A concise introduction to measure theory 9783030032401, 9783030032418

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Satish Shirali

A Concise Introduction to Measure Theory

A Concise Introduction to Measure Theory

Satish Shirali

A Concise Introduction to Measure Theory

123

Satish Shirali Formerly of Panjab University Chandigarh, India and The University of Bahrain Zallaq, Bahrain

ISBN 978-3-030-03240-1 ISBN 978-3-030-03241-8 https://doi.org/10.1007/978-3-030-03241-8

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Library of Congress Control Number: 2018960220 Mathematics Subject Classification (2010): 28-01, 28A05, 28A10, 28A12, 28A20, 28A25, 28A35 © Springer Nature Switzerland AG 2018 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, express 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

The concept of measure in an abstract space, and of the integral with respect to it, has played a fundamental role in analysis and probability in the decades after the pathbreaking work of H. Lebesgue in 1902. It has therefore been an essential component of a mathematics curriculum, accessible to any student who has acquired a facility with basic analysis, including the Heine–Borel theorem, the theory of Riemann integration, infinite series, and also with the use of sets and quantifiers (“for all” and “there exists”) in definitions and proofs. This book assumes such a facility on the part of the reader and also an understanding of the notions of countability and Cartesian product. Equivalence classes and the axiom of choice are needed only once. It is well known that the abstract integral can be interpreted as an improper Riemann-type integral of a related function on the real line, thereby rendering it relatively more concrete. The related function is one that closely resembles what is called the cumulative distribution in probability and statistics and may reasonably be called simply the distribution. This book introduces the concept of the integral of a nonnegative measurable function on a measure space via the improper integral of the related distribution function, without the intermediacy of simple functions. Doing so brings out more vividly how the Lebesgue approach begins by partitioning the range rather than the domain of the integrand. Also, the concept is then amenable to a quick extension to “fuzzy” measures. It also helps to appreciate how the monotone convergence property is unrelated to the additivity of measure.

v

vi

Preface

It is natural to begin by recasting the Riemann integral of a step function as the integral of its distribution function, which is essentially the Abel summation formula. This is what Chap. 1 starts with. For other kinds of functions, it is not always clear that a distribution function exists at all. A countably additive measure as the “size” of a set is then presented as motivated by the need for a limit of functions having distributions again to have a distribution. The concept of a countably additive measure on a r-algebra and measurability of functions are introduced in Chap. 2 against the backdrop of the motivation described earlier. The integral of a nonnegative measurable function is defined as the improper integral of its distribution function, and the monotone convergence theorem is proved. The integral of a function that may take negative values is introduced, but its properties are dealt with in the next chapter. Simple functions are defined in Chap. 3 and used to establish the additivity of the integral. The dominated convergence theorem, for which the additivity of the integral is used, is also treated in this chapter. There is a discussion of the extension to subadditive fuzzy measures, but this material is optional. Chapter 4 is about constructing Lebesgue measure. After defining Lebesgue outer measure, Carathéodory’s ideas are applied to an abstract outer measure in the usual manner. It is shown that the classical Riemann integral agrees with the Lebesgue integral for Riemann integrable functions, and the existence of a nonmeasurable subset of the reals is discussed. The chapter ends with induced measures, which find application in connection with product measures and in identifying certain improper integrals as being Lebesgue integrals. The counting measure and interchanging the order of summation in a repeated sum as amounting to interchanging the order of integration are treated at length in Chap. 5. In this connection, the unconditional sum is identified with the integral respect to the counting measure. Product measures and the theorems of Tonelli and Fubini are taken up in Chap. 6. In Chap. 7, the relation to differentiation is discussed in some detail. There are many approaches to the topic, the one adopted in this book being via the Vitali covering theorem and Tonelli’s theorem. While the concept of total variation is presented ad hoc in most texts, here it is presented as a natural outcome of attempts to decompose a function as the sum of an increasing and a decreasing function. The relation between differentiation and Lebesgue integration is not as straightforward as in the case of Riemann integration, and no discussion of the matter can be complete without the Cantor set and function. Their essentials are discussed in Chap. 8. The symbols R, Q, Z and N are commonly understood these days, and no explanation of them has been given. For matters that are treated differently or not treated at all in most other works covering measure and integration, a prospective reader having prior experience of the subject may wish to check out the following:

Preface

vii

1.1.3, 1.1.6 (motivation), 1.2.4, 3.1.9, 3.1.10, 3.1.P1, 3.1.P7, 3.1.P8, 3.1.P9, Figures in Sect. 3.2, Sect. 3.3, 4.1.2(b), 4.2.P6, 4.2.P8, 4.2.P9, 4.4.1, 5.1.2, 5.1.4, 5.1.10, 5.2.5, 5.2.7, 5.2.8, 6.2.P4, 6.6.4, 7.1.3(b), 7.1.P3, beginning of Sect. 7.6 (motivation), 7.8.2, Sect. 8.2 (manner of constructing the Cantor function), 8.2.P3, 8.2.P4, 8.2.P5.

The author is grateful to Dr. H. L. Vasudeva for his valuable comments on the presentation, which contributed significantly to the improvement in readability. Gurugram, India September 2018

Satish Shirali

Contents

1 Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.1 The Riemann Integral Revisited . . . . . . . . . . . . . . . . . . . . . . . . . . 1.2 Improper Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

1 1 10

2 Measure Space and Integral . . . . . . . . . . 2.1 Measure and Measurability . . . . . . . . 2.2 Measurability and the Integral . . . . . . 2.3 The Monotone Convergence Theorem

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3 Properties of the Integral . . . . . 3.1 Simple Functions . . . . . . . . 3.2 Other Measurable Functions 3.3 Subadditive Fuzzy Measures

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39 39 47 56

4 Construction of a Measure . . . . . . . . . . . 4.1 Lebesgue Outer Measure . . . . . . . . . . 4.2 Measure from Outer Measure . . . . . . 4.3 Lebesgue Measure . . . . . . . . . . . . . . 4.4 Induced Measure and an Application .

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65 65 71 78 85

5 The Counting Measure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.1 Interchanging the Order of Summation . . . . . . . . . . . . . . . . . . . . 5.2 Integration with the Counting Measure . . . . . . . . . . . . . . . . . . . .

91 91 98

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6 Product Measures . . . . . . . . . . . . . . . . . . . . 6.1 Algebras and Monotone Classes of Sets . 6.2 Defining a Product r-Algebra . . . . . . . . 6.3 Sections of a Subset and of a Function . 6.4 Defining a Product Measure . . . . . . . . . 6.5 The Tonelli and Fubini Theorems . . . . . 6.6 Borel Sets . . . . . . . . . . . . . . . . . . . . . .

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103 103 105 108 111 116 122

ix

x

7 Differentiation . . . . . . . . . . . . . . . . . . . . . . . 7.1 Integrability and Step Functions . . . . . . 7.2 The Vitali Covering Theorem . . . . . . . . 7.3 Dini Derivatives . . . . . . . . . . . . . . . . . . 7.4 The Lebesgue Differentiability Theorem . 7.5 The Derivative of the Integral . . . . . . . . 7.6 Bounded Variation . . . . . . . . . . . . . . . . 7.7 Absolute Continuity . . . . . . . . . . . . . . . 7.8 The Integral of the Derivative . . . . . . . .

Contents

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127 127 133 138 142 148 151 158 162

8 The Cantor Set and Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 169 8.1 A Surjection from the Cantor Set to [0, 1] . . . . . . . . . . . . . . . . . . 169 8.2 A Bijection from the Cantor Set to [0, 1] and the Cantor Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 173 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 179 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 267 Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 269

Chapter 1

Preliminaries

1.1 The Riemann Integral Revisited It is elementary from the basic theory of the Riemann integral, for example as in Rudin [7] or Shirali and Vasudeva [9], that if a function on a closed bounded interval [α, β] is constant on (α, β), its integral is the product of that constant with the length of the interval. The values of the function at the endpoints make no difference. Each term in a lower or upper sum of a bounded function f : [a, b] → R is a product of this form. To put it more precisely, denote the lower sum of a bounded function f : [a, b] → R over a partition P: a  x0 < x1 < · · · < xn  b of [a, b] as usual by L( f, P) 

n 

m j (x j − x j−1 )

j1

where mj means the infimum of f over the subinterval [x j−1 , x j ] of P. The term mj (x j − x j−1 ) is the Riemann integral over [x j−1 , x j ] of the function that equals mj on the subinterval (x j−1 , x j ), regardless of what value the function takes at the endpoints. Taking into account that the sum of the integrals of a function on contiguous intervals [α, β] and [β, γ] is the integral of that function over [α, γ], it is seen that the lower b sum is equal to the integral a m of any function m: [a, b] → R which, for each j, equals the constant mj on the interval (x j−1 , x j ). The values of m at the points x j make no difference; but in order to be specific, we shall choose m to be mj on the left closed subinterval [x j−1 , x j ) of P and m(b) to be 0. The argument why refining the partition increases the lower sum can then be interpreted in terms of such a function m as proving that refining increases the function m. Similar observations apply to the upper sum, which will be denoted as usual by U ( f, P) 

n 

M j (x j − x j−1 )

j1

© Springer Nature Switzerland AG 2018 S. Shirali, A Concise Introduction to Measure Theory, https://doi.org/10.1007/978-3-030-03241-8_1

1

2

1 Preliminaries

where M j means the supremum of f over the interval [x j−1 , x j ]. The notation L(f , P) for the lower sum and U(f , P) for the upper sum will be used in the sequel without explanation. Functions that remain constant on the open intervals of a partition are lurking within the definition of upper and lower sums, although one may not make explicit mention of them. As we intend to consider them at some length now, we formally introduce the standard name by which such functions are known. 1.1.1. Definition. A function f : [a, b] → R is called a step function if there is some partition P: a  x0 < x1 < · · · < xn  b of [a, b] such that f is constant on each of the open intervals (x j−1 , x j ). It is said to be constant on the subintervals of P. It should be borne in mind that such a function is constant only on the open subintervals of P, although we shall usually suppress the word “open” for convenience. Evidently, the function is also constant on the subintervals of any refinement of P; thus the partition P mentioned in the definition is far from being unique. Although we shall generally denote the Riemann integral over [a, b] of a function b b φ by a φ, it will be convenient to use the more customary notation a φ(x)d x when the function is known by its expression φ(x) without any symbol such as φ having been introduced for it.  b nThe Riemann integral a f of a step function f is easy to compute: It is simply j1 c j (x j − x j−1 ), where cj denotes the value on (x j−1 , x j ). In the language of elementary calculus, the “area under the graph” of a nonnegative step function is the sum of the areas of rectangles stretching downward from the graph all the way to the horizontal axis. The terms in the sum represent the areas of the rectangles. If the graph is drawn with the axes interchanged, then the rectangles stretch leftward from the graph all the way to the vertical axis. 1.1.2. Exercise. For each of the functions below, determine whether it is a step function, and describe in terms of intervals the set S mentioned along with it: (a) f : [0, 4] → R, where f (x)  6 if 0 ≤ x < 1, f (x)  2 if 1 ≤ x ≤ 3, f (x)  6 if 3 < x ≤ 4; S  {x ∈ [0, 4]: f (x) > 2}. (b) f : [0, 2] → R, where f (x)  x 2 ; S  {x ∈ [0, 2]: f (x) > 41 }. (c) f : [0, π] → R, where f (x)  sin x; S  {x ∈ [0, π]: f (x) > t}, where 0 ≤ t. (d) f : [0, π] → R, where f (x)  1 + sin x; S  {x ∈ [0, π]: f (x) > t}, where 0 ≤ t. Solutions: (a) Consider the partition P: 0 < 1 < 3 < 4. On the (open) subintervals (0, 1), (1, 3) and (3, 4), the function has the constant values 6, 2 and 6 respectively. Therefore f is a step function. S  [0, 1)∪(3, 4]. (b) The function takes infinitely many different values and therefore cannot be a step function. S  ( 21 , 2]. (c) The function takes infinitely many different values and therefore cannot be a step function. If t < 1, we have S  (sin−1 t, π − sin−1 t), as a graph would reveal; we dispense with the formal argument. When t ≥ 1, the set S is empty because of the inequality sin x ≤ 1.

1.1 The Riemann Integral Revisited

3

(d) The function takes infinitely many different values and therefore cannot be a step function. S  [0, π] if 0 ≤ t < 1 and S  (sin−1 (t −1), π−sin−1 (t −1)) if 1 ≤ t < 2. For t ≥ 2, the set S is empty. In view of what has been noted in parts (c) and (d) of the foregoing Exercise, we shall henceforth consider the empty set as an interval having length 0. With this convention, we can say that the set S has turned out to be a finite union of disjoint intervals in each case. In part (b), if the number 41 is replaced by an arbitrary t ≥ 0, √ then S is the interval ( t, 2] if 0 ≤ t < 4 and ∅ if 4 ≤ t. Since the length of the interval ∅ is finite, we shall count it among bounded intervals; moreover, just as in the case of R, it will be considered as closed and open at the same time. This changes the meaning of “interval” slightly, but all that is needed is to exercise some care in using results about intervals. In part (a) of the exercise, one can ask for S  {x ∈ [0, 4]: f (x) > t} for an arbitrary t ≥ 0. The reader will find it helpful to draw a graph of the step function f to see that S varies with the value of t in the following manner: ⎧ ⎪ [0, 4] if 0 ≤ t < 2 ⎨ if 2 ≤ t < 6 S  [0, 1) ∪ (3, 4] ⎪ ⎩ ∅ if 6 ≤ t Thus the subset of the domain on which the function exceeds any given number t ≥ 0 turns out to be a finite union of intervals in each of the instances in the exercise. There are many other functions for which this happens, for example, those that are monotone. When it does happen, the function has a distribution, to be denoted herein by f˜, which is defined to be the function whose value at any t ≥ 0 is the sum of the lengths of all the finitely many disjoint intervals whose union is the subset S of the domain on which the function exceeds t. If the subset is empty, we take f˜(t) to be 0. In particular, when t ≥ sup f , the set S is empty and f˜(t)  0. A significant point to note is that the distribution, if it exists, is a decreasing function (which means nonincreasing in this book), and therefore Riemann integrable (see Theorem 9.3.3. of Shirali and Vasudeva [9] or 6.9 of Rudin [7]). It is decreasing because t 1 > t 2 implies {x: f (x) > t1 } ⊆ {x: f (x) > t2 }. The reader may note however that disjoint intervals comprising the subset need not be unique—Example: [0, 3] is the union of the disjoint intervals [0, 1), [1, 3] as well as [0, 1], (1, 2], (2, 3]—and therefore one could legitimately ask whether their total length is unique, and consequently whether the concept of the distribution f˜ is unambiguous. Some authors may not share these doubts (see Rudin [7]) and in any case, the matter can be settled in the affirmative. However, we prefer to leave the technical details to the problems and proceed by taking the affirmative answer for granted. Readers who do not feel any need for the technical details of the matter would do well to avoid engaging with Problems 1.1.P6, 1.1.P7, 1.1.P8 and 1.1.P9. For the function f in part (a) of the exercise above, the set {x ∈ [0, 4]: f (x) > t} has been described at the beginning of the preceding paragraph and one can conclude therefrom that the distribution f˜: [0, ∞) → R is given by

4

1 Preliminaries

⎧ ⎪ ⎨ 4 if 0 ≤ t < 2 ˜ f (t)  2 if 2 ≤ t < 6 ⎪ ⎩ 0 if 6 ≤ t. Since sup f  6, the last line in this description of f˜ reflects the earlier observation that, in general, f˜(t)  0 when t ≥ sup f . We draw attention to the fact that 4

sup f

f  6 · 1 + 2 · 2 + 6 · 1  16 and 0

f˜ 

0

6

f˜  4 · 2 + 2 · 4  16,

0

4  sup f so that 0 f and 0 f˜ are equal to each other. The figures below are intended to illustrate how the two sums lead to the area of the same region broken up into rectangles in two different ways. The figure on the left shows the “area under the

f(x)

f(x)

t

x

x

f (t)

graph” of f in the usual manner as a union of rectangles. Adding up the areas of 4 the rectangles is the same computation as in the computation of 0 f above. The middle figure shows the same area broken up into rectangles differently. The figure on the right shows the graph of f˜ with the axes interchanged and the area that would have been under the graph if the axes had not been interchanged. The darker shaded rectangle has been broken up into two by a vertical line so as to show the relation to the rectangles in the middle figure. Again, adding up the areas of the rectangles is the  sup f f˜ above. One can visually confirm same computation as in the computation of 0 that the total shaded areas are the same in all three figures, which is a reflection of the equality of integrals that was obtained above by computation. The forthcoming proposition shows that the equality of the integrals of the nonnegative step function and its distribution is not a coincidence. 1.1.3. Proposition. Any nonnegative-valued step function f : [a, b] → R has a distribution f˜ and b

sup f

f  a

0

f˜.

1.1 The Riemann Integral Revisited

5

Proof: The case when f is 0 everywhere is trivial because sup f  0 in this situation. Therefore we consider only the case when f has at least one positive value. The set {x ∈ [a, b]: f (x) > t}, even if empty, is the union of a finite family of disjoint bounded intervals. By Problem 1.1.P9, the total length of disjoint intervals comprising the set is then uniquely determined, so that the distribution indeed exists. Let P: a  x0 < x1 < · · · < xm  b be a partition of [a, b] such that f has constant value cj on the open interval (x j−1 , x j ). Then b f 

m 

c j (x j − x j−1 ).

j1

a

By grouping together the terms for which the values cj agree but are not 0, we rewrite the above sum as b f  a

n 

Tk λk ,

(1.1)

k1

where the Tk (1 ≤ k ≤ n) are the distinct positive values among c1 , . . . , cm and λk is the total length of the intervals on which f takes the value T k . Any values of f that are not among the T k are taken only at one of the points x j of the partition P and therefore the total length of the intervals where such values are taken is 0. Thus we can include them in the sum with 0 as the corresponding λk . The numbers T k then represent all the distinct positive values of f . By renumbering if necessary, we may assume that they are in increasing order. For reasons that will become clear in the next paragraph, it will be convenient to set T 0  0, although it does not appear in the sum. Since T1 , . . . , Tn are the distinct positive values of f arranged in increasing order, we have Tn  sup f and f˜(Tn )  0.

(1.2)

Consider any integer k such that 1 ≤ k ≤ n. Surely, f (x) > Tk_1 ⇔ f (x) ≥ Tk . In other words, {x ∈ [a, b]: f (x) > Tk−1 }  {x ∈ [a, b]: f (x) > Tk } ∪ {x ∈ [a, b]: f (x)  Tk }. (1.3) Denote by F1 and F2 the families of disjoint intervals comprising the two sets on the right side. Since the sets are disjoint, we deduce that the intervals in F1 are disjoint from those in F2 , so that F1 ∪ F2 is a family of disjoint intervals comprising the set on the left side of (1.3). In particular, F1 and F2 can have no nonempty interval in

6

1 Preliminaries

common, so that the total length of the intervals in F1 ∪ F2 is the sum of the separate total lengths. This means f˜(Tk−1 )  f˜(Tk ) + λk , and hence by (1.1), b f 

n 

Tk ( f˜(Tk−1 − f˜(Tk )).

(1.4)

k1

a

If n  1, then by (1.2), f˜(T1 )  0; therefore, keeping in mind that we have set T 0  0, we find that the sum in (1.4) becomes T1 ( f˜(T0 ) − f˜(T1 ))  (T1 − T0 ) f˜(T0 ) 

n 

f˜(Tk−1 )(Tk − Tk−1 ).

k1

Hence, b f  a

n 

f˜(Tk−1 )(Tk − Tk−1 )

(1.5)

k1

when n  1. Suppose n ≥ 2. By (1.2), f˜(Tn )  0, and therefore the equality (1.4) can, by separating the last term in the sum, be recast as b f  a

n−1 

Tk ( f˜(Tk−1 ) − f˜(Tk )) + Tn f˜(Tn−1 ).

k1

Also, Tk  (Tk − Tk−1 ) + · · · + (T1 − T0 ), remembering that T 0  0. Using the Abel summation formula of Problem 1.1.P5 (with ak  f˜(Tk−1 ) and bk  T k − T k −1 ), we find that (1.5) holds in the present case too, namely, n ≥ 2.  sup f  It remains to show only that 0 f˜  nk1 f˜(Tk−1 )(Tk − Tk−1 ).  sup f T In view of the first equality in (1.2), we have 0 f˜  0 n f˜. Since f does not take any value between T k −1 and T k , it is true of any t ∈ [T k −1 , T k ) that f (x) > t ⇔ f (x) > Tk−1 . It follows that f˜(t)  f˜(Tk−1 ) for every t ∈ [T k −1 , T k ), which is to say, the restriction of f˜ to [0, sup f ] is a step function having constant value f˜(Tk−1 ) on each interval [T k −1 , T k ). The equality that remained to be shown is now an immediate consequence.  For a monotone function f , the set {x: f (x) > t} is an interval and therefore the function has a distribution f˜. The function of Exercise 1.1.2(d) is not even monotone and yet the aforementioned set is an interval. It is natural to ask whether the equality proved above for step functions is valid for other (nonnegative) functions that have a distribution. We shall not prove this yet, because more technical hassles are involved than proving that the total length of a disjoint union of intervals is determined uniquely, i.e. independently of the particular representation as such a union. One cannot imitate the proof for the case of step functions because the integral is no

1.1 The Riemann Integral Revisited

7

longer a finite sum. Although the proof is deferred until a proper formulation and argument are possible (appearing as Theorem 4.3.4 later on), we shall illustrate some cases now. 1.1.4. Example. (a) As noted in the paragraph following Exercise 1.1.2, for the 2 function f : [0, 2] √ → R such that f (x)  x , the set S  {x ∈ [0, 2]: f (x) > t} ∅ if 4 ≤ t. Also sup f  4. Therefore is the interval ( t, 2] if 0 ≤ t < 4 and √ f˜: [0, 4] → R is given by f˜(t)  (2 − t) for 0 ≤ t ≤ 4, and thus b

2 f 

a

sup f 2

x d x and 0

0

f˜ 

4 (2 −

√

t)dt.

0

Elementary computations show that both are 83 . (b) As seen earlier in Exercise 1.1.2(d), for the function f : [0, π] → R, where f (x)  1 + sin x, the set S  {x ∈ [0, π]: f (x) > t} is the interval [0, π] when 0 ≤ t < 1, interval (sin−1 (t − 1), π − sin−1 (t − 1)) when 1 ≤ t < 2 and ∅ when 2 ≤ t. Of course, sup f  2. Therefore f˜: [0, sup f ] → R is given on [0, sup f ) by

π if 0 ≤ t < 2 f˜(t)  −1 π − 2 sin (t − 1) if 1 ≤ t < 2. Thus b

π f 

a

sup f

(1 + sin x)d x and 0

0

f˜  π(1 − 0) +

2

(π − 2 sin−1 (t − 1))dt.

1

Elementary computations show that both are π + 2. M When f is bounded, the distribution f˜ is 0 on [sup f , ∞) and the integral 0 f˜ is the same for all M ≥ sup f . Thus the restriction of f˜ to [0, sup f ] can “do everything for us” that the unrestricted function can and therefore will also be called the distribution f˜. The Exercise about to be presented will be used to obtain a glimpse into the technical hassles mentioned above. The concept of countability will play an essential role in the exercise as well as further on. In the terminology of this book, “countable” includes finite. The elements of a countably infinite set A can by definition be arranged in a sequence with no repetitions and range equal to the entire set A. Any such sequence will be called an enumeration of A.

8

1 Preliminaries

1.1.5. Exercise. Let {r k }k ≥1 be an enumeration of the set of all rational numbers in (0, 1). For each n ∈ N, let f n : [0, 1] → R be the function which vanishes at the first n numbers of the sequence {r k }k ≥1 and also at 0 and 1, while having the value 1 at every other point of its domain. Also, let f : [0, 1] → R be the function for which f (x)  inf{ f n (x): n ∈ N} for each x ∈ [0, 1]. Obviously, 0 ≤ f (x) ≤ 1. Show that (i) (ii) (iii) (iv)

for each x ∈ [0, 1] and n ∈ N, the inequality f n (x) ≥ f n+1 (x) holds; for each x ∈ [0, 1], f (x)  limn→∞ f n (x); each f n is a step function with f˜n (t)  1 when 0 ≤ t < 1; S  {x ∈ [0, 1]: f (x) > t} is not a countable union of intervals when 0 ≤ t < sup f .

Solution: The functions f n and f n+1 differ only at r n+1 and f n (r n+1 )  1 > 0  f n+1 (r n+1 ). This proves (i). It now follows from elementary facts about monotone sequences that (ii) holds (see Theorem 3.3.3 of Shirali and Vasudeva [9] or 3.14 of Rudin [7]). The first n numbers of the sequence {r k }k ≥1 provide a partition of [0, 1] when arranged in increasing order. By definition, f n is 1 on each open subinterval of the partition, so that it is a step function. Besides, for any t such that 0 ≤ t < 1, the set S n  {x ∈ [0, 1]: f n (x) > t} is the union of the open subintervals of the partition, which means its total length is 1. This proves (iii). To prove (iv), we first note that f (x)  0 if x is rational and 1 otherwise. This is because, on the one hand, since the range of {r k }k ≥1 is the set of all rational numbers in (0, 1), any rational x equals r k for some k and hence we have f n (x)  f n (r k )  0 for every n ≥ k; on the other hand, since any irrational x differs from r k for every k, we have f n (x)  1 for every n ∈ N. From the foregoing description of f , we observe that sup f  1. Moreover, when 0 ≤ t < 1, the set S  {x ∈ [0, 1]: f (x) > t} consists of all the irrational numbers in [0, 1]. Thus S  ∩∞ n1 Sn . If S were to be a countable union of intervals, then each interval would have to consist of at most one point (because any interval consisting of more than one point contains a rational number, which cannot be in S), thereby making their union a countable set, which S is not. The above exercise illustrates that it can happen for a sequence {f n }n≥1 of step functions converging at each point to a limit function f that all functions of the sequence have perfectly simple distributions and yet the limit function has none. This jarring situation has arisen because although (in the notation of the solution presented above) S  ∩∞ n1 Sn , where each S n is a finite union of intervals, the set S is not even a countable union of intervals. Of course, one has the option of switching over to the complement of S in [0, 1], which consists of all the rational numbers in [0, 1] and is thus a countable union of the single point intervals [0, 0], [1, 1] and [r n , r n ], whose total length can be taken as 0 + 0 + …  0. This makes it reasonable to think of the “total length” of S as being 1, although it has not been arrived at as a sum of lengths. Whatever we understand by total length of a more general kind of a set has to go beyond a mere totaling process. The obstacle will eventually be surmounted in Chap. 4, but only after we have looked at some other benefits that will accrue from considering distributions, something we begin doing in the next section.

1.1 The Riemann Integral Revisited

9

1.1.6. Remark. In summary: In order to arrange for a more general class of functions to have distributions that are sure to be decreasing functions, what we would like is that the total length idea be extended—if necessary by going beyond a totaling process—so as to ensure that: (1) it is applicable to a class of sets that is broad enough to include (i) unions of countable families of sets in the class and (ii) complements of sets in the class; (intersections of countable families of sets in the class will then necessarily be included); (2) the total length (in the extended sense) of a set is never less than that of a subset, so that a distribution is sure to be a decreasing function. The complement of the set S occurring in Exercise 1.1.5 was seen to be the union of a countable number of disjoint single point intervals, and the totaling process to obtain its length led to a series with its nth partial sum representing the sum of the lengths of the first n intervals. This would have been so even if the lengths of the intervals were not 0. This suggests that the total length in the extended sense should also have the property that (3) the (extended) total length of a set S  ∪∞ n1 Sn , where any two among the sets S n are disjoint, agrees with the sum of the series formed by (extended) total lengths of S n . The other concepts that we shall develop by using distributions have far- reaching consequences. For one of them, the reader may go through the statement of Exercise 3.1.10 right away. Problem Set 1.1 1.1.P1. For each t ≥ 0, find S  {x ∈ [0, 4]: f (x) > t}, where f : [0, 4] → R is the function given by: f (x)  6 if 0 ≤ x < 1, f (1)  5, f (x)  2 if 1 < x < 3, f (3)  1, f (x)  6 if 3 < x ≤ 4. 1.1.P2. For f : [1, 2] → R given by f (x)  ex , find the distribution f˜: [0, sup f ] → R  sup f 2 and show that 0 f˜ agrees with 1 f . 1.1.P3. Let A > 0. For the function f : [0, A] → R given by f (x)  e−x , find the  sup f A distribution f˜: [0, sup f ] → R and show that 0 f˜ agrees with 1 f . Hint: Use an appropriate substitution and integrate by parts. 2

1.1.P4. Abel summation formula. If a1 , . . . , an and b1 , . . . , bn are two finite sequences of n terms each (n ≥ 2) and Sk  b1 + · · · + bk , then bk  S k − S k −1 for k  2, . . . , n and therefore a1 b1 + · · · + an bn  a1 S1 + a2 (S2 − S1 ) + a3 (S3 − S2 ) + · · · + an (Sn − Sn−1 )  (a1 − a2 )S1 + (a2 − a3 )S2 + (a3 − a4 )S3

10

1 Preliminaries

+ · · · + (an−1 − an )Sn−1 + an Sn . In  notation: Let {ak }nk=1 and {bk }nk=1 be finite sequences of numbers, where n ≥ 2, and let  Sk  kj1 b j for 1 ≤ k ≤ n. Then n 

ak bk 

k1

n−1 

(ak − ak+1 )Sk + an Sn .

k1

Prove this in  notation without resorting to algebraic manipulations concealed behind an  T k be defined in order to obtain the alternative  ellipsis · · · How should version nk1 ak bk  a1 T1 + nk2 (ak − ak−1 )Tk ? 1.1.P5. (Relevant for Problem 1.1.P7 and to be used in Theorem 4.3.1) If one visualizes two disjoint intervals on the real line, one of them lies entirely on the left of the other. Prove the following precise formulation of this idea for bounded intervals: If two bounded nonempty intervals are disjoint, then the right endpoint of one of them is less than or equal to the left endpoint of the other. Deduce that, if the union of two disjoint intervals is an interval, then the sum of their lengths is equal to the length of the union. 1.1.P6. Suppose F is a finite family of disjoint bounded intervals. Show that there exists a nonempty finite family G of bounded intervals such that (a) a union of two distinct intervals in G is never an interval; (b) ∪G  ∪F (Here “∪H” means union of all the intervals in H); (c) the total length of the intervals in G is the same as for F. [Note that the intervals in G are necessarily disjoint in view of (a).] 1.1.P7. If an interval I has an endpoint that belongs to an interval J, show that the union I ∪ J is an interval. (Only a left endpoint need be considered.) 1.1.P8. Let G be a finite nonempty family of bounded intervals such that a union of two distinct intervals of G is never an interval. Suppose F is a finite family of disjoint bounded intervals such that ∪F  ∪G. Show that every interval of F is contained in some interval of G. What more can be said if F too is nonempty and has the property that a union of two distinct intervals of it is never an interval? 1.1.P9. Let F1 and F2 be finite families of bounded intervals, each having the property that its intervals are disjoint. If ∪F1  ∪F2 , show that the total length of the intervals of each family is the same.

1.2 Improper Integrals Towards the end of the preceding section, we encountered some difficulty in assigning a size, or total length, to certain subsets of R. What we did have was an instance

1.2 Improper Integrals

11

of a family F of subsets of an interval, with each set A ∈ F having an associated nonnegative real number; in effect, a function μ: F → R such that μ(A) ≥ 0 for every A ∈ F. It is easy to conceive of other instances of a family F of subsets of some set X with a nonnegative-valued function μ: F → R. In fact, here is a trivial example: Let X consist of two distinct elements a and b, and let F consist of all subsets. Define μ by setting μ(∅)  0, μ({a})  2, μ({b})  3 and μ(X)  5. The fact that so trivial an example, lacking any ostensible purpose, is possible testifies to the wide sweep of the idea. It turns out however that examples that do have an ostensible purpose and are equally far removed from the original one in which X is an interval are also possible. Readers who have worked with probability will recognize that, in what is called a sample space, the family F of subsets called “events” has a probability μ: F → R defined on it. In recent years, there has been great interest in what are called belief measure, plausibility measure and possibility measure, all of which are cases of a number μ(A) being assigned to every subset A of some set. For details, the reader may consult the book “Fuzzy Measure Theory” by Wang and Klir [11]. A function whose domain consists of some or all subsets of a set is sometimes called a set function. All the functions in the above three paragraphs are set functions. What is of immediate interest against the background of the preceding section is that the concept of distribution makes sense in such a general framework. If a nonnegative-valued function f : X → R has the property that {x ∈ X : f (x) > t} belongs to the family Ffor every t ≥ 0, then μ({x ∈ X : f (x) > t}) is meaningful and we can take it as f˜(t) for t ≥ 0. If f takes only finitely many values, then as observed before for step functions, the set {x ∈ X : f (x) > t} remains unchanged as t varies in the open interval between two consecutive values; thus f˜: [0, ∞) → R is a step function and therefore has a Riemann integral on any bounded subinterval of its domain. Since it is 0 on [sup f , M ∞), the integral 0 f˜ is the same for all M ≥ sup f . Therefore M lim

M→∞ 0

f˜ 

sup f

f˜.

0

M The limit here is of course the improper integral 0 f˜. To accommodate functions that may take infinitely many values, in which case the distribution may fail to be a step function, we would like μ to be monotone in the following sense: A ⊆ B ⇒ μ(A) ≤ μ(B); this ensures that f˜ is decreasing, because t1 > t2 ⇒ {x ∈ X : f (x) > t1 } ⊆ {x ∈ X : f (x) > t2 }. M Since f˜ is decreasing, we know that the integral 0 f˜ exists for all M (see Theorem 9.3.3 of Shirali and Vasudeva [9] or 6.9 of Rudin [7]) and increases with M in

12

1 Preliminaries

view of the nonnegativity of f˜. Consequently, the limit

∞ 0

f˜  lim

either as a real number or as ∞.

M

M→∞ 0

f˜ exists

 1.2.1 Remark. We have not ruled out the possibility that f˜  ∞. We also do not wish to rule out the possibility that the size of a subset is ∞, so that we can take X to be R or [0, ∞); this means allowing f˜ to take ∞ as a value. We shall therefore need some conventions about ∞ when it appears in a computation, and we state them below. If one or more of the terms in a statement or computation turn out to be ∞, then the validity of the assertion will generally have to be checked separately by applying the conventions we are about to articulate. Doing so is usually straightforward and details will be left to the reader. In adopting these conventions, we are introducing ∞ as a mathematical object and no longer regarding it as just convenient shorthand. x < ∞ for every x ∈ R; ∞ ≮ ∞; x + ∞  ∞ + x  ∞ for every x ∈ R; ∞ + ∞  ∞; ∞ · ∞  ∞; ∞ p  ∞ for every p > 0; x · ∞  ∞ · x  ∞ for every positive x ∈ R; ∞ · 0  0 · ∞  0.   Because of (a), the statement f˜  ∞ [or f˜(t) < ∞ ] is equivalent to f˜  ∞  [or f˜(t)  ∞], which is also expressed by saying that f˜ [or f˜(t)] is finite. Furthermore, 0 < ∞ and therefore ∞ is regarded as positive but only “extended real”. Thus, a function taking values in {x ∈ R: x ≥ 0} ∪ {∞}, a set to be denoted henceforth by R∗+ , is said to be nonnegative extended real-valued. Note that such a function need not actually take the value ∞. The symbol R+ will denote the set {x ∈ R: x ≥ 0} of nonnegative real numbers; a function taking values in R+ will be described as nonnegative real-valued. We also adopt the convention that the supremum of a nonempty set that is not bounded above is ∞. Moreover, for a sequence {sn }n≥1 in R∗+ , we take limn→∞ sn  ∞ to mean that for every A > 0 there exists a natural number n0 such that n > n 0 ⇒ sn > A. Then it is easy to see that, as far as an increasing sequence {sn }n≥1 is concerned, the statements (a) (b) (c) (d) (e) (f)

lim sn  ∞ and sup{sn : n ≥ 1}  ∞

n→∞

both mean that the sequence is unbounded above. Therefore, for such a sequence, we have lim sn  sup{sn : ≥ 1}.

n→∞

It is elementary that this equality holds good for a bounded increasing sequence (see Theorem 3.3.3 of Shirali and Vasudeva [9] or 3.14 of Rudin [7]). Thus it holds for any increasing sequence, whether bounded or not. In particular, the sum of a series

1.2 Improper Integrals

13

of nonnegative terms is the supremum of its partial sums, and the sum is ∞ if and only if the partial sums are unbounded. We shall later need the consequence of the “rearrangement theorem” (see Theorem 4.2.4 of Shirali and Vasudeva [9] or 3.55 of Rudin [7]) that the sum of a series of nonnegative terms is independent of the order of terms even if the sum is ∞. For increasing sequences, we have limn→∞ (sn + tn )  limn→∞ sn + limn→∞ tn whether the limits are finite or not; similarly, for nonempty sets, we have sup {s+t: s ∈ S, t ∈ T }  sup S + sup T , whether the suprema are finite or not (obvious when one supremum is ∞; for finite suprema, see Problem 1.6.P6 of Shirali and Vasudeva [9]). The first part can be summarized as “the limit of a finite sum of increasing sequences is the sum of their separate limits”. If one term in a series of nonnegative terms is ∞, then every partial sum up to that term and beyond is ∞ and hence so is the sum of the series. Suppose that for each n ∈ N, we have a series ∞ k1 an,k of nonnegative terms in sums form a sequence of nonnegative terms, using which one could set R∗+ . Then their  ∞ a . What this means is that we have a function N×N → up a series ∞ n,k n1 k1 ∞ a: ∞ R∗+ , called a double sequence, and an associated repeated sum ∞ k1 an,k . By  n1 reversing the roles of n and k, we get another repeated sum ∞ k1 n1 an,k . It will be useful to note that values of the two sums are the same. This is of course trivial it is true when every finite, when one of the terms an,k is ∞. To see why is  Kterm N K N a  consider any N, K ∈ N. It is elementary that n1 n,k k1 k1 n1 an,k . Therefore ∞  N  k1 n1

an,k  lim

K →∞

K  N  k1 n1

an,k  lim

K →∞

N  K  n1 k1

an,k 

N  ∞ 

an,k ,

n1 k1

where the final  K equality is based on the fact that the limit as K → ∞ of the sum of the N an,k is the sum of their separate limits. Since each an,k is nonnegative, sequences k1  ∞ ∞  N  N ∞ we can now argue that ∞ k1 n1 an,k ≥ k1 n1 an,k  n1 k1 an,k . ∞ ∞ ∞ Upon taking the limit as N → ∞, we obtain ∞ k1 n1 an,k ≥ n1 k1 an,k . The reverse inequality can be established by an analogous argument and therefore ∞ ∞ ∞ ∞ n1 k1 an,k  k1 n1 an,k (“interchanging the order of summation”). For a detailed discussion of this topic, particularly when an,k is real or complex and not restricted to be nonnegative when real, the reader may consult the article “Double sequences and double series” by Habil [4]. The concepts of liminf and limsup carry over quite smoothly and we shall encounter the liminf in Theorem 3.2.3. We have not introduced −∞, although we could have done so by extending appropriately  the list (a)–(e) above. Therefore, care must be taken to avoid such expressions as − f˜ or − f˜ without ensuring first that the integral or function concerned is finitevalued. ∞ We have encountered 0 f˜ before and now we need to lay down what such an “improper” integral means when the integrand can have an infinite value and need not vanish beyond some M > 0. However, we need consider only functions that are

14

1 Preliminaries

decreasing and take nonnegative values, ∞ included; in other words, decreasing nonnegative extended real-valued functions. They need not be distributions of any functions, and accordingly, we shall dispense with the tilde in the rest of this section. 1.2.2. Definition. For a decreasing nonnegative extended real-valued function f on [0, ∞), we define the integral from 0 to ∞ as ∞

b f  sup{

0

g: 0 < a < b < ∞; g: [a, b] → R Riemann integrable a

and 0 ≤ g ≤ f on [a, b]}.

(1.6)

It is understood here that g is real-valued, because otherwise it would not be Riemann integrable. This is similar to the “extended integral” defined by Munkres [5, p. 121], except that the domain of integration here is always [0, ∞) and the integrand must be decreasing, but the integrand as well as the integral may take ∞ as a value. Consider the case when f is real-valued except possibly at 0. Since it is decreasing, its restriction to any [a, b], where 0 < a < b < ∞, is Riemann integrable. Moreover, b b a f ≥ a g for any g of the kind mentioned in (1.6). It follows that ∞ f real-valued on (0, ∞) ⇒

b f  sup{

0

f : 0 < a < b < ∞}.

(1.7)

a

If f is not real-valued on (0, ∞), that is to say, f (α)  ∞ for some real α > 0, then by considering a sequence of nonnegative functions gn ≤ f such that gn (x)  n on [0, α], ∞ we find that 0 f  ∞. Therefore ∞ f < ∞ ⇒ f real-valued on (0, ∞).

(1.8)

0

Now suppose f is real-valued on (0, ∞). If f (0) is also real, then f is Riemann integrable over [0, A] for every A > 0, because it is decreasing. The usual meaning of ∞ A the improper integral 0 f is then lim A→∞ 0 f which agrees with the supremum in (1.6) because f is nonnegative-valued. If f (0)  ∞, then f is Riemann integrable over [x, B] whenever 0 < x < B < ∞. Two possibilities arise. One is that f vanishes beyond ∞ B some B > 0; the usual meaning of the improper integral 0 f is then lim x→0 x f , which again agrees with the supremum in (1.6). The other possibility is that  ∞there is no such B. In this situation, the usual meaning of the improper integral 0 f is B A lim x→0 x f + lim A→∞ B f , which is independent of the chosen B > 0. With a little effort, one can see that this too agrees with the supremum in (1.6). Here, the result that the sum of the integrals of a function on contiguous intervals [α, β] and [β, γ] equals the integral of that function over [α, γ] is relevant.

1.2 Improper Integrals

15

We conclude from the above paragraph that when f is real-valued on (0, ∞), the ∞ integral 0 f in the sense of Definition 1.2.2 can be computed as usual. 1.2.3. Lemma. If {gn }n≥1 is a sequence of real-valued functions on an interval [a, b] such that (a) gn is a decreasing function on [a, b] for each n ∈ N, (b) gn (x) ≤ gn+1 (x) for each x ∈ [a, b] and each n ∈ N, (c) limn→∞ gn (x)  g(x) for each x ∈ [a, b] and g(a) < ∞, then g is Riemann integrable and b

b g  lim

gn .

n→∞

a

a

Proof: Since (a) and (c) together imply that g is a real-valued decreasing function on [a, b], it follows that it is Riemann integrable. We need prove only the limit. In view b of (b) and (c), we already know that limn→∞ a gn exists and b

b g ≥ lim

gn .

n→∞

a

(1.9)

a

The proof will be complete as soon as we prove the reverse inequality. For this purpose, consider any ε > 0. There exists a partition P: a  x0 < x1 < . . . < xm  b of [a, b] such that the lower sum L(g, P) of g over P satisfies b g − L(g, P)
0, there exist a and b such that 0 < a ≤ b < ∞ and ∞ 0≤

b f −

f
0, there exist a and b such that 0 < a ≤ b < ∞ and ∞ > a f > K . The foregoing lemma yields an integer N ∈ N such that b

b f −

n≥N ⇒0≤ a

∞

b fn
K . a

∞

Therefore a f  limn→∞ a f n also when a f  ∞ and f is real-valued on (0, ∞). ∞ Finally, suppose a f  ∞ but f is not real-valued on (0, ∞). Then f (α)  ∞ for some real α > 0. From (b) and (c) of the hypothesis, we deduce that for any M > 0, there exists an N ∈ N such that n ≥ N ⇒ f n (α) ≥ M. It follows from (a) that ∞ ∞ n ≥ N ⇒ f n ≥ M on [ α2 , α], so that 0 f n ≥ α/2 M  α2 M. Since such an N exists ∞ ∞  for any M > 0, we have limn→∞ a f n  ∞  a f . Problem Set 1.2 1.2.P1. For the double sequence a: N × N → R given by an,k  1 if n  k, an,k  −1 if k  n + 1 and an,k  0 in all other cases, ∞ ∞ ∞ ∞ show thatthe repeated sums  n1 k1 an,k and k1 n1 an,k are unequal, ∞ ∞ ∞ ∞ although n1 | k1 an,k | and k1 | n1 an,k | are convergent. Hint: One can think of an,k as representing an infinite matrix with first three rows: 1 −1 0 0 . . . 0 1 −1 0 0 . . . 0 0 1 −1 0 0 . . . 1.2.P2. Let b > 0 and g be a decreasing nonnegative function on [0, b] that is realvalued except perhaps at 0. Denote by f the decreasing nonnegative function on [0, ∞) obtained by extending g to be equal to 0 to the right of b. Show that ∞

b f  lim

g.

x→0

0

x

Note: If g(0) < ∞, the function g is Riemann integrable on [0, b] and the limit here b is equal to a g (see Problem 9.3.P2 of Shirali and Vasudeva [9]).

18

1 Preliminaries

1.2.P3. In Lemma 1.2.3, show that the condition g(a) < ∞ does not follow from the remaining conditions in the hypothesis even if we impose the additional condition that g is real-valued on (a, b]. If g(a)  ∞ but g (which must be decreasing) is realb b valued on (a, b], then the improper integral a g in the usual sense is lim x→0 x g, ∞ b though it may be ∞; show that 0 g  limn→∞ a gn . If g is not real-valued on (a, b b], which to say, g(α)  ∞ for some α > a, show that limn→∞ a gn  ∞.

Chapter 2

Measure Space and Integral

2.1 Measure and Measurability The intent behind the definitions that are about to follow may be easier to appreciate if Remark 1.1.6 is kept in view. Symbol-free descriptions of the properties in the definitions are given alongside. The complement of a set will be indicated by a superscript “c”; thus, the symbol Ac will mean the complement of the set A. 2.1.1. Definition. A family F of subsets of a nonempty set X is called a σ-algebra (or a σ-field) if (a) ∅ ∈ F (contains the empty set), (b) for every sequence {Ak }k ≥1 of sets, each belonging to F, ∪∞ k1 Ak ∈ F (closed under countable unions), (c) Ac ∈ F whenever A ∈ F (closed under complements). Since X  ∅c , conditions (a) and (c) in the definition together imply that X ∈ F. Also, conditions (b) and (c) together imply closure under countable intersections, considering that ∞  k1

Ak  (

∞ 

Ack )c .

k1

From conditions (a) and (b), we infer that a σ-algebra is closed under finite unions as well: if Ak ∈ F for 1 ≤ k ≤ n, then

n 

Ak ∈ F.

k1

© Springer Nature Switzerland AG 2018 S. Shirali, A Concise Introduction to Measure Theory, https://doi.org/10.1007/978-3-030-03241-8_2

19

20

2 Measure Space and Integral

This is true because we may take Ak  ∅ for k > n. The same goes for finite intersections as well. In particular, if A and B belong to F, then so does the set-theoretic difference A\B, which is defined to be A ∩ Bc . These facts and their simple consequences will be used without further ado. We shall speak of three or more sets as being disjoint when no two of them have an element in common. This is the same as what many authors call “mutually disjoint” or “pairwise disjoint”. 2.1.2. Definition. A nonnegative extended real-valued set function μ: F → R∗+ , where F is a σ-algebra of subsets of a set X, is called a measure if (a) μ(A) ≥ 0 for every A ∈ F (nonnegative), (b) μ(∅)  0 (vanishes at the empty set), (c) for every sequence {An }n≥1 of disjoint sets, each belonging to F,

μ

∞ 

 An

n1



∞ 

μ(An )(countably additive).

n1

The triplet (X, F, μ) is called a measure space. The backdrop for countable additivity is requirement (3) of Remark 1.1.6. Given any finite sequence A1 , …, An of disjoint sets belonging to F, it is posible to extend it to an infinite sequence of disjoint sets belonging to F by appending ∅, ∅, …, and hence properties (b) and (c) of Definition 2.1.2 justify the assertion that  n  n   Ak  μ(An )(finitely additive). μ k1

k1

This has the further consequence that A ⊆ B ⇒ B  A ∪ (B\A) ⇒ μ(B)  μ(A) + μ(B\A) by finite additivity, because A ∩ (B\A)  ∅; monotonicity (which means A ⊆ B ⇒ μ(A) ≤ μ(B)) now follows from property (a) of Definition 2.1.2. In the terminology of these definitions, Remark 1.1.6 says that we would like the total length idea to be extended so as to become a measure μ on a σ-algebra F that includes all those subsets of [a, b] that are finite unions of disjoint intervals. By “extend” we mean that for such finite unions, μ must agree with total length. Elements of a σ-algebra F are called F-measurable, or just measurable especially when the σ-algebra is understood and has not been denoted by any symbol. When this is so, it is easier to speak of a measure on a set X, with the tacit understanding that some σ-algebra is intended but may or may not have been named. Similarly, it is often convenient to speak of X as a measure space, with the tacit understanding that some σ-algebra and measure are intended but may or may not have been named. It will be convenient to abbreviate μ(A) as μA when there is no risk of confusion.

2.1 Measure and Measurability

21

2.1.3. Example. (a) Consider the example mentioned at the beginning of Sect. 1.2, in which X consists of two distinct elements a and b, and F consists of all subsets. The set function μ is defined by setting μ(∅)  0, μ({a})  2, μ({b})  3 and μ(X)  5. It is easy to check that μ is a measure. Let the function f on X be defined as f (a)  1, f (b)  3. Then the set S  {x ∈ X: f (x) > t} is seen to be as described in the next three lines: X if 0 ≤ t < 1 {b} if 1 ≤ t < 3  sup f ∅

if 3 ≤ t.

Since F consists of all subsets of X, there is such a thing as μS whatever nonnegative number t may be. Therefore there exists a distribution f˜, as explained in Sect. 1.2. (b) Let X consist of three distinct elements a, b, c and let F consist of all subsets. Define μ by setting μ∅  0, μ{a}  2, μ{b}  3, μ{c}  4, μ{b, c}  6, μ{c, a}  5, μ{a, b}  4 and μX  8. Obviously, μ is nonnegative and vanishes at the empty set; one can easily check case by case that it is also monotone. However, it is not a measure, as it fails to be finitely additive: μ{a}  2, μ{b}  3 but μ{a, b}  4  2 + 3. Next, consider the function f on X defined as f (a)  1, f (b)  3, f (c)  4. For this function, the set S  {x ∈ X: f (x) > t} is seen to be as described in the next four lines: X {b, c} {c} ∅

if 0 ≤ t < 1 if 1 ≤ t < 3 if 3 ≤ t < 4  sup f if 4 ≤ t.

Since F consists of all subsets of X, there is such a thing as μS whatever nonnegative number t may be. Therefore there is a distribution f˜. If F were to consist of selected subsets, then f would have to have the property that S is always one of the selected subsets before we could speak of its distribution. 2.1.4. Definition. Let F be a σ -algebra of subsets of a set X. A function f : X → R ∪ {∞} is said to be F-measurable (or just measurable) if , for every t ∈ R, the set S  {x ∈ X: f (x) > t} is F−measurable, i.e. belongs to the σ-algebra F. If there is also an extended real-valued function μ: F → R∗+ , then the distribution of a measurable function f on X is the function ˜f : [0, ∞) → R∗+ such that f˜(t)  μ({x ∈ X : f (x) > t}).

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2 Measure Space and Integral

The distribution of a sum f + g will be denoted by ( f + g ) or (  f + g), depending on convenience. When we speak of a distribution, μ will usually be understood to be a measure but will occasionally be just nonnegative, monotone and vanishing at ∅, as was the case in Example 2.1.3(b). As was indicated just before Remark 1.2.1, this is enough to ensure that the distribution is nonnegative and decreasing. If f takes only negative values, then {x ∈ X: f (x) > t} is empty for any t ≥ 0 and hence f˜ is 0 everywhere; however, we shall have no occasion to consider such a situation. In fact, we shall be dealing with distributions only of measurable nonnegative functions although measurability has been defined to be applicable to functions that may take negative values. For reasons set forth in Sect. 1.1, when f is nonnegative and bounded, the restriction of f˜ to [0, sup f ] will also be called the distribution. The set {x ∈ X: f (x) > t}, sometimes called the “t-cut” of f , will be denoted by X(f > t) when convenient. Without explicit mention of F, one can state that an R ∪ {∞}valued function f on a set X is measurable if its t-cut is measurable for every real number t. Thus, when we speak of a measurable function on a set X, it is understood that some σ-algebra of subsets of X is intended. 2.1.5. Exercise. Let f , g be measurable functions on a space X with measure μ, and f (x)  g(x) for all x ∈ A, where A is a measurable subset of X satisfying μ(Ac )  0. Show that f˜  g. ˜ Solution: Consider any real t and the t-cuts B and C of f and g respectively. Since a measure is monotone and μ(Ac )  0, it follows that μ(B ∩ Ac )  0. Now B  (B ∩ Ac ) ∪ (B ∩ A) and additivity of measure implies μ(B)  μ(B ∩ Ac ) + μ(B ∩ A)  μ(B ∩ A). This means f˜(t)  μ(B ∩ A). Similarly, g(t) ˜  μ(C ∩ A). By hypothesis, f and g agree on the set A, which implies that the t-cuts B and C satisfy B ∩ A  C ∩ A. 2.1.6. Definition. For any subset A ⊆ X, the characteristic function χ A is the function with domain X, for which  1 if x ∈ A χ A (x)  0 if x ∈ / A. is 1, unless For the characteristic function χ A of any subset A ⊆ X, the supremum A is empty, in which case the supremum is 0. Thus the interval 0, sup χ A is [0, 1] if A is nonempty and is the single point interval [0, 0] when A is empty. It is easy to verify that (i) if A ⊆ B, then χA ≤ χB (ii) if A ∩ B  ∅, then χA χB  0 everywhere and χA +χB  χA∪B . In Exercise 1.1.5, the relation between the function f n and the complement S n of the set formed by the first n terms of the sequence {r k }k ≥1 is described. In the language of the above definition, the relation is that f n  χ Sn . The relation described

2.1 Measure and Measurability

23

there between the function f  limn→∞ f n and the set S  ∩∞ n1 Sn is that f  χS . On the basis of the inclusion S n ⊇ S n+1 , we obtain χ Sn ≥ χ Sn+1 , which is simply a restatement of the inequality f n ≥ f n+1 of that exercise (cf. Problem 2.1.P1). 2.1.7. Proposition. The characteristic function of a subset A ⊆ X is measurable if and only if the subset A is measurable. When there is a measure μ on X, the characteristic function has a distribution χ A , whose restriction to [0, 1] is a step function having value A on [0, 1) and 0 at 1. When μA  0, this means χ A is 0 everywhere. Moreover, ∞

sup χA

χ A  0

χ A  μA. 0

Proof: For t < 0, the t-cut X(χA > t) is X, which is always measurable (because any σ-algebra has X as an element). For 0 ≤ t < 1, the set X(χ A > t) is A and for t ≥ sup χ A , the set X(χ A > t) is ∅, which is always measurable (because any σ-algebra has ∅ as an element). One consequence of this observation is that χ A is measurable if and only if A is. Another consequence is that χ A (t) is μA on [0, 1) and 0 at 1. Thus χ A is a step function on [0, 1].

sup χ A

1 If A  ∅, then sup χ A  1 and hence 0 χ A  0 χ A  μA; if A  ∅, then

sup χ A

0 χ A  0 χ A  0  μA.  sup χ A  0 and hence 0 It is worth noting that the above proposition is valid for any set function μ: F → R∗+ that vanishes at the empty set. 2.1.8. Example. (a) Let X consist of two elements a and b and F consist of only ∅ and X. It is trivial that F is a σ-algebra. The function f for which f (a)  1, f (b)  3, is not measurable because X ( f > 1)  {b} ∈ / F. Therefore it has no distribution. (b) Let X, F and μ be as in Example 2.1.3(a). Define the nonnegative functions f and g as below: f (a)  1, f (b)  3 and g(a)  5, g(b)  1. Then (f + g)(a)  6, (f + g)(b)  4. Also, ⎧ ⎧ ⎪ ⎪ ⎪ ⎪ 5 5 0 ≤ t < 1 ⎪ ⎪ ⎪ ⎪ ⎨ ⎨ f˜(t)  3 1 ≤ t < 3 , g(t) ˜  2 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩0 3≤t t)  (X ( f > r ) ∩ X (g > t − r )). r ∈Q

The sets X(f > r) ∩ X(g > t − r) on the right side of this equality are measurable and are countable in number (i.e. can be arranged in a sequence). Therefore their union, which has been shown to be X(f + g > t), must be measurable. (c) Here we shall prove only the case when both functions are real-valued. The case when both are nonnegative extended real-valued is left as Problem 2.2.P11. First we shall prove that measurability of f implies that of f 2 . If t < 0, then X(f 2 > t)  2 X, which is measurable. If t ≥ 0, then √ f (x) > t is equivalent to2the assertion that √ either f (x) > t√or f (x) < − t and hence we have X ( f > t)  X ( f > √ t) ∪ X ( f > − t), which is measurable by Proposition 2.2.2(a). To prove measurability of the product fg, we note that fg 

 1 ( f + g)2 − f 2 − g 2 2

when f and g are both real-valued, and apply parts (a) and (b). (d) For an arbitrary real number t, we have max{ f, g}(x) > t if and only if f (x) > t or g(x) > t. Therefore X (max{ f, g} > t)  X ( f > t) ∪ X (g > t).

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2 Measure Space and Integral

Since both sets on the right are measurable, the one on the left is measurable. It follows that max{f , g} is measurable. The argument for min{f , g} is similar.  We need the following simple fact to proceed further. 2.2.4. Proposition. For any x ∈ R ∪ {∞}, let x +  max{x, 0}, x −  max{−x, 0} i f x ∈ R, and ∞+  ∞, ∞−  0. (a) Then x + ≥ 0, x − ≥ 0, x  x + − x − and |x|  x + + x − , where |∞| means ∞. In particular, x + ≤ |x| and x − ≤ |x|. (b) If x ∈ R, x+ 

1 1 (|x| + x) and x −  (|x| − x). 2 2

(c) Moreover, if x  y − z, where y ≥ 0 and z ≥ 0, then y ≥ x + and z ≥ x − . (Obviously, z cannot be ∞ here.) Proof: (a) Since this is trivial when x  ∞, we need only consider x ∈ R. If x ≥ 0, then x +  max{x, 0}  x ≥ 0, x −  max{− x, 0}  0 ≥ 0 and |x|  x. Therefore x + − x −  x − 0  x and x + + x −  x + 0  x  |x|. In the contrary case when x < 0, we have x +  max{x, 0}  0 ≥ 0, x −  max{− x, 0}  −x ≥ 0 and |x|  −x. Therefore x + − x −  0 − (−x)  x and x + + x −  0 + (−x)  −x  |x|. Thus the two equalities hold in both cases. (b) This follows from (a) by an elementary computation. (c) Since z cannot be ∞ here (otherwise “y − z” would make no sense), x and y are both ∞ or both finite. If both are ∞, the truth of the assertion is trivial. In the contrary case when x, y, z are all finite, y  x + z ≥ x and y ≥ 0, whence y ≥ max{x, 0}  x + . Also, z  y − x ≥ −x and z ≥ 0, whence z ≥ max{− x, 0}   x− .

2.2 Measurability and the Integral

31

Using Proposition 2.2.4, for any real-valued function f on any set X, one can form the associated functions f +  max{f ,0} and f _  max{_ f ,0}, which then have the property that f  f + − f − and | f |  f + + f − . It When f is nonnegative, we have f +  f everywhere and f −  0 everywhere.

+  f and follows that if X is a measure space and f ≥ 0 is measurable, then f



− f  0, so that f  f + − f − . For a general measurable f that need not be nonnegative, it is still true in view of Theorem 2.2.3 that the nonnegative functions

f + and f − are measurable and f + − f − makes sense at least when both integrals are real numbers. 2.2.5. Definition. Suppose μ is a measure on a set X. For any measurable realvalued function f : X → R, the measure space integral is defined to be f −, f  f+ −

provided that f − is finite (i.e.

is not ∞). The function f is said to be summable or if f is finite. integrable if both f + and f − are finite, or equivalently,

f, f dμ, or X f dμ(x) in place Often it makes for easier reading if we write X X

of f , especially if two or more measure spaces are involved in the discussion. Until we reach Sect. 3.2, we shall be dealing mainly with integrals of nonnegative (extended real-valued) measurable functions. Definition 2.2.5 has been stated here only for completeness. An R∗+ -valued measurable function with a finite integral will not be called integrable unless it is known to take only finite values. The restriction that a function be finite-valued in order to be regarded as integrable can be removed but will be retained here so that we can consider −f without having to introduce −∞ and handle the attendant complications with “∞−∞”. At this juncture, it would be well to recall that for a bounded function f : [a, b] → R, the supremum (resp. infimum) of lower (resp. upper) sums, taken over all partitions of [a, b], is called the lower (resp.upper) integral of the function over the interval

b

b

b

b  f resp. ¯ f . In general, f ≤ ¯ f . If equality holds, the and is denoted by _a

_a

a

a

function is said to be Riemann integrable and the common value of the lower and

b upper integrals is the Riemann integral a f . 2.2.6. Proposition. Let f : [a, b] → R be Riemann integrable and c > 0. Then the function g: [ca, cb] → R given by g(u)  f ( uc ) is Riemann integrable and cb

b gc

ca

which is to say,

f, a

32

2 Measure Space and Integral

cb

  f uc du  c

ca

b f (u)du. a

Proof: Let P: a  x 0 < x 1 < ··· < x n  b be any partition of [a, b]. Since c > 0, Q: ca  cx0 < cx1 < · · · < cxn  cb is a partition of [ca, cb]. For any x ∈ [x j−1 , x j ], we have u  cx ∈ [cx j−1 , cx j ] and f (x)  f ( cxc )  g(cx)  g(u). Conversely, for any u ∈ [cx j−1 , cx j ], we have

x  uc ∈ x j−1 , x j and g(u)  f ( uc )  f (x). Therefore { f (x): x ∈ [x j−1 , x j ]}  {g(u): u ∈ [cx j−1 , cx j ]}, so that inf{ f (x): x ∈ [x j−1 , x j ]}  inf{g(u): u ∈ [cx j−1 , cx j ]} and similarly for the suprema. Upon multiplying by cx j − cx j-1  c(x j − x j−1 ), and taking the summation over all j from 1 to n, we get

cb

cL( f, P)  L(g, Q) ≤

g and cU ( f, P)  U (g, Q) ≥

¯

cb

g.

ca

_ ca

Upon taking the supremum and infimum respectively over all P, it follows from here that

b

c

f ≤

_a

Since

cb _ ca

cb

g and c

¯

b

f ≥

¯

a

_ ca

cb

g.

ca

cb

b

b

b g ≤ ¯ ca g and _ a f  ¯ a f  a f by hypothesis, we conclude that c a

b

f 

cb

_ ca

g

¯

cb

g.  ca

2.2.7. Proposition. Suppose μ is a measure on a set X. For a measurable f : X → R∗+ and for 0 ≤ c < ∞, we have (c f )  c f. Proof: For c  0, the function cf is 0 everywhere and equality is seen to hold trivially, keeping in mind that ∞ · 0  0 · ∞  0 (Remark 1.2.1(f)). So, suppose c > 0.

2.2 Measurability and the Integral

33

We have (cf )(x) > t ⇔ f (x) > ct , so that X(cf > t)  X(f > ct ), which implies that cf is measurable and that  (c f )(t)  f˜( ct ). In particular, f˜ is real-valued wherever  (c f ) is. If f˜ is not real-valued on (0, ∞), that is to say, f˜(α)  ∞ for some real α > 0, then the same is true of  (c f ), in which case the required equality holds because both sides are equal to ∞. Consider the case when f˜ is real-valued except possibly at 0. The same must be true of  (c f ), and it follows by equality (1.7) of Sect. 1.2 and from Proposition 2.2.6 that the required equality holds.  It is an immediate consequence of Proposition 2.1.7 and Definition 2.2.1 that

for any measurable set, χ A  μA; it now follows from Proposition 2.2.7 that (cχ A )  c(μA) when 0 ≤ c < ∞. 2.2.8. Proposition. Suppose μ is a measure on a set X and the nonnegative

measurable functions f and g on X satisfy f (x) ≤ g(x) for every x ∈ X. Then f ≤ g Proof: From the hypothesis that f (x) ≤ g(x) for every x ∈ X, it follows that f˜(t) ≤ g(t) ˜ for every t. The required inequality follows directly from this.  2.2.9. Proposition. Suppose μ is a measure on a set X and f : X → R∗+ is mea surable. If f < ∞, then there exists a (measurable) set A such that μA  0 and f (x) < ∞ for x ∈ / A. Proof: Let A  {x ∈ X: f (x)  ∞}, so that f (x) < ∞ for x ∈ / A. The set A is meaX ( f > n). We need only prove that μA  0. Now, surable because it equals ∩∞ n1 x ∈ X and every positive integer n. It follows in view nχA (x) ≤ f (x) for every

of

and therefore n(μA) ≤ f. Proposition 2.2.8 that nχ A ≤ f . But nχ A  n(μA)

Since this must hold for every positive integer n and f < ∞, we get a contradiction unless μA  0.  2.2.10. Proposition. Suppose μ is a measure on a set X and f : X → R∗+ is mea surable. If f < ∞, then there exists a sequence {An }n≥1 of measurable sets such that (a) μAn < ∞ for every n, (b) f (x)  0 for x ∈ / ∪∞ n1 An , (c) f  f χA , where A  ∪∞ n1 An . Proof: Put An  X ( f > n1 ), so that n1 χ An (x) ≤ f (x) for every x ∈ X. By Proposition

2.2.8, we obtain ( n1 χ An ) ≤ f . But ( n1 χ An )  n1 (μAn ) and therefore n1 (μAn ) ≤

f . Since f < ∞, it follows that n1 (μAn ) < ∞, and hence (μAn ) < ∞. Thus (a) must hold. / An for every n, which means f (x) ≤ If x ∈ / ∪∞ n1 An , then x ∈ f (x) ≥ 0, it follows that f (x)  0. So, (b) must hold.

1 n

for every n. Since

Consider any x ∈ X. If x ∈ A  ∪∞ n1 An , we have χA (x)  1 and hence f (x)  f (x)χA (x)  (f χA )(x). In the contrary case, χA (x)  0 by definition of a characteristic function and f (x)  0 by (b), which leads to f (x)  0  f (x)χA (x)  (f χA )(x). Thus (c) must also hold. 

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2 Measure Space and Integral

Problem Set 2.2 2.2.P1. Prove Proposition 2.2.2(b). 2.2.P2. If f : X → R∗+ has the property that X(f ≥ t) is measurable for every t ∈ Q, show that f is measurable. 2.2.P3. Let {f n }n≥1 be a sequence of extended real-valued measurable functions on X. Then the function f defined on X as f (x)  inff n (x), i.e. inf{ f n (x): n ∈ N}, is measurable. 2.2.P4. In Problem 2.1.P3, show that the measurability of the limit function f holds without the hypothesis that f n (x) ≤ f n+1 (x). 2.2.P5. If f : X → R is measurable and φ: R → R is increasing, show that the composition φ ◦ f : X → R is measurable. 2.2.P6. Suppose μ is a measure on a set X. Prove that (a) For a measurable function f : X → R+∗ and any real p > 0, the nonnegative function f p : X → R∗+ is measurable; moreover, f  0 if and only if f˜(t)  0 for every t > 0, which, in turn, is equivalent to

f p  0 for any p > 0.

(b) If f : X → R ∗+ and g: X → R∗+ are measurable functions satisfying 0  g, then ( f + g)  0.

f 

2.2.P7. (a) Let f , g be nonnegative measurable functions on a space X with measure μ, and f (x)  g(x) for all x ∈ A, where A is a measurable subset of X satisfying μ(Ac )  0. Show that f  g. (b) Let f , g be measurable functions on a space X with measure μ, and f (x)  g(x) for all x ∈ A, where A is a measurable subset of X satisfying μ(Ac )  0. If g is

integrable, show that f is also integrable and f  g. 2.2.P8. Let f , g be nonnegative measurable functions on a space X with measure μ, c and f (x) ≤ g(x)

for all

x ∈ A, where A is a measurable subset of X satisfying μ(A )  0. Show that f ≤ g. 2.2.P9. For a function f : X → R ∪ ∞, let A be the set X(f < ∞). Show that (a) (b) (c) (d)

If t < 0, then X(f χA > t)  Ac ∪(A∩X(f > t)). If t ≥ 0, then X(f χA > t)  A∩X(f > t). The product f χA is real-valued. If f is measurable, then A is measurable and f χA is measurable.

2.2.P10. Let the function f : X → R ∪ ∞ have the property that X(t < f < ∞) is measurable for every t ∈ R. Show that f is measurable. (The converse is trivial.)

2.2 Measurability and the Integral

35

2.2.P11. If f : X → R+∗ and g: X → R+∗ are both measurable, show that their product f g: X → R+∗ is measurable. 2.2.P12. For an integrable function f : X → R∗ , show that α · μ(X (| f | > α)) ≤ | f | for all α > 0. 2.2.P13. Suppose f : X → R∗+ is a measurable function such that, for some α > 0, the function min{f , α} has a finite integral. Show that f has a finite integral if and ˜ only if the series ∞ k1 f (k) converges.

2.3 The Monotone Convergence Theorem It follows from the result of Problem 2.1.P2 that, for any sequence {f n }n≥1 of measurable functions such that f n ≤ f n+1 everywhere, we have f˜n ≤ f˜n+1 everywhere. If limn→∞ f n (x) is a real-valued function f , then it follows from Problem 2.1.P3 that it is measurable, so that it has a distribution f˜. Since f n ≤ f everywhere, it further follows from the result of Problem 2.1.P2 that the distribution satisfies f˜ ≥ f˜n . The fact that a measure has a property called inner continuity makes it possible to obtain something more satisfying; see Theorem 2.3.2 further below. 2.3.1. Proposition. Let μ be a measure. Then for a sequence of measurable sets {S n }n≥1 , we have μ(∪∞ n1 Sn )  lim n→∞ μ(Sn ) provided that S n ⊆ S n+1 for every n. (inner continuous) Proof: As observed immediately after the definition of measure (Definition 2.1.2), μ is finitely additive. Now consider a sequence {S n }n≥1 of measurable sets such that S n ⊆ S n+1 for every n ∈ N. Let A1  S 1 and An  S n \S n−1 for n > 1. Then each An is measurable and the condition S n ⊆ S n+1 has the consequence that the sets An are disjoint and that

Sn 

n 

Aj.

(2.3)

μ( A j ).

(2.4)

j1

Therefore by finite additivity, we have μ(Sn ) 

n  j1

Using countable additivity and then using (2.4), we get

36

2 Measure Space and Integral

⎛ μ⎝

∞ 

⎞ Aj⎠ 

j1

∞ 

μ( A j )

j1

 lim μ(Sn ). n→∞

(2.5)

But ∞  n1

Sn 

∞ 

Aj

j1

in view of (2.3). It follows from this equality and from (2.5) that ∞   μ Sn  lim μ(Sn ). n1

n→∞

 The functions μ in all our examples so far are inner continuous, including the one in Example 2.1.3(b), although it is not even a measure (it fails to be countably additive). The proof below does not use additivity directly but depends crucially on inner continuity. 2.3.2. Monotone Convergence Theorem. Let μ be a measure on a set X. Suppose f is a nonnegative extended real-valued function on X and {f n }n≥1 a sequence of measurable nonnegative extended real-valued functions such that (i) f n ≤ f n+1 (x) and (ii) f x  lim f n (x) f or ever y x ∈ X. n→∞

Then f is measurable, f˜(t)  limn→∞ f˜n (t) for every t ≥ 0, and also f (t)  limn→∞ f n (t). Proof: Measurability of f is a simple consequence of Problem 2.2.P4. In the sequence { f˜n }n≥1 of nonnegative extended real-valued functions, each term f˜n is a decreasing function on [0, ∞). By Problem 2.1.P3, we know that 0 ≤ f˜n (t) ≤ f˜n+1 (t) for each t ∈ [0, ∞) and each n ∈ N. By inner continuity (Proposition 2.3.1), in conjunction ˜ with Problem 2.1.P3, we also know that limn→∞ f˜n (t)

∞ f (t) for each t ∈ [0, ∞). ∞ ˜ ˜ Therefore Proposition 1.2.4 yields 0 f  limn→∞ 0 f n . But by Definition 2.2.1,

 this means f  limn→∞ f n . 2.3.3. Exercise. Show that the “monotonicity” hypothesis (i) of the Monotone Convergence Theorem 2.3.2 cannot be dropped.    Solution: Let X  N and μ(A)  j∈A 21j . Then μ is a measure. One way to see this is to note that it is by applying Problem 2.1.P10 with X j  { j} ⊂ N  obtained  and μj defined by μ j X j  21j . Define f n to be 2n times the characteristic function of {n}. Then f n  1 for each n. Also, for each j ∈ X  N, n ≥ j + 1 ⇒ f n ( j)  0; so limn→∞ f n ( j)  0 for each j ∈ X. Thus limn→∞ f n  0 although limn→∞ f n  0.

2.3 The Monotone Convergence Theorem

37

Problem Set 2.3 2.3.P1. (Needed in Problem 3.2.P1 and Theorem 4.3.4) Let μ be a measure on X and f a nonnegative measurable function such that f  0. Prove μ(X(f > 0))  0. 2.3.P2. Let {f n }n≥1 be a sequence of real-valued functions on an arbitrary nonempty set X. Suppose f is a real-valued function on X such that, for every x ∈ X, some subsequence of the real sequence {f n (x)}n≥1 converges to f (x). Show for any real t that X(f > t) ⊆ ∪∞ n=1 X(f n > t). 2.3.P3. In a measure space, suppose that every set of infinite measure contains a subset of finite positive measure. Show that every set of infinite measure contains a subset of arbitrarily large finite measure. Hint: For a given set of infinite measure, show that there exists a subset whose measure is the supremum of the measures of all subsets of finite measure. If its measure is finite, then the measure of its complement in the given set must be infinite. 2.3.P4. Let μ be a measure. For a sequence of measurable sets {S n }n≥1 with μ(S 1 ) < ∞, show that μ(∩∞ n1 Sn )  lim n→∞ μ(Sn ) provided that S n ⊇ S n+1 for every n. (outer continuous).

Chapter 3

Properties of the Integral

3.1 Simple Functions So far, we have proved the Monotone    Convergence Theorem but not the seemingly simpler property ( f + g)  f + g. In order to prove this, we need to make full use of the finite additivity of measure, which we did not have to do for the Monotone Convergence Theorem, because inner continuity of measure, a consequence of its countable additivity, was sufficient. In this section, we shall establish the property in question for a restricted class of functions and extend it to a broader class only later. 3.1.1. Definition A measurable function f : X → R is said to be simple if its range is finite. Simple functions are a generalization of step functions, which also take finitely many values, each value being taken on a union of finitely many intervals; thus, step functions are linear combinations of characteristic functions of intervals. For example, if s : [1, 3] → R is the step function that equals 3 on [1, 2), 4 on [2, 3) and s(3)  3, then s  3χ[1,2)∪[3,3] + 4χ[2,3) . The generalization consists in replacing unions of finitely many intervals by measurable sets. 3.1.2. Examples. (a) The characteristic function χ A of a nonempty measurable set A ⊂ X is a simple function with range {0, 1}. If the measurable set A is empty or equals X, the range of χ A consists of only 0 or only 1. (b) Suppose A and B are nonempty measurable sets, both distinct from X. As observed in (a), each of χ A and χ B has range {0, 1}; but the range of the sum χ A + χ B may consist of 1, 2 or 3 numbers. Indeed, the range of χ A + χ B is {1} if A ∩ B  ∅ and A ∪ B  X {0, 1} if A ∩ B  ∅ and A ∪ B  X {0, 2} if A  B (in this event, A ∩ B  A  X ) © Springer Nature Switzerland AG 2018 S. Shirali, A Concise Introduction to Measure Theory, https://doi.org/10.1007/978-3-030-03241-8_3

39

40

3 Properties of the Integral

{1, 2} if A ∪ B  X and A ∩ B  ∅ {0, 1, 2} if A ∪ B  X, A  B and A ∩ B  ∅. (c) The sum of two functions with finite range has a finite range. Since the sum of two measurable functions is measurable (Theorem  2.2.3(b)), the sum of two simple functions is simple. Therefore a finite sum nj1 α j χ A j , where the sets A1 , . . . , An are measurable and α1 , . . . , αn are real numbers, is always a simple function. It can have a value which is none of the numbers α j , as illustrated in part (b), where χ A +χ B  1· χ A +1·χ B  α1 ·χ A +α2 ·χ B with α1  1  α2  2, but χ A + χ B can have 2 as a value. 3.1.3. Remarks. (a) Let α1 , . . . , αn be the distinct values assumed by a simple function s.  Then the sets A j  X (s  α j ) are nonempty and disjoint, and have union nj1 A j  X ; moreover, each of them is measurable by virtue of Proposition 2.2.2(a), and s(x) 

n 

α j χ A j (x).

j1

. , An are any n measurable sets that are nonempty Conversely, suppose A1 , . . X , and let α1 , . . . , αn be any n distinct and disjoint, having union ∞ j1 A j  numbers. Then the simple function s  nj1 α j χ A j has the numbers α1 , . . . , αn as its distinct values and satisfies X (s  α j )  A j . In particular, the range s(X) of the function is {α j : 1 ≤ j ≤ n} . If the n distinct values α1 , . . . , αn are arranged in increasing order, we shall refer to the sum as the canonical representation of the simple function s. (b) The canonical representations of χ A + χ B in the five illustrations in Example 3.1.2(b) are respectively 1 · χ X , 0 · χ( A∪B)c + 1 · χ A∪B , 0 · χ Ac + 2 · χ A , 1 · χ( A∩B)c + 2 · χ A∩B , 0 · χ( A∪B)c + 1 · χ AB + 2 · χ A∩B . Here A  B means the “symmetric difference” ( A ∩ B c ) ∪ ( Ac ∩ B). 3.1.4. Exercise. When X  {a, b, c, d} and A  X, B  {c, d}, C  {d}, find the canonical form of the simple function s  χ A + χ B + χC . Solution: Since a and b belong to precisely one among the three sets A, B, C (namely, A), we have s(a)  s(b)  1. Since c belongs to precisely two of the sets (namely, A and B), s(c)  2. Lastly, since d belongs to all three sets, s(d)  3. To summarize, s(a)  s(b)  1, s(c)  2 and s(d)  3. Thus s takes the three values 1, 2, 3. Now,

3.1 Simple Functions

41

A1  X (s  1)  {a, b}, A2  X (s  2)  {c} and A3  X (s  3)  {d}. So, the canonical form is s  1 · χ A1 + 2 · χ A2 + 3 · χ A3  1 · χ{a,b} + 2 · χ{c} + 3 · χ{d} . The next proposition involving a canonical representation will be used to prove the stronger result that the same conclusion is true with a representation that is not canonical. The latter will make it transparent why integrals of simple nonnegative functions add up as expected. 3.1.5. Proposition. Suppose (X, F, μ) is a measure space and s : X → R a nonnegative simple function with canonical representation s(x) 

n 

α j χ A j (x).

j1

Then  s

n 

α j μ( A j ).

j1

Proof: The case when s takes only one value is trivial. Therefore we consider only the case when s has at least two values, i.e. n ≥ 2. Since the numbers α1 , . . . , αn are in increasing order (by definition of canonical), sup s  αn . Therefore 

sup s  αn s˜  s˜ . s 0

(3.1)

0

Another consequence of the numbers α1 , α2 . . . , αn being in increasing order is that, for j > 1 and α j−1 ≤ t < α j , the inequality s(x) > t holds if and only if s(x)  αk for some k ≥ j, i.e. x ∈ A j ∪ · · · ∪ An , which is to say, {x ∈ X : s(x) > t}  s(t)  μ({x ∈ X : s(x) > t})  μ( A j ∪ · · · ∪ An ) when A j ∪ · · · ∪ An . Therefore  α j−1 ≤ t < α j . Since the sets Aj are disjoint, the additivity of μ yields μ( A j ∪ · · · ∪ An )  μ( A j ) + · · · + μ( An ). Thus the distribution  s satisfies  s(t)  μ( A j ) + · · · + μ( An ) for t ∈ [α j−1 , α j ) and 1 < j ≤ n, and if α1 > 0, also satisfies

(3.2)

42

3 Properties of the Integral

 s(t)  μ( A1 ) + · · · + μ( An ) for t ∈ [0, α1 ).

(3.3)

If each μ(Aj ) is finite,  s is a step function and it follows from (3.2) and (3.3) that, whether or not α1 > 0, we have αn  s  α1 (μ( A1 ) + · · · + μ( An )) +

n 

 (α j − α j−1 ) μ( A j ) + · · · + μ( An ) . (3.4)

j2

0

One can transform the right side here by using the Abel summation formula of Problem 1.1.P5 with b1  α1 , b j  α j − α j−1 , ( j ≥ 2) and a j  μ( A j ) + · · · + μ( An ), thereby obtaining αn  s 0

n 

α j μ( A j ).

(3.5)

j1

Now  (3.1) and (3.5) together imply the required equality for the case when each μ A j is finite.  s(α)  ∞ for α ∈ Next, suppose μ A j  ∞ for some j > 1. In this event,  sides of the required equality are ∞. [α j−1 , α j ) and hence both  Finally, suppose μ A j is finite for all j > 1 but μ(A1 )  ∞. If α1 > 0, then  s(t)  ∞ for t ∈ [0, α1 ) and again both sides of the equality in question are ∞. If s is a step function and (3.4) holds with μ(A1 ) replaced α1  0, then (3.2) shows that by any real number. As before, we arrive at (3.5) but with μ(A1 ) replaced by a real number. However, (3.5) continues to hold if μ(A1 ) is changed back to ∞ because  now α1  0. Once again, (3.1) and (3.5) together imply the required equality. When μ(A) is the number of elements in A, the function μ is a measure and is called the counting measure; it is understood that μ(A)  ∞ when A is an infinite set. In Example 3.1.4, if we use the counting measure, s can be computed on the strength of the preceding proposition as  s  1 · 2 + 2 · 1 + 3 · 1  2 + 2 + 3  7.    χ A + χ B + χC  μ( A) + μ(B) + μ(C)  We note that s  χ A + χ B + χC and  4 + 2 + 1  7, which agrees with s. 3.1.6. Exercise.   Let X  {a, b, c, d} and A, B, C be subsets of X such that s  χ A + χ B + χC works out to be the 3.1.4.   If A  {b, c, d}  sameas in Exercise and B  {a, d}, find C and compare χ A + χ B + χC with s when μ is the counting measure.

3.1 Simple Functions

43

Solution: Since s(a)  s(b)  1, the elements a and b must belong to precisely one among the three sets A, B, C. It is given that they belong to B and A respectively. Therefore they do not belong to C. Since s(c)  2, c must belong to precisely two among the sets A, B, C. It is given that it belongs to A but not B. It follows that it belongs to C. A similar argument shows that d too belongs to C. Thus C  {c, d}. Therefore    χ A + χ B + χC  μ( A) + μ(B) + μ(C)  3 + 2 + 2  7,  which agrees with the value of s computed above (just before this exercise).  Before proceeding, we remind the reader (see Remark 3.1.3(a)) that nj1 α j χ A j , where α1 , α2 , . . . , αn are in increasing order, is a canonical representation of a simple function on a measurable set X if and only if (a) the measurable sets Aj are disjoint with union X, while (b) the real numbers α j are distinct and the sets Aj nonempty. We shall say that the sets Aj form a partition of X when they satisfy (a), regardless of whether some of them are empty. It will be understood of course that the sets are measurable. 3.1.7. Proposition. Let (X, F, μ) be a measure space and let s: X → R be a nonnegative simple function such that s

m 

αi χ Ai 

i1

n 

β j χB j ,

j1

where the sets Ai (1 ≤ i ≤ m) form a partition of X and so do the sets B j (1 ≤ j ≤ n), which is to say, Ai ∩ Ai1  ∅ when i  i 1 ,

B j ∩ B j1  ∅ when j  j1

(3.6)

and m

Ai 

i1

n

Bj  X.

(3.7)

j1

Then m  i1

αi μ( Ai ) 

n 

 β j μ(B j ) 

s.

(3.8)

j1

Proof: In view of (3.6), the sets Ai ∩ B j , (1 ≤ j ≤ n), are disjoint for each fixed i, and (3.7) shows that their union is

44

3 Properties of the Integral n

( Ai ∩ B j )  Ai ∩

j1

n

B j  Ai ∩ X  Ai .

j1

Therefore by finite additivity of μ, μ( Ai ) 

n 

μ( Ai ∩ B j ) for each i.

(3.9)

j1

Similarly, μ(B j )  m  i1

αi μ(Ai ) 

m 

μ( Ai ∩ B j ) for each j, and together with (3.9), this leads to

i1 n m  

αi μ( Ai ∩ B j ) and

i1 j1

n 

β j μ(B j ) 

j1

m n  

β j μ( Ai ∩ B j ).

j1 i1

(3.10) Now consider any i (1 ≤ i ≤ m) and any j (1 ≤ j ≤ n). If Ai ∩ B j  ∅, the function s takes the value αi as well as β j on this nonempty set and therefore αi  β j , so that αi μ( Ai ∩ B j )  β j μ( Ai ∩ B j ). In the contrary case, i.e. Ai ∩ B j  ∅, the equality still holds, because both sides are then zero. Thus it holds in both cases for all i and j. Using this in (3.10), we arrive at the first equality in (3.8).  Finally, since (3.6) and (3.7) are fulfilled when nj1 α j χ A j is a canonical representation of s, the second equality in (3.8) follows from Proposition 3.1.5.  3.1.8. Proposition. Let (X, F, μ) be a measure space and suppose s : X → R and t : X → R are nonnegative simple functions. Then    (s + t)  s + t.  m αi χ Ai and t  nj1 β j χ B j , where the sets Ai (1 ≤ i ≤ m) as Proof: Let s  i1 well as the sets B j (1 ≤ j ≤ n) form partitions of X. Then the sets Ai ∩ B j also form a partition of X, and moreover, s + t takes the value αi  β j on Ai ∩ B j , so that ⎞ ⎛ m n    ⎝ s+t  (αi + β j )χ Ai ∩B j ⎠. i1

j1

By Proposition 3.1.7,  s

m  i1

 αi μ( Ai ),

t

n  j1

β j μ(B j )

(3.11)

3.1 Simple Functions

45

and 

⎞ ⎛ m n    ⎝ (s + t)  (αi + β j )μ( Ai ∩ B j ) ⎠ i1



m  i1



m 

j1

⎞   m n n     ⎝ αi μ( Ai ∩ B j ) ⎠ + β j μ( Ai ∩ B j ) ⎛

j1

⎛ αi ⎝

i1

n 

⎞ μ( Ai ∩ B j )⎠ +

j1 n 

j1

βj

i1

 m 

j1

For each i, the set Ai is the disjoint union of μ, n 

n j1

 μ( Ai ∩ B j ) .

(3.12)

i1

( Ai ∩ B j ) and therefore by additivity

μ( Ai ∩ B j )  μ( Ai ).

j1

Similarly, for each j, m 

μ( Ai ∩ B j )  μ(B j ).

i1

These equalities together with (3.12) and (3.11) yield  (s + t) 

m  i1

αi μ( Ai ) +

n 

 β j μ(B j ) 

 s+

t.



j1

3.1.9. Exercise. Three junior managers in a company submit lists of three cities each. When the lists are compiled, it turns out that they together mention only five different cities, a, b, c, d, e. The cities a and b appear in 2 lists each and c, d in 1 list each. In how many lists does the city e appears? Solution: Let the sets of cities submitted by the junior managers be A, B, C. With the counting measure as μ, we have μ(A)  μ(B)  μ(C)  3. Therefore by Proposition 3.1.8, the simple function s  χ A + χ B + χC satisfies s  μ( A) + μ(B) + μ(C)  3 + 3 + 3  9. Note that s(x) tells us how many among the sets A, B, C contain the element x, i.e. how many lists the city x appears in. Now, it is given that a, b ∈ X(s  2) and c, d ∈ X(s  1). If e appears in only one list, then X(s  1)  {c, d, e} and hence μ(X (s  1))  3; also, X(s  2)  {a, b} and hence μ(X (s  2))  2. Therefore s  1 · 3 + 2 · 2  7  9. Consequently, e cannot appear in only one list. By similar considerations, we deduce that if e were  to appear in precisely two lists, then we would get s  1 · 2 + 2 · 3  8  9.

46

3 Properties of the Integral

Consequently, e cannot appear in precisely two lists. This shows that e must appear in all three lists. (Indeed, when e appears in all three lists,  we have μ(X (s  1))  2, μ(X (s  2))  2 and μ(X (s  3))  1, leading to s  1 · 2 + 2 · 2 + 3 · 1  9, as required. Problem 3.1.P3 is relevant here.) 3.1.10. Exercise. Each of four junior managers in a company submits a list of three cities to the manager, because the senior manager wants them to. The manager goes through the lists and informs the senior manager that there are seven cities in all the four lists taken together, and that three cities appear in two lists each and the remaining four in one list each. When the senior manager (who had studied a certain measure theory book very carefully) hears this, she is furious—Why? Solution: 2 · 3 + 1 · 4  3 + 3 + 3 + 3—that’s why! Problem Set 3.1 3.1.P1. Let μ be a measure on X and A1 ,…,An be measurable subsets of X such that μ( A1 ∪ . . . ∪ An ) < ∞. Show that the Inclusion-Exclusion Principle holds: ⎞ ⎛   ⎝(−1)r −1 μ( A1 ∪ · · · ∪ An )  μ( A j1 ∩ · · · ∩ A jr )⎠. 1≤r ≤n

1≤ j1 g))  0. 3.2.P2. Let X  {a, b} with the counting measure. Let {f n }n≥1 be the sequence of functions on X defined by f n  χ{a} if n is even and χ{b} if n is odd. Find   fn . (lim inf f n ) and lim inf n→∞

n→∞

3.2.P3. In this problem, (X, F, μ) is a measure space and f , g are measurable realvalued functions on X. Then the nonnegative real-valued function | f g| is measurable by Theorem 2.2.3(c) and Proposition 2.2.2(b); also, for any real p > 0 and q > 0, the real-valued functions | f + g|, | f + g| p , | f | p and |g|q are measurable by Theorem 2.2.3(b), Proposition 2.2.2(b) and Problem 2.2.P6(a). If p > 1, then | f + g| p−1 is also measurable.

3.2 Other Measurable Functions

55

(a) Suppose p, q are positive real numbers such that 

 | f g| ≤

1 p

 1  | f |p

p

1 q

+

 1. Show that

1 |g|q

q

.

This is known as Hölder’s Inequality. (b) Suppose p is a real number such that p ≥ 1. Show that 

1 | f + g|

p

p

1

 ≤

|f|

p

p

1

 +

|g|

p

p

.

This is known as Minkowski’s Inequality. 3.2.P4. (Needed in Proposition 7.7.5) Let f be an integrable function on X, where (X, F, μ) is a measure space. Given any ε> 0, show  that there exists a δ > 0 such that every A ∈ F with μ(A) < δ satisfies  (χ A f ) < ε. 3.2.P5. Show that the “dominated” hypothesis of the Dominated Convergence Theorem 3.2.6, namely that, for some integrable g, the inequality | f n (x)| ≤ g(x) holds for every x ∈ X, cannot be dropped.   1 n j−1 3.2.P6. For each fixed n ∈ N, the series ∞ converges to n+1 . Show j1 n n+1 n  ∞ 1 n j−1 for every n ∈ N and every j ∈ N that any series j1 a j satisfying a j > n n+1 diverges. 3.2.P7. Show that the following sequence converges to 0: ⎧ ⎫ ∞ ⎨ ⎬ n ⎩ j(n 2 + j 2 ) ⎭ j1

n∈N

3.2.P8. Given any measure space, show that a nonnegative measurable function f satisfies    f  sup s : 0 ≤ s ≤ f, where s is simple . Hint: Use Theorem 3.2.1 and the Monotone Convergence Theorem 2.3.2. 3.2.P9. Given any measure space (X, F, μ), let S denote the class of nonnegative simple functions s vanishing outside a set of finite measure, i.e. μ(X(s > 0)) < ∞.  This is obviously equivalent to s < ∞. (a) If every set of infinite measure contains a subset of finite positive measure, show for any nonnegative measurable function f that    f  sup s : 0 ≤ s ≤ f, s ∈ S

56

3 Properties of the Integral

(b) Prove the converse of (a). Hint: (a) Use Problem 2.3.P3. (b) Take f to be the characteristic function of any given set of infinite measure;  the condition yields s ∈ S with s > 0. Now consider any set of positive measure on which s takes a positive value. 3.2.P10. (a) Let μ be a measure on a set X. Suppose f is a nonnegative real-valued function on X and { f n }n≥1 a sequence of measurable nonnegative real-valued functions such that (i) f n (x) ≥ f n+1 (x) and (ii) f (x)  lim f n (x) for every x ∈ X. n→∞

 f N < ∞ for some N, then f  lim f n . n→∞  (b) Show that in part (a), the hypothesis that f N < ∞ for some N cannot be dropped even if μ(X ) is finite and the only subset of measure 0 is the empty set. Then f is measurable, and if



1

3.2.P11. Let the function f : [0, 1] → R be given by f (x)  x 2 . In the notation of the proof of Theorem 3.2.1, find the set E 3,9 and the number s3 43 .

3.3 Subadditive Fuzzy Measures This section can be omitted without loss of continuity. Example 2.1.3(b) exhibits a nonnegative-valued function μ with a σ-algebra F as its domain, vanishing at the empty set and satisfying monotonicity: A ⊆ B ⇒ μ(A) ≤ μ(B), where it is understood of course that A, B ∈ F. Recall that μ is not a measure because it is not even finitely additive. It was noted just before Definition 2.2.1 that the concept of the integral of a nonnegative measurable function makes sense when μ is merely nonnegative and monotone. In this section, we shall mostly be concerned with a monotone nonnegative-valued function with a σ-algebra F as its domain and vanishing at the empty set; such an object will be referred to as a fuzzy measure on X, it being understood that the domain is a σ-algebra F of subsets of X. Thus what was called “μ” in Example 2.1.3(b) is a fuzzy measure. To avoid confusion, we μ and integration with respect to it  shall denote a fuzzy measure by   ˜ the latter being more useful when there are two or more fuzzy by ∼ or by ∼ d μ, measures in the discussion.    Observe that Example 2.1.8(c) computes ∼ ( f + g), ∼ f and ∼ g for certain functions f and g, and demonstrates that the analog of Minkowski’s Inequality of Problem 3.2.P3(b) need not hold for a fuzzy measure when p  1. Similarly, Problem 2.1.P6 shows that it need not hold when p  2. A suitably modified version of Minkowski’s Inequality will be proved below in Theorem 3.3.5 for fuzzy measures that are finitely subadditive in the sense of Problem 2.1.P5. In this context, we shall drop the qualification “finitely” in the present section.

3.3 Subadditive Fuzzy Measures

57

The fuzzy measure of Example 2.1.3(b) is subadditive. Besides, the outer measures to be studied in Chap. 4 will be subadditive fuzzy measures. From a given subadditive fuzzy measure  μ, one can easily generate other subadditive fuzzy measures ν by choosing any increasing function φ : R∗+ → R∗+ such that φ(0)  0, φ(u + v)) ≤ φ(u) + φ(v) and setting  ν(E) φ( μ(E)). 3.3.1. Exercise Let  μ be a subadditive fuzzy measure with domain a σ-algebra F. If A, B ∈ F and  μ(B)  0, show that  μ( A\B)   μ(A). Solution: On the one hand, we have  μ( A) ≥  μ( A\B) and  μ( A ∩ B)  0 by monotonicity, while on the other hand, μ( A\B) +  μ( A ∩ B) by subadditivity.  μ(A) ≤  Proposition 2.2.7 and Proposition 2.2.8 about the integral with respect to a measure μ do not use the full force of the hypothesis that μ is a measure. The reader can check that their analogs for any subadditive fuzzy measure  μ are valid. We record the analogs here for convenience. 3.3.2. Proposition. Suppose  μ is a subadditive fuzzy measure on a set X. For a measurable f : X → R∗+ and for 0 ≤ c < ∞, we have   (c f )  c f. ∼

∼

Proof: The proof of Proposition 2.2.7 carries over.



3.3.3. Proposition. Suppose  μ is a subadditive fuzzy measure on a set X and the nonnegative measurable functions f and g on X satisfy f (x) ≤ g(x) for every x ∈ X.   Then ∼ f ≤ ∼ g. Proof: The proof of Proposition 2.2.8 carries over.



3.3.4. Proposition. Suppose  μ is a subadditive fuzzy measure on a set X. Then (a) For a measurable function f : X → R∗+ and any real p > 0, the nonnegative function f p : X → R∗+ is measurable; moreover,  f  0 if and only if f˜(t)  0 for every t > 0, ∼

 which, in turn, is equivalent to ∼ f p  0 for any p > 0. ∗+ ∗+ (b) If f : X →  R and g : X → R are measurable functions satisfying ∼ f  0  ∼ g, then ∼ ( f + g)  0.

58

3 Properties of the Integral



Proof: The solution of Problem 2.2.P6 carries over.

The result of the next theorem has already been illustrated in Example 2.1.8(c) with p  1 (but cf. Problem 3.3.P2 and Problem 3.3.P6) and in Problem 2.1.P6 with p  2. In both illustrations, the subadditive fuzzy measure was not a measure. 3.3.5. Theorem. Suppose  μ is a subadditive fuzzy measure on a set X and f : X → R∗+ , g : X → R∗+ are measurable functions. Then for any real p > 0, 

 ( f + g)

p

1 p+1

∼



 ≤

f

1 p+1

p



 +

∼

g

p

1 p+1

∼

and 

 ( f + g) ≤ 2 ( p

∼



p

p

g p ).

f + ∼

∼

  Proof: If either ∼ f p or ∼ g p is ∞, there is nothing to prove. So, we may assume both to be finite. If both are 0, then ∼ ( f + g) p  0 by Proposition 3.3.4 and the inequalities hold. So, we proceed only with the case ∼ f p > 0. Consider an arbitrary λ satisfying 0 < λ < 1. For any such λ, we have       1 1 1  X ( f + g) p > t  X f + g > t p ⊆ X f > λt p ∪ X g > (1 − λ)t p . Therefore the monotonicity and subadditivity of  μ imply that   + g) p (t) ≤ f p λ p t + gp (1 − λ) p t . ( f It follows from here that   ( f + g) p ≤ ∼

∞



∞

f p (λ p t)dt +

0

 gp (1 − λ) p t dt.

0

On the basis of Proposition 2.2.6, we know that  ∞   ∞ 1 1 fp f p (λ p t)dt  p f p (t)dt  p λ 0 λ ∼ 0 and 

∞ 0

Therefore

gp ((1 − λ) p t)dt 

1 (1 − λ) p

 0

∞

gp (t)dt 

1 (1 − λ) p

 ∼

g p.

3.3 Subadditive Fuzzy Measures

59

 ∼

( f + g) p ≤

1 λp

 fp+ ∼

1 (1 − λ) p

 ∼

g p.

(3.17)

Since  p this holds for all λ satisfying 0 < λ < 1, we may take the limit as λ → 1. If ∼g  0, taking the limit yields the first one of the desired inequalities. However, if ∼ g p > 0, it is permissible to select 1 p p+1 ∼ f 1 1  p p+1 + p p+1 ∼g 

λ

 ∼

f

 because thisnumber λ satisfies 0 < λ < 1, considering that we now have ∼ f p > 0 as well as ∼ g p > 0. Upon using it in (3.17), we obtain the first of the desired inequalities after a brief but elementary computation.  The second inequality follows from (3.17) by taking λ  21 . 3.3.6. Exercise. Let μ be the counting measure on the two-element set X  {a, b}. Choose f to be the characteristic function of {a} and g to be the characteristic function of {b}. Show that for p  21 , Minkowski’s Inequality does not hold but the inequality of Theorem 3.3.5 does.  Solution: Being  a measure, μ is also a subadditive fuzzy measure and means the same thing as ∼ .     p 1  p 1 Now, f p  1  g p and ( f + g) p  2. Therefore f p 1 g p 1  and ( f + g) p p  4. Since 4 > 1 + 1, Minkowski’s Inequality does not hold. How 1  p 1  p 1  2 ever, p + 1  23 and therefore, f p+1  1  g p+1 , ( f + g) p p+1  2 3 . 2 Since 2 3 < 1 + 1, the inequality of Theorem 3.3.5 does hold. 3.3.7. Exercise. Suppose  μ is a subadditive fuzzy measure on a set X and φk : X → R∗+ are measurable functions for 1 ≤ k ≤ n, where n ≥ 2. For p > 0, show that  p n−1     n  p φk ≤ 2 pk φk + 2 p(n−1) φnp . ∼

k1

∼

k1

∼

 Solution: Use induction on n. For n  2, the assertion is that ∼ (φ1 + φ2 ) p ≤  p p 2 p ∼ φ1 +2 p ∼ φ1 , which is true by virtue of the second inequality of Theorem 3.3.5. Assume as induction hypothesis that the result is true for some n ≥ 2. Then by virtue of the same inequality again, p p      n+1 n+1 p p p φk ≤2 φ1 + 2 φk . ∼

k1

∼

∼

k2

60

3 Properties of the Integral

The induction hypothesis further yields p  n       n+1  p p p p p p(k−1) p(n−1) φk ≤2 φ1 + 2 2 φk + 2 φn+1 ∼

k1



≤ 2p ≤

n  k1

p

∼

φ1 +

 n 

∼

∼



p

2 pk ∼

φk + 2 pn

∼

φk + 2 pn

∼





p

2 pk

k2



∼

k2



∼

φn+1 p

φn+1 p ,

which shows that the result is true for n + 1. Readers who are aware of pseudometric spaces and completeness will note that 1  | f − g| p p , where p ≥ 1 and the integral is taken with respect setting ρ( f, g)  to a measure, gives  rise to a pseudometric on the space of all real-valued measurable functions f with | f | p < ∞, the crucial triangle inequality being a consequence of Minkowski’s Inequality. It is well known that the corresponding metric space is complete (see 5.6.6 of Craven [1, p. 157]). If we take a subadditive fuzzy measure  μ 1  p p+1 , where p > 0, can be instead of a measure, then setting ρ( f, g)  ∼ | f − g| seen to give rise to a pseudometric on the space of all real-valued measurable functions  f with ∼ | f | p < ∞, the crucial triangle inequality now being a consequence of Theorem 3.3.5. Completeness of the corresponding metric space can be proved with the additional hypothesis that inner continuity as described in Sect. 2.3 holds for  μ. By Proposition 2.3.1, inner continuity holds for any arbitrary measure, but this is not so for subadditive fuzzy measures. For a counterexample, consider the set N with  μ defined on the σ-algebra of all subsets by putting  μ(∅)  0,  μ(N)  2 and  μ(E)  1whenever ∅  E  N; it is easy to check that  μ is a subadditive fuzzy measure but is not inner continuous. Given inner continuity for a subadditive fuzzy measure, the proof of completeness can be carried out by taking advantage of Exercise 3.3.7. Details may be found in Shirali [8]. Theorem 3.3.5 is an analog of Minkowski’s Inequality, valid for subadditive fuzzy measures. The next result is an analog of Hölder’s Inequality of Problem 3.2.P3(a), valid for subadditive fuzzy measures. 3.3.8. Theorem. Suppose  μ is a subadditive fuzzy measure on a set X and f : X → R∗+ , g : X → R∗+ are measurable functions. Suppose p, q are positive real numbers such that 1p + q1  1. Then  ∼

fg ≤



√1 p

+

√1 q

2 

 1  f

∼

p

p

1 g

∼

q

q

.

 Proof: First we dispose of the case when ∼ f p  0. Proposition 3.3.4 implies that  f (t)  0 for every t > 0. Take an arbitrary t > 0. Then for any x ∈ X,

3.3 Subadditive Fuzzy Measures

61

f (x)g(x) > t ⇒ f (x) > 0 ⇒ f (x) > s for some s > 0. In other words, X ( f g > t) ⊆ X ( f > s) for some s > 0. By monotonicity of  μ, it f g  0 by Proposition follows that f g(t)  0. Since t > 0 is arbitrary, this implies ∼   3.3.4. Thus the inequality holds when ∼ f p  0. Similarly, it holds when ∼ g q  0.   It remains to establish the inequality when ∼ f p > 0 and ∼ g q > 0. If one of them is ∞, the right side is ∞, and there is nothing to prove. So, we also assume both of them to befinite.  p 1/ p and G  g/( ∼ g q )1/q . Then by Proposition 3.3.2,  Setp F  f /( ∼q f ) ∼ F  1  ∼ G . By the inequality ab ≤

1 p 1 q a + b for any nonnegative a and b p q

(easily proved by considering the derivative of φ(ξ)  1/q + ξ/ p − ξ1/ p for ξ ∈ (0, ∞)), we have F G ≤ F p / p + G q /q. Employing Proposition 3.3.2, Proposition 3.3.3 and Theorem 3.3.5 (wherein we select ‘p’ to be 1), we obtain 2   √1 + √1 . This leads to the required inequality upon using Proposition ∼ FG ≤ p q 3.3.2 once again.   3.3.9. Example. We illustrate the set of nonnegative functions with ∼ f  1 in the particular case when X consists of two distinct elements a and b, the σ-algebra F consists of all subsets, and  μ is defined by setting  μ∅  0,  μ{a}  2,  μ{b}  3 and  μ{a, b}  4. It is easy to check that  μ is a subadditive fuzzy measure. Since X consists of two points, a nonnegative real-valued function on it can be represented by a point in the first quadrant of the coordinate plane. Let f (a)  x and f (b)  y, where x and y are both positive. Its representation in the coordinate plane is the point with coordinates (x, y). Now, ⎧ ⎧ ⎪ ⎪ ⎪ 4 0≤t 0, there exists a sequence {In }n≥1 of open intervals I n such that A ⊆ ∪∞ n1 In and ∞ 

(In ) ≤ m ∗ ( A) + ε.

n1

When it is finite, i.e. when m ∗ (A) < ∞, one may replace “≤” in the above inequality by “ 0, the sequence {In }n≥1 of open intervals with I1  (a − ε, b∞+ ε) and In  ∅ for n > 1 covers any subset of [a, b], and satisfies 0 ≤ n1 (In )  b − a + 2ε. So by Definition 4.1.3, 0 ≤ m*(A) ≤ b − a + 2ε for every ε > 0. (b) The empty set  ∅ is covered by the sequence {I n }n≥1 of open intervals with every ∗ In  ∅. Since ∞ n1 (In )  0, it follows that m (∅)  0. (c) Any sequence of open intervals that covers B also covers A and hence the set of numbers of which m ∗ ( A) is the infimum contains the corresponding set for m ∗ (B). Consequently, m ∗ ( A) ≤ m ∗ (B). (d) For every ε > 0, the sequence {In }n≥1 of openintervals with I 1  (x − ε, x + ε) and I n  ∅ for n > 1 covers {x} and satisfies ∞ n1 (In )  2ε. (e) Whenever {In }n≥1 is a sequence of open intervals that covers A, {I n + x}n≥1 is a sequence of open intervals that covers A + x and (In )  (In + x) for each n. It follows that m ∗ (A) ≥ m ∗ (A + x). Similarly, m ∗ (A + x) ≥  m ∗ (( A + x) + (−x))  m ∗ ( A). The above proposition says essentially that Lebesgue outer measure m ∗ is defined on the σ-algebra of all subsets of R and it has two special properties: (i) m ∗ of any set containing a single element is 0; and (ii) no set has m ∗ greater than the length of any interval it may be contained in. The same is true of the restriction of m ∗ to all subsets of an interval [a, b]. The rest of this section is devoted primarily to establishing the two further properties that (iii) m ∗ of an interval agrees with its length (Theorem 4.1.8) and (iv) m ∗ is “countably subadditive” (Proposition 4.1.9). 4.1.6. Example. Suppose that A ⊆ R is countable and infinite. Then m ∗ ( A) can be shown to be 0 as follows: The supposition means that the points of A can be arranged in a sequence {x k }k ≥1 . Given ε > 0, let I k be an open interval of length ε/2k with x k . The sequence {I k }k ≥1 of open intervals then covers A and satisfies ∞ midpoint  ∞ k ∗ ∗ k1 (Ik )  k1 ε/2  ε. So, m ( A) ≤ ε. A particular case is that m (Q)  0. ∗ It is of course trivial to show that m ( A)  0 if A is finite. It follows from what has just been noted that a set A ⊆ R for which m ∗ ( A) > 0 must be uncountable.

68

4 Construction of a Measure

4.1.7. Proposition. When a nonempty closed bounded interval is covered by a finite family of open intervals, the total length of the intervals in the family exceeds that of the closed interval. Proof: When there is only a single interval in the family, there is nothing to prove. Assume as induction hypothesis that whenever a nonempty closed bounded interval I is covered by a family of n open intervals, their total length exceeds that of I. Now consider a family of n + 1 open intervals covering a nonempty closed bounded interval [a, b]. One of them, call it (α, β), must contain a. Then α < a < β and it follows that b − α > b − a. Consequently, it is sufficient to argue that the total length of all the n + 1 intervals exceeds or equals b − α. If β ≥ b, it follows that β − α ≥ b − α, which means the length of the interval (α, β) alone exceeds or equals b − α. If β < b, we have [β, b] ⊆ [a, b] but [β, b] ∩ (α, β)  ∅. It follows that the other n intervals in the family must cover I  [β, b], which is a nonempty closed bounded interval. Therefore by the induction hypothesis, their total length exceeds b − β and hence the total length of all the n + 1 intervals exceeds (b − β) + (β − α)  b − α.  4.1.8. Theorem. If A is an interval, then its Lebesgue outer measure is the same as its length: m ∗ (A)  (A). Proof: As there is nothing to prove when A  ∅, we shall consider only nonempty A. First suppose that A is a bounded interval with endpoints a and b, where a ≤ b. Then A ⊆ [a, b] and (A)  b − a. Therefore it follows from Proposition 4.1.5(a) that m ∗ (A) ≤ (A). We shall show that m ∗ (A) ≤ (A) by considering the separate cases when A is closed and when A is not closed. Consider the case when the nonempty bounded interval A is closed; call it [α, β]. Let {I n }n≥1 be a sequence of open intervals that covers A and the family B be the range of the sequence. Since A is closed and bounded, the Heine–Borel Theorem provides a finite subfamily A ⊆ B that covers A. By Proposition 4.1.7, the total length of the intervals in A is greater than β − α.Now the length of each interval of A appears as a term at least once in the series ∞ n1 (In ). It follows that the sum of this series is greater than or equal to the total length of the intervals in A, and is therefore greater than β − α. This establishes that m ∗ ( A) ≥ β − α  ( A) in the case when A is closed. Next, consider the case when the nonempty bounded interval A is not closed. When this is so, A must contain more than one point and, in fact, for every ε > 0, there must exist a closed interval I ⊆ A such that (I ) > ( A) − ε. Since I is closed, we know from the case already established above that m ∗ (I )  (I ), and since I ⊆ A, we also know that m ∗ (A) ≥ m ∗ (I ). It follows that m ∗ (A) > ( A)−ε.

4.1 Lebesgue Outer Measure

69

This inequality has been shown to hold for any ε > 0, and therefore we have m ∗ (A) ≥ (A). Finally, suppose A is not a bounded interval, which is to say, it is an unbounded interval and (A)  ∞. Then it contains bounded intervals of arbitrarily great length and hence of arbitrarily great Lebesgue outer measure. From Proposition 4.1.5(c), it  follows that m ∗ (A)  ∞. The next proposition provides finite subadditivity of m ∗ upon taking A j  ∅ for j ≥ 3. In presenting its proof, it is customary to dispense with the song and dance of Remark 4.1.2(b), regarding it as obvious that the repeated sum in (4.2) below is equal to the sum of lengths of a suitable sequence of open intervals. 4.1.9. Proposition. The Lebesgue outer measure m ∗ is countably subadditive; that is, for every sequence {An }n≥1 of subsets of R, m∗(

∞ 

An ) ≤

n1

∞ 

m ∗ ( An ).

(4.1)

n1

Proof: Consider any ε > 0. In accordance with Remark 4.1.4, corresponding to each n ∈ N, there is a sequence {I n,k }k ≥1 of open intervals I n,k such that An ⊆ ∪∞ k1 In,k and ∞ 

(In,k ) ≤ m ∗ ( An ) +

k1

ε . 2n

Then the family {In,k : (n, k) ∈ N × N} covers ∪∞ n1 An and ∞  ∞ 

(In,k ) ≤

n1 k1

∞ 

m ∗ ( An ) + ε.

(4.2)

n1

Next, let φ: N → N × N be a bijection. Then the sequence {Iφ( p) } p≥1 of open intervals covers ∪∞ n1 An and hence m∗(

∞ 

An ) ≤

n1

∞ 

(Iφ( p) ).

p1

It follows from Remark 4.1.2(b) and from (4.2) that m∗(

∞  n1

An ) ≤

∞ 

m ∗ ( An ) + ε.

n1

Since this holds for every ε > 0, the desired inequality (4.1) follows.



70

4 Construction of a Measure

4.1.10. Exercise. If B is the set of all irrational numbers in [a, b], show that m ∗ (B)  b − a. Solution: By Proposition 4.1.5(a), m ∗ (B) ≤ b − a. Let A be the set of all rational numbers in [a, b]. Then A is countable and hence m ∗ (A)  0, as shown in Example 4.1.6. By Proposition 4.1.9, we have m ∗ ( A ∪ B) ≤ m ∗ ( A) + m ∗ (B)  m ∗ (B) ≤ b − a. On the other hand, A ∪ B  [a, b], so that m ∗ ( A ∪ B)  b − a by Theorem 4.1.8. Therefore the inequality displayed above yields b − a ≤ m ∗ (B) ≤ b − a. Problem Set 4.1 4.1.P1. The outer measure m ∗ on [a, b] is defined on the σ-algebra F of all subsets of [a, b] and therefore all functions are measurable and have a distribution. Find the distribution f˜ of the function f : [a, b] → R such that f (x)  1 if x is irrational  sup f f˜. (The reader is reminded that the Riemann and 0 otherwise and hence find 0 b integral a f does not exist.) 4.1.P2. For nonempty A ⊆ R, the intersection I A of all intervals containing A is a nonempty interval. Show that m ∗ (A) ≤ sup{|x − y|: x, y ∈ A}  (I A ). 4.1.P3. For any subset A of [a, b] and any ε > 0, show that there exists a subset B of ∗ ∗ [a, b] such that A ⊆ B  ∪∞ j1 J j , m (B) < m ( A)+ε and each J j is the intersection of an open interval with [a, b]. ∞ 4.1.P4. Consider   2 the set C, known as the Cantor set, defined as ∩n1 Cn , where 1 C1  0, 3 ∪ 3 , 1 , and C n+1 is obtained from C n by removing the open middle         thirds of its closed intervals. Thus, C2  0, 19 ∪ 29 , 13 ∪ 23 , 79 ∪ 89 , 1 , and so on.

0

1/3

1

2/3

C1 1/9

2/9

7/9

8/9

C2

Show that m ∗ (C)  0. 4.1.P5. Prove the reverse of the inequality  established in Remark 4.1.2(b): Suppose ∗+ that for each n ∈ N, we have a series ∞ k1 an,k of nonnegative terms in R . Let φ: N → N × N be a bijection. Prove that ∞  p1

aφ( p) ≥

∞  ∞ 

an,k .

n1 k1

Hint:  N First  N show that,  P given any N , K ∈ N, there exists a P ∈ N such that a ≤ n1 k1 n,k p1 aφ( p) . Also use the fact that the limit of a finite sum of sequences is the sum of their separate limits.

4.2 Measure from Outer Measure

71

4.2 Measure from Outer Measure The outer measure m ∗ as defined in Sect. 4.1 is an example of a nonnegative, monotone, countably subadditive set function that is defined on all subsets of R and satisfies m ∗ (∅)  0. However, it does not provide the extension we wanted in Remark 1.1.6 because we have not proved that m ∗ satisfies requirement (3) recorded there. It may be noted that we did not necessarily want the extended total length to be defined on all subsets. What we shall do now is select a σ-algebra of subsets such that the restriction of m ∗ to the σ-algebra is a measure satisfying (1), (2) as well as (3) of that remark. Of course, the selected σ-algebra will be shown to contain intervals, so that the restriction is indeed an extension of total length. The process of obtaining a σ-algebra such that the restriction of m ∗ to it is a measure is actually quite general and can be carried out with a general set X instead of R or [a, b] as the underlying set. The reader may recall that before Sect. 4.1, our σ-algebras consisted of subsets of a general set X. We return to that wider context. Instead of m ∗ , we consider a nonnegative countably subadditive function defined on the σ-algebra of all subsets of some (nonempty) set X, vanishing at ∅ and taking “bigger values on bigger sets”. Such an entity is called an outer measure. 4.2.1. Definition. An outer measure μ∗ on a nonempty set X is an R∗+ -valued function defined on every subset of X and having the properties that (a) μ∗ (∅)  0 (vanishes at the empty set); (b) A ⊆ B ⇒ μ∗ (A) ≤ μ∗ (B)(monotone); (c) for every sequence {An }n≥1 of subsets of X, μ∗ (

∞  n1

An ) ≤

∞ 

μ∗ ( An ) (countably subadditive).

n1

Finite subadditivity follows exactly as it does for Lebesgue outer measure. In the preceding section, we have shown that m ∗ has all the properties of Definition 4.2.1 and it is therefore an outer measure on R. When X is finite, any sequence of subsets must eventually repeat terms. So, a finitely subadditive nonnegative extended real-valued function defined on all subsets is perforce countably subadditive and is therefore an outer measure if it is monotone and vanishes at ∅. Examples of these have been presented before and can now be marshaled to serve as illustrations of outer measures. It is trivial to see that when an outer measure on a set X is restricted to the subsets of some Y ⊆ X, it becomes an outer measure on Y . Usually it is convenient to avoid introducing a new name or symbol for a restriction. Thus we have Lebesgue measure m ∗ on R as well as on its subsets, particularly intervals. If there is a σ-algebra of subsets such that the restriction of μ∗ to the σ-algebra is countably additive, then it is a trivial deduction that the restriction is a measure. One such σ-algebra is the one consisting of only the two subsets ∅ and X. We shall

72

4 Construction of a Measure

discuss how to construct one that has a chance of being larger, and in the case of Lebesgue outer measure on an interval, is large enough to include all subintervals. 4.2.2. Definition. When μ∗ is an outer measure on X, a subset E ⊆ X is said to be μ∗ -measurable if μ∗ ( A)  μ∗ ( A ∩ E) + μ∗ ( A ∩ E c ) f or ever y A ⊆ X. An informal description of μ∗ -measurability of a set E is that E breaks up every subset A of X in such a way as to make μ∗ look like it were additive. A set which is m ∗ -measurable, where m ∗ is Lebesgue outer measure, is said to be Lebesgue measurable. 4.2.3 Remarks. (a) By finite subadditivity, we have μ∗ ( A) ≤ μ∗ ( A ∩ E) + μ∗ ( A ∩ E c ) for every A ⊆ X. Thus, to prove μ∗ -measurability of E, it is enough to show the reverse of this inequality, which is known as the Carathéodory condition. It holds trivially when A ⊆ E or A ⊆ E c and also when μ∗ ( A)  0 or ∞. (b) If E is μ∗ -measurable, so is its complement E c . (c) If μ∗ (E)  0, then E is μ∗ -measurable. This is because, for any A ⊆ X , nonnegativity and monotonicity of μ∗ yield 0 ≤ μ∗ ( A ∩ E) ≤ μ∗ (E)  0 and μ∗ ( A ∩ E c ) ≤ μ∗ (A). So, μ∗ (A) ≥ μ∗ ( A ∩ E c ) + 0  μ∗ ( A ∩ E c ) + μ∗ ( A ∩ E), which implies E is μ∗ -measurable by (a) above. By monotonicity, every subset of E also has outer measure 0 and is therefore μ∗ -measurable. Moreover by (b), it follows that the complement is μ∗ -measurable. In particular, ∅ and X are always μ∗ -measurable. (d) A countable subset of R is m ∗ -measurable. This follows from (c) above and Example 4.1.6. 4.2.4. Example. Let X be the set consisting of two elements a and b, and μ∗ be the outer measure given by μ∗ (∅)  0, μ∗ ({a})  2, μ∗ ({b})  3 and μ∗ (X)  4. The subset E  {a} is not μ*-measurable, because when A  X, we find that μ∗ ( A ∩ E)  μ∗ (E)  2, μ∗ ( A ∩ E c )  μ∗ ({b})  3 and μ∗ (A)  μ∗ (X )  4  2 + 3. It follows that the complement E c  {b} is also not μ∗ -measurable. Thus the only μ∗ -measurable subsets are ∅ and X. 4.2.5. Definition. The collection of all μ∗ -measurable subsets of X will be denoted by M. We propose to show that M is a σ-algebra and that the restriction of μ∗ to M is countably additive and hence a measure. This will be achieved in several stages. 4.2.6. Proposition. M is closed under finite unions and finite intersections. Proof: It is sufficient to prove that E 1 , E 2 ∈ M ⇒ E 1 ∪ E 2 ∈ M. The rest follows by induction and the fact that M is closed under complements.

4.2 Measure from Outer Measure

73

So, let E 1 , E 2 ∈ M. In order to prove that E 1 ∪ E 2 ∈ M, consider any subset A ⊆ X. Since E 1 ∈ M (which means it is μ∗ -measurable), we have μ∗ ( A)  μ∗ ( A ∩ E 1 ) + μ∗ ( A ∩ E 1c ).

(4.3)

Since E 2 ∈ M, applying Definition 4.2.2 with A ∩ E 1 and A ∩ E 1c in place of A, we get μ∗ ( A ∩ E 1 )  μ∗ ( A ∩ E 1 ∩ E 2 ) + μ∗ ( A ∩ E 1 ∩ E 2c )

(4.4)

μ∗ ( A ∩ E 1c )  μ∗ ( A ∩ E 1c ∩ E 2 ) + μ∗ ( A ∩ E 1c ∩ E 2c ).

(4.5)

and

Now, (E 1 ∩ E 2 ) ∪ (E 1 ∩ E 2c ) ∪ (E 1c ∩ E 2 )  E 1 ∪ E 2 and hence ( A ∩ E 1 ∩ E 2 ) ∪ ( A ∩ E 1 ∩ E 2c ) ∪ ( A ∩ E 1c ∩ E 2 )  A ∩ (E 1 ∪ E 2 ).

(4.6)

Also, ( A ∩ E 1c ∩ E 2c )  A ∩ (E 1 ∪ E 2 )c .

(4.7)

On using (4.6), (4.7) and subadditivity of outer measure, we obtain μ∗ ( A ∩ (E 1 ∪ E 2 )) + μ∗ ( A ∩ (E 1 ∪ E 2 )c ) ≤ μ∗ ( A ∩ E 1 ∩ E 2 ) + μ∗ ( A ∩ E 1 ∩ E 2c ) + μ∗ ( A ∩ E 1c ∩ E 2 ) + μ∗ ( A ∩ E 1c ∩ E 2c ). On using (4.4), (4.5) and then (4.3), we further obtain μ∗ ( A ∩ (E 1 ∪ E 2 )) + μ∗ ( A ∩ (E 1 ∪ E 2 )c ) ≤ μ∗ ( A ∩ E 1 ) + μ∗ ( A ∩ E 1c )  μ∗ ( A). In view of Remark 4.2.3(a), this completes the proof that E 1 ∪ E 2 ∈ M.



4.2.7. Exercise. (a) Let B ⊆ A∪E and μ∗ (E)  0. Show that Ac ∩ B ∈ M. (b) Let B ⊆ A ∪ E, B ∈ M and μ∗ (E)  0. Show that A ∩ B ∈ M. (c) Let B ⊆ A ∪ E, A ⊆ B ∪ E, B ∈ M and μ∗ (E)  0. Show that A ∈ M. Solutions: (a) Ac ∩ B ⊆ Ac ∩ ( A ∪ E)  ( Ac ∩ A) ∪ ( Ac ∩ E)  ∅ ∪ ( Ac ∩ E) ⊆ E. Since μ∗ (E)  0, monotonicity implies μ∗ ( Ac ∩ B)  0 and hence Ac ∩ B ∈ M by Remark 4.2.3(c).

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4 Construction of a Measure

(b) A∩ B  ( A∩ B)∪(B c ∩ B)  ( A∪ B c )∩ B  ( Ac ∩ B)c ∩ B. Since Ac ∩ B ∈ M by part (a) and B ∈ M by hypothesis, we have A ∩ B ∈ M by Proposition 4.2.6. (c) A  A ∩ (B c ∪ B)  ( A ∩ B c ) ∪ ( A ∩ B). Since A ∩ B c ∈ M by part (a) and A ∩ B ∈ M by part (b), we get A ∈ M by Proposition 4.2.6. To prove closure under countable unions, we shall use one more result, namely, that the restriction of μ∗ to M is finitely additive. 4.2.8. Proposition. If E 1 , E 2 ,…, E n is a finite sequence of disjoint μ∗ -measurable subsets of X and A ⊆ X is any subset whatsoever, then μ∗ ( A ∩

n 

E j) 

j1

n 

μ∗ ( A ∩ E j ).

(4.8)

j1

In particular, if A  X, then μ∗ (

n 

E j) 

j1

n 

μ∗ (E j ).

j1

Proof: We shall prove this by induction on n. For n  1, the equality is obvious. Suppose for some n ∈ N that (4.8) holds whenever E 1 , E 2 ,…, E n is a finite sequence of disjoint μ∗ -measurable subsets of X and A ⊆ X is arbitrary (induction hypothesis). In order to arrive at the validity of (4.8) with n + 1 in place of n, first note that μ∗ (E)measurability of E n+1 entails that μ∗ ( A ∩

n+1 

E j )  μ∗ (( A ∩

j1

n+1 

E j ) ∩ E n+1 ) + μ∗ (( A ∩

n+1 

j1

c E j ) ∩ E n+1 ). (4.9)

j1

On the basis of the disjointness of E 1 , E 2 ,…, E n+1 , we have (A ∩

n+1 

E j ) ∩ E n+1  A ∩ E n+1 and ( A ∩

j1

n+1 

E j) ∩

c E n+1

 A∩

j1

n 

E j.

j1

In conjunction with (4.9), this implies μ∗ ( A ∩

n+1  j1

E j )  μ∗ ( A ∩ E n+1 ) + μ∗ ( A ∩

n 

E j ).

j1

The induction hypothesis now yields equality (4.8) with n + 1 in place of n.



4.2 Measure from Outer Measure

75

4.2.9. Theorem. M is a σ-algebra. Proof: From Remark 4.2.3(b) and (c), we know that M is closed under complements and that ∅ is an element of M. It remains only to show that, if E j ⊆ X are all μ∗ (E)∗ measurable ( j ∈ N), then ∪∞ j1 E j is also μ -measurable. c  Let F 1  E 1 and Fk  E k ∩ ∪k−1 E  E k \(E 1 ∪ · · · ∪ E k−1 ) for k > 1. Then j1 j ∗ ∗ each F k ⊆ E k and hence μ (Fk ) ≤ μ (E k ). Furthermore, the sets F k are all disjoint and μ∗ -measurable. The definition of F k also implies that ∪nj1 E j  ∪nj1 F j for all n ∈ N and that ∞ ∪ j1 E j  ∪∞ j1 F j . Therefore, if we show μ∗ ( A) ≥ μ∗ ( A ∩ (

∞ 

F j )) + μ∗ ( A ∩ (

j1

∞ 

F j )c )

(4.10)

j1

for every A ⊆ X, the desired μ∗ -measurability of ∪∞ j1 E j will follow. p For any positive integer p, the finite union ∪ j1 F j is known to be μ∗ -measurable by Proposition 4.2.6. Therefore we know that ∗

∗

μ ( A)  μ ( A ∩ (

p 

∗

F j )) + μ ( A ∩ (

j1

p 

F j )c ).

(4.11)

j1

Since the sets F j are disjoint and μ∗ -measurable, μ∗ ( A ∩ (

p  j1

F j )) 

p 

μ∗ ( A ∩ F j )

(4.12)

j1

c  p by finite additivity of μ∗ on M (Proposition 4.2.8). Also, since ∪ j1 F j ⊇  c ∪∞ and μ∗ is monotone, we have j1 F j μ∗ ( A ∩ (

p 

F j )c ) ≥ μ∗ ( A ∩ (

j1

∞ 

F j )c ).

(4.13)

j1

Now, (4.11), (4.12) and (4.13) together lead to μ∗ ( A) ≥

p 

μ∗ ( A ∩ F j ) + μ ∗ ( A ∩ (

j1

This is valid for all p while the sequence {  ∞ ∗

j1 μ A ∩ F j . Therefore

∞ 

F j )c ).

j1

∞ j1

 μ∗ A ∩ F j } p≥1 has supremum

76

4 Construction of a Measure

μ∗ ( A) ≥

∞ 

μ∗ ( A ∩ F j ) + μ ∗ ( A ∩ (

j1

≥ μ∗ (

∞ 

∞ 

F j )c )

j1

( A ∩ F j )) + μ∗ ( A ∩ (

j1

∞ 

F j )c )

j1



 ∞ by countable subadditivity of μ∗ . Since ∪∞ j1 A ∩ F j  A∩ ∪ j1 F j , the inequality just derived asserts the same thing as (4.10).  As was mentioned in Sect. 2.1, a set is said to be F-measurable in the context of a σ-algebra F if it belongs to F. Being an M-measurable set is the same as being a μ∗ -measurable set. 4.2.10. Exercise. Let g:X → R be an M-measurable function (which means X (g > t) ∈ M for every real t) and suppose f : X → R satisfies μ∗ (X (g  f ))  0. Show that f is M-measurable. Solution: Let E  X (g  f ). Then μ∗ (E)  0 by hypothesis. For any t, let A  X(f > t) and B  X(g > t). Then B ∈ M by hypothesis. Also, x∈A

⇒

f (x) > t

⇒

g(x) > t or g(x)  f (x)

⇒

x ∈ B or x ∈ E,

so that A ⊆ B ∪ E. A similar argument shows that B ⊆ A ∪ E. It follows by Exercise 4.2.7(c) that A ∈ M, i.e. X ( f > t) ∈ M. Thus f is M-measurable. Recall that when μ∗ is m ∗ , whether on all subsets of R or of [a, b], an Mmeasurable function is said to be Lebesgue measurable. Thus f : [a, b] → R is Lebesgue measurable if and only if, for every real t, the set {x ∈ [a, b]: f (x) > t} is m*-measurable. Theorem. 4.2.11. The outer measure μ∗ is countably additive on the σ-algebra M of μ∗ -measurable subsets of X and its restriction to M is therefore a measure. Proof: Let E j j≥1 be a sequence of disjoint μ∗ -measurable subsets. By monotonicity of μ∗ and its finite additivity proved in Proposition 4.2.8, it is true for an arbitrary positive integer p that ∗

μ (

∞ 

∗

E j) ≥ μ (

j1

p 

E j) 

j1

p 

μ∗ (E j ).

j1

Therefore ∗

μ (

∞  j1

E j) ≥

∞ 

μ∗ (E j ).

j1

The reverse inequality follows from the countable subadditivity of μ∗ .



4.2 Measure from Outer Measure

77

Problem Set 4.2 4.2.P1. Let μ∗ be a finite outer measure on a set X and A ⊆ B be subsets of X. Show that (a) if μ∗ (B\A)  0, then μ∗ (B)  μ∗ (A); (b) if μ∗ (B)  μ∗ (A) and A ∈ M, then μ∗ (B\A)  0. State the analog for three sets. 4.2.P2. Show that, if E 1 and E 2 are μ∗ -measurable, where μ∗ is an outer measure, then μ∗ (E 1 ∪ E 2 ) + μ∗ (E 1 ∩ E 2 )  μ∗ (E 1 ) + μ∗ (E 2 ). 4.2.P3. What has been called “μ” in Example 2.1.3(b) is easily seen to be an outer measure μ∗ . Which subsets are μ∗ -measurable? 4.2.P4. Let X  {a, b, c} and F be the σ-algebra consisting of the subsets {a, b},{c} besides ∅ and X. Then μ: F → R, where μ∅  μ{a, b}  0, μ{c}  μX  1, is clearly a measure X. For an arbitrary subset A ⊆ X, define μ∗ ( A) to be the infimum on ∞ of all numbers n1 μ(In ), where {I n }n≥1 is a sequence of sets in F that covers A (the infimum is taken over all such sequences, but only finitely many need be used). Find μ∗ ( A) for each of the eight subsets of X. It can be checked easily that μ∗ is an outer measure; find the μ∗ -measurable subsets. 4.2.P5. Let H be a family of subsets of a nonempty set X such that ∅ ∈ H and suppose α: H → R is nonnegative-valued, satisfying α∅ = 0. Assume that some sequence {H n }n≥1 of sets belonging to H covers X, hence also every subset of X. Show that α∗ A  inf{

∞ 

α(In ): every In ∈ H and A ⊆



n1

n≥1

In }

defines an outer measure on X. 4.2.P6. For X  {a, b, c, d}, a set of four elements, let H  {∅, {d}, {a, b}, {a, c}, {a, b, c}, X } and α: H → R be defined as α∅  0, α{d}  3, α{a, b}  0, α{a, c}  3, α{a, b, c}  2 and α X  2. Find α ∗ ( A) for all the sixteen subsets of X and determine which ones are α ∗ measurable.

78

4 Construction of a Measure

4.2.P7. Let H be a σ-algebra of subsets of X and (X, H, α) be a measure space. Any sequence of sets in H having X as one of its terms covers X. Show that for the outer measure α ∗ obtained as in Problem 4.2.P5, every set in H is α ∗ -measurable. Is the converse true? 4.2.P8. For infinite A ⊆ N, let α(A)  1 and for finite A ⊆ N, let α( A)   n j j1 1/2 , where n is the number of elements in A. (If A  ∅, the sum becomes “empty” and is to be interpreted as usual to be 0.) Show that α is an outer measure on N and find all α-measurable subsets of N. 4.2.P9. (Cf. Problem 4.3.P7) Let F be a σ-algebra of subsets of a nonempty set X and suppose the set function α: F → R∗+ vanishes at the empty set, is monotone and countably subadditive. (In other words α has all the properties of an outer measure except that its domain is F.) For any A ∈ F, put μ( A)  sup{

∞  j1

α(E j ):

∞ 

E j  A, E j ∩ E k  ∅ whenever j  k, every E j ∈ F}.

j1

(a) Show that μ is a measure and that, for all A ∈ F, we have (i) μ(A) ≥ α(A)(ii) μ(A)  0 ⇔ α(A)  0. (b) If α is as in Problem 4.2.P8, what is μ?

4.3 Lebesgue Measure In the light of what has been shown so far, the restriction of Lebesgue outer measure m ∗ on an interval I to the σ-algebra M of m ∗ -measurable sets of the interval is a measure on that σ-algebra. It is known as Lebesgue measure (on I) and we shall denote it by the letter m. Thus the triplet (I, M, m) is a measure space in accordance with Definition 2.1.2. The measure space integral with reference to Lebesgue measure is called the Lebesgue integral. We shall mostly be concerned with the cases when I is either R or a closed bounded interval [a, b]. We have yet to prove that the σ-algebra of Lebesgue measurable sets contains all subintervals of [a, b], failing which, Lebesgue measure m is not an extension of the total length concept. This is all that remains to be done for achieving what was envisaged in Remark 1.1.6 and we are now in a position to bestow the idea with entelechy. 4.3.1. Theorem. When m ∗ is Lebesgue outer measure on R or on an interval [a, b], every subinterval is m ∗ -measurable. Furthermore, it has Lebesgue measure equal to its length. Proof: It is sufficient to prove that any subinterval I of the form (c, b] is m ∗ measurable. Let A be any subset of [a, b] and let A1  A ∩ I and A2  A ∩ I c  A ∩ [a, c]. We must show that

4.3 Lebesgue Measure

79

m ∗ ( A) ≥ m ∗ ( A1 ) + m ∗ ( A2 ).

(4.14)

We need consider only the case when m ∗ ( A) is finite. The definition of m ∗ guarantees that, for every ε > 0, there exists a sequence {I n }n≥1 of open intervals such that A ⊆ ∪∞ n1 In and ∞ 

(In ) ≤ m ∗ ( A) + ε.

n1

Let I  n  I n ∩ I and I n  I n ∩ I c . Then I  n and I n are subintervals of [a, b] (some of which may be empty) and by Problem 1.1.P6, (In )  (In ) + (In ). Now, the length of a subinterval is the same as its Lebesgue outer measure according to Theorem 4.1.8. Therefore (In )  m ∗ (In ) + m ∗ (In ). ∞ ∞   Since A ⊆ ∪∞ n1 In , we have A1 ⊆ ∪n1 In and A2 ⊆ ∪n1 In . Therefore

m ∗ ( A1 ) ≤ m ∗ (

∞ 

In ) ≤

n1

∞ 

m ∗ (In )

n1

and ∗

∗

m ( A2 ) ≤ m (

∞ 

In )

n1

≤

∞ 

m ∗ (In ).

n1

Hence m ∗ ( A1 ) + m ∗ ( A2 ) ≤

∞  n1

m ∗ (In ) +

∞ 

m ∗ (In ) 

n1

∞ 

[m ∗ (In ) + m ∗ (In )]

n1



∞ 

(In ) ≤ m ∗ ( A) + ε.

n1

Since this is true for every ε > 0, we conclude that (4.14) holds. It follows by Theorem 4.1.8 that the Lebesgue measure of an interval equals its length.  Note that it is a consequence of the preceding Theorem and Proposition 4.2.8 that there is an unambiguously determined total length of a union E of a finite family of disjoint bounded intervals, namely, m(E). Thus, we have another proof of the result of Problem 1.1.P9.

80

4 Construction of a Measure

4.3.2. Corollary. A nonnegative-valued step function on [a, b] is Lebesgue measurable and its Riemann integral equals its Lebesgue integral. Proof: The Lebesgue measurability follows from the first assertion of Theorem 4.3.1. The second assertion thereof implies that the distribution function as understood in the context of the Lebesgue measure space (see the paragraph immediately after Definition 2.1.4) is the same as the distribution as understood for the purpose of Proposition 1.1.3. Therefore the equality claimed in the present corollary is just a restatement of Proposition 1.1.3.   f n of Exercise 1.1.5 and their limit  1f , show that 4.3.3. Exercise. For the functions f exists and equals limn→∞ f n , although the Riemann integral 0 f does not exist. Solution: Consider any n. As noted in Exercise 1.1.5, when 0 ≤ t < 1, the t-cut S n  {x ∈ [0,1]: f n (x) > t} is a finite union of disjoint intervals with total length 1; obviously, it is empty when 1 ≤ t. So, m(S n ) equals 1 for all t ∈ [0, 1) and equals 0 for all t ∈ [1, ˜ ∞). Hence f n equals 1 on [0, 1) and equals 0 on [1, ∞) for every n, which implies f n  1 for every n. It was also noted that the t-cut S  {x ∈ [0, 1]: f (x) > t} is the set of all irrational numbers in [0, 1] when 0 ≤ t < 1; obviously, it is empty when 1 ≤ t. By finite additivity of m and Example 4.1.6, it follows that m(S) equals 1 for ˜ ˜ all  t ∈ [0, 1) and equals  0 for all t ∈ [1, ∞). Thus f  f n for all n, which implies f  1. Therefore f  limn→∞ f n trivially. Since f (x) equals 1 if x is irrational 1 and equals 0 otherwise, it follows that the Riemann integral 0 f does not exist. In Example 1.1.4, we considered two Riemann integrable functions f : [a, b] → R b  sup f and computed a f as well as the integral 0 f˜, which was what we can now call  b  the Lebesgue integral f . In both cases, we found that a f  f . We shall now show that this must always happen for a Riemann integrable function. In particular, a continuous function is Lebesgue measurable. 4.3.4. Theorem. Let f : [a, b] → R be Riemann integrable. Then it is Lebesgue measurable and its Lebesgue integral agrees with its Riemann integral: 

 f  [a,b]

b f 

f. a

Proof: It is sufficient to consider only nonnegative f . For each n ∈ N, there exists a partition Pn of [a, b] such that the lower sum L(f , Pn ) and upper sum U(f , Pn ) satisfy b

b f −

a

1 n

< L( f, Pn ) ≤ U ( f, Pn )
0, there exist Lebesgue measurable sets B and C such that B ⊆ A ⊆ C and m(C ∩ B c ) < ε. (Here m denotes Lebesgue measure as usual.) Show that A is Lebesgue measurable. 4.3.P2. Let A ⊆ [a, b] and m denote Lebesgue measure. Show that there exists a Lebesgue measurable set B ⊆ [a, b] such that m(B)  m ∗ (B)  m ∗ (A) and A ⊆ B. The same is true with R in place of [a, b]. 4.3.P3. Show that the Lebesgue outer measure m ∗ is inner continuous.

In other  words, if {Sn }n≥1 is a sequence of sets such that S1 ⊆ S2 ⊆ … , then m ∗ ∪∞ n1 Sn  limn→∞ m ∗ (Sn ). 4.3.P4. Let E ⊆ [a, b]. Show that E is a Lebesgue measurable subset of [a, b] if and only if it is a Lebesgue measurable subset of R. 4.3.P5. For any S ⊆ R and any x ∈ R, let S + x denote the set {s + x: s ∈ S}. Prove the following:

4.3 Lebesgue Measure

85

(a) If E is a Lebesgue measurable subset of R, then E + x is also a Lebesgue measurable subset of R and m(E + x)  m(E). (b) If E is a Lebesgue measurable subset of [a, b] and E + x ⊆ [a, b], then E + x is also a Lebesgue measurable subset of [a, b]. 4.3.P6. Let {r n }n≥1 be an enumeration of the rationals in [0, 1]. For each i ∈ N, let gi be a continuous nonnegative function on [0,1] such that 1 gi (ri ) ≥ i and

gi (x)d x  1/2i . 0

n Define f n  i1 gi . Then each f n is nonnegative and continuous. Besides, the sequence {f n }n≥1 is increasing and therefore has a limit function f that is nonnegative extended real-valued and measurable.  Show that the function f is unbounded on every subinterval of [0, 1] and satisfies [0,1] f  1. 4.3.P7. For any interval I ⊆ R of positive length and any M > 0, show that there  exists ∞ ∗ E > M. a sequence {E j }j≥1 of disjoint sets such that ∪∞ E  I and m j j j1 j1 4.3.P8. For an increasing nonnegative real-valued function φ on R satisfying φ(0)  0 and φ(u + υ) ≤ φ(u) + φ(υ) whenever u, υ ∈ R, show that φ(1)1/2

⎛ 1 ⎞1/2 ⎛ 1 ⎞1/2   √ √ ≤ ⎝ φ(1 − t)dt ⎠ + ⎝ φ( 1 − t)dt ⎠ , 0

0

where the integrals are understood in the sense of Riemann.

4.4 Induced Measure and an Application Let (X, F, μ) be a measure space and ∅  E ∈ F. Then the family of subsets F E  {F: F ∈ F and F ⊆ E} ⊆ F is clearly the same as the family {G ∩ E: G ∈ F}. It is easy to verify that this is a σ-algebra of subsets of E and that the restriction of μ to the σ-algebra is a measure. We shall denote the restriction by μE ; also, we shall denote the restriction to E of any function φ on X by φ|E . If it happens that the function has been described only by its expression φ(x) without any symbol such as φ having been introduced for it, then the restriction may be denoted by φ(x)|E . The σ-algebra F E is called the σ-algebra induced by F on E and the measure μE is called the measure induced by μ on E. It is clear what is meant by measure space induced on E. It is an immediate consequence of Problem 4.3.P4 that the measure induced on an interval [a, b] by Lebesgue measure on R is nothing but Lebesgue measure on [a, b]. This renders it possible to apply Proposition 4.4.1 below in the proof of the

86

4 Construction of a Measure

subsequent Proposition 4.4.2, which makes the latter an application of the concept of induced measure. 4.4.1. Proposition. Let (X, F, μ) be a measure space and ∅  E ∈ F. For any function φ: X → R∗ , the restriction φ|E is F E -measurable if and only if the product function φ χ E : X → R∗ is F-measurable, in which case the equality   φ χ E dμ  φ| E dμ E (4.19) X

E

holds for nonnegative extended real-valued φ. For any real-valued function φ: X → R, the restriction φ|E is integrable if and only if the product function φχ E : X → R is integrable, in which case (4.19) holds. Proof: Given a function φ: X → R∗ , assume that φ|E is F E -measurable. Then the F-measurability of φχ E follows from the observation that, for α ≥ 0, we have X (φχ E > α)  {x ∈ E: φ(x) > α}  {x ∈ E: φ| E (x) > α}  E( φ| E > α} ∈ F E ⊆ F and for α < 0, we have X (φχ E > α)  E c ∪ {x ∈ E: φ| E (x) > α}  E c ∪ E( φ| E > α} ∈ F. Now assume conversely that φ χ E is F-measurable. Then the F E -measurability of φ|E follows from the observation that, for α ≥ 0, we have E( φ| E > α)  X (φ χ E > α) ∈ F and E( φ| E > α) ⊆ E and for α < 0, we have E( φ| E > α)  E ∩ {x ∈ X : φ(x) > α}  E ∩ X (φ χ E > α) ∈ F E . We proceed to establish the equality (4.19) for a function φ: X → R∗+ when φ|E is F E -measurable, or equivalently, when φχ E is F-measurable. As is easily verified, for any F ∈ F, we have (χ F∩E )| E  (χ F χ E )| E  χ F | E and all three are equal to the characteristic function of F ∩ E defined on the sub set E. Therefore E χ F | E dμ E  μ E (F ∩ E)  μ(F ∩ E)  X (χ F χ E )dμ. This shows that (4.19) holds when φ is the characteristic function χ F of an arbitrary F ∈ F. It follows that it holds for any simple function φ on X and hence for any nonnegative F-measurable function φ on X. Now suppose instead of the F-measurability of φ, we assume only the Fmeasurability of φχ E . Applying what has been proved in the preceding paragraph to φχ E , we get   (φ χ E )χ E dμ  (φ χ E )| E dμ E . X

E

4.4 Induced Measure and an Application

87

Upon combining this with the straightforward equalities φχ E  (φχ E )χ E and (φχ E )| E  φ| E , we find that (4.19) holds when φχ E is F-measurable. The last part about a real-valued function φ: X → R now follows from the above by using the obvious equalities φ+ | E  φ| E + , (φχ E )+  φ+ χ E , φ− | E  φ|−E and  (φχ E )−  φ− χ E .  Remark. So far we have used the symbol I f , where I is an interval, to mean the integral of f with respect to Lebesgue measure m on I. It means the same thing as I f dm. Since Lebesgue measure on I isthe same as the measure mI induced  on I by Lebesgue measure m on R, the symbol I f also means the same thing as I f dm I . It is worth bearing this in mind for the next proposition, which is an application of induced measures. 4.4.2. Proposition. Suppose f : [β, ∞) → R is Riemann integrable over [β, B] for every B > β. If the Lebesgue integral ∫[β,∞) f over [β, ∞) exists, then the improper ∞ integral β f also exists and agrees with the Lebesgue integral. Proof: Consider an arbitrary sequence {Bn }n≥1 in [β, ∞) tending to ∞. For each n ∈ N, let f n  f χ[β,Bn ] and gn  f |[β,Bn ] . Then each f n is Lebesgue integrable and each gn is Riemann integrable by hypothesis, with Bn

Bn gn 

β

f.

(4.20)

β

Moreover, f n → f everywhere and | f n |≤ | f | everywhere. By the Dominated Convergence Theorem 3.2.6, we find that   lim f n exists and equals f. (4.21) n→∞ [β,∞)

[β,∞)

By Proposition 4.4.1 and Theorem 4.3.4 (in that order), for each n ∈ N, we have 

 fn 

[β,∞)

[β,Bn ]

Bn gn 

gn β

Bn 

f β

by (4.20). B  It follows from here in the light of (4.21) that limn→∞ β n f exists and equals [β,∞) f . Since {Bn }n≥1 is an arbitrary sequence tending to ∞, the contention of the proposition stands proved. 

88

4 Construction of a Measure

   The integral  E φ| E dμ E is usually abbreviated as E φ. Recall that X φ is usually 4.4.1 can then  as φ. The equality (A) of Proposition   be stated as  written φχ E  E φ. If E  ∅, we shall understand E φ to be 0 because φχ E  0. c 4.4.3. Proposition. Let (X, F, μ) be a measure space  and H ∈ F satisfy μ(H )  0. Then for any measurable f : X → R∗+ , we have f  H f . The same holds for any integrable f : X → R.

Proof: If H  ∅, it follows that μ(X )  0 and all integrals are 0; so there is nothing to prove. If H c  ∅, again there is nothing to prove. So, assume both H and H c  G to be nonempty. Then there are induced measures on both H and G. However the induced measure on G is the zero measure and therefore all integrals on it are 0; in  particular, G f  0. Now,      f (1 − χ H )  f χH + f χG f  f χH +    f + f by Proposition 4.4.1 H

G





f. H

 with a proposition that justifies such computations as  this section   We close f + f  f , where a ≤ b ≤ c. However, no reference to it will be [a,b] [b,c] [a,c] given when it is used. 4.4.4. Proposition. Let (X, F, μ) be a measure space and the subsets E, F ∈ F satisfy μ(E ∩ F)  0 and E ∪ F  X . Then for any measurable f : X → R∗+ , we have    f + f  f. (4.22) E

F

X

For any real-valued function f : X → R, the restrictions to E and to F are both integrable if and only if f is integrable, in which case (4.22) holds. Proof: Since E ∪ F  X , we have χ X + χ E∩F  χ E + χ F . Consider a measurable function f : X → R∗+ . By the above equality, additivity of the integral for nonnegative functions (Theorem 3.2.2) and Proposition 4.4.1,     f + f  f + f. X

E∩F

E

F

4.4 Induced Measure and an Application

But

 E∩F

89

f  0 because μ(E ∩ F)  0 and hence (4.22) holds.

Now consider an integrable function f : X → R. Since | f χ E |≤ | f |, | f χ F |≤ | f | and | f χ E∩F |≤ | f |, all the three functions f χ E , f χ F and f χ E∩F are integrable and the above equality regarding characteristic functions, taken in conjunction with additivity of the integral for integrable functions (Theorem 3.2.5) and Proposition 4.4.1, leads to (4.22) as above. Finally, suppose the restrictions to E and F of the real-valued function f : X → R are both integrable. By Proposition 4.4.1, the products f χ E and f χ F are both integrable and the inequality | f χ E∩F |≤ | f χ E | implies that f χ E∩F is also integrable. The above equality regarding characteristic functions can be written in the form χ X  χ E + χ F − χ E∩F . When taken in conjunction with the additivity and homogeneity of the integral for integrable functions (Theorem 3.2.5) and Proposition 4.4.1, the equality leads to     f  f + f − f. X

E

F

E∩F

 Problem Set 4.4 4.4.P1. (a) Let g: [1, ∞) → R be given by g(x)  sinx and I k be the interval [2kπ, (2k + 1)π], where k ∈ N. Show that Ik g(x)d x  2. (b) Let f : [1, ∞) → R be given by f (x)  x1 sinx and I k be the interval [2kπ,  2 . (2k + 1)π], where k ∈ N. Show that Ik f (x)d x ≥ (2k+1)π 4.4.P2. Let (X, F, μ) be ameasure  space and {f k }k ≥1 be a sequence of integrable functions on X such that ∞ k | < ∞. Show that there exists an integrable k1 | f function f on X such that the series ∞ converges k1 f k   to f on a measurable subset fk . H ⊆ X satisfying μ(H c )  0, and that f  ∞ k1 4.4.P3. Let F be a σ-algebra of subsets of X and A, B ∈ F be nonempty subsets satisfying A ∪ B  X . Show that a function φ: X → R∗ is F-measurable if and only if the restrictions φ|A and φ|B are F A -measurable and F B -measurable respectively.

Chapter 5

The Counting Measure

5.1 Interchanging the Order of Summation Since a sequence a  {aj}j≥1 is nothing but a function with domain N, it makes sense to speak of the integral a with reference to some measure on the σ-algebra of all subsets of N, in particular the counting measure. It was noted in Problem 3.1.P5 that  when each aj is nonnegative, the integral a is just the sum ∞ j1 aj . be nonnegative but require For this paragraph, drop the requirement that every a j  + a , the sum of the series formed by the nonevery aj to be finite. Then a+  ∞ j1 j  −  ∞ negative terms of j1 aj ; correspondingly for a . By Definition 2.2.5, a is then  ∞ +  ∞ −   − + aj − j1 aj provided ∞ aj− is convergent; thus a  ∞ j1 j1 j1 (aj − aj )  ∞ reference to the counting measure, what we need j1 aj . For a to be integrable with  ∞ − + according to Definition 2.2.5 is that ∞ j1 aj and j1 aj should both be convergent. It is well known from the elementary theory of series that convergence of both is  a . In any event, the equivalence can equivalent to the absolute convergence of ∞ j j1 − x+ + x− . be easily derived from the fact that any x ∈ R satisfies x+ ≤ |x|, x∞ ≤|x| ∞  It  was established in Sect. 1.2 that the equality n1 k1 an,k ∞ ∞ a , called interchanging the order of summation, is valid for any funcn,k k1 n1 tion a : N × N → R∗+ . We know from Problem 1.2.P1 that this inequality can fail if a is not required to be nonnegative even if it is restricted to being real. However, for a : N × N → R, an additional hypothesis does permit the interchange (see Problem 5.1.P4). To reinterpret these repeated sums in terms of integrals with respect to counting measures, it will be convenient  to denote an,k by a(n, k) and introduce some further notation. In taking the sum ∞ k1 an,k , we are holding n fixed and regarding a(n, k) as a function of k only. The function will be denoted by an . Similarly, in taking the  a , we are holding k fixed and regarding a(n, k) as a function of n other sum ∞ n,k n1 only. It will be denoted by ak .

© Springer Nature Switzerland AG 2018 S. Shirali, A Concise Introduction to Measure Theory, https://doi.org/10.1007/978-3-030-03241-8_5

91

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5 The Counting Measure

5.1.1. Definition. Given a function a: N × N → R ∪ {∞} and n ∈ N, the function with domain N such that each k ∈ N is mapped into a(n, k) will be denoted by an : an (k)  a(n, k) for every k ∈ N. Likewise, for any k ∈ N, the function with domain N such that each n ∈ N is mapped into a(n, k) will be denoted by ak : ak (n)  a(n, k) for every n ∈ N.   an  terms of the notation just introduced, ∞ k1 an,k    5.1.2. Remark. In  k a  a  a(n, k)d μ(n) when a is nonnegativea(n, k)d ν(k) and ∞ n,k n1 valued, μ and ν both being the counting measure on N. Moreover, interchanging the order of summation says that     ( an )  ( ak ), or in standard and more readable notation,     ( a(n, k)d ν(k))d μ(n)  ( a(n, k)d μ(n))d ν(k) (for nonnegative-valued a). This is interchanging the order of integration in repeated integrals, the integration being with respect to the counting measure on N every time. It is natural to ask whethersuch interchangingis valid for any function f : X × Y → R ∪ {∞}, at least when f (x, y)d ν(y) and f (x, y)d μ(x) describe integrable (or nonnegative measurable) functions of x and y respectively. In standard and more traditional notation, we are asking whether     d ν(y) f (x, y)d μ(x)  d μ(x) f (x, y)d ν(y), where μ and ν are measures on some σ-algebras of subsets of X and Y respectively, and it is assumed that all integrals involved make sense. We shall first give an example to show that interchanging the order of integration need not always be valid even if f takes only the values 0 and 1. 5.1.3. Example. Let X  Y  [0, 1], F  G  the σ-algebra of Lebesgue measurable subsets of [0, 1]; suppose μ is the Lebesgue measure on F and ν is the counting measure on G. Let V  {(x, y) ∈ [0, 1] × [0, 1]: x  y}. Then as argued below,     d μ(x) χV (x, y)d ν(y)  1 but d ν(y) χV (x, y)d μ(x)  0.

5.1 Interchanging the Order of Summation

93

Indeed,  given x ∈ X, we have χV (x, y)  1 for y  x and 0 for y  x, which given y ∈ Y , we have χV (x, y)  1 for x  y implies χV (x, y)d ν(y)  1, whereas  and 0 for x  y, which implies χV (x, y)d μ(x)  0. Readers who may be conversant with the Tonelli Theorem 6.5.2 and Fubini Theorem 6.5.5 for what are called σ-finite measures may note that the upcoming theorem does not assume the measure μ to be σ-finite and is therefore not a special case of those theorems. 5.1.4. Theorem. Let μ be a measure on a set X. Consider N with the counting measure (all subsets measurable) and let f : X × N → R∗+ be a function such that, for each k ∈ N, the function f k on X described by f k (x)  f (x, k) is measurable. Then interchanging the order of integration is valid, which is to say,    ∞ ∞   f (x, k) d μ(x)  f (x, k)d μ(x) . k1

k1

Proof: For brevity, weshall write dμ(x) as simply dx. Define a sequence of functions φn on X by φn (x)  nk1 f (x, k). Then each φn is nonnegative-valued and measurable. Besides, φn (x) ≤φn+1 (x) for every x ∈ X and every n ∈ N. By the Monotone Convergence Theorem 2.3.2,   

  ∞  f (x, k) dx  lim φn (x) dx  lim φn (x)dx n→∞

k1

 lim

n→∞

n→∞

n   k1

f (x, k)dx 

∞  

f (x, k)dx .

k1

 ∞

5.1.5. Remark. Let us return to the context of the sum j1 aj being the integral  ∗+ a with respect to the counting measure on N, where ∞ a takesvalues in R . It is clear that for any finite subset A ⊆ N, we have j1 aj ≥ j∈A aj . Therefore ∞  j1 aj ≥ sup{ j∈A aj : A ⊆ N finite}. It is quite easy to see that actually equality holds: ∞  j1

aj  sup{



aj : A ⊆ N finite}

j∈A

Indeed, the right side here is greater than or equal to the supremum taken over the special finite subsets of the form {1, 2, . . . , n}, i.e. sup{ nj1 aj : n ∈ N}, which  is the same as ∞ j1 aj . It makes no difference whether we allow A  ∅, because  a  0 by convention. j j∈∅

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5.1.6. Exercise. Let S denote a class of finite subsets of N such that every finite subset is contained in some F ∈ S. (For example, S may consist of all subsets of the form {1, 2, . . . , n}  or of the form  {1, 2, . . . , n} ∪ {n + 3}.) For any R∗+ -valued ∞ sequence a, show that j1 aj  sup{ j∈F aj : F ∈ S}.   Solution: Since each F ∈ S is finite, sup{ j∈A aj : A ⊆ N finite} ≥ sup{ j∈F aj : F ∈ S}. Consider any finite A ⊆ N. By some F ∈ S such  hypothesis,  there exists  ≥ 0, it follows that j∈A aj ≤  a ≤ sup{ that A ⊆ F. Since aj  j∈F j j∈F aj : F ∈ S}. Consequently, sup{ j∈A aj : A ⊆ N finite} ≤ sup{ j∈F aj : F ∈ S}. Thus the two  suprema are equal. But one of them is known to be ∞ j1 aj (Remark 5.1.5). The advantage of looking at the sum as the supremum is that the latter makes sense even when N is replaced by an arbitrary set. 5.1.7. Definition. For any function a : X → R∗+ , the sum of all a(x) over x ∈ X is   a(x)  sup{ a(x) : A ⊆ X finite}. x∈X

x∈A

It should be noted that this is not a measure-theoretic definition. Situations in which the need to interchange the order of summations in repeated sums of this kind arises naturally will not be discussed here but the interested reader is referred to Dunford and Schwartz [2, p. 1010]. However, the matter of interchanging the order of summation will be taken up.  5.1.8. of Remark 5.1.5, when X  N, the sum x∈X a(x), or  Remark. In view ∞ . Recall once again from Problem 3.1.P5 that this j1 a j∈N aj , is the same as j  sum is the same as the integral a or a(x)d μ(x), where μ is the counting measure on X  N. Therefore interchange of the order of summation in repeated sums that was proved in Sect. 1.2 can be reformulated in terms of integrals  respect to  with counting measures. In other words, what was expressed as i∈N j∈N a(i, j)    a(i, j), or equivalently, j∈N i∈N   a(x, y)  a(x, y) x∈A y∈B

y∈B x∈A

for countable sets A, B and a : A × B → R∗+ , can also be expressed as interchanging the order of integration:     ( a(x, y)d ν(y))d μ(x)  ( a(x, y)d μ(x))d ν(y), μ and ν being the counting measure s on A and B respectively. We are going to justify interchanging the order of summation when the function is defined on the Cartesian product of sets that are not assumed countable, but first a preparatory result.

5.1 Interchanging the Order of Summation

95

5.1.9. Proposition. (a) If a subset A ⊆ X has the property  that a(x)  0 whenever x∈ / A, i.e. a(x)  0 unless x ∈ A, then x∈A a(x)  x∈X a(x). (b) If x∈X a(x) < ∞, then there is a countable subset A ⊆ X with the property that a(x)  0 whenever x ∈ / A. Any subset B of X that contains A has the same property. Proof: (a) The assertion is clearly true when X is finite. Any  finite subset F of A is also a finite subset of X and therefore x∈A a(x) ≤ x∈X a(x). To prove the reverse inequality, consider an arbitrary finite subset The intersection  F of X.  B  F ∩ A is a finite subset of A and therefore x∈A a(x) ≥ x∈B a(x). Now, a(x)  0 unless x ∈A. Therefore,for x ∈ F, we have a(x)  0 unless x ∈ F ∩ A  B. Consequently, x∈B a(x)  x∈F a(x), considering that F is finite. It follows   a(x) ≥ x∈Fa(x). As this holds for every finite subset F of X, we that x∈A infer that x∈A a(x) ≥ x∈X a(x), which is the desired reverse inequality. (b) Left to the reader as Problem 5.1.P2.  With the above preparatory result in hand, we come to the theorem that justifies interchanging the order of summation when the function is defined on the Cartesian product of sets that are not assumed countable. There are authors who regard the matter as being too obvious to be worthy of a proof (Dunford and Schwartz [2, p. 1010]), presumably because the first assertion in part (b) of the above Proposition 5.1.9 “reduces” the matter to the countable case. But the reduction is not quite as straightforward as may seem at first sight, something that will be clear from the argumentation below. The reduction idea will be oppugned further in Remark 6.5.4(a). 5.1.10. Theorem. For any function a : X × Y → R∗+ ,   a(x, y)  a(x, y). x∈X y∈Y

y∈Y x∈X

  Proof: First suppose S  x∈X y∈Y a(x, y) < ∞. By Proposition 5.1.9(b), there exists a countable A ⊆ X such that  a(x, y)  0 for every x ∈ / A. y∈Y

Hence a(x, y)  0 for every x ∈ / A and every y ∈ Y . Together with Proposition 5.1.9(a), these statements imply  a(x, y) S x∈A y∈Y

and

(5.1)

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5 The Counting Measure



a(x, y) 



a(x, y) for every y ∈ Y .

(5.2)

x∈X

x∈A

 Using the finiteness of S again, we have y∈Y a(x, y) < ∞ for every x ∈ X. Therefore Proposition 5.1.9(b) yields for each x ∈ X a countable subset Bx ⊆ Y such that a(x, y)  0 for every y ∈ / Bx .

(5.3)

Let B  ∪x∈A Bx . Then B ⊆ Y is countable, and from (5.3) we have a(x, y)  0 for every y ∈ / B and x ∈ A. Hence by (5.2), 

a(x, y)  0 for every y ∈ / B.

x∈X

Together with Proposition 5.1.9(a), these statements imply   a(x, y)  a(x, y) for every x ∈ A y∈Y

(5.4)

y∈B

and 

a(x, y) 



a(x, y).

(5.5)

Since A and B are countable sets, we know that   a(x, y)  a(x, y).

(5.6)

y∈Y x∈X

y∈B x∈X

x∈A y∈B

y∈B x∈A

Now, S



a(x, y) by (5.1)

x∈A y∈Y





a(x, y) by (5.4)

x∈A y∈B





a(x, y) by (5.6)

y∈B x∈A





a(x, y) by (5.2)

y∈B x∈X



 

y∈Y x∈X

a(x, y) by (5.5).

5.1 Interchanging the Order of Summation

97

  This proves the equality in question  when x∈X y∈Y a(x, y) < ∞. A similar argument goes through when y∈Y x∈X a(x, y) < ∞. In the remaining case, the two are equal because both are ∞.  It was noted in Remark 5.1.8 that interchanging the order of summation for a nonnegative function a with domain A × B with both A and B countable can be interpreted as interchanging the order of integration with respect to the counting measures on A and B. We intend to show in Sect. 5.2 (see Remark 5.2.2) that interchanging the order of summation for a nonnegative function a with domain X × Y with general X and Y —asserted by Theorem 5.1.10—can also be interpreted as interchanging the order of integration with respect to the counting measures on X and Y . It will later be shown in Theorem 5.2.3 that the sums in Theorem 5.1.10 are both equal to (x,y)∈X ×Y a(x, y). Problem Set 5.1 5.1.P1. Let X be any nonempty set. Define F to consist of those subsets of X that are either countable or have a countable complement. (It is easy to argue that F is a σalgebra, sometimes called the“co-countable” σ-algebra). Now let φ : X → R∗+ be any function and set μ(A)  x∈A φ(x) for any A ∈ F, where the sum is understood in the sense of Definition 5.1.7. Show that μ is a measure on F. (Note: If φ > 0 everywhere, then the only subset of measure 0 is the empty set.) 5.1.P2. Prove Proposition 5.1.9(b). 5.1.P3. Let μ be a measure on a set X. Consider N with the counting measure (all subsets measurable) and let f : X × N → R be a real-valued function such that, for each k ∈ N, the function f k on X described by f k (x)  f (x, k) is measurable. Suppose  also that there is an integrable function φ : X → R+ such that | nk1 f (x, k)|≤ φ(x) for each x ∈ X and each n ∈ N. If the series ∞ k1 f (x, k) converges, show that interchanging the order of integration is valid, which is to say,    ∞ ∞   f (x, k) d μ(x)  f (x, k)d μ(x) . k1

k1

5.1.P4. Suppose the real-valued function  a : N × N → R has nthe property that ∞ there exists a sequence {bm }m≥1 such that  m1 bm < ∞ and | k1 am,k |≤ bm for ∞ each and each m. Show ∞ if the series k1 am,k converges for each m, then ∞ that, ∞ n  ∞ . (Note: (1) The hypothesis about {bm }m≥1 is m1 k1 am,k  k1 m1  am,k ∞ stronger than convergence of ∞ m1 | k1 am,k |. The latter is not sufficient for the purpose at hand, in view of Problem 1.2.P1. (2) A proof of the result of the present problem by using metric spaces is found in Feldman [3, p. 2]. What has been stated there has a stronger ∞ palatable version of the ∞hypothesis about {bm }, in that but more it is assumed that ∞ m1 k1 |am,k | < ∞, so that k1 am,k not only converges for each m but does so absolutely.)

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5 The Counting Measure

5.2 Integration with the Counting Measure With a view to the earlier stated purpose of interpreting Theorem 5.1.10 as interchanging the order of integration in repeated integrals, we show that the sum defined by Definition 5.1.7 is actually the integral with respect to the counting measure. 5.2.1. Theorem. For any function a : X → R∗+ ,   a(x)  a, x∈X

where the integral is taken with respect to the counting measure on X. Proof: For any function a : X → R∗+ that takes the value ∞, this is obvious because the sum and integral are both ∞. So we need consider only real-valued a. Suppose there is a simple  function s such that 0 ≤ s ≤ a and s is nonzero on an infinite set. Then a ≥ s  ∞. Moreover, s takes some positive value σ on an infinite subset B ⊆ X. Therefore, for any M > 0, there exists a finite subset F ⊆ B with a(x) ≥ more than M /σ elements. It follows that x∈X a(x) ≥ x∈F x∈F s(x) >  a(x)  ∞. Thus σ · (M /σ)  M . Since this holds for every M > 0, we have x∈X   a  x∈X a(x). Now consider the case when there is no simple function s such that 0 ≤ s ≤ a and s is nonzero on an infinite set. In other words, any simple function s such that0 ≤ s ≤ a is nonzero only set F. An application of Proposition 3.1.7 yields s  on a finite     s(x). But s(x) ≤ a(x) ≤ a(x) and hence s ≤ a(x). x∈F x∈F x∈F x∈X x∈X By Theorem 3.2.1 and the Monotone Convergence Theorem 2.3.2, it follows that   a(x). a≤ x∈X

To derive the reverse inequality, consider any finite subset  A ⊆ X.  The product aχA is a nonnegative simple function not exceeding a and hence a ≥ aχA . As Proposition   3.1.7 implies aχA  x∈A a(x), the desired reverse inequality now follows right away.  5.2.2. Remark. In the light of Theorem 5.2.1, the interchange of the order of summation in Theorem 5.1.10 can be restated as an interchange of the order of integration:     d μ(x) f (x, y)d ν(y)  d ν(y) f (x, y)d μ(x), where μ,ν are both counting measures and the function f is R∗+ -valued. For other measures, we have seen in Example 5.1.3 that interchanging the order of integration need not hold. We can nevertheless ask under what conditions it does hold.

5.2 Integration with the Counting Measure

99

 Next, we show that in the context of Theorem 5.1.10, the two sums are equal to (x,y)∈X ×Y a(x, y). 5.2.3. Theorem. For any function a : X × Y → R∗+ ,   a(x, y)  a(x, y). (x,y)∈X ×Y

x∈X y∈Y

Proof: The argument that 

a(x, y) ≤

(x,y)∈X ×Y



a(x, y)

x∈X y∈Y

is straightforward and is left as Problem 5.2.P1. We shall show that strict inequality cannot hold. This is trivially true if the left side is ∞. So we assume  S a(x, y) < ∞. (x,y)∈X ×Y

 The finiteness immediately that of y∈Y a(x, y) for each x. To get a   implies a(x, y) > S. Then there exists a finite set contradiction, suppose  x∈X y∈Y a(x, y) > S. The finiteness of A and the finiteA ⊆ X such that x∈A y∈Y    x together imply that ness of y∈Y a(x, y)for each x∈A y∈Y a(x, y) < ∞.  a(x, y) − S, we have 0 < α < ∞. Suppose A conSo, if we set α  x∈A y∈Y  a(x, y) yields tains N elements. For each x ∈ X, finiteness of y∈Y    a finite subset α ⊆ Y such that a(x, y) > a(x, y) − . So, B x y∈Bx y∈Y x∈A y∈Bx a(x, y) > N   a(x, y) − α  S. If we select F ⊆ X × Y to be {(x, y) ∈ X × Y : x ∈ A, y x∈A y∈Y  and the foregoing inequality leads to (x,y)∈F a(x, y) > S. ∈ Bx }, then F is finite  However, this implies (x,y)∈X ×Y a(x, y) > S, contrary to the definition of S. The left side in the equality asserted by Theorem 5.2.3 can be represented as the integral of a with respect to the counting measure on the Cartesian product X × Y . With this as a backdrop, we are led to the question of whether there is a measure on X × Y such that the integral with respect to that measure, a “double integral”, agrees with the repeated integrals even when the measures on X and Y are not counting measures. These matters will be taken up in the next chapter. The reader is cautioned that the integral with respect to the counting measure on X × Y is not the same as what is called the sum of a “double series” (see Shirali and Vasudeva [10, pp. 70–71] or Habil [4, pp. 4, 20–21]); the latter takes into account only a limited class of finite subsets of X × Y . 5.2.4. Remark. It is an easy consequence for any real of −Theorem 5.2.1 that  + f (x) < ∞, we have valued function f : X → R such that x∈X x∈X f (x) −   − f , where the integral is taken with respect to the counting measure x∈X f (x)  on X.

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5 The Counting Measure

 We proceed to discuss how f , when it is finite and f is real-valued, can be described as a sum over the set X without resort to f + and f − . This kind of sum arises naturally in some contexts where f takes vectors or complex numbers as values and there is no such thing as f + and f − . +  Let f : X → R be a nonnegative real-valued function on a set X. If σ  x∈X f (x) < ∞, then it is the unique real number with the property that for any ε > 0, there exists a finite F ⊆ X such that  f (x) − σ|< ε whenever F ⊆ G ⊆ X and G is finite. (5.7) | x∈G

 Conversely, if sucha real number σ exists, then ∞ > x∈X f (x)  σ. This reformulation of what x∈X f (x) is remains meaningful even if f is permitted to take negative values. With this in mind, we enunciate the following definition. 5.2.5. Definition. If f : X → R is a real-valued function, an unconditional sum  un x∈X f (x) is a real number σ having the property (5.7), if there is any. 5.2.6. Remarks. (a) The resemblance to the definition of a limit of a sequence suggests how to prove by an 2ε -argument, using the fact that the union of two finite sets is a finite set, that there cannot be two different real numbers σ satisfying the unconditional sum, un if it exists, is unique. (5.7). In otherwords, un g(x) both exist, then Also, if un x∈X f (x) and x∈X x∈X (f + g)(x) exists and un f (x) + g(x). equals un x∈X x∈X (b) In view of the observation preceding the above definition, when  the function f  f (x) exists if and only if is nonnegative real-valued, un x∈X x∈X f (x) is finite, f (x)  f (x). in which case un x∈X x∈X (c) In particular, interchanging the order of summation in Theorem 5.1.10 continues to hold when the function a is nonnegative real-valued and the sums are understood as unconditional sums. Similarly for Theorem   5.2.3. + − (d) If f : X → R is any real-valued function and either x∈X f (x), un −  x∈X+ f (x) un + f (x), f (x) are both finite, then f (x)  are both finite or x∈X x∈X  unx∈X − un + − f (x), f (x)  f (x) by (b) above. When this is so, once x∈X x∈X x∈X again the resemblance between Definition 5.2.5 and the definition of a limit of a sequence suggests how to prove the following statement by an 2ε -argument, using the fact that the union of two finite sets is a finite set: un un un   f (x)  f + (x)− f − (x)  f + (x)− f − (x). x∈X

x∈X

x∈X

In particular, the unconditional sum

 un

x∈X

x∈X

f (x) exists.

x∈X

The next result is a sort of a converse. 5.2.7. Theorem. Suppose f : X → R is a real-valued  function on a set X and its  + − unconditional sum un x∈X f (x), x∈X f (x) are both finite x∈X f (x) exists. Then and

5.2 Integration with the Counting Measure



f + (x)−

x∈X

101



f − (x) 

x∈X

un

f (x),

x∈X

the unconditional sum. Moreover, un

f (x) 

x∈X

un

 un f + (x)− f − (x)  f ,

x∈X

x∈X

the integral with respect to the counting measure. In particular,



f exists.

Proof: Taking ε  1 in (5.7), we find that X has a finite subset F for which  | x∈G f (x)|< | un x∈X f (x)|+1 whenever F ⊆ G ⊆ X and G is finite. Since F can have only finitely many subsets, some real α > 0 satisfies  f (x)|< α whenever G ⊆ F or F ⊆ G ⊆ X and G is finite. (5.8) | x∈G

 +  Now x∈X f (x)  ∞. Then there exists a finite subset G ⊆ X for which  suppose +  + 0, so that x∈G  f (x) > 2α. Let H be the subset of G on which  f (x) >  f + f on H +  + and f  0 on G \H . Then we have x∈H f (x)  x∈H f (x)  x∈G f (x) and hence  f (x) > 2α. x∈H

Let G1  F\H and G2  H ∪ F. By (5.8), it follows that  f (x)|< α | x∈G1

because G1 is a subset of F, and also that  f (x)|< α | x∈G2

because G2 is a finite subset containing F. But H and G1 are disjoint with union G2 , all of them being finite. Therefore    f (x)  f (x) + f (x). x∈G2

x∈H

x∈G1

This  is +plainly in contradiction with the precedingthree −inequalities. Therefore x∈X f (x) is finite. A similar argument shows that x∈X f (x) is also finite.

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5 The Counting Measure

  + − Once we know that both finite, it  follows from  x∈X+ f (x),  x∈X f − (x) are  un un + f (x) − f (x)  f (x)  Remark 5.2.6(d) that x∈X x∈X x∈X x∈X f (x) − un − x∈X f (x). The last part about the integral now follows by virtue of Theorem 5.2.1.  5.2.8. Theorem. Suppose f : X → R is a real-valued function on  a set X and  + the integral f with respect to the counting measure is finite. Then x∈X f (x),     − + − x∈X f (x) are both  finite and x∈X f (x) − x∈X f (x)  f . Moreover, the unconditional sum un x∈X f (x) exists and  un un un f (x)  f + (x)− f − (x)  f . x∈X

x∈X

x∈X

Furthermore,    un un un  |f (x)| f + (x) + f − (x)  f + (x)+ f − (x)  |f (x)|. |f |  x∈X

x∈X



x∈X





x∈X

x∈X

x∈X



Proof: Since f is finite, f  f +− f − and both integrals on the right are  the  + + − −  f (x) and f  f finite. But by Theorem 5.2.1, f x∈X     x∈X − (x). Therefore  + − + f . The x∈X f (x), x∈X f (x) are both finite and x∈X f (x) − x∈X f (x)  rest is a consequence of Theorem 5.2.1, the finiteness and Remark 5.2.6(d), (a).  Problem Set 5.2 5.2.P1. Complete the proof of Theorem 5.2.3 by showing for any function a : X × Y → R∗+ that   a(x, y) ≤ a(x, y). (x,y)∈X ×Y

x∈X y∈Y

5.2.P2. Let f : X ×Y → R be a real-valued  function on the Cartesian product X × Y finite and suppose that the unconditional sum un (x,y)∈X ×Y f (x, y) exists (necessarily un  f (x, y) exists for each x ∈ X, (ii) f (x, y) by definition). Show that (i) un x∈X un un y∈Y un f (x, y)  f (x, y). exists for each y ∈ Y , and (iii) un x∈X y∈Y y∈Y x∈X 5.2.P3. (Cf. Problem 5.1.P4) Suppose the real-valued function f :  X ×Y → R + has the property that there exists a map φ : X → R such that x∈X φ(x) <  ∞ and | y∈F f (x, y)|≤ φ(x) for each finite subset F ⊆ Y and each x ∈ X.   for each y ∈ Y , un Show that (i) un x∈X f (x, y) exists y∈Y f (x, y) exists for each un un f (x, y) exists, and (iii) if the other x ∈ X, (ii) the repeated sum x∈X y∈Y  un f (x, y) also exists, then the two repeated sums are equal: repeated sum un y∈Y x∈X  un un un un x∈X y∈Y f (x, y)  y∈Y x∈X f (x, y).

Chapter 6

Product Measures

6.1 Algebras and Monotone Classes of Sets After Theorem 5.2.3, the question was raised whether there is a measure on X × Y such that the integral with respect to that measure, a “double integral”, agrees with the repeated integrals even when the measures on X and Y are not counting measures. In this section, we begin to lay out the background for obtaining a σ-algebra in the Cartesian product X × Y , which would be suitable for such a measure. The σ-algebra will only be set up in Sect. 6.2 and the measure in Sect. 6.4. 6.1.1. Definition. A family A of subsets of a nonempty set X is called an algebra if (a) ∅ ∈ A (contains the empty set), (b) A ∪ B ∈ A whenever A ∈ A, B ∈ A (closed under unions), (c) Ac ∈ A whenever A ∈ A (closed under complements). Since X  ∅c , conditions (a) and (c) in the definition together imply that X ∈ A. Also, conditions (b) and (c) together imply closure under intersections, considering that A ∩ B  (Ac ∪ Bc )c . It is immediate that an algebra is closed under finite unions and intersections as well: if Ak ∈ A for 1 ≤ k ≤ n, then

n  k1

Ak ∈ A and

n 

Ak ∈ A.

k1

Moreover, if A and B belong to A, then so does the set-theoretic difference A\B, which is defined to be A ∩ Bc . These facts and their simple consequences will be used without further ado. © Springer Nature Switzerland AG 2018 S. Shirali, A Concise Introduction to Measure Theory, https://doi.org/10.1007/978-3-030-03241-8_6

103

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Since the intersection of any family of σ-algebras is again a σ-algebra (this is easy to prove), it is clear that given any family S of subsets of X, there always exists a smallest σ-algebra containing S. Here “smallest” means contained in every σ-algebra containing S. The intersection of the family of all σ-algebras containing S serves the purpose. It is usually called the σ-algebra generated by S. When S is an algebra, there is an alternative description of the σ-algebra generated by it. 6.1.2. Definition. A monotone class is a family M of subsets of a nonempty set X such that:  A ∈ M, (a) A1 ⊆ A2 ⊆ · · · and each Aj ∈ M implies ∞ j1 j (b) B1 ⊇ B2 ⊇ · · · and each Bj ∈ M implies ∞ j1 Bj ∈ M. Any σ-algebra is a monotone class. When A ⊆ X, where X is any nonempty set, the family M  {A} is a monotone class which is not a σ-algebra. As with σalgebras, there is always a monotone class generated by a family of subsets of X. The alternative description promised above is as follows. 6.1.3. Proposition. The σ-algebra generated by an algebra is also the smallest monotone class generated by the algebra. Proof: Let A be an algebra of subsets of X and M0 be the monotone class generated by A, which is to say, the smallest monotone class containing A. Since any σ-algebra is a monotone class, the σ-algebra generated by A certainly contains M0 . The converse will follow if we show that M0 is a σ-algebra. First we argue that M0 is closed under complementation. Let M0  {A ∈ M0 : A ∈ M0 }, the family of those subsets in M0 whose complements also belong to M0 . Since A is an algebra and M0 ⊇ A, we have M0 ⊇ A. Now, it is obvious that ∞ c ∞ ∞ c c c M0 is also a monotone class, because (∪∞ j1 Aj )  ∩j1 Aj and (∩j1 Aj )  ∪j1 Aj . Therefore M0 ⊇ M0 , which shows that M0 is closed under complementation. We proceed to argue that M0 is closed under unions. As a preliminary step, we show that A ∪ F ∈ M0 for every A ∈ A and every F ∈ M0 . Consider any A ∈ A and set U (A)  {F ∈ M0 : A∪F ∈ M0 }. We shall first see why U(A) is a monotone class containing A. Since A is an algebra and M0 ⊇ A, we have U (A) ⊇ A. Now consider sets E1 ⊆ E2 ⊆ · · · with each E j ∈ U(A). Since U (A) ⊆ M0 by definition and M0 is a monotone class, we have ∪∞ j1 Ej ∈ M0 . The definition of U(A) yields A ∪ Ej ∈ M0 for each j; also, A ∪ E1 ⊆ A ∪ E2 ⊆ · · ·. As ∞ M0 is a monotone class, it follows that ∪∞ j1 (A ∪ Ej ) ∈ M0 , i.e. A ∪ (∪j1 Ej ) ∈ M0 . ∞ Therefore ∪j1 Ej ∈ U (A). The other property required for U(A) to be a monotone class can be established analogously. Thus U(A) is a monotone class containing A, with the result that U (A) ⊇ M0 . But the reverse inclusion holds by definition of U(A). Therefore U (A)  M0 . As this has been shown to hold for every A ∈ A, it follows that A ∪ F ∈ M0 for every A ∈ A and every F ∈ M0 . Now consider any F ∈ M0 , and set V (F)  {G ∈ M0 : F ∪ G ∈ M0 }. We shall demonstrate that V (F) is a monotone class containing A. From the previous paragraph, we know that V (F) ⊇ A and the same argument as before enables us to c

6.1 Algebras and Monotone Classes of Sets

105

infer that V (F) is a monotone class. Since V (F) ⊆ M0 by definition, it follows that V (F) M0 . As this has been shown to hold for every F ∈ M0 , we have F ∪ G ∈ M0 for all F, G ∈ M0 . This shows that M0 is closed under unions. We have so far shown that M0 is an algebra and it remains to prove that M0 is a σ-algebra. Let Fn ∈ M0 for every n. For each n, consider En  ∪nk1 Fk . Then E 1 ∞ ⊆ E 2 ⊆ ··· and ∪∞ n1 E n  ∪n1 F n . Since M0 has been shown to be an algebra, each E n belongs to it, which implies ∪∞ n1 En ∈ M0 because M0 is a monotone class.  Problem Set 6.1 6.1.P1. Suppose the set X consists of three elements a, b, c and B is the family of subsets {a} and {b, c}. List all the subsets in (i) the smallest algebra of sets that contains B and (ii) the monotone class generated by B. 6.1.P2. Let X consist of four elements a, b, c, d and S consist of the subsets of X described in unorthodox notation as ∅, a, bcd, bc, ad, abcd. Determine (i) whether S is an algebra, (ii) whether S is a monotone class, (iii) what the monotone class generated by S is, and (iv) what the σ-algebra generated by S is. 6.1.P3. Let X be an uncountable set and B be the family of all subsets that are either finite or have finite complement. Show that B is an algebra of subsets. Is it a σ-algebra? 6.1.P4. Let A be an algebra of subsets of a nonempty set X. If there are finitely many sets in A, show that the number of sets in it is 2n , where n is the number of nonempty sets B in the algebra (called ‘atoms’) having the property that the only set B1 ∈ A satisfying ∅ ⊂ B1 ⊆ B must be B. 6.1.P5. Let F be a σ-algebra of subsets of X generated by a subfamily H and let A ∈ F. Show that the induced σ-algebra FA is the same as the σ-algebra generated by the family HA  {H ∩ A : H ∈ H} of intersections of the sets of H with A.

6.2 Defining a Product σ-Algebra In what follows, F denotes a σ-algebra of subsets of X and G denotes a σ-algebra of subsets of Y . We shall define the anticipated σ-algebra in X × Y and also give an alternative description of it in terms of monotone classes (Corollary 6.2.9). 6.2.1. Definition. A rectangle is a subset A × B ⊆ X × Y , where A ∈ F, B ∈ G. It is trivial that an intersection of two rectangles is a rectangle. 6.2.2. Exercise. Show that the representation of a rectangle in the form A × B is unique if and only if it is nonempty.

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Solution: The representation of ∅ in the form A × B is not unique because ∅  X × ∅  ∅ × Y . However, if A × B  P × Q is nonempty, then we can show that A  P and B  Q by arguing thus: Since A × B is nonempty, B is also nonempty; so choose y ∈ B. Then x ∈ A ⇒ (x, y) ∈ A × B  P × Q ⇒ (x, y) ∈ P × Q ⇒ x ∈ P, thereby validating the inclusion A ⊆ P. One may validate the inclusions B ⊆ Q, P ⊆ A and Q ⊆ B analogously. 6.2.3. Definition. The product σ-algebra of F and G is the σ-algebra generated by rectangles and is denoted by F × G [the same symbol as for the Cartesian product although it is not the Cartesian product]. Our next objective is to extend the set function defined on the family of all rectangles A × B as μ(A)ν(B) to a measure on the product σ-algebra F × G. We begin by defining a class of sets that provides a stepping stone. 6.2.4. Definition. A finite union of disjoint rectangles is called an elementary set. The family of elementary sets will be denoted by E. Clearly, any σ-algebra containing all rectangles must also contain all elementary sets and therefore F × G is the same as the σ-algebra generated by elementary sets. 6.2.5. Proposition. A union of finitely many disjoint elementary sets is an elementary set. Proof: Consider finitely many disjoint elementary sets E 1 , E 2 , …, E n . The rectangles comprising any one of the sets E i are disjoint not only from each other but also from the rectangles comprising any other E j , j  i. Thus the finitely many rectangles occurring in all the n unions forming the respective elementary sets E i are all disjoint from each other. This means their union F is an elementary set. But F is the same as  the union of E 1 , E 2 , …, E n . 6.2.6. Proposition. If P1 , P2 , …, Pm are disjoint rectangles and Q is a rectangle, then (P1 \Q) ∪ (P2 \Q) ∪ ··· ∪ (Pm \Q) is an elementary set. 

Proof: Relegated to Problem 6.2.P2. 6.2.7. Proposition. E is closed under intersections and differences. That is, E ∈ E, F ∈ E ⇒ E ∩ F ∈ E, F\E ∈ E.

(6.1)

Proof: Let E  P1 ∪ P2 ∪ · · · ∪ Pm and F  Q1 ∪ Q2 ∪ · · · ∪ Qn , where P1 , P2 , …, Pm are disjoint rectangles and Q1 , Q2 , …, Qn are also disjoint rectangles. Then E∩F 

m  n  i1 j1

(Pi ∩ Qj )

6.2 Defining a Product σ-Algebra

107

and Pi ∩ Qj (1 ≤ i ≤ m, 1 ≤ j ≤ n) are disjoint rectangles. So, E ∩ F ∈ E. To complete the proof of (6.1), it remains to show that E\F ∈ E. We do so by induction on n. For n  1, it is immediate from Proposition 6.2.6 that E\F ∈ E. So, assume as induction hypothesis that E\F ∈ E whenever F is a finite disjoint union of k rectangles. When F is a finite disjoint union F of k + 1 rectangles Qj (1 ≤ j ≤ k +1), we have c ) E\F  E\(Q1 ∪ Q2 ∪ · · · ∪ Qk+1 )  E ∩ (Q1c ∩ Q2c ∩ · · · ∩ Qkc ∩ Qk+1   c c c c c  E ∩ (Q1 ∩ Q2 ∩ · · · ∩ Qk ) ∩ Qk+1  (E\(Q1 ∪ Q2 ∪ · · · ∪ Qk )) ∩ Qk+1 .

By the induction hypothesis, the set E\(Q1 ∪ Q2 ∪ ··· ∪ Qk ) on the right side here is an elementary set. Therefore the already established case when n  1 leads to the conclusion that the entire right side is an elementary set and hence so is the left side E\F. This completes the induction argument.  6.2.8. Corollary. E is closed under taking finite unions and under complementation. That is, E ∈ E, F ∈ E ⇒ E ∪ F ∈ E, E c ∈ E. Thus E is an algebra. Proof: Since E∪F  E∪(F\E), the desired conclusion is a straightforward deduction from Proposition 6.2.5 and Proposition 6.2.7.  6.2.9. Corollary. F × G is the monotone class generated by E. Proof: The result follows from Corollary 6.2.8 and Proposition 6.1.3.



Problem Set 6.2 6.2.P1. Let F and G be σ-algebras of subsets of X and Y respectively. Suppose ∞ X  ∪∞ n1 Xn and Y  ∪n1 Yk , where {X n } and {Y k } are disjoint sequences of sets. Then Fn  {F ∩ Xn : F ∈ F} and Gk  {G ∩ Yk : G ∈ G} are σ-algebras of subsets of X n and Y k respectively. (If X n and Y k are all measurable, Fn and Gk are the induced σ-algebras in the sense of Sect. 4.4.) Show that every E ∈ F × G satisfies E ∩ (Xn × Yk ) ∈ Fn × Gk for every n and k. 6.2.P2. Prove Proposition 6.2.6. 6.2.P3. Let F and G be the σ-algebras consisting of all subsets of X and Y respectively, where X and Y are both countable. Show that F × G consists of all subsets of X ×Y.

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6.2.P4. Let F and G be σ-algebras of subsets of X and Y respectively. Suppose A ∈ F and B ∈ G. Then the σ-algebra (F × G)A×B induced by the product σ-algebra F × G on A × B is the same as the product FA × GB of the induced σ-algebras FA and GB . In particular, if f : X × Y → R∗+ is (F × G)-measurable then the restriction of f to A × B is (FA × GB )-measurable.

6.3 Sections of a Subset and of a Function The reader may recall the notation an , ak introduced in Definition 5.1.1 and f k in Theorem 5.1.4. The superscript or subscript there represented an element of N. We now introduce the same concept again but with the superscript or subscript representing an element of a more general set. Since measurability, which was not an issue in Sect. 5.1, is now going to be of major concern, it is necessary to have a corresponding concept for sets as well. However, the concept itself does not depend on the availability of a σ-algebra and therefore the same is true of some of the results in this section. 6.3.1. Definition. For any x ∈ X, the x-section of a subset E ⊆ X × Y is the subset Ex  {y ∈ Y : (x, y) ∈ E} of Y . Similarly, for any y ∈ Y , the y-section is the subset Ey  {X ∈ X : (x, y) ∈ E} of X. 6.3.2. Examples. (a) Let A = S×T , where S ⊆ X and T ⊆ Y . Then   S if y ∈ T T if x ∈ S y Ax  and A  ∅ if y ∈ /T ∅ if x ∈ /S (b) Suppose X  Y  R and A  {(x, y) : x2 + 4y2 ≤ 4} ⊆ R × R. Then  [− 21 (4 − x2 )1/2 , 21 (4 − x2 )1/2 ] if |x| ≤ 2 Ax  ∅ otherwise and  A  y

[−2(1 − y2 )1/2 , 2(1 − y2 )1/2 ] if |y| ≤ 1 ∅

otherwise.

 (c) Let A  (x, y) : 0 ≤ y ≤ x2 and x ≥ 0 ⊆ R × R. Then  [0, x2 ] if x ≥ 0 Ax  ∅ otherwise and

6.3 Sections of a Subset and of a Function

 A  y

109

(y1/2 , ∞) if y ≥ 0 ∅

otherwise.

6.3.3. Proposition. (Ex )c  (E c )x and (E y )c  (E c )y . Moreover, for any sequence of ∞ ∞ ∞ sets E 1 , E 2 , …, we have ∪∞ j1 (Ej )x  (∪j1 Ej )x and ∩j1(Ej )x  (∩j1 Ej )x . Similarly for y-sections and for finite sequences. In particular, when E and F are disjoint, E x and F x are disjoint and E y and F y are disjoint. If A ⊆ X and B ⊆ Y , then χ(A×B)x (y)  χA (x)χB (y)  χ(A×B)y (x). Proof: The assertions that y ∈ (Ex )c and that y ∈ (E c )x are both equivalent to (x, y) ∈ / E. The argument that (E y )c  (E c )y is analogous. ∞ The assertions that y ∈ ∪∞ j1 (E j )x and that y ∈ (∪j1 E j )x are both equivalent to ∞ ∞ (x, y) ∈ E j for some j. This means ∪j1 (E j )x  (∪j1 E j )x . ∞ The assertions that y ∈ ∩∞ j1 (E j )x and that y ∈ (∩j1 E j )x are both equivalent to ∞ ∞ (x, y) ∈ E j for every j. This means ∩j1 (E j )x  (∩j1 E j )x . The argument for y-sections is similar. The equalities for finite sequences follow because ∅x and ∅y are both empty. In particular, (E x ∩ F x )  (E ∩ F)x and (E y ∩ F y )  (E ∩ F)y , which justifies the assertion regarding disjointness. The argument for the last part is that

χ(A×B)x (y)  1 ⇔ y ∈ (A × B)x ⇔ (x, y) ∈ A × B ⇔ x ∈ A as well as y ∈ B ⇔ χA (x)  1 as well as χB (y)  1 ⇔ χA (x)χB (y)  1, while at the same time, (x, y) ∈ A × B ⇔ x ∈ (A × B)y ⇔ χ(A×B)y (x)  1.  6.3.4. Lemma. Suppose E = A × B, where A ⊆ X and B ⊆ Y . Then E x  B whenever / A; also, E y  A whenever y ∈ B and E y  ∅ whenever x ∈ A and E x  ∅ whenever x ∈ y∈ / B. If A ∈ F and B ∈ G, then ν(E x )  ν(B)χA (x) and μ(E y )  μ(A)χB (y). Proof: (The first part has been seen in Example 6.3.2(a) but we nevertheless give a detailed argument here.) By definition, y ∈ E x if and only if (x, y) ∈ E. Consider any x ∈ A. Since E  A × B, therefore (x, y) ∈ E if and only if y ∈ B. Thus y ∈ E x if and / A. Then (x, y) ∈ / E A ×B only if y ∈ B. This means E x  B. Now consider any x ∈ for every y ∈ Y , which means Ex  ∅. The proof for E y is analogous. The last part is now trivial.  6.3.5. Proposition. Let E ∈ F × G. Then Ex ∈ G for every x ∈ X, and E y ∈ F for every y ∈ Y .

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Proof: Let H be the family of all E ∈ F × G for which Ex ∈ G for every x ∈ X and E y ∈ F for every y ∈ Y. It is a consequence of Proposition 6.3.3 that E ∈ H ⇒ E c ∈ H and that for any sequence of sets E 1 , E 2 , … in H we have ∪∞ j1 Ej ∈ H. Thus H is a σ-algebra. Moreover, in view of Lemma 6.3.4, it contains all rectangles. By definition of the σ-algebra F × G, it follows that F × G ⊆ H. On the other hand, H ⊆ F × G by the definition of H. Therefore H  F × G.  6.3.6. Definition. Given x ∈ 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 such that fx (y)  f (x, y) for all y ∈ Y . Similarly for the y-section f y with a given y ∈ Y . As the y-section and the yth power are denoted by the same symbol, one has to judge from the context which one is meant. When fx is integrable or has integral ∞, we shall denote Y fx by the more elaborate the notation used in Calculus. Similarly, symbol Y f (x, y)d ν(y), which resembles we shall denote the integral X f y by X f (x, y)d μ(x). 6.3.7. Exercise. If E ⊆ X × Y , show that χEx (y)  (χE )x (y)  (χE )y (x)  χE y (x)  χE (x, y) for all x ∈ X and y ∈ Y . If moreover E ∈ F × G, then show that

χE (x, y) d ν(y)  (χE )x  χEx  ν(Ex ) for all x ∈ X Y

Y

Y

and

χE (x, y)d μ(x)  X

(χE ) 

χE y  μ(E y ) for all y ∈ Y .

y

X

X

Solution: From the definition of section, we have / Ex χEx (y)  1 if y ∈ Ex and 0 if y ∈  1 if(x, y) ∈ E and 0 if(x, y) ∈ / E, and / E. (χE )x (y)  (χE )y (x)  χE (x, y)  1 if(x, y) ∈ E and 0 if(x, y) ∈ Similarly, / E. χE y (x)  1 if(x, y) ∈ E and 0 if(x, y) ∈ This proves the first part.

6.3 Sections of a Subset and of a Function

111

For the second part, the equality of the integrals follows from the first part. We need to show only that Y χEx  ν(Ex ) and X χE y  μ(E y ). This follows from the fact that the integral of the characteristic function of a measurable set is always equal to measure of that set. 6.3.8. Remark. It is trivial that (φ + ψ)x  φx + ψx and (αφ)x  α(φx ), where α is a real number; correspondingly for y-sections. It is a simple consequence of Exercise 6.3.7 and Lemma 6.3.4 that, if E is a rectangle A × B, then (χE )x  χEx  χB when x ∈ A and (χE )y  χE y  χA when y ∈ B. 6.3.9. Proposition. If the extended real-valued function f on X × Y is measurable, then its sections fx and f y are also measurable. Proof: Let α be any real number; we must show that {y ∈ Y : fx (y) > α} ∈ G. This set is nothing but the section E x , where E  {(x, y) ∈ X × Y : f (x, y) > α}. Since f is measurable, E ∈ F × G. It follows by Proposition 6.3.5 that EX ∈ G. Thus fx is a  measurable function. A similar argument shows that f y is also measurable. Problem Set 6.3 6.3.P1. Let X,Y and V be as in Example 5.1.3. For each x ∈ X, describe the set V x and for each y ∈ Y , describe the set V y . If f  χV , describe the function fx for each x ∈ X.

6.4 Defining a Product Measure We now proceed to set forth a measure on X × Y that is intended to serve the purpose described in the opening paragraph of this chapter, but we do so under a certain restriction called “σ-finiteness”. The question of whether it actually serves the intended purpose will be taken up in Sect. 6.5. The definition of the measure in question presupposes a certain result involving sections of sets, which will be taken up shortly. 6.4.1. Definition. A measure space (X, F, μ) is called σ-finite if there exists a sequence {X n }n≥1 of subsets such that ∪∞ n1 Xn  X and each X n is F-measurable with μ(Xn ) < ∞. One also speaks of the measure μ itself as being σ-finite. The measure space (R, M, m) is σ-finite [here m means Lebesgue measure] because the sets X n  [−n, n], where n ∈ N, satisfy ∪∞ n1 Xn  R, Xn ∈ M and m(X n ) < ∞ for each n. Clearly, the same is true if R is replaced by an interval. Let μ be the counting measure on the σ-algebra P(N) of all subsets of N. Then the sets X n  {n}, consisting of a single point each, satisfy the requirements that ∪∞ n1 Xn  N

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and μ(Xn )  1 < ∞ for each n. So, (N, P(N), μ) is a σ-finite measure space. Of course, σ-finiteness holds for the counting measure on any countable set. In the σ-finiteness condition, we are free to take the X n to be disjoint because each X n with n > 1 can be replaced by Xn \∪n−1 k1 Xk . 6.4.2. Proposition. Let (X, F, μ) and (Y , G,ν) be σ-finite measure spaces and E ∈ F × G. Then (a) the function that maps each x ∈ X into ν(E x ) is measurable, (b) the each y ∈ Y into μ(E y ) is measurable, function that maps (c) X ν(Ex )d μ(x)  Y μ(E y )d ν(y). Proof: We first consider finite measures μ and ν. Let  be the family of all subsets E ∈ F × G such that assertions (a), (b) and (c) hold. We shall show that   F × G. It is enough to show that  contains every elementary set and that it is a monotone class contained in F × G, because Corollary 6.2.9 will then imply   F × G. Consider a rectangle E  A × B, where A ∈ F and B ∈ G. By Lemma 6.3.4, ν(Ex )  ν(B)χA (x) and therefore ν(E x ) is a measurable function on X satisfying

ν(Ex )d μ(x)  ν(B)μ(A). (6.2) X

Similarly, μ(E y ) is a measurable function on Y and its integral over Y equals the right side of (6.2). Therefore a rectangle always belongs to . Now consider an elementary set E  ∪Nn1 (An × Bn ), a finite union of disjoint rectangles. By Proposition 6.3.3, we have χEx (y) 

N 

χ(An ×Bn )x (y) 

n1

N 

χAn (x)χBn (y),

n1

and hence ν(Ex ) 

N 

⎛ ⎝χAn (x)

n1

⎞ χBn (y)d ν(y)⎠ 

N 

ν(Bn )χAn (x).

n1

Y

This shows that ν(E x ) is a sum of measurable functions and therefore measurable. Moreover,

ν(Ex )d μ(x)  X

N 

ν(Bn )μ(An ).

n1

Similarly, μ(E y ) is a measurable function on Y and

6.4 Defining a Product Measure

113

μ(E y )d ν(y)  Y

N 

ν(Bn )μ(An ).

n1

Thus the assertions (a), (b), (c) of the proposition hold for every elementary set, which means  contains every elementary set. We proceed to show that  is a monotone class. Let En ∈ , n ≥ 1, satisfy y y En ⊆ En+1 for every n, and put E  ∪∞ n1 En . Then (E n )x ⊆ (E n+1 )x and (E n ) ⊆ (E n+1 ) for every x ∈ X and y ∈ Y . Hence the sequences of nonnegative measurable functions {ν((En )x )}n≥1 and {μ((En )y )}n≥1 are increasing, and limn→∞ ν((En )x )  ν(Ex ), limn→∞ μ((En )y )  μ(E y ). This means ν(E x ) and μ(E y ) are limits of measurable functions and are therefore measurable, i.e. E satisfies (a) and (b). We have to show that E satisfies (c) as well. By the Monotone Convergence Theorem 2.3.2, we have

ν(Ex )d μ(x)  lim ν((En )x )d μ(x) n→∞

X

X

and

μ(E y )d ν(y)  lim

μ((En )y )d ν(y).

n→∞

Y

Y

Since En ∈  for each n ≥ 1, we have

ν((En )x )d μ(x)  μ((En )y )d ν(y). X

Y

Therefore, it follows from the limits noted above that

ν(Ex )d μ(x)  μ(E y )d ν(y). X

Y

Consequently, E  ∪∞ n1 En satisfies (c) as well, so that E ∈ . Similarly, if E n ∈  and E n ⊇ E n+1 , n ≥ 1, we can conclude that E  ∩∞ n1 En ∈  by using the Dominated Convergence Theorem 3.2.6 and the hypothesis that μ and ν are both finite measures. This completes the proof that  is a monotone class, thereby establishing the contention of the proposition for the case of finite measures. ∞ For the infinite but σ-finite case, we have X  ∪∞ n1 Xn , Y  ∪k1 Yk , where {X n } and {Y k } are disjoint sequences of nonempty sets of finite measure. Denote by (X n , Fn , μn ) and (Y k , Gk , νk ) the measure spaces induced on X n and Y k respectively. Consider an arbitrary E ∈ F × G and take En,k  E ∩ (Xn × Yk ). Then En,k ∈ F × G and hence by Proposition 6.3.5, its sections are measurable. Moreover, considering that the sets E n,k are disjoint and their union is E, we infer on the basis of Proposition

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6 Product Measures

6.3.3 that the x-sections (E n,k )x are disjoint and their union is E x ; correspondingly for y-sections. Therefore ν(Ex ) 

∞  ∞ 

(ν(En,k )x ) and μ(E y ) 

n1 k1

∞  ∞ 

μ((En,k )y ).

(6.3)

n1 k1

By Problem 6.2.P1, we have En,k ∈ Fn × Gk , and hence by what has been proved in the case of finite measures, it follows that the functions of x and y defined by νk ((E n,k )x ) and μn ((E n,k )y ) on X n and Y k respectively are measurable, and that

ν((En,k )x )d μn (x)  μ((En,k )y )d νk (y). (6.4) Xn

Yk

Now, the functions defined on the whole of X and the whole of Y by ν((E n,k )x ) and μ((E n,k )y ) respectively agree with the aforementioned functions on X n and Y k while being zero elsewhere. In other words, they agree with their own respective products with the characteristic functions of X n and Y k and their restrictions to the latter are precisely the aforementioned functions. Upon invoking Proposition 4.4.1, we find that they are measurable and

ν((En,k )x )d μ(x)  ν((En,k )x )d μn (x) X

Xn

and

μ((En,k )y )d ν(y)  Y

μ((En,k )y )d νk (y). Yk

However, the last recorded equality of integrals above, (6.4), says that the right sides here are equal. It therefore follows that the left sides here are equal:

ν((En,k )x )d μ(x)  μ((En,k )y )d ν(y). (6.5) X

Y

∞

 y Since the terms in the series k1 ν((En,k )x ) and ∞ k1 μ((En,k ) ) have been shown to be measurable functions on X and Y respectively, the same is true of the functions represented their sums and hencealso of the functions represented by their repeated  by  ∞ ∞ ∞ y ν((E ) ) and sums ∞ n,k x n1 k1 n1 k1 μ((En,k ) ). However, the repeated sums are precisely the right sides in (6.3). Therefore the functions represented by the left sides ν(E x ) and μ(E y ) in (6.3) are measurable, showing that assertions (a) and (b) of the proposition hold. Finally, in view of (6.3), (6.5) and the Monotone Convergence Theorem,

6.4 Defining a Product Measure

ν(Ex )d μ(x) 

115

∞  ∞ 

  ν (En,k )x d μ(x)

n1 k1 X n

X



∞  ∞ 

  μ (En,k )y d ν(y) 

n1 k1 Y k

μ(E y )d ν(y), Y

showing that assertion (c) of the proposition also holds.  Example 5.1.3 shows that the σ-finiteness condition in the above proposition is essential, because the set V there can be  to be measurable as follows.  shown j and V n  ∪nj1 (I j × I j ). Clearly, each , For any positive integer n, set Ij  j−1 n n V n is a union of measurable rectangles and therefore measurable, and hence so is ∩∞ n1 V n . But this intersection is precisely V . In the definition about to be presented, the equality of the integrals is assured by Proposition 6.4.2. 6.4.3. Definition. Let (X, F, μ) and (Y ,G, ν) be σ-finite measure spaces. Then the set function μ × ν defined on F × G by

(μ × ν)(E)  ν(Ex ) d μ(x)  μ(E y )d ν(y) X

Y

is called the product measure of μ and ν. 6.4.4. Theorem. Let (X,F, μ) and (Y , G, ν) be σ-finite measure spaces. Then the product μ × ν defined on the σ-algebra F as

(μ × ν)(E)  ν(Ex )d μ(x)  μ(E y )d ν(y) X

Y

is a σ-finite measure. Proof: Relegated to Problem 6.4.P1.





6.4.5. Remark. In view of Exercise 6.3.7, Y χE (x, y)d ν(y)  Y (χE )x  ν(Ex ) while X χE (x, y)d μ(x)  X (χE )y  μ(E y ). Parts (a) and (b) of Proposition 6.4.2 assert that these two functions of x and y respectively are measurable; part (c) asserts that ⎛ ⎞ ⎛ ⎞



⎝ (χE )x ⎠  ⎝ (χE )y ⎠, X

or in alternate notation,

Y

Y

X

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6 Product Measures



d μ(x) X

χE (x, y)d ν(y) 

Y

d ν(y)

Y

χE (x, y)d μ(x) X

Since μ × ν is a measure by Theorem 6.4.4, we have

(μ × ν)(E)  χE d (μ × ν)  χE (x, y)d (μ × ν)(x, y). X ×Y

X ×Y

Therefore on the basis of Definition 6.4.3, ⎛ ⎞ ⎛ ⎞





⎝ (χE )x ⎠  χE d (μ × ν)  ⎝ (χE )y ⎠, X

X ×Y

Y

which may also be written as



d μ(x) X

χE (x, y)d ν(y)  Y

Y

X

χE (x, y)d (μ × ν)(x, y) 

X ×Y

d ν(y)

Y

χE (x, y)d μ(x). X

In other words, the “double integral” of the characteristic function of a measurable subset of the product space equals both repeated It should integrals of that function. be noted that measurability of the functions Y χE (x, y)d ν(y) and X χE (x, y)d μ(x) is not assumed but claimed. In the next section, we shall prove the equality of double and repeated integrals for a larger class of measurable functions on the product space. 6.4.6 Remark. When (X, F, μ) is a measure space, a measurable subset of X is said to be σ-finite if the measure space induced on it is σ-finite. Consider an integrable function f :X → R∗ . By Problem 2.2.P12, for each n ∈ N, we 1 have μ(X (|f |> 1n )) ≤ n |f |< ∞. Therefore the set X (|f |> 0)  ∪∞ n1 X (|f |> n ) is σ-finite. This is the analog of Proposition 5.1.9(b) for a general measure that need not be a counting measure. Problem Set 6.4 6.4.P1. Prove Theorem 6.4.4.

  j . For the set V of Example 5.1.3, , 6.4.P2. For any positive integer n, set I j  j−1 n n ∞ n show that V  ∩n1 Vn , where Vn  ∪j1 (Ij × Ij ).

6.5 The Tonelli and Fubini Theorems We now have the requisite machinery for formally defining what we have informally called a “double integral” so far, and for relating it to repeated integrals. There are two well-known theorems in this regard. Both of them validate interchanging the order of integration under certain conditions provided the measures are σ-finite.

6.5 The Tonelli and Fubini Theorems

117

6.5.1. Definition. Given σ-finite measure spaces (X, F, μ) and (Y , G, ν), any integral with respect to the product measure μ × ν is called a double integral. 6.5.2. Tonelli’s Theorem. Let (X, F, μ) and (Y , G, ν) be σ-finite measure spaces and f be a nonnegative extended real-valued (F × G)-measurable function on X × Y . Then Y f (x, y)d ν(y) is an F-measurable function on X, X f (x, y)d μ(x) is a G-measurable function on Y, and the double integral of f agrees with each of its repeated integrals:



d μ(x) f (x, y)d ν(y)  f (x, y)d (μ × ν)(x, y)  d ν(y) f (x, y)d μ(x). X

X ×Y

Y

Y

X

Proof: It is sufficient to prove the first equality. By Proposition 6.3.9, the integral f (x, y)d ν(x) exists for each x ∈ X. We need to prove that it provides a measurable Y function on X with integral equal to X ×Y fd (μ × ν). As observed in Remark 6.4.5, this is true when f is the characteristic function of an (F × G)-measurable set. It follows immediately from Remark 6.3.8 that the same is true also when f is any simple function. Now let f be an arbitrary nonnegative extended real-valued (F × G)-measurable function on X × Y . There exists an increasing sequence of nonnegative simple functions sk , k ∈ N, having limit f (see Theorem 3.2.1). Since the required conclusion is known to be valid for simple functions, each Y sk (x, y)d ν(y) is an F-measurable function on X and

d μ(x) sk (x, y)d ν(y)  sk (x, y)d (μ × ν)(x, y). (6.6) Y

X

X ×Y

By applying the Monotone Convergence Theorem 2.3.2, we obtain

lim sk (x, y)d ν(y)  f (x, y)d ν(y), k→∞

Y

Y

from which we know that Y f (x, y)d ν(y) is an F-measurable function on X. Moreover, the F-measurable functions Y sk (x, y)d ν(y) form an increasing sequence, and hence by another application of the Monotone Convergence Theorem, we obtain



lim d μ(x) sk (x, y)d ν(y)  d μ(x) f (x, y)d ν(y), k→∞

X

Y

X

Y

which is the same as

lim sk (x, y)d (μ × ν)(x, y)  d μ(x) f (x, y)d ν(y) k→∞ X ×Y

X

Y

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6 Product Measures

in the light of (6.6). Finally, one further application of the Monotone Convergence Theorem yields

lim sk (x, y)d (μ × ν)(x, y)  f (x, y)d (μ × ν)(x, y). k→∞ X ×Y

X ×Y

The required equality now springs forth from the preceding two.



6.5.3. Remark. Given any measure space (X, F, μ) and an integrable function f on X, the subset E ⊆ X on which f takes the value ∞, obviously measurable, + must have 0. This is because f ≥ nχE for every positive integer n, which measure + implies X f ≥ n X χE ≥ nμ(E) for every positive integer, thereby contradicting the finiteness of X f unless μ(E)  0. It is a consequence that there exists a measurable subset F (e.g. E c ) on which f is finite and μ(F c )  0. Such an F need not be precisely the complement of E and could be a smaller set instead, because “removing” a nonempty subset of measure 0 from any such F leads to a smaller subset with the property that f is finite at each point of it and its complement has measure 0. It is often convenient to avoid explicit mention of the set of measure 0 by describing the relevant property, such as being finite, that holds at each point of its complement as holding almost everywhere. Phrased in this terminology, the remark in the preceding paragraph asserts that an integrable function is finite almost everywhere. If the property has been formulated in terms of an explicitly named point x ∈ X, then one can express the same thing by saying that the property holds for almost all x. The standard abbreviation for “almost everywhere” is a.e. Thus for instance, the assertion that f  g a.e. means the same thing as “f (x)  g(x) a.e.” or “f (x)  g(x) for almost all x”. 6.5.4. Remarks. (a) Before stating Theorem 5.1.10 about interchanging the order of summation, i.e. integration with the counting measure, over uncountable sets, we had mentioned the idea of reducing the matter to the countable case by exploiting Proposition 5.1.9(b). Now that we have the analog of Proposition 5.1.9(b) for a general measure (see Remark 6.4.6), the same idea would suggest reducing the matter of interchanging the order of integration with non-σ-finite measures to the σ-finite case, which has already been established in Tonelli’s Theorem 6.5.2. Accepting such a reduction would lead to the conclusion that interchanging the order of integration with respect to measures that may not be σ-finite is valid. However, we know from Example 5.1.3 that this conclusion is definitely erroneous. This shows that reduction to the countable case is a fallacious line of reasoning. In particular, interchanging the order of summation over uncountable sets is not a special case of Tonelli’s Theorem. (b) Suppose φ and ψ are integrable nonnegative extended real-valued functions. It makes no sense to speak of a function φ − ψ, because ψ may take the value ∞. By Remark 6.5.3, the functions φ and ψ are finite a.e. That is, there exist measurable sets E and F such that μ(E c )  μ(F c )  0, and the

6.5 The Tonelli and Fubini Theorems

119

functions φ and ψ are finite-valued on E and F respectively. Then they are both finite-valued on the set E ∩ F and μ((E ∩ F)c )  0. Thus, it is true almost everywhere that both are finite-valued. There is a function φ − ψ on the domain E ∩ F, that is to say, φ − ψ is defined a.e. If we extend it to all of X by setting it equal to 0 outside E ∩ F, the resulting function  is is a measurable function on X. Now, the same as φχE ∩F − ψχE ∩F , which φχ  φ and ψχ  E∩F E∩F X X X X ψ by Problem 2.2.P7. Since φ and ψ and ψχ are integrable, so are φχ E ∩ F E ∩F , and therefore also ; moreover,   (φχ − ψχ )  φχ − ψχ  E∩F E∩F E∩F E∩F X X X X X φ − X ψ. We thus see that there exists an integrable function  that agrees with φ − ψ almost everywhere and satisfies X   X φ − X ψ. (c) Let {fn }n≥1 be a sequence of measurable nonnegative functions on a measure space X and f a measurable function such that limn→∞ fn (x)  f (x) for almost all x ∈ X. It follows from Fatou’s Lemma 3.2.3, as we shall presently show, that

f ≤ lim inf fn . n→∞

By definition of almost all, there exists a measurable set A ⊆ X such that μ(Ac )  0 and limn→∞ fn (x)  f (x) for all x ∈ A. It follows that  f (x)χA (x) for all x ∈ A. Therefore by Fatou’s Lemma, limn→∞ fn (x)χA (x) f χA ≤ lim inf fn χA . But also, f (x)χA (x)  f (x) and fn (x)χA (x)  fn (x) for n→∞ all x ∈ A. Therefore by Problem 2.2.P7, f χA  f and fn χA  fn . (d) Now suppose  and 1 are integrable functions and g is some function defined almost everywhere such that both  and 1 agree with g almost everywhere. Then  and 1 agree with each other almost everywhere, that is, they agree on some set G with μ(Gc )  0. By Problem 2.2.P7,





  + − −  + χG − − χG  +1 χG − − 1 χG X

X

X



+1 − X

X

X

− 1 

X

X

X

1 . X

By X g we shall understand X φ, where  is any integrable function that agrees with g almost everywhere. In part (c) of Fubini’s Theorem below, the first and third integrals are understood in this sense, keeping part (b) of the theorem in mind. 6.5.5. Fubini’s Theorem. Let (X, F, μ) and (Y , G, ν) be σ-finite measure spaces and the real-valued function f on X × Y be integrable. Then (a) fx is integrable for almost all x, and f y is integrable for almost all y; (b) there exist integrable functions on X and Y , which agree almost everywhere with Y f (x, y)d ν(y) and X f (x, y)d μ(x) respectively;

120

6 Product Measures

(c) the double integral of f equals both its repeated integrals:



d μ(x) X



f (x, y)d ν(y) 

f (x, y)d (μ × ν)(x, y) 

X ×Y

Y

d ν(y) Y

f (x, y)d μ(x). X

Proof: By Proposition 6.3.9, the extended real-valued function fx on Y is measurable for each x∈X. Correspondingly for f y . Since f + ≥ 0 is measurable (see the observation just before Definition 2.2.5), funcit follows by Tonelli’s Theorem 6.5.2 that Y f + (x, y)d ν(y) is a measurable + d μ(x) f (x, y)d ν(y)  tion on X and that its integral over X satisfies X Y + f d (μ × ν)(x, y). Since f is integrable, it follows that X ×Y

d μ(x) f + (x, y)d ν(y) < ∞. (6.7) X

Y

By Remark 6.5.3, this has the consequence that the measurable function on X given by Y f + (x, y)d ν(y) is finite almost everywhere. Since Y f + (x, y)d ν(y) means the same thing as Y (f + )x , the preceding assertion is precisely that (f + )x is integrable for almost all x. By a similar argument,

d μ(x) f − (x, y)d ν(y) < ∞, (6.8) X

Y

and the measurable function on X given by Y f − (x, y)d ν(y) is finite almost everywhere, which means (f − )x is integrable for almost all x. From the obvious equality (see Proposition 2.2.4) fx  (f + − f − )x  (f + )x − (f − )x ,

(6.9)

we deduce that fx is integrable for almost all x. This proves the first part of (a) and the second part follows by an analogous argument. By (6.7) and (6.8), Y f + (x, y)d ν(y) and Y f − (x, y)d ν(y) represent integrable functions on X. By Remark 6.5.4(b), there is an integrable function  on X such that

+ (x)  f (x, y)d ν(y) − f − (x, y)d ν(y) (6.10) Y

Y

almost everywhere and has integral equal to



  d μ(x) f + (x, y)d ν(y) − d μ(x) f − (x, y)d ν(y). X

X

Y

X

Y

From (6.9) we also deduce that, for almost all x, we have

(6.11)

6.5 The Tonelli and Fubini Theorems

121





fx  Y

(f + )x − Y

(f − )x ,

Y

which can be expressed as

+ f (x, y)d ν(y)  f (x, y)d ν(y) − f − (x, y)d ν(y). Y

Y

Y

Together with (6.10), this leads to

(x)  f (x, y)d ν(y) for almost all x.

(6.12)

Y

Thus there exists an integrable function on X that agrees with Y f (x, y)d ν(y) almost everywhere. By a similar argument, there exists an integrable function on Y that agrees with X f (x, y)d μ(x) almost everywhere. This proves (b). It follows from Remark 6.5.4(d), (6.11) and (6.12) that



d μ(x) f (x, y)d ν(y)  d μ(x) f + (x, y)d ν(y) − d μ(x) f − (x, y)d ν(y). X

Y

X

Y

X

Y

(6.13) By definition of integral (applied to μ × ν on X × Y ), we have



f (x, y)d (μ × ν)(x, y)  X ×Y

f + (x, y)d (μ × ν)(x, y) − X ×Y

f − (x, y)d (μ × ν)(x, y).

X ×Y

By Tonelli’s Theorem, the right side here agrees with that of (6.13). Therefore the left side here agrees with that of (6.13), which means the first equality of part (c) is valid. The second equality is also valid for analogous reasons.  Problem Set 6.5 6.5.P1. Let X  Y  [1, ∞) with Lebesgue measure m. Suppose f :X × Y → R satisfies f (0, 0)  0 and f (x, y) 

x2 − y2 . (x2 + y2 )2

Show that f cannot be integrable even if it is measurable. 6.5.P2. If X and Y are both countable, show that the product measure of the counting measures on them is the counting measure on X × Y. 6.5.P3. Show that repeated integrals with respect to counting measures can be unequal even if the function on the product space is measurable. Hint: See Problem 1.2.P1.

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6 Product Measures

6.5.P4. Give an example of two functions on a measure space that are equal almost everywhere, one is measurable but the other is not. 6.5.P5. Suppose {fn }n≥1 is a sequence of measurable functions such that lim fn (x) n→∞ exists for all x in some measurable set of convergence A. Show that the function defined to agree with the limit on A and to be 0 elsewhere is measurable. Show also that two different measurable functions obtained in this manner by using two different sets of convergence, both having complement of measure 0, agree a.e. 6.5.P6. Let (X, F, μ) and (Y , G, ν) be measure spaces and f : X → R, φ : X → R be measurable functions such that f  φ a.e. Suppose g:Y → X has the property that F ∈ F ⇒ g −1 (F) ∈ G, so that the compositions f ◦ g:Y → R and φ ◦ g:Y → R are both measurable. Is it necessary that f ◦ g  φ ◦ g a.e.? 6.5.P7. Let f :[0, 1] × [0, 1]→ R be the function given by 1 1 f (x, y)  (x − )−3 if 0 < y < |x − | and f (x, y)  0 otherwise. 2 2

(1,1)

0 in this pentagon _1_

(1,_12_ )

(0, 2 ) (x

_ _1_

_3

2)

(x

_ _1_

_3

2)

Show that the repeated integral [0,1] dx [0,1] f (x, y)dy does not even exist but the repeated integral [0,1] dy [0,1] f (x, y)dx exists.

6.6 Borel Sets Although a product σ-algebra F × G is easy to define, it can be intractable in the concrete situation when F and G are the σ-algebra of Lebesgue measurable sets in R or an interval. This can make it difficult to verify measurability of a function on the product space R × R for applying the Fubini or Tonelli Theorem. In order to alleviate this difficulty, we introduce a smaller but more tractable σ-algebra. The benefits will be harvested in Chap. 7. So far, we have used a symbol such as “(a, b)” to mean either the open interval from a to b or the ordered pair with a as the first element and b the second, as there was no possibility of confusion. However, in the present section, confusion can arise

6.6 Borel Sets

123

and we shall denote the ordered pair by “a, b”. But we shall continue to denote the value of a function f at a, b by the symbol “f (a, b)” as usual. 6.6.1. Definition. The σ-algebra of Lebesgue measurable subsets of R will be denoted by M. The σ-algebra generated by all open intervals is called the Borel σ-algebra and will be denoted by B. A set belonging to B is called a Borel set. Since every interval is a Lebesgue measurable set, it is immediate that B ⊆ M and hence that B × B ⊆ M × M. Since a Borel set B is a Lebesgue measurable set, it has a Lebesgue measure m(B), which is sometimes referred to as its Borel measure. In other words, the restriction to B of the Lebesgue measure m is sometimes called the “Borel measure”. It is left to the reader as Problem 6.6.P1 to show that every interval is a Borel set. 6.6.2. Proposition. Let I ⊆ R be an open interval. Then {x, y ∈ R × R : x + y ∈ I } ∈ B × B. Proof: Denote the set in question by A and consider an arbitrary point ξ, η ∈ A. Then ξ + η ∈ I. Therefore there exists a δ > 0 such that (ξ + η − 2δ, ξ + η + 2δ) ⊆ I . It follows that x, y ∈ (ξ − δ, ξ + δ) × (η − δ, η + δ) ⇒ −δ < x − ξ < δ and − δ < y − η < δ ⇒ −2δ < (x + y) − (ξ + η) < 2δ ⇒ (ξ + η) − 2δ < x + y < (ξ + η) + 2δ ⇒ x + y ∈ (ξ + η − 2δ, ξ + η + 2δ) ⇒x+y ∈I ⇒ x, y ∈ A. Thus, (ξ − δ, ξ + δ) × (η − δ, η + δ) ⊆ A. Now let p, q, r, s be rational numbers such that ξ − δ < p < ξ < q < ξ + δ and η − δ < r < η < s < η + δ. We then have ξ, η ∈ (p, q) × (r, s) ⊆ (ξ − δ, ξ + δ) × (η − δ, η + δ) ⊆ A. This shows that each ξ, η ∈ A belongs to a Cartesian product (p, q) × (r, s) ⊆ A of intervals with rational endpoints p, q, r, s. The union of all these Cartesian products is therefore precisely A. Now all such Cartesian products (p, q) × (r, s) belong to B × B and there can be only countably many of them, as their endpoints p, q, r, s are rational. Thus, A is a countable union of sets belonging to B × B and hence must belong to B × B.  6.6.3. Exercise. Let D be a σ-algebra of subsets of R × R. Show that the family E of subsets E of R such that

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6 Product Measures

{x, y ∈ R × R : x + y ∈ E} ∈ D is a σ-algebra. Solution: For any E ⊆ R, denote the associated set {x, y ∈ R × R : x + y ∈ E} by E*. We have to show that the family E of subsets E of R such that E ∗ ∈ D is a σ-algebra. First we argue that ∅∗  ∅. When E  ∅, there cannot be any x, y ∈ R × R for which x + y ∈ E, which means ∅∗  E ∗  {x, y ∈ R × R : x + y ∈ E}  ∅. Next, let {E k }k ≥1 be a sequence of subsets of R. Then  ∞ ∗ ∞   Ek  Ek∗ , k1

k1

because for any x, y ∈ R × R, we have  ∞ ∗ ∞   x, y ∈ Ek ⇔ x + y ∈ Ek ⇔ x + y ∈ Ek for some k ∈ N k1

k1

⇔ x, y ∈ Ek∗ for some k ∈ N ⇔ x, y ∈

∞ 

Ek∗ .

k1

It is equally easy to see that (E ∗ )c  (E c )∗ . In fact, / E∗ ⇔ x + y ∈ /E x, y(E ∗ )c ⇔ x, y ∈ c ⇔ x + y ∈ E ⇔ x, y ∈ (E c )∗ . It follows from the equalities established in the foregoing three paragraphs that E is a σ-algebra because D is. 6.6.4. Proposition. Let B ⊆ R be any Borel set. Then {x, y ∈ R × R : x + y ∈ B} ∈ B × B. Proof: Let E be the family of subsets E of R such that {x, y ∈ R × R : x + y ∈ E} ∈ B × B.

6.6 Borel Sets

125

By the result of Exercise 6.6.3, the family E is a σ-algebra, and by Proposition 6.6.2, it contains every open interval. By Definition 6.6.1 of the σ-algebra B, we have E ⊇ B. This is precisely the conclusion we wanted to arrive at.  Recall the terminology introduced in Remark 6.5.3 that in the context of a measure space (X, F, μ), a property involving a point of X is said to hold almost everywhere, abbreviated as “a.e.”, if it holds at each point of X outside some set of measure 0. Suppose there is a finite or infinite sequence of properties, all of which hold a.e., but each with its own set of measure 0. Using the fact that a union of a finite or infinite sequence of sets of measure 0 is again a set of measure 0, we can deduce that all the properties hold simultaneously outside a single set of measure 0, and thus hold a.e. Such deductions will often be carried out with no express mention of the sets of measure 0, as is customary. In the lemma below, the hypothesis that the measurable set under consideration has finite measure can be removed but we shall not need to. 6.6.5. Lemma. Let E ⊆ R be a Lebesgue measurable set of finite measure. Then for ∞ any  η > 0, there exists a sequence {I n }n≥1 of open intervals such that E ⊆ ∪n1 I n and I )\E ≤ η. m (∪∞ n1 n 

Proof: Left to the reader as Problem 6.6.P2.

6.6.6. Proposition. Any Lebesgue measurable set E ⊆ R is contained in a Borel set F such that m(F\E)  0; in particular, m(F)  m(E). Proof: First suppose that E has finite measure. By Lemma 6.6.5, for every p ∈ N, there exists a Borel set Bp ⊇ E such that m(Bp \E) ≤ p1 . The intersection F  ∩∞ p1 Bp is then a Borel set such that F ⊇ E. Moreover, F\E ⊆ Bp \E for every p ∈ N and therefore m(F\E) ≤ m(Bp \E) ≤ p1 for every p ∈ N, so that m(F\E)  0. Now consider a Lebesgue measurable set E of infinite measure. For each n ∈ N, let E n  E ∩ [−n, n], so that ∪∞ n1 E n  E. Furthermore, each E n has finite measure and therefore the result of the preceding paragraph yields a Borel set F n such that Fn ⊇ En and m(Fn \En )  0. It follows that ∞  n1

Fn ⊇

∞ 

En  E

(6.14)

n1

∞ ∞ ∞ and m(∪∞ n1 (Fn \En ))  0. But ∪n1 Fn \∪n1 En ⊆ ∪n1 (Fn \En ), as is easy to check. ∞ F \∪ E )  0. By (6.14), this means m((∪∞ Therefore m(∪∞ n1 n n1 n n1 Fn )\E)  0. F satisfies m(F\E)  0; moreover, it satisfies F Therefore the Borel set F  ∪∞ n1 n ⊇ E by virtue of (6.14).  We can now assert that a property holds outside some Lebsgue measurable set of measure 0 if and only if it holds outside some Borel set of measure 0. Indeed, the “if”

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6 Product Measures

part is obvious because B ⊆ M, and the converse is a consequence of Proposition 6.6.6 applied to a Lebesgue measurable set of measure 0. Any function that is F-measurable in the sense of Definition 2.1.4 with F  B is said to be Borel measurable (or B-measurable or just Borel). 6.6.7. Proposition. Given a real-valued Lebesgue measurable function f defined on R, there exists a real-valued Borel function g on R such that f  g a.e. Proof: We begin by proving this when f  χE , the characteristic function of a Lebesgue measurable set E. By Proposition 6.6.6, there exists a Borel set F such that E ⊆ F and m(F\E)  0. Select g  χF . We claim that f  g a.e., that is χE  χF a.e. More precisely, we claim that χE  χF outside F\E. To demonstrate why this is so, we first split R as     R  F c ∪ F  F c ∪ F ∩ (E c ∪ E)  F c ∪ (F ∩ E c ) ∪ (F ∩ E)  F c ∪ (F\E) ∪ E.

Next, we note that it follows from the inclusion E ⊆ F that χE and χF are both 0 on F c and are both 1 on E. Upon combining this with the above splitting of R, we find that f (x)  g(x) as long as x ∈ / (F\E). But m(F\E)  0 and therefore f  g a.e. So far, we have proved that the result in question holds for the characteristic function of a Lebesgue measurable set. It is a straightforward consequence that it holds for all simple Lebesgue measurable functions. Now consider any nonnegative Lebesgue measurable function f . By Theorem 3.2.1, there exists a sequence {sn }n≥1 of Lebesgue measurable simple functions converging pointwise to f . According to what has been established in the preceding paragraph, each sn has a Borel function gn and a Borel set Gn of measure 0 such / Gn . The Borel set G  ∪∞ that sn (x)  gn (x) whenever x ∈ n1 Gn then has the property / G. So, gn (x)χGc (x) → f (x)χGc (x) that m(G)  0 and also gn (x) → f (x) for each x ∈ for each x ∈ R. Since each gn χGc is a Borel function, so is their limit f χGc . However / G. Taking into account that m(G)  0, we have f  f (x)  f (x)χGc (x) for each x ∈ f χGc a.e. Thus the function g  f χGc is a Borel function such that f  g a.e. Since f is real-valued, the product f χGc is also real-valued. For a Lebesgue measurable function f that need not be nonnegative-valued, we  obtain the result by considering the related functions f + and f − . We note in passing that Proposition 6.6.7 holds even if f is extended real-valued but with the proviso that g may also be extended real-valued. Problem Set 6.6 6.6.P1. (Cf. Problem 6.5.P4) Suppose f and g are functions on [a, b] or R such that f  g a.e. If f is Lebesgue measurable, show that g is also Lebesgue measurable. 6.6.P2. Prove Lemma 6.6.5. 6.6.P3. Let F be a σ-algebra of subsets of a nonempty set X and F:X→R be an F-measurable map. Show that any Borel set A satisfies F −1 (A) ∈ F.

Chapter 7

Differentiation

7.1 Integrability and Step Functions The two fundamental theorems of calculus are about the derivative of the Riemann integral and Riemann integral of the derivative. There are corresponding results about the Lebesgue integral addressing the same matters, but they require a considerable amount of spadework. The foundational result of this section is Theorem 7.1.5 and its proof requires the following fact: The characteristic function on [a, b] of a subset which is a finite union of disjoint subintervals is a step function. It is fairly easy to convince oneself of this via a figure, and furthermore that the same is true even if the subintervals are not assumed disjoint because the union can be rewritten as the finite union of disjoint subintervals. The reader who is prepared to accept this assertion as being too obvious to need a proof is welcome to proceed directly to Theorem 7.1.5. The more skeptical reader may want to ratiocinate over the material leading up to Proposition 7.1.4, wherein the assertion is proved but on the assumption that the subintervals involved are of positive length, which is adequate for our purpose. However, the material can profitably be postponed until after the rest of the section has been read. 7.1.1. Remark. If the intersection of an open interval I with a closed interval J of positive length is nonempty, we shall show that it is an interval of positive length. That the intersection is an interval is trivial. Suppose u ∈ I ∩ J . Since I is open, some interval of the form (u − δ, u + δ), where δ > 0, must be contained in I; and since J is closed with positive length, some interval of the form either [u, u + δ ) or (u − δ , u], where δ > 0, must be contained in J. It now follows easily that an interval of one of the latter two forms must be contained in I ∩ J . 7.1.2. Exercise. It is known from elementary Analysis that a subset of R is an interval if and only if any number lying between two numbers belonging to the subset also belongs to the subset. Using this, prove that the union of two nondisjoint intervals is an interval. © Springer Nature Switzerland AG 2018 S. Shirali, A Concise Introduction to Measure Theory, https://doi.org/10.1007/978-3-030-03241-8_7

127

128

7 Differentiation

Solution: Consider intervals I and J with u ∈ I ∩ J and x, y ∈ I ∪ J satisfying x < y. We shall show that any z ∈ R such that x < z < y lies in I ∪ J . If z  u, there is nothing to prove. If x, y both belong to I or both belong to J, then again there is nothing to prove. So, suppose x ∈ I , y ∈ J . If z < u, we have x < z < u with x, u ∈ I , which implies z ∈ I ⊆ I ∪ J , and if z > u, we have u < z < y with u, y ∈ J , which implies z ∈ J ⊆ I ∪ J . The remaining possibility that x ∈ J , y ∈ I can be argued analogously. 7.1.3. Proposition. (a) A finite union of intervals is also a finite union of disjoint intervals. (b) A finite union of intervals of positive length is a finite union of disjoint intervals of positive length. Proof: (a) When the union is empty or when there is only one interval, the matter is trivial. First consider a union of two intervals. If they are not disjoint, then Exercise 7.1.2 ensures that their union is an interval, which makes it a union of disjoint intervals. Now assume as induction hypothesis that any union of n + 1 intervals (n ≥ 1) is a union of disjoint intervals, and consider a union of n + 2 intervals I 1 , …, I n+2 . If I n+2 is disjoint from ∪n+1 k1 Ik , then the induction hypothesis yields the desired conclusion immediately. If on the contrary I n+2 is not disjoint from ∪n+1 k1 Ik , then we may take In+1 ∩ In+2 to be nonempty, whereupon it follows by using to the consequence that Exercise 7.1.2   n that In+1 ∪ In+2 is an interval  n J, leading ∪ ∪ J , which is a union of n + 1 I  ∪ I ∪ I I ∪n+2  ∪ (I ) n+1 n+2 K1 k K1 k K1 k intervals, and thus a union of disjoint intervals by the induction hypothesis. (b) This follows upon merely reinterpreting the word “interval” in the reasoning above to mean “interval of positive length”.  In what follows, an interval named as [a, b] will be assumed to be of positive length, i.e. a < b, but not in Sect. 7.6. 7.1.4. Proposition. The characteristic function on [a, b] of a finite union of subintervals thereof , each having positive length, is a step function. Proof: On the basis of Proposition 7.1.3(b), we may take the subintervals to be disjoint and with positive length. We shall refer to them as given subintervals, because there will be occasion to mention other subintervals too. Let the left endpoints of the given subintervals be a1 , …, an in increasing order and the corresponding right endpoints be b1 , …, bn . If n  1, we have a ≤ a1 < b1 ≤ b; this gives rise to the four cases a  a1 < b1  b, a  a1 < b1 < b, a < a1 < b1  b, a < a1 < b1 < b. It is easy to verify in each case that the characteristic function of any subinterval with endpoints a1 , b1 is a step function.

7.1 Integrability and Step Functions

129

Now assume as induction hypothesis that the result in question holds whenever there are n given subintervals, and suppose we have n + 1 given subintervals at hand with endpoints aj , bj , 1 ≤ j ≤ n + 1. We claim that b1 ≤ a2 . If not, then a2 < min{b1 , b2 } and any u ∈ (a2 , min{b1 , b2 }) satisfies a1 < u< b1 as well as a2 < u< b2 , contradicting disjointness. Consider the characteristic function of the union of the n + 1 given subintervals at hand. Its restriction to the interval [a, b1 ] agrees with the characteristic function of the given subinterval with endpoints a1 , b1 , except perhaps at b1 if b1  a2 , and is therefore seen to be a step function on [a, b1 ] on the grounds of what has been proved in the paragraph above. As for the restriction to [b1 , b], since it agrees with the characteristic function of the union of the remaining n given subintervals, except perhaps at b1 , it must also be a step function by the induction hypothesis. Thus the restrictions to [a, b1 ] and [b1 , b] are both step functions. It follows that the characteristic function under consideration is a step function, as was to be proved.  For reasons that need not delay us here, the content of the next theorem and its corollary will be summarized by saying that simple functions are “dense”. Note that, whenever p, q and M are real numbers such that p ≤ M, we can show that |p − min{q, M }| ≤ |p − q| in the following manner: If q ≤ M, then min{q, M}  q, so that |p − min{q, M }|  |p − q|, whereas if q > M, then p ≤ M  min{q, M } < q, so that |p − min{q, M }| < |p − q|. 7.1.5. Theorem. Given an integrable function  f : [a, b] → R and any ε > 0, there exists a step function s: [a, b] → R such that [a,b] |f − s| < ε. Moreover, if |f | ≤ M for some M ≥ 0, then s can be chosen to satisfy |s| ≤ M. Proof: First we prove this when f  χE , the characteristic function of a measurable set E ⊆ [a, b]. Set η  2ε . By Lemma 6.6.5, there exists a sequence {I n }n≥1 of open intervals such that E ⊆ ∪∞ n1 In and  ∞   m ( In )\E ≤ η. (7.1) n1

Consider the intersections Jn  In ∩ [a, b]. They are all subintervals of [a, b] and the nonempty ones have positive length (see Remark 7.1.1 above). Moreover, they ∞ ∞ Jn ⊇ E, which ensures that satisfy ∪ n1 (Jn ∩ E)   ∪   E. By Proposition 2.3.1, we  n1 ∞ N have m ∪n1 (Jn ∩ E)  limN →∞ m ∪ n1 (Jn ∩ E) . Hence there exists an N ∈ N such that 0 ≤ m(E) − m ∪Nn1 (Jn ∩ E) < η. This implies  N   m E\ (Jn ∩ E) < η. (7.2) n1

Denoting the finite union ∪Nn1 Jn ⊆ [a, b] by G, we have E\G  E\∪Nn1 (J  n ∩ E) and hence m(E\G) < η by (7.2). On the other hand, G\E  ∪Nn1 Jn \E ⊆

130

7 Differentiation

 ∞     ∪n1 In \E and therefore m(G\E) ≤ m ∪∞ n1 In \E ≤ η by (7.1). Consequently, the symmetric difference EG  (E\G) ∪ (G\E) satisfies m(EG) < 2η. The  easily proven equality χEG  |χE − χG | now leads to [a,b] |χE − χG | < 2η  ε. But G is a finite union of subintervals of [a, b] of positive length, and hence Proposition 7.1.4 shows that the function χG : [a, b] → R is a step function. The assertion of the theorem has thus been proved when the function f is the characteristic function of a measurable set E ⊆ [a, b]. This extends to all simple functions because they are finite linear combinations of characteristic functions of measurable sets. Next, consider an integrable nonnegative f . By Theorem 3.2.1 and the Monotone Convergence Theorem 2.3.2, function σ : [a, b] → R such   thereexists a simple that 0 ≤ σ ≤ f and 0 ≤ [a,b] f − [a,b] σ  [a,b] |f − σ| < 2ε . Since the assertion of the theorem has been shown to hold for simple  functions, the simple function σ yields a step function s : [a, b] → R such that [a,b] |σ − s| < 2ε . This step function  s then satisfies [a,b] |f − s| < ε. Thus the assertion in question is found to hold for any integrable nonnegative f . Extending this to an arbitrary integrable function is now a matter of employing the usual kind of 2ε -argument after writing f  f + − f − . For the last part, first suppose f is nonnegative and satisfies f ≤ M for some M ≥ 0. Then min{s, M} is a step function which serves our purpose because |f − min{s, M }| ≤ |f − s|. For an arbitrary integrable f , the required step function is obtained by considering f + and f − separately.  Since a step function, by definition, has a closed bounded interval as domain, the above theorem does not carry over directly to the case when the domain is R. However, if we agree to extend a step function to be zero outside its domain, it does carry over, as we shall see in Corollary 7.1.7. 7.1.6. Notation. Given a function g: [a, b] → R, its extension to all of R obtained by setting it equal to 0 outside [a, b] will be denoted by g* . It is clear that g ∗ |[a,b]  g and g ∗ χ[a,b]  g ∗ . Recall from Sect. 4.4 that the measure induced on [a, b] by Lebesgue measure on R is nothing but Lebesgue measure on [a, b]. It follows upon using Proposition 4.4.1 that g : [a, b] → R is measurable if when this is so, g is integrable if and only if and only if g ∗ : R →  R is. Moreover,  g ∗ is, in which case [a,b] g  R g ∗ .   It is trivial to see that |g1 − g2 |∗  g1∗ − g2∗ . 7.1.7. Corollary. Given an integrable function f : R → R and  any ε > 0, there exists an interval [a, b] and a step function s: [a, b] → R such that R |f − s∗ | < ε. Moreover, if |f | ≤ M for some M ≥ 0, then s∗ can be chosen to satisfy |s∗ | ≤ M. Proof: The sequence of functions {fn }n≥1 , where fn  f χ[−n,n] , converges pointwise to f on R and it follows from the Dominated Convergence Theorem 3.2.6 that  |f | R n − f → 0 as n → ∞. Therefore there exists a k ∈ N such that ε |fk − f | < . 2 R

7.1 Integrability and Step Functions

131

Now fk is integrable and hence, so is its restriction g to [−k, k]; moreover, since fk vanishes outside [−k, k], we have g∗  fk . By Theorem 7.1.5, there  exists a step function s: [−k, k] → R such that [−k,k] |g − s| < 2ε . However, [−k,k] |g − s|    |g ∗ − s∗ |  R |fk − s∗ |, so that R   fk − s ∗  < ε , 2 R

which leads to

  f − s ∗  ≤

R



|fk − f | +

R

  fk − s∗  < ε.

R

The interval [a, b]  [−k, k] has thus been found to be of the required kind. If |f | ≤ M for some M ≥ 0, then the same holds for fk  f χ[−k,k] , and for its restriction g. Therefore by the last part of Theorem 7.1.5, the step function s : [−k, k] → R can be chosen to satisfy the same inequality; the extension s∗ then also satisfies the inequality.  7.1.8. Mean Continuity Theorem. For an integrable function f : R → R, lim |f (x + h) − f (x)|dx  0. h→0

R

Proof: First suppose f is the characteristic function of an interval [a, b]. Then f (x + h) is the characteristic function of the interval [a − h, b − h] and consequently, |f (x + h) − f (x)| is the characteristic function of the symmetric difference of the two intervals. Consider h such that 0 < h < b − a; then a − h < a < b − h < b and the symmetric difference is the union [a − h, a) ∪ (b − h, b], the measure of which is 2h. Therefore |f (x + h) − f (x)|dx  2h R

and the desired limit must hold from the right. A similar argument shows that it also holds from the left. Thus the limit in question holds for the characteristic function of a closed bounded interval. If the interval is not closed but is bounded, then its characteristic function differs from that of the corresponding closed interval at finitely many points whereas the integral involved has the same value for both. So no separate argumentation is required in this case.

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7 Differentiation

It now follows in the obvious manner that the limit holds for any step function. Its validity for an arbitrary integrable function f now follows from the fact that step functions s are dense (Corollary 7.1.7) and that |f (x + h) − f (x)|dx ≤ |f (x + h) − s(x + h)|dx R



R



|s(x + h) − s(x)|dx +

+ R

|s(x) − f (x)|dx R

and

|f (x + h) − s(x + h)|dx  R

|s(x) − f (x)|dx. R

 Problem Set 7.1 7.1.P1. A subset U ⊆ R is said to be open if for every u ∈ U, there exists a δ > 0 such that (u − δ, u + δ) ⊆ U . (Obviously, any open interval is an open subset; an interval that is an open subset is an open interval; any union of open subsets (or open intervals) is an open subset.) Show that any open subset U ⊆ R is a countable union of disjoint open intervals.

 j 7.1.P2. Let X  [0, 1] and consider the intervals j−1 , 1 ≤ j ≤ k, k ∈ N, , k k arranged in some order (without repetition). Show that the sequence {fn }n≥1 of characteristic functions on X of the respective intervals does not converge at any point, although their integrals converge to zero. 7.1.P3. (a) Both the figures below show a stick and a second stick with the overlap of their shadows underneath. They also show a third stick whose shadow overlaps with some part of the existing overlap of the shadows of the first two sticks (i.e. all three shadows together have some overlap). If in the figure on the left, the second stick is removed, the combined shadow of the remaining two sticks is the same as that of all three sticks. If in the figure on the right, the first stick is removed, the combined shadow of the remaining two sticks is the same as that of all three sticks. Question: Can you place three sticks in such a way that all three shadows together have some overlap but if any one among the three sticks is removed, the combined shadow of the remaining two sticks is no longer the same as that of all three sticks?

7.1 Integrability and Step Functions

third stick

133

third stick second stick

second stick stick

shadow overlap

combined shadow of all three sticks

stick

shadow overlap

combined shadow of all three sticks

(b) Suppose F is a finite nonempty family of bounded intervals. Show that there exists a subfamily G ⊆ F such that ∪G  ∪F and no interval of G is contained in the union of two other intervals of G. (c) Can the finiteness hypothesis of part (b) be dropped?

7.2 The Vitali Covering Theorem The theorem of the title is an important stepping stone in our approach to the remarkable theorem of Lebesgue (next section) that a function that is increasing or decreasing (monotone) is differentiable almost everywhere. However, there are several other approaches that do not use the theorem. A “Vitali covering” of a subset A ⊆ R is a family of closed intervals such that every number in A belongs to an arbitrarily short interval of the family, but with positive length. A trivial example would be A consisting of a single point and the family of all closed intervals centered at x with lengths 1n , where n ∈ N. A less trivial example would be A  [0, 1] and the family of all possible closed intervals with rational centers in A and rational lengths. One variation on this is to drop the requirement that the centers and lengths be rational; another is to restrict the lengths to be less than some specified positive number. The concept can be described formally keeping in mind that the length of an interval is the same as its Lebesgue measure: 7.2.1. Definition. A family C of closed intervals is a Vitali covering of a subset A ⊆ R if, for any given x ∈ A and ε > 0, there exists an I ∈ C for which x ∈ I and 0 < m(I) < ε.

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7 Differentiation

Since closed intervals of length 0 or ∞ do not serve any purpose in a Vitali covering, it will be tacitly assumed that all intervals in a Vitali covering have finite positive length. The substance of the Vitali Covering Theorem to be proved below, and to be invoked later in the proof of the Lebesgue Differentiability Theorem, can be summarized as follows: Any Vitali covering of a set of finite (Lebesgue) outer measure contains finitely many disjoint intervals such that removing them all from the set results in a subset of arbitrarily small outer measure. The proof involves some facts about intervals that the reader will easily believe on the basis of a figure. However, we prefer to spell out the precise reasons, which undoubtedly make for dreary reading. It is possible to proceed directly to Exercise 7.2.5 from here and decide at a later stage whether to suffer through the intervening material. 7.2.2. Remarks. (a) Recall that the interval obtained by deleting the endpoints (if any) of a given interval is called its interior. It will be convenient to denote the interior of an interval I by Iº. To belong to the interior of an interval is to belong to the interval while being at a positive distance from any endpoint(s) it may have, which is the same as being the center of a suitable closed subinterval of positive length. Surely, an interval must have positive length if its interior is nonempty. (b) A point which is outside a closed interval I belongs to the interior of a closed bounded interval that is disjoint from I: First suppose I  [a,

∞) and consider x∈ / [a, ∞). Then x < a and for δ  a − x > 0, we have y ∈ x − 21 δ, x + 21 δ ⇒ y−x < δ  a−x ⇒ y < a ⇒ y ∈ / [a, ∞), so that [x− 21 δ, x+ 21 δ]∩[a, ∞)  ∅.

Thus, x belongs to the interior of the closed bounded interval x − 21 δ, x + 21 δ , which is disjoint from [a, ∞). Similar reasoning works when I is a closed interval of any other type. (c) The intersection of a finite number of closed intervals of positive length whose interiors all have one given point in common can be shown to be a closed interval which has that point in its interior. In this connection, it is easy to see that the intersection of a finite number of closed intervals is a closed interval, perhaps empty. (However, it need not be of positive length even if all the intersecting intervals have positive length and have a common point, as is illustrated by the example [0, 1] ∩ [1, 2]  [1, 1].) Suppose now that their interiors have a common point. By (a), each of them contains a closed subinterval of positive length with that point as center. The intersection of the finitely many subintervals so obtained is then a closed interval of positive length contained in the intersection of the original intervals; moreover, it has the common point as its center. Thus the intersection of the original intervals has the common point in its interior. (d) If x ∈ (a, b), then any interval to which x belongs and has length less than the positive number min{x − a, b − x} can be shown to be contained in (a, b). To wit, let [a0 , b0 ] be such an interval. Then

7.2 The Vitali Covering Theorem

135

b0 − a0 < x − a and a0 ≤ x ≤ b0 . The second of these implies x − a0 ≤ b0 − a0 , which when combined with the first yields x − a0 < x − a. This further yields a < a0 . The reason why b0 < b is similar. 7.2.3. Exercise. Show that a point which lies outside each of a finite number of closed intervals belongs to the interior of a closed bounded interval that is disjoint from each one of the finitely many closed intervals. Solution: By Remark 7.2.2(b), to each of the finitely many closed intervals corresponds a closed bounded interval of positive length that is disjoint from the former and having the point in its interior. The intersection of the intervals so obtained is then disjoint from each of the given intervals and is bounded. By Remark 7.2.2(c), it is a closed interval whose interior contains the point. 7.2.4. Exercise. Suppose the distance of a point x ∈ R from the center of a closed interval does not exceed α times the length  of the interval (here α > 0 of course). Show that x lies in a closed interval with the same center but having length 2α.

Solution: Call the center y, so that the interval is y − 21 , y + 21  . Then |y − x| ≤ α.

Therefore x ∈ y − α, y + α , which is precisely the closed interval with the same center but length 2α. 7.2.5. Exercise. Let C be a Vitali covering of a subset A ⊆ R and U be a union of open intervals (i.e. is an open subset as in Problem 7.1.P1) containing A. Show that the family of those intervals belonging to C that are contained in U is a Vitali covering of A. Solution: Consider any x ∈ A and ε > 0. We have to conclude that there exists an I ∈ C for which x ∈ I , 0 < m(I ) < ε and also I ⊆ U . Since x ∈ U, a union of open intervals, there exists a δ > 0 such that (x − δ, x + δ) ⊆ U .

(7.3)

Take η  min{ε, δ} > 0. Now C is a Vitali covering of A and hence there exists an I ∈ C for which x ∈ I and 0 < m(I ) < η ≤ ε. All we need to show is that I ⊆ (x − δ, x + δ), because the required conclusion will then follow from (7.3). To this end, let I  [a, b]. Since b − a  m(I) < η ≤ δ, we have

(7.4)

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7 Differentiation

b < a + δ, equivalently, a > b − δ.

(7.5)

Since x ∈ I  [a, b], we also have a ≤ x ≤ b. Hence by (7.5), b − δ < x < a + δ. This implies x − δ < a and b < x + δ. These two inequalities together yield [a, b] ⊆ (x − δ, x + δ). Since I  [a, b], this is the same as (7.4), which has already been noted as all that needs to be shown. 7.2.6. Lemma. Suppose C is a Vitali covering of A ⊆ R and I 1 , …, I n are finitely many disjoint intervals belonging to C. Then for any x ∈ A\∪nj1 Ij , there exists an interval belonging to C that has x as element of it and is disjoint from each of the Ij , 1 ≤ j ≤ n. Proof: Consider any x ∈ A\∪nj1 Ij . It lies outside each of the finitely many closed intervals I j , 1 ≤ j ≤ n, and therefore belongs to an interval (a, b) that is disjoint from each of the Ij , 1 ≤ j ≤ n (Exercise 7.2.3). Now, every interval to which x belongs and has sufficiently small length must be contained in (a, b) (Remark 7.2.2(d)). Since C is a Vitali covering of A, there certainly exists such an interval belonging to C. Being contained in (a, b), it is disjoint from each of the I j , 1 ≤ j ≤ n. Thus, we obtain an interval belonging to C that has x as an element of it and is disjoint from each of the  I j , 1 ≤ j ≤ n. 7.2.7. Vitali Covering Theorem. Suppose C is a Vitali covering of A ⊆ R and m∗ (A) < ∞. Then for any ε > 0, there exist finitely many disjoint intervals I 1 , …, I n belonging to C such that m∗ (A\

n 

Ij ) < ε.

j1

Proof: In the event that either m∗ (A)  0 or there exist finitely many disjoint intervals I 1 , …, I n belonging to C such that A ⊆ ∪nJ 1 Ij , there is nothing to prove. So we may assume that m∗ (A) > 0 and that whenever I 1 , …, I n are finitely many disjoint intervals belonging to C, the difference A\∪nj1 Ij is nonempty. By definition of Lebesgue outer measure, there exists a countable union G of open intervals such that A ⊆ G and m(G) < ∞. By Exercise 7.2.5, we may legitimately assume that every interval belonging to C is contained in G. The consequence that the sum of the measures of any sequence of disjoint intervals belonging to C is finite will be used below. First consider an arbitrary finite sequence {Ij }1≤j≤n of disjoint intervals belonging to C. The difference A\∪nj1 Ij is nonempty and therefore there exists some x ∈ A\∪nj1 Ij . By Lemma 7.2.6, we obtain an interval belonging to C that has x as an element of it and is disjoint from each of the I , 1 ≤ j ≤ n. Let σn be the supremum of the lengths of all intervals belonging to C that are disjoint from each of the I j , 1 ≤ j

7.2 The Vitali Covering Theorem

137

≤ n, but not necessarily having x as an element. Then 0 < σn < ∞ and there exists an interval In+1 ∈ C that is disjoint from ∪nj1 Ij and satisfies 2 · m(In+1 ) > σn . A straightforward inductive construction now leads to an infinite sequence {I n }n≥1 of disjoint intervals belonging to C and a sequence {σn }n≥1 of positive real numbers such that I ∈ C and I ∩ Ij  ∅ for 1 ≤ j ≤ n ⇒ σn ≥ m(I ) for every n ∈ N

(7.6)

and 2 · m(In+1 ) > σn for every n ∈ N.

(7.7)

We shall show by contradiction that ∗

m (A\

n 

Ij ) < ε for some n ∈ N.

(7.8)

Ij ) ≥ ε for every n ∈ N.

(7.9)

j1

Suppose not. Then m∗ (A\

n  j1

As noted at the end of the second paragraph, one hand that

∞ j1

  m Ij < ∞. This implies on the

m(Ij ) → 0 as j → ∞

(7.10)

and on the other hand that some N ∈ N has the property that ∞ 

m(Ij )
0. Thus there must be some j > N for which I ∩ Ij  ∅. Denote the smallest such integer j by k. Then k>N and I is disjoint from each of the Ij , 1 ≤ j ≤ k − 1, and therefore by (7.6) and (7.7),

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7 Differentiation

m(I ) ≤ σk−1 < 2 · m(Ik ).

(7.12)

At the same time, I ∩ Ik  ∅ and therefore the distance of x ∈ I from the center of I k is at most m(I ) + 21 m(Ik ) (see Problem 7.2.P2 below), and hence by (7.12) is less than 2 · m(Ik ) + 21 m(Ik )  25 m(Ik ). It follows that the closed interval Ik with the same center as I k but length 5m(I k ) has x as an element (Exercise 7.2.4, wherein α  25 ). Taking into account that x ∈ A\∪Nj1 Ij was arbitrary and k > N, we obtain A\∪Nj1 Ij ⊆ ∪j>N Ij . On the basis of (7.9) and (7.11), we further obtain ⎛ ⎞ N ∞ ∞    ε ≤ m∗ ⎝A\ Ij ⎠ ≤ m(Ij )  5 m(Ij ) < ε. j1

jN +1

jN +1

This contradiction proves (7.8) and hence also the theorem.



Problem Set 7.2 7.2.P1. Let C be a Vitali covering of a subset A ⊆ R and J be a family of closed intervals such that the union of their interiors contains A. Show that the family C0  {I ∩ J : I ∈ C, J ∈ J } of closed intervals consisting of the intersections of intervals belonging to C with intervals belonging to J is a Vitali covering of A. 7.2.P2. If I and J are intervals of finite positive lengths and I ∩ J  ∅, show that the distance of any point in I from the center of J is at most m(I ) + 21 m(J ).

7.3 Dini Derivatives So far, we have not spoken of −∞, at least not as a mathematical entity subject to any computational rules. In the present section we shall speak of both ∞ and −∞ as possible limits, the latter as an entity not to compute with but only to compare, as for instance, −∞ < a < ∞ for every real number a. It then follows that the supremum of any set to which ∞ belongs is ∞ and the infimum of any set to which −∞ belongs is −∞. Since every element of R ∪ {∞, −∞} is “vacuously” a lower bound as well as an upper bound of ∅, it follows that inf ∅  ∞ and sup ∅  −∞. The empty set is the only one for which the infimum exceeds the supremum. The implication that A ⊆ B

⇒

inf A ≥ inf B and sup A ≤ sup B

continues to hold even if A  ∅ or B  ∅. Suppose the domain of the real-valued function f is some interval I ⊆ R of (c) , positive length and let c ∈ I . Then the so-called “difference quotient at c”, f (c+h)−f h makes sense for all sufficiently small h unless c is an endpoint. We speak of right

7.3 Dini Derivatives

139

and left difference quotients depending on whether h is positive or negative. At the left endpoint of I, only right difference quotients make sense and we therefore take any set of left difference quotients there to be ∅. Similarly, we take any set of right difference quotients at the right endpoint of I to be ∅. When δ ∈ R∗+ , the set   f (c + h) − f (c) : 0 0, the difference quotient works out to be h− 2 cos 1h , which has 1 the value (2πn) 2 when h  (2πn)−1 and n ∈ N. Given any K > 0, one can choose n 1 such that the inequalities (2πn)−1 < δ and (2πn) 2 > K both hold. So, sup Qδ  ∞. 1 1 On the other hand, the infimum is −∞, because h− 2 cos 1h has the value −(π(2n+1)) 2 when h  (π(2n + 1))−1 and n ∈ N. 1

In what follows, any number denoted by δ or a variant thereof will be understood to be positive. 7.3.2. Remark. If δ ≤ δ , then Qδ ⊆ Qδ and therefore inf Qδ ≤ inf Qδ and sup Qδ ≤ sup Qδ . When c is not an endpoint, Qδ  ∅ and therefore we also have inf Qδ ≤ sup Qδ , with the consequence that inf Qδ ≤ sup Qδ as well as inf Qδ ≤ sup Qδ . We go on to consider infima of sets of the form {sup Qδ : δ < δ1 }, where δ1 is given. Note that these sets are never empty even though Qδ can be. 7.3.3. Proposition. For any δ1 and δ2 ∈ R*+ , inf{sup Qδ : δ < δ1 }  inf{sup Qδ : δ < δ2 }. Proof: We need prove the equality only for δ1 < δ2 . When this is so, {sup Qδ : δ < δ1 } ⊆ {sup Qδ : δ < δ2 }. It is immediate from this inclusion that inf{sup Qδ : δ < δ1 } ≥ inf{sup Qδ : δ < δ2 }.

(7.13)

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7 Differentiation

To arrive at the reverse inequality, it is sufficient to show that for any q in the bigger of the two sets in (7.13), there exists a p in the smaller set such that p ≤ q. Accordingly, consider any q in the bigger set. There exists δ < δ2 for which q  sup Qδ .   Put   min δ , 21 δ1 and p  sup Q . Since  ≤ δ , we have Q ⊆ Qδ and hence sup Q ≤ sup Qδ , which is the same as p ≤ q. Since  < δ1 , we have sup Q ∈ {sup Qδ : δ < δ1 }, which means exactly that p belongs to the smaller of the two sets in (7.13).  The above Proposition 7.3.3 allows for δ2  ∞. Therefore, in taking the infimum of the set of suprema, it makes no difference whether we impose any such restriction as δ < δ1 or not. Accordingly, in Definition 7.3.4 below, we do not impose a restriction but it can be introduced if needed. Similar considerations apply with inf and sup interchanged and/or h > 0 replaced by h < 0. Thus four possibilities can occur. When h < 0, we work with   f (c + h) − f (c) : 0 < −h < δ . Q−δ  h When c is not an endpoint of the domain of f , the sets Qδ and Q−δ are never empty and therefore the infimum of either of them cannot exceed its supremum. 7.3.4. Definition. Let f be a real-valued function on an interval I of positive length. The four Dini derivatives of f at an interior point c ∈ I are defined as D+ f (c)  inf{sup Qδ : 0 < δ}, D+ f (c)  sup{inf Qδ : 0 < δ}, D− f (c)  inf{sup Q−δ : 0 < δ}, and D− f (c)  sup{inf Q−δ : 0 < δ}, where Qδ 



f (c + h) − f (c) : 0 0 such that D+ f (c) ≤ sup Qδ < M. By Remark 7.3.2, any smaller positive δ satisfies the same inequality. It follows that there is a sequence {hn }n≥1 of positive numbers such that hn → 0 and

f (c + hn ) − f (c) → D+ f (c). hn

If D+ f (c)  ∞, the same conclusion can be arrived at even more easily by simply dropping the reference to M. Consequence: for any α < D+ f (c), there is a sequence {hn }n≥1 of positive numbers such that hn → 0 and

f (c + hn ) − f (c) > α for every n ∈ N. hn

Analogously, when D− f (c) > −∞, for any M < D− f (c), there is some δ > 0 such that M < inf Q−δ ≤ D− f (c). By the obvious analog of Remark 7.3.2, any smaller

142

7 Differentiation

positive δ satisfies the same inequality. As before, we have the following consequence whether or not D− f (c) > −∞: for any β > D− f (c), there is a sequence {hn }n≥1 of negative numbers such that hn → 0 and

f (c + hn ) − f (c) < β for every n ∈ N. hn

Two other such assertions can of course be proved regarding D+ f (c) and D− f (c). 7.3.7. Exercise. If g  −f , show that D+ g(c)  −D+ f (c). Solution: For any δ > 0, it is immediate from the definition that Qδ (g)  −Qδ (f ) and hence that sup Qδ (g)  − inf Qδ (f ). Therefore inf{sup Qδ (g) : 0 < δ}  − sup{inf Qδ : 0 < δ}. By Definition 7.3.4, this means D+ g(c)  −D+ f (c). Problem Set 7.3 7.3.P1. Prove the following partial converse of the “consequence” in Remark 7.3.6 in the case of D+ f (c) : If α ∈ R ∪ {∞, −∞} and there exists a sequence {hn }n≥1 of positive numbers as in the remark (in which case α  ∞ of course), then α ≤ D+ f (c). 7.3.P2. Show that α < D+ f (c) if and only if for some β > α there is a sequence {hn }n≥1 of positive numbers such that hn → 0 and

f (c + hn ) − f (c) > β for every n ∈ N. hn

7.3.P3. Give an example of continuous functions f and g on an interval containing 0 such that D+ f (0) + D+ g(0)  D+ (f + g)(0). 7.3.P4. If g  −f , show that D− g(c)  −D− f (c). 7.3.P5. Suppose f and g are defined on a common interval I and c ∈ I ◦ . Show that D+ (f + g)(c) ≥ D+ f (c) + D+ g(c), provided that the sum on the right side is meaningful. What about D+ (f + g)(c) ≤ D+ f (c) + D+ g(c)?

7.4 The Lebesgue Differentiability Theorem Recall the notions of left and right derivatives from elementary Analysis (see e.g., Shirali and Vasudeva [9, p. 200]). The reader is called upon to make the obvious modification to permit ±∞ as their values. A little reflection will then show that D+ f (c)  D+ f (c) if and only if f has a right derivative f+ (c) at c, in which case D+ f (c)  D+ f (c)  f+ (c). Similarly, D− f (c)  D− f (c) if and only if f has a left derivative f− (c) at c, in which case D− f (c)  D− f (c)  f− (c). Moreover, f has a (two sided) derivative f  (c) at c if and only if f+ (c)  f− (c), in which case f+ (c)  f− (c)  f  (c). In terms of Dini derivatives, this says that f has a (two sided)

7.4 The Lebesgue Differentiability Theorem

143

derivative f  (c) at c if and only if D+ f (c)  D+ f (c)  D− f (c)  D− f (c), in which case D+ f (c)  D+ f (c)  D− f (c)  D− f (c)  f  (c). Differentiability at c means f  (c) is finite, which is equivalent to all four Dini derivatives at c being equal and finite. Differentiability here is understood as two sided. So endpoints of the interval are automatically out of the running. 7.4.1. Lebesgue Differentiability Theorem. If a real-valued function on an interval is monotone, then it is differentiable almost everywhere. Proof: It is sufficient to prove the result either for an increasing function or for a decreasing function. So, let f be an increasing function on an interval I. We may assume that I has positive length because otherwise its measure is 0 and there is nothing to prove. It is sufficient to consider the case when I is bounded, because in the unbounded case, I can be represented as the countable union of the disjoint bounded intervals I ∩ (n, n + 1], n ∈ Z, to each of which the bounded case applies; the required conclusion then follows for I. So, suppose I is bounded. Since there cannot be two sided differentiability at endpoints (if any), we need consider only interior points, which means we can take I to be open. Let A  {x ∈ I : D+ f (x) > D− f (x)}. We shall demonstrate that m∗ (A)  0. Now A is the union of the countably many sets Ap,q  {x ∈ I : D+ f (x) > p > q > D− f (x)}, where p, q ∈ Q. It is sufficient therefore to show that each Ap,q has outer measure 0. The increasing nature of the function f guarantees that all Dini derivatives are nonnegative and hence Ap,q  ∅ only if p > q > 0. So we need consider only those p, q for which this inequality holds. With this in view, take any Ap,q for which p > q > 0. and consider an arbitrary ε > 0. Being a subset of the bounded interval I, the set Ap,q has finite outer measure and therefore   there exists a countable union G of open intervals such that Ap,q ⊆ G and m∗ Ap,q ≤ m(G) < m∗ (Ap,q ) + ε. By Remark 7.3.6, to each x ∈ Ap,q correspond intervals [x + h, x] ⊆ G with h < 0 and arbitrarily small in absolute value such that f (x+h)−f (x) < q, i.e. such that h f (x + h) − f (x) > qh. Since this holds for each x ∈ Ap,q , the intervals constitute a Vitali covering of Ap,q . By

Covering Theorem 7.2.7, there exist finitely many disjoint intervals the Vitali xj + hj , xj , 1 ≤ j ≤ J , in the Vitali covering such that m∗ (Ap,q \

J  j1

[xj + hj , xj ]) < ε.

(7.14)

144

7 Differentiation

The choice of the intervals of the Vitali covering ensures that f (xj + hj ) − f (xj ) > qhj for every j.

(7.15)

Now, observe that since the disjoint intervals are contained in G, (remember hj < 0)  we have Ji1 hj ≥ −m(G). Since q > 0, taken in tandem with (7.15), the preceding observation leads to J J       f (xj + hj ) − f (xj ) > q · hj ≥ −q · m(G) > −q · m∗ (Ap,q ) + ε . j1

(7.16)

j1

In conjunction with (7.14), the obvious equality ⎞ ⎞ ⎛ ⎛ J J   m∗ ⎝Ap,q \ (xj + hj , xj )⎠  m∗ ⎝Ap,q \ [xj + hj , xj ]⎠, j1

j1

gives rise to ⎛ m∗ ⎝Ap,q \

J 

⎞ (xj + hj , xj )⎠ < ε.

(7.17)

j1

  By Remark 7.3.6, to each y ∈ Ap,q ∩ ∪Jj1 xj + hj , xj correspond intervals

   (y) y, y + h ⊆ ∪Jj1 xj + hj , xj with arbitrarily small h > 0 such that f (y+hh)−f > p,  i.e. such that f (y + h ) − f (y) > ph . Since this holds for each y ∈ Ap,q ∩ ∪Jj1 (xj + hj , xj ), the intervals constitute a Vitali covering of Ap,q ∩ ∪Jj1 (xj + hj , xj ). By the Vitali Covering Theorem 7.2.7, there exist finitely many disjoint intervals [yk ,yk +hk ], 1 ≤ k ≤ K, in the Vitali covering such that ⎞ ⎛ K J   m∗ ⎝(Ap,q (xj + hj , xj ))\ [yk , yk + hk ]⎠ < ε. (7.18) j1

k1

The choice of the intervals of the Vitali covering ensures that f (yk + hk ) − f (yk ) > phk for every k.

(7.19)

Since the Vitali covering consists of intervals contained in ∪Jj1 (xj + hj , xj ), we have

7.4 The Lebesgue Differentiability Theorem K 

145

J 

yk , yk + hk ⊆ (xj + hj , xj ).

(7.20)

j1

k1

We claim that K 

hk > m∗ (Ap,q ) − 2ε.

(7.21)

k1

 

To prove (7.21), let us denote ∪Jj1 xj + hj , xj by J and ∪Kk1 yk , yk + hk by K. Then m∗ (K)  m(K) 

K 

hk .

(7.22)

k1

Also, K ⊆ J in view of (7.20), so that J c  J c ∩ Kc . Together with (7.17), this implies m∗ (Ap,q ∩ J c ∩ Kc )  m∗ (Ap,q ∩ J c ) < ε. From (7.18), we have m∗ ((Ap,q ∩ J ) ∩ Kc ) < ε. Since       (Ap,q ∩ J c ∩ Kc ) ∪ (Ap,q ∩ J ) ∩ Kc  (Ap,q ∩ Kc ) ∩ J c ∪ (Ap,q ∩ Kc ) ∩ J  Ap,q ∩ Kc , the preceding two inequalities yield m∗ (Ap,q ∩ Kc ) < 2ε. In combination with the fact that Ap,q  (Ap,q ∩K)∪(Ap,q ∩Kc ), the above inequality leads us to m∗ (Ap,q ) ≤ m∗ (Ap,q ∩ K) + m∗ (Ap,q ∩ Kc ) ≤ m∗ (K) + m∗ (Ap,q ∩ Kc ) < m∗ (K) + 2ε    and hence m∗ Ap,q < Kk1 hk + 2ε by (7.22).

our claim (7.21). This proves It follows from (7.20) that each interval yk , yk + hk on the left side therein is contained in the union of finitely many disjoint intervals on the right side. The nature

146

7 Differentiation

of an interval necessitates that each interval yk , yk + hk is actually contained in a single interval on the right side. This fact and the increasing nature of the function f together justify the inequality (recall that hj < 0) J K       f (xj ) − f (xj + hj ) ≥ f (yk + hk ) − f (yk ) . j1

k1

By (7.19) and (7.21), this implies J K       f (xj ) − f (xj + hj ) > p · hk > p · m∗ (Ap,q ) − 2ε . j1

k1

Therefore by (7.16),     q · m∗ (Ap,q ) + ε > p · m∗ (Ap,q ) − ε .   Since this has been established for an arbitrary ε > 0, it follows that q · m∗ Ap,q ≥  p · m∗ Ap,q , which is only possible if m∗ (Ap,q )  0, because p > q. This completes the demonstration that m∗ (A)  0, where A  {x ∈ I : D+ f (x) > D− f (x)}. By similar reasoning, we can show that the set on which any pair of Dini derivatives differ has outer measure 0. Thus the four Dini derivatives agree almost everywhere. It remains to show that the set on which any one of them is ∞ has outer measure 0. We set B  {x ∈ I : D+ f (x)  ∞} and proceed to argue that m∗ (B)  0. Any nonempty interval is a countable union of closed subintervals, and it is sufficient therefore to prove the result for the subintervals. In effect, we may replace I by a closed subinterval [α, β]. Note that m∗ (B) ≤ β − α < ∞. Assuming m∗ (B) > 0 for the purpose of deriving a contradiction, let η

1 ∗ m (B) > 0 2

and ∞>M >

2(f (β) − f (α)) ≥ 0. m∗ (B)

(7.23)



By Remark 7.3.6, to each z ∈ B correspond intervals z, z + h ⊆ [α, β] with  arbitrarily small h > 0 such that f (z+hh)−f (z) > M , i.e. such that f (z + h ) − f (z) > Mh .

7.4 The Lebesgue Differentiability Theorem

147

Since this holds for each y ∈ B, the intervals constitute a Vitali covering of the set. By

Covering Theorem 7.2.7, there exist finitely many disjoint intervals the Vitali zr , zr + hr , 1 ≤ r ≤ R, in the Vitali covering such that m∗ (B\

R 

[zr , zr + hr ]) < η.

(7.24)

r1

The choice of the intervals of the Vitali covering ensures that f (zr + hr ) − f (zr ) > Mhr for every r.

(7.25)

We claim that R 

hr > m∗ (B) − η.

(7.26)

r1

To prove (7.26), let us denote ∪Rr1 [zr , zr + hr ] by L. Then m∗ (L)  m(L) 

R 

hr ,

(7.27)

r1

and by (7.24), m∗ (B ∩ Lc ) < η. In combination with the fact that B  (B ∩ L) ∪ (B ∩ Lc ), the above inequality leads to m∗ (B) ≤ m∗ (B ∩ L) + m∗ (B ∩ Lc ) ≤ m∗ (L) + m∗ (B ∩ Lc ) < m∗ (L) + η,  which implies m∗ (B) < Rr1 hr + η by (7.27).

This proves our claim (7.26). The disjointness of the intervals zr , zr + hr and the increasing nature of f together justify the inequality f (β) − f (α) ≥

R    f (zr + hr ) − f (zr ) . r1

By (7.25) and (7.26), this implies

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7 Differentiation

f (β) − f (α) > M ·

R 

  hr > M · m∗ (B) − η .

r1

Recalling that η was defined to be 21 m∗ (B), we deduce from the above inequality   that f (β) − f (α) > M · 21 m∗ (B) . However, this conflicts with (7.23) and thereby  completes the contradiction proof that m∗ (B)  0. 7.4.2. Proposition. If a real-valued function f on an interval is a sum of an increasing and a decreasing function, then there is a Borel function on that interval which agrees with the derivative f  a.e. Any two such functions are equal a.e. Proof: It is sufficient to prove the result for an increasing function f . By the Lebesgue Differentiability Theorem 7.4.1, there is a subset A of the domain on which f has a derivative f  and the complement of which has Lebesgue measure 0. By Proposition 6.6.6, we may take the subset A to be a Borel set. Define g to agree with the derivative on A and to be 0 on Ac . We shall show that g is a Borel function. An increasing function is certainly a Borel function because the set on which f (x) > t is an interval and therefore a Borel set. So, the functions   f x ± 1n − f (x) , n∈N ± 1n are Borel functions. On the set A, the limit of the sequence of these functions exists and equals g. Since g is 0 outside the Borel set A, the deduction is implemented by appealing to Problem 6.5.P5. That any two such functions must be equal a.e. is trivial.  Problem Set 7.4 7.4.P1. For an increasing function f on a bounded open interval I, show by a direct argument, as in the proof of the Lebesgue Differentiability Theorem 7.4.1, that the set A  {x ∈ I : D+ f (x) < D− f (x)} has Lebesgue outer measure 0. 7.4.P2. Assuming that the result m∗ {x ∈ I : D+ f (x) > D− f (x)}  0 established in the course of proving the Lebesgue Differentiability Theorem 7.4.1 is valid for decreasing functions, deduce the result of Problem 7.4.P1. 7.4.P3. Suppose f and g are defined on a common interval I and c ∈ Iº is such that g has a finite right derivative g+ (c) at c. Show that D+ (f + g)(c)  D+ f (c) + g+ (c).

7.5 The Derivative of the Integral One of the fundamental theorems of calculus says that the derivative of the “indefinite” Riemann integral of a continuous function exists everywhere and agrees with

7.5 The Derivative of the Integral

149

that function. The analog for the Lebesgue integral is that the derivative of the indefinite integral of an integrable function exists almost everywhere and agrees with that function. We begin by showing that the indefinite integral of an integrable function is differentiable almost everywhere. 7.5.1. Proposition. Suppose f : [a, b] → R is integrable and F : [a, b] → R is the function given by f for all x ∈ [a, b]. F(x)  [a,x]

Then F is differentiable almost everywhere and there exists a Borel function on [a, b] that agrees with the derivative a.e. Moreover, any two such Borel functions are equal a.e. Proof: By definition of integrability, the functions G : [a, b] → R and H : [a, b] → R given by G(x)  f + and H (x)  f − for all x ∈ [a, b] [a,x]

[a,x]

are real-valued and F  G − H. Since f + and f − are nonnegative-valued, G and H are increasing functions. It follows by the Lebesgue Differentiability Theorem 7.4.1 that both are differentiable a.e. and hence their difference F is also differentiable a.e. The existence of a Borel function of the kind claimed in the theorem follows from Proposition 7.4.2, considering that F is the sum of the increasing function G and the decreasing function −H. So does the assertion about any two functions of the kind being equal a.e.  In order to simplify the notation, any Borel function of the kind in the above proposition will be denoted by F  . Any two such functions agree a.e. and the kind of thing we have to say about any one of them is equally true of the others. So, there is no harm in speaking of such a function as though it is unique. We do so in the impending proof. 7.5.2. Theorem (Derivative of the Integral). Suppose f : [a, b] → R is integrable and F : [a, b] → R is the function given by f for all x ∈ [a, b]. F(x)  [a,x]

Then for almost all x ∈ [a, b], the derivative F  (x) exists and equals f (x). Proof: That the derivative exists for almost all x ∈ [a, b] is known from Proposition 7.5.1. It suffices to show that the function F  (understood to be a Borel function

150

7 Differentiation

as in Proposition 7.5.1) agrees a.e. with some function, which in turn, agrees a.e. with f . By Proposition 6.6.7, there exists a real-valued Borel function g on R that agrees a.e. with the extension f * (see 7.1.6 for the meaning of f * ). By Problem 3.2.P1, it is sufficient to establish that    F − g   0. (7.28) [a,b]

Let {hn }n≥1 be a sequence of negative numbers tending to zero and {F n }n≥1 be the sequence of functions on R for which −1 Fn (x)  −hn g for all x ∈ R. [x+hn ,x]

Then for all x ∈ (a, b) and sufficiently large n, the number F n (x) agrees with h−1 n (F(x+ hn ) − F(x)). Hence by Proposition 7.5.1, lim Fn  F  a.e. on [a, b].

(7.29)

n→∞

We shall first prove that lim

n→∞ [a,b]

|Fn − g|  0.

(7.30)

For all x ∈ R and n ∈ N, we have −1 −1 Fn (x) − g(x)  −hn (g(t) − g(x))dt  −hn (g(x + τ) − g(x))d τ. [x+hn ,x]

[hn ,0]

Therefore −1 |Fn − g| ≤|hn | dx |g(x + τ) − g(x)|d τ for all n ∈ N. R

R

(7.31)

[hn ,0]

Now, the real-valued function on R × R given by g(x + τ) is (B × B)-measurable by Proposition 6.6.4 and hence (M × M)-measurable. It follows by Tonelli’s Theorem 6.5.2 that  −1   −1    h  dx |g(x + τ) − g(x)|d τ  hn d τ |g(x + τ) − g(x)|dx. n R

[hn ,0]

[hn ,0]

R

7.5 The Derivative of the Integral

151

Therefore by (7.31), −1 |Fn − g| ≤|hn | d τ |g(x + τ) − g(x)|dx for all n ∈ N. R

[hn ,0]

(7.32)

R

Now, consider any ε > 0. Since hn → 0, by the Mean Continuity Theorem 7.1.8, there exists an N ∈ N such that n ≥ N implies |g(x + τ) − g(x)|dx < ε whenever hn ≤ τ < 0, R

and hence by (7.32) implies

   |Fn − g| ≤h−1 n

R

εd τ  ε.

[hn ,0]

Thus, (7.30) has been shown to hold. Now it emerges from (7.29), Remark 6.5.4(c) (a consequence of Fatou’s Lemma) and (7.30) that equality (7.28) must hold. As already noted when (7.28) was formulated above, this is all that needed to be established. 

7.6 Bounded Variation By the Lebesgue Differentiability Theorem, a sum of an increasing and a decreasing function is differentiable a.e. Although an increasing or a decreasing function is usually easy to identify, their sum may not be. In order to get a feel for what may be involved in seeking to identify a function f as a sum of an increasing function g and a decreasing function h, we enter into a heuristic discussion with a concrete example. Of course, g and h cannot be unique. f (x)

a

c

_ h(x) = f (x) g(x)

g(x)

b

x

a

c

b

x

a

c

b

x

Consider the function f : [a, b] → R with a graph as in the figure above, extreme left. To keep matters simple, we take f (a) to be 0. In seeking to represent it as f  g + h, with increasing g and decreasing h, it is natural to try first to take g as in the middle of the figure. What g “does” is to mimic f as long as f is increasing but freeze into a

152

7 Differentiation

constant as soon as f starts decreasing. The function g that does this will serve our purpose provided that h  f − g turns out to be decreasing. The graph on the extreme right in the above figure shows the function h  f − g; the graph of g is included there in a lighter and thinner line in order to facilitate understanding why the graph of f − g is indeed as shown. Unsurprisingly, h “does” the exact opposite of what g does, which is to say, it remains frozen as a constant as long as f is increasing and begins to mimic (run parallel to) f as soon as the latter starts decreasing. V(x)

a

c

b

x

Now consider the related function V as in the figure just above this line. An informal description of it is that it is obtained by reversing even the minutest decrease in f into an increase of the same magnitude, but retaining even the minutest increase in f exactly as it is. Another way to put it is that V (x) is the accumulated absolute variation in f from a to x. In particular, V (a)  0 because there is no change in the value of f from a to a. On the interval of increase, which is [a, c], the function V mimics f and happens to agree with f , because f (a)  0  V (a). It can now be read off the figure quite easily that g is the mean between V and f on the entire domain. On the interval of decrease, which is [c, b], the function V is readily described as being given by V (x)  f (c) + |f (x) − f (c)|. On this interval, g(x)  f (c), a constant, and V (x)  2f (c) − f (x). As a result, a simple computation confirms what had been read off the figure earlier: g(x)  21 (V (x) + f (x)) for x ∈ [c, b]. This is trivially true on the interval [a, c] of increase, because g  V  f there. For the above equality to hold even when f (a)  0, which does not alter the fact that V (a)  0, we have to have g(a)  21 f (a). This only entails that the description of what g “does” be modified to say that g begins at a by being equal to 21 f (a) and then mimics f on [a, c]. 7.6.1. Example. Let f : [0, 3] → R be given by f (x)  3 − x for x ∈ [0, 2] and f (x)  2x − 4 for x ∈ (2, 3]. Note that f is left continuous at 2 but not right continuous. We identify the function V (x) on the basis of its informal description as the accumulated absolute variation in f from 0 to x.

7.6 Bounded Variation

153

If x ∈ [0, 2], then V (x)  |f (x) − f (0)|  f (0) − f (x)  3 − f (x)  x. Across x  2, the function jumps up by −1 because f (2)  1 and the right limit f (2+)  0. Therefore, if x ∈ (2, 3], then V (x)  V (2) + |−1|+(f (x) − f (2+))  2x − 1. Note that the function g : [0, 3] → R given by g(x)  21 (V (x) + f (x)) works out to be 23 on [0, 2] and to be 2x − 25 on (2, 3]. This is an increasing function. Moreover, h(x)  f (x) − g(x)  f (x) − 21 (V (x) + f (x))  21 (f (x) − V (x)) works out to be 23 − x on [0, 2] and to be − 23 on (2, 3]. This is a decreasing function. It should be noted how the jump was taken into account in the description of V over (2, 3]. In this connection, the reader will have no difficulty agreeing that V (x) can be approximated on (2, 3] by |f (2 − δ) − f (0)| + |f (2 + δ) − f (2 − δ)| + |f (x) − f (2 + δ)| with “small” positive δ. In fact, drawing a figure will suggest that the supremum of such sums over δ > 0 is nothing but V (x) when x ∈ (2, 3]. 7.6.2. Exercise. Let f : [0, 4] → R be given by f (x)  x 2 + 1 for x ∈ [0, 1) and f (x)  5 − x for x ∈ [1, 4]. If 0 < 1 − δ < 1 + δ < x0 ≤ 4, compute the sum S  |f (1 − δ) − f (0)| + |f (1 + δ) − f (1 − δ)| + |f (x0 ) − f (1 + δ)| and find its supremum over δ > 0.

4

1 1-δ

0

1+δ

1

x0

4

154

7 Differentiation

Solution: Since 0 and 1 − δ both belong to the interval [0, 1), we have f (1 − δ) − f (0)  (1 − δ)2 + 1−f (0)  (1 − δ)2 ≥ 0. Therefore the first term in the sum S is |f (1 − δ) − f (0)|  1 − 2δ + δ2 .

(7.33)

Now, 1 − δ ∈ [0, 1) and 1 + δ ∈ [1, 4]; therefore f (1 + δ) − f (1 − δ)  5 − (1 + δ) − ((1 − δ)2 + 1)  2 + δ − δ2  2 + δ(1 − δ). Since 0 < δ < 1, we have δ(1 − δ) > 0 and consequently, the second term in the sum S is |f (1 + δ) − f (1 − δ)|  2 + δ − δ2 .

(7.34)

Next, 1 + δ and x 0 both belong to the interval [1, 4]. So, we have f (x0 ) − f (1 + δ)  5 − x − (5 − (1 + δ))  (1 + δ) − x0 < 0 because it is given that 1 + δ < x0 . Therefore the third (and final) term in the sum S is |f (x0 ) − f (1 + δ)|  −1 − δ + x0 .

(7.35)

It follows from (7.33), (7.34) and (7.35) that S  (1 − 2δ + δ2 ) + (2 + δ − δ2 ) + (−1 − δ + x0 )  x0 + 2 − 2δ. Hence the required supremum is x 0 + 2. The reader may note how the “jump” at 1 is “captured” by 1 − δ and 1 + δ in the process of taking the supremum, because arbitrarily small positive values of δ are taken into account thereby. With the foregoing considerations as backdrop, it would seem reasonable that a mathematically precise way to encapsulate what V is would be as follows, the possibility that a  b being allowed. 7.6.3. Definition. For a real-valued function f : [a, b] → R, the total variation from a to x, where x ∈ [a, b], is V xa

n    f (xj ) − f (xj−1 ) (f )  sup j1

over all possible partitions P : a  x0 < x1 < · · · < xn  x of [a, x]. The function f is said to be of bounded variation over [a, x] if Vax (f ) < ∞. The sum on the right side above will be denoted by S(f , P).

7.6 Bounded Variation

155

It is clear that any monotone function f : [a, b] → R is of bounded variation over [a, x] with Vax (f )  ±(f (x) − f (a)) according as the function is increasing or decreasing. Also, since the only partition P of [a, a] consists of a alone and S(f , P)  |f (a) − f (a)|  0, we have Vaa (f )  0, whatever the function f may be. 7.6.4. Example. Let such that f (0)  0 and, for any  f : [0, 1] → R be any function   k ∈ N, we have f k1  k1 if k is even and f k1  0 if k is odd. An example of a continuous function of this kind is given by f (x)  x cos2 (π/2x) for x  0 and f (0)  0. To see that f is not of bounded variation over [0, 1], we consider the partitions P: 0
0. There exists a partition P  of [a, y] such that S(f , P  ) > M . The points of P  and the point z together constitute a partition P of [a, z] and this partition satisfies S(f , P) ≥ S(f , P  ) > M . Since this has been proved for any M > 0 and some partition P of [a, z], we conclude that V za (f )  ∞, as required. The case when Vyz (f )  ∞ is entirely analogous. y Now suppose Va (f ) and Vyz (f ) are both finite. y We shall prove first that Vaz (f ) ≥ Va (f ) + Vyz (f ). With this in view, consider any ε > 0. There exist partitions Q and R of [a, y] and [y, z] respectively such that y S(f , Q) > Va (f ) − 2ε and S(f , R) > Vyz (f ) − 2ε . The points of Q and R together y form a partition P of [a, z] and S(f , P)  S(f , Q) + S(f , R) > Va (f ) + Vyz (f ) − ε. Since this has been proved for any ε > 0 and some partition P of [a, z], we y conclude that Vaz (f ) ≥ Va (f ) + Vyz (f ), as required. y It remains to prove the reverse inequality, that is, Vaz (f ) ≤ Va (f ) + Vyz (f ). Consider an arbitrary partition P of [a, z]. If necessary, refine it by adding the point y. Then the refinement P0 satisfies S(f , P) ≤ S(f , P0 ) by Exercise 7.6.5. The points of P0 lying in [a, y] form a partition Q of [a, y] and those lying in [y, z] form a partition R of [y, z]. Besides, S(f , P0 )  S(f , Q) + S(f , R) ≤ y y Va (f )+Vyz (f ). It follows that S(f , P) ≤ S(f , P0 ) ≤ Va (f )+Vyz (f ). Since this has y been proved for any partition P of [a, z], we conclude that Vaz (f ) ≤ Va (f )+Vyz (f ), as required. (e) (e) It follows from parts (a) and (d) that Vax (f ) ≤ Vab (f ) for every x ∈ [a, b]. 

7.6 Bounded Variation

157

If f : [a, b] → R is of bounded variation, V xa (f ) describes a nonnegative real-valued function on [a, b] in view of Proposition 7.6.6(e). It will be convenient to introduce the notation V a f or Va f  for this function. Thus, (Va f )(ξ), Va f (ξ) and V a f (ξ) all ξ mean the same thing as Va (f ). It is clear from the definition of total variation that Vax (−f )  Vax (f ) for all x, which is to say, Va −f   Va f . 7.6.7. Exercise. Show that if f : [a, b] → R is of bounded variation, the nonnegative real-valued function V a f is increasing and that Vax (Va f )  Vax (f ) for all x ∈ [a, b]. y

y

Solution: Let a ≤ x ≤ y ≤ b. By Proposition 7.6.6(d), Va (f )  Vax (f ) + Vx (f ). Since y V a f is nonnegative-valued, it follows that Va (f ) ≥ Vax (f ). Therefore V a f is increasing. Hence Vax (Va f )  (Va f )(x) − (Va f )(a)  Vax (f ) − Vaa (f )  Vax (f ) for all x ∈ [a, b]. 7.6.8. Theorem. A real-valued function on a closed bounded interval is of bounded variation if and only if it is the sum of an increasing function and a decreasing function. Proof: Since a monotone function is always of bounded variation, the “if” part is trivial by Proposition 7.6.6(b). For the converse, consider any function f : [a, b] → R such that Vab (f ) < ∞. Put g  21 (Va f + f ), which is to say, g(x)  21 (Vax (f ) + f (x)). Then g : [a, b] → R is real-valued. If a ≤ y ≤ z ≤ b, then  1 z Va (f ) − Vay (f ) + f (z) − f (y) 2  1 z  Vy (f ) + f (z) − f (y) by Proposition 7.6.6(d) 2 ≥ 0 by Definition 7.6.3.

g(z) − g(y) 

Thus g is an increasing function. As this has to hold even if we replace f by −f , the function 21 (Va −f  − f )  21 (Va f − f ) is also increasing, so that the function h  21 (f − Va f ) is decreasing. Besides, g + h  21 (Va f + f ) + 21 (f − Va f )  f .  7.6.9. Remark. It now follows from the Lebesgue Differentiability Theorem 7.4.1 that a function of bounded variation is differentiable a.e. Moreover, by Proposition 7.4.2, its derivative agrees a.e. with a Borel function.  The first paragraph in the proof of Proposition 7.5.1 begins by arguing that F(x)  [a,x] f describes a function of bounded variation. Problem Set 7.6 7.6.P1. Show that the product of two functions of bounded variation is of bounded variation.

158

7 Differentiation

7.6.P2. Suppose a < b and the bounded function f : [a, b] → R is decreasing on (a, b], so that the right limit f (a+) exists. Show that Vab (f )  f (a+) − f (b) + |f (a+) − f (a)|. 7.6.P3. If f : [a, b] → R has a bounded derivative on the open interval (a, b), show that it is of bounded variation on the closed interval [a, b]. 7.6.P4. If f : [a, b] → R is of bounded variation, show that it has one sided limits at each point and that, for any c ∈ [a, b), the equality Va (f )(c+) − Va (f )(c)  |f (c+) − f (c)| holds. (In particular, f is right continuous at c if and only if V a (f ) is.)

7.7 Absolute Continuity A theorem about the integral of the derivative can be expected to say something like f (x) − f (a)  [a,x] f  for a ≤ x ≤ b, albeit under appropriate conditions on the derivative f  . For the equality to make sense, the derivative should exist almost everywhere and the function resulting from extending the derivative function to be 0 elsewhere should be integrable. So, this could be one appropriate condition to assume. Additionally, properties that the indefinite integral of an integrable function must necessarily possess, such as continuity, could also be appropriate to assume about f . Actually, an indefinite integral always possesses a stronger property than continuity, called “absolute continuity”. If we take f to be absolutely continuous, it turns out that it also has to be of bounded variation, and the Lebesgue Differentiability Theorem 7.4.1 renders it redundant to assume the existence of the derivative almost everywhere. Now, assuming absolute continuity, the strategy could be to show that the left side of the equality we seek to establish has the same derivative as the right side. The hurdle in this strategy is that equality of derivatives can be expected to hold only almost everywhere, thereby making the Mean Value Theorem inapplicable. However, absolute continuity comes to the rescue, because vanishing of the derivative almost everywhere does indeed imply that the function is constant provided that the function is absolutely continuous. It may be pertinent to note here as an aside that even the requirement that the function f be increasing does not warrant the desired equality. The best that can be extracted from the assumption that the function is increasing will be given later in Theorem 7.8.1. 7.7.1. Definition. A real-valued function f : [a, b] → R is said to be absolutely continuous if , for every ε > 0, there exists a δ > 0 such that whenever (a1 , b1 ), (a2 , b2 ), . . . , (an , bn ) are disjoint subintervals of [a, b],

7.7 Absolute Continuity n 

159

(bj − aj ) < δ implies

j1

n    f (bj ) − f (aj ) < ε. j1

It is clear what it means for a function f : [a, b] → R to be absolutely continuous on a closed subinterval of the domain. 7.7.2. Exercise. Let (a1 , b1 ), (a2 , b2 ), . . . , (an , bn ) be subintervals of [a, b] and k be an integer such that 1 ≤ k ≤ n. Suppose the intervals (α1 , β1 ), (α2 , β2 ), . . . , (αn+1 , βn+1 ) are obtained by replacing (ak , bk ) by the two retaining (ai , bi ) for which i  k. intervals (a k , c) and (c, bk ), while    all the intervals Show that ni1 |f (bi ) − f (ai )| ≤ nj1 f (βj ) − f (αj ). Solution: In view of the manner in which the intervals (αj , βj ) are obtained from the intervals (ai , bi ), n+1 n     f (βj ) − f (αj ) − |f (bi ) − f (ai )| j1

i1

 |f (bk ) − f (c)| + |f (c) − f (ak )| − |f (bk ) − f (ak )| ≥ 0 by the triangle inequality. 7.7.3. Theorem. An absolutely continuous function f : [a, b] → R is continuous and of bounded variation. In particular, it is differentiable a.e. and its derivative agrees a.e. with a Borel function. Proof: That it is continuous is trivial from the definition of absolute continuity. Corresponding to ε  1, choose δ > 0 satisfying the condition in the definition of absolute continuity. Let N be some positive integer such that Nδ > b − a. We shall show that Vab (f ) ≤ N . Observe that by choice of δ, any subinterval [α, β] ⊆ [a, b] of length less than δ β satisfies Vα (f ) ≤ ε  1. If N  1, then δ > b − a and it is immediate from the observation in the preceding paragraph that Vab (f ) ≤ 1  N . So, we need consider only N > 1. Let δ be the number such that N δ  b − a. Then δ > 0 and N δ  b − a < N δ, so that δ < δ. Thus, 0 < δ < δ. Now consider the points x0 < x1 < . . . < xN in [a, b] given by xr  a + rδ , 0 ≤ r ≤ N . Obviously, x 0  a and x N  b. Since each interval [xr , xr+1 ] has length δ < δ, it follows by the observation above that each Vxxrr+1 (f ) ≤ 1. Therefore, on applying Proposition 7.6.6(d) repeatedly N − 1 times, we obtain Vab (f ) 

N −1  r0

as promised.

Vxxrr+1 (f ) ≤ N ,

160

7 Differentiation

The rest follows from Remark 7.6.9.  In Example 7.6.4, we have exhibited a continuous function that is not of bounded variation. By Theorem 7.7.3, it cannot be absolutely continuous; continuity therefore does not imply absolute continuity. It will be seen later in Example 7.8.2 that there exists a function of bounded variation, in fact a strictly increasing one, which is continuous but not absolutely continuous. Now for the result that lets us hurtle over the hurdle of the Mean Value Theorem not being available. 7.7.4. Theorem. If an absolutely continuous function f : [a, b] → R has the property that f   0 a.e. on [a, b], then it is constant on [a, b]. Proof: It is sufficient to show that f (a)  f (b). Given any ε > 0, we shall prove that |f (b) − f (a)| ≤ ε(1 + (b − a)). Choose δ > 0 satisfying the condition in the definition of absolute continuity. Let N ⊆ [a, b] consist of b and all the points where f  is either nonzero or does not exist. Then N has measure 0. To each α ∈ [a, b]\N correspond intervals [α, β] ⊆ [a, b] with β−α arbitrarily small such that |f (β) − f (α)| < ε(β − α). Since this holds for each α ∈ [a, b]\N, the intervals constitute a Vitali covering of the set. By the Vitali Covering Theorem 7.2.7, there exist finitely many disjoint intervals [αj , βj ], 1 ≤ j ≤ n, in the Vitali covering such that n  m(N0 ) < δ, where N0  ([a, b]\N )\ [αj , βj ]. j1

Since N0 ∪ N ⊇ [a, b]\∪nj1 [αj , βj ], we obtain m([a, b]\

n 

[αj , βj ]) ≤ m(N0 ∪ N ) ≤ m(N0 ) + m(N )  m(N0 ) < δ.

j1

By renumbering the intervals [αj , βj ] if necessary, we may assume that their centers are in increasing order. Set β0  a and αn+1  b. Then we have a  β0 ≤ α1 < β1 ≤ · · · < βn ≤ αn+1  b. It follows that

7.7 Absolute Continuity

161

|f (b) − f (a)| ≤

n n+1       f (βj ) − f (αj ) + f (αj ) − f (βj−1 ). j1

j1

The choice of the Vitali covering ensures that n n     f (βj ) − f (αj ) ≤ ε(βj − αj ) ≤ ε(b − a). j1

j1

Now the intervals (βj−1 , αj ) are all disjoint from ∪nj1 [αj , βj ] and therefore their union is a subset of [a, b]\∪nj1 [αj , βj ], which has been shown to have measure less than δ. As they are disjoint from each other, their total length is the same as the measure of their union and hence is less than δ. Consequently, in view of the condition that δ was chosen to satisfy with reference to the definition of absolute continuity, n+1    f (αj ) − f (βj−1 ) < ε. j1

When this is combined with the preceding two inequalities, the desired inequality springs forth.  7.7.5. Proposition. Suppose f : [a, b] → R is integrable and F : [a, b] → R is the function given by F(x)  f for all x ∈ [a, b]. [a,x]

Then F is absolutely continuous. Proof: Let ε > 0. By Problem 3.2.P4,  there exists a δ > 0 such that for any measurable A with m(A) < δ, we have A f < ε. Now  consider disjoint subintervals n A  (a1 , b1 ), (a2 , b2 ), . . . , (an , bn ) of [a, b] such that j1 (bj − aj ) < δ. Put 

 n n ∪j1 aj , bj . Then m(A)  j1 (bj − aj ) < δ, with the consequence that A f < ε. But f  A

n 

f 

j1 [a ,b ] j

j

n  

 F(bj ) − F(aj ) .

j1

 Problem Set 7.7 7.7.P1 (a) Show that a continuous function on [0, 1] which is absolutely continuous on [η, 1] for every positive η less than 1 need not be absolutely continuous on [0, 1].

162

7 Differentiation

(b) Show that a continuous function on [0, 1] which is absolutely continuous on [η, 1] for every positive η less than 1 and is of bounded variation must be absolutely continuous on [0, 1]. 1

7.7.P2. (a) Show that the function g : [0, 1] → R given by g(x)  x 2 is absolutely continuous. (b) Show that the function f : [0, 1] → R given by f (x)  x2 cos4 (π/2x) for x  0 and f (0)  0 is absolutely continuous. (c) Show that a composition of absolutely continuous functions need not be absolutely continuous. 7.7.P3. (a) Suppose f : [a, b] → [α, β] and g : [α, β] → R are absolutely continuous. If f is increasing, show that the composition g ◦ f : [a, b] → R is absolutely continuous. Does the same conclusion hold if the function f is decreasing instead? (b) Give an example of a function that has an unbounded derivative, is not monotone, but is absolutely continuous. 7.7.P4. If f : [a, b] → R is absolutely continuous on each of the subintervals [a, c] and [c, b], where a < c < b, show that it is absolutely continuous on [a, b]. 7.7.P5. Given a finite number of points in (a, b), show that there exists an absolutely continuous function on [a, b] that fails to be differentiable at the given points.

7.8 The Integral of the Derivative As recorded just before the statement of Theorem 7.5.2, a Borel function that agrees a.e. with the derivative of a function f is denoted by f  and we speak of it as though it is unique. By Proposition 7.4.2, this may be done with a monotone function, as we shall do in Theorem 7.8.1. By virtue of Theorem 7.7.3, Theorem 7.6.8 and Proposition 7.4.2, the same can also be done with an absolutely continuous function, and we shall do so in Theorem 7.8.3. The reader is reminded that, for a real-valued monotone function f : [a, b] → R one sided limits f (x+) and f (x−) exist at every interior point x ∈ (a, b) and so do f (a+) and f (b−). 7.8.1. Theorem. If f : [a, b] → R is increasing, then f  is integrable on [a, b] and f (b−) − f (a) ≥ f . [a,b]

Proof: The result will follow by the Dominated Convergence Theorem 3.2.6 if we prove it with b replaced by an arbitrary number c ∈ (a, b) on the smaller side of the 1 < n ∈ N. inequality. The integral [a,c+ 1 f then makes sense when b−c n

7.8 The Integral of the Derivative

163

Let the functions fn : [a, c] → R be given by fn (x) 

f (x + 1n ) − f (x) 1 n

1 < n ∈ N. b−c

,

Then limn→∞ fn  f  a.e. on [a, c]. In tandem with Remark 6.5.4(c) (a consequence of Fatou’s Lemma), this leads to ⎞ ⎛ 1 f  ≤ lim inf fn  lim inf ⎝n f (x + ) − f (x)dx⎠ n→∞ n→∞ n [a,c] [a,c] [a,c] ⎛ ⎞ ⎜ ⎟ ⎟  lim inf ⎜ n f (x)dx − n f (x)dx ⎝ ⎠ n→∞

1 [c,c+ ] n

⎛

⎜ ≤ lim inf ⎜ ⎝n

1 [a,a+ ] n



f (c−)dx − n

n→∞

1 [c,c+ ] n

⎞ ⎟ f (a)dx⎟ ⎠

1 [a,a+ ] n

≤ f (c−) − f (a).  The inequality in the above theorem cannot be replaced by equality. There actually exists a continuous function from [0, 1] to [0, 1], which is surjective, has derivative 0 a.e., and is also strictly increasing to boot. 7.8.2. Example. Define a sequence {fn }n≥0 of functions on the domain [0, 1] as follows. The first one is f0 (x)  x, a strictly increasing function whose values on any with 0 ≤ k ≤ 20 − 1 (yes, there is only one!) are linear interpolations interval 2k0 , k+1 20 of its values at the endpoints (straight line graph on the interval). Suppose fn is with a strictly increasing function on [0, 1] whose values on any interval 2kn , k+1 2n 0 ≤ k ≤ 2n −1 are linear interpolations of its values at the endpoints. It is pertinent to note for later purposes that linear interpolation implies that the value at the midpoint is the mean of the values at the endpoints: fn

 2k+1  2n+1



1  k  1  k+1  fn n + fn 2 n . 2 2 2

(7.36)

The function fn+1 is obtained from fn by (see figure below) first setting its values at k , where 0 ≤ k ≤ 2n+1 , to be the 2n+1 + 1 points 2n+1 fn+1

fn+1



2k 2n+1

 2k+1  2n+1





 fn+1

k 2n

 fn

k 2n

for 0 ≤ k ≤ 2n ,

1  k  2  k+1  fn n + fn 2 n for 0 ≤ k ≤ 2n − 1, 3 2 3

(7.37)

(7.38)

164

7 Differentiation

and then extending it by linear interpolation to the interiors of the 2n+1 intervals k n+1 , 2k+1 − 1. formed by the points. Note that the 2n+1 intervals are 2n+1 n+1 , 0 ≤ k ≤ 2

fn+1 fn

2k  1 2 n 1

k 2n

k 1 2n

Equation (7.38) mandates the value of fn+1 at the midpoint of [ 2kn , k+1 ] to be a 2n convex combination of the values of fn at the endpoints but “biased” in favor of the value at the right endpoint, which is the larger one. Therefore, intuition would lead one to expect that the convex combination must work out to be greater than the mean. The expectation is confirmed by the computation     1 2 1 1 1 α+ β − α + β  (β − α). 3 3 2 2 6 Accordingly, fn+1

 2k+1  2n+1

>

1  k  1  k+1  fn n + fn 2 n , 2 2 2

and hence by (7.36), fn+1

 2k+1  2n+1

> fn

 2k+1  . 2n+1

  The stipulation about linear interpolation now implies that fn+1 > fn on 2kn , k+1 and 2n hence that fn+1 ≥ fn on all of [0, 1]. Since these functions have values lying between 0 and 1, there is a function F on [0, 1] with values lying between 0 and 1 such that F  limp→∞ fp ≥ fn for every n. From (7.37), we infer that F

k 2n

 fn

k 2n

In particular, F(0)  0 and F(1)  1. The computations

for 0 ≤ k ≤ 2n and n ≥ 0.

(7.39)

7.8 The Integral of the Derivative



165

   1 2 2 2 1 1 α + β − α  (β − α) and β − α + β  (β − α), 3 3 3 3 3 3

when viewed in the light of (7.37) and (7.38), confirm that fn+1

k 2n

< fn+1

 2k+1  2n+1

< fn+1

 k+1  . 2n

The stipulation about linear interpolation now implies that fn+1 is strictly increasing. Consequently, every function fn in the inductively constructed sequence is strictly increasing. It follows that the limit function F is increasing. We shall show that it is in fact strictly increasing. To this end, suppose x < y, where x, y ∈ [0, 1]. Some k and n must satisfy x < 2kn < y. Using the already established facts that (a) F is     increasing, (b) F 2kn  fn 2kn always, (c) every fn is strictly increasing, and (d) F ≥ f n always, in that order, we find that F(x) ≤ F

k

 fn

2n

k 2n

< fn (y) ≤ F(y).

The computations at the beginning of the preceding paragraph also confirm that fn+1

 2k+1  2n+1

− fn+1

k 2n



  2   k+1  fn 2n − fn 2kn 3

(7.40)



  1   k+1  fn 2n − fn 2kn . 3

(7.41)

and fn+1

 k+1  2n

− fn+1

 2k+1  2n+1

Now consider an arbitrary x ∈ [0, 1]. There exists a nested sequence of intervals , each containing x. From (7.39), (7.40) and (7.41), it follows [αn , βn ]  2knn , kn2+1 n for n ≥ 0 that F(βn+1 ) − F(αn+1 )  λn (F(βn ) − F(αn )), where each λn 

2 1 or . 3 3

Repeated application of this equality leads to F(βn+1 ) − F(αn+1 ) 

n 

λk (F(β0 ) − F(α0 )) 

k0

n 

λk .

k0

Since each |λk | ≤ 23 , the right side here has limit 0 as n → ∞. This proves that F is continuous at x. 1 , the derivative F  (x), if it exists, must be the limit of Since βn+1 − αn+1  2n+1 2n+1 (F(βn+1 ) − F(αn+1 )) 

n  k0

τk , where each τk 

2 4 or . 3 3

166

7 Differentiation

We shall demonstrate that the product on the right side cannot have a finite positive limit, no matter which among the τk are 23 and which ones are 43 . Suppose to the contrary that it has a finite limit τ > 0. Then there exists an n ∈ N such that   14 14 16 16 τ< τ< τ. τk < τ as well as τk < 15 15 15 15 n

n+1

k0

k0

If τn+1  23 , the second part of the first double inequality implies n+1 

τk


16 4 14 τ> τ, 3 15 15

contradicting the second part of the second double inequality. The possibility that F  (x) is finite and nonzero is thereby ruled out. Since F must have a derivative a.e. by the Lebesgue Differentiability Theorem 7.4.1, we conclude that the strictly increasing continuous function F has derivative 0 a.e.  Thus F(1−) − F(0)  F(1) − F(0)  1 > 0  [a,b] F  , showing that equality need not hold in Theorem 7.8.1. Furthermore, by the next theorem, F cannot be absolutely continuous, although it is continuous and strictly increasing, in particular, of bounded variation. The above function F is surprising because it is a continuous surjection from [0, 1] to itself with derivative 0 a.e. but is strictly increasing nonetheless. At the opposite extreme, there are continuous surjections from [0, 1] to itself not only with derivative 0 a.e. but also constant on all the middle thirds that are removed in the process of forming the Cantor set (see Problem 4.1.P4). The standard example of such a continuous surjection, called the Cantor function, will be presented in the next chapter. These continuous surjections illustrate the same thing regarding Theorem 7.8.1 as F does. 7.8.3. Theorem (Integral of the Derivative). If f : [a, b] → R is absolutely continuous, then f  is integrable on [a, b] and f (x) − f (a)  f  for allx ∈ [a, b]. [a,x]

Proof: Since f is absolutely continuous, it is of bounded variation and therefore a sum of an increasing function g and a decreasing function h. By Theorem 7.8.1, both g

7.8 The Integral of the Derivative

167

and h are integrable on [a, b]. Since f   g  − h , the derivative f  is integrable. The right side of the equality sought to be proved has derivative f  a.e. by Theorem 7.5.2 (Derivative of the Integral) and is absolutely continuous by Proposition 7.7.5. The same is true of the left side. Upon invoking Theorem 7.7.4, we find that the left side and right side differ by a constant. As they agree at x  a, we can assert that they are equal everywhere.  7.8.4. Corollary. A function f : [a, b] → R is absolutely continuous if and only if there exists an integrable function g : [a, b] → R such that g for allx ∈ [a, b]. f (x) − f (a)  [a,x]

Proof: Immediate from Proposition 7.7.5 and Theorem 7.8.3.



Problem Set 7.8 7.8.P1. (a) Let f be a real-valued function on [a, b] such that its values on (a, b) are linear interpolations between its values at the endpoints. If one among f (a) and f (b) is positive and the other nonnegative, show that f (x) is positive for all x ∈ (a, b). (b) Let f and g be real-valued functions on [a, b] such that f (a) ≥ g(a), f (b) ≥ g(b), one of the inequalities being strict, and let the values of both functions on (a, b) be linear interpolations between their values at the endpoints. Show that f (x) > g(x) for all x ∈ (a, b). 7.8.P2. (a) Let f be a real-valued function on [a, b] such that its values on (a, b) are linear interpolations between its values at the endpoints and let a < c < b. Show that the function has the same interpolation property over the intervals (a, c) and (c, b). (b) Let f and g be real-valued functions on [a, b] such that f (a) ≥ g(a), f (b) ≥ g(b), and the values of g on (a, b) are linear interpolations between its values at the endpoints. Also, let a < c < b and the values of f on each of (a, c) and (c, b) be linear interpolations between its values at the endpoints of the respective intervals. If f (c) > g(c), show that f (x) > g(x) for all x ∈ (a, b). a sequence of numbers such that, for 7.8.P3. Suppose 0 < α < 1 < β and {τj }j≥1 is  each j, either 0 < τj ≤ α or τj ≥ β. Let σn  nj1 τj . Show that σn cannot have a finite positive limit. 7.8.P4. Given that f : [a, b] → R is absolutely continuous, show that the function ξ Va f : [a, b] → R defined by (Va f )(ξ)  Va (f ) for all ξ ∈ [a, b] is also absolutely continuous. Can one draw the inference that an absolutely continuous function is the sum of an absolutely continuous increasing function and an absolutely continuous decreasing function?

Chapter 8

The Cantor Set and Function

8.1 A Surjection from the Cantor Set to [0, 1] Let b be an integer greater than 1. If {αk }k≥1 is a sequenceof integers such that −k converges 0 ≤ αk ≤ b − 1 for each k, it is easy to show that the series ∞ k1 αk b to a sum x ∈ [0, 1] by comparing it with the geometric series with common . It is well known that every x ∈ [0, 1] has at least one representation ratio b−1 ∞ −k as x  k1 αk b , where the αk are integers satisfying 0 ≤ αk ≤ b − 1 (see Theorem 4.1.9 in Shirali and Vasudeva [9]). We shall refer to any such representation as a b-expansion (or b-representation) of x. The easily verified fact that 0 and 1 have unique b-expansions will matter but lack of uniqueness in general is an inconvenience. However, the inconvenience can be managed. The integer b in this context is called the base of the expansion. The standard name for a 2-expansion is binary expansion (representation). The terms of the sequence {αk }k≥1 are called the digits in the b-expansion. One learns in early encounters with decimal expansions (base 10) how to compare two numbers by spotting the first decimal place where they differ. The procedure needs to be formulated with due care to adjust for absence of uniqueness, and then validated. We shall do so for an arbitrary base b although we shall use it, with one exception, only for base 2 afterwards. The exception is Problem 8.1.P3, which concerns base 3 and is a standalone with no purpose to serve in anything else relating to the Cantor set that we discuss. ∞ −k −k α b and are b-expansions, then we have 0 ≤ αk ≤ b − 1 If ∞ k1 k k1 βk b as well as 0 ≤ βk ≤ b − 1 for each k. Two simple consequences of these inequalities will be used in the next proof: For each k, (1) |αk − βk | ≤ b − 1 and (2) αk −β k  −(b − 1) ⇔ αk  0, βk  b − 1. ∞ ∞ −k −k and y  are b8.1.1. Proposition. Suppose x  k1 αk b k1 βk b expansions and there exists a positive integer j such that α j > β j and αk  βk for k < j (vacuously true in case j  1). Then (1) x ≥ y and (2) x  y ⇔ not only α j −β j  1 but also every k > j satisfies αk  0, βk  b − 1. © Springer Nature Switzerland AG 2018 S. Shirali, A Concise Introduction to Measure Theory, https://doi.org/10.1007/978-3-030-03241-8_8

169

170

8 The Cantor Set and Function

Proof: First note that ∞  k j

(αk − βk )b−k ≥ b− j +

∞ 

(αk − βk )b−k because α j − β j ≥ 1

k j+1

≥ b− j − (b − 1)

∞ 

b−k because αk − βk ≥ −(b − 1) for each k.

k j+1

Now, equality holds in the first line above ⇔ α j −β j  1 and holds in the second line ⇔ every k > j satisfies αk −βk  −(b − 1) ⇔ every k > j satisfies αk = 0, βk  b − 1. So, equality holds in both lines ⇔ not only α j −β j  1 but also every k > j satisfies  −k  x − y and αk = 0, βk  b − 1. And now the sockdolager: ∞ k j (αk − βk )b  ∞ −j −k  b − (b − 1) k j+1 b  0. In the above proposition, the integer j is the smallest k such that αk  βk and is unique if it exists, which it certainly does unless αk  βk for all k. 8.1.2. Example. Let b  2 and x  1 · 2−1 + 0 · 2−2 + 0 · 2−2 + · · · ,

y  0 · 2−1 + 1 · 2−2 + 1 · 2−2 + · · · .

Then the smallest k such that αk  βk is k  1, and α1 > β1 . From Proposition 8.1.1, we obtain x ≥ y, which can independently be seen to be true because x  21  y. Besides, α1 − β1  1 − 0  1 and every k > 1 satisfies αk  0, βk  1  b − 1. 8.1.3. Remarks. (a) When the open middle third of a closed bounded interval is removed, what remains is a union of two disjoint closed subintervals, which we may distinguish as the left and right thirds. If two numbers in the closed interval belong to different ones among the left and right thirds, then the bigger number must be in the right third (and the smaller in the left). (b) When the open middle third of a closed bounded interval is removed, the left endpoint not only belongs to the left third but is also its left endpoint. In particular, when the open middle third of the left third is removed, it again belongs to the left third of the left third. Analogously for the right endpoint.  1 2   Recall that the Cantor set is defined as ∞ n1 C n , where C 1  0, 3 ∪ 3 , 1 , and Cn+1 is obtained  C n by  removing   the  open  from  middle thirds of its closed intervals. Thus, C2  0, 19 ∪ 29 , 13 ∪ 23 , 79 ∪ 89 , 1 , and so on. In general, Cn+1 is the union of the closed intervals obtained by removing the open middle thirds from each of the closed intervals whose union    disjoint  is C n . Thus C 1 is the union of the left third 0, 13 of [0, 1] and the right third 23 , 1 . By induction,  n C n is the union of 2n closed intervals, each of length 13 . So the total length of the  2 n closed intervals in C n is 3 . It will be convenient to refer to the 2n intervals as “comprising” C n or as “components” of C n and distinguish them as left and right components. They are left and right thirds respectively of the components of C n–1 if

8.1 A Surjection from the Cantor Set to [0, 1]

171

n > 1 and of [0, 1] if n  1. In this terminology, half of the 2n intervals comprising C n are left components and half are right components and they are all disjoint from each other. The disjointness ensures that, for each n, a given number in the Cantor set belongs to precisely one among the components of C n . The interval may be a left component or a right component. Accordingly, we can map each number u in the Cantor set into a uniquely determined sequence {αk }k≥1 of 0s and 1s by setting αk  0 or 1 according as the number is in a left component or a right component of C k . We have thus described a mapping from the Cantor set into the set of all sequences {αk }k≥1 of 0s and 1s.   8.1.4. Example. The number 13 belongs to the left component 0, 13 among the two components of C 1 . On the other hand, it is the right endpoint of that left component   and therefore by Remark 8.1.3(b), it is the right endpoint of the right third of 0, 13  1  1  3 when the open middle third of 0, 3 is removed. In fact, 0, 3  0, 9 and its right   third is 29 , 13 . So, 13 is the right endpoint of a right component of C 2 . By repeated application of Remark 8.1.3(b), we find that 13 is in a right component of C n for all n > 1. This shows not only that 13 is in the Cantor set but also that the sequence {αk }k≥1 obtained by applying the above mapping to 13 is 01111…. In contrast, the number   2 is in the right component 23 , 1 among the two components of C 1 but is the left 3 endpoint of it and therefore belongs to a left component of C n for all n > 1. Hence not only is 23 in the Cantor set but also the sequence {αk }k≥1 obtained by applying the above mapping to 23 is 10000….     1 Let m be a positive integer. The interval 0, 3m+1 is the left third of 0, 31m , which shows by induction that the latter is a component interval of C m . On the other hand, 1 is the right endpoint of that left component and therefore by Remark 8.1.3(b), 3m+1   1 it is the right endpoint of the right third of 0, 3m+1 when the open middle third   1 of 0, 3m+1 is removed. Applying Remark 8.1.3(b) repeatedly as in the preceding 1 paragraph, we deduce that 3m+1 is in a right component of C n for all n > m + 1. This 1 shows that 3m+1 is in the Cantor set and hence that the Cantor set contains arbitrarily small positive numbers. We note certain properties of the mapping that we shall need. It is manifest from the Nested Interval Theorem (see Theorem 3.3.6 of Shirali and Vasudeva [9]) that the mapping is surjective. That it is injective will follow from something stronger that we need and are about to prove. The inverse of the mapping will subsequently be denoted by φ. 8.1.5. Proposition. Suppose u ∈ C is mapped into {αk }k≥1 and v ∈ C is mapped into {βk }k≥1 . If αk  βk for 1 ≤ k ≤ n, then u and v belong to the same component of C n . If u > v, then there is some k for which αk  βk and if j denotes the smallest such k, we have α j > β j . In particular, the mapping is injective. Proof: If n  1, this is obvious. Assume it true for some n and let αk  βk f for 1 ≤ k ≤ n + 1. Then u and v belong to the same component I of C n . Since αn+1  βn+1 , the numbers u and v both belong to a left third among the components of Cn+1 or both

172

8 The Cantor Set and Function

belong to a right third among the components of Cn+1 . However, I contains only one left third and one right third. Therefore u and v belong to the same component of Cn+1 . Now suppose u > v. If we were to have αk  βk for all k, then u and v would be in the same component of C k for every k, so that the distance between them would  k be at most 13 for every k. This is not possible because u  v. Therefore αk  βk for some k and there must be a smallest such  it by  j. If j  1, then u and  k. Denote v belong to different subintervals among 0, 13 and 23 , 1 and the fact that u > v     implies that u ∈ 23 , 1 and v ∈ 0, 13 , which further implies that α1  1 and β1  0, so that αj > βj , as desired. Now suppose j > 1. Then αk  βk for 1 ≤ k ≤ j–1 and therefore u and v are in the same component of C j–1 . Since αj  βj , one among u and v belongs to the left third of a component of C j–1 and the other belongs to the right third. The fact that u > v implies by Remark 8.1.3(a) that u must belong to the right third and v to the left third. So, αj  1 and βj  0; this confirms that αj > βj , as desired.  8.1.6. Remark. It is trivial to deduce from the second assertion of Proposition 8.1.5 that its converse is also true. Thus if the sequences {αk }k≥1 , {βk }k≥1 fulfill the condition that there is a k for which αk  βk and αj > βj , where j is the smallest k with αk  βk , then φ({αk }k≥1 ) > φ({βk }k≥1 ).  −k Now, the mapping which takes a sequence {αk }k≥1 into ∞ has range [0, k1 αk 2 1] because every number in [0, 1] has at least one binary expansion. We make no claim that it is injective (see Example 8.1.2.). By composing this mapping with the one discussed in Proposition 8.1.5 above, namely φ−1 , we obtain a surjective map κ0 from the Cantor set to [0, 1]. Although an application of the Bernstein–Schröder Theorem immediately yields the conclusion that there is a bijection between the Cantor set and [0, 1], we shall derive this conclusion independently of that theorem in the next section (see Theorem 8.2.3), wherein we  shall use the function κ0 . −k A more formal definition of κ0 is that κ0 (u)  ∞ k1 αk 2 , where {αk }k≥1  −1 φ (u). It is worth bearing in mind that κ0 (u) is the number in [0, 1] having φ−1 (u) as its sequence of digits in at least one binary expansion of it.   A rephrasing of Example 8.1.4 in terms of φ−1 is that the sequence φ−1 13 has   first term 0 and all subsequent terms 1, whereas φ−1 23 has first term 1 and all   subsequent terms 0. This leads to κ0 13  0 · 2−1 + 1 · 2−2 + 1 · 2−2 + · · · and 2     κ0 3  1 · 2−1 + 0 · 2−2 + 0 · 2−2 + · · ·. So, κ0 13  21  κ0 21 , showing that κ0 is not injective. 8.1.7. Proposition. The function κ0 is increasing (not strictly of course) and κ0 (u)  0 implies u  0. −1 Proof: Consider u > v and let φ−1 (u) k≥1 , φ (v)  {β k }k≥1 . By definition of   {αk }−k ∞ −k α 2 and κ (v)  the function κ0 , we have κ0 (u)  ∞ 0 k1 k k1 βk 2 . We want to show that

8.1 A Surjection from the Cantor Set to [0, 1]

κ0 (u) ≥ κ0 (v), i.e.

∞  k1

173

αk 2−k ≥

∞ 

βk 2−k .

k1

From Proposition 8.1.5 and the inequality u > v, we know that there is a k for which αk  βk and that αj > βj , where j is the smallest such k; hence (1) of Proposition 8.1.1 yields just what we want. Suppose κ0 (u)  0. Then 0 has φ−1 (u) as its sequence of digits in at least one binary expansion of it. However, 0 has only one binary expansion and every digit in it is 0; therefore φ−1 (u) is the sequence with every term equal to 0, which is to say,  φ−1 (u)  φ−1 (0). Consequently, u  0. Problem Set 8.1   −k −k and y  ∞ are b-expansions and there 8.1.P1. Suppose x  ∞ k1 αk b k1 βk b exists a positive integer j such that αj > βj and αk  βk for k < j (vacuously true in case j  1). If there does not exist any positive integer J such that αk  0 for every k > J, show that x > y. 8.1.P2. If y, z ∈ [a, b] and |y−z|  b − a, show that one among y and z is a and the other is b. (Ignore the trivial case when a  b.) 8.1.P3. Let {βk }k≥1 be a sequence in which each term is either 0 or 1.  −k ∈ [0, 1] and the series is a 3-expansion (called ternary) (a) Then ∞ k1 2 · βk 3 of the sum. (b) If the  first n terms of the sequence {βk }k≥1 of digits in any one ternary expansion −k of the number x agree with the first n terms of φ−1 (u), x  ∞ k1 2 · βk 3 where u ∈ C, then x is in the same component of C n as u is. ∞  (c) If φ−1 (u)  {αk }k≥1 , then 2 · αk 3−k is a ternary expansion of u. k1

(d) A number in [0, 1] belongs to C if and only if it has a ternary expansion in which every digit is either 0 or 2. (e) If a number in [0, 1] has a ternary expansion in which every digit is either 0 or 2, then it has no other ternary expansion of the same kind.

8.2 A Bijection from the Cantor Set to [0, 1] and the Cantor Function For the purposes of this section, a sequence {ξk }k≥1 of any kind of terms will be called recurring if all the terms from some k onward are equal to each other. It could also have been called ‘eventually constant’. A b-expansion will be called recurring if the sequence of digits in it is recurring. 8.2.1. Proposition. If a number x ∈ [0, 1] has two distinct b-expansions, then both are recurring.

174

8 The Cantor Set and Function

Proof: Left to the reader as Problem 8.2.P1(a).



8.2.2. Proposition. If a number x ∈ [0, 1] has a recurring b-expansion, then it is rational. Proof: Left to the reader as Problem 8.2.P1(b).



8.2.3. Theorem. There exists a bijection from the Cantor set to [0, 1]. Proof: Let C 0 be the subset of the Cantor set C consisting of those u for which the associated sequence φ−1 (u)  {αk }k≥1 is recurring. Such sequences are countably infinite in number, and therefore C 0 is countably infinite. Consider also the infinite subset I 0 of [0, 1] consisting of numbers having a recurring binary expansion. Any number having such an expansion is rational by Proposition 8.2.2 and therefore I 0 too is countably infinite. Therefore there is a bijection from C 0 to I 0 . Suppose u, v ∈ C\C0 , where u > v and let φ−1 (u)  {αk }k≥1 and φ−1 (v)  {βk }k≥1 . Both these sequences are nonrecurring. By definition of the function κ0 (see 8.1), the numbers x  κ0 (u) and y  κ0 (v) are the ones that satisfy x   ∞ Sect. −k −k and y  ∞ k1 αk b k1 βk b . By Proposition 8.1.7, x ≥ y. Now, by Proposition 8.1.5, there is a smallest k such that αk  βk and if j denotes this smallest k, then αj > βj . Since x has a nonrecurring binary expansion, it is a consequence of Proposition 8.2.1 that x > y, i.e. κ0 (u) > κ0 (v). This shows that κ0 is strictly increasing on C\C0 and is therefore injective on C\C0 . We shall show that κ0 maps C\C0 into [0, 1]\I0 surjectively. For any u ∈ C\C0 , the sequence φ−1 (u) is nonrecurring and therefore κ0 (u) has a nonrecurring binary expansion. By Proposition 8.2.1, it has no other binary expansion and therefore κ0 (u) ∈ [0, 1]\I0 , showing that κ0 maps C\C0 into [0, 1]\I0 . It is trivial to see that it does so surjectively. Since we have already shown that κ0 is injective on C\C0 , it follows that it is a bijection from C\C0 to [0, 1]\I0 . But there is also a bijection from C 0 to I 0 , and the two bijections together provide a bijection from C to [0, 1].  8.2.4. Remark. Let g be an increasing real-valued function on a closed interval. At an interior point x of the closed interval, either g(x−)  g(x+), in which case the function is continuous at x, or g(x−) < g(x+). In the latter case, at least one of the intervals (g(x−), g(x)) and (g(x), g(x+)) is nonempty and forms a gap in the range of the function. Consequently, an increasing real-valued function on a closed interval whose range is an interval must be continuous in the interior of its domain. Similar reasoning shows that continuity holds at an endpoint as well. 8.2.5. Proposition. Let the function κ0 :C → [0, 1] be as in Sect. 8.1. Define the function κ : [0, 1] → R by setting κ(x)  sup{κ0 (u) : u ∈ C, u ≤ x}. Then κ0 is the restriction of κ to C. Moreover, κ is a continuous increasing function that maps [0, 1] surjectively to [0, 1] and is constant on any interval disjoint from C.

8.2 A Bijection from the Cantor Set to [0, 1] and the Cantor Function

175

Proof: Since 0 ∈ C and the range of κ0 is [0, 1], the set {κ0 (u): u ∈ C, u ≤ x} is always a nonempty subset of [0, 1] and hence its supremum always exists and belongs to [0, 1]. Thus κ maps [0, 1] to itself. By Proposition 8.1.7, the function κ0 is increasing and therefore κ(x)  κ0 (x) whenever x ∈ C. This means κ0 is the restriction of κ to C. Now, the range of κ0 is [0, 1] and therefore the range of κ is also [0, 1]. Thus κ is surjective. For x, y ∈ [0, 1] such that x < y, we have {κ0 (u): u ∈ C, u ≤ x} ⊆ {κ0 (u): u ∈ C, u ≤ y} and hence κ(x) ≤ κ(y). Consequently, κ is increasing. Since κ is an increasing real-valued function whose domain and range are both intervals, it follows by Remark 8.2.4 that it is continuous. Lastly, let I be a subinterval of [0, 1] that is disjoint from C. We consider only nonempty I, because it is trivial that κ is constant on ∅. Suppose x, y ∈ I and x ≤ y. Since κ has been shown to be increasing, we know that κ(x) ≤ κ(y). Consider any u ∈ C such that u ≤ y. Since [x, y] ⊆ I and is therefore disjoint from C, it follows that u < x. This shows that {κ0 (u): u ∈ C, u ≤ y} ⊆ {κ0 (u): u ∈ C, u ≤ x} and hence κ(y) ≤ κ(x). Consequently, κ(x)  κ(y).  The function κ of Proposition 8.2.5 is called the Cantor function. 8.2.6. Remark. As seen in Example 8.1.4 (second paragraph), the Cantor set contains arbitrarily small positive numbers. It therefore follows from Proposition 8.1.7 that 0 < u ≤ 1 implies κ(u) > 0. It is possible to use the Cantor function to set up, though less explicitly, a function of the kind in Example 7.8.2: strictly increasing and continuous with derivative 0 almost everywhere. To do so, we shall need another well known theorem of Fubini, towards which we first discuss a preliminary result. Lemma. 8.2.7 that such  Let {φk }k≥1 be a sequence of increasing functions on [a, b] ∞

φ converges everywhere. Then the differentiated series φ the series ∞ k1 k k1 k is defined a.e. and its partial sums form an increasing sequence bounded above by the  ∞ φ of the sum; in particular, the differentiated series is convergent derivative k1 k a.e. Proof: n Let s denote the sum of the undifferentiated series and sn its partial sum k1 φk . All of them are increasing functions on [a, b]. By the Lebesgue Differentiability Theorem 7.4.1, the sum s and all φk are differentiable a.e. and therefore there is a single set E ⊆ [a, b] on which s, all sn and all φk are differentiable and whose complement in [a, b] has measure 0. It follows that the differentiated series is defined on all of E and hence almost everywhere. As for convergence, we note that the increasing nature of the functions involved

≤ s everywhere on E. implies regarding derivatives that they satisfy sn ≤ sn+1 Consequently, the partial sums of the differentiated series, at any point of E, form a bounded increasing sequence, so that its convergence is ensured. Thus the differentiated series converges everywhere on E, hence almost everywhere. 

176

8 The Cantor Set and Function

8.2.8. Fubini Series Theorem.  Let {φk }k≥1 be a sequence of increasing functions Then the differentiated on [a, b] such that the series ∞ k1 φk converges everywhere.  ∞ ∞ series k1 φk is converges a.e. to the derivative k1 φk of the sum. Proof: It is sufficient to prove the theorem with the additional hypothesis that φk (a)  0 for each k.  Let s denote the sum of the undifferentiated series and sn its partial sum nk1 φk . All of them are increasing functions on [a, b]. By Lemma 8.2.7, there exists an E ⊆[a, b] such that the complement of E in [a, b] has measure 0 and the partial sums sn form an increasing sequence on E with s as an upper bound. The assertion of the theorem is that sn → s on some subset of E whose complement in [a, b] has measure 0. Since the sn form an increasing sequence, the assertion will follow if we construct a subsequence sn k such that s − sn k → 0 on some subset of E whose complement in [a, b] has measure 0. Taking into account the additional hypothesis that φk (a)  0 for each k, we know for every n and every x ∈ [a, b] that 0  s(a) − sn (a) ≤ s(x) − sn (x) ≤ s(b) − sn (b). Since sn (b) → s(b), there exists a sequence {n k }k≥1 such that s(b) − sn k (b) < 2−k for each k. Taken  in conjunction with the inequality displayed above, this implies that the series ∞ k1 (s − sn k ) converges everywhere. Now, the terms s − sn k of this series are increasing functions. Therefore,by appealing to Lemma 8.2.7 once again,

we deduce that the differentiated series ∞ k1 (s − sn k ) converges on some subset of [a, b] whose complement has measure 0. The intersection of this subset with E is of E with the property that its complement has measure 0 and the series a∞subset



 k1 (s − sn k ) converges on it, so that the kth term (s − sn k ) tends to 0 on it. 8.2.9. Example. We now present another example of a continuous strictly increasing function on [0, 1] with derivative 0 almost everywhere taking the value 0 at 0 and 1 at 1. Let g be the extension to R of the Cantor function κ obtained by setting it equal to 1 for x > 1 and equal to 0 for x < 0. That is, ⎧ ⎪ if x < 0 ⎨0 g(x)  κ(x) if 0 ≤ x ≤ 1 ⎪ ⎩1 if 1 < x. Then g is continuous and increasing, and it has derivative 0 almost everywhere. Let {rk }k≥1 be a sequence, the range of which is the set of all rational numbers in [0, 1], and set F(x) 

∞  1 g(x − rk ). 2k k1

8.2 A Bijection from the Cantor Set to [0, 1] and the Cantor Function

177

Since 0 ≤ g ≤ 1 everywhere, the series is uniformly convergent and therefore F is continuous. Moreover, F  0 a.e. by the Fubini Series Theorem 8.2.8. To see why the function F is strictly increasing on [0, 1], consider any x, y ∈ [0, 1] satisfying x < y. Then x−r k < y−r k and therefore g(x − rk ) ≤ g(y − rk ) for every k. As the range of the sequence {rk }k≥1 is the set of all rational numbers in [0, 1], there exists a j such that x < r j < y. This implies x−r j < 0 < y−r j < 1. The definition of g therefore yields g (x−r j )  0 and g (y−r j )  κ(y−r j ) ≥ 0. But κ (y−r j ) > 0 by Remark 8.2.6, so that g (x−r j ) < g (y−r j ). This inequality, when combined with the fact that g (x−r sk ) ≤ g (y−r k ) for every k, implies via the definition of F that F (x) < F(y). Problem Set 8.2 8.2.P1. (a) Prove Proposition 8.2.1. (b) Prove Proposition 8.2.2. Ordinarily, the “post inverse” (or “left inverse”) g of a function G is understood to have the range of G as its domain and to satisfy the condition that g(G(x))  x for every x in the domain of G. However, for the present purpose, it will be necessary to broaden the concept somewhat. If G : [a, b] → R is strictly increasing, then any function g defined on [G(a), G(b)] and satisfying the aforementioned condition will be called a post inverse. Its values at points of [G(a), G(b)] that are outside the range of G are not uniquely determined unless some additional condition is imposed. 8.2.P2. (Rubel’s Lemma [6]) Let G : [a, b] → R be a strictly increasing function. Then there is a continuous increasing function g: [G(a), G(b)] → R which is a “post inverse” of G, i.e. satisfies g(G(x))  x for each x ∈ [a, b], and which is constant on any open subinterval of [G(a), G(b)] contained in the complement of the range of G. 8.2.P3. Let G be a strictly increasing real-valued function on the domain [a, b] and g0 be an increasing function with domain containing the range of G and satisfying g0 (G(x))  x for each x ∈ [a, b]. (In particular, g0 could be an increasing post inverse of G.) Show the following. (a) For each y ∈ [G(a), G(b)] that belongs to the domain of g0 , g0 (y)  sup{u ∈ [a, b] : G(u) ≤ y}. (b) The increasing post inverse is unique and g0 agrees with the unique increasing post inverse at points of [G(a), G(b)] that belong to its domain. Hint: Argue for any y ∈ [G(a), G(b)] that belongs to the domain of g0 that g0 (y) lies between the numbers sup{u ∈ [a, b]: G(u) ≤ y} and inf{u ∈ [a, b]: G(u) ≥ y} and that these two numbers are equal.

178

8 The Cantor Set and Function

8.2.P4. Let G be a strictly increasing real-valued function on the domain [a, b]. Show that the function g : [G(a), G(b)] → R defined by g(y)  inf{u ∈ [a, b] : G(u) ≥ y} is an increasing post inverse of G. 8.2.P5. Let G1 and G2 be strictly increasing real-valued functions on the domain [a, b] and let g1 and g2 be their respective (unique) increasing post inverses. If G1  G2 almost everywhere, show that g1  g2 everywhere on the intersection of their domains, i.e. on [G 1 (a), G 1 (b)] ∩ [G 2 (a), G 2 (b)]. 8.2.P6. In the course of proving Theorem 8.2.3, it was established that the mapping κ0 is strictly increasing on C\C0 , where C 0 is the subset of the Cantor set C consisting of those u for which the associated sequence φ−1 (u)  {αk }k≥1 is recurring. Use this fact in conjunction with Problem 6.6.P3 to show that there exists a measurable set that is not a Borel set. Hint: Consider the inverse image by κ0 of the intersection of a nonmeasurable subset of [0, 1] with the range of κ0 .

Solutions

1.1

The Riemann Integral Revisited

1.1.P1. For each t
0, find S ¼ fx 2 ½0; 4: f ð xÞ [ tg, where f : ½0; 4 ! R is the function given by: f ð xÞ ¼ 6 if 0  x\1; f ð1Þ ¼ 5; f ð xÞ ¼ 2 if 1\x\3; f ð3Þ ¼ 1; f ð xÞ ¼ 6 if 3\x  4: Solution:

8 ½0; 4 > > > > < ½0; 3Þ [ ð3; 4 S ¼ ½0; 1 [ ð3; 4 > > ½0; 1Þ [ ð3; 4 > > : ; © Springer Nature Switzerland AG 2018 S. Shirali, A Concise Introduction to Measure Theory, https://doi.org/10.1007/978-3-030-03241-8

if0  t\1 if1  t\2 if2  t\5 if5  t\6 if6  t: 179

180

Solutions

1.1.P2. For f : ½1; 2 ! R given by f ð xÞ ¼ ex , find the distribution ~f : ½0; sup f  ! R R sup f R2 and show that 0 ~f agrees with 1 f . Solution: The set fx 2 ½1; 2: f ð xÞ [ tg is [1, 2] if 0  t\e; ðln t; 2 if e  t\e2 and is ; if e2  t. Also, sup f ¼ e2 . Therefore ~f ðtÞ ¼



1 2  ln t

if 0  t\e if e  t\e2 :

So, Zsup f

~f ¼ 1  ðe  0Þ þ

Ze2

  ð2  ln tÞdt ¼ e þ ð3e2  e2 ln e2 Þ  ð3e  e ln eÞ

e

0

¼ e2  e ¼

Z2

ex dx:

1

1.1.P3. Let A [ 0. For the function f : ½0; A ! R given by f ð xÞ ¼ ex , find the RA R sup f distribution ~f : ½0; sup f  ! R and show that 0 ~f agrees with 0 f . Hint: Use an appropriate substitution and integrate by parts. 2 Solution: Here sup f ¼ 1 and inf f¼ eA . Also, fx 2 ½0; A: f ð xÞ [ tg is [0, A] if qffiffiffiffiffiffiffiffiffiffi 2 2 t ¼ 0  t\eA and is 0; ðln1t Þ if eA  t  1. Consequently, 2



 t\eA pAffiffiffiffiffiffi if 0A 2 1 ln t if e  t  1: qffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffi R1 R sup f 2 So, 0 ~f ¼ AeA þ expðA2 Þ ðln1t Þdt. Upon substituting for ðln1t Þ and then RA 2 integrating by parts, we find that this agrees with 0 ex dx. 1.1.P4. Abel summation formula. If a1 ; . . .; an and b1 ; . . .; bn are two finite sequences of n terms each (n
2) and Sk ¼ b1 þ    þ bk , then bk ¼ Sk  Sk1 for k ¼ 2; . . .; n and therefore ~f ðtÞ ¼

2

a1 b1 þ    þ an bn ¼ a1 S1 þ a2 ðS2  S1 Þ þ a3 ðS3  S2 Þ þ    þ an ðSn  Sn1 Þ ¼ ða1  a2 ÞS1 þ ða2  a3 ÞS2 þ ða3  a4 ÞS3 þ    þ ðan1  an ÞSn1 þ an Sn : In R notation: Let fak gnk¼1 and fbk gnk¼1 be finite sequences of numbers, where n
2, and let P Sk ¼ kj¼1 bj for 1  k  n. Then

Solutions

181 n X

ak bk ¼

k¼1

n1 X

ðak  ak þ 1 ÞSk þ an Sn :

k¼1

Prove this in R notation without resorting to algebraic manipulations concealed behind an Pellipsis   . How should P Tk be defined in order to obtain the alternative version nk¼1 ak bk ¼ a1 T1 þ nk¼2 ðak  ak1 ÞTk ? Solution: For an induction proof, see Shirali and Vasudeva [9; pp. 133–4]. Here is a proof without induction. Set S0 ¼ 0 ¼ an þ 1 . Then Sk  Sk1 ¼ bk for 1  k  n. Also, n X

ak Sk1 ¼

k¼1

n1 X

ak þ 1 Sk

k¼0

¼ ¼

n1 X k¼1 n X

ak þ 1 Sk

because S0 ¼ 0

ak þ 1 Sk

because an þ 1 ¼ 0:

k¼1

Pn

Subtracting this from

k¼1

n X

ak Sk , we get

ak ðSk  Sk1 Þ ¼

k¼1

n X

ðak  ak þ 1 ÞSk :

k¼1

In view of the observation that Sk  Sk1 ¼ bk for 1  k  n, the above equality becomes n X

ak bk ¼

k¼1

n X k¼1

¼

n1 X

ðak  ak þ 1 ÞSk ¼

n1 X

ðak  ak þ 1 ÞSk þ ðan  an þ 1 ÞSn

k¼1

ðak  ak þ 1 ÞSk þ an Sn

because an þ 1 ¼ 0:

k¼1

P For the alternative version, Tk ¼ nj¼k bj . 1.1.P5. (Relevant for Problem 1.1.P6 and to be used in Theorem 4.3.1) If one visualizes two disjoint intervals on the real line, one of them lies entirely on the left of the other. Prove the following precise formulation of this idea for bounded intervals: If two bounded nonempty intervals are disjoint, then the right endpoint of one of them is less than or equal to the left endpoint of the other. Deduce that, if the union of two disjoint intervals is an interval, then the sum of their lengths is equal to the length of the union.

182

Solutions

Solution: Let the left and right endpoints of an interval I be a and b and those of J be a and b respectively; suppose I \ J ¼ ;. We must prove that either b  a or b  a. If a ¼ b, the interval I consists of a (= b) alone and, because of the disjointness, the inequality a\a\b cannot hold; therefore either b  a or a  a, which in this case means b  a. The case a ¼ b is similar and we need argue only the case when a\b as well as a\b. Suppose it is not true that either b  a or b  a. Then a\b as well as a\b. Thus we have the four inequalities a\b;

a\b;

a\b;

a\b:

Therefore a\minfb; bg and a\minfb; bg, so that there exist x and y such that a\x\minfb; bg and a\y\minfb; bg. If a  a, then x 2 I \ J, while if a  a, then y 2 I \ J, thus contradicting the hypothesis that I and J are disjoint. This contradiction shows that either b  a or b  a. If one of the intervals is empty, there is nothing to prove. Continuing with the notation of the above paragraph, suppose b  a. If I [ J is an interval, we claim that b ¼ a. If not, then any number between b and a fails to belong to I [ J, although it lies between the midpoints of I and J which are in I [ J. This cannot happen when I [ J is an interval. Consequently, b ¼ a as claimed. It follows that I [ J has length b  a ¼ ðb  aÞ þ ðb  aÞ. 1.1.P6. Suppose F is a finite family of disjoint bounded intervals. Show that there exists a nonempty finite family G of bounded intervals such that (a) a union of two distinct intervals in G is never an interval; (b) [ G ¼ [ F (Here “ [ H” means union of all the intervals in H); (c) the total length of the intervals in G is the same as for F . (Note that the intervals in G are necessarily disjoint in view of (a).) Solution: Induction on the number of intervals in F . If there is only one interval in F , then G ¼ F serves the purpose; if F is empty, then take G ¼ f;g. Assume that such a family G exists whenever F is a family of k disjoint bounded intervals (induction hypothesis). Consider a family F of (k + 1) disjoint bounded intervals. If it is the case that a union of two distinct intervals in F is never an interval, then G ¼ F serves the purpose. Otherwise there exist two distinct intervals in F whose union is an interval; if one of them is empty, then the length of the union is the sum of the lengths of the two intervals; if neither is empty, then the same holds in view of Problem 1.1.P5. Let F 0 be the family obtained from F by replacing the two intervals by their union. Then first, [ F 0 ¼ [ F ; second, F 0 is a family of k disjoint bounded intervals, and third, the total length of the intervals in F 0 is the same as for F . Therefore by the induction hypothesis, there is a family G of bounded intervals such that (a) a union of two distinct intervals in G is never an interval (b) [ G ¼ [ F 0 ¼ [ F and (c) the total length of the intervals in G is the same as for F 0 , which is the same as for F .

Solutions

183

1.1.P7. If an interval I has an endpoint that belongs to an interval J, show that the union I [ J is an interval. (Only a left endpoint need be considered.) Solution: Let a 2 J be the left endpoint of I. Consider any x, u, y such that x\u\y, where x; y 2 I [ J. All that need be shown is that u 2 I [ J. First suppose u  a. Then x\a. Since a is the left endpoint of I, it follows that x 62 I, so that x 2 J. Thus x\u  a with x 2 J and a 2 J, which implies u 2 J. Next, suppose instead that a\u. If y 2 J, we have a\u\y with a 2 J and y 2 J, which implies u 2 J. If y 62 J, we have y 2 I and hence a\u\y with a the left endpoint of I and y 2 I, which implies u 2 I. 1.1.P8. Let G be a finite nonempty family of bounded intervals such that a union of two distinct intervals of G is never an interval. Suppose F is a finite family of disjoint bounded intervals such that [ F ¼ [ G. Show that every interval of F is contained in some interval of G. What more can be said if F too is nonempty and has the property that a union of two distinct intervals of it is never an interval? Solution: If G ¼ f;g, necessarily F ¼ ; or f;g. If G contains a nonempty interval, then the hypothesis implies that ; 62 G. So we may assume that all the intervals in G are nonempty, and therefore freely use results about intervals from elementary Analysis, wherein ; is usually not considered to be an interval. Let I1 be any nonempty interval in F and x1 2 I1 . Then x1 2 [ F ¼ [ G and hence x1 2 I for some interval I belonging to G. We shall show that I1 I. Let the left and right endpoints of I be a and b respectively (the possibility that a = b, in which case I ¼ fag ¼ ½a; a is not excluded). Then x1 2 I1 and x1
a: We shall establish that the left endpoint of I1 is greater than or equal to a and that, if a 62 I, then a 62 I1 . A similar argument can then establish the corresponding assertion about right endpoints and it will follow that I1 I. Case 1. a 62 I. Consider any interval J of G other than I, if there is any. By hypothesis, I [ J is not an interval. It follows by Problem 1.1.P7 that a 62 J. This is true for every J belonging to G (including for I) and therefore a 62 [ G. But [ G ¼ [ F and therefore a 62 [ F . It follows that a 62 I1 . Since x1 2 I1 and x1 [ a 62 I1 , the left endpoint of I1 must be greater than or equal to a. This establishes what we want in this case. Case 2. a 2 I. In the subcase that every x 2 [ G satisfies x
a, we have x 2 I1 ) x
a, because I1  [ F ¼ [ G. Hence the left endpoint of I1 is greater than or equal to a. In the contrary subcase, [ G does contains a number less than a and hence there must be an interval in G which contains that number. By hypothesis, whenever J is an interval of this kind, I [ J is not an interval. It follows by Problem 1.1.P7 that neither I nor J can contain an endpoint of the other and hence that a 62 J and that a cannot be an endpoint of J. Combined with the fact that J contains a number less than a, this implies that the right endpoint of J must be less than a. But there are only finitely many such intervals (G is given to be a finite) and there must therefore be a greatest right endpoint c for them. We claim that

184

Solutions

c\x\a ) x 62 [ G. Indeed, suppose there is some x 2 [ G satisfying c\x\a; then x 2 J0 for some J0 belonging to G and it therefore follows as before that the right endpoint of J0 is less than a while being greater than c, thereby contradicting the maximality of c. Since c\x\a ) x 62 [ G and I1  [ F ¼ [ G, it follows that I1 cannot contain numbers between c and a. But I1 contains the number x1
a. Consequently, the left endpoint of I1 is greater than or equal to a. When F too is nonempty and has the property that a union of two distinct intervals of it is never an interval, it follows from what has been proved above that every interval of G is also contained in some interval of F . Since the intervals in each family are disjoint among themselves, it follows that F ¼ G. 1.1.P9. Let F 1 and F 2 be finite families of bounded intervals, each having the property that its intervals are disjoint. If [ F 1 ¼ [ F 2 , show that the total length of the intervals of each family is the same. Solution: In accordance with Problem 1.1.P6, let Gi ði ¼ 1; 2Þ be a nonempty finite family of bounded intervals such that (a) a union of two distinct intervals in Gi is never an interval; (b) [ Gi ¼ [ F i ; (c) the total length of the intervals in Gi is the same as for F i . By (b) and the hypothesis that [ F 1 ¼ [ F 2 , we have [ [ [ [ G1 ¼ F1 ¼ F2 ¼ G2 : By (a) and the last part of Problem 1.1.P8, we have G1 ¼ G2 . The required conclusion now follows from (c).

1.2 Improper Integrals 1.2.P1. For the double sequence a: N  N ! R given by an;k ¼ 1 if n ¼ k;

an;k ¼ 1 if k ¼ n þ 1 and

an;k ¼ 0 in all other cases,

P1 P1 P1 P1 show that the repeated sums n¼1 P k¼1 an;k and k¼1 n¼1 an;k are unequal,   P1 P1 P  and 1  1 an;k  are convergent. although n¼1 a n;k k¼1 k¼1 n¼1 Hint: One can think of an;k as representing an infinite matrix with first three rows: 1 1 0 1 0 0

0 0 1 0 1 1

 0  0 0



Solution: It is clear from the given description Pof an;k that, for each n, there always exists a k such that k ¼ n þ 1, and therefore 1 k¼1 an;k ¼ 0. For the other sum, the

Solutions

185

possibility that k ¼ n þ 1 arises if and only if k > 1 and therefore P when k = 1 and 1 n¼1 an;k ¼ 1 when k > 1. Consequently, 1 X 1 X n¼1 k¼1

an;k

   1 X 1 X   ¼ an;k  ¼ 0    n¼1 k¼1

but

1 X 1 X k¼1 n¼1

an;k

P1

n¼1

an;k ¼ 1

   1 X 1 X   ¼ an;k  ¼ 1:    k¼1 n¼1

1.2.P2. Let b > 0 and g be a decreasing nonnegative function on [0, b] that is real-valued except perhaps at 0. Denote by f the decreasing nonnegative function on [0, ∞) obtained by extending g to be equal to 0 to the right of b. Show that Z1

f ¼ lim

Zb g:

x!0

0

x

Note: If g(0) < ∞, the function g is Riemann integrable on [0, b] and the limit Rb here is equal to 0 g (see Problem 9.3.P2 of Shirali and Vasudeva [9]). Solution: Consider any x 2 (0, b). On the interval [x, b], the function g is Riemann R1 Rb integrable and satisfies 0  g  f . It follows by Definition 1.2.2 that x g  0 f . Rb R1 Consequently, limx!0 x g  0 f . We proceed to prove the reverse inequality. Let 0\a\b\1 and g0 be any Riemann integrable function on ½a; b such that 0  g0  f . Keeping in mind how f is obtained from g, we see that g0  g on [0, b] Rb Rb and g0 vanishes on ðb; 1Þ. We claim that a g0  limx!0 x g. For b\a, the inequality claimed must hold because the left side is 0. Suppose a  b. If b  b, the Rb Rb validity of the inequality g0  g on [0, b] leads to a g0  a g and the nonnegaRb Rb tivity of g leads to a g  limx!0 x g. Thus the inequality claimed holds if b  b. Rb Rb In the contrary case b\b, the vanishing of g0 on ðb; 1Þ leads to a g0 ¼ a g0 , the Rb Rb validity of the inequality g0  g on [0, b] leads to a g0  a g and the nonnegaRb Rb tivity of g leads to a g  limx!0 x g. Thus the inequality claimed holds also if b\b. This completes the justification of the claim in all cases. Since g0 is an arbitrary Riemann integrable function on ½a; bð0; 1Þ satisfying 0  g0  f , Definition 1.2.2 now yields the desired reverse inequality. 1.2.P3. In Lemma 1.2.3, show that the condition gðaÞ\1 does not follow from the remaining conditions in the hypothesis even if we impose the additional condition that g is real-valued on (a, b]. If gðaÞ ¼ 1 but g (which must be decreasing) is realRb Rb valued on (a, b], then the improper integral a g in the usual sense is limx!0 x g, Rb Rb though it may be 1; show that a g ¼ limn!1 a gn . If g is not real-valued on Rb (a, b], which to say, gðaÞ ¼ 1 for some a [ a, show that limn!1 a gn ¼ 1.

186

Solutions

Solution: Let gn ð0Þ ¼ n and gn ð xÞ ¼ minfn; ðx  aÞ1 g for x 2 (a, b]. Also, let gðaÞ ¼ 1 and gð xÞ ¼ ðx  aÞ1 for x 2 (a, b]. Then all other conditions in the hypothesis are fulfilled but gðaÞ ¼ 1 and g is real-valued on (a, b]. Without loss of generality, we may assume a = 0. Since gð xÞ
gn ð xÞ
gn ðbÞ
g1 ðbÞ for all x and all n, we may, by subtracting g1(b) from every function, assume that all functions are nonnegative-valued. (This lays the groundwork for applying Proposition 1.2.4.). Extend all functions gn and g to the right of b by setting them equal to 0 there and call them fn and f. RThen the hypothesis R 1 of 1 Proposition 1.2.4. holds and hence we can infer that 0 f ¼ limn!1 0 fn . It remains only to note that by virtue of Problem 1.2.P2, Z1

fn ¼ lim

Zb

x!0

0

x

gn ¼

Zb

Z1 gn

and

0

f ¼ lim

Zb

x!0

0

g¼

x

Zb g: 0

For the last part, suppose gðaÞ ¼ 1, where a [ 0. Then for any real M [ 0, there exists an integer N 2 N such that n
N ) gn ðaÞ
M ) gn
M on Rb Rb ½0; a ) a gn
aM. Consequently, limn!1 a gn ¼ 1.

2.1 Measure and Measurability 2.1.P1. Let fSn gn
1 be a sequence of subsets of a set X, such that Sn Sn þ 1 for every n 2 N (no r-algebra intended) and let S ¼ [ 1 n¼1 Sn . Denote the characteristic functions of Sn and of S respectively by fn and f. For each x 2 X, show that ði) fn ðxÞ  fn þ 1 ðxÞ

and

ðii) f ðxÞ ¼ lim fn ðxÞ: n!1

Solution: (i) If x 2 Sn , then fn ð xÞ ¼ 1 and x 2 Sn þ 1 , so that fn þ 1 ð xÞ ¼ 1 and consequently, fn ð xÞ  fn þ 1 ð xÞ. If x 62 Sn, then fn ð xÞ ¼ 0  fn þ 1 ð xÞ. (ii) Let x 2 S. Then f (x) = 1. Also, since S ¼ [ 1 n¼1 Sn , there exists some N such that x 2 SN . But Sn Sn þ 1 for every n 2 N, and therefore n
N ) Sn SN ) fn ð xÞ ¼ 1, so that limn$1 fn ð xÞ ¼ 1 ¼ f ð xÞ. Now let x 62 S. In this case, f ð xÞ ¼ 0. Also, since S ¼ [ 1 n¼1 Sn , we have x 62 Sn for every n 2 N, which has the consequence that fn ð xÞ ¼ 0, so that limn$1 fn ð xÞ ¼ 0 ¼ f ð xÞ. 2.1.P2. If f and g are functions such that f  g everywhere, show for any real t that the t-cut of f is a subset of the t-cut of g. If f and g are also measurable and l is a measure on their domain, show that ~f ðtÞ  ~ gðtÞ for every t. Solution: f ð xÞ [ t ) gð xÞ [ t; so, Xðf [ tÞX ðg [ tÞ. By monotonicity of measure, lðXðf [ tÞÞ  lðX ðg [ tÞÞ, i.e. ~f ðtÞ  ~ gðtÞ.

Solutions

187

2.1.P3. Let ffn gn
1 be a sequence of functions on X such that, for every x 2 X and n 2 N, the inequality fn ð xÞ  fn þ 1 ð xÞ holds. For any t 2 R, denote the t-cut of fn by Sn ðtÞ. Show that, Sn ðtÞSn þ 1 ðtÞ for every n 2 N. If furthermore f is a function on X such that f ð xÞ ¼ limn!1 fn ð xÞ for every x 2 X, show that Xðf [ tÞ ¼ [ 1 n¼1 Sn ðtÞ. Hence conclude that if each fn is measurable, then f is measurable. (The latter part is true without the hypothesis that fn ð xÞ  fn þ 1 ð xÞ, as is to be proved in Problem 2.2.P4.) Solution: x 2 Sn ðtÞ ) fn ð xÞ [ t ) fn þ 1 ð xÞ [ t ) x 2 Sn þ 1 ðtÞ. This proves Sn ðtÞSn þ 1 ðtÞ for every n 2 N. For the next part, first note that, since f ð xÞ ¼ limn!1 fn ð xÞ and also fn ð xÞ  fn þ 1 ð xÞ for every n 2 N, it follows when f ðxÞ\1 that every N 2 N satisfies 0  f ðxÞ  fN ð xÞ and hence satisfies jf ðxÞ  fN ðxÞj ¼ f ðxÞ  fN ð xÞ. Now, let f (x) > t. Since f ð xÞ ¼ limn!1 fn ð xÞ, there exists some N 2 N such that either jf ðxÞ  fN ðxÞj\f ðxÞ  t or fN ð xÞ [ t according as f ð xÞ\1 or f ð xÞ ¼ 1. In the former case, as noted in the preceding paragraph, jf ðxÞ  fN ðxÞj ¼ f ðxÞ  fN ð xÞ and therefore f ðxÞ  fN ð xÞ\f ðxÞ  t and hence fN ð xÞ [ t, just as in the latter case. Thus x 2 SN ðtÞ in either case. Hence Xðf [ tÞ [ 1 n¼1 Sn ðtÞ. Conversely, consider S ð t Þ. There exists some N 2 N such that x 2 SN ðtÞ, which means x 2 [1 n¼1 n fN ð xÞ [ t. Since fN ð xÞ  f ð xÞ, it follows that f ð xÞ [ t. Thus [ 1 n¼1 Sn ðtÞXðf [ tÞ. Finally, if each fn is measurable, then Sn ðtÞ is measurable for every n and every t. By definition of r-algebra, it follows that the union [ 1 n¼1 Sn ðt Þ is measurable, whereby Xðf [ tÞ is measurable. Thus the function f is measurable. 2.1.P4. (Note: This problem is of interest in Combinatorics rather than Measure Theory, because it relates to the Inclusion-Exclusion Principle there; see Problem 3.1.P1.) Let A1 ; . . .; An be subsets of X. Using the symbol v½B to denote the characteristic function of a set B, show that v½A1 [    [ An  ¼

X

X

ðð1Þr1

1rn

v½Aj1 \    \ Ajr Þ:

1  j1 \\jr  n

Solution: First we note the two (obvious) equalities v½B1     v½Bn  ¼ v½B1 \    \ Bn  and

1  v½B ¼ v½Bc :

Now we invoke the purely algebraic identity (justified for the sake of the skeptical reader at the end of this solution), 1

Y 1jn

ð1  aj Þ ¼

X 1rn

ð1Þ

r1

X 1  j1 \\jr  n

! aj1    ajr :

188

Solutions

Upon substituting aj ¼ v½Aj  and using the first of the two equalities noted above, we find that the right side becomes X X ðð1Þr1 v½Aj1 \    \ Ajr Þ: 1rn

1  j1 \\jr  n

Also, each factor 1  aj on the left side is 1  v½Aj , which is v½Acj  by the second of the two equalities noted above. Using the first of the two equalities once again, we find that the product on the left side equals the characteristic function of the intersection Ac1 \    \ Acn . Therefore the left side equals the characteristic function of the complement of Ac1 \    \ Acn . However, this complement is nothing but the union A1 [    [ An . For the justification of the algebraic identity used above, it is sufficient to demonstrate that Y

X

1rn

1  j1 \\jr  n

ð1 þ aj Þ ¼ 1 þ

1jn

!

X

aj 1    aj r :

For n = 2, the equality becomes ð1 þ a1 Þð1 þ a2 Þ ¼ 1 þ a1 þ a2 þ a1 a2 , which is true. Assume it to hold for some n
2 and we shall show that it holds for n + 1. Since it has been assumed to hold for n, we deduce that Y Y ð1 þ aj Þ ¼ ½ ð1 þ aj Þ  ð1 þ an þ 1 Þ ¼ S þ T; 1jnþ1

1jn

where S ¼ 1þ

!

X

X

1rn

1  j1 \\jr  n

aj1    ajr

and T ¼ an þ 1 þ

!

X

X

1rn

1  j1 \\jr  n

aj1    ajr an þ 1 :

It is therefore sufficient to prove SþT ¼ 1þ

X

X

1rnþ1

1  j1 \\jr  n þ 1

! aj1    ajr :

ðAÞ

Solutions

189

We rewrite S by separating out the term with r = 1 as S ¼ 1 þ ða1 þ    þ an Þ þ U;

where U ¼

!

X

X

2rn

1  j1 \\jr  n

aj1    ajr ;

and also rewrite T by separating out the term with r = n as T ¼ an þ 1 þ V þ a1    an þ 1 ; X X where V ¼ 1  r  n1

! aj 1    aj r an þ 1 :

1  j1 \\jr  n

Now U can be expressed as U¼

!

X

X

2rn

1  j1 \\jr  n þ 1;jr 6¼n þ 1

aj1    ajr

and V can be expressed as V¼ ¼

!

X

X

2rn

1  j1 \\jr1  n

aj1    ajr1 an þ 1 !

X

X

2rn

1  j1 \\jr  n þ 1; jr ¼n þ 1

aj1    ajr :

These expressions for U and V show that UþV ¼

!

X

X

2rn

1  j1 \\jr  n þ 1

aj1    ajr :

It follows that S þ T ¼ 1 þ ða1 þ    þ an þ 1 Þ þ

X

X

2rn

1  j1 \\jr  n þ 1

! aj1    ajr

þ a1    an þ 1 :

But this is the same as (A) with the terms corresponding to r = 1 and r = n + 1 separated out. 2.1.P5. If l: F ! R is monotone as well as finitely additive, then show that it is also finitely subadditive in the sense that lðA [ BÞ  lðAÞ þ lðBÞ for measurable sets A and B, disjoint or not. Solution: Whether the measurable sets A and B are disjoint or not, lðA [ BÞ ¼ lðAÞ þ lðB \ Ac Þ because the sets A and B \ Ac are disjoint and measurable, with union equal to A [ B, and lðB \ Ac Þ  lðBÞ by virtue of monotonicity.

190

Solutions

2.1.P6. Let l be as in Example 2.1.3(b) and the functions f, g be defined on X = {a, b, c} as f ðaÞ ¼ 1; f ðbÞ ¼

pffiffiffi 3; f ðcÞ ¼ 2;

gð aÞ ¼

pffiffiffi 2; gðbÞ ¼ 1; gðcÞ ¼ 2:

Show that vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi u u sup f 2 u sup g2 u supZðf þ gÞ2 uZ uZ u 3 3 3 u u u fe2 þ t ge2 ; ðf þ gÞ2 \t t

g

0

0

0

but vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi u u sup f 2 u sup g2 u supZðf þ gÞ2 uZ uZ u u u u ge2 : ðf þ gÞ2 [ t fe2 þ t t

g

0

pffiffiffi Solution: ðf þ gÞðaÞ ¼ 1 þ 2; 8 8 > > < 6 e 2 f ðtÞ ¼ >4 > : 0

0

0

pffiffiffi ðf þ gÞðbÞ ¼ 1 þ 3;

0  t\1 1  t\3 3  t\4 4  t\1;

8 8 > > < 5 e 2 g ðtÞ ¼ >4 > : 0

ðf þ gÞðcÞ ¼ 4. Also, 0  t\1 1  t\2 2  t\4 4  t\1

8 pffiffiffi > 8 0p  ffiffit\3 þ 2 2 pffiffiffi > ffi < 6 3 þ 2 2p ffiffiffit\4 þ 2 3 and ðf þ gÞ2 ðtÞ ¼ > 4 4 þ 2 3  t\16 > : 0 16  t\1:

g

Hence 2 sup Z f

fe2 ¼ 8 þ 12 þ 4 ¼ 24;

0

2 sup Z g

ge2 ¼ 8 þ 5 þ 8 ¼ 21;

0

and 2 supðf Z þ gÞ

0

pffiffiffi pffiffiffi pffiffiffi pffiffiffi ðf þ gÞ2 ¼ 8ð3 þ 2 2Þ þ 6ð1 þ 2 3  2 2Þ þ 4ð12  2 3Þ

g

pffiffiffi pffiffiffi ¼ 78 þ 4 2 þ 4 3:

Solutions

191

It can be verified (without abject subservience to a calculator) that pffiffiffi pffiffiffi pffiffiffi pffiffiffi 78 þ 4 2 þ 4 3\ð241=3 þ 211=3 Þ3 but 78 þ 4 2 þ 4 3 [ ð241=2 þ 211=2 Þ2 as follows: pffiffiffi pffiffiffi 78 þ 4 2 þ 4 3\78 þ 4ð2 þ 3Þ ¼ 98. On the other hand, 21 [ 15:625 ¼ 2:53 . Therefore 211=3 [ 2:5 and hence 241=3 [ 2:5. It follows that 241=3 þ 211=3 [ 5, pffiffiffi pffiffiffi  3 which leads to 241=3 þ 211=3 [ 125. Thus 78 þ 4 2 þ 4 3\98\125\ ð241=3 þ 211=3 Þ3 . pffiffiffi pffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffi pffiffiffi 78 þ 4 2 þ 4 3 [ 78 þ 4ð 1:96 þ 2:89Þ ¼ 78 þ ð1:4 þ 1:7Þ [ 78 þ 4ð3Þ ¼ 90. pffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffi On the other hand, ð241=2 þ 211=2 Þ2 ¼ 45 þ 2 504\90, because 2 504\45, as can be seen from the inequality 4ð504Þ\452 , which is the same as 2016 < 2025. 2.1.P7. Show that a measure is countably subadditive in the following sense: If

ðX; F ; lÞ is a measure space and Aj j2N is a sequence of measurable sets, then P1   lð [ 1 j¼1 Aj Þ  j¼1 l Aj . Is the corresponding statement true regarding a finite sequence of measurable sets? c Solution: Let B1 ¼ A1 and, for k > 1, let Bk ¼ Ak \ ð [ k1 j¼1 Aj Þ . In other words, Bk ¼ Ak nðA1 [    [ Ak1 Þ. Then each Bk Ak and is measurable (by definition of r-algebra), so that lðBk Þ  lðAk Þ. Furthermore, the sets Bk are all disjoint and therefore by countable additivity of measure, we have l

1 [ j¼1

! Bj

¼

1 X j¼1

lðBj Þ 

1 X

lðAj Þ:

j¼1

But the definition of Bk also implies that [ nj¼1 Aj ¼ [ nj¼1 Bj for all n 2 N and hence 1 [1 j¼1 Aj ¼ [ j¼1 Bj . The same argument shows that the corresponding statement is true for a finite sequence. 2.1.P8. Let F be a r-algebra of subsets of a set X. Suppose l: F ! R þ is finitely additive and vanishes at ;. If it is also countably subadditive (as defined in Problem 2.1.P7), show that it is a measure. Solution: We need show only that l is countably additive. Since it is nonnegative and finitely additive, it is monotone. Let fAk gk
1 be a sequence of disjoint sets, A Þ
lð [ m 2 N by monoeach belonging to F . Then lð [ 1 k¼1 Ak Þ for each m P Pm k¼1 k m m A Þ
tonicity. But lð [ k¼1 Ak Þ ¼ k¼1 lðAk Þ. Therefore lð [ 1 k¼1 k k¼1 lðAk Þ for P m each m. It follows that lð [ 1 A Þ
lðA Þ. The reverse inequality is assured k k¼1 k k¼1 by the given countable subadditivity.

192

Solutions

2.1.P9. Suppose that, for each j 2 N, F j is a r-algebra of subsets of Xj . Define the family F of sets to consist of those subsets A of the union X ¼ [ 1 j¼1 Xj for which A \ Xj 2 F j for each j 2 N. Show that F is a r-algebra. Solution: Clearly, ;; X 2 F . For any AX, the intersection Ac \ Xj is the complement in Xj of the intersection A \ Xj . Therefore A 2 F ) Ac 2 F . Finally, let fAk gk
1 be a sequence of sets in F ; then for any j, we have 1 ð[1 k¼1 Ak Þ \ Xj ¼ [ k¼1 ðAk \ Xj Þ 2 F j . 2.1.P10. For each j 2 N, let ðXj ; F j ; lj Þ be a measure space and denote the union [1 j¼1 Xj by X. Set F ¼ fAX: A \ Xj 2 F j for each j 2 N} and for each A 2 F , set P lð A Þ ¼ 1 j¼1 lj ðA \ Xj Þ. Show that l is a measure. Solution: Obviously, l(A) is nonnegative and vanishes at ;. From Problem 2.1.P9, we know that F is a r-algebra. We shall first prove that l is finitely additive. It is sufficient to show that lðA [ BÞ ¼ lðAÞ þ lðBÞ for any two disjoint sets A; B 2 F . In view of the fact that each lj is finitely additive, we have lj ððA [ BÞ \ Xj Þ ¼ lj ðA \ Xj Þ þ lj ðB \ Xj Þ. Therefore lðA [ BÞ ¼

1 1  X   X  lj ðA [ BÞ \ Xj ¼ lj ðA \ Xj Þ þ lj ðB \ Xj Þ j¼1

¼

1 X

j¼1

lj ðA \ Xj Þ þ

j¼1

1 X

lj ðB \ Xj Þ ¼ lðAÞ þ lðBÞ:

j¼1

This proves the finite additivity of l, and therefore by Problem 2.1.P8, it remains only to prove its countable subadditivity. Consider a sequence fAk gk
1 of sets, each belonging to F . Thus Ak \ Xj 2 F j P for each j 2 N. Since lðAk Þ ¼ 1 j¼1 lj ðAk \ Xj Þ, what we want to prove is that P P 1 1 1 lð [ k¼1 Ak Þ  k¼1 ð j¼1 lj ðAk \ Xj ÞÞ. Since each lj is countably subadditive (by Problem 2.1.P7), we have lð

1 [

Ak Þ ¼

k¼1



1 X j¼1 1 X j¼1

lj ðð

1 [

Ak Þ \ Xj Þ ¼

k¼1

ð

1 X

1 X j¼1

lj ð

1 [

ðAk \ Xj ÞÞ

k¼1

lj ðAk \ Xj ÞÞ:

k¼1

When the order of summation is interchanged, the required inequality emerges. 2.1.P11. Let F be a r-algebra of subsets of a nonempty set X and Y any set with a map F: X!Y. Take G to be the family of subsets fBY: F 1 ðBÞ 2 F g of Y. Here, F−1(B) means {x 2 X : F(x) 2 B}, as usual. Show that G is a r-algebra. If l is a measure on F , show that the map mðBÞ: G ! R þ defined by mðBÞ ¼ lðF 1 ðBÞÞ is a measure on G.

Solutions

193

Solution: It is trivial why ;, Y 2 G. Suppose BY and Bc is the complement of B in Y. Then x 2 F 1 ðBc Þ , FðxÞ 2 Bc , FðxÞ 62 B , x 62 F 1 ðBÞ , x 2 F 1 ðBÞc and hence F 1 ðBc Þ ¼ F 1 ðBÞc : Consequently, B 2 G ) F 1 ðBÞ 2 F ) F 1 ðBÞc 2 F ) F 1 ðBc Þ 2 F ) Bc 2 G. Thus G is closed under complements. We shall show that it is also closed under countable unions. Let fBk gk
1 be a sequence of subsets of Y. Then x 2 F 1 ð

1 [

Bk Þ , FðxÞ 2

k¼1

k,x2

1 [

Bk , FðxÞ 2 Bk for some k , x 2 F 1 ðBk Þ for some

k¼1 1 [ 1

F ðBk Þ:

k¼1 1 1 Consequently, F 1 ð [ 1 k¼1 Bk Þ ¼ [ k¼1 F ðBk Þ. Now suppose each Bk 2 G. This 1 1 means each F ðBk Þ 2 F and it follows that [ 1 k¼1 F ðBk Þ 2 F , and hence that 1 1 1 F ð [ k¼1 Bk Þ 2 F ; i:e: [ k¼1 Bk 2 G. We proceed to argue that m is a measure. Certainly, m is nonnegative and vanishes at ;. We shall show that it is countably additive. Let fBk gk
1 be a sequence of 1 because for disjoint sets belonging   to1G. Then the sets F ðBk Þ are also disjoint, 1 j 6¼ k, x 2 F Bj \ F ðBk Þ ) F ð xÞ 2 Bj \ Bk ¼ ;: Since F 1 ð [ 1 k¼1 Bk Þ ¼ 1 1 [ k¼1 F ðBk Þ as proved above,

mð

1 [

1

Bk Þ ¼ l F ð

k¼1

1 [ k¼1

! Bk Þ

¼l

1 [ k¼1

! 1

F ðBk Þ

¼

1 1 X   X l F 1 ðBk Þ ¼ mðBk Þ: k¼1

k¼1

2.1.P12. Let F be a r-algebra of subsets of a set X, and E 2 F have the property that every subset of it belongs to F . (a) If BA [ E, show that Ac \ B 2 F . (b) If BA [ E, B 2 F , show that A \ B 2 F . (c) If BA [ E, AB [ E, B 2 F , show that A 2 F . Solution: (a) Ac \ BAc \ ðA [ EÞ ¼ ðAc \ AÞ [ ðAc \ EÞ ¼ ; [ ðAc \ EÞE. Since every subset of E belongs to F , we have Ac \ B 2 F . (b) A \ B ¼ ðA \ BÞ [ ðBc \ BÞ ¼ ðA [ Bc Þ \ B ¼ ðAc \ BÞc \ B. Since Ac \ B 2 F by part (a) and B 2 F by hypothesis, we have A \ B 2 F .

194

Solutions

(c) A ¼ A \ ðBc [ BÞ ¼ ðA \ Bc Þ [ ðA \ BÞ. Since A \ Bc 2 F by part (a) and A \ B 2 F by part (b), we get A 2 F . 2.1.P13. Prove that the result of Exercise 2.1.5 holds even if the set function l: F ! R þ is assumed to be only finitely additive provided it is monotone and vanishes at ;. Solution: Consider any real t and the t-cuts B and C of f and g respectively. Since l is monotone and nonnegative the hypothesis lðAc Þ ¼ 0 implies that lðB \ Ac Þ ¼ 0: Now B ¼ ðB \ Ac Þ [ ðB \ AÞ and finite subadditivity of l implies lðBÞ  lðB \ Ac Þ þ lðB \ AÞ ¼ lðB \ AÞ: But lðB \ AÞ  lðBÞ by monotonicity, so that lðB \ AÞ ¼ lðBÞ. This means ~f ðtÞ ¼ lðB \ AÞ. Similarly, ~gðtÞ ¼ lðC \ AÞ. By hypothesis, f and g agree on the set A, which implies that the t-cuts B and C satisfy B \ A ¼ C \ A.

2.2 Measurability and the Integral 2.2.P1. Prove Proposition 2.2.2(b). Solution: Suppose f is real-valued so that f is defined. As X ðf [ tÞ ¼ Xðf \  tÞ, measurability of f implies that of f in view of (ii) of part (a) of this proposition. Now consider any f : X ! R [ f1g. While f may not be defined, it is true nevertheless that jf ðxÞj [ t , f ðxÞ [ t or f ðxÞ\  t; and hence X ðjf j [ tÞ ¼ Xðf [ tÞ [ Xðf \  tÞ. It now follows from (ii) of part (a) of this proposition and the fact that a r-algebra is closed under finite unions that measurability of f implies that of j f j. 2.2.P2. If f : X ! R þ has the property that Xðf
tÞ is measurable for every t 2 Q, show that f is measurable. Solution: Consider any s 2 R. Then f ð xÞ [ s , f ð xÞ
t for some rational t > s. This means Xðf [ sÞ is the union of all sets Xðf
tÞ with t 2 Q and t > s. But these sets are measurable and are also countable in number (i.e. can be arranged in a sequence). Therefore their union, which has been shown to be Xðf [ sÞ, must be measurable. 2.2.P3. Let ffn gn
1 be a sequence of extended real-valued measurable functions on X. Then the function f defined on X as f ð xÞ ¼ inf fn ð xÞ, i.e. infffn ð xÞ: n 2 Ng, is measurable. Solution: Let x 2 X and n 2 N. Then f ð xÞ
t , fn ð xÞ
t for every n 2 N. So, Xðf
tÞ ¼ \ 1 n¼1 Xðfn
tÞ. By Proposition 2.2.2, each set Xðfn
tÞ is measurable and hence, so is their intersection. Thus Xðf
tÞ is measurable. This has been shown to be the case for all real t. Using Proposition 2.2.2 once again, we find that f is measurable.

Solutions

195

2.2.P4. In Problem 2.1.P3, show that the measurability of the limit function f holds without the hypothesis that fn ð xÞ  fn þ 1 ð xÞ. Solution: We are given a sequence ffn gn
1 of measurable functions and a function f such that f ð xÞ ¼ limn!1 fn ð xÞ for every x 2 X. We must show that the limit function f is measurable. For each x 2 X, f ðxÞ ¼ lim inf fn ðxÞ ¼ lim gn ðxÞ; n!1

n!1

where gn ð xÞ ¼ infffm ð xÞ: n  m 2 Ng. By Problem 2.2.P3, each function gn is measurable, and obviously, gn ð xÞ  gn þ 1 ð xÞ. It follows by Problem 2.1.P3 that f is measurable. 2.2.P5. If f : X ! R is measurable and /: R ! R is increasing, show that the composition /  f : X ! R is measurable. Solution: Xð/  f [ tÞ ¼ fx 2 X: f ð xÞ 2 fu 2 R: /ðuÞ [ tgg. Since / is increasing, fu 2 R: /ðuÞ [ tg is an interval (perhaps empty). For any interval I, the set fx 2 X: f ð xÞ 2 Ig is measurable by Proposition 2.2.2. Hence Xð/  f [ tÞ is measurable for any t. 2.2.P6. Suppose l is a measure on a set X. Prove that (a) For a measurable function f : X ! R þ and any real p > 0, the nonnegative function f p : X ! R þ is measurable; moreover, Z f ¼ 0 if and only if ~f ðtÞ ¼ 0 for every t [ 0; which, in turn, is equivalent to

R

f p ¼ 0 for any p > 0.

þ are measurable functions satisfying (b) RIf f : X !RR þ and R g: X ! R f ¼ 0 ¼ g, then ðf þ gÞ ¼ 0. 1

Solution: (a) The measurability of f p follows from the fact that Xðf p [ tÞ ¼ Xðf [ tp Þ. For the rest, the “if” part is clear. For the “only if” part, suppose for some t > 0 that ~f ðtÞ [ 0. Since ~f is a decreasing function, we have ~f ðsÞ
~f ðtÞ for 0  s  t and R R1 Rt R R therefore f ¼ 0 ~f
0 ~f
t~f ðtÞ [ 0. The equivalence of f ¼ 0 and f p ¼ 0 follows from the fact that ~f ðtÞ ¼ lðXðf [ tÞÞ ¼ 0

for every t [ 0

if and only if fep ðtÞ ¼ lðXðf p [ tÞÞ ¼ 0

for every t [ 0:

196

Solutions

R R (b) Suppose f ¼ 0 ¼ g. Then by (a), we have ~f ðsÞ ¼ 0 ¼ ~gðsÞ for every s > 0. By definition of a distribution, this means lðXðf [ sÞÞ ¼ 0 ¼ lðXðg [ sÞÞ for every s > 0. Now, consider an arbitrary t > 0. For any x 2 X, ðf þ gÞðxÞ [ t

)

either f ðxÞ [

t t or gðxÞ [ : 2 2

Therefore Xðf þ g [ tÞXðf [ 2t Þ [ Xðg [ 2t Þ. It follows by monotonicity and finite subaditivity proved in Problem 2.1.P5 Rthat lðXðf þ g [ tÞÞ ¼ 0. This is so for an arbitrary t > 0 and hence by (a), ðf þ gÞ ¼ 0. 2.2.P7. (a) Let f, g be nonnegative measurable functions on a space X with measure l, and f ð xÞ ¼ gð xÞ for all Rx 2 A,R where A is a measurable subset of X satisfying lðAc Þ ¼ 0. Show that f ¼ g. (b) Let f, g be measurable functions on a space X with measure l, and f ð xÞ ¼ gð xÞ for all x 2 A, where A is a measurable subset of lðAc Þ ¼ 0. If g is R X satisfying R integrable, show that f is also integrable and f ¼ g. Solution: (a) By Example 2.1.5, ~f ¼ ~ g. (b) Under the given hypothesis, f+ and g+ are nonnegative measurable functions þ þ all satisfying R x 2 A. It follows from part (a) that R þ R f þ ð xÞ ¼ g ð xÞR for f ¼ g . Similarly, f  ¼ g . 2.2.P8. Let f, g be nonnegative measurable functions on a space X with measure l, and f (x)  g(x) for Rall x 2R A, where A is a measurable subset of X satisfying lðAc Þ ¼ 0. Show that f  g. Solution: Let v be the characteristic function of A. Then f ð xÞ ¼ ðf vÞð xÞ for all x 2 A and similarly for g. Also, for any t
0, the t-cut of f v is the intersection of the t-cut of f with A; therefore f v is a measurable nonnegative function. Similarly, Rgv is Ra measurable nonnegative function. It follows by Problem 2.2.P7 that f ¼ ðf vÞ and similarly for g. But ðf vÞð xÞ  ðgvÞð xÞ for all x 2 X. Therefore by R R Proposition 2.2.8, ðf vÞ  ðgvÞ. 2.2.P9. For a function f : X ! R [ f1g, let A be the set Xðf \1Þ. Show that (a) (b) (c) (d)

If t < 0, then Xðf vA [ tÞ ¼ Ac [ ðA \ Xðf [ tÞÞ. If t
0, then Xðf vA [ tÞ ¼ A [ Xðf [ tÞ. The product f vA is real-valued. If f is measurable, then A is measurable and f vA is measurable.

Solution: (a) First consider any t 2 R. Suppose ðf vA ÞðxÞ [ t. Either x 2 Ac or x 2 A. If x 2 A, we have ðf vA ÞðxÞ ¼ f ðxÞ and therefore also f ðxÞ [ t, so that x 2 A \ Xðf [ tÞ. This proves the inclusion Xðf vA [ tÞAc [ ðA \ Xðf [ tÞÞ for all t 2 R. For t < 0, we shall prove the reverse inclusion. With this in view, consider any x 2 Ac [ ðA \ Xðf [ tÞÞ. If x 2 Ac , then ðf vA ÞðxÞ ¼ 0 [ t. If x 2 A \ Xðf [ tÞ, then ðf vA ÞðxÞ ¼ f ðxÞ [ t. This proves the reverse inclusion for t < 0.

Solutions

197

(b) Now suppose t
0 and ðf vA ÞðxÞ [ t. Then ðf vA ÞðxÞ [ 0 and therefore vA ðxÞ 6¼ 0. It follows that vA ðxÞ ¼ 1, so that x 2 A, and hence f ðxÞ ¼ ðf vA ÞðxÞ [ t. This proves the inclusion Xðf vA [ tÞA \ Xðf [ tÞ for t
0. To prove the reverse inclusion, consider any x 2 A \ Xðf [ tÞ. Since x 2 A, we have ðf vA ÞðxÞ ¼ f ðxÞ and since x 2 Xðf [ tÞ, we also have f (x) > t. Therefore ðf vA ÞðxÞ [ t, thereby proving the reverse inclusion. (c) If x 2 A, then ðf vA ÞðxÞ ¼ f ðxÞ and f (x) < 1. If x 2 Ac , then ðf vA ÞðxÞ ¼ 0\1. c (d) Since A ¼ ð \ 1 n¼1 Xðf [ nÞÞ and f is measurable, the set A is measurable. That f vA is real measurable now follows from parts (a) and (b). 2.2.P10. Let the function f : X ! R [ f1g have the property that X(t < f < 1) is measurable for every t 2 R. Show that f is measurable. (The converse is trivial.) Solution: It is true for any x 2 X that f ð xÞ\1 , n\f ðxÞ\1 for some n 2 Z. This means Xðf \1Þ ¼ [ n2Z Xðn\f \1Þ. From the given property of f, it follows that Xðf \1Þ is measurable and therefore its complement Xðf ¼ 1Þ is also measurable. Since Xðf [ tÞ ¼ Xðt\f \1Þ [ Xðf ¼ 1Þ, it further follows that X (f > t) is measurable. 2.2.P11. If f : X ! R þ and g: X ! R þ are both measurable, show that their product fg: X ! R þ is measurable. Solution: Let A = X(f < 1) and B = X(g < 1), both measurable by Problem 2.2.P9 (d). By Problem 2.2.P10, it is sufficient to show that X(t < fg < 1) is measurable for every t 2 R. First consider the possibility that t < 0. Since fg
0 > t everywhere, the set in question is then the same as X(fg < 1). Now f (x)g(x) < 1 if and only if either f (x) < 1 as well as g(x) < 1 or one among f (x) and g(x) is 0. Consequently, Xðfg\1Þ ¼ ðA \ BÞ [ Xðf ¼ 0Þ [ Xðg ¼ 0Þ; which is measurable. So, X(t < fg < 1) is measurable when t < 0. Next consider the possibility that t
0. When this is so, Xðt\fg\1Þ ¼ Xð0\f \1Þ \ Xð0\g\1Þ \ Xðf vA gvB [ tÞ: The first two sets on the right side are measurable. Since f vA and gvB are real-valued measurable functions by Problem 2.2.P9(c) and (d), their product is measurable by Theorem 2.2.3(c), as a consequence of which the third set on the right is also measurable. It follows that Xðt\fg\1Þ is again measurable when t
0. 2.2.P12. For an integrable function f : X ! R , show that Z a  lðXðj f j [ aÞÞ  j f j for all a [ 0:

198

Solutions

Solution: It is sufficient to prove the inequality for a nonnegative measurable function. Since ~f is a decreasing function, we have ~f ðtÞ
~f ðaÞ for 0  t  a. This Ra R1 yields a  ~f ðaÞ  0 ~f  0 ~f for all a > 0. However, ~f ðaÞ ¼ lðXðf [ aÞÞ and R R1 ~ f. 0 f ¼ 2.2.P13. Suppose f : X ! R þ is a measurable function such that, for some a > 0, the function min{f,a} has a finite integral. Show that f has a finite integral if and P ~ only if the series 1 k¼1 f ðkÞ converges. Solution: Denote min{f,a} by g. The distribution ~g agrees with ~f on the interval [0, Ra R1 R Ra a) and is 0 on [a, 1). Therefore 0 ~f ¼ 0 ~ g ¼ 0 ~g ¼ g, which is given to be R1 R1 finite. One consequence of this is that 0 ~f is finite if and only if a ~f is finite. Ra R1 Another consequence is that ~f ðaÞ is finite and hence 1 ~f ðor a ~f Þ is finite, so that R1 R1 ~ ~ a fR is finite if and only if 1 fR is finite. These two consequences together show 1~ 1 that 0 f is finite if and only if 1 ~f is finite. The equivalence of the finiteness of P1 this integral and the series k¼1 ~f ðkÞ is now just the integral test for the convergence of a series of decreasing nonnegative terms (Theorem 10.2.4 of Shirali and Vasudeva [9] or Example 8 on pp.138–9 of Rudin [7]).

2.3 The Monotone Convergence Theorem 2.3.P1. (Needed in Problem 3.2.P1 and Theorem 4.3.4) Let l beR a measure on X and f a nonnegative measurable function such that f ¼ 0. Prove lðXðf [ 0ÞÞ ¼ 0. Solution: It follows by Problem 2.2.P6 that lðXðf [ tÞÞ ¼ 0 for every t > 0. Now, f(x) > 0 if and only if f ð xÞ [ 1n for some n 2 N, which means 1 1 1 Xðf [ 0Þ ¼ [ 1 n¼1 Xðf [ nÞ. Since Xðf [ nÞXðf [ n þ 1Þ, inner continuity yields  1 lðXðf [ 0ÞÞ ¼ lim l Xðf [ Þ : n!1 n But each lðXðf [ n þ1 1ÞÞ ¼ 0 according to what has been proved earlier. 2.3.P2. Let ffn gn
1 be a sequence of real-valued functions on an arbitrary nonempty set X. Suppose f is a real-valued function on X such that, for every x 2 X, some subsequence of the real sequence ffn ð xÞgn
1 converges to f (x). Show for any real t that Xðf [ tÞ [ 1 n¼1 Xðfn [ tÞ. Solution: Consider any x 2 X(f > t), which means f (x) > t. Since the real sequence ffn ð xÞgn
1 has some subsequence converging to the real number f (x), there exists an n 2 N such that jfn ðxÞ  f ðxÞj\f ðxÞ  t. This inequality implies f ðxÞ  fn ð xÞ\ f ðxÞ  t, which in turn implies fn(x) > t. Thus there exists an n 2 N such that fn ð xÞ [ t, i.e. x 2 Xðfn [ tÞ.

Solutions

199

2.3.P3. In a measure space, suppose that every set of infinite measure contains a subset of finite positive measure. Show that every set of infinite measure contains a subset of arbitrarily large finite measure. Hint: For a given set of infinite measure, show that there exists a subset whose measure is the supremum of the measures of all subsets of finite measure. If its measure is finite, then the measure of its complement in the given set must be infinite. Solution: Denote the measure by l. Consider any measurable set A with l(A) = 1 and set a ¼ supflðBÞ: BA; 0\lðBÞ\1g. Then a > 0 and we have to prove that a = 1. Assume otherwise. For each n 2 N, there exists a Bn A such that a  1n \lðBn Þ\1. The set B ¼ [ 1 n¼1 Bn is measurable and l(B) = a by Proposition 2.3.1. Since lð AÞ ¼ lðBÞ þ lðAnBÞ ¼ a þ lðAnBÞ and a has been assumed finite we have lðAnBÞ ¼ 1. According to the hypothesis, AnB contains a subset C such that 0 < l(C) < 1. Since B and C are disjoint, lðC [ BÞ ¼ lðCÞ þ lðBÞ ¼ lðCÞ þ a. Since C [ BA and 0\lðCÞ þ a\1, this is contrary to the definition of a. 2.3.P4. Let l be a measure. For a sequence of measurable sets fSn gn
1 with lðS1 Þ\1, show that lð \ 1 n¼1 Sn Þ ¼ limn!1 lðSn Þ provided that Sn Sn þ 1 for every n. (outer continuous) Solution: Let Tn ¼ S1 \ Scn . Since Sn S1 , the set S1 is the disjoint union Sn [ Tn , where Tn is also measurable. Therefore lðS1 Þ ¼ lðTn Þ þ lðSn Þ. Since lðS1 Þ\1, we can rewrite this as lðTn Þ ¼ lðS1 Þ  lðSn Þ:

ðSol:1Þ

Now, 1 [ n¼1

Tn ¼

1 [

ðS1 \ Scn Þ ¼ S1 \

n¼1

1 [

Scn ¼ S1 \ ð

n¼1

1 \

Sn Þc

n¼1

and 1 \

Sn S1 :

n¼1

An argument similar to that leading to (Sol.1) now leads to lð

1 [

n¼1

Tn Þ ¼ lðS1 Þ  lð

1 \ n¼1

Sn Þ:

ðSol:2Þ

200

Solutions

Moreover, T1 T2     and hence by inner continuity (Proposition 2.3.1) and (Sol.1), lð

1 [

Tn Þ ¼ lim lðTn Þ ¼ lðS1 Þ  lim lðSn Þ: n!1

n¼1

n!1

ðSol:3Þ

The required equality follows from (Sol.2), (Sol.3) and the finiteness of lðS1 Þ.

3.1 Simple Functions 3.1.P1. Let l be a measure on X and A1 ; . . .; An be measurable subsets of X such that lðA1 [    [ An Þ\1. Show that the Inclusion-Exclusion Principle holds: ! X X r1 ð1Þ lðAj1 \    \ Ajr Þ : lðA1 [    [ An Þ ¼ 1rn

1  j1 \\jr  n

If it is not given that lðA1 [    [ An Þ\1, what is the best one can say? Solution: In the equality established in Problem 2.1.P4, transpose all terms with even r to other side, so that all powers of −1 can be taken to be 1. Then apply Proposition 3.1.8. Convert the integrals of characteristic functions to measures of the corresponding sets. If lðA1 [    [ An Þ\1 all terms are finite and one may therefore transpose them back to their original positions. If it is not given that lðA1 [    [ An Þ\1, the transposing back is impermissible and one can only say that X X lðA1 [    [ An Þ þ ð lðAj1 \    \ Ajr ÞÞ ¼

X

1  r  n and r even 1  j1 \\jr  n

ð

X

lðAj1 \    \ Ajr ÞÞ:

1  r  n and r odd 1  j1 \\jr  n

3.1.P2. When X ¼ fa; b; c; d g and A ¼ X; B ¼ fc; d g; C ¼ fd g, find the canonical form of the simple function s ¼ vA þ 2vB þ vC . Solution: Since a and b belong only to A, we have sðaÞ ¼ sðbÞ ¼ 1. Since c belongs to only A and B, we have sðcÞ ¼ 1 þ 2 ¼ 3. Lastly, since d belongs to all three sets, sðd Þ ¼ 1 þ 2 þ 1 ¼ 4. To summarize, sðaÞ ¼ sðbÞ ¼ 1; sðcÞ ¼ 3

and

sðdÞ ¼ 4:

Thus s takes the three values 1, 3, 4. Now, A1 ¼ Xðs ¼ 1Þ ¼ fa; bg; A2 ¼ Xðs ¼ 3Þ ¼ fcg and So, the canonical form is

A3 ¼ Xðs ¼ 4Þ ¼ fdg:

Solutions

201

s ¼ 1  vA1 þ 3  vA2 þ 4  vA3 ¼ 1  vfa;bg þ 3  vfcg þ 4  vfdg : 3.1.P3. Does the data of Exercise 3.1.9 determine the sets A, B, C uniquely? Solution: No. There are at least two possibilities consistent with the data: (in abbreviated notation) abe, ace, bde and also abe, ade, bce. 3.1.P4. Each of three junior managers in a company submits a list of three cities to the manager, because the senior manager wants them to. The manager goes through the lists and informs the senior manager that there are five cities in all the three lists taken together, and that two cities appear in two lists each and two others in three lists each. When the senior manager hears this, she is furious—Why? Solution: Let l denote counting measure on the set of all five cities and let A, B, C R If we take s ¼ vA þ vB þ vC , then by Proposition 3.1.8, R denote R the Rthree lists. s ¼ vA þ vB þ vC ¼ lðAÞ þ lðBÞ þ lðCÞ ¼ 3 þ 3 þ 3 ¼ 9. But according to the information provided by the manager, lðXðs ¼ 2ÞÞ
2 and lðXðs ¼ 3ÞÞ
2. R Therefore s
2  2 þ 3  2 ¼ 10 [ 9. P 3.1.P5. Let aj be a series of nonnegative terms. Show thatRthere is a measure l P P on N, every subset measurable, such that lðAÞ ¼ j2A aj and f ¼ 1 j¼1 aj f ðjÞ for every nonnegative function f on N. (Note that when each aj is 1, this is the counting measure.)   Solution: Consider the sequence of measure spaces Xj ; F j ; lj ; j 2 N, in which Xj ¼ f jg 2 N; F j is the only possible r-algebra of subsets of a set containing a   single element and lj Xj ¼ aj . Then [ 1 X ¼ N. By Problem 2.1.P10, there is a j¼1 j

measure l on N with domain F ¼ AX: A \ Xj 2 F j for each j 2 N and for  P1  each A 2 F ; lðAÞ ¼ j¼1 lj A \ Xj . However, in the present situation, F is  P1  seen to consist of all subsets of N, and moreover, j¼1 lj A \ Xj ¼ P P1 j2A aj . j¼1 lj ðA \ f jgÞ ¼ Now let f be any nonnegative function on N but real-valued. For any n 2 N, define fn to agree with f on f1; 2; . . .; ng and vanish outside the set f1; 2; . . .; ng. each fn is a simple nonnegative-valued function with integral RThen P fn ¼ nj¼1 aj f ðjÞ by Proposition 3.1.7 or by Proposition 3.1.8. But we have fn ðjÞ  fn þ 1 ðjÞ and f ðjÞ ¼ limRn!1 fn ðjÞ for every j 2 N. It follows by the Monotone R P Convergence Theorem that f ¼ limn!1 fn ¼ 1 j¼1 aj f ðjÞ. Next, consider the possibility when f takes 1 as a value. If for some j, we have R P f ðjÞ ¼ 1 with aj [ 0, then 1 a f ðjÞ ¼ 1 and also f ¼ 1, the latter because j Rj¼1 f [ Kvf jg for any K > 0, and Kvf jg ¼ Klðf jgÞ ¼ Kaj . If there is no such j, then aj ¼ 0 whenever f ðjÞ ¼ 1. This means lðAc Þ ¼ 0, where AN is the subset on c which f is finite-valued. Define R R/: N ! N to agree with f on A and be zero on A . Then by Problem 2.2.P7, f ¼ /. But / is nonnegative real-valued and therefore R P1 P1 P /¼ 1 j¼1 aj /ðjÞ. We shall argue that j¼1 aj /ðjÞ ¼ j¼1 aj f (j). In fact, it is true that aj /ðjÞ ¼ aj f (j) or each j, the reason being that, if f ðjÞ\1, then /ðjÞ ¼ f ðjÞ, and if f ðjÞ ¼ 1, thenR aj ¼ R0 and so, aj /ðjÞ ¼ 0  /ðjÞ ¼ 0 ¼ 0  1 ¼ P P1 aj f ðjÞ. We have thus shown that f = / ¼ 1 j¼1 aj /ðjÞ ¼ j¼1 aj f (j).

202

Solutions

3.1.P6. Show that the following sequence in R diverges: (

1 X j¼1

n jðn þ jÞ

) n2N

1 Solution: Let l be the measure on N such R that lðf jgÞ ¼ j2 . By Problem 3.1.P5, such a measure is possible and the integral / of any nonnegative function / on N is R P1 1 P nj n fn ¼ 1 j¼1 j2 /ðjÞ. Therefore the function fn ðjÞ ¼ n þ j has integral j¼1 jðn þ jÞ. Now, fn ðjÞ  fn þ 1 ðjÞ and n!1 fn ðjÞ ¼ j for every j 2 N, and the limit function f ðjÞ ¼ j R lim P 1 has integral f ¼ 1 j¼1 j ¼ 1. By the Monotone Convergence Theorem, R R limn!1 fn ¼ f ¼ 1. 3.1.P7. Let l be a measure on X and let A1 ; . . .; An , where n
5, be measurable subsets of X such that whenever 1  j1 \j2 \j3 \j4 \j5  n, the intersection Aj1 \ Aj2 \ Aj3 \ Aj4 \ Aj5 of the corresponding subsets is empty. (In symbol-free terminology, the intersection of sets with any five distinct indices is empty.) Show that

X

X

lðAi Þ  lðA1 [    [ An Þ þ

  l Aj1 \ Aj2 :

1  j1 \j2  n

i

Solution: If lðA1 [    [ An Þ ¼ 1, there is nothing to prove. So, assume the contrary. Then by the Inclusion-Exclusion Principle of Problem 3.1.P1, we have lðA1 [    [ An Þ ¼ þ

X

lðAi Þ 

  l Aj1 \ Aj2

1  j1 \j2  n

i

X

X

X

  l Aj1 \ Aj2 \ Aj3 

1  j1 \j2 \j3  n

  l Aj1 \ Aj2 \ Aj3 \ Aj4 :

1  j1 \j2 \j3 \j4  n

The sets Aj1 \ Aj2 \ Aj3 \ Aj4 are disjoint and their union is contained in [



 Aj1 \ Aj2 \ Aj3 :

1  j1 \j2 \j3  n

It follows that X

  l Aj1 \ Aj2 \ Aj3 \ Aj4  l

1  j1 \j2 \j3 \j4  n

[



1  j1 \j2 \j3  n



X

1  j1 \j2 \j3  n

Aj1 \ Aj2 \ Aj3



  l Aj1 \ Aj2 \ Aj3 :

!

Solutions

203

Hence the sum of the last two summations in the above expression for lðA1 [    [ An Þ is greater than or equal to 0. Consequently, X X   lðA1 [    [ An Þ
lðAi Þ  l Aj1 \ Aj2 ; 1  j1 \j2  n

i

which leads at once to the inequality sought to be proved. 3.1.P8. (a) If A1 ; . . .; A4 are respectively {2, 3, 4}, {1, 3, 4}, {1, 2, 4}, {1, 2, 3}, then the intersection of all four sets is obviously empty. With counting   P measure l, compute 4j¼1 l Aj and lð [ 4j¼1 Aj Þ. Do the same when A1 ; . . .; A4 are respectively {1, 2, 3, 4}, {1, 2, 4}, {1, 3}, {2, 3, 4}, which also have empty intersection. of (b) Give an example of five distinct sets A1 ; . . .; A5 such that

the intersection  P5   5 any four among them is empty and j¼1 l Aj ¼ 3l [ j¼1 Aj , where l again means the counting measure. Solution: (a) When A1 ; . . .; A4 are respectively {2, 3, 4}, {1, 3, 4}, {1, 2, 4}, {1, 2, 3}, P4   4 we have j¼1 l Aj ¼ 3 þ 3 þ 3 þ 3 ¼ 12 and lð [ j¼1 Aj Þ ¼ 4. When A ; . . .; A are respectively {1, 2, 3, 4}, {1, 2, 4}, {1, 3}, {2, 3, 4}, we have P1 4  4  4 j¼1 l Aj ¼ 4 þ 3 þ 2 þ 3 ¼ 12 and lð [ j¼1 Aj Þ ¼ 4. (b) Let A1 ; . . .; A5 be respectively {1, 2, 5}, {1, 2, 3}, {3, 4, 5}, {2, 4, 5}, {1, 3, 4 }. It is a bit laborious to check that the intersection of any four among them is P empty. It is simple to verify that 5j¼1 lðAj Þ ¼ 3 þ 3 þ 3 þ 3 þ 3 ¼ 15 and lð [ 5j¼1 Aj Þ ¼ 5. 3.1.P9. Let l be a measure on X and let A1 ; . . .; An , where n
4, be measurable subsets of X such that whenever 1  j1 \j2 \j3 \j4  n, the intersection Aj1 \ Aj2 \ Aj3 \ Aj4 of the corresponding subsets is empty. (In symbol-free terminology, the intersection of sets with any four distinct indices is empty.) Suppose also that lðA1 [    [ An Þ\1. Show that X

lðAj Þ ¼ lð

j

[

Aj Þ þ l

1  j1 \j2  n





Aj 1 \ Aj 2



!



l Aj1 \ Aj2 ¼ l

[



Aj 1 \ Aj 2

X

þ

1  j1 \j2  n

j

X

[



!

1  j1 \j2 \j3  n

þ2

1  j1 \j2  n

j

lðAj Þ  3lð

X 1  j1 \j2 \j3  n

and that X

  l Aj 1 \ Aj 2 \ Aj 3 ;

[ j

Aj Þ:

  l Aj 1 \ Aj 2 \ Aj 3

204

Solutions

Solution: All summations, unions and intersections will be understood to be taken over all distinct indices. By the Inclusion-Exclusion Principle of Problem 3.1.P1, we have [ X   X   X   lð Aj Þ ¼ l Aj1 \ Aj2 þ l Aj1 \ Aj2 \ Aj3 : l Aj  The sets Aj1 \ Aj2 are indexed by two-element subsets fj1 ; j2 gf1; 2; . . .; ng and may be denoted by Ba , where the index a ranges over the set of two-element subsets fj1 ; j2 gf1; 2; . . .; ng. Then X X   lðBa Þ has the same meaning as l Aj1 \ Aj2 and lð

[

Ba Þ has the same meaning as l

[ 

Aj1 \ Aj2

 ;

facts which will be used below. Now, the number of two-element subsets of a set of 3 elements is 3 C2 ¼ 3 and therefore the union of four distinct two-element subsets of f1; 2; . . .; ng must contain at least four elements. It follows that the intersection of sets Ba with four distinct a must be empty. Therefore, upon applying the Inclusion-Exclusion Principle of Problem 3.1.P1 to the sets Aj1 \ Aj2 ¼ Ba , we get

[  X X X l Ba ¼ lð B a 1 \ B a 2 Þ þ lðBa1 \ Ba2 \ Ba3 Þ: lðBa Þ  This equality may be written as X   X l Aj1 \ Aj2 ¼ lðB Þ

[ a X X ¼l Ba þ lðBa1 \ Ba2 \ Ba3 Þ: lðBa1 \ Ba2 Þ  The summations on the right will be examined in this paragraph and the next. P Consider any one term in the summation lðBa1 \ Ba2 Þ. We have a1 ¼ fj1 ; j2 g and a2 ¼ fj3 ; j4 g for some j1 ; j2 ; j3 ; j4 2 f1; 2; . . .; ng. If fj1 ; j2 ; j3 ; j4 g consists of four distinct elements, then Ba1 \ Ba2 is the intersection of sets Aj with four distinct indices and is therefore empty. Therefore the summation need extend only over those a1 ; a2 whose union consists of precisely three distinct indices from f1; 2; . . .; ng. If a1 [ a2 ¼ fj1 ; j2 ; j3 g, then Ba1 \ Ba2 ¼ Aj1 \ Aj2 \ Aj3 . This means   that the term lðBa1 \ Ba2 Þ equals l Aj1 \ Aj2 \ Aj3 . Moreover, there are precisely two other pairs of distinct a1 ; a2 whose union is fj1 ; j2 ; j3 g. We deduce from this that

Solutions

205

X

lðBa1 \ Ba2 Þ ¼ 3

X   l Aj1 \ Aj2 \ Aj3 :

P Next, consider any one term in the summation lðBa1 \ Ba2 \ Ba3 Þ. It can be nonzero only if a1 [ a2 [ a3 ¼ fj1 ; j2 ; j3 g for some distinct j1 ; j2 ; j3 2 f1; 2; . . .; ng. When this so, Ba1 \ Ba2 \ Ba3 ¼ Aj1 \ Aj2 \ Aj3 . We need consider only such terms. So, a term in the summation under discussion is equal to a uniquely determined P  term in the sum l Aj1 \ Aj2 \ Aj3 . Besides, the converse can be shown to be true. The converse is the assertion is that, given distinct j1 ; j2 ; j3 2 f1; 2; . . .; ng, the three distinct two-element sets a1 ; a2 ; a3 such that a1 [ a2 [ a3 ¼ fj1 ; j2 ; j3 g are uniquely determined. This is true because, being a three-element set, fj1 ; j2 ; j3 g has only three distinct two-element subsets (i.e. 3 C2 ¼ 3). Consequently, X X   lðBa1 \ Ba2 \ Ba3 Þ ¼ l Aj1 \ Aj2 \ Aj3 : Combining the preceding three equalities, we obtain

[  X  X    Ba þ 2 l Aj1 \ Aj2 ¼ l l Aj1 \ Aj2 \ Aj3

[  X   ¼l ðAj1 \ Aj2 Þ þ 2 l Aj1 \ Aj2 \ Aj3 : This is the second of the inequalities asked for in the problem. Combining it with the very first equality derived above from the Inclusion-Exclusion Principle, we further obtain

[ 

[  X    X   Aj þ l l Aj ¼ l Aj1 \ Aj2 þ l Aj1 \ Aj2 \ Aj3 : j

This proves the first equality that was asked for in the problem. To arrive at the inequality, we need only show that the second and third terms on the right side in the equality that has just been proved are each bounded above by the first. For the second term, this is obvious because the union therein is a subset of the union in the first term. The third term is handled as follows. The sets whose measures are being taken are disjoint and therefore

[  X    l Aj1 \ Aj2 \ Aj3 ¼ l Aj1 \ Aj2 \ Aj3 : However, [

 [ Aj1 \ Aj2 \ Aj3  Aj :

206

Solutions

3.2 Other Measurable Functions 3.2.P1. Let l beRa measure on X and f, g be integrable functions such that f
g R everywhere and f ¼ g. Prove lðXðf [ gÞÞ ¼ 0. / ¼ f  g is nonnegative and, by Theorem 3.2.5, RSolution: The R function R ðf  gÞ ¼ f  g, which is given to be 0. The result now follows by Problem 2.3.P1. 3.2.P2. Let X ¼ fa; bg with counting measure. Let ffn gn
1 be the sequence of functions on X defined by fn ¼ vfag if n is even and vfbg if n is odd. Find Z  lim inf fn n!1

Z and

fn :

lim inf n!1

Solution: For each x 2 X, the sequence ffn ðxÞgn
1 alternates between 0 and 1, starting with 0 or 1 depending on whether x = aRor x = b. Hence lim inf n!1 fn is 0 everywhere and therefore has integral 0. But fn ¼ 1 for every n and therefore R lim inf n!1 fn ¼ 1. 3.2.P3. In this problem, ðX; F ; lÞ is a measure space and f, g are measurable real-valued functions on X. Then the nonnegative real-valued function jfgj is measurable by Theorem 2.2.3(c) and Proposition 2.2.2(b); also, for any real p > 0 and q > 0, the real-valued functions jf þ gj; jf þ gjp ; j f jp and jgjp are measurable by Theorem 2.2.3(b), Proposition 2.2.2(b) and Problem 2.2.P6(a). If p > 1, then jf þ gjp1 is also measurable. (a) Suppose p, q are positive real numbers such that 1p þ 1q ¼ 1. Show that Z

jfgj 

Z

jfj

p

1p Z

j gj

q

1q

:

This is known as Hölder's Inequality. (b) Suppose p is a real number such that p
1. Show that Z

j f þ gj

p

1p



Z

jfj

p

1p

þ

Z

jgj

p

1p

:

This is known as Minkowski's Inequality. Solution: (a) If one of the integrals on the right side is 0, then by Problem 2.2.P6, the integral on the left side is also 0; therefore we may assume that both the integrals on the right side are positive. Then if one of them is 1, the right side is 1, and there is nothing to prove. So, we also assume both of them to be finite. R 1=p R 1=q Set F ¼ j f j= j f jp and G ¼ jgj= jgjp . Then by homogeneity (secR q R p ond part of Theorem 3.2.2), F ¼ 1 ¼ G . By the inequality

Solutions

207

ab 

1 p 1 q a þ b for any nonnegative a and b p q

(easily proved by considering the derivative of /ðnÞ ¼ 1=q þ n=p  n1=p for n 2 ð0; 1Þ), we have FG  F p =p þ Gq =q.R Employing Proposition 2.2.8 and both parts of Theorem 3.2.2, we obtain FG  1p þ 1q ¼ 1. This leads to the required inequality upon using the second part of Theorem 3.2.2 once again. (b) If p = 1, then the required inequality is a straightforward consequence of Proposition 2.2.8. So, suppose p > 1. We need argue only the case when both integrals on the right side are finite. When this is so, the inequality   ða þ bÞp þ 1  2p ap þ 1 þ bp þ 1 for all a; b; p
0 (easily proved by considering the derivative of /ðxÞ ¼ ðx þ bÞp þ 1  2p ðxp þ 1 þ bp þ 1 Þ; x 2 ½0; b) shows that jfRþ gjp  2p1 ðj f jp þ jgjp Þ and hence by Proposition 2.2.8 and Theorem 3.2.2, jf þ gjp is also finite, a fact needed later in this proof. Since jf þ gjp ¼ jf þ gjp1 jf þ gj  jf þ gjp1 ðj f j þ jgjÞ; R R R we have jf þ gjp  jf þ gjp1 j f j þ jf þ gjp1 jgj. Upon applying p , so that 1p þ 1q ¼ 1 and Hölder’s Inequality of part (a) here, with q ¼ p1 pq  q ¼ p, we get Z 1p Z 1q Z 1p Z 1q Z p1 p qðp1Þ p p ¼ j f þ gj j f þ gj : j f þ gj j f j  jfj jfj R 1 R 1 jf þ gjp1 jgj  jgjp p jf þ gjp q . These two inequalities R R R R 1 1 1 jointly imply that jf þ gjp  ðð j f jp Þp þ ð jgjp Þp Þð jf þ gjp Þq . As we have already noted that the integral on the left side is finite, we may divide both R 1 sides by jf þ gjp q , whereupon the desired inequality follows. Similarly,

R

3.2.P4. (Needed in Proposition 7.7.5) Let f be an integrable function on X, where ðX; F ; lÞ is a measure space. Given any e > 0, show  that there exists a d > 0 such R that every A 2 F with lðAÞ\d satisfies  ðvA f Þ\e. R  R Solution: Since  f   j f j, we may assume f
0. By Theorem 3.2.1, there exists an increasing sequence fsn gn
1 of simple functions such that sn ðxÞ ! f ðxÞ as n!1 for every x 2 X. By the Monotone Theorem 2.3.2, R Convergence R e f ¼ limn!1 sn . Fix any N 2 N such that 0  ðf  sN Þ\ 2. Then Z Z Z e for any A 2 F ; ðvA f Þ  ðvA sN Þ  ðf  sN Þ\ 0 2 because 0  vA ðf  sN Þ  f  sN everywhere. Being a simple function, sN is e bounded above by some positive real number K. Set d ¼ 2K . Then any A 2 F with

208

Solutions

R R lðAÞ\d satisfies 0  ðvA sN Þ  K vA ¼ KlðAÞ\Kd\ 2e. Together with the R inequality displayed above, this leads to 0  ðvA f Þ\e. 3.2.P5. Show that the “dominated” hypothesis of the Dominated Convergence Theorem 3.2.6, namely that, for some integrable g, the inequality jfn ðxÞj  gðxÞ holds for every x 2 X, cannot be dropped. Solution: The same example as in Example 2.3.3 for the Monotone Convergence Theorem will serve the purpose. P 1 n j1 3.2.P6. For each fixed n 2 N, the series 1 converges to n þn 1. Show that j¼1 n n þ 1  j1 P1 1 n any series for every n 2 N and every j 2 N j¼1 aj satisfying aj
n n þ 1 diverges. Solution: Consider the measure l on N such that lðf jgÞ ¼ j12 . By Problem 3.1.P5, R such a measure is possible and the integral / of any nonnegative function / on N is   R P1 1 2 1 n j1 has integral fn ¼ n þn 1 and the j¼1 j2 /ðjÞ. Therefore the function fn ðjÞ ¼ j n n þ 1 R P function gðjÞ ¼ j2 aj has integral g ¼ 1 j¼1 aj . If this series were to be convergent,  j1 then g would be integrable and it would follow from the inequality aj
1n n þn 1 and R the Dominated Convergence Theorem 3.2.6 that n þn 1 ! limn!1 fn . However, this not theP case, because limn!1 fn is 0 everywhere. This contradiction shows that the series 1 j¼1 aj cannot converge if the given inequality holds. 3.2.P7. Show that the following sequence converges to 0: (

1 X j¼1

n 2 jðn þ j2 Þ

) n2N

Solution: Consider the same measure as in the solution of Problem 3.2.P6 but with functions fn and g given by fn ðjÞ ¼ j2

n jðn2 þ j2 Þ

and

gðjÞ ¼ 1:

R Then limn!1 fn is 0 everywhere. The integral / ofR any nonnegative function / on R P1 P 1 P1 1 N is j¼1 j2 /ðjÞ. Therefore fn ¼ j¼1 jðn2 nþ j2 Þ and g ¼ 1 j¼1 j2 . In particular, g is integrable. Also, 0  fn ðjÞ  gðjÞ for all n and j. It follows by theR Dominated Convergence Theorem 3.2.6 that the given sequence converges to limn!1 fn , which is 0. 3.2.P8. Given any measure space, show that a nonnegative measurable function f satisfies Z

f ¼ sup

Z

 s: 0  s  f ; where s is simple :

Hint: Use Theorem 3.2.1 and the Monotone Convergence Theorem 2.3.2.

Solutions

209

R Solution: Clearly, f cannot be less than the supremum. By Theorem 3.2.1 and the Monotone Convergence Theorem 2.3.2, there of simple functions R exists a sequence R R sn such that 0  sn  f and f ¼ limn!1 sn . Therefore f cannot be greater than the supremum. 3.2.P9. Given any measure space ðX; F ; lÞ, let S denote the class of nonnegative simple functions s vanishing outside a set of finite measure, i.e. lðXðs [ 0ÞÞ\1. R This is obviously equivalent to s\1. (a) If every set of infinite measure contains a subset of finite positive measure, show for any nonnegative measurable function f that Z  Z f ¼ sup s: 0  s  f ; s 2 S: (b) Prove the converse of (a). Hint: (a) Use Problem 2.3.P3. (b) Take f to be the characteristic R function of any given set of infinite measure; the condition yields s 2 S with s [ 0. Now consider any set of positive measure on which s takes a positive value. R R Solution: (a) If f \1, the equality is trivial. So, suppose f ¼ 1. We need R only show that there exists some s 2 S such that 0  s  f and s is arbitrarily large. R Consider any M > 0. Since f ¼ 1, we know from Problem 3.2.P8 that there R R exists a simple function s satisfying 0  s  f and s [ M. If s\1, then R s 2 S and we already have the required s. Suppose s ¼ 1. Then for some finite a > 0, we have lðXðs ¼ aÞÞ ¼ 1. By Problem 2.3.P3, the set must contain a subset A for which 1 [ lðAÞ [ M=a. Then svA ¼ avA is aR nonnegative simple function not exceeding s and hence not exceeding f. Also, svA ¼ alðAÞ [ M and is finite, so that svA 2 S. Thus svA serves our purpose. (b) For the converse, consider a measurable set A with lðAÞ ¼ 1. Apply the givenR condition with vA as f. We obtain a function s 2 S such that 0  s  vA and s [ 0. Now, s must take a positive value on some set of positive measure and that set must not only be a subset of A because 0  s  vA but must also have finite measure because s 2 S. 3.2.P10. (a) Let l be a measure on a set X. Suppose f is a nonnegative real-valued function on X and ffn gn
1 a sequence of measurable nonnegative real-valued functions such that ðiÞ fn ðxÞ
fn þ 1 ðxÞ and ðii) f ðxÞ ¼ lim fn ðxÞ for every x 2 X: n!1

Then f is measurable, and if

R

fN \1 for some N, then

R

f ¼ limn!1

R

fn .

210

Solutions

R (b) Show that in part (a), the hypothesis that fN \1 for some N cannot be dropped even if lðXÞ is finite and the only subset of measure 0 is the empty set. Solution: (a) Measurability of f is a simple consequence of Problem 2.2.P4. We shall assume, as is legitimate, that N = 1. The functions fn  f are nonnegative R f1 \1, the Dominated real-valued and satisfy jfn  f j  f1 . Since Convergence Theorem 3.2.6 yields the desired conclusion. (b) Consider the measure l on N such that lðf jgÞ ¼ j12 . By Problem 3.1.P5, such a R measure is possible and the integral / of any nonnegative function / on N is P1 1 j¼1 j2 /ðjÞ. Moreover, lðXÞ is finite and the only subset of measure 0 is the 1

empty set. Define fn on N by fn ðjÞ ¼ j2 ðjn  1Þ. Then each fn is nonnegative real-valued and measurable. Moreover, each j satisfies fn ðjÞ
fn þ 1 ðjÞ for each n as well as limn!1 fn ðjÞ ¼ 0. So, f ¼ limn!1 fn is 0 everywhere and has P 1 n integral 0. However, the integral of fn is 1 j¼1 ðj  1Þ which is 1 because, for 1

n each n, Rthe jth term R ðj  1Þ of the series tends to 1 as j ! 1. So, limn!1 fn ¼ 1 6¼ f . 1

3.2.P11. Let the function f : ½0; 1 ! R be given by f ðxÞ ¼ x2. In  the notation of the 3 proof of Theorem 3.2.1, find the set E3;9 and the number s3 4 .   Solution: Since 23 ¼ 8 and 9  1 ¼ 8, the set E3;9 is the inverse image f 1 88 ; 98 , which is {1}, because   f ð1Þ ¼ 1 and f ð xÞ\1 when 1 6¼ x 2 ½0; 1. To evaluate s3 34 , we need to know the integer k for which 1  k  3  23 and 312 1 1 3 1 2 4 2 E3;k i.e. 8 ðk  1Þ  4 \ 8 k. This inequality is equivalent to k  1  4  3 \k. 1 Therefore k  1 is the integer part of 4  32 , which can be determined to be 6. Thus  1 s3 34 ¼ 68 ¼ 34. To determine the integer part of 4  32 , we need only note that 1

ð4  32 Þ2 ¼ 16  3 ¼ 48, which lies between 62 and 72 .

3.3 Subadditive Fuzzy Measures 3.3.P1. Let X ¼ N and F be the r-algebra of all subsets of X. Take ~m to be defined P by ~ mðEÞ ¼ ki¼1 bi , where b 2 ð0; 1Þ is fixed and k is the number of elements in E. (If E = ;, then k = 0 and the summation becomes “empty”, in which case its value is understood to be 0 as usual.) Show that ~m is a subadditive fuzzy measure that is not a measure. Is it inner continuous? Solution: It is immediate from the definition of ~ m that ~m; ¼ 0 and ~mðEÞ
0 for all E. If EF, then F has at least as many elements in it as E, which implies that ~mðEÞ  ~mðFÞ. Therefore ~ m is monotone and hence a fuzzy measure. Now consider any two subsets E and F. If one of them contains infinitely many elements, there is nothing to prove. Suppose they contain k and ‘ elements respectively. Then E [ F contains at most Pj P P j ¼ k þ ‘ elements and therefore ~ mðE [ F)  i¼1 bi ¼ ki¼1 bi þ ‘i¼1 bi þ k  P Pk ‘ i i mðEÞ þ ~ mðFÞ. This shows that ~m is subadditive. i¼1 b þ i¼1 b ¼ ~

Solutions

211

Let E and F consist of one element each, E 6¼ F. Then E \ F ¼ ;, m~ðEÞ ¼ b ¼ ~ mðEÞ þ ~ mðFÞ. This establishes that ~m is not mðFÞ and ~ mðE [ FÞ ¼ b þ b2 6¼ b þ b ¼ ~ a measure. We shall argue that ~ m is inner continuous. Consider a sequence of subsets En of X such that En En þ 1 for each n and denote their union by E. If E is finite,  then  there is some p such that Ep ¼ E and it is apparent that ~mðEÞ ¼ ~m Ep ¼ P1 i mðEn Þ. If on the other hand E is infinite, then ~mðE) = limn!1 ~ i¼1 b , while at the P1 i same time, limn!1 ~ mðEn Þ ¼ i¼1 b because the number of elements in En must tend to 1 as n!1. ~ is a subadditive fuzzy measure on a set 3.3.P2. (Cf. Problem 3.3.P6) Suppose l X and f : XR! R þ , g:RX ! RR þ are measurable functions. If fg = 0 everywhere, show that f þ g  f þ g. R R Solution: We begin as in the proof of Theorem 3.3.5. If either f or g is 1, there is nothing to prove. So, we may assume both to be finite. The hypothesis that fg = 0 everywhere provides that f and g cannot both be nonzero at the same point, and therefore Xðf þ g [ tÞXðf [ tÞ [ Xðg [ tÞ: ~ now lead to The monotonicity and subadditivity of l g ðf þ gÞðtÞ  ~f ðtÞ þ ~ gðtÞ: Upon integrating from 0 to 1, we arrive at the required inequality. ~ on a r-algebra F of subsets of a set 3.3.P3. Given a subadditive fuzzy measure l X and a nonnegative measurable function f : X ! R þ , prove that the function ~m defined on all sets E 2 F by setting ~ mðEÞ ¼

Z

f vE d~ l

~ðEÞ ¼ 0 is a subadditive fuzzy measure. Prove also that (i) ~mðEÞ ¼ 0 whenever l ~ is. and (ii) ~ m is a measure (i.e. is countably additive) whenever l Solution: That ~ mðEÞ
0 and ~ mð;Þ ¼ 0 need no argument. Since R R AB implies l  f vB d~ l when vA  vB , which further implies f vA  f vB , leading to f vA d~ A; B 2 F , we find that ~ m is monotone. As regards subadditivity, first consider disjoint sets A; B 2 F . In this case, vA þ vB ¼ vA [ B and the product ðf vA Þðf vB Þ is 0 everywhere. It follows that

212

Solutions

~ mðA [ BÞ ¼ 

Z

Z Z

f vA [ B d~ l¼ ðf vA + f vB Þd~ l Z f vA d~ lþ f vB d~ l by Problem 3:3:P2

~ mðAÞ þ ~ mðBÞ: When A, B are not disjoint, the sets A\B and B are surely disjoint and have union A [ B. Therefore we can reason as follows: ~ mðA [ BÞ ¼ ~ mððAnBÞ [ BÞ  ~ mðAnBÞ þ ~mðBÞ  ~mðAÞ þ ~mðBÞ; where the final step invokes the already established motonicity of m~. ~ðEÞ ¼ 0 implies Since f vE can take a nonzero value only on E, the hypothesis l that fg vE ¼ 0 everywhere on [0, 1) and hence that ~mðEÞ ¼ 0. This proves (i). ~ is a measure and consider a sequence fAn gn
1 of disjoint sets, Now suppose l A by A and [ Nn¼1 An by BN . To prove each belonging to F . Denote the union [ 1 n¼1 P1 n mðAn Þ. On account of the disjoint(ii), we have to demonstrate that ~ mðAÞ ¼ n¼1 ~ PN and hence ness of the sets An, we have n¼1 f vAn ¼ f vBN  R R  PN ~ d~ l . Since l is a measure, the integral is additive by f v d~ l ¼ f v BN An n¼1 Theorem 3.2.2, and therefore we further have Z

f vBN d~ l¼

N Z X n¼1



f vAn d~ l:

The definition of A andBN also has the consequence that f vBN  f vBN þ 1 for every N, and that limN!1 f vBN ðxÞ ¼ ðf vA ÞðxÞ for every x 2 X (cf. Problem 2.1.P1), so that the Monotone Convergence Theorem 2.2.3 yields Z lim

N!1



f vBN d~ l¼

Z

f vA d~ l:

Upon combining this with the equality displayed further above, we obtain R P1 R f v d~ l ¼ f v l, which, by the definition of ~m, is the same as what we A An d~ n¼1 had to demonstrate. ~ and ~ 3.3.P4. Let l m be finite subadditive fuzzy measures on a r-algebra F of ~ðE) = 0. MustR there exist a nonsubsets of a set X such that ~ mðE) = 0 whenever l negative measurable function f : X ! R þ satisfying ~mðEÞ ¼ f vE d~ l for each E 2 F? ~ on the r-algebra Solution: No. Let c 2 ð0; 1Þ be fixed and consider the measure l of all subsets of N as in Problem 3.1.P5 with every aj ¼ c j . This measure satisfies P j ~ðE) = 0 ) E ¼ ;. It follows for any fuzzy ~ðN) = l j2N c \1 and also satisfies l

Solutions

213

~ðE) = 0. If a function f : X ! R þ of the kind measure ~ m that ~ mðE) = 0 whenever l in question were to exist, it would follow from Problem 3.3.P3 that ~m must be a measure. However, the example in Problem 3.3.P1 shows that ~m need not be a measure. 3.3.P5. Let a; b [ 0 and s > 1. For any x such that maxfa; bg  x  a þ b, show that

14    1    1 ð1 þ sÞ3 x  x þ s3  1 a 4 þ x þ s3  1 b 4 : ~ on a two element set {a, b} as l ~; ¼ 0, l ~fag ¼ a, l ~fbg ¼ b and Solution: Define l ~fa; bg ¼ x. Then l ~ is a subadditive fuzzy measure. Consider the functions f and l g on {a, b} such that f ðaÞ ¼ 1; f ðbÞ ¼ s and gðaÞ ¼ s; gðbÞ ¼ 1. Then f + g takes R the value 1 + s at both points and hence ðf þ gÞ3 ¼ ð1 þ sÞ3 x. Also, 8 0 and a + b
1 (subadditivity numbers a ¼ l ~). Denote f (a) by x
0 and f(b) by y
0. Also, denote g(a) by n
0 and g(b) by of l g
0. Then 8 8 < 1 0  t\x < 1 0  t\y ~f ðtÞ ¼ b x  t\y if x  y and ~f ðtÞ ¼ a y  t\x if y  x : : 0 y  t\1 0 x  t\1: (When y = x, both descriptions of ~f hold, the middle line being vacuous.) Similarly for g, but with x and y replaced by n and η respectively. From the above computation of ~f , we obtain Z

f ¼



x þ bðy  xÞ y þ aðx  yÞ

if x  y if x
y

Z and

similarly for

g:

RSuppose x  y.R If also Rn  g and hence x þ n  y þ g, then it is straightforward that ðf þ gÞ ¼ f þ g. So, suppose n
g instead. Then we have

214

Solutions

Z

f ¼ x þ bðy  xÞ

Z and

g ¼ g þ aðn  gÞ:

There are two cases to consider: x þ n  y þ g and x þ n
y þ g. First consider the case when x þ n  y þ g. This inequality has the consequence that Z

ðf þ gÞ ¼ x þ n þ b ðy þ g  x  nÞ:

Keeping in view that a þ b
1, i.e. a
1  b, and that n  g
0, we reason that Z Z fþ g ¼ x þ bðy  xÞ þ g þ aðn  gÞ




x þ bðy  xÞ þ g þ ð1  bÞðn  gÞ and the smaller side of this inequality simplifies to x þ bðy  xÞ þ g þ ðn  gÞ  bðn  gÞ ¼ x þ n þ bððy  xÞ  ðn  gÞÞ ¼ x þ n þ bðy þ g  x  nÞ Z ¼ ðf þ gÞ:

Next, consider the case when x þ n
y þ g. Analogously to the previous case, this inequality has the consequence that Z ðf þ gÞ ¼ y þ g þ aðx þ n  y  gÞ:

Keeping in view that a þ b
1, i.e. b
1  a, and that y  x
0, we reason that Z

fþ

Z

g ¼ x þ bðy  xÞ þ g þ aðn  gÞ
x þ ð1  aÞðy  xÞ þ g þ aðn  gÞ

and the smaller side of this inequality simplifies to x þ ðy  xÞ  aðy  xÞ þ g þ aðn  gÞ ¼ y þ g þ aððn  gÞ  ðy  xÞÞ ¼ y þ g þ aðx þ n  y  gÞ Z ¼ ðf þ gÞ:

Solutions

215

This proves the required inequality when x  y. In the event that x
y, an entirely analogous argument goes through.

4.1 Lebesgue Outer Measure 4.1.P1. The outer measure m on [a, b] is defined on the r-algebra F of all subsets of [a, b] and therefore all functions are measurable and have a distribution. Find the distribution ~f of the function f : ½a; b ! R such that f (x) = 1 if x is irrational and 0 R sup f otherwise and hence find 0 ~f . (The reader is reminded that the Riemann integral Rb a f does not exist.) Solution: The function f is the characteristic function of the set A of all irrational numbers in [a, b]. By Exercise 4.1.10, its outer measure is b  a. Therefore ~f ðtÞ ¼ R sup f b  a if 0  t\1 and 0 if 1  t. Lastly, 0 ~f ¼ b  a. 4.1.P2. For nonempty AR, the intersection IA of all intervals containing A is a nonempty interval. Show that m ðAÞ  supfjx  yj: x; y 2 Ag ¼ ‘ðIA Þ. Solution: Since AIA , we have m ðAÞ  m ðIA Þ ¼ ‘ðIA Þ. We need show only that ‘ðIA Þ ¼ supfjx  yj: x; y 2 Ag ¼ s, say. Observe that s = 0 (resp. 1) if and only if ‘ðIA Þ ¼ 0 (resp. 1). So we take 0\s\1. Then IA contains more than one number and is bounded. Denote its left and right endpoints by a and b respectively. Then ½a; b IA A. So, we have ‘ðIA Þ  jx  yj for all x; y 2 A. This yields ‘ðIA Þ
s. In order to prove the reverse inequality, consider an arbitrary e [ 0 and let g ¼ minfe; b  ag [ 0. The interval ½a þ g; b does not contain IA and hence does not contain A, but [a, b] does. Therefore there exists an x 2 A such that a  x\a þ g. Similarly, there exists a y 2 A such that b  g\y  b. Consequently, s [ b  a  2g ¼ ‘ðIA Þ  2g
‘ðIA Þ  2e. Since e [ 0 is arbitrary, this proves the required reverse inequality s
‘ðIA Þ. 4.1.P3. For any subset A of [a, b] and any e [ 0, show that there exists a subset B of



[a, b] such that AB ¼ [ 1 j¼1 Jj ; m ðBÞ\m ðAÞ þ e and each Jj is the intersection of an open interval with [a, b].

Solution: By Definition 4.1.3 of m , there exists a sequence Ij j
1 of open intervals such that A [ 1 j¼1 Ij and 1 X

‘ðIj Þ\m ðAÞ þ e:

ðSol:4Þ

j¼1

Let Jj ¼ Ij \ ½a; b; then each Jj is the intersection of an open interval with [a, b].

 1 1 Take B ¼ [ j¼1 Jj . Since A½a; b, therefore A ¼ A \ ½a; b [ j¼1 Ij \ ½a; b ¼   [1 j¼1 Ij \ ½a; b ¼ B. By countable subadditivity of m* (see Proposition 4.1.9),   P1   P

m ðBÞ  1 j¼1 m Jj  j¼1 m Ij . But since each Ij is an interval, we have by

216

Solutions

      P Theorem 4.1.8 that m Ij ¼ ‘ Ij . It follows that m ðBÞ  1 j¼1 ‘ Ij . By (Sol.4), we have m ðBÞ\m ðAÞ þ e. 4.1.P4. Consider the set C, known as the Cantor set, defined as \ 1 n¼1 Cn , where     C1 ¼ 0; 13 [ 23 ; 1 , and Cn þ 1 is obtained from Cn by removing the open middle         thirds of its closed intervals. Thus, C2 ¼ 0; 19 [ 29 ; 13 [ 23 ; 79 [ 89 ; 1 , and so on. Show that m ðC Þ ¼ 0. Solution: Every Cn is the union of 2n closed intervals of length 1=3n each. Therefore the total length of the closed intervals comprising Cn is ð2=3Þn . From Theorem 4.1.8 and Proposition 4.1.9, it follows that m ðCn Þ  ð2=3Þn . Considering that CCn , we obtain m ðCÞ ¼ 0. 4.1.P5. Prove the reverse of the inequality established in Remark 4.1.2(b): Suppose P

þ that for each n 2 N, we have a series 1 . Let k¼1 an;k of nonnegative terms in R /: N ! N  N be a bijection. Prove that 1 X p¼1

a/ðpÞ


1 X 1 X

an;k :

n¼1 k¼1

Hint: First show that, given any N; K 2 N, there exists a P 2 N such that PP PN P K n¼1 k¼1 an;k  p¼1 a/ðpÞ . Also use the fact that the limit of a finite sum of sequences is the sum of their separate limits.

Solution: Consider any N; K 2 N. Let P ¼ max /1 ðn; kÞ: 1  n  N; 1  k  K . Then 1  n  N; 1  k  K ) /1 ðn; kÞ  P ) ðn; kÞ ¼ /ðpÞ for some p  P ) P P P P an;k ¼ a/ðpÞ for some p  P. Therefore Nn¼1 Kk¼1 an;k  Pp¼1 a/ðpÞ  1 p¼1 a/ðpÞ . This inequality has been shown to be true for any N; K 2 N. Since the limit of a finite sum of sequences is the sum of their separate limits, for any N 2 N, we further P P P1 PN PK have Nn¼1 1 k¼1 an;k  limK!1 n¼1 k¼1 an;k  p¼1 a/ðpÞ . Now take the limit as N!1.

4.2 Measure from Outer Measure 4.2.P1. Let l be a finite outer measure on a set X and AB be subsets of X. Show that (a) if l (B\A) = 0, then l (B) = l (A); (b) if l (B) = l (A) and A 2 M, then l (B\A) = 0. Solution: (a) By the subadditivity of outer measure, l ðBÞ ¼ l ðA [ ðBnAÞÞ  l ðAÞ þ l ðBnAÞ ¼ l ð AÞ since l ðBnAÞ ¼ 0. The reverse inequality follows from the monotonicity of outer measure. (b) Since A is l -measurable,

Solutions

217

l ðBÞ ¼ l ðB \ AÞ þ l ðB \ Ac Þ ¼ l ðAÞ þ l ðBnAÞ: 4.2.P2. Show that, if E1 and E2 are l -measurable, where l is an outer measure, then l ðE1 [ E2 Þ þ l ðE1 \ E2 Þ ¼ l ðE1 Þ þ l ðE2 Þ: State the analog for three sets. Solution: The restriction l of l to l -measurable sets is a measure (Theorem 4.2.11) and we only need to prove lðE1 [ E2 Þ þ lðE1 \ E2 Þ ¼ lðE1 Þ þ lðE2 Þ: Although on proceeding as in the solution of Problem 3.1.P1 with n = 2 we get the required equality forthwith, we offer a direct proof. For i = 1, 2, the set Ei is the union of the disjoint sets Ei nðE1 \ E2 Þ and E1 \ E2 , and therefore lðEi Þ ¼ lðEi nðE1 \ E2 ÞÞ þ lðE1 \ E2 Þ. Adding these for i = 1, 2 results in lðE1 Þ þ lðE2 Þ ¼ lðE1 nðE1 \ E2 ÞÞ þ lðE2 nðE1 \ E2 ÞÞ þ 2lðE1 \ E2 Þ ¼ ðlðE1 nðE1 \ E2 ÞÞ þ lðE2 nðE1 \ E2 ÞÞ þ lðE1 \ E2 ÞÞ þ lðE1 \ E2 Þ ¼ lðE1 [ E2 Þ þ lðE1 \ E2 Þ; the justification of the last step being that E1 [ E2 is the union of the disjoint sets E1 \ E2 and Ei nðE1 \ E2 Þ; i ¼ 1; 2. The analog for three sets is that l ðE1 [ E2 [ E3 Þ þ l ðE1 \ E2 Þ þ l ðE2 \ E3 Þ þ l ðE1 \ E3 Þ ¼ l ðE1 Þ þ l ðE2 Þ þ l ðE3 Þ þ l ðE1 \ E2 \ E3 Þ for l -measurable sets E1 ; E2 ; E3 . A justification can be given by using the two sets case twice. 4.2.P3. What has been called “l” in Example 2.1.3(b) is easily seen to be an outer measure l . Which subsets are l -measurable? Solution: Only ; and X. It is enough to show that {a}, {b} and {c} are not l measurable. It will follow that their complements are also not l -measurable (Remark 4.2.3(b)). For E ={a}, the set A = {c, a} does not satisfy the equality required for l -measurability of E, because l ðAÞ ¼ 5; l ðA \ EÞ ¼ 2 and l ðA \ E c Þ ¼ 4. For E ={b}, the set A = {b, c} does not satisfy the equality required for l -measurability of E, because l ðAÞ ¼ 6; l ðA \ EÞ ¼ 3 and l ðA \ E c Þ ¼ 4. For E ={c}, the set A = {b, c} does not satisfy the equality required for l measurability of E, because l ðAÞ ¼ 6; l ðA \ EÞ ¼ 4 and l ðA \ E c Þ ¼ 3. 4.2.P4. Let X ¼ fa; b; cg and F be the r-algebra consisting of the subsets {a, b},{c} besides ; and X. Then l: F ! R, where l; ¼ lfa; bg ¼ 0; lfcg ¼ lX ¼ 1, is

218

Solutions

clearly a measure P1 on X. For an arbitrary subset AX, define l A to be the infimum of all numbers n¼1 lðIn Þ, where fIn gn
1 is a sequence of sets in F that covers A (the infimum is taken over all such sequences, but only finitely many need be used). Find l A for each of the eight subsets of X. It can be checked easily that l is an outer measure; find the l -measurable subsets. Solution: l fag ¼ l fbg ¼ 0; l fb; cg ¼ l fa; cg ¼ 1; the other four subsets are in F , and l agrees with l on each of them. All subsets are l -measurable. 4.2.P5. Let H be a family of subsets of a nonempty set X such that ; 2 H and suppose a: H ! R is nonnegative-valued, satisfying a; ¼ 0. Assume that some sequence fHn gn
1 of sets belonging to H covers X, hence also every subset of X. Show that

(

a A ¼ inf

1 X

aðIn Þ: every In 2 H and A

[

) In

n
1

n¼1

defines an outer measure on X. Solution: It is obvious that a is R þ -valued. Consider ABX. Any sequence fIn gn
1 of sets In belonging to H that covers B also covers A. Therefore a A  a B. Thus a is monotone. Since ; is covered by the sequence ;; ;; . . .; of sets in H and a; ¼ 0, we have a ; ¼ 0. It remains only to prove countable subadditivity. P

Let fAn gn
1 be a sequence of subsets of X and e [ 0. If 1 n¼1 a ðAn Þ ¼ 1, there is nothing to prove. So, we may assume this sum to be finite and hence also

every a ðA n 2 N, there is a n Þ. In accordance with the definition of a , for each I sequence In;k k
1 of sets belonging to H such that An  [ 1 k¼1 n;k and 1 X   e a In;k  a ðAn Þ þ n : 2 k¼1



Then the family In;k : ðn; kÞ 2 N  N covers [ 1 n¼1 An and 1 X 1 1 X   X a In;k  a ðAn Þ þ e: n¼1 k¼1

ðSol:5Þ

n¼1



Let /: N ! N  N be a bijection. The sequence I/ðpÞ p
1 of sets belonging to H covers [ 1 n¼1 An and hence a ð

1 [

n¼1

An Þ 

1 X   a I/ðpÞ : p¼1

It follows from (Sol.5) and Remark 4.1.2(b) that

Solutions

219

a ð

1 [

n¼1

An Þ 

1 X

a ðAn Þ þ e:

n¼1

Since this holds for every e [ 0, countable subadditivity must hold. 4.2.P6. For X ¼ fa; b; c; dg, a set of four elements, let H ¼ f;; fd g; fa; bg; fa; cg; fa; b; cg; X g and a: H ! R be defined as a; ¼ 0; afd g ¼ 3; afa; bg ¼ 0; afa; cg ¼ 3; afa; b; cg ¼ 2

and aX ¼ 2:

Find a A for all the sixteen subsets of X and determine which ones are a measurable. Solution: a ; ¼ 0; a fag ¼ a fbg ¼ a fa; bg ¼ 0, all other subsets E of X have a E ¼ 2. The sets ;; fag; fbg; fa; bg and their complements are a -measurable. The remaining eight sets fail to be a -measurable. The subset E ¼ fcg is not a -measurable, because when A ¼ fc; d g, we find that

a ðA \ EÞ ¼ a ðEÞ ¼ 2; a ðA \ E c Þ ¼ a ðfd gÞ ¼ 2 and a ð AÞ ¼ 2 6¼ 2 þ 2. The subset E ¼ fd g is not a -measurable, because when A ¼ fc; d g, we find that a ðA \ EÞ ¼ a ðEÞ ¼ 2; a ðA \ E c Þ ¼ a ðfcgÞ ¼ 2 and a ð AÞ ¼ 2 6¼ 2 þ 2. The subset E ¼ fa; cg is not a -measurable, because when A ¼ fc; d g, we find that a ðA \ EÞ ¼ a fcg ¼ 2; a ðA \ Ec Þ ¼ a ðfd gÞ ¼ 2 and a ð AÞ ¼ 2 6¼ 2 þ 2. The subset E ¼ fa; d g is not a -measurable, because when A ¼ fc; d g, we find that a ðA \ EÞ ¼ a fd g ¼ 2; a ðA \ E c Þ ¼ a ðfcgÞ ¼ 2 and a ð AÞ ¼ 2 6¼ 2 þ 2. The subset E ¼ fb; cg is not a -measurable, because its complement fa; d g has been shown not to be a -measurable. The subset E ¼ fb; d g is not a -measurable, because its complement fa; cg has been shown not to be a -measurable. The subset E ¼ fa; b; d g is not a -measurable, because its complement fcg has been shown not to be a -measurable. The subset E ¼ fa; b; cg is not a -measurable, because its complement fd g has been shown not to be a -measurable. 4.2.P7. Let H be a r-algebra of subsets of X and ðX; H; aÞ be a measure space. Any sequence of sets in H having X as one of its terms covers X. Show that for the outer measure a obtained as in Problem 4.2.P5, every set in H is a -measurable. Is the converse true? Solution: Let E 2 H and AX be an arbitrary subset. We need prove only the Carathéodory condition that a ð AÞ
a ðA \ EÞ þ a ðA \ E c Þ. So, let a ð AÞ\1 and consider an arbitrary sequence fEn gn
1 of sets in H that covers A. Then fEn \ Egn
1 and fEn \ E c gn
1 are sequences of sets in H that cover A \ E and A \ E c respectively. It follows that

220

Solutions 1 X

aðEn \ EÞ
a ðA \ EÞ and

n¼1

1 X

aðEn \ E c Þ
a ðA \ Ec Þ:

n¼1

Since a is a measure, we have aðEn \ EÞ þ aðEn \ E c Þ ¼ aðEn Þ for each n. Therefore 1 X n¼1

aðEn Þ ¼

1 X n¼1

aðEn \ EÞ þ

1 X

aðEn \ Ec Þ
a ðA \ EÞ þ a ðA \ E c Þ:

n¼1

Since fEn gn
1 is an arbitrary sequence of sets in H that covers A, we may take the infimum on the left side, thereby obtaining the desired Carathéodory condition and completing the proof that every set in H is a -measurable. The converse is false. Let X consist of three points a; b; c and H consist of ;; fa; bg; fcg and X. Consider the measure a on H satisfying að;Þ ¼ afa; bg ¼ 0 and afcg ¼ aX ¼ 1. Since a fag ¼ 0, the set {a} is a -measurable but fag 62 H. P 4.2.P8. For infinite AN, let að AÞ ¼ 1 and for finite AN, let að AÞ ¼ nj¼1 ð1=2 j Þ, where n is the number of elements in A. (If A ¼ ;, the sum becomes “empty” and is to be interpreted as usual to be 0.) Show that a is an outer measure on N and find all a-measurable subsets of N. Solution: The set function a is certainly nonnegative real-valued and bounded above by 1. Besides, að AÞ\1 whenever A is finite. Suppose ABN. If B is infinite, then aðBÞ ¼ 1 and hence að AÞ  aðBÞ. If B is finite, then so is A and P P að AÞ ¼ nj¼1 ð1=2 j Þ; aðBÞ ¼ pj¼1 ð1=2 j Þ, where n is the number of elements in A and p is the number of elements in B. But n  p, which implies að AÞ  aðBÞ. This shows that a is monotone. For countable subadditivity, consider a sequence fAn gn
1 of subsets of N. For the purpose at hand, we may take every An to be nonempty. If [ 1 n¼1 An is infinite, then infinitely many among the sets An must contain one or more elements and P 1 therefore have a equal to 12 or greater. This implies 1 n¼1 aðAn Þ ¼ 1. If [ n¼1 An is 1 finite, then it can have only finitely many distinct subsets and therefore [ n¼1 An ¼ [ Nn¼1 An for some N. So, we need prove only finite subadditivity with finite sets, which will follow by induction if we prove it for two finite sets. With this in view, let A and B be finite subsets having n and p elements respectively. Then A [ B has at P þp P most n + p elements and hence aðA [ BÞ  nj¼1 ð1=2 j Þ ¼ nj¼1 ð1=2 j Þ þ P P Pp jþn Þ  nj¼1 ð1=2 j Þ þ pj¼1 ð1=2 j Þ ¼ aðAÞ þ aðBÞ. j¼1 ð1=2 Let E be a subset that is neither empty nor N. Let A be a subset consisting of any one element belonging to E and any one element belonging to E c . Then A contains precisely two elements and each of A \ E and A \ E c contains a single element. Therefore að AÞ ¼ 34 and aðA \ EÞ ¼ aðA \ E c Þ ¼ 12, so that að AÞ\aðA \ EÞ þ aðA \ Ec Þ. So E is not a-measurable. This shows that the only a-measurable subsets of N are ; and N.

Solutions

221

4.2.P9. (Cf. Problem 4.3.P7) Let F be a r-algebra of subsets of a nonempty set X and suppose the set function a: F ! R þ vanishes at the empty set, is monotone and countably subadditive. (In other words a has all the properties of an outer measure except that its domain is F .) For any A 2 F , put ( lðAÞ ¼ sup

1 X j¼1

aðEj Þ:

1 [

) Ej ¼ A; Ej \ Ek ¼ ; whenever j 6¼ k; every Ej 2F :

j¼1

(a) Show that l is a measure and that, for all A 2 F , we have (i) lð AÞ
að AÞ (ii) lð AÞ ¼ 0 , að AÞ ¼ 0. (b) If a is as in Problem 4.2.P8, what is l?

Solution: (a) Given any A 2 F , one possible sequence Ej j
1 such that [ 1 j¼1 Ej ¼ A; Ej \ Ek ¼ ; whenever j 6¼ k and every Ej \ F is given by E1 ¼ A; Ej ¼ ;   P for j > 1. For this sequence, 1 j¼1 a Ej ¼ að AÞ and therefore lð AÞ
að AÞ. This proves (i). It follows from (i) that l is nonnegative-valued. For A ¼ ;, the only sequence Ej j
1 of the required kind has every Ej ¼ ;, and therefore lð;Þ ¼ 0. Before proceeding to countable additivity, we establish monotonicity. Suppose AB, both sets being in F , i.e. measurable. Consider an arbitrary sequence Ej j
1 of disjoint measurable sets with [ 1 j¼1 Ej ¼ A. Then the

sequence of sets obtained by prefixing B\A to the sequence Ej j
1 consists of disjoint measurable and has union equal to B. So, lðBÞ
  sets P P1   aðBnAÞ þ 1
a E j j¼1 a Ej . Since this holds for an arbitrary j¼1

sequence Ej j
1 of disjoint measurable sets with union A, we have lðBÞ
lð AÞ. Regarding countable additivity, consider A ¼ [ 1 n¼1 An , where fAn gn
1 is a

sequence of disjoint measurable sets. Let Ej j
1 be an arbitrary sequence of

disjoint measurable sets such that [ 1 j¼1 Ej ¼ A. Then for each n, An \ Ej j
1   is a sequence of disjoint measurable sets such that [ 1 An \ Ej ¼ An . P1 j¼1 Hence the definition of l yields j¼1 aðAn \ Ej Þ  lðAn Þ. P1 P1 P1 Consequently, n¼1 j¼1 aðAn \ Ej Þ  n¼1 lðAn Þ. By Remark 4.1.2(b) and P1 P1 P1 Problem 4.1.P5, we can assert that j¼1 n¼1 aðAn \ Ej Þ  n¼1 lðAn Þ.

However, the sequence An \ Ej n
1 consists of disjoint measurable sets satisfying [ 1 n¼1 ðAn \ Ej Þ ¼ Ej and therefore countable subadditivity of a leads   P aðA \ Ej Þ. In tandem with the preceding inequality, this to a Ej  1 P1 n¼1  n P1 a E lðAn Þ. This has been shown to hold for an leads to j  j¼1 n¼1 arbitrary sequence Ej j
1 of disjoint measurable sets having union A. By P definition of l, it now follows that lð AÞ  1 n¼1 lðAn Þ.

222

Solutions

To arrive at the reverse inequality, we may assume lð AÞ\1. Then monotonicity provides lðAn Þ\1 for each n. Take

an arbitrary e > 0. Corresponding to each n 2 N, there is a sequence En;j j
1 of disjoint measurable sets such P1   e that An ¼ [ 1 j¼1 En;j and j¼1 a En;j
lðAn Þ  2n . Then the sets of the family

are measurable, are disjoint and have union En;j : ðn; jÞ 2 N  N P1 P1   A ¼ A. Besides, they satisfy the inequality [1 n¼1 n n¼1 j¼1 a En;j
P1 n¼1 lðAn Þ  e. Next, let /: N ! N  N be a bijection. Then the sequence fE/ðpÞ gp
1 of disjoint measurable sets has union A and hence P1 p¼1 aðE/ðpÞ Þ  lð AÞ: Upon combining this with the previous inequality and P Problem 4.1.P5, we obtain 1 n¼1 lðAn Þ  e  lð AÞ. Since e > 0 is arbitrary, we arrive at the desired reverse inequality. It remains to prove (ii). It is trivial from (i) that lð AÞ ¼ 0 ) a ð AÞ ¼ 0. If að AÞ ¼ 0, the monotonicity of a ensures that every sequence Ej j
1 of P1   measurable sets satisfying [ 1 j¼1 Ej ¼ A also satisfies j¼1 a Ej ¼ 0, thereby rendering lð AÞ ¼ 0.

(b) Consider a subset A consisting of a single point x. In any sequence Ej j
1 of disjoint measurable sets having union A, all sets must be ; except one and the P 1 exceptional one must be equal to A. Therefore 1 j¼1 a Ej ¼ að AÞ ¼ 2. It follows that lf xg ¼ 12 for every x and hence that l is 12 times the counting measure.

4.3 Lebesgue Measure 4.3.P1. Suppose that A is a subset of R with the property that, for every e [ 0, there exist Lebesgue measurable sets B and C such that BAC and mðC \ Bc Þ\e. (Here m denotes Lebesgue measure as usual.) Show that A is Lebesgue measurable. Solution: For n 2 N, there exist Lebesgue measurable Bn and Cn such that 1 Bn ACn and mðCn \ Bcn Þ\ 1n. Let B0 ¼ [ 1 n¼1 Bn and C0 ¼ \ n¼1 Cn . Then B0 and c c C0 are Lebesgue measurable and mðC0 \ B0 Þ  mðCn \ Bn Þ  1n for each n 2 N. This implies mðC0 \ Bc0 Þ ¼ 0. Moreover, A \ Bc0 C0 \ Bc0 , so that m ðA \ Bc0 Þ  m ðC0 \ Bc0 Þ ¼ 0. Hence A \ Bc0 2 M and so, A ¼ ðA \ Bc0 Þ [ B0 2 M. 4.3.P2. Let A½a; b and m denote Lebesgue measure. Show that there exists a Lebesgue measurable set B½a; b such that mðBÞ ¼ m ðBÞ ¼ m ð AÞ and AB. The same is true with [a, b] replaced by R. Solution: First suppose A½a; b. By Problem 4.1.P3, for each n 2 N, there exists a



1 subset Bn of [a, b] such that ABn ¼ [ 1 j¼1 Jj;n such that m ðBn Þ  m ðAÞ þ n and each Jj;n is the intersection of an open interval with [a, b]. Applying Theorem 4.3.1 and then Theorem 4.2.9, we find that each Bn is Lebesgue measurable. Appealing to Theorem 4.2.9 again, we deduce that the set B ¼ \ 1 n¼1 Bn ½a; b is also measurable. Now AB and by monotonicity of m , we have m ð AÞ  m ðBÞ  m ðBn Þ  m ðAÞ þ 1n for each n 2 N. Hence m ðBÞ ¼ m ð AÞ. Since B is Lebesgue measurable, it is trivial that mðBÞ ¼ m ðBÞ.

Solutions

223

When [a, b] is replaced by R, the appeal to Problem 4.1.P3 can be replaced by an appeal to Remark 4.1.4; each Jj;n is an open interval and m ðBn Þ can be 1. (The matter is trivial if m ð AÞ ¼ 1.) 4.3.P3. Show that the Lebesgue outer measure m is inner continuous. In other words, if fSn gn
1 is a sequence of sets such that S1 S2    , then

m ð [ 1 n¼1 Sn Þ ¼ limn!1 m ðSn Þ. Solution: Since the sequence fm ðSn Þgn
1 is increasing, we know that limn!1 m ðSn Þ exists, perhaps as 1. Also, m ðSk Þ  m ð [ 1 n¼1 Sn Þ for each k, and hence lim m ðSn Þ  m ð

n!1

1 [

Sn Þ:

ðSol:6Þ

n¼1

By Problem 4.3.P2, there exist Lebesgue measurable Tn Sn such that mðTn Þ ¼ m ðTn Þ ¼ m ðSn Þ:

ðSol:7Þ

Setting An ¼ \ j
n Tj , we have A1 A2     ; with each An Lebesgue measurable

ðSol:8Þ

(by Theorem 4.2.9) and also Sn An Tn . Therefore, we have 1 [

Sn 

n¼1

1 [

An

ðSol:9Þ

n¼1

as well as m ðSn Þ  m ðAn Þ  m ðTn Þ. Hence from (Sol.7), we get m ðAn Þ ¼ m ðSn Þ:

ðSol:10Þ

From (Sol.8) and Proposition 2.3.1 (inner continuity of measure), we know mð [ 1 n¼1 An Þ ¼ limn!1 mðAn Þ, i.e. m ð

1 [ n¼1

An Þ ¼ lim m ðAn Þ n!1

¼ lim m ðSn Þ by ðSol.10Þ n!1

Together with (Sol.9), this leads to the reverse of inequality (Sol.6). 4.3.P4. Let E½a; b. Show that E is a Lebesgue measurable subset of [a, b] if and only if it is a Lebesgue measurable subset of R. Solution: We shall denote complements in R by a superscript ‘c’ as usual; however, we shall denote the complement of a set F½a; b by ½a; bnF.

224

Solutions

First suppose E is a Lebesgue measurable subset of R. To show that it is a Lebesgue measurable subset of [a, b], consider an arbitrary subset A½a; b. Note that A \ E c ¼ fx 2 A: x 62 Eg ¼ fx 2 A: x 2 ½a; b; x 62 Eg ¼ A \ ð½a; bnEÞ. Since E is a Lebesgue measurable subset of R, we have m ðAÞ ¼ m ðA \ EÞ þ m ðA \ Ec Þ ¼ m ðA \ EÞ þ m ðA \ ð½a; bnEÞÞ: So, E is a Lebesgue measurable subset of [a, b]. Next, suppose for the converse that E is a Lebesgue measurable subset of [a, b]. To show that it is a Lebesgue measurable subset of R, consider an arbitrary subset AR. From the inclusion E½a; b, we know that A \ ½a; b \ E ¼ A \ E

and ½a; bc ¼ E c \ ½a; bc :

Now, since [a, b] is a Lebesgue measurable subset of R, m ðAÞ ¼ m ðA \ ½a; bÞ þ m ðA \ ½a; bc Þ ¼ m ðA \ ½a; bÞ þ m ðA \ E c \ ½a; bc Þ: Also, since E is a Lebesgue measurable subset of [a, b] and A \ ½a; b½a; b, we have m ðA \ ½a; bÞ ¼ m ðA \ ½a; b \ EÞ þ m ðA \ ½a; b \ ð½a; bnEÞÞ ¼ m ðA \ EÞ þ m ðA \ ½a; b \ ð½a; bnEÞÞ; as E½a; b: But ½a; bnE½a; b and hence A \ ½a; b \ ð½a; bnE Þ ¼ A \ ð½a; bnEÞ. So, the preceding equality can be restated as m ðA \ ½a; bÞ ¼ m ðA \ EÞ þ m ðA \ ð½a; bnEÞÞ: It follows that m ðAÞ ¼ m ðA \ EÞ þ m ðA \ ð½a; bnEÞÞ þ m ðA \ E c \ ½a; bc Þ ¼ m ðA \ EÞ þ m ðA \ Ec \ ½a; bÞ þ m ðA \ E c \ ½a; bc Þ: However, m ðA \ Ec \ ½a; bÞ þ m ðA \ E c \ ½a; bc Þ ¼ m ðA \ E c Þ because [a, b] is a Lebesgue measurable subset of R. Hence m ðAÞ ¼ m ðA \ EÞ þ m ðA \ E c Þ: 4.3.P5. For any SR and any x 2 R, let S + x denote the set fs þ x: s 2 Sg. Prove the following: (a) If E is a Lebesgue measurable subset of R, then E + x is also a Lebesgue measurable subset of R and mðE þ xÞ ¼ mðE Þ.

Solutions

225

(b) If E is a Lebesgue measurable subset of [a, b] and E þ x½a; b, then E + x is also a Lebesgue measurable subset of [a, b]. Solution: (a) Note that, for any AR, ER and any x 2 R, the following equalities hold: A \ ðE þ xÞ ¼ ½ðA  xÞ \ E þ x and

A \ ðE þ xÞc ¼ ½ðA  xÞ \ E c  þ x:

Moreover, in view of Proposition 4.1.5(e), m ðS þ xÞ ¼ m ðSÞ for any SR and any x 2 R: Now let AR be arbitrary and E be a Lebesgue measurable subset of R. It follows from the three equalities noted in the above paragraph and the measurability of E that 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: The equality mðE þ xÞ ¼ mðEÞ now follows from the fact that it holds for m . (b) By Problem 4.3.P4, the sets E and E + x are Lebesgue measurable of subsets of [a, b] if and only if they are Lebesgue measurable subsets of R. Therefore the required conclusion is immediate from (a). 4.3.P6. Let frn gn
1 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 gi ðri Þ
i and

Z1

gi ðxÞdx ¼

1 : 2i

0

P Define fn ¼ ni¼1 gi . Then each fn is nonnegative and continuous. Besides, the sequence ffn gn
1 is increasing and therefore has a limit function f that is nonnegative extended real-valued and measurable. Show R that the function f is unbounded on every subinterval of [0, 1] and satisfies ½0;1 f ¼ 1. Solution: Consider an arbitrary subinterval ½a; b½0; 1, where a < b, and an arbitrary M > 0. The interval [a, b] must contain infinitely many rational numbers and hence must contain some rn with n > M. The given definition of fn leads to fn ðrn Þ
gn ðrn Þ
n [ M. By continuity, [a, b] has some subinterval [a, b] of positive length on which fn [ M. It follows that f
fn [ M on ½a; b. This shows that f is unbounded on every subinterval of [0, 1].

226

Solutions

R R1 By Theorem 4.3.4, each gn satisfies ½0;1 gi ¼ 0 gi ðxÞdx ¼ 1=2i . It follows that R Pn i n i¼1 ð1=2 Þ ¼ 1  ð1=2 Þ. Since the sequence ffn gn
1 is increasing, we ½0;1 fn ¼ R infer from the Monotone Convergence Theorem 2.3.2 that ½0;1 f ¼ limn!1 R ½0;1 fn ¼ 1. 4.3.P7. For any interval IR

of positive length and any M > 0, show that there exists a sequence Ej j
1 of disjoint sets such that [ 1 j¼1 Ej ¼ I and P1   i¼1 m Ej [ M. Solution: In Problem 4.2.P9, choose X to be R, F to be the family of all subsets of R and a to be m . We get a measure l defined on all subsets of R such that the first inequality in (i) of Proposition 4.3.7 is satisfied and (ii) holds. It follows by an application of that proposition that the second inequality in (i) cannot hold. Therefore there exists a bounded interval J such that lðJ Þ ¼ 1. Being bounded, J is a subset of the union finitely many translates of any interval I of positive length. Therefore such an interval I must satisfy lðI Þ ¼ 1. Considering the manner in which l is set up in the aforementioned problem as a supremum, the assertion that lðI Þ ¼ 1 is precisely equivalent to what is to be proved here. 4.3.P8. For an increasing nonnegative real-valued function / on R satisfying /ð0Þ ¼ 0 and /ðu þ tÞ  /ðuÞ þ /ðtÞ whenever u,t 2 R, show that 0 1=2

/ð1Þ

@

Z1

11=2 0 1 11=2 Z pffiffiffiffiffiffiffiffiffiffiffiffi pffi /ð1  tÞdtA þ @ /ð 1  tÞdtA ;

0

0

where the integrals are understood in the sense of Riemann. ~A ¼ /ðmð AÞÞ, where m is Lebesgue measure, we obtain a Solution: By setting l subadditive fuzzy measure on any Lebesgue measurable subset of real numbers, in particular, on the interval [0, 1]. Consider the functions f and g on [0, 1] defined as f ðxÞ ¼ x2 and gð xÞ ¼ 1  x2 . Their distributions are given by ~f ðtÞ ¼ /ð1 

pffi tÞ

for

0  t\1 and 0 for t
1

and ~ gðtÞ ¼ /ð

pffiffiffiffiffiffiffiffiffiffiffiffi 1  tÞ

for

0  t\1

and

0

for t
1:

/ð

pffiffiffiffiffiffiffiffiffiffiffiffi 1  tÞdt:

Therefore Z

f ¼

Z1

pffi /ð1  tÞdt

0

R

Z and

g¼

Z1 0

~ Since f þ g ¼ 1 everywhere, ðf þ gÞ ¼ l½0; 1 ¼ /ð1Þ. Theorem 3.3.5 with p = 1, we arrive at the desired inequality.

On

applying

Solutions

227

4.4 Induced Measure and an Application 4.4.P1. (a) Let g: ½1; 1Þ ! R be given by gRð xÞ ¼ sin x and Ik be the interval ½2kp; ð2k þ 1Þp, where k 2 N. Show that Ik gðxÞdx ¼ 2. (b) Let f : ½1; 1Þ ! R be given by f ð xÞ ¼ 1x sin x and Ik be the interval R ½2kp; ð2k þ 1Þp, where k 2 N. Show that Ik f ðxÞdx
ð2k þ2 1Þp. R R R Solution: (a) Ik gðxÞdx ¼ ½1;1Þ gð xÞvIk ðxÞdx ¼ Ik gjIk ðxÞdx by Proposition 4.4.1. Now, by Proposition 4.3.4, Z

gjIk ðxÞdx ¼

ð2kZþ 1Þp

Ik

gjIk ðxÞdx ¼

ð2kZþ 1Þp

2kp

sin xdx 2kp

¼ ½cosðð2k þ 1ÞpÞ  cosð2kpÞ ¼ 2: R

R ð2k þ 1Þp f ðxÞdx ¼ 2kp f jIk ðxÞdx by Proposition 4.4.1 R ð2k þ 1Þp 1 R ð2k þ 1Þp 1 f jIk ðxÞdx ¼ 2kp and Proposition 4.3.4. Now, 2kp x sin xdx
ð2k þ 1Þp R ð2k þ 1Þp sin xdx ¼ ð2k þ2 1Þp. 2kp

(b) As in part (a),

Ik

4.4.P2. Let ðX; F ; lÞ be a measure space and ffk gk
1 be a sequence of integrable P1 R functions on X such that k j\1. Show that there exists an integrable k¼1 jfP converges to f on a measurable subset function f on X such that the series R1 k¼1 fk P R . HX satisfying lðH c Þ ¼ 0, and that f ¼ 1 f k P P1 R k¼1 Solution: Let g ¼ 1 k¼1 jfk j. Since R k¼1 jfk j\1, it follows by the Monotone Convergence Theorem 2.3.2 that g\1. By Proposition 2.2.9, there exists a measurable set HX such that g is finite on H and lðH c Þ ¼ 0. Thus the series P1 for all k¼1 fk is convergent on H, and its sum f0 on H satisfies jf0 ð xÞj  gðxÞ P x 2 H. Since H is measurable and restrictions of the partial sums of 1 k¼1 fk to H are measurable with reference to the induced r-algebra F H on H, therefore f0 is an F H -measurable function on H. Hence by Proposition 4.4.1, its extension f to X obtained by setting it equal to 0 on H c is real-valued and F -measurable (considering that f jH ¼ f0 and f vH ¼ f ). Moreover, the extension satisfies jf j  g on P P X and is therefore integrable. Now, j nk¼1 fk ðxÞj  gð xÞ and limn!1 nk¼1 fk ðxÞ ¼ f ðxÞ for every x 2 H. The Dominated Convergence Theorem 3.2.6, the fact that lðH c Þ ¼ 0 and Proposition 4.4.3 together lead to lim

n!1

n Z X k¼1

fk ¼ lim

n!1

n Z X k¼1

fk ¼ lim

Z X n

n!1

H

H

k¼1

fk ¼

Z H

f ¼

Z

f:

228

Solutions

4.4.P3. Let F be a r-algebra of subsets of X and A; B 2 F be nonempty subsets satisfying A [ B ¼ X. Show that a function /: X ! R is F -measurable if and only if the restrictions /jA and /jB are F A -measurable and F B -measurable respectively. Solution: Proposition 4.4.1 will be used repeatedly without reference. If /: X ! R

is F -measurable, then so are the products /vA and /vB and it follows that the restrictions /jA and /jB are F A -measurable and F B -measurable respectively. For the converse, suppose the restrictions /jA and /jB are F A -measurable and F B -measurable respectively. Then the products /vA and uvB are F -measurable and it follows that the products ð/vA ÞvB , ð/vA ÞvAnB and ð/vB ÞvBnA are also F -measurable. But ð/vA ÞvB ¼ /ðvA vB Þ ¼ /vA \ B ; ð/vA ÞvAnB ¼ /ðvA vAnB Þ ¼ /vAnB and ð/vB ÞvBnA ¼ /ðvB vBnA Þ ¼ /vBnA : Thus the products /vA \ B , /vAnB and /vBnA are F -measurable. The sum of these three products is /vA [ B , which is the same as / because it is given that A [ B ¼ X.

5.1 Interchanging the Order of Summation 5.1.P1. Let X be any nonempty set. Define F to consist of those subsets of X that are either countable or have a countable complement. (It is easy to argue that F is a r-algebra, sometimes called the “co-countable” r-algebra). Now let /: X ! R þ P be any function and set lð AÞ ¼ x2A /ð xÞ for any A 2 F , where the sum is understood in the sense of Definition 5.1.7. Show that l is a measure on F . (Note: If / [ 0 everywhere, then the only subset of measure 0 is the empty set.) Solution: Countable additivity is all that needs to be proved. Let fAn gn
1 be a sequence of disjoint sets in F and A be their union. For any finite subset BA, each of the intersections B \ An is finite and, in view of thePdisjointness P of the P An , is / ð x Þ ¼ empty for all except finitely many n. Hence x2B n x2B \ An P /ð xÞ  Pn lðAn Þ. Since this holds for every finite subset BA, it follows that lð AÞ  n lðAn Þ. To prove the reverse inequality, we first note that it is trivial when lð AÞ ¼ 1, so that we may assume lð AÞ\1. Consider an arbitrary e > 0. P For each n, there exists a finite subset Bn An such that x2Bn /ð xÞ [ lðAn Þ  2en . For each positive integer N, let CN ¼ [ Nn¼1 Bn . Then CN APand is finite. Combined P with the disjointness of the Bn , this implies lð AÞ
x2CN /ð xÞ ¼ 1  n  N P P /ð xÞ
1  n  N lðAn Þ  e. As this holds for every positive integer N, we x2BP n get n lðAn Þ  lðAÞ þ e. Since e > 0 is arbitrary, we conclude that P l ð A Þ  lð AÞ, which is the desired reverse inequality. n n 5.1.P2. Prove Proposition 5.1.9(b). Solution: Let A ¼ fx 2 X: að xÞ [ 0g. Then any subset B of X containing A has the property that a(x) = 0 whenever x 62 B. For each n 2 N, consider the set

Solutions

229

P An ¼ fx 2 X: að xÞ [ 1ng. It contains at most n  x2X að xÞ elements and therefore must be finite. The set [ 1 n¼1 An is therefore countable. But it is precisely the same set as A. 5.1.P3. Let l be a measure on a set X. Consider N with counting measure (all subsets measurable) and let f : X  N ! R be a real-valued function such that, for each k 2 N, the function f k on X described by f k ð xÞ ¼ f ðx; kÞ is measurable. Suppose also that there is an integrable function /: X ! R þPsuch that Pn j k¼1 f ðx; kÞj  /ð xÞ for each x 2 X and each n 2 N. If the series 1 k¼1 f ðx; kÞ converges, show that interchanging the order of integration is valid, which is to say, Z

1 X

! f ðx; kÞ dlðxÞ ¼

k¼1

1 Z X

f ðx; kÞdlðxÞ :

k¼1

Solution: For brevity, we shall write P dlð xÞ as simply dx. Define a sequence of functions /n on X by /n ð xÞ ¼ nk¼1 f ðx; kÞ. Then each /n is real-valued and measurable. Besides, j/n ð xÞj  /ð xÞ for every x 2 X and every n 2 N. By the Dominated Convergence Theorem 3.2.6, Z

1 X

! f ðx; kÞ dx ¼

Z



lim /n ðxÞ dx ¼ lim

n!1

k¼1

¼ lim

n!1

n Z X k¼1

Z

n!1

f ðx; kÞdx ¼

1 Z X

/n ðxÞdx



f ðx; kÞdx :

k¼1

5.1.P4. Suppose the real-valued function P a: N  N ! R has Pnthe property that there exists a sequence fbm gm
1 such that 1 b \1 and j m m¼1 k¼1 am;k j  bm for each P1 n and each m. Show that, if the series a converges for each m, then k¼1 m;k P1 P1 P1 P1 m¼1 k¼1 am;k ¼ k¼1 m¼1 ak;m. (Note: (1) The hypothesis about fbm gm
1 is P P1 am;k . The latter is not sufficient for the stronger than convergence of 1 m¼1 k¼1 purpose at hand, in view of Problem 1.2.P1. (2) A proof of the result of the present problem by using metric spaces is found in Feldman [3; p. 2]. What has been stated there has a stronger P1 palatable version of the P1hypothesis about fbm g, in that Pbut more ja j\1, so that it is assumed that 1 m¼1 k¼1 m;k k¼1 am;k not only converges for each m but does so absolutely.) Solution: This is an immediate consequence of Problem 5.1.P3: Choose X in that problem to be N, replace “x 2 X” by “m 2 N”, choose l to be counting measure, the function f to be a and /ð xÞ to be bm . Then / is integrable because P 1 bm \1, and what has been denoted by /n ð xÞ in the solution there is now Pm¼1 n k¼1 am;k .

230

Solutions

5.2 Integration with the Counting Measure 5.2.P1. Complete the proof of Theorem 5.2.3 by showing for any function a: X  Y ! R þ that X

aðx; yÞ 

ðx;yÞ2XY

XX

aðx; yÞ:

x2X y2Y

Solution: Let FX  Y be finite. Then the subsets AX and BY given by A ¼ fx 2 X: ðx; yÞ 2 F for some y 2 Yg; B ¼ fy 2 Y: ðx; yÞ 2 F for some x 2 Xg (“projections” on X and Y) are both finite and FA  B, a finite subset of X  Y. Now, X ðx;yÞ2F

aðx; yÞ 

X

aðx; yÞ ¼

ðx;yÞ2AB

XX

aðx; yÞ 

x2A y2B

XX

aðx; yÞ:

x2X y2Y

Since F is an arbitrary finite subset of X  Y, we obtain XX X aðx; yÞ  aðx; yÞ: ðx;yÞ2XY

x2X y2Y

5.2.P2. Let f : X  Y ! R be a real-valued function on the Cartesian product X  Y P and suppose that the unconditional sum un ðx;yÞ2XY f ðx; yÞ exists (necessarily finite P Pun by definition). Show that (i) y2Y f ðx; yÞ exists for each x 2 X, (ii) un x2X f ðx; yÞ Pun Pun Pun Pun exists for each y 2 Y, and (iii) x2X y2Y f ðx; yÞ ¼ y2Y x2X f ðx; yÞ. Pun Solution: Since the unconditional sum ðx;yÞ2XY f ðx; yÞ exists, it follows from P P Theorem 5.2.7 that the sums ðx;yÞ2XY f þ ðx; yÞ and ðx;yÞ2XY f  ðx; yÞ are both finite. Now, by Theorem 5.2.3, X XX X XX f þ ðx; yÞ ¼ f þ ðx; yÞ and f  ðx; yÞ ¼ f  ðx; yÞ: ðx;yÞ2XY

x2X y2Y

ðx;yÞ2XY

x2X y2Y

Therefore P the repeated sums P on the right sides are both finite. Consequently, the sums y2Y f þ ðx; yÞ and y2Y f  ðx; yÞ are finite for each x. Together with Remark P 5.2.6(d), this implies that un y2Y f ðx; yÞ exists for each x 2 X. This completes the argument for (i). The argument for (ii) is analogous.

Solutions

231

By Theorem 5.2.7, we have X X un f ðx; yÞ ¼ f þ ðx; yÞ  ðx;yÞ2XY

ðx;yÞ2XY

¼

XX

¼

un

x2X

¼

f ðx; yÞ 

X

XX

f ðx; yÞ 

X

¼ ¼ ¼

Z X Z

un

x2X

f þ ðx; yÞdx 

Z X

y2Y

¼

X

X

f  ðx; yÞ

by Remark 5:2:6ðbÞ

y2Y

f  ðx; yÞdx

by Theorem 5:2:7

y2Y

un þ

f ðx; yÞdx 

Z X

y2Y

un 

y2Y

X

un þ

f ðx; yÞ 

y2Y

Z

f  ðx; yÞ by Theorem 5:2:3

x2X y2Y þ

y2Y

Z X

f  ðx; yÞ

ðx;yÞ2XY

þ

x2X y2Y

X

X

X

! un

X

f ðx; yÞdx

by Remark 5:2:6ðbÞ

!

un 

f ðx; yÞ dx

y2Y

f ðx; yÞ dx by ðiÞ and Theorem 5:2:7

y2Y un

x2X

By a similar argument,

X

un

f ðx; yÞ

by Theorem 5:2:8:

y2Y

Pun

ðx;yÞ2XY

f ðx; yÞ ¼

Pun Pun y2Y

x2X

f ðx; yÞ.

5.2.P3. (Cf. Problem 5.1.P4) Suppose the real-valued function f : X  Y ! R has P the property that there exists a map /: X ! R þ such that x2X /ð xÞ\1 and P j y2F f ðx; yÞj  /ð xÞ for each finite subset FY and each x 2 X. Show that P Pun (i) un x2X f ðx; yÞ exists for each y 2 Y; y2Y f ðx; yÞ exists for each x 2 X, (ii) the Pun Pun repeated sum f ðx; yÞ exists, and (iii) if the other repeated sum x2X y2Y Pun Pun x2X f ðx; yÞ also exists, then the two repeated sums are equal: Py2Y P Pun un Pun ¼ un x2X y2Y f ðx; yÞ y2Y x2X f ðx; yÞ. P Solution: Since j y2F f ðx; yÞj  /ð xÞ\1 for each finite subset FY, it follows P P for each y 2 Y upon choosing F = {y} that x2X jf ðx; yÞj  x2X /ð xÞ\1 and Pun hence via Theorem 5.2.1 and Theorem 5.2.8 that x2X f ðx; yÞ exists and also that it R equals f ðx; yÞdx. P Next, consider an arbitrary x 2 X. Since j y2F f ðx; yÞj  /ð xÞ\1 for each finite subset FY, it is immediate that the P Psame is true when f is replaced either by f þ or by f  . Therefore y2Y f þ ðx; yÞ; y2Y f  ðx; yÞ  /ð xÞ\1. It now follows Pun on the basis of Remark 5.2.6(d) that y2Y f ðx; yÞ exists. This shows that Pun f ðx; yÞ exists for each x 2 X. This proves (i). y2Y

232

Solutions

From the second assertion in (i), it follows that, for each x 2 X and each n 2 N, there exists a finite Fn Y such that whenever Gn Y is a finite subset containing Fn, P P 1 the inequality j y2Gn f ðx; yÞ  un y2Y f ðx; yÞj\ n holds. Define a sequence of functions /n on X in terms of an arbitrary such sequence fGn gn
1 of sets by taking P /n ð xÞ ¼ y2Gn f ðx; yÞ. Then f/n gn
1 is a sequence of real-valued functions measurable with respect to the r-algebra of all subsets of X, which converges P everywhere to un y2Y f ðx; yÞ. Moreover,        X  X  X X       þ  þ  f ðx; yÞ  f ðx; yÞ   f ðx; yÞ þ  f ðx; yÞ  2/ðxÞ j/n ðxÞj ¼     y2G   y2G y2G y2G n

n

n

n

for every x 2 X and every n 2 N. Hence the Dominated Convergence P Theorem 3.2.6 implies that un y2Y f ðx; yÞ ¼ limn!1 /n ð xÞ is integrable with respect to counting measure on X, so that by virtue of Theorem 5.2.8, it follows that Pun Pun x2X y2Y f ðx; yÞ exists. Thus, (ii) stands proved. In order to justify (iii), we begin by noting that by virtue of Theorem 5.2.8, it also follows that X x2X

un

X

un

f ðx; yÞ ¼

y2Y

Z

X

! un

f ðx; yÞ dx:

y2Y

Moreover, the Dominated Convergence Theorem also implies that Z

Z X Z Z X un ð f ðx; yÞÞdx ¼ ð lim /n ðxÞÞdx ¼ lim ð /n ðxÞdxÞ ¼ lim f ðx; yÞdx n!1

y2Y

¼ lim

n!1

XZ y2Gn

n!1

f ðx; yÞdx ¼ lim

n!1

n!1

XX

un

y2Gn

f ðx; yÞ:

y2Gn x2X

Taking advantage of the fact that this equality holds for an arbitrary sequence fGn gn
1 of finite sets Gn such that Fn Gn Y, we can now choose Gn , pursuant to P Pun the existence of un y2Y x2X f ðx; yÞ, in such a manner as to satisfy    1 X X X X   un un un f ðx; yÞ  f ðx; yÞ\   n y2G x2X y2Y x2X

for every n 2 N:

n

R P P Pun Then the foregoing equality leads to ð un f ðx; yÞÞdx ¼ R un f ðx; yÞ. y2Y Pun Pun y2Y Pun x2X Since it has already been established that x2X y2Y f ðx; yÞ ¼ ð y2Y f ðx; yÞÞdx, the required equality of repeated sums shines forth.

Solutions

233

6.1 Algebras and Monotone Classes of sets 6.1.P1. Suppose the set X consists of three elements a, b, c and B is the family of subsets {a} and {b, c}. List all the subsets in (i) the smallest algebra of sets that contains B and (ii) the monotone class generated by B. Solution: (i) ;, {a},{b, c},X (ii) {a},{b, c}. 6.1.P2. Let X consist of four elements a, b, c, d and S consist of the subsets of X described in unorthodox notation as ;, a, bcd, bc, ad, abcd. Determine (i) whether S is an algebra, (ii) whether S is a monotone class, (iii) what the monotone class generated by S is, and (iv) what the r-algebra generated by S is. Solution: (i) S is not closed under unions and is therefore not an algebra. (ii) Being a finite family, it is certainly a monotone class. (iii) Consequently, the monotone class generated by it is itself. (iv) When S is enlarged by including abc and its complement d, what we obtain can be verified to be an algebra and therefore a r-algebra in view of its finiteness. Hence this enlarged family is the r-algebra generated by S. 6.1.P3. Let X be an uncountable set and B be the family of all subsets that are either finite or have finite complement. Show that B is an algebra of subsets. Is it a r-algebra? Solution: Surely B contains ; and is closed under complements. To show that it is closed under unions, consider any A; B 2 B. If both are finite, then so is A [ B and therefore A [ B 2 B. If one of them is not finite, then its complement must be finite, which implies that the complement of the bigger set A [ B is finite and hence that A [ B 2 B. Thus B is an algebra. However, it is not a r-algebra, because a countably infinite subset fails to be in B despite being a countable union of subsets belonging to B. 6.1.P4. Let A be an algebra of subsets of a nonempty set X. If there are finitely many sets in A, show that the number of sets in it is 2n, where n is the number of nonempty sets B in the algebra (called ‘atoms’) having the property that the only set B1 2 A satisfying ; B1 B must be B. Solution: Our first step will be show that every nonempty set A in the algebra contains an atom. Suppose A 2 A contains no atom. Then there exists an A1 2 A satisfying ; A1 A. Now, A1 also contains no atom and hence there exists an A2 2 A satisfying ; A2 A1 . Proceeding in this manner, we obtain infinitely many sets A1 ; A2 ; . . . in A, contradicting the finiteness of A. This contradiction establishes what was promised in the first step. The second step is to note that any two distinct atoms must be disjoint. The reason is that nonemptiness of their intersection would, by definition of an atom, make it equal to each one of the atoms, contradicting their distinctness. The third step is to show that every nonempty set A in the algebra is the union of all the atoms it contains. In accordance with the first step, A does contain at least one atom and they must be finitely many because A is finite. Let B be their union. Then A B 2 A and the contention is that A = B. If not, then C ¼ AnBA is a nonempty set in A and is disjoint from B. In accordance with the first step, there exists an atom A0 CA. This implies that A0 is an atom contained in A but disjoint from B, which is not possible because of the way B was set up.

234

Solutions

The fourth and final step is to show that there is a bijection between the family of all subsets = of the n-element set T of all atoms and the algebra A. As this family of subsets has 2n elements, it will follow that A also has 2n elements. Consider the map from the family into A given by = ! [ =. Recall that = ¼ ; ) [ = ¼ ;. Therefore we know from the third step that the map is surjective. It remains only to show that it is injective. Let =1 and =2 be distinct subsets of T , i.e. distinct sets of atoms. One of them must contain an atom A0 that does not belong to the other. To be specific, suppose A0 2 =1 but A0 62 =2 . Our second step guarantees that A0 is disjoint from every atom belonging to =2 and hence also from [ =2 . Since [ =1 A0 6¼ ;, it follows that [ =1 6¼ [ =2 . 6.1.P5. Let F be a r-algebra of subsets of X generated by a subfamily H and let A 2 F . Show that the induced r-algebra F A is the same as the r-algebra generated by the family HA ¼ fH \ A: H 2 Hg of intersections of the sets of H with A. Solution: Let G be the r-algebra of subsets of A generated by HA . We have to show that G ¼ F A . Since HF , it is clear that HA F A and hence that GF A . We shall show that G F A. For any family S consisting of subsets of X, the symbol ‘S A ’ will, in conformity with the symbols ‘HA ’ and ‘F A ’, denote the family consisting of subsets of A that are intersections with A of sets belonging to S. In symbols, S A ¼ fS \ A: S 2 Sg: Note that for any collection M of families S consisting of subsets of X, \

SA

S2M

\

! S

S2M

: A

Take M to be the collection of all r-algebras of subsets of X that contain H. Then \ S2M S is given to be the same as F and therefore the right side of the above inclusion is the induced r-algebra F A . Consequently, the inclusion can be stated as \

SA F A:

S2M

It remains only to be shown that the intersection on the left side here equals G. We first show that it contains G. Let S be an arbitrary family belonging to M, i.e. an arbitrary r-algebra of subsets of X that contains H. Then S A is clearly a r-algebra of subsets of A and contains HA . Therefore S A G. Since this is true for any S 2 M whatsoever, it follows that \ S2M

S A G:

Solutions

235

With the aim of proving the reverse inclusion, consider an arbitrary r-algebra K of 0 subsets of A that contains HA . Let S be the family of those subsets of X whose 0 0 intersection with A belongs to K, so that S A ¼ K. Also, S is a r-algebra of subsets of X that contains H, which is to say, S 0 2 M. Therefore the intersection of all such K must contain the intersection on the left side of the foregoing inclusion. However, the intersection of all such K is nothing but G, and therefore the preceding statement means that the reverse of the foregoing inclusion holds. Thus, \

S A ¼ G;

S2M

which is all that remained to be shown.

6.2 Defining a Product r-algebra 6.2.P1. Let F and G be r-algebras of subsets of X and Y respectively. Suppose 1 X ¼ [1 n¼1 Xn and Y ¼ [ k¼1 Yk , where fXn g and fYk g are disjoint sequences of sets. Then F n ¼ fF \ Xn : F 2 F g and Gn ¼ fG \ Yk : G 2 Gg are r-algebras of subsets of Xn and Yk respectively. (If Xn and Yk are all measurable, F n and Gk are the induced r-algebras in the sense of Sect. 4.4.) Show that every E 2 F  G satisfies E \ ðXn  Yk Þ 2 F n  Gk for every n and k Solution: Let H ¼ fEX  Y: E \ ðXn  Yk Þ 2 F n  Gk g. By Problem 2.1.P9, this is a r-algebra. First we note that H contains every rectangle AB. This is because, by definition of F n and Gk , A \ Xn 2 F n ;

B \ Yk 2 Gk

and hence ðA  BÞ \ ðXn  Yk Þ ¼ ðA \ Xn Þ  ðB \ Yk Þ 2 F n  Gk for every n and k. Since H is a r-algebra containing every rectangle, it follows by definition of F  G that H F  G. By definition of H, this inclusion implies the assertion we seek to prove. 6.2.P2. Prove Proposition 6.2.6. Solution: It is sufficient to prove that P\Q is an elementary set whenever P and Q are rectangles, because the rest will then follow by Proposition 6.2.5. Let P = A  B and Q = CD, where A,C 2 F and B, D 2 G. Then

236

Solutions

PnQ ¼ ððAnCÞ  BÞ [ ððA \ CÞ  ðBnDÞÞ; which is an elementary set. (A∩C)×(B\ D)

A×B

(A\ C)×B

C×D

Illustration of the set-theoretic equality in the proof of Proposition 6.2.6 6.2.P3. Let F and G be the r-algebras consisting of all subsets of X and Y respectively, where X and Y are both countable. Show that F  G consists of all subsets of XY. Solution: Let (a, b) be any point in XY. Since F and G consist of all subsets of X and Y respectively, the sets fagX and fbgY belong to F and G respectively. Therefore {a}{b} = {(a, b)} is a rectangle. By Definition 6.2.3, it must belong to F  G. But X  Y is countable and therefore so is any subset of it. Therefore any subset of it is a countable union of sets of the form {(a, b)}. It follows that it must be in F  G. 6.2.P4. Let F and G be r-algebras of subsets of X and Y respectively. Suppose A 2 F and B 2 G. Then the r-algebra ðF  GÞAB induced by the product r-algebra F  G on A  B is the same as the product F A  GB of the induced r-algebras F A and GB . In particular, if f : X  Y ! R þ is ðF  GÞ-measurable then the restriction of f to A  B is ðF A  GB Þ-measurable. Solution: Let H be the family of rectangles R in XY. By definition, F  G is the r-algebra generated by H. By the result of Problem 6.1.P5, the r-algebra ðF GÞAB is the same as the r-algebra generated by the family HA ¼ fR \ ðA  BÞ: R 2 Hg of intersections of the sets of H with AB. Now, F A  GB is, by definition, the r-algebra generated by the rectangles in A  B. It is sufficient therefore to show that HA is precisely the family of rectangles in AB. It is obvious that a rectangle in A  B is a rectangle in X  Y, which means it equals R \ ðA  BÞ, where R 2 H. Now consider any rectangle P  Q in X  Y. By definition, R = P  Q, where P 2 F and Q 2 G. Then P \ A 2 F A and Q \ B 2 GB and therefore ðP \ AÞ  ðQ \ BÞ is a rectangle in A  B. However, ðP \ AÞ  ðQ \ BÞ ¼ ðP  QÞ \ ðA  BÞ ¼ R \ ðA  BÞ 2 HA .

6.3 Sections of a Subset and of a Function 6.3.P1. Let X, Y and V be as in Example 5.1.3. For each x 2 X, describe the set Vx and for each y 2 Y, describe the set V y . If f ¼ vV , describe the function fx for each x 2 X.

Solutions

237

Solution: By Definition 6.3.1, Vx ¼ fy 2 Y: ðx; yÞ 2 Vg and by the definition of V in Example 5.1.3, V ¼ fðx; yÞ 2 ½0; 1  ½0; 1: x ¼ yg. Therefore Vx ¼ fy 2 Y: x ¼ yg ¼ f xg. Similarly, V y ¼ fyg. By Definition 6.3.6, the function fx has domain Y and satisfies fx ð yÞ ¼ f ðx; yÞ for all y 2 Y. Since f ðx; yÞ ¼ vV ðx; yÞ, we have fx ð yÞ ¼ vV ðx; yÞ ¼ 1 if (x, y) 2 V and 0 otherwise. Taking into account what V is given to be, we find that fx ð yÞ ¼ 1 if y = x and 0 otherwise. (In other words, fx ¼ vfxg .)

6.4 Defining a Product Measure 6.4.P1. Prove Theorem 6.4.4. Solution: Since the function mðEx Þ defined on X is nonnegative and measurable by Proposition 6.4.2, it has an integral, though not necessarily finite. We need only show that l  m is a measure, because the r-finiteness will then follow trivially. Clearly, ðl  mÞð;Þ ¼ 0 and ðl  mÞðEÞ
0 for every E 2 F  G. Consider

a sequence fEn gn
1 of disjoint sets of F  G. The sets in the sequence ðEn Þx n
1 of their sections are disjoint and consequently, on the basis of the Monotone Convergence Theorem 2.3.2, ðl  mÞð

1 [ n¼1

¼

Z X 1 X

n¼1

En Þ ¼

Z mðð X

1 [

En Þx ÞdlðxÞ ¼

n¼1

mððEn Þx ÞdlðxÞ ¼

1 Z X n¼1

X

Z mð X

1 [

ðEn Þx ÞdlðxÞ

n¼1

mððEn Þx ÞdlðxÞ ¼

1 X

ðl  mÞðEn Þ:

n¼1

Hence l  m is countably additive and therefore a measure. 6.4.P2. For any positive integer n, set Ij ¼ ½j1 ; j . For the set V of Example 5.1.3,  n n  1 n show that V ¼ \ n¼1 Vn , where Vn ¼ [ j¼1 Ij  Ij . Solution: Consider any point (x, x) 2 V. For each n, we have ½0; 1 ¼ [ nj¼1 Ij and therefore there exists a j such that x 2 Ij . This j satisfies ðx; xÞ 2 Ij  Ij , so that x 2 Vn . Since this holds for every n, it follows that x 2 \ 1 n¼1 Vn . This shows that V \ 1 n¼1 Vn . Now consider any ðx; yÞ 2 ½0; 1  ½0; 1 such that x 6¼ y and we shall argue that ðx; yÞ 62 \ 1 n¼1 Vn . We may take x [ y. Choose any integer n [ 1=ðx  yÞ. If there were to exist a j such that ðx; yÞ 2 Ij  Ij , then we would have x 2 Ij and y 2 Ij , which implies x  y\1=n, a contradiction. This means there can be no such j. In other words, for this particular n, there is no j such that ðx; yÞ 2 Ij  Ij , i.e.   ðx; yÞ 62 [ nj¼1 Ij  Ij ¼ Vn . The existence of such an integer n means that ðx; yÞ 62 \ 1 n¼1 Vn .

238

Solutions

6.5 The Tonelli and Fubini Theorems 6.5.P1. Let X = Y = [1, 1) with Lebesgue measure m. Suppose f : X  Y ! R satisfies f ð0; 0Þ ¼ 0

and

f ðx; yÞ ¼

x2  y2 ðx2 þ y2 Þ2

:

Show that f cannot be integrable even if it is measurable. Solution: The function that maps y 2 [1, 1) into y12 has Riemann integral 1  B1 over any interval [1, B] and therefore has the same Lebesgue integral by Theorem 4.3.4. It follows on applying the Monotone Convergence Theorem 2.3.2 that it is has Lebesgue integral 1 over [1, 1). Next, the section fx satisfies jfx j  y12 , which makes it Lebesgue integrable over [1, 1). Therefore by Theorem 4.4.2, the Lebesgue integral may be computed as the improper integral, which can be done by noting 2 2 that @y@ x2 þy y2 ¼ ðxx2 þyy2 Þ2 , thereupon obtaining the value 1 þ1 x2 . It follows that one repeated integral is given by Z

Z dmðxÞ X

Y

Z  f ðx; yÞdmðyÞ ¼  X

Z1  1 1 p  dmðxÞ ¼ dx ¼  : 2 2 1þx 1þx 4 1

For the other repeated integral, an analogous computation leads to a value that turns out to be the negative of what has just been computed. Thus the two repeated integrals do not agree, and therefore Fubini's Theorem 6.5.5 yields the conclusion that f cannot be integrable. 6.5.P2. If X and Y are both countable, show that the product measure of the counting measures on them is the counting measure on X  Y. Solution: Denote by l and m the counting measures on X and Y respectively. Their domains F and G are the r-algebras of all subsets of X and Y respectively. We know from Problem 6.2.P3 that the product r-algebra F  G on X  Y consists of all subsets. It is sufficient to show for each ða; bÞ 2 X  Y that ðl  mÞfða; bÞg ¼ 1. For any x 2 X, the section fða; bÞgx is {b} if x = a and ; if x 6¼ a. Consequently, mfða; bÞgx ¼ mfbg ¼ 1 if x = a and 0 if x 6¼ a. In other words, the R function on X given by x ! mfða; bÞgx is the characteristic function of {a}. Hence X mfða; bÞgx dlð xÞ ¼ lfag ¼ 1. By Definition 6.4.3, this means ðl  mÞfða; bÞg ¼ 1. 6.5.P3. Show that repeated integrals with respect to counting measures can be unequal even if the function on the product space is measurable. Hint: See Problem 1.2.P1. Solution: Consider the function a: N  N ! R of Problem 1.2.P1. As seen in that problem, the repeated sums are unequal. However, the repeated sums are precisely the repeated integrals with respect to counting measure. Since the product r-algebra consists of all subsets by Problem 6.2.P3, the function a is measurable. (Fubini's Theorem 6.5.5 now implies that a cannot be integrable with respect to counting

Solutions

239

measure on N  N; however, the lack of integrability can also be obtained directly from the definition of a by taking into account the infinitude of the points (p, p) where it takes the value 1, or alternatively, the infinitude of the points (p + 1, p).) 6.5.P4. Give an example of two functions on a measure space that are equal almost everywhere, one is measurable but the other is not. Solution: Let X = {a, b, c}, F ¼ f;; fa; bg; fcg; Xg and the measure l on F satisfy lðfa; bgÞ ¼ 0. Consider the functions f and g on X such that f ðaÞ ¼ f ðbÞ ¼ 1; f ðcÞ ¼ 2 and gðaÞ ¼ 0; gðbÞ ¼ 1; gðcÞ ¼ 2. Then f is measurable but g is not (because X ðg ¼ 0Þ ¼ fag 62 F ). But f ð xÞ ¼ gð xÞ for x 62 fa; bg and lðfa; bgÞ ¼ 0. (Note that it is crucial here that the set {a, b} has measure 0 and has a nonmeasurable subset {a}, a phenomenon that cannot occur with Lebesgue measure by Remark 4.2.3(c).) 6.5.P5. Suppose ffn gn
1 is a sequence of measurable functions such that limn!1 fn ð xÞ exists for all x in some measurable set of convergence A. Show that the function defined to agree with the limit on A and to be 0 elsewhere is measurable. Show also that two different measurable functions obtained in this manner by using two different sets of convergence, both having complement of measure 0, agree a.e. Solution: Suppose A is a measurable set such that limn!1 fn ð xÞ exists for all x 2 A. Denote by f the function that is defined to agree with the limit on A and to be 0 on Ac . We have to show that f is measurable. But f ð xÞ ¼ f vA ð xÞ ¼ limn!1 fn vA ð xÞ for all x, whether in A or in Ac . Each fn vA is measurable, and hence the limit f is measurable. If B is another such measurable set and lAc ¼ lBc ¼ 0, then the function obtained from B in the same manner agrees with f on C ¼ A \ B and lC c ¼ lðAc [ Bc Þ ¼ 0. 6.5.P6. Let ðX; F ; lÞ and ðY; G; mÞ be measure spaces and f : X ! R; /: X ! R be measurable functions such that f ¼ / a.e. Suppose g: Y ! X has the property that F 2 F ) g1 ðFÞ 2 G, so that the compositions f  g: Y ! R and /  g: Y ! R are both measurable. Is it necessary that f  g ¼ /  g a.e.? Solution: No. It can happen that ðf  gÞð yÞ 6¼ ð/  gÞð yÞ for all y 2 Y. Let f be the characteristic function of a nonempty subset A of X having measure 0 and g be any measurable function with range contained in A. Then f ¼ / a.e., where / ¼ 0 everywhere, but ðf  gÞð yÞ ¼ 1 6¼ 0 ¼ ð/  gÞð yÞ for all y 2 Y. 6.5.P7. Let f : ½0; 1  ½0; 1 ! R be the function given by (1,1)

0 in this pentagon _ (0, 12 )

_

(1, 12 )

(x

_ 1_

_

2)

3

(x

_ 1_

_

2)

3

f ðx; yÞ ¼ ðx  12Þ3 if 0\y\jx  12 j and f(x, y) = 0 otherwise.

240

Solutions

R R Show that the repeated integral ½0;1 dx ½0;1 f ðx; yÞdy does not even exist but the R R repeated integral ½0;1 dy ½0;1 f ðx; yÞdx exists.   Solution: Suppose x 2 ½0; 1Þ. Then jx  1 j ¼ 1  x [ 0 and therefore, y\x  1 , 2

2

2

2

x  12 \y\ 12  x , y\ 12  x, considering that x  12 \0 and y [ 0. Consequently, Z

f ðx; yÞdy ¼

½0;1

Z

1 3 2

ðx  Þ dy ¼ ðx  Þ2 1 2

1 2

for x 2 ½0; Þ:

½0;12x

    Now suppose x 2 ð12 ; 1. Then x  12 ¼ x  12 [ 0 and therefore, y\x  12 , 1 1 1 1 2  x\y\x  2 , y\x  2, considering that 2  x\0 and y > 0. Consequently, Z

f ðx; yÞdy ¼

½0;1

Z

1 3 2

ðx  Þ dy ¼ ðx  Þ2 1 2

1 2

for x 2 ð ; 1:

½0;x12

From the last two computations of integrals, we obtain 0 B @

Z

1þ

0

1 C B f ðx; yÞdyA ¼ ðx  Þ2 v½12;1 ðxÞ and @ 2

½0;1

1

Z

C f ðx; yÞdyA

½0;1

¼ ðx  Þ2 v½0;12 ðxÞ: 1 2

R R R R It is now elementary that ½0;1 dxð ½0;1 f ðx; yÞdyÞ þ and ½0;1 dxð ½0;1 f ðx; yÞdyÞ are both 1.   For all x 2 [0, 1], we have jx  12 j  12, and therefore 12  y  1 implies y
x  12 R and hence f ðx; yÞ ¼ 0. It follows that ½0;1 f ðx; yÞdx ¼ 0 for 12  y  1 So, consider 0\y\ 12. We write Z

f ðx; yÞdx ¼

½0;1

Z

f ðx; yÞdx þ

½0; 

Z

f ðx; yÞdx:

½ ;1

1 2

1 2

Once again, suppose x 2 ½0; 12Þ. Then jx  12 j ¼ 12  x [ 0 and therefore, y\jx  12 j , x  12 \y\ 12  x , y\ 12  x, considering that x  12  0 and y > 0. However, y\ 12  x , x\ 12  y and we know that 12 [ 12  y [ 0. Consequently, Z ½0;12

f ðx; yÞdx ¼

Z

12y  1 3 1 1 1 ðx  Þ dx ¼  ðx  Þ2  ¼  y2 þ 2: 2

½0;12y

2

2

0

2

Solutions

241

Now suppose x 2 ð12 ; 1. Then jx  12 j ¼ x  12 [ 0 and therefore, y\jx  12 j , 1 1 1 1 2  x\y\x  2 , y\x  2, considering that 2  x  0 and y > 0. However, 1 1 1 1 y\x  2 , x [ 2 þ y and, we know that 2 \ 2 þ y\1. Consequently, Z

Z

f ðx; yÞdx ¼

½12;1

12 þ y  1 3 1 1 1 ðx  Þ dx ¼ ðx  Þ2  ¼ y2  2: 2

2

2

2

1

½12 þ y;1

From the last two computations R Rof integrals, we obtain 0\y\ 12. It follows that ½0;1 dy ½0;1 f ðx; yÞdx ¼ 0.

R

½0;1

f ðx; yÞdx ¼ 0 also for

6.6 Borel Sets 6.6.P1. (Cf. Problem 6.5.P4) Suppose f and g are functions on [a, b] or R such that f = g a.e. If f is Lebesgue measurable, show that g is also Lebesgue measurable. Solution: Let E½a; b ¼ X, or ER ¼ X, have measure 0 and f ð xÞ ¼ gð xÞ for x 62 E. For any real t, we have Xðg [ tÞ ¼ ðXðg [ tÞ \ EÞÞ [ ðXðg [ tÞ \ E c Þ ¼ ðXðg [ tÞ \ E Þ [ ðXðf [ tÞ \ Ec Þ: Now ðX ðf [ tÞ \ E c Þ is Lebesgue measurable because f is Lebesgue measurable. The set ðX ðg [ tÞ \ EÞ is also Lebesgue measurable because it is a subset of E and therefore has measure 0, which makes it Lebesgue measurable by Remark 4.2.3(c). 6.6.P2. Prove Lemma 6.6.5. Solution: Since E is measurable, its Lebesgue measure m(E) is the same as its Lebesgue outer measure m (E), which is given to be finite. As noted in Remark 4.1.4, there exists a sequence fIn gn
1 of open intervals such that E

1 [

In

ðSol:11Þ

n¼1

and

P1

n¼1

‘ðIn Þ  m ðEÞ þ g; i.e. mð

1 [ n¼1

In Þ 

P1

n¼1

1 X

mðIn Þ  mðEÞ þ g, which implies

mðIn Þ  mðEÞ þ g\1:

n¼1

   Together with (Sol.11), this inequality leads to m [ 1 n¼1 In nE  g. 6.6.P3. Let F be a r-algebra of subsets of a nonempty set X and F: X ! R be an F -measurable map. Show that any Borel set A satisfies F 1 ð AÞ 2 F .

242

Solutions

Solution: Take G to be the family of subsets fBR: F 1 ðBÞ 2 F g of R. We have to show that every Borel set belongs to G. By Problem 2.1.P11, in which we take Y ¼ R, the family G is a r-algebra. By Definition 2.1.4 of F -measurability, every interval of the form (t, 1) belongs to G. Since G is a r-algebra, every open interval belongs to G. By Definition 6.6.1 of Borel sets, every Borel set belongs to G.

7.1 Integrability and Step Functions 7.1.P1. A subset UR is said to be openif for every u 2 U, there exists a d [ 0 such that ðu  d; u þ dÞU. (Obviously, any open interval is an open subset; an interval that is an open subset is an open interval; any union of open subsets (or open intervals) is an open subset.) Show that any open subset UR is a countable union of disjoint open intervals. Solution: This is trivial for U ¼ ; because f;g is a countable family of disjoint open intervals having union ;. So, assume U 6¼ ;. For each u 2 U, let Ku be the union of all open intervals that are contained in U and have u as an element. The definition of open subset shows that some such interval must always exist. Then u 2 Ku U. In order to see why Ku must be an interval, suppose x\z\y, where x; y 2 Ku . By definition, there is exist open intervals IKu and JKu such that u 2 I \ J and x 2 I; y 2 J. Now it follows as in Exercise 7.1.2 that z 2 I [ JKu . Thus Ku is always an interval. Being a union of open subsets of R, it must again be an open subset and hence an open interval. It is straightforward that U is the union of the family fKu : u 2 Ug of open intervals. We shall show that that the intervals in the family are disjoint. In fact, if Ku and Kt are not disjoint, then the open set Ku [ Kt is an interval by Example 7.1.2 and hence an open interval; but it is contained in U and has both u and t as elements, which renders it a subset of Ku as well as of Kt , which is possible only if it equals both Ku and Kt . That the intervals are countable in number is an immediate consequence of the observation that every nonempty open interval has to  contain a rational number. j 7.1.P2. Let X = [0, 1] and consider the intervals j1 ; k k ; 1  j  k; k 2 N, arranged in some order (without repetition). Show that the sequence ffn gn
1 of characteristic functions on X of the respective intervals does not converge at any point, although their integrals converge to zero. Solution: The number of intervals of length 1k or greater is finite (in fact, 12 k ðk þ 1Þ). Therefore, regardless of the order in which the intervals may be arranged, for every k 2 N, there exists an N 2 N such that all intervals from the Nth one onwards have length less than 1k, so that the integral of fn is less than 1k when n
N. This shows that the integrals converge to zero. On the other hand, for any x 2 [0, 1] and any k > 2, j one or two of the intervals j1 k ; k , 1  j  k, must include x while others must exclude x, which shows that fn ð xÞ ¼ 1 for some arbitrarily large n while fn ð xÞ ¼ 0 for another arbitrarily large n.

Solutions

243

7.1.P3. (a) Both the figures below show a stick and a second stick with the overlap of their shadows underneath. They also show a third stick whose shadow overlaps with some part of the existing overlap of the shadows of the first two sticks (i.e. all three shadows together have some overlap). In the figure on the left, if the second stick is removed, the combined shadow of the remaining two sticks is the same as that of all three sticks. In the figure on the right, if the first stick is removed, the combined shadow of the remaining two sticks is the same as that of all three sticks. Question: Can you place three sticks in such a way that all three shadows together have some overlap but if any one among the three sticks is removed, the combined shadow of the remaining two sticks is no longer the same as that of all three sticks?

third stick

third stick second stick

second stick stick

shadow overlap

combined shadow of all three sticks

stick

shadow overlap

combined shadow of all three sticks

(b) Suppose F is a finite nonempty family of bounded intervals. Show that there exists a subfamily GF such that [ G ¼ [ F and no interval of G is contained in the union of two other intervals of G. (c) Can the finiteness hypothesis of part (b) be dropped? Solution: (a) The interpretation of the question in terms of intervals is as follows. Can there be three bounded intervals such that the intersection of all three is nonempty but none is contained in the union of the other two? We shall argue that there cannot be. In other words, given three bounded intervals with a nonempty intersection, one of the three must be contained in the union of the other two. To begin with observe that, for a bounded interval of positive length, a number a is its left endpoint if and only if it contains no number less than a but contains infinitely many numbers of the form a þ 1n, where n 2 N. Analogously for the right endpoint.

244

Solutions

If one of the three intervals in question has length zero, then the matter is trivial because then the interval must be equal to the intersection and therefore be contained in both the other intervals. We may therefore assume that all three intervals are of positive length. On applying the result of Exercise 7.1.2 twice, we find that the union of all the three intervals is also an interval, which must of course be bounded and have positive length. Let its left endpoint be a and right endpoint be b > a. In view of the observation in the preceding paragraph, the union then contains no number less than a but contains infinitely many numbers of the form a þ 1n, each of which must of course belong to one or the other of the three intervals. It follows that at least one among the three must contain infinitely many such numbers but none less than a, and in view of the above observation once again, must have left endpoint a. Name one such interval as A. When the union contains a, the interval A can be so chosen as to contain a. First consider the case when the other endpoint of A is b. Then the interiors of both the other intervals are contained in A. In the event that neither of them contains b, both are contained in A and the required conclusion follows. In the event that one of them contains b, call that one B, the remaining interval C is contained in A [ B and the required conclusion is found to hold. Next consider the case when the other endpoint of A is not b. Then, as argued regarding a, the number b must be the right endpoint of some other interval, which we shall call B. As before, when the union contains b, the interval B can be so chosen as to contain b. Now, A [ B is an interval. Its left endpoint must be a and its right endpoint must be b. Thus the interior of A [ B is (a, b). Moreover, if a [resp. b] is in the union of all three intervals, then it is in A [resp. B]. This implies that A [ B coincides with the union A [ B [ C of all three intervals. Therefore CA [ B [ C ¼ A [ B. This shows that C is contained in the union of the other two. (b) Induction on the number of intervals in F . If the number of intervals in F is less than three, there is nothing to prove. If there are precisely three intervals in F , then the result about intervals proved in (a) immediately yields the required subfamily G. Now suppose the result is true for every family containing k intervals (induction hypothesis), and consider a family F containing k + 1 intervals. If F already has the property that no interval of F is contained in the union of two other intervals of F , then we may choose G ¼ F . Otherwise, one of the intervals of F is contained in the union of two others. The subfamily F 0 obtained by deleting that interval from F has the property that it contains k intervals and [ F 0 ¼ [ F . By the induction hypothesis, F 0 contains a subfamily G such that no interval of it is contained in the union of two others and [ G ¼ [ F 0 . Then G is a subfamily of F such that no interval of it is contained in the union of two others and [ G¼ [ F 0¼ [ F . (c) No. Let F be the family of open intervals 1n ; 3  1n . Then [ F ¼ ð0; 3Þ. Moreover, this interval cannot be the union of any finite subfamily of F while any infinite subfamily has the property that each interval of the subfamily is contained in another one, à fortiori in the union of two others.

Solutions

245

7.2 The Vitali Covering Theorem 7.2.P1. Let C be a Vitali covering of a subset AR and J be a family of closed intervals such that the union of their interiors contains A. Show that the family C0 ¼ fI \ J: I 2 C; J 2 J g of closed intervals consisting of the intersections of intervals belonging to C with intervals belonging to J is a Vitali covering of A. Solution: Consider any x 2 A and any e [ 0. Since C is a Vitali covering of A, there exists an I 2 C for which x 2 I and 0\mðIÞ\e. And since the union of the interiors of the intervals belonging to J contains A, there also exists a J 2 J such that x belongs to the interior J o of J. Then x 2 I \ J o I \ J 2 C0 and mðI \ JÞ  mðIÞ\e. On the other hand, I \ J o is an intersection of open intervals and is nonempty because it contains x. Therefore it is an interval of positive length, which means it has positive measure. Since I \ J o I \ J, it follows that mðI \ JÞ [ 0. 7.2.P2. If I and J are intervals of finite positive lengths and I \ J 6¼ ;, show that the distance of any point in I from the center of J is at most mðIÞ þ 12 mðJÞ. Solution: Denote the center of J by c. Since I \ J 6¼ ;, there exists some z 2 I \ J. As z 2 J, we have jz  cj  12 mðJÞ. As z 2 I, any x 2 I satisfies jx  zj  mðIÞ. It follows for any x 2 I that jx  cj  jx  zj þ jz  cj  mðIÞ þ 12 mðJÞ.

7.3 Dini Derivatives 7.3.P1. Prove the following partial converse of the “consequence” in Remark 7.3.6 in the case of D þ f ðcÞ: If a 2 R [ f1; 1g and there exists a sequence fhn gn
1 of positive numbers as in the remark (in which case a 6¼ 1 of course), then a  D þ f ðcÞ. Solution: We may assume that fhn gn
1 is decreasing. Consider any d > 0. Since there exists an n with 0\hn \d, the set Qd contains numbers greater than a, so that supQd [ a. Since this inequality holds for all d > 0, it follows that a  inf fsup Qd : 0\dg ¼ D þ f ðcÞ. 7.3.P2. Show that a\D þ f ðcÞ if and only if for some b [ a there is a sequence fhn gn
1 of positive numbers such that hn ! 0 and

f ðc þ hn Þ  f ðcÞ [ b for every n 2 N: hn

Solution: Suppose a\D þ f ðcÞ. Then there exists a b [ a such that b\D þ f ðcÞ. Remark 7.3.6 now implies that the required sequence fhn gn
1 of positive numbers exists. The converse follows trivially from Problem 7.3.P1.

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Solutions

7.3.P3. Give an example of continuous functions f and g on an interval containing 0 such that D þ f ð0Þ þ D þ gð0Þ 6¼ D þ ðf þ gÞð0Þ. Solution: Let f be the function on R defined by f ð0Þ ¼ 0 and f ðxÞ ¼ j xjcos 1x for x 6¼ 0. When h [ 0, the difference quotient works out to be cos 1h, which has the value 1 when h ¼ ð2pnÞ1 and n 2 N. Also cos 1h  1 for all h 6¼ 0. One can always choose n such that ð2pnÞ1 \d. So, sup Qd ðf Þ ¼ 1. It follows that D þ f ð0Þ ¼ inffsup Qd ðf Þ: 0\dg ¼ 1. On the other hand, the difference quotient cos 1h has the value −1 when h ¼ ðpð2n þ 1ÞÞ1 and n 2 N. Also cos 1h
 1 for all h 6¼ 0. One can always choose n such that ðpð2n þ 1ÞÞ1 \d. So, inf Qd ðf Þ ¼ 1. It follows that D þ f ð0Þ ¼ supfinf Qd : 0\dg ¼ 1. Now let g ¼ f . By Exercise 7.3.7, we have D þ gð0Þ ¼ D þ f ð0Þ ¼ 1. Therefore D þ f ð0Þ þ D þ gð0Þ ¼ 2. However, f þ g ¼ 0 and therefore D þ ðf þ gÞð0Þ ¼ 0. 7.3.P4. If g ¼ f , show that D gðcÞ ¼ D f ðcÞ. Solution: It is immediate from the definition that Qd ðgÞ ¼ Qd ðf Þ. It follows for any d [ 0 that sup Qd ðgÞ ¼ inf Qd ðf Þ. Therefore inf fsup Qd ðgÞ: 0\dg ¼ supfinf Qd ðf Þ: 0\dg. By Definition 7.3.4, this means D gðcÞ ¼ D f ðcÞ. 7.3.P5. Suppose f and g are defined on a common interval I and c 2 I o . Show that D þ ðf þ gÞðcÞ
D þ f ðcÞ þ D þ gðcÞ, provided that the sum on the right side is meaningful. What about D þ ðf þ gÞðcÞ  D þ f ðcÞ þ D þ gðcÞ? Solution: Obviously, we may assume that D þ f ðcÞ 6¼ 1 6¼ D þ gðcÞ. Consider arbitrary a and b such that a\D þ f ðcÞ and b\D þ gðcÞ. Then there exists a D [ 0 such that inf QD ðf Þ [ a and inf QD ðgÞ [ b. It follows for 0\h\D that f ðc þ hÞ  f ðcÞ [a h

and

gðc þ hÞ  gðcÞ [ b; h

which implies that every number in the set QD ðf þ gÞðcÞ is greater than a þ b. Therefore inf QD ðf þ gÞðcÞ
a þ b. The existence of such a D [ 0 implies that supfinf Qd ðf þ gÞðcÞ: d [ 0g
a þ b. Thus, D þ ðf þ gÞðcÞ
a þ b. Since this holds for arbitrary a and b such that a\D þ f ðcÞ and b\D þ gðcÞ, the desired inequality follows. The inequality asked about holds provided that the sum on the right side is meaningful.

7.4 The Lebesgue Differentiability Theorem 7.4.P1. For an increasing function f on a bounded open interval I, show by a direct argument, as in the proof of the Lebesgue Differentiability Theorem 7.4.1, that the set A ¼ fx 2 I: D þ f ð xÞ\D f ð xÞg has Lebesgue outer measure 0. Solution: The set A is the union of the countably many sets Ap;q ¼ fx 2 I: D þ f ð xÞ\ p\q\D f ð xÞg, where p; q 2 Q. It is sufficient therefore to show that each Ap;q has

Solutions

247

outer measure 0. The increasing nature of the function f guarantees that all Dini derivatives are nonnegative and hence Ap;q 6¼ ; only if p > q > 0. So we need consider only those p, q for which this inequality holds. With this in view, take any Ap;q with p > q > 0 and consider an arbitrary e > 0. Being a subset of the bounded interval I, the set Ap;q has finite outer measure and therefore there exists a countable union G of open intervals such that Ap;q G and   m Ap;q  mðGÞ\m ðAp;q Þ þ e. By Remark 7.3.6, to each x 2 Ap;q correspond intervals ½x; x þ hG with h > 0 and arbitrarily small in absolute value such that f ðx þ hÞf ðxÞ \p, i.e. such that h f ðx þ hÞ  f ðxÞ\ph: Since this holds for each x 2 Ap;q , the intervals constitute a Vitali covering of Ap;q . By Exercise 7.2.5, we may restrict ourselves to only those intervals that are contained in G. Covering Theorem 7.2.7, there exist finitely many disjoint  By the Vitali  intervals xj ; xj þ hj ; 1  j  J, in the Vitali covering such that m ðAp;q n

J [

½xj ; xj þ hj Þ\e:

ðSol:12Þ

j¼1

The choice of the intervals of the Vitali covering ensures that f ðxj þ hj Þ  f ðxj Þ\phj

for every j:

ðSol:13Þ

Now, observe that since the disjoint intervals are contained in G, we have PJ j¼1 hj  mðGÞ. Since p > q > 0, taken in tandem with (Sol.13), the preceding observation leads to J  X

J X    f ðxj þ hj Þ  f ðxj Þ \p  hj  p  mðGÞ\p  m ðAp;q Þ þ e :

j¼1

ðSol:14Þ

j¼1

In conjunction with (Sol.12), the obvious equality m ðAp;q n

J [

ðxj ; xj þ hj ÞÞ ¼ m ðAp;q n

j¼1

gives rise to m ðAp;q n

J [

½xj ; xj þ hj Þ;

j¼1

J [

ðxj ; xj þ hj ÞÞ\e:

ðSol:15Þ

j¼1

  By Remark 7.3.6, to each y 2 Ap;q \ [ Jj¼1 xj ; xj þ hj correspond intervals   0 ½y þ h0 ; y [ Jj¼1 xj ; xj þ hj with arbitrarily small h0 \0 such that f ðy þ hh0Þf ðyÞ [ q, i.e. such that

248

Solutions

f ðy þ h0 Þ  f ðyÞ\qh0 :   Since this holds for each y 2 Ap;q \ [ Jj¼1 xj ; xj þ hj , the intervals constitute a Vitali covering of Ap;q \ [ Jj¼1 xj ; xj þ hj . By the Vitali Covering Theorem 7.2.7,   there exist finitely many disjoint intervals yk þ h0k ; yk ; 1  k  K, in the Vitali covering such that J [

m ððAp;q \

ðxj ; xj þ hj ÞÞn

j¼1

K [

½yk þ h0k ; yk Þ\e:

ðSol:16Þ

k¼1

The choice of the intervals of the Vitali covering ensures that f ðyk þ h0k Þ  f ðyk Þ\qh0k

for every k:

ðSol:17Þ

  Since the Vitali covering consists of intervals contained in [ Jj¼1 xj ; xj þ hj , we have K [

½yk þ h0k ; yk 

J [

ðxj ; xj þ hj Þ:

ðSol:18Þ

j¼1

k¼1

We claim that 

K X

h0k [ m ðAp;q Þ  2e:

ðSol:19Þ

k¼1

    To prove (Sol.19), let us denote [ Jj¼1 xj ; xj þ hj by J and [ kk¼1 yk þ h0k ; yk by K. Then m ðKÞ ¼ mðKÞ ¼ 

K X

h0k :

ðSol:20Þ

k¼1

Also, KJ in view of (Sol.18), so that J c ¼ J c \ Kc . Together with (Sol.15), this implies m ðAp;q \ J c \ Kc Þ ¼ m ðAp;q \ J c Þ\e: From (Sol.16), we have   m ðAp;q \ J Þ \ Kc \e:

Solutions

249

Since 

          Ap;q \ J c \ Kc [ Ap;q \ J \ Kc ¼ Ap;q \ Kc \ J c [ Ap;q \ Kc \ J ¼ Ap;q \ Kc ;

the preceding two inequalities yield m ðAp;q \ Kc Þ\2e: In combination with the fact that Ap;q ¼ ðAp;q \ KÞ [ ðAp;q \ Kc Þ, the above inequality leads us to m ðAp;q Þ  m ðAp;q \ KÞ þ m ðAp;q \ Kc Þ  m ðKÞ þ m ðAp;q \ Kc Þ \m ðKÞ þ 2e   P and hence m Ap;q \  Kk¼1 h0k þ 2e by (Sol.20). This proves our claim (Sol.19).   It follows from (Sol.18) that each interval yk þ h0k ; yk on the left side therein is contained in the union of finitely many disjoint intervals on the   right side. The nature of an interval necessitates that each interval yk þ h0k ; yk is actually contained in a single interval on the right side. This fact and the increasing nature of the function f together justify the inequality ½recall that h0k \0 J  X

K   X  f ðxj þ hj Þ  f ðxj Þ
f ðyk Þ  f ðyk þ h0k Þ :

j¼1

k¼1

By (Sol.17) and (Sol.19), this implies J  X

K X    f ðxj þ hj Þ  f ðxj Þ [  q  h0k [ q  m ðAp;q Þ  2e :

j¼1

k¼1

Therefore by (Sol.14), p  ðm ðAp;q Þ þ eÞ [ q  ðm ðAp;q Þ  2eÞ: Since  this has been for an arbitrary  e [ 0, it follows that  established  q  m Ap;q  p  m Ap;q , which is only possible if m Ap;q ¼ 0, because p < q. This completes the demonstration that m ðAÞ ¼ 0, where A ¼ fx 2 I: D þ f ð xÞ\D f ð xÞg. 7.4.P2. Assuming that the result m fx 2 I: D þ f ð xÞ [ D f ð xÞg ¼ 0 established in the course of proving the Lebesgue Differentiability Theorem 7.4.1 is valid for decreasing functions, deduce the result of Problem 7.4.P1.

250

Solutions

Solution: Let g ¼ f . Then g is decreasing and therefore m fx 2 I: D þ gð xÞ [ D gð xÞg ¼ 0. By Exercise 7.3.7, D þ gð xÞ ¼ D þ f ð xÞ and by Problem 7.3.P4, D gð xÞ ¼ D f ð xÞ. Therefore fx 2 I: D þ gð xÞ [ D gxÞg ¼ fx 2 I: D þ f ð xÞ\ D f ð xÞg. 7.4.P3. Suppose f and g are defined on a common interval I and c 2 I o is such that g has a finite right derivative g0þ ðcÞ at c. Show that D þ ðf þ gÞðcÞ ¼ D þ f ðcÞ þ g0þ ðcÞ. Solution: By hypothesis, D þ gðcÞg0þ ðcÞ. Therefore from Problem 7.3.P5, we know that D þ ðf þ gÞðcÞ  D þ f ðcÞ þ g0þ ðcÞ. We shall prove the reverse inequality. Consider any a\D þ f ðcÞ. By Remark 7.3.6, there exists a sequence fhn gn
1 of positive numbers such that hn ! 0 and

f ðc þ hn Þ  f ðcÞ [ a for every n 2 N: hn

Given any b\g0þ ðcÞ ¼ D þ gðcÞ, the sequence fhn gn
1 must also satisfy gðc þ hn Þ  gðcÞ [ b for sufficiently large n: hn The sequence obtained by deleting the first few terms of fhn gn
1 then has limit 0 and satisfies ðf þ gÞðc þ hn Þ  ðf þ gÞðcÞ [ a þ b for all n: hn By Problem 7.3.P1, we obtain D þ ðf þ gÞðcÞ
a þ b. Since this is valid for any a\D þ f ðcÞ and b\D þ gðcÞ ¼ g0þ ðcÞ, the reverse inequality we seek is an immediate consequence.

7.6 Bounded Variation 7.6.P1. Show that the product of two functions of bounded variation is of bounded variation. Solution: First of all, a function f : ½a; b ! R of bounded variation is bounded. This is because, for any x 2 ½a; b, the two points a and x constitute a partition of [a, x] and therefore, by definition of total variation, Vax ðf Þ
jf ðxÞ  f ðaÞj, which implies f jf ðxÞj  jf ðaÞj þ Vax ðf Þ  jf ðaÞj þ Vab ðf Þ. Now let f : ½a; b ! R and g: ½a; b ! R both be of bounded variation and be bounded in absolute value by A and B respectively. Consider an arbitrary partition P: af ¼ x0 \x1 \    \xn ¼ bof [a, b]. Then for any integer j satisfying 1  j  n, we have

Solutions

251

      ðfgÞðxj Þ  ðfgÞðxj1 Þ  Agðxj Þ  gðxj1 Þ þ Bf ðxj Þ  f ðxj1 Þ and therefore Sðfg; PÞ  A  Sðg; PÞ þ B  Sðf ; PÞ  A  Vab ðgÞ þ B  Vab ðf Þ. Since this   holds for every partition P, it follows that Vab ðfgÞ  AVab gÞ þ BVab ðf \1. 7.6.P2. Suppose a < b and the bounded function f : ½a; b ! R is decreasing on (a, b], so that the right limit f(a+) exists. Show that Vab ðf Þ ¼ f ða þ Þ  f ðbÞ þ jf ða þ Þ  f ðaÞj: Solution: If f ðaÞ
f ða þ Þ, then the function is decreasing on the closed interval [a, b] and the equality in question is seen to be true, both sides being equal to f ðaÞ  f ðbÞ. So, we assume that f ðaÞ\f ða þ Þ. Then the equality in question becomes Vab ðf Þ ¼ f ða þ Þ  f ðbÞ þ f ða þ Þ  f ðaÞ: Consider an arbitrary x 2 (a, b]. In view of Proposition 7.6.6(d), we have Vab ðf Þ ¼ Vax ð f Þ þ Vxb ðf Þ. Since f is decreasing on (a, b], we also have Vxb ðf Þ ¼ f ðxÞ  f ðbÞ. Therefore Vab ðf Þ ¼ Vax ð f Þ þ f ðxÞ  f ðbÞ ¼ ðVa f ÞðxÞ þ f ðxÞ  f ðbÞ. As this holds for an arbitrary x 2 (a, b], we have Vab ðf Þ ¼ ðVa f Þða þ Þ þ f ða þ Þ  f ðbÞ. It remains to show that ðVa f Þða þ Þ ¼ f ða þ Þ  f ðaÞ. Take any positive e\f ða þ Þ  f ðaÞ. There exists some d [ 0 such that 0\x  a\d ) 0  f ða þ Þ  f ðxÞ\ 2e. For any n 2 ða; b such that n  a\d, we shall show that Van ðf Þ ¼ ðVa f ÞðnÞ satisfies ðf ða þ Þ  f ðaÞÞ  e\Van ðf Þ\ðf ða þ Þ  f ðaÞÞ þ e: Consider an arbitrary partition P: a\x1 \    \xn ¼ n of [a, n]. Since a\x1  n, therefore 0\x1  a  n  a\d and hence e 2

ðSol:21Þ

e 0  f ða þ Þ  f ðnÞ\ : 2

ðSol:22Þ

0  f ða þ Þ  f ðx1 Þ\ and

As e\f ða þ Þ  f ðaÞ, the second inequality in (Sol.21) yields f ðx1 Þ [ f ða þ Þ 

e 2

[ f ða þ Þ  ðf ða þ Þ  f ðaÞÞ ¼ f ðaÞ:

ðSol:23Þ ðSol:24Þ

252

Solutions

Moreover, f ðnÞ  f ðx1 Þ  f ða þ Þ because a\x1  n. Together with (Sol.22), this implies e 0  f ðx1 Þ  f ðnÞ  f ða þ Þ  f ðnÞ\ : 2

ðSol:25Þ

Since f is decreasing on ½x1 ; n if x1 \n, we have Sðf ; PÞ ¼ jf ðx1 Þ  f ðaÞj þ ðf ðx1 Þ  f ðnÞÞ ¼ ðf ðx1 Þ  f ðaÞÞ þ ðf ðx1 Þ  f ðnÞÞby ðSol:24Þ:

ðSol:26Þ

Therefore by (Sol.25) and the first inequality in (Sol.21), Sðf ; PÞ\ðf ðx1 Þ  f ðaÞÞ þ

e e  ðf ða þ Þ  f ðaÞÞ þ : 2 2

This holds for an arbitrary partition P of [a, n] and consequently, e ðVa f ÞðnÞ ¼ Van ðf Þ  ðf ða þ Þ  f ðaÞÞ þ \ðf ða þ Þ  f ðaÞÞ þ e: 2 On the other hand, it also follows from (Sol.26), (Sol.23) and the first inequality in (Sol.22) that



 e e f ða þ Þ   f ðaÞ þ f ða þ Þ   f ða þ Þ 2 2 ¼ ðf ða þ Þ  f ðaÞÞ  e:

Sðf ; PÞ [

This leads to ðVa f ÞðnÞ ¼ Van ðf Þ [ ðf ða þ Þ  f ðaÞÞ  e: 7.6.P3. If f : ½a; b ! R has a bounded derivative on the open interval (a, b), show that it is of bounded variation on the closed interval [a, b]. Solution: Let M be an upper bound for the absolute value of the derivative. Take any c 2 (a, b). Then by the Mean Value Theorem, jf ðxÞj  jf ðcÞj þ Mðb  aÞ for any x 2 (a, b). Denote jf ðcÞj þ Mðb  aÞ by K and maxfjf ðaÞj; jf ðbÞjg by L. Now, let P: a ¼ x0 \x1 \    \xn ¼ b be any partition of [a, b] with n
3. By what has been proved in the preceding paragraph,       f ðxj Þ  f ðxj1 Þ  f ðxj Þ þ f ðxj1 Þ  L þ K

for j ¼ 1 and j ¼ n:

Solutions

253

Another consequence of the Mean Value Theorem is that a\y\z\b implies jf ðzÞ  f ðyÞj  Mðz  yÞ, so that n1  X  f ðxj Þ  f ðxj1 Þ  Mðb  aÞ: j¼1

Adding the two inequalities displayed above yields Sðf ; PÞ  L þ K þ Mðb  aÞ. This has been derived on the assumption that P contains at least four points (i.e. n
3). Since any partition has a refinement consisting of at least four points, the inequality must actually be valid for all partitions. Upon taking the supremum over all partitions, we arrive at Vab ðf Þ  L þ K þ Mðb  aÞ. 7.6.P4. If f : ½a; b ! R is of bounded variation, show that it has one sided limits at each point and that, for any c 2 [a, b), the equality Va h f iðc þ Þ  Va h f iðcÞ ¼ jf ðc þ Þ  f ðcÞj holds. (In particular, f is right continuous at c if and only if Va hf i is.) Solution: Every monotone function has one sided limits at each point and therefore the same is true of a sum of two monotone functions. To simplify the notation, let us denote Va hf i simply by V. First we shall prove that Vðc þ Þ  V ðcÞ  jf ðc þ Þ  f ðcÞj. Let Vðc þ Þ  V ðcÞ [ 0 and consider an arbitrary a\Vðc þ Þ  V ðcÞ. The inequality in question will follow if we can show that jf ðc þ Þ  f ðcÞj
a. To achieve this, it is sufficient to prove that, given any b\a and any d [ 0, there exists a t 2 ðc; c þ dÞ such that jf ðtÞ  f ðcÞj [ b. With this goal in view, consider an arbitrary b\a and an arbitrary d [ 0. Since V(c+) exists, there must exist an s 2 (c, b) satisfying c\x  s

)

VðsÞ  VðxÞ\a  b:

And since a\Vðc þ Þ  V ðcÞ, the number s can be so chosen so as to satisfy also VðsÞ  VðcÞ [ a: This inequality implies that there is a partition P of [c, s] such that Sðf ; PÞ [ a. By refining P if necessary, we may assume that P has at least three points in it and that the point t immediately to the right of c lies in ðc; c þ dÞ. Let S0 ðf ; PÞ denote the sum of all terms in Sðf ; PÞ except jf ðtÞ  f ðcÞj. Then S0 ðf ; PÞ ¼ Sðf ; PÞ  jf ðtÞ f ðcÞj [ a  jf ðtÞ  f ðcÞj. But on the other hand, the inequality S0 ðf ; PÞ  VðsÞ  V ðtÞ\a  b must obtain. Consequently, we have a  b [ a  jf ðtÞ  f ðcÞj, which yields jf ðtÞ  f ðcÞj [ b. The existence of the required kind of t 2 ðc; c þ dÞ has thus been established, thereby completing the argument why Vðc þ Þ  V ðcÞ  jf ðc þ Þ  f ðcÞj.

254

Solutions

We shall also prove the reverse inequality, which is easy to do. Recall from the proof of Theorem 7.6.8 that the functions g ¼ V þ f and h ¼ V  f are both increasing, Therefore gðc þ Þ  gðcÞ
0 and hðc þ Þ  hðcÞ
0. However, gðc þ Þ  gðcÞ ¼ ðVðc þ Þ  VðcÞÞ þ ðf ðc þ Þ  f ðcÞÞ and hðc þ Þ  hðcÞ ¼ ðVðc þ Þ  VðcÞÞ  ðf ðc þ Þ  f ðcÞÞ: The reverse inequality we seek now rolls in.

7.7 Absolute Continuity 7.7.P1. (a) Show that a continuous function on [0, 1] which is absolutely continuous on [η, 1] for every positive η less than 1 need not be absolutely continuous on [0, 1]. (b) Show that a continuous function on [0, 1] which is absolutely continuous on [η, 1] for every positive η less than 1 and is of bounded variation must be absolutely continuous on [0, 1]. Solution: (a) The function f given by f ðxÞ ¼ x cos2 ðp=2xÞ for x 6¼ 0 and f(0) = 0 is continuous on [0, 1]. Since it has a bounded derivative on [η, 1] for every positive η less than 1, it is absolutely continuous on every such interval. But as seen in Example 7.6.4, it is not even of bounded variation on [0, 1] and therefore by Theorem 7.7.2 certainly not absolutely continuous. (b) Call the function f and consider any e [ 0. By Problem 7.6.P4, V0 hf i is continuous at 0, and therefore there exists some positive g\1 such that 0  V0 hf iðgÞ\ 2e. It follows for any finitely many disjoint intervals  P  ða1 ; b1 Þ; ða2 ; b2 Þ; . . .; ðak ; bk Þ½0; g that kj¼1 f ðbj Þ  f ðaj Þ  V0 hf iðgÞ\ 2e. The given function f is absolutely continuous on ½g; 1 and therefore there exists a d [ 0 satisfying the condition in the definition of absolute continuity but with 2e in place of e and ½g; 1 in place of [a, b]. Consider finitely many disjoint open subintervals of [0, 1] with total length less than d. If one of them contains η, we may break up the corresponding closed subinterval into two parts, one containing η and the points to the left of it and the other containing η and the points to the right of it. By doing so, the kind of sum we need to consider may increase or remain the same but not decrease. Also, the corresponding open subintervals continue to be disjoint and each is contained either in ½0; g or in ½g; 1. Denote the former, if any, by ða1 ; b1 Þ; ða2 ; b2 Þ; . . .; ðak ; bk Þ and

Solutions

255

the latter, if any, by ða1 ; b1 Þ; ða2 ; b2 Þ; . . .; ðap ; bp Þ. If intervals of one of these kinds do not actually occur, then the sums relating to them are empty and therefore  P  understood to be 0. With this proviso, we have kj¼1 f ðbj Þ  f ðaj Þ\ 2e, and since P we started with subintervals with total length d, we also have pr¼1 ðbr  ar Þ\d Pp The choice of d now leads to r¼1 jf ðbr Þ  f ðar Þj\ 2e: It follows that p k  X  X f ðbj Þ  f ðaj Þ þ jf ðbr Þ  f ðar Þj\e: r¼1

j¼1

Consequently, f is seen to be absolutely continuous on [0, 1]. 7.7.P2. (a) Show that the function g: ½0; 1 ! R given by gðxÞ ¼ x12 is absolutely continuous. (b) Show that the function f : ½0; 1 ! R given by f ðxÞ ¼ x2 cos4 ðp=2xÞ for x 6¼ 0 and f (0) = 0 is absolutely continuous. (c) Show that a composition of absolutely continuous functions need not be absolutely continuous. Solution: (a) The function is increasing on [0, 1] and is absolutely continuous on [η, 1] for every positive η less than 1, by reason of having a bounded derivative on such an interval. We deduce absolute continuity on [0, 1] by invoking Problem 7.7.P1(b). (b) The function in question has a bounded derivative on [0, 1]. (c) Consider the functions f and g above. The composition g  f is given by ðg  f Þð xÞ ¼ x cos2 ðp=2xÞ for x 6¼ 0 and f(0) = 0. This function is not of bounded variation, as seen in Example 7.6.4, and hence by Theorem 7.7.2, it cannot be absolutely continuous. 7.7.P3. (a) Suppose f : ½a; b ! ½a; b and g: ½a; b ! R are absolutely continuous. If f is increasing, show that g  f : ½a; b ! R is absolutely continuous. Does the same conclusion hold if the function f is decreasing instead? (b) Give an example of a function that has an unbounded derivative, is not monotone, but is absolutely continuous. Solution: (a) Let e [ 0 be arbitrary. Choose d0 [ 0 in accordance with the definition of absolute continuity for g on ½a; b. Then choose d [ 0 in accordance with the definition of absolute continuity for f on [a, b] but with d0 in place of e. Consider finitely many disjoint open subintervals ða1 ; b1 Þ; ða2 ; b2 Þ; . . .; ðak ; bk Þ P of [a, b] with total length less than d. That is, kj¼1 ðbj  aj Þ\d. By choice of  Pk  f ðbj Þ  f ðaj Þ\d0 . Since f is increasing, f ðaj Þ  f ðbj Þ d, it follows that j¼1

for every j, and besides, the intervals ðf ðaj Þ; f ðbj ÞÞ (some of which may be empty because f ðaj Þ can be equal to f ðbj Þ) must be disjoint. Considering that  Pk Pk  0 0  j¼1 f ðbj Þ  f ðaj Þ \d , the choice of d forces j¼1 jðg  f Þðbj Þ  ðg  f Þ ðaj Þj\e.

256

Solutions

Consequently, g  f : ½a; b ! R is seen to be absolutely continuous. The same conclusion holds if the function f is decreasing instead; the intervals ðf ðaj Þ; f ðbj ÞÞ have to be replaced by the intervals ðf ðbj Þ; f ðaj ÞÞ. (b) The function on [0, 1] that maps x 2 ð0; 1 into x2 cosðp=2xÞ and maps 0 into 0 is absolutely continuous, because it has a bounded derivative and the function 1 described by x2 is also absolutely continuous by Problem 7.7.P2(a). Moreover, the latter is increasing. It follows by part (a) that the composition, which maps 1 x 2 ð0; 1 into x cosðp=2x2 Þ and maps 0 into 0, is absolutely continuous. However, its derivative maps x 2 ð0; 1 into cosðp=2x2 Þ þ ðp=4Þx2 sinðp=2x2 Þ: 1

1

1

Therefore the value of the derivative when x ¼ ð4k þ 1Þ2 is ðp=4Þð4k þ 1Þ, and when x ¼ ð4k þ 3Þ2 , it is ðp=4Þð4k þ 3Þ. Thus the derivative takes large positive values as well as large negative values as x approaches 0, with the consequence that the function has an unbounded derivative and is not monotone. 7.7.P4. If f : ½a; b ! R is absolutely continuous on each of the subintervals [a, c] and [c, b], where a < c < b, show that it is absolutely continuous on [a, b]. Solution: Take any e [ 0. Let d1 be as in the definition of absolute continuity for f on the subinterval [a, c] but with 2e in place of e, and d2 the same on the subinterval [c, b]. Set d ¼ minfd1 ; d2 g [ 0. Consider finitely many disjoint open subintervals of [a, b] with total length less than d. If none of them contains c, then each one is contained either in [a, c] or in [c, b]. Denote the former, if any, by ða1 ; b1 Þ; ða2 ; b2 Þ; . . .; ðak ; bk Þ and the latter, if any, by ða1 ; b1 Þ; ða2 ; b2 Þ; . . .; ðap ; bp Þ. If intervals of one of these kinds do not actually occur, then the sums relating to them are empty and therefore understood to be 0. With this Pk Pp proviso, we have j¼1 ðbj  aj Þ\d1 and also j¼1 ðbr  ar Þ\d2 , because we started with subintervals with total length less than d. The choice of d1 and d2 now P P leads to kj¼1 jf ðbj Þ  f ðaj Þj\ 2e and pr¼1 jf ðbr Þ  f ðar Þj\ 2e. It follows that p k  X  X f ðbj Þ  f ðaj Þ þ jf ðbr Þ  f ðar Þj\e: j¼1

r¼1

Now consider finitely many disjoint open subintervals of [a, b] with total length less than d, one of them containing c. We may replace the subinterval by two parts of it, one containing all its points that are to the left of c and the other containing all its points that are to the right of c. By doing so, the kind of sum we need to consider may increase or remain the same but not decrease (see Exercise 7.7.2). Also, the open subintervals continue to be disjoint and each is contained either in [a, c] or in [c, b]. We can now arrive at the desired inequality by proceeding as in the preceding paragraph. Thus f is seen to be absolutely continuous on [a, b].

Solutions

257

7.7.P5. Given a finite number of points in (a, b), show that there exists an absolutely continuous function on [a, b] that fails to be differentiable at the given points. Solution: Define a function to be 1 and 0 alternately at a, the given points in increasing order and at b; extend it by linear interpolation to the interiors of the intervals formed by the points at which it has just been defined.. Then it is absolutely continuous on each of the corresponding closed intervals, considering that it has a constant derivative in each interior. It is not differentiable at the given points but is absolutely continuous on [a, b] by Problem 7.7.P4.

7.8 The Integral of the Derivative 7.8.P1. (a) Let f be a real-valued function on [a, b] such that its values on (a, b) are linear interpolations between its values at the endpoints. If one among f (a) and f (b) is positive and the other nonnegative, show that f (x) is positive for all x 2 ða; bÞ. (b) Let f and g be real-valued functions on [a, b] such that f ðaÞ
gðaÞ, f ðbÞ
gðbÞ, one of the inequalities being strict, and let the values of both functions on (a, b) be linear interpolations between their values at the endpoints. Show that f ðxÞ [ gðxÞ for all x 2 ða; bÞ. Solution: (a) By the linear interpolation hypothesis, f ðxÞ  f ðaÞ ¼ ðf ðbÞ  f ðaÞÞðb  aÞ1 ðx  aÞ for all x 2 ½a; b: Let x 2 ða; bÞ, so that x  a [ 0 and 0\ðb  aÞ1 ðx  aÞ\1. Then if f ðbÞ  f ðaÞ [ 0, we have f ðxÞ [ f ðaÞ
0. If f ðbÞ  f ðaÞ ¼ 0, we have f ðxÞ ¼ f ðaÞ ¼ f ðbÞ [ 0. If f ðbÞ  f ðaÞ\0, then in view of the fact that 0\ðb  aÞ1 ðx  aÞ\1, we have ðf ðbÞ  f ðaÞÞðb  aÞ1 ðx  aÞ [ f ðbÞ f ðaÞ, and hence f ðxÞ  f ðaÞ [ f ðbÞ  f ðaÞ, so that f ðxÞ [ f ðbÞ
0. (b) It is easy to check that f  g also has the property that its values on (a, b) are linear interpolations between its values at the endpoints. The desired conclusion is straightforward from the result of part (a). 7.8.P2. (a) Let f be a real-valued function on [a, b] such that its values on (a, b) are linear interpolations between its values at the endpoints and let a\c\b. Show that the function has the same interpolation property over the intervals (a, c) and (c, b). (b) Let f and g be real-valued functions on [a, b] such that f ðaÞ
gðaÞ, f ðbÞ
gðbÞ, and the values of g on (a, b) are linear interpolations between its values at the endpoints. Also, let a\c\b and the values of f on each of (a, c) and (c, b) be linear interpolations between its values at the endpoints of the respective intervals. If f ðcÞ [ gðcÞ, show that f ðxÞ [ gðxÞ for all x 2 ða; bÞ.

258

Solutions

Solution: (a) By the linear interpolation hypothesis, f ðxÞ  f ðaÞ ¼ ðf ðbÞ  f ðaÞÞðb  aÞ1 ðx  aÞ for all x 2 ½a; b:

ðSol:27Þ

Taking x = c, we get f ðcÞ  f ðaÞ ¼ ðf ðbÞ  f ðaÞÞðb  aÞ1 ðc  aÞ;

ðSol:28Þ

and hence ðf ðbÞ  f ðaÞÞðb  aÞ1 ¼ ðf ðcÞ  f ðaÞÞðc  aÞ1 : Using this in (Sol.27), we obtain an equality valid for x 2 [a, b] and whose validity on [a, c] asserts that the values of f on (a, c) are linear interpolations between its values at the endpoints of (a, c). Upon subtracting (Sol.28) from (Sol.27), we obtain f ðxÞ  f ðcÞ ¼ ðf ðbÞ  f ðaÞÞðb  aÞ1 ðx  cÞ for all x 2 ½a; b: Taking x = b in this, we get f ðbÞ  f ðcÞ ¼ ðf ðbÞ  f ðaÞÞðb  aÞ1 ðb  cÞ; and hence ðf ðbÞ  f ðaÞÞðb  aÞ1 ¼ ðf ðbÞ  f ðcÞÞðb  cÞ1 : Using this in (Sol.27), we obtain an equality valid for x 2 ½a; b and whose validity on [c, b] asserts that the values of f on (c, b) are linear interpolations between its values at the endpoints of (c, b). (b) By part (a), the values of g on (a, c) are linear interpolations between its values at the endpoints. Since f ðaÞ
gðaÞ and f (c) > g(c), it follows from Problem 7.8.P1(b) that f (x) > g(x) for all x 2 ða; cÞ. Similarly, f (x) > g(x) for all x 2 ðc; bÞ. It is given that f (c) > g(c). Thus f (x) > g(x) for all x 2 ða; bÞ. 7.8.P3. Suppose 0 < a < 1 < b and fsj gj
1 is a sequence of numbers such that, for each j, either 0\sj  a or sj
b: Let rn ¼ Pnj¼1 sj : Show that rn cannot have a finite positive limit. Solution: Suppose to the contrary that limn!1 rn ¼ r, a finite positive number.

The hypothesis on a and b implies 0\max a; b1 \1. Call it c. Let c0 be any number such that c\c0 \1 and set k ¼ c0 =c [ 1. Then c0 r\r\kr. By our supposition, there must exist an n 2 N such that

Solutions

259

c0 r\rn \kr

as well as

c0 r\rn þ 1 \kr:

If sn þ 1  a, then the second part of the first double inequality implies rn þ 1 \kra  kra ¼ rc0 ; contradicting the first part of the second double inequality, whereas if sn þ 1
b, the first part of the first double inequality implies rn þ 1 [ c0 rb
c0 rc1 ¼ kr; contradicting the second part of the second double inequality. The contradiction thus obtained shows that our supposition is untenable. 7.8.P4. Given that f : ½a; b ! R is absolutely continuous, show that the function Va f : ½a; b ! R defined by ðVa f ÞðnÞ ¼ Vax ðf Þ for all n 2 ½a; b is also absolutely continuous. Can one draw the inference that an absolutely continuous function is the sum of an absolutely continuous increasing function and an absolutely continuous decreasing function? Solution: By Corollary 7.8.4, there exists an integrable g: ½a; b ! R such that f ðxÞ  f ðaÞ ¼

Z g

for all x 2 ½a; b

½a;x

and the function /: ½a; b ! R defined by /ðxÞ  /ðaÞ ¼

Z

jgj

for all x 2 ½a; b

½a;x

is absolutely continuous. Since jf ðx2 Þ  f ðx1 Þj  j/ðx2 Þ  /ðx1 Þj ¼ /ðx2 Þ  /ðx1 Þ whenever ½x1 ; x2 ½a; b, it follows that we also have jðVa f Þðx1 Þ  ðVa f Þðx2 Þj  j/ðx1 Þ  /ðx2 Þj whenever ½x1 ; x2 ½a; b. Therefore the absolute continuity of / yields the absolute continuity of Va f . The inference can indeed be drawn, because a sum of absolutely continuous functions is absolutely continuous.

8.1 A Surjection from the Cantor Set to [0, 1] P P k k 8.1.P1. Suppose x ¼ 1 and y ¼ 1 are b-expansions and there k¼1 ak b k¼1 bk b exists a positive integer j such that aj [ bj and ak ¼ bk for k < j (vacuously true if j = 1). If there does not exist any positive integer J such that ak ¼ 0 for every k > J, show that x > y.

260

Solutions

Solution: By (1) of Proposition 8.1.1, the inequality x
y must hold. If x were to be equal to y, then by (2) of Proposition 8.1.1, j would have the property that ak ¼ 0 for every k > j. This would contradict the present hypothesis. So, x > y. 8.1.P2. If y; z 2 ½a; b and jy  zj ¼ b  a, show that one among y and z is a and the other is b. (Ignore the trivial case when a = b.) Solution: Since jy  zj [ 0, we know y 6¼ z. Suppose y < z. Then the hypothesis is that b  a ¼ z  y. In conjunction with the inequality a  y, this yields b  z. But z  b and therefore z = b. In combination with the equality b  a ¼ z  y, this yields y = a. 8.1.P3. Let fbk gk
1 be a sequence in which each term is either 0 or 1. P k 2 ½0; 1 and the series is a 3-expansion (called ternary) (a) Then 1 k¼1 2  bk 3 of the sum. (b) If the first n terms of the sequence fbk gk
1 of digits in any one ternary P k of the number x agree with the first n terms of expansionx ¼ 1 k¼1 2  bk 3 1 / ðuÞ, where u 2 C, then x is in the same component of Cn as u is. P k is a ternary expansion of u. (c) If /1 ðuÞ ¼ fak gk
1 , then 1 k¼1 2  ak 3 (d) A number in [0, 1] belongs to C if and only if it has a ternary expansion in which every digit is either 0 or 2. (e) If a number in [0, 1] has a ternary expansion in which every digit is either 0 or 2, then it has no other ternary expansion of the same kind.     P P k k Solution: (a) 0  1 2 1 ¼ 2 13 1= 1  13 ¼ 1. Since 2bk k¼1 2  bk 3 k¼1 3 is an integer satisfying 0  2bk  ð3  1Þ, the series is a ternary expansion of its sum. (b) Regardless of the matter of ternary expansions, we note to begin with that x 2 ½a; b is in the left third ½a; a þ 13 ðb  aÞ of [a, b] if and only if   x  a  13 ðb  aÞ; similarly, x 2 ½a; b is in the right third a þ 23 ðb  aÞ; b if and only if x  a
23 ðb  aÞ. Let /1 ðuÞ ¼ fak gk
1 : If b1 ¼ a1 , then by separating the first term in the given ternary expansion of x, we obtain the double inequality 2  a1 31 ¼ 2  b1 31  x  2  b1 31 þ

1 X k¼2

1 3

2  3k ¼ 2  a1 31 þ :

If u is in the component ½0; 13 of C1, then a1 ¼ 0 and the second half of the above double inequality shows that x  13, so that x is found to be in the same component as u. If on the other hand u is in the component ½23 ; 1 of C1, then a1 ¼ 1 and the first half of the above double inequality shows that x
23, so that x is once again found to be in the same component as u. The assertion at issue has thus been proved to hold for n = 1.

Solutions

261

Now assume as induction hypothesis that the assertion at issue holds for some n, and let the first n + 1 terms of the sequence fbk gk
1 of digits in the ternary P k expansion x ¼ 1 agree with the first n + 1 terms of /1 ðuÞ. k¼1 2  bk 3 Then certainly the first n terms of the sequence fbk gk
1 of digits in the ternary P k expansion x ¼ 1 agree with the first n terms of /1 ðuÞ. Note that k¼1 2  bk 3 the same happens to be true regarding y¼

n X

2  ak 3k

and

k¼1

z¼

n X k¼1

2  ak 3k þ

1 X

2  3k :

k¼n þ 1

The induction hypothesis therefore implies that not only x but also y and z are in P k ¼ 3n , which the same component of Cn as u is. Now, z  y ¼ 1 k¼n þ 1 2  3 is the length of the component. It follows from Problem 8.1.P2 that y is the left endpoint of the component. So, the component is [a, b], where a ¼ y ¼ P Pn k k and b  a ¼ 3n . Also, x  a ¼ 1 k¼1 2  ak 3 k¼n þ 1 2  bk 3 , where bn þ 1 ¼ an þ 1 . Separating the first term in the series for x − a, we obtain the double inequality

 1 2  an þ 1 3n1  x  a  2  an þ 1 3n1 þ 2  ð3n2 Þ 1=ð1  Þ ¼ 2  an þ 1 3n1 þ 3n1 :

3

The component of Cn þ 1 that u is in is either the left third of the component [a, b] of Cn , in which case an þ 1 ¼ 0, or the right third of the component [a, b] of Cn , in which case an þ 1 ¼ 1. If u is in the left third of the component [a, b] of Cn , then an þ 1 ¼ 0 and  the second half of the above double inequality shows that x  a  3n1 ¼ 13 3n ¼ 13 ðb  aÞ, which implies that x is also in the left third and hence is in the same component of Cn þ 1 as u. If u is in the right third of the component [a, b] of Cn , then an þ 1 ¼ 1 and the first half of the above double inequality shows that x  a
2  3n1 ¼ 2  ð13Þ3n ¼ 23 ðb  aÞ, which implies that x is also in the right third and hence is again in the same component of Cn þ 1 as u. This shows that in both the possible cases, x is in the same component of Cn þ 1 as u. P k as one ternary expansion has (c) In view of (b), a number having 1 k¼1 2  ak 3 the property that, for every n, it is in the same component of Cn as u. Therefore its distance from u is at most 3n for every n. (d) It was shown in (c) that every number in C has a ternary expansion in which every digit is either 0 or 2. Conversely, a ternary expansion in which every P k digit is either 0 or 2 is of the form 1 k¼1 2  ak 3 , where every ak is either 0 or 1 1. Set u ¼ /ðfak gk
1 Þ 2 C. Then / ðuÞ ¼ fak gk
1 , and hence by (c), P1 3k is a ternary expansion of u. Therefore the number with ternary k¼1 2  akP k expansion 1 is the same as u, which belongs to C. k¼1 2  ak 3 P P k 0 k (e) Consider two distinct ternary expansions 1 and 1 of a k¼1 ak b k¼1 ak b 0 number and suppose every ak as well as every ak is either 0 or 2. Since the

262

Solutions

expansions are distinct, there must be some k such that ak 6¼ a0k . Let j be the smallest among such k. Then ak 6¼ a0k for k < j (vacuously true if j = 1) and aj 6¼ a0j . We may assume aj [ a0j . Now, (2) of Proposition 8.1.1 yields aj  a0j ¼ 1, which is not possible because each of aj and a0j is either 0 or 2.

8.2 A Bijection from the Cantor Set to [0, 1] and the Cantor Function 8.2.P1. (a) Prove Proposition 8.2.1. (b) Prove Proposition 8.2.2. P P k 0 k Solution: (a) Let 1 and 1 be two distinct b-expansions of the k¼1 ak b k¼1 ak b same x 2 ½0; 1. Since they are distinct, there must exist some k such that ak 6¼ a0k . It follows that there is a smallest k with this property. Call it j. Then either ak [ a0k or ak \a0k . To be specific, suppose ak [ a0k . It then follows by statement (2) of Proposition 8.1.1 that every k > j satisfies ak ¼ 0; a0k ¼ b  1. Thus the sequences fak gk
1 ; fa0k gk
1 are both recurring. P Pj k (b) Suppose x ¼ 1 and ak ¼ a for k > j. Then x ¼ k¼1 ak bk þ k¼1 ak b P1 P P P j 1 j k k k k j ¼ k¼1 ak b þ a k¼j þ 1 b ¼ k¼1 ak b þ ab ðb  1Þ1 , k¼j þ 1 ak b which is a rational number. Ordinarily, the “post inverse” (or “left inverse”) g of a function G is understood to have the range of G as its domain and to satisfy the condition that g(G(x)) = x for every x in the domain of G. However, for the present purpose, it will be necessary to broaden the concept somewhat. If G: ½a; b ! R is strictly increasing, then any function g defined on [G(a), G(b)] and satisfying the aforementioned condition will be called a post inverse. Its values at points of [G(a), G(b)] that are outside the range of G are not uniquely determined unless some additional condition is imposed. 8.2.P2. (Rubel's Lemma [6]) Let G: ½a; b ! R be a strictly increasing function. Then there is a continuous increasing function g: ½GðaÞ; GðbÞ ! R which is a “post inverse” of G, i.e. satisfies g(G(x)) = x for each x 2 ½a; b, and which is constant on any open subinterval of [G(a), G(b)] contained in the complement of the range of G. Solution: Let g: ½GðaÞ; GðbÞ ! R be the function such that gðyÞ ¼ supfu 2 ½a; b: GðuÞ  yg: Then it is obvious that the range of g is contained in [a, b]. Let us show first that g(G(x)) = x for each x 2 ½a; b. With this in mind, consider any x 2 ½a; b. Since G is strictly increasing, we have GðuÞ  Gð xÞ , u  x. Therefore fu: GðuÞ  Gð xÞg ¼ fu: u  xg; hence supfu: GðuÞ  Gð xÞg ¼ x. By definition of g, this is the same as saying that g(G(x)) = x.

Solutions

263

Next, we show that g is increasing. Suppose y; g 2 ½GðaÞ; GðbÞ and y  g. Then GðuÞ  y ) GðuÞ  g, and therefore fu: GðuÞyg  fu: GðuÞ  gg. This implies supfu: GðuÞ  yg  supfu: GðuÞ  gg, which means gðyÞ  gðgÞ by definition of g. Thus g is an increasing function. We proceed to prove continuity. Since g(G(x)) = x for each x 2 ½a; b, the range of g is the interval [a, b]. As we have shown that the function is increasing, its continuity now follows by Remark 8.2.4. Let I be a subinterval of [G(a), G(b)] that is contained in the complement of the range of G. We consider only nonempty I, because it is trivial that g is constant on ;. Suppose y1 ; y2 2 I and y1  y2 . Since g has been shown to be increasing, we know that gðy1 Þ  gðy2 Þ. Consider any u 2 ½a; b such that GðuÞ  y2 . Since ½y1 ; y2 I and is therefore contained in the complement of the range of G, it follows that GðuÞ\y1 . This shows that fu: GðuÞ  y2 gfu: GðuÞ  y1 g, which implies gðy2 Þ  gðy1 Þ by definition of g. Consequently, gðy1 Þ ¼ gðy2 Þ. 8.2.P3. Let G be a strictly increasing real-valued function on the domain [a, b] and g0 be an increasing function with domain containing the range of G and satisfying g0 ðGð xÞÞ ¼ x for each x 2 ½a; b. (In particular, g0 could be an increasing post inverse of G.) Show the following. (a) For each y 2 ½GðaÞ; GðbÞ that belongs to the domain of g0 , g0 ðyÞ ¼ supfu 2 ½a; b: GðuÞ  yg: (b) The increasing post inverse is unique and g0 agrees with the unique increasing post inverse at points of ½GðaÞ; GðbÞ that belong to its domain. Hint: Argue for any y 2 ½GðaÞ; GðbÞ that belongs to the domain of g0 that g0 ð yÞ lies between the numbers supfu 2 ½a; b: GðuÞ  yg and inffu 2 ½a; b: GðuÞ
yg and that these two numbers are equal. Solution: (a) Let y 2 ½GðaÞ; GðbÞ belong to the domain of g0 and A ¼ fu 2 ½a; b: GðuÞ  yg: We have to prove that g0 ðyÞ ¼ sup A: Let B = {u 2 [a, b] : G(u)
y}. Then A [ B ¼ ½a; b. Both A and B are nonempty because a 2 A and b 2 B. So, sup A and inf B both exist and belong to [a, b]. Since it is given that g0 is increasing, that its domain contains the range of G and that y is in the domain of g0 , we can assert that

264

Solutions

u 2 A ) u ¼ g0 ðGðuÞÞ  g0 ðyÞ u 2 B ) u ¼ g0 ðGðuÞÞ
g0 ðyÞ: Hence sup A  g0 ð yÞ  inf B. But if sup A < inf B, then any x 2 ðsup A; inf BÞ must belong to [a, b] and also satisfy x 62 A as well as x 62 B, which is not possible because A [ B ¼ ½a; b. Therefore sup A = inf B, which implies that g0 ð yÞ ¼ sup A, as was to be proved. (b) This is an easy consequence of (a). 8.2.P4. Let G be a strictly increasing real-valued function on the domain [a, b]. Show that the function g: ½GðaÞ; GðbÞ ! R defined by gðyÞ ¼ inffu 2 ½a; b: GðuÞ
yg is an increasing post inverse of G. Solution: It is obvious that the range of g is contained in [a, b]. Let us show first that gðGðxÞÞ ¼ x for each x 2 ½a; b. With this in mind, consider any x 2 ½a; b. Since G is strictly increasing, we have GðuÞ  Gð xÞ , u  x. Therefore fu: GðuÞ
Gð xÞg ¼ fu: u
xg; hence inffu: GðuÞ
Gð xÞg ¼ x. By definition of g, this is the same as saying that gðGðxÞÞ ¼ x. It remains to show that g is increasing. Suppose y1 ; y2 2 ½GðaÞ; GðbÞ and y1  y2 . Then GðuÞ
y2 ) GðuÞ
y1 , and therefore fu: GðuÞ
y2 gfu: GðuÞ
y1 g. This implies inffu: GðuÞ
y2 g
inffu: GðuÞ
y1 g, which means gðy1 Þ  gðy2 Þ by definition of g. Thus g is an increasing function. 8.2.P5. Let G1 and G2 be strictly increasing real-valued functions on the domain [a, b] and let g1 and g2 be their respective (unique) increasing post inverses. If G1 ¼ G2 almost everywhere, show that g1 ¼ g2 everywhere on the intersection of their domains, i.e. on ½G1 ðaÞ; G1 ðbÞ \ ½G2 ðaÞ; G2 ðbÞ. Solution: Let y 2 ½G1 ðaÞ; G1 ðbÞ \ ½G2 ðaÞ; G2 ðbÞ. Then fu 2 ½a; b: G1 ðuÞ  yg and fu 2 ½a; b: G2 ðuÞ  yg are both nonempty. Consider x1 ¼ supfu 2 ½a; b: G1 ðuÞ  yg; x2 ¼ supfu 2 ½a; b: G2 ðuÞ  yg. By Problem 8.2.P3, we need only prove that x1 ¼ x2 . This will follow if we derive a contradiction from the assumption that x1 \x2 . So, assume x1 \x2 . By definition of x2 , the interval ðx1 ; x2  contains some u0 such that G2 ðu0 Þ  y. Since G2 is increasing, it follows that every u 2 ðx1 ; u0 Þ satisfies G2 ðuÞ  y. Since G1 ¼ G2 almost everywhere, some u 2 ðx1 ; u0 Þ must satisfy G1 ðuÞ ¼ G2 ðuÞ  y. Since x1 \u, this contradicts the definition of x1 . 8.2.P6. In the course of proving Theorem 8.2.3, it was established that the mapping j0 is strictly increasing on CnC0 , where C0 is the subset of the Cantor set C consisting of those u for which the associated sequence /1 ðuÞ ¼ fak gk
1 is recurring. Use this fact in conjunction with Problem 6.6.P3 to show that there exists a measurable set that is not a Borel set. Hint: Consider the inverse image by j0 of the intersection of a nonmeasurable subset of [0, 1] with the range of j0 . Solution: Denote the restriction of j0 to CnC0 by j1 . We know from the proof of Theorem 8.2.3 that j1 is strictly increasing and has range ½0; 1nI0 , where I0 is the

Solutions

265

subset of [0, 1] consisting of numbers having a recurring binary expansion. Here C0 and I0 are both countably infinite. Since j1 is strictly increasing, so is its inverse j1 1 , which has domain ½0; 1nI0 . Let N1 be a subset of R that is not Lebesgue measurable (Corollary 4.3.8). Then at least one among the countably many intersections N1 \ ½n; n þ 1, where n 2 Z, must fail to be Lebesgue measurable. It follows that N ¼ fx 2 R: x þ n 2 N1 \ ½n; n þ 1g is a subset of [0, 1] that is not Lebesgue measurable. Since I0 is countable, the same is true of N \ I0 , which implies that it is measurable. Since N ¼ ðN \ I0 Þ [ ðNnI0 Þ, we infer that NnI0 is not Lebesgue measurable. [(Since N  [0,1], the set-theoretic difference NnI0 is the same as the intersection N \ ð½0; 1nI0 Þ: Since j1 is a strictly increasing and therefore measurable function on the 1 Lebesgue measurable subset ½0; 1nI0 of R, the inverse image by j1 1 of any Borel set is measurable by Problem 6.6.P3. Now the inverse image by j1 1 of j1 ðNnI0 Þ is precisely NnI0 , which is not measurable. Therefore j1 ðNnI0 Þ is not a Borel set. However, it is a Lebesgue measurable set because it is a subset of C and the latter has measure 0.

References

1. 2. 3. 4.

5. 6. 7. 8. 9. 10. 11.

Craven, B.D.: Lebesgue Measure & Integral. Pitman Publishing Inc, Marshfield, MA (1982) Dunford, N., Schwartz, J.T.: Linear Operators, Part II. Interscience Publishers, New York (1963) Feldman, J.: Interchanging the order of summation. http://www.math.ubc.ca/*feldman/ m321/twosum.pdf. Last accessed 19 June 2016 Habil, E.D.: Double sequences and double series. IUG J. Nat. Stud. 14(1), 1–32 (2006). http://www2.iugaza.edu.ps/ar/periodical/articles/volume%2014-%20Issue%201%20-studies %20-16.pdf. Last accessed 19 June 2016 Munkres, J.R.: Analysis on Manifolds. Westview Press, Boulder, Colorado (1991) Rubel, L.A.: Differentiability of monotonic functions. Colloq. Math. 10, 277–279 (1963) Rudin, W.: Principles of Mathematical Analysis, 3rd edn. McGraw-Hill, New York (1976) Shirali, S.: Analog of Lp with a subadditive measure. Ricerche Mat. 57, 43–54 (2008) Shirali, S., Vasudeva, H.L.: Introduction to Mathematical Analysis. Narosa, New Delhi (2014) Shirali, S., Vasudeva, H.L.: Multivariable Analysis. Springer, London (2010) Wang, Z., Klir, G.J.: Fuzzy Measure Theory. Plenum Press, New York (1992)

© Springer Nature Switzerland AG 2018 S. Shirali, A Concise Introduction to Measure Theory, https://doi.org/10.1007/978-3-030-03241-8

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Index

A Abel summation formula, 6, 9, 42 Additive, 26 Additivity, 41 r-algebra, 19 co-countable, 97, 99 generated by, 104 induced, 85 Algebra of subsets, 103 Almost all, 118–120 Almost everywhere, 118–121, 125 Axiom of Choice, 83 B Bernstein–Schröder Theorem, 172 Binary expansion, 169, 172 Borel function, 126, 149, 150 Borel measurable function, 126 Borel measure, 123 Borel set, 122–124, 148, 178 Bounded derivative, 158 Bounded variation, 151, 154, 158 C Calculator, 191 Canonical representation, 40 Cantor function, 166, 173, 175 Cantor set, 70, 84, 169, 170, 172 Carathéodory condition, 72 Characteristic function, 22 Closed under complements, 19, 103 Closed under countable unions, 19 Closed under finite intersections, 103 Closed under finite unions, 19, 103

Closed under unions, 103 Closure under countable intersections, 19 Co-countable, 97, 99 Combinatorics, 26 Complement, 19 Composition, 34, 122, 162 Countability, 7 Countable families of sets, 9 Countably additive, 69, 76 Countably infinite, 174 Countably subadditive, 26 Counting measure, 42, 46, 91, 93, 94, 96, 98, 100 Covered by, 65 Craven, 27, 60 D Dense, 129 Derivative of the integral, 148, 149 Dini derivatives, 138, 140 Disjoint intervals, 127, 128 Distribution, 3, 11 Distribution of a measurable function, 21 Dominated Convergence Theorem, 53, 81, 113, 130 Double integral, 99, 101, 103, 117, 120 Dunford, 94–97 E Extended real-valued, 14 F Fatou’s Lemma, 51, 119, 151, 163 Feldman, 97, 99 r-field, 19

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270 r-finite, 93, 95, 111 Finite family, 10, 133 Finitely subadditive, 26 F -measurable, 20, 21 Form a partition, 43 Fubini Series Theorem, 176, 177 Fubini Theorem, 93, 95, 116, 119, 122 Furious, 46 Fuzzy measure subadditive, 56–60 H Habil, 13, 99, 101 Heine–Borel Theorem, 68 Hölder’s Inequality, 55, 60 I 1, 12 Inclusion-Exclusion Principle, 26 Induced r-algebra, 85 measure, 85 Inner continuity, 35, 60 Integrable, 31, 52 Integral Lebesgue, 78, 80 improper, 10, 13, 81, 87 measure space, 27 Riemann, 1, 11, 80 Integral of the derivative, 162, 166 Interchanging order of integration, 92, 94, 96–100, 116, 118 order of integration, not valid, 92, 94, 118 order of summation, 13, 91, 93, 94, 96–100 Interior of an interval, 134 J Junior manager, 46 K Klir, 11 L Lebesgue Differentiability Theorem, 134, 142, 143, 149 Lebesgue integral, 78 Lebesgue measurable function, 76 Lebesgue measurable set, 72 Lebesgue measure, 71, 78 Lebesgue outer measure, 66 List of three cities, 46

Index M Mean Continuity Theorem, 131, 151 Mean Value Theorem, 158 l -measurable, 72 Measurable, 20, 21 Measure, 19, 20 induced, 85 Measure space, 20 Measure space integral, 27, 31, 52 Measure zero, 37, 72, 118, 143, 146, 148 uncountable set of, 84 Monotone, 3, 11, 71 Monotone class generated by, 104 Monotone Convergence Theorem, 35, 36, 51, 93, 95, 98, 100, 113, 114, 117, 130 Munkres, 14 N Nested Interval Theorem, 84, 171 Nonmeasurable set, 84, 178 O Open subset, 132 Order of integration interchange of, 92, 94, 97, 99 Order of summation interchange of, 13, 94, 96, 97, 99 Outer measure, 71 P Probability, 11 Product r-algebra, 105 Product measure, 103, 111, 115, 117 R Rearrangement Theorem, 13 Recurring, 173 Repeated integral, 92, 94, 98–101, 103, 116, 117, 120 sum, 13, 69, 94, 96 Riemann integral, 11, 80 Rubel, 177 Rudin, 3 S Sample space, 11 Schwartz, 94–97 Sections of a function, 108, 110 of a subset, 108 Senior manager, 46

Index Sequence of disjoint measurable sets, 19, 20, 69, 71 Set function, 11 Set-theoretic difference, 20, 103 Shirali, 60, 99, 101, 169 Simple function, 39 Step function, 2, 11, 39, 80, 127, 128 Subadditive, 57 countably, 26, 69 finitely, 26 fuzzy measure, 56–60 Subadditivity, 25 Sum of the lengths, 3 Summable, 31, 52 Supremum, 12 Symbol free descriptions, 19 Symmetric difference, 40 T Ternary expansion, 173 Third left, 170

271 middle, 70, 170 right, 170 Tonelli Theorem, 93, 95, 116, 117, 121, 122, 150 Total length existence of, 9 uniqueness of, 3, 5, 6, 10 U Unconditional sum, 100, 102 Uncountable set of measure zero, 84 V Vacuously, 138 Vasudeva, 99, 101, 169 Vitali covering, 133 Vitali Covering Theorem, 133, 134, 136 W Wang, 11