\lðXk Þ þ elðXnXk Þ\lðXk Þ þ elðXÞ: meas

Since fk ! f , we have l(Xk) ! 0, and therefore qðfk ; f Þ\eð1 þ lðXÞÞ for all large enough k; that is, q(fk, f) ! 0. (d) The argument is similar to (c). (e) Let f be the constant function 1. Then q(f, 0) = 12l(X) but q(2f, 0) = 2q(f, 0) when l(X) 6¼ 0. 6.2.P6. For any two measurable functions f and g on X, define

2 3

l(X) 6¼

qðf ; gÞ ¼ inffc þ lðXðjf  gj [ cÞÞ : c [ 0g: Show that, if either l(X) < ∞ or f, g, fk 2 L∞, then (a) ∞ > q(f, g)
0, and q(f, g) = 0 if and only if f = g a.e.; also, q(f, g) = q(g, f); (b) q(f, h)  q(f, g) + q(g, h) [triangle inequality]; Remark This means q is a pseudometric in the sense of Definition 1.3.3. meas (c) fk ! f if and only if q(fk, f) ! 0 as k ! ∞; (d) a sequence {fk}k
1 of measurable functions is Cauchy in measure if and only if it is Cauchy in the sense of the pseudometric q; (e) if A X is measurable and a > 0, then q(avA, 0) = min{a, l(A)}; (f) setting k f k ¼ q(f, 0) does not provide a norm on the space of equivalence classes of measurable functions that agree a.e., except in the trivial case that no subset A X satisfies 0 < l(A) < ∞. Hint: (a) Clearly, q(f, g)
0 and q(f, g) = q(g, f). Now, q(f, g) < ∞ if either l(X) < ∞ or f, g 2 L∞; in fact, q(f, g)  kf  gk1 if f, g 2 L∞. The rest of the arguments require only that q(f, g) always be finite and therefore remain valid when

8 Hints

571

either l(X) < ∞ or all functions are in L∞. Assume q(f, g) = 0 and consider any e > 0. There is a sequence {ck}k
1 of positive numbers such that ck \

e e and lðXðjf  gj [ ck ÞÞ\ k for each k: k 2 2

The inequality ck \ 2ek for each k implies that ck! 0. Therefore X(f 6¼ g) = 1 P X(| f − g| > 0) = [ k
1X(| f − g| > ck) and hence lðXðf 6¼ gÞÞ  lðXðj f  gj [ ck ÞÞ 

1 P k¼1

k¼1 e 2k

¼ e. Consequently, l(X(f 6¼ g)) = 0.

Conversely, assume l(X(f 6¼ g)) = 0, i.e. l(X(| f − g| > 0)) = 0, and consider any e > 0. Then l(X(| f − g| > 0)) < 2e and therefore 2e + l(X(| f − g| > 0)) < e. But this implies c + l(X(| f − g| > c)) < e when c = 2e : Therefore inf{c + l(X(| f − g| > c)): c > 0} < e, which means q(f, g) < e, by definition of q(f, g). Since this holds for any e > 0, it follows that q(f, g) = 0. (b) It is obvious from the definition that q(f, g) = q(f − g, 0). Therefore the triangle inequality will follow if we can show that q(f + g, 0)  q(f, 0) + q(g, 0). To establish this, consider any e > 0. Then there exist c, c′ > 0 such that e e c þ lðXðj f j [ cÞÞ\qðf ; 0Þ þ and c0 þ lðXðjgj [ c0 ÞÞ\qðg; 0Þ þ : 2 2 This implies ðc þ c0 Þ þ lðXðj f j [ cÞÞ þ lðXðjgj [ c0 ÞÞ\qðf ; 0Þ þ qðg; 0Þ þ e: Now, | f + g|(x) > c + c′ ) [| f |(x) > c or |g|(x) > c′] and therefore lðXðjf þ gj [ c þ c0 ÞÞ  lðXðj f j [ cÞÞ þ lðXðjgj [ c0 ÞÞ; so that ðc þ c0 Þ þ lðXðjf þ gj [ c þ c0 ÞÞ\qðf ; 0Þ þ qðg; 0Þ þ e: It follows that

qðf þ g; 0Þ\qðf ; 0Þ þ qðg; 0Þ þ e:

Since this holds for any e > 0, we have q(f + g, 0)  q(f, 0) + q(g, 0), as required. (c) To begin with, observe that by definition of q(f, g), the inequality q(fk, f) < e means inf{c + l(X(| fk − f| > c)): c > 0} < e, and by definition of infimum, this means

572

8 Hints

c0 þ lðXðjfk  f j [ c0 ÞÞ\e for some c0 [ 0:

ð8:72Þ

meas

Assume fk ! f . We must show that, for any e > 0, there exists an N such that k
N implies that (8.72) holds. With this in view, consider any e > 0. The meas assumption that fk ! f means l(X(| fk − f| > c) ! 0, no matter which positive number c may be. In particular, this is so when c = 2e : Hence there exists an N such that k
N implies l(X(| fk − f| > 2eÞ < 2e ; which in turn implies e e þ lðXðjfk  f j [ ÞÞ\e; 2 2 that is, (8.72) holds with c′ = 2e : Conversely, assume q(fk, f) ! 0. We must show that, for any c > 0 and any η > 0, there exists an N such that k
N implies l(X(| fk − f| > c) < η. Accordingly, consider any c > 0 and any η > 0. The assumption that q(fk, f) ! 0 means that for e = min{c, η}, there exists an N such that k
N implies that (8.72) holds, that is, c0 þ lðXðjfk  f j [ c0 ÞÞ\ minfc; gg for some c0 [ 0: The positive number c′ must satisfy c′ < min{c, η}  c and also l(X(| fk − f| > c′)) < min{c, η}  η. However, l(X(| fk − f| > c)  l(X(| fk − f| > c′) because c′ < c. Therefore l(X(| fk − f| > c) < η. Remark Since q(f, g)  kf  gk1 , we have another proof that L∞ convergence implies convergence in measure. (d) The argument is similar to (c). (e) Denote avA by f. Let c
a. Then X(| f | > c) is empty and c + l(X (| f | > c)) = c. Therefore inffc þ lðXðj f j [ cÞÞ : c
ag ¼ a: But if c < a, then X(| f | > c) = A and c + l(X(| f | > c)) = c + l(A). Therefore inffc þ lðXðj f j [ cÞÞ : c\ag ¼ lðAÞ: It follows that q(f, 0) = inf{c + l(X(| f | > c)): c > 0} = min{a, l(A)}. (f) Consider any subset A X that satisfies 0 < l(A) < ∞ and let f(x) = 12l(A)vA. Then q(f, 0) = 12l(A) and q(4f, 0) = l(A) in view of part (e), which means q(4f, 0) = 2q(f, 0) 6¼ 4q(f, 0), considering that l(A) > 0.

8 Hints

573 meas

meas

6.2.P7. If fk ! f and g 2 L∞(X), then show that fk g ! fg. Hint: Let M denote the essential sup of g. Then the set XM = X(|g| > M) has measure 0. Xðjfn g  fgj [ aÞ fx 2 XnXM : jfn g  fgj [ ag [ fx 2 XM : jfn g  fgj [ ag a fx 2 XnXM : jfn  f j [ g [ fx 2 XM : jfn g  fgj [ ag: M 6.2.P8. If f and {fk}k
1 are measurable functions defined on X with l(X) < ∞, show that the following are equivalent: meas

(i) fk ! f ; (ii) every subsequence of {fk}k
1 has a subsequence converging to f a.e. meas

Hint: (i) ) (ii). Since fk ! f ; the same is true of every subsequence. By Corollary 6.2.7, any subsequence has a subsequence converging to f a.e. (ii) ) (i). Let a > 0 be given. Let Ek = X(|fk − f|
a}; we then have to show that l(Ek) ! 0. If this fails to hold, then l(Ekj ) > d for some d > 0 and some subsequence {fkj }j
1. In view of the hypothesis, by selecting a subsequence of {kj} if ae

necessary, we may assume that fkj ! f , so that l(limsupEkj ) = 0. Hence by Proposition 2.3.21(d), which can be shown to be valid for a general measure,     0 ¼ l lim supEkj
lim supl Ekj
d [ 0; and this is a contradiction. 6.2.P9. For Lebesgue measure on [0, 1], show that there cannot exist a q satisfying (a) and (b) of Problems 6.2.P5 and 6.2.P6, but satisfying ae

fk !f if and only if qðfk ; f Þ ! 0 as k ! 1 instead of (c). Hint: Suppose such a q exists. The sequence {fk}k
1 of Example 6.2.20 converges to 0 in measure and by Problem 6.2.P8, we conclude that every subsequence of it has a subsequence converging to 0 a.e. We can use q to show that the sequence converges to 0 a.e., which was observed not to be the case in Example 6.2.20. If the sequence were not to converge to 0 a.e., then there would exist some η > 0 and some subsequence ffkj gj
1 of {fk}k
1 satisfying qðfkj ; 0Þ [ g for every j. It follows that every subsequence of the subsequence ffkj gj
1 also satisfies the same and therefore fails to converge to 0 a.e., contradicting the conclusion obtained above by applying Problem 6.2.P8.

574

8 Hints meas

ae

6.2.P10. (a) If fk ! f ; and fk ! g, then g = f a.e. meas ae (b) If fk ! f ; and fk  fk+1 a.e. for each k, then fk ! f . meas Hint: (a) Since fk ! f ; by Corollary 6.2.7, there exists a subsequence converging ae to f a.e. Since fk ! g, the subsequence converges to g as well. Therefore g = f a.e. Note that this works even if g is extended real-valued. (b) The hypothesis fk  fk+1 a.e. for each k means that for each k, there exists a set Ek of measure 0 such that fk  fk+1 on the complement Eck. The union E = [ k
1Ek then has measure 0 and, every x 2 Ec satisfies the inequality fk  fk+1 for every k. Thus {fk(x)}k
1 is an increasing sequence for every x 2 Ec and therefore has an extended real limit for every x 2 Ec. Let g be the function defined to be this limit for ae x 2 Ec and 0 on E. Then fk ! g and by part (a), it follows that g = f a.e. ae Consequently, fk ! f . 6.2.P11. Show that if X = Z with the counting measure, then convergence in measure is equivalent to uniform convergence. Does this equivalence hold for the counting measure on an arbitrary set? Hint: l(X(| fk − f |
a)) ! 0. There exists a positive integer K such that 1 k
K ) lðXðjfk  f j
aÞÞ\ : 2 Since each nonempty subset of Z has measure at least 1, it follows that X(| fk − f |
a) = £ for k
K. That is to say, k
K ) jfk ðxÞ  f ðxÞj\a for all x 2 Z: unif

Thus fk ! g. Yes, the equivalence holds for the counting measure on an arbitrary set. 6.2.P12. Let (X,F ; l) be a measure space with l(X) < ∞ and suppose f and {fk}k
1 are measurable functions on X. For e > 0 and integer k
1, put Eke ¼ Xðjfk  f j
eÞ: ( Prove that fk! f (pointwise) if and only if, for all e [ 0;

S k
1

) decreases to £.

Eke n
1

Hint: First suppose fk! f and consider any e > 0. Observe that [ k
n+1Eek [ k
nEek. Let f e = \ n
1( [ k
nEek). We have to show that f e = £. Consider any x 2 f e. By supposition, there exists an n0 such that | fk(x) − f(x)| < e for all k
n0. Since x 2 f e, we have x 2 [ k
nEek for all n. In particular, x 2 [ k
n0 Eke . So there exists a k
n0 such that x 2 Eek, which implies | fk(x) − f(x)|
e. This contradiction shows that f e = £, proving the “only if” part. Now, suppose that for any e > 0, { [ k
nEek}n
1 decreases to £ and consider any x 2 X. We must show that fk(x) ! f(x). Consider any η > 0. There exists an n0

8 Hints

575

such that x 62 [ k
n0 Ekg , which implies x 62 Eek for every k
n0. Thus for every k
n0, we have | fn(x) − f(x)| < e. So, fk(x) ! f(x). 6.2.P13. Let (X,F ; l) be a measure space with l(X) < ∞ and suppose f and {fk}k
1 are measurable functions on X. For e > 0 and integer k
1, put Eke ¼ Xðjfk  f j
eÞ:   ae Prove that fk ! f if and only if lim l [ k
n Eke ¼ 0 for all e > 0. n!1

ae

Hint: First suppose fk ! f . By definition, this means there exists a measurable set G with l(Gc) = 0 and fk(x) ! f(x) for every x 2 G. Consider any e > 0. Observe that [ k
n+1Eek [ k
nEek. Therefore by Proposition 3.1.8(b), we need only prove that \ [ ð Eke ÞÞ ¼ 0: lð n
1 k
n

Let f e = \ n
1( [ k
nEek). We have to show that l(f e) = 0. Consider any x 2 G \ f e. By supposition, there exists an n0 such that | fk(x) − f(x)| < e for all k
n0. Since x 2 f e, we have x 2 [ k
nEek for all n. In particular, x 2 [ k
n0 Eke . So there exists a k
n0 such that x 2 Eek, which implies | fk(x) − f(x)|
e. This contradiction shows that G \ f e = £. Therefore f e = (G \ f e) [ (Gc \ f e) = Gc \ e f e Gc. Since l(Gc) = 0, it follows that  l(f ) = 0. e Now, suppose that lim l [ k
n Ek ¼ 0 for any e > 0. n!1

Since [ k
n+1Eek [ k
nEek, we have l( \ n
1( [ k
nEek)) = 0 for any e > 0 in view of Proposition 3.1.8(b). Upon considering e = 1/p for p = 1, 2, …, we find that l( [ p
1( \ n
1( [ k
nE1/p So, the set G = ( [ p
1( \ n
1 k ))) = 0. c c ( [ k
nE1/p ))) satisfies l(G ) = 0. Consider any x 2 G. We claim that fk(x) ! f(x). k 1=p Consider an arbitrary integer p > 0. There exists an n0 such that x 2 [ k
n0 Ek , 1/p which implies x 62 Ek for every k
n0. Thus for every k
n0, we have | fk(x) − f(x)| < 1/p. Note that we have shown such an n0 to exist for an arbitrary integer p > 0. Consequently, fk(x) ! f(x). 6.2.P14. Let (X, F ; l) be a measure space with l(X) < ∞ and suppose f and {fk}k
1 are measurable functions on X. For e > 0 and integer m
1, let Eme ¼ Xðj fm
n  f j
eÞ: ae

Prove that fk ! f if and only if lim l n!1

S m
n

! Eme

¼ 0 for all e > 0.

Hint: This is a direct consequence of Problem 6.2.P13 and Egorov’s Theorem 6.2.21. Problem Set 7.1 7.1.P1. Let X = Y = [0, 1], F = G be the r-algebra of measurable subsets of [0, 1] and l = m be Lebesgue measure on [0, 1]. Let

576

8 Hints

f ðx; yÞ ¼

x2  y2 ðx2 þ y2 Þ2

;

ðx; yÞ 2 ð0; 1Þ ð0; 1Þ:

Prove that each of the iterated integrals of f exists. The function f is, however, not in L1([0, 1] [0, 1]). Hint: For each fixed x, the function f(x, y), though undefined at y = 0 and y = 1, becomes Riemann (and hence Lebesgue) integrable on [0, 1], when defined in any manner at y = 0 and y = 1. It is easy to verify that the Riemann integral is R1 R 1 0 f ðx; yÞdy ¼ 1 þ x2 . It follows that the Lebesgue integral ð0;1Þ f(x, y)dm(y) is also R R equal to 1 þ1 x2 . Hence ð0;1Þ ð0;1Þ f ðx; yÞdmðyÞ dlðxÞ ¼ p4 : Similarly, it can be R R checked that ð0;1Þ ð0;1Þ f ðx; yÞdlðxÞ dmðyÞ ¼  p4. Since the iterated integrals are unequal, it follows by Fubini’s Theorem that f 62 L1([0, 1] [0, 1]). Alternatively, by Tonelli’s Theorem 7.1.23, Z Z Z ð jf ðx; yÞjdðl mÞ ¼ jf ðx; yÞjdmðxÞÞdlðyÞ ð0;1Þ ð0;1Þ ð0;1Þ ð0;1Þ Z Z x2  y2 dmðxÞÞdlðyÞ ð ¼ 2 ð0;1Þ ð0;1Þ ðx2 þ y2 Þ Z Z x2  y2
dmðxÞÞdlðyÞ ð 2 2 2 ð0;1Þ ð0;yÞ ðx þ y Þ Z ¼ Z ¼

ð0;1Þ

ð0;1Þ

Z ¼

Z ð

y2  x2

dmðxÞÞdlðyÞ ðx2 þ y2 Þ2 ! x x  dlðyÞ x2 þ y2 x¼y x2 þ y2 x¼0 ð0;yÞ

1 dlðyÞ ¼ 1: ð0;1Þ 2y

Thus f 62 L1((0, 1) (0, 1)). Computing the last integral above as an improper integral is justified by the Monotone Convergence Theorem, keeping in mind that the integrand is nonnegative. 7.1.P2. Let X = Y = [−1, 1] and F = G be the r-algebra of measurable subsets of [−1, 1] and l = m be Lebesgue measure on [−1, 1]. Let

8 Hints

577

f ðx; yÞ ¼

if ðx; yÞ 6¼ ð0; 0Þ otherwise:

xy ðx2 þ y2 Þ2

0

Show that Z ½1;1

Z ð

Z ½1;1

f ðx; yÞdmðyÞÞdlðxÞ ¼ 0 ¼

½1;1

Z ð

½1;1

f ðx; yÞdlðxÞÞdmðyÞ;

_ but the function isR not Lebesgue integrable over [_1, 1] R [ 1, 1]. Hint: For y 6¼ 0, [–1,1] f(x, y)dl(x) = 0 and for x 6¼ 0, [–1,1] f(x, y)dm(y) = 0; so each of the iterated integrals is zero. However, f is not integrable on [−1, 1] [−1, 1], as the following argument shows. Since Z Z jf ðx; yÞjdðl mÞ  jf ðx; yÞjdðl mÞ\1;

½0;1 ½0;1

½1;1 ½1;1

it will follow that f is integrable on [0, 1] [0, 1] if it is integrable on [−1, 1] [−1, 1]. By Tonelli’s Theorem 7.1.23 ! Z Z Z xy dmðyÞ dlðxÞ: jf ðx; yÞjdðl mÞ ¼ 2 ½0;1 ½0;1 ð0;1Þ ð0;1Þ ðx2 þ y2 Þ Simple calculations show that Z ð0;1Þ ðx2

xy þ y2 Þ 2

dmðyÞ ¼

1 x  : 2x 2ð1 þ x2 Þ

R However, ð0;1Þ 2x1  2ð1 þx x2 Þ dlðxÞ ¼ 1. Computing this integral as an improper integral is justified by the Monotone Convergence Theorem, keeping in mind that the integrand is nonnegative. 7.1.P3. Let f 2 L1(X, F ; l) and g 2 L1(Y, G; m), and suppose /(x, y) = f(x)g(y), x 2 X, y 2 Y. Show that / 2 L1(X Y, F G; l m) and Z Z Z /ðx; yÞdðl mÞ ¼ ð f ðxÞdlðxÞÞð gðyÞdmðyÞÞ: X Y

X

Y

Hint: The function f*(x, y) = f(x) is F G-measurable. Indeed,

578

8 Hints

fðx; yÞ 2 X Y : f  ðx; yÞ [ ag ¼ fx 2 X : f ðxÞ [ ag Y: Similarly, the function g*(x, y) = g(y) is F G-measurable. The function f(x)g (y) = f*(x, y)g*(x, y), being the product of two F G-measurable functions, is F G-measurable. Since the iterated integral Z Z



Z

jf ðxÞgðyÞjdmðyÞ dlðxÞ ¼ X

Y

 Z jgðyÞjdmðyÞ

Y

 jf ðxÞjdlðxÞ \1;

X

Fubini’s Theorem 7.1.25 is applicable and the desired result follows. 7.1.P4. Let X = Y = [0, 1], F = G = the r-algebra of Lebesgue measurable subsets of [0, 1] and l = m = Lebesgue measure. Suppose that either f 2 L1([0, 1] [0, 1]) or f is a measurable nonnegative function on [0, 1] [0, 1]. Prove that Z ½0;1

Z ½0;x

! f ðx; yÞdy dx ¼

Z ½0;1

Z ½y;1

! f ðx; yÞdx dy:

Hint: Depending on whether f 2 L1([0, 1] [0, 1]) or f is a measurable nonnegative function, apply either Fubini’s Theorem 7.1.25 or Tonelli’s Theorem 7.1.23 to the function g given on the domain [0, 1] [0, 1] by g = fvA, where A = {(x, y) 2 [0, 1] [0, 1]: y 2 [0, x]} after arguing that A is (i) measurable and (ii) the same set as {(x, y) 2 [0, 1] [0, 1]: x 2 [y, 1]}. Here, (ii) is obvious because the conjunction of inequalities 0  x  1 and 0  y  x is plainly equivalent to the conjunction 0  y  1 and y  x  1:    j j Regarding (i), for any positive integer n, set Ij ¼ j1 and n ; n ; Jj ¼ 0; n Vn = [ 1  j  n(Ij Jj). Clearly, each Vn, being a union of measurable rectangles, is measurable, and hence so is \ n
1Vn. However, this intersection is A. R 1 7.1.P5. Let f 2 L1((0, a)) and gðxÞ ¼ ½x;a f ðtÞ t dt; 0\x  a: Show that g 2 L ((0, a)) R R and that [0,a]g(x)dx = [0,a] f(t)dt. Hint: We wish to show that

8 Hints

579

Z f ðtÞ dt dx\1: jgjdm ¼ ½0;a ½0;a ½x;a t

Z

Z

It is enough to show that

R

R

½0;a ½x;a

j f ðtÞj t dtdx\1.

The functions given by

f ðtÞ j f ðtÞj t and t

on (0, a) (0, a) are both measurable. Interchanging the order of integration and using the obvious analogue of the result of 7.1.P4 above, we transform the last integral as Z

Z

jf ðtÞj dxdt ¼ t ½0;t

½0;a

Z ½0;a

jf ðtÞjdt\1:

Thus g 2 L1((0, a)). Now, by the definition of g, Z

Z

Z gðxÞdx ¼

½0;a

½0;a

f ðtÞ dtdx: ½x;a t

Using the obvious analogue of the result of 7.1.P4 above once again, Z

Z

Z ½0;a

gðxÞdx ¼ Z ¼

f ðtÞ dxdt ½0;t t

½0;a

½0;a

f ðtÞdt:

7.1.P6. Let X = Y = [0, 1], F = G = the r-algebra of Lebesgue measurable subsets of [0, 1] and l = m = Lebesgue measure. Define for x, y 2 [0, 1],

f ðx; yÞ ¼

x2Q x 62 Q:

1 2y

Compute Z ½0;1

Z ð

½0;1

Z f ðx; yÞdmðyÞÞ dlðxÞ and

Does f 2 L1([0, 1] [0, 1])?

½0;1

Z ð

½0;1

f ðx; yÞdlðxÞÞdmðyÞ:

580

8 Hints

Hint: The function can be expressed as f(x, y) = vI \ Q (x)⋅1 + vI/ Q (x)⋅2y. Since each of the two terms is a measurable function on [0, 1] [0, 1], the sum f is a measurable function too. Its values lie between 0 and 2 and it is therefore integrable. Moreover, by Tonelli’s Theorem 7.1.23, its integral equals each of its repeated integrals. It is sufficient therefore to compute either one of the two repeated integrals asked for. For each x 2 [0, 1], we have f(x, y) = 1 or 2y according as x 2 Q or x 62 Q; so that f (x, y) = 2y for almost all x 2 [0, 1]. Therefore Z ½0;1

f ðx; yÞdmðyÞ ¼ 1 for almost all x 2 ½0; 1:

Hence !

Z

Z ½0;1

½0;1

f ðx; yÞdmðyÞ dlðxÞ ¼ 1:

Alternatively, we may compute the other repeated integral: For each y 2 [0, 1], we have f(x, y) = 1 or 2y according as x 2 Q or x 62 Q; so that f(x, y) = 2y for almost all x 2 [0, 1]. Therefore Z ½0;1

f ðx; yÞdlðxÞ ¼ 2y for all y 2 ½0; 1:

Hence Z

!

Z ½0;1

½0;1

f ðx; yÞdlðxÞ dmðyÞ ¼ 1; as expected:

7.1.P7. Suppose that {am,n}m,n
1 is a double sequence of nonnegative real numbers. Then show that 1 X 1 X m¼1 n¼1

am;n ¼

1 X 1 X

am;n :

n¼1 m¼1

Hint: Let (X, F ; l) = (Y, G; m) = (N; P (NÞ, c), where c is the counting measure on N; obviously r-finite. The product r-algebra P(NÞ P(NÞ is easily verified to be P(N NÞ, so that every function on N N is measurable. Define f: N N!R by f(m, n) = am,n. Then by Tonelli’s Theorem,

8 Hints 1 X 1 X m¼1 n¼1

581

am;n ¼

1 Z X

 Z Z f ðm; nÞdcðnÞ ¼ f ðm; nÞdcðnÞdcðmÞ

N

m¼1

N

N

Z Z ¼

N

N

f ðm; nÞdcðmÞdcðnÞ ¼

1 X 1 X

am;n :

n¼1 m¼1

7.1.P8. Let X = Y = [0, 1], F = G = the r-algebra of Borel subsets of [0, 1]; suppose l is Lebesgue measure on Borel subsets of X and let c be the counting measure on Y. Show that V = {(x, y) 2 [0, 1] [0, 1]: x = y} is F G-measurable and Z

Z

Z

Z

vV dl ¼ 0 but

dc Y

vV dc ¼ 1:

dl

X

Y

X

  j Hint: For any positive integer n, set Ij ¼ j1 n ; n and Vn = [ 1  j  n(Ij Ij). Clearly, each Vn, being a union of measurable rectangles, is measurable, and so is V = \ n
1Vn. Now, Z

Z

Z vV dl ¼

dc Y

and

lðV y Þdc ¼ 0

X

Z

Y

Z

Z vV dc ¼

dl X

Y

cðVx Þdl ¼ 1: X

Fubini’s and Tonelli’s Theorem both fail because c is not r-finite. 7.1.P9. Let X = Y = N and l = m = counting measure on N: Let 8