Integral and Measure

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MATHEMATICS AND STATISTICS SERIES

In Part 1, the definition of the integral of a one-variable function is different (not essentially, but rather methodically) from traditional definitions of Riemann or Lebesgue integrals. Such an approach allows us, on the one hand, to quickly develop the practical skills of integration as well as, on the other hand, in Part 2, to pass naturally to the more general Lebesgue integral. Based on the latter, in Part 2, the author develops a theory of integration for functions of several variables. In Part 3, within the same methodological scheme, the author presents the elements of theory of integration in an abstract space equipped with a measure; we cannot do without this in functional analysis, probability theory, etc. The majority of chapters are complemented with problems, mostly of the theoretical type.

Vigirdas Mackevičius is Professor of the Department of Mathematical Analysis in the Faculty of Mathematics of Vilnius University in Lithuania. His research interests include stochastic analysis, limit theorems for stochastic processes, and stochastic numerics.

www.iste.co.uk

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Integral and Measure From Rather Simple to Rather Complex

Integral and Measure

The book is mainly devoted to students of mathematics and related specialities. However, Part 1 can be successfully used by any student as a simple introduction to integration calculus.

Vigirdas Mackevičius

This book is devoted to integration, one of the two main operations in calculus.

Vigirdas Mackevičius

Integral and Measure

To my beloved wife Eugenija

Series Editor Nikolaos Limnios

Integral and Measure From Rather Simple to Rather Complex

Vigirdas Mackevičius

First published 2014 in Great Britain and the United States by ISTE Ltd and John Wiley & Sons, Inc.

Apart from any fair dealing for the purposes of research or private study, or criticism or review, as permitted under the Copyright, Designs and Patents Act 1988, this publication may only be reproduced, stored or transmitted, in any form or by any means, with the prior permission in writing of the publishers, or in the case of reprographic reproduction in accordance with the terms and licenses issued by the CLA. Enquiries concerning reproduction outside these terms should be sent to the publishers at the undermentioned address: ISTE Ltd 27-37 St George’s Road London SW19 4EU UK

John Wiley & Sons, Inc. 111 River Street Hoboken, NJ 07030 USA

www.iste.co.uk

www.wiley.com

© ISTE Ltd 2014 The rights of Vigirdas Mackevičius to be identified as the author of this work have been asserted by him in accordance with the Copyright, Designs and Patents Act 1988. Library of Congress Control Number: 2014945514 British Library Cataloguing-in-Publication Data A CIP record for this book is available from the British Library ISBN 978-1-84821-769-0

Contents

P REFACE

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

ix

N OTE FOR THE T EACHER OR W HO IS BETTER , R IEMANN OR L EBESGUE ? . . . . . . . . . . . . . . . . . . . . . . . . . .

xi

N OTATION

. . . . . . . . . . . . . . . . . . . . . . . . . . . . .

xiii

PART 1. I NTEGRATION OF O NE -VARIABLE F UNCTIONS . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

1

C HAPTER 1. F UNCTIONS WITHOUT S ECOND - KIND D ISCONTINUITIES . . . . . . . . . . . . . . . . . . . . . . . .

3

P.1. Problems . . . . . . . . . . . . . . . . . . . . . . . . . .

9

C HAPTER 2. I NDEFINITE I NTEGRAL . . . . . . . . . . . . .

11

P.2. Problems . . . . . . . . . . . . . . . . . . . . . . . . . .

16

C HAPTER 3. D EFINITE I NTEGRAL

. . . . . . . . . . . . . .

19

3.1. Introduction . . . . . . . . . . . . . . . . . . . . . . . . P.3. Problems . . . . . . . . . . . . . . . . . . . . . . . . . .

19 38

C HAPTER 4. A PPLICATIONS OF THE I NTEGRAL . . . . . .

43

4.1. Area of a curvilinear trapezium . . . . . . . . . . . . . . 4.2. A general scheme for applying the integrals . . . . . .

43 51

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Integral and Measure

. . . .

52 53 54 56

C HAPTER 5. OTHER D EFINITIONS : R IEMANN AND S TIELTJES I NTEGRALS . . . . . . . . . . . . . . . . . . . . .

59

5.1. Introduction . . . . . . . . . . . . . . . . . . . . . . . . P.5. Problems . . . . . . . . . . . . . . . . . . . . . . . . . .

59 75

C HAPTER 6. I MPROPER I NTEGRALS . . . . . . . . . . . . .

79

P.6. Problems . . . . . . . . . . . . . . . . . . . . . . . . . .

88

PART 2. I NTEGRATION OF S EVERAL - VARIABLE F UNCTIONS . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

91

C HAPTER 7. A DDITIONAL P ROPERTIES OF S TEP F UNCTIONS . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

93

7.1. The notion “almost everywhere” . . . . . . . . . . . . . P.7. Problems . . . . . . . . . . . . . . . . . . . . . . . . . .

97 104

C HAPTER 8. L EBESGUE I NTEGRAL . . . . . . . . . . . . . .

105

4.3. 4.4. 4.5. P.4.

Area of a surface of revolution Area of curvilinear sector . . . Applications in mechanics . . Problems . . . . . . . . . . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

8.1. Proof of the correctness of the definition of integral . . . . . . . . . . . . . . . . . . . . . 8.2. Proof of the Beppo Levi theorem . . . . . . 8.3. Proof of the Fatou–Lebesgue theorem . . . P.8. Problems . . . . . . . . . . . . . . . . . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

106 114 119 133

C HAPTER 9. F UBINI AND C HANGE - OF -VARIABLES T HEOREMS . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

139

P.9. Problems . . . . . . . . . . . . . . . . . . . . . . . . . .

157

C HAPTER 10. A PPLICATIONS OF M ULTIPLE I NTEGRALS . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

161

10.1. Calculation of the area of a plane figure . . . . . . . . 10.2. Calculation of the volume of a solid . . . . . . . . . . .

161 162

Contents

vii

10.3. Calculation of the area of a surface . . . . . . . . . . . 10.4. Calculation of the mass of a body . . . . . . . . . . . . 10.5. The static moment and mass center of a body . . . . .

162 165 166

C HAPTER 11. PARAMETER - DEPENDENT I NTEGRALS . . .

169

11.1. Introduction . . . . . . . . . . . . . . . . . . . . . . . . 11.2. Improper PDIs . . . . . . . . . . . . . . . . . . . . . . . P.11. Problems . . . . . . . . . . . . . . . . . . . . . . . . . .

169 177 187

PART 3. M EASURE AND I NTEGRATION IN A M EASURE S PACE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

191

C HAPTER 12. FAMILIES OF S ETS . . . . . . . . . . . . . . .

193

12.1. Introduction . . . . . . . . . . . . . . . . . . . . . . . . P.12. Problems . . . . . . . . . . . . . . . . . . . . . . . . . .

193 197

C HAPTER 13. M EASURE S PACES

. . . . . . . . . . . . . . .

199

P.13. Problems . . . . . . . . . . . . . . . . . . . . . . . . . .

206

C HAPTER 14. E XTENSION OF M EASURE . . . . . . . . . . .

209

P.14. Problems . . . . . . . . . . . . . . . . . . . . . . . . . .

220

C HAPTER 15. L EBESGUE –S TIELTJES M EASURES ON THE R EAL L INE AND D ISTRIBUTION F UNCTIONS . . . . . . . .

223

P.15. Problems . . . . . . . . . . . . . . . . . . . . . . . . . .

229

C HAPTER 16. M EASURABLE M APPINGS AND R EAL M EASURABLE F UNCTIONS . . . . . . . . . . . . . . . . . . .

233

P.16. Problems . . . . . . . . . . . . . . . . . . . . . . . . . .

239

C HAPTER 17. C ONVERGENCE A LMOST E VERYWHERE AND C ONVERGENCE IN M EASURE . . . . . . . . . . . . . .

241

P.17. Problems . . . . . . . . . . . . . . . . . . . . . . . . . .

246

C HAPTER 18. I NTEGRAL

. . . . . . . . . . . . . . . . . . . .

249

P.18. Problems . . . . . . . . . . . . . . . . . . . . . . . . . .

263

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Integral and Measure

C HAPTER 19. P RODUCT OF T WO M EASURE S PACES . . .

267

P.19. Problems . . . . . . . . . . . . . . . . . . . . . . . . . .

275

B IBLIOGRAPHY . . . . . . . . . . . . . . . . . . . . . . . . . .

277

I NDEX . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

279

Preface

This textbook is devoted to integration, an important part of calculus. In Part 1, the definition of the integral of a one-variable function is different (not essentially, but rather methodically) from traditional definitions of Riemann or Lebesgue integrals. Such an approach allows us, on the one hand, to quickly develop practical skills of integration and, on the other hand, later, in Part 2, to pass naturally to the more general Lebesgue integral. Based on the latter, in Part 2, we develop a theory of integration for functions of several variables. In Part 3, within the same methodological scheme, we present the elements of theory of integration in an abstract space equipped with a measure; we cannot do without this in functional analysis, probability theory, etc. The majority of the chapters are complemented with problems, mostly of the theoretical type. Although the three parts of the book are methodically related to each other, they are somewhat independent. For example, any reader accustomed to the Riemann integral and wishing to get into the theory of the Lebesgue integral is encouraged to begin with Part 2. Those who feel they are lacking in the basics of general theory of measure and integration should open the book from Part 3. The book is mainly devoted to students of mathematics and related specialities. However, Part 1 can be successfully used by any students as a simple introduction to integration calculus.

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Integral and Measure

We use the double numbering of statements (theorems, propositions, etc.): the first number denotes the number of the chapter, and the second indicates the number of the statement within the chapter. Though a large number of books and papers have influenced the contents of this book, the short reference list includes only those directly used by the author. The book is essentially a revised translation of the author’s book Integral and Measure (TEV, Vilnius, 1998) from Lithuanian. The author would like to thank Vilijandas Bagdonaviˇcius and a large number of students of the Faculty of Mathematics and Informatics of Vilnius University. Thanks to them, the book contains significantly fewer misprints. Vigirdas Mackeviˇcius Vilnius, July 2014

Note for the Teacher or Who is better, Riemann or Lebesgue?

Which integral, that of Riemann or Lebesgue, is preferable in teaching calculus to first-year students of mathematics? In discussions on this question, the opponents present serious arguments in favor of “their” integral and, as a rule, never change their opinion. The author of this textbook, although a supporter of the Lebesgue integral, proposes the third approach to teaching integration (first, for the first-year students). It is mentioned in some mathematical books; however, I did could not find it in textbooks for university students. In my opinion, the main advantages of the approach are the following: (1) a relatively simple definition of the integral of one-variable functions and the proof of its existence and main properties, (2) a natural passage to the Lebesgue integral (of multi-variable functions and in abstract measure spaces) when the students already have the basic skills of integration and its applications. To comprehend the approach, let us recall, in a few words, the Daniel construction of the Lebesgue integral. First, the integral is defined for some class of “simple” functions, say step functions, and the main its properties are proved. Then, by passing to the limit, the integral is extended to the limits of step functions. In which sense are these limits taken? Actually, the Lebesgue integral is defined for the functions that can be expressed as the difference of two functions that

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Integral and Measure

are the limits (almost everywhere) of monotonic sequences of step functions. What if we consider only uniform limits (i.e. the functions that are limits of uniformly converging sequences of step functions)? Applying the Cauchy criterion of uniform convergence, we immediately get the convergence of the corresponding sequence of integrals and call its limit the integral of the limit function. The correctness of such a definition is almost obvious. The main properties of the integral (linearity, additivity, the principal formula of integration, etc.) are proved very simply. The class of functions integrable in this sense consists of all functions without discontinuities of the second kind; although this class is narrower than the Riemann-integrable functions, it is sufficient for acquiring the main skills of integration and its applications in other areas (differential equations, mechanics, etc.). Later, moving to the Lebesgue integral is extremely natural by considering – instead of uniformly converging – pointwise (almost everywhere) converging sequences of step functions that are Cauchy sequences in the mean. Matching this scheme with the Daniel scheme is rather simple since the pointwise (a.e.) limit of a Cauchy sequence can be represented as the difference of two limits of monotonic sequences of step functions. Having indicated the advantages of our approach, it is also worth mentioning one “disadvantage”. When the integral is defined in the way mentioned, the students need to know the relatively difficult notion of uniform convergence of a series of functions, which must be taught perhaps earlier than usual.

Notation N

The set of positive integers {1, 2, . . .}

N

N ∪ {+∞}

N+

N ∪ {0}

R

Real line (−∞, +∞)

R

Extended real line R ∪ {−∞, +∞}

R+

The set of non-negative real numbers [0, +∞)

∀

“for all”, “for each”, “for every”

∃... : ...

There exists . . . such that. . .

:=

“Denote”, “is equal by definition to”

≡

Identically equal

=⇒ ⇒

“implies”

⇐⇒ , ⇔

“if and only if”, “necessary and sufficient”, “equivalent”

∅

Empty set

x∈A

An element x belongs to a set A

xA

An element x does not belong to a set A

A

c

The complement of a set A

A ⊂ B, B ⊃ A

A set A is a subset of a set B

{xn } ⊂ A

A sequence {xn , n ∈ N} of elements of a set A

xiv

Integral and Measure

A ∪ B, A ∩ B ∞   An , ∞ n=1 An

n=1 ∞  n=1

An ,

∞

n=1 An

The union and intersection of sets A and B The union of a sequence of sets An , n ∈ N The intersection of a sequence of sets An , n ∈ N

f:X → Y

A function (mapping) from X to Y

fn ⇒ f

A sequence of functions { fn } uniformly converges to a function f (in some given set A)

[x]

The integer part of a number x (the maximal integer not exceeding x)

x ∨ y, x ∧ y

max{x, y}, min{x, y}

1lA

The indicator of a set A: 1lA (x) = 1 for x ∈ A; 1lA (x) = 0 for x ∈ Ac

Part 1

Integration of One-Variable Functions

1 Functions without Second-kind Discontinuities

D EFINITION 1.1.– A function f : I → R, where I ⊂ R is an interval, is said to be continuous at a point x0 ∈ I if lim x→x0 f (x) = f (x0 ). Otherwise, f is said to be discontinuous (or has a jump) at x0 or that x0 is a discontinuity point of f . A discontinuity point x0 of a function f is called: 1) a first-kind (or simple) discontinuity point if there exist finite onesided limits f (x0 − 0) := lim f (x) x→x0 −0

and

f (x0 + 0) := lim f (x); x→x0 +0

2) a second-kind discontinuity point in the remaining cases, where at least one of the limits f (x0 ± 0) is infinite (infinite discontinuity) or does not exist (oscillating discontinuity). The functions f : I → R that do not have second-kind discontinuities in I are called regular. We denote the set of all such functions by D(I) (the letter D from “discontinuous”). In the case of a concrete interval, we will omit the parentheses: D[a, b] := D([a, b]), D[a, b) := D([a, b)), etc. It is obvious that C(I) ⊂ D(I) (where, as usual, C(I) denotes the set of all continuous functions f : I → R). R EMARK 1.1.– Because of its generality, the term regular is not perfect for functions without second-order discontinuities. As we will

4

Integral and Measure

see further, mathematically, a more appropriate term is uniformly measurable; however, it is rather long and inconvenient.

Figure 1.1. A regular function f ∈ D[a, b]

D EFINITION 1.2.– A function ϕ: [a, b] → R is called a step function if the interval [a, b] can be partitioned into finitely many intervals Ik , k =  1, 2, . . . , m ( m k=1 Ik = [a, b]) so that, in each interval Ik , ϕ is constant (i.e. ∃ yk ∈ R, k = 1, 2, . . . , m : ϕ(x) = yk for x ∈ Ik ). We denote the set of all such functions by S [a, b]. Note that here we consider one-point sets {c} as closed intervals [c, c].

Figure 1.2. A step function ϕ ∈ S [a, b]

P ROPOSITION 1.1.– Let f ∈ C[a, b]. Define the sequence of step functions {ϕn } by ⎧ 

n n n ⎪ ⎪ ⎨ f xk f or x ∈ xk , xk+1 , k = 0, 1, . . . , n − 1, ϕn (x) := ⎪ ⎪ ⎩ f (b) f or x = xnn = b; xnk := a +

b−a k, n

k = 0, 1, 2, . . . , n.

Then ϕn ⇒ f in the interval [a, b], that is, the sequence {ϕn } converges to f uniformly in [a, b].

Functions without Second-kind Discontinuities

5

Figure 1.3. Approximation of a continuous function by step functions

P ROOF.– Take arbitrary ε > 0. Since f is uniformly continuous in [a, b] (Cantor theorem), there exists δ > 0 such that    f (x) − f (y) < ε if |x − y| < δ, x, y ∈ [a, b]. Taking N ∈ N such that |ϕn (x) − f (x)| < ε,

b−a N

< δ, we get that, for all n > N,

x ∈ [a, b].

Indeed, let n > N. For any x ∈ [a, b), let [xkn , xnk+1 ) be the interval to which the point x belongs. Then |xnk − x| < δ, and therefore, |ϕn (x) − f (x)| = | f (xnk ) − f (x)| < ε. Finally, for x = b, we always have ϕn (b) − f (b) = 0 < ε. This means that, as required, ϕn ⇒ f in [a, b] (the choice of N depends only on ε > 0 and not on x ∈ [a, b]).   R EMARK 1.2.– From the proof of proposition 1.1, we can easily see that the uniform convergence ϕn ⇒ f in the interval [a, b] remains true if (see Figure 1.4): 1) instead of the points xnk = a + b−a n that partition the interval [a, b] into n intervals [xnk , xnk+1 ], k = 0, 1, . . . , n − 1, of equal size b−a n , we take n n n arbitrary points a = x0 < x1 < · · · < xkn = b, n ∈ N, satisfying the condition  n  max  xk+1 − xkn  → 0

0kkn −1

as n → 0;

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Integral and Measure

Figure 1.4. A more general approximation of a continuous function by step functions

2) we define the values of the step functions ϕn as the values of f taken not at the left points of the partition interval [xkn , xnk+1 ], but rather n ], that is, at arbitrary points ξkn ∈ [xnk , xk+1 ⎧  n

n n ⎪ ⎪ ⎨ f ξk for x ∈ xk , xk+1 , k = 0, 1, . . . , kn − 1, ϕn (x) := ⎪ ⎪ ⎩ f (b) for x = xnk = b. n T HEOREM 1.1.– 1) For any function f ∈ D[a, b], there exist a sequence of step functions {ϕn } ⊂ S [a, b] converging to f uniformly in the interval [a, b]. 2) The set of discontinuity points of any f ∈ D[a, b] is finite or countable. 3) Every function f ∈ D[a, b] is bounded. R EMARK 1.3.– The converse of the first statement is also true: the limit of a uniformly converging sequence of step functions is a function without second-kind discontinuities. Therefore, the class D[a, b] can be characterized as the functions that are uniform limits of sequences of step functions in [a, b]. P ROOF.– [of theorem 1.1]

Functions without Second-kind Discontinuities

7

1) Let f ∈ D[a, b]. Take arbitrary ε > 0. Then, for every x ∈ [a, b], there exists δ x > 0 such that   ε  f (t) − f (x + 0) < for t ∈ (x, x + δ x ) ∩ [a, b] 2 and   ε  f (t) − f (x − 0) < 2 Since



Uδx (x) =

x∈[a,b]

for t ∈ (x − δx , x) ∩ [a, b].

(x − δ x , x + δx ) ⊃ [a, b],

x∈[a,b]

by the compactness of the interval [a, b] (finite subcovering property), there exist x1 , x2 , . . . , xn ∈ [a, b] such that Uδx1 (x1 ) ∪ Uδx2 (x2 ) ∪ · · · ∪ Uδxn (xn ) ⊃ [a, b]. Let {a0 , a1 , . . . , ak } be the set consisting of the numbers a, b, xi , xi − δxi , xi + δ xi belonging to the interval [a, b] and written in the increasing order: a = a0 < a1 < · · · < ak = b. Define the function ϕ ∈ S [a, b] by ϕ(ai ) := f (ai ), ϕ(x) := f (c j )

i = 0, 1, . . . , k; for x ∈ (a j−1 , a j ),

with arbitrary c j ∈ (a j−1 , a j ) (say, c j = (a j−1 + a j )/2). Then |ϕ(x) − f (x)| < ε for all x ∈ [a, b]. Indeed, let x ∈ [a, b]. If x = a j for some j, then |ϕ(x) − f (x)| = 0 < ε. If x ∈ (a j−1 , a j ), then there is i such that (a j−1 , a j ) ⊂ (xi , xi + δ xi ) or (a j−1 , a j ) ⊂ (xi − δ xi , xi ).

8

Integral and Measure

Suppose, for example, that case (1) is true. Then,     ϕ(x) − f (x) =  f (c j ) − f (x)       f (c j ) − f (xi + 0) +  f (xi + 0) − f (x) <

ε ε + =ε 2 2

since 0 < c j − xi < a j − xi < δ xi and 0 < x − xi < δ xi . Thus, for arbitrary ε > 0, we have a function ϕ ∈ S [a, b] such that |ϕ(x) − f (x)| < ε for all x ∈ [a, b]. By taking ε = 1n , n = 1, 2, . . . , we get a sequence {ϕn } ⊂ S [a, b] such that 1 |ϕn (x) − f (x)| < , n

x ∈ [a, b], n ∈ N.

Clearly, ϕn ⇒ f in [a, b]. 2) Let f ∈ D[a, b], and let {ϕn } ⊂ S [a, b] be a sequence constructed in the previous item (ϕn ⇒ f ). The set T n of discontinuity points of each ϕn is finite (since it consists of the ends of constancy intervals of  ϕn ). All the functions ϕn are continuous on the set A = [a, b]\( ∞ n=1 T n ). Therefore, f is continuous on the set A.1 This means that f may have  discontinuities only at the points of the set T = ∞ n=1 T n . Since the set T is finite or countable (as a countable union of finite sets), the statement is proved. 3). For any f ∈ D[a, b], take ϕ ∈ S [a, b] such that    f (x) − ϕ(x)  1,

x ∈ [a, b].

1 If all the functions in a uniformly converging sequence are continuous at some point, so is the limit function.

Functions without Second-kind Discontinuities

9

Denoting M := max x∈[a,b] |ϕ(x)| (ϕ ∈ S [a, b] can take only finitely many distinct values), we get      f (x)  ϕ(x) + 1  M + 1,

x ∈ [a, b].

 

R EMARK 1.4 (for a lecturer).– If the latter proof seems to be too difficult for the first-year students, we may omit it (and the whole chapter as well) and later (in Chapter 3) define the integral for functions that are uniform limits of sequences of step functions (that is, in fact, for all functions from D[a, b]), without proving that such functions exhaust the whole class D[a, b]. Of course, in such a case, we need to prove that at least all continuous functions are integrable (proposition 1.1) and, as an easy consequence, so are the functions with finitely many first-kind discontinuities. Such functions are sufficient for developing practical integration skills. P.1. Problems P ROBLEM 1.1.– Give an example of a function f ∈ D[a, b] with an infinite (countable) set of discontinuity points. P ROBLEM 1.2.– May the set of discontinuity points of a function f ∈ D[a, b] be the rationals of the interval [a, b]? P ROBLEM 1.3.– Does the Dini lemma hold in the class D[a, b], that is, does the monotone convergence D[a, b]  fn ↓ 0 imply that fn ⇒ 0? P ROBLEM 1.4.– Show that if f ∈ D[a, b] is right continuous (that is, f (x + 0) = f (x), x ∈ [a, b)), then the sequence{ϕn } in theorem 1.1 can be chosen to be of right-continuous functions.

2 Indefinite Integral

D EFINITION 2.1.– A function F: I → R,1 is called a primitive (or an antiderivative) of a function f : I → R if F  (x) = f (x),

x ∈ I.

P ROPOSITION 2.1.– 1) If F1 and F 2 are two primitives of a function f : I → R, then F1 − F 2 is a constant (i.e. ∃C ∈ R : F1 (x) − F 2 (x) = C, x ∈ I). 2) If F is a primitive of a function f : I → R, then for any C ∈ R, the function F1 = F +C is also a primitive of f . In other words, all primitives of f only differ by constants. P ROOF.– 1) F1 = F2 = f ⇒ (F1 − F2 ) = F 1 − F 2 = f − f = 0 ⇒ F 1 − F2 = const. 2) F  = f ⇒ (F + C) = F  + C  = f + 0 = f .

 

R EMARK 2.1.– If I is not an interval, the proposition is not true: consider the example F(x) = sgn x, x ∈ I := (−1, 1) \ {0}.

1 Here and below in this section, I ⊂ R is an interval.

12

Integral and Measure

D EFINITION 2.2.– The indefinite integral

of a function f : I → R is the family of its primitives. It is denoted as f (x) dx. By proposition 2.1,  f (x) dx = {F + C : C ∈ R}, where F is any primitive of f (provided that it does exist). It is common to write for short  f (x) dx = F(x) + C. Table of main integrals   1) dx = 1 · dx = x + C;  xa+1 2) xa dx = + C (a  −1); a+1  dx 3) = ln |x| + C; x 1  x dx a arctg a + C, 4) = 1 − a arcctg ax + C1 ; a2 + x2   dx arcsin ax + C, 5) = (a > 0); √ − arccos ax + C1 2 − x2 a   ax x 6) a dx = + C (a > 0, a  1); e x dx = e x + C; ln a   7) sin x dx = − cos x + C; cos x dx = sin x + C;   dx dx 8) = −ctg x + C; = tg x + C; 2 cos2 x  sin x   dx 9) = ln x + x2 + a + C; √ 2 x +a  dx 1  x − a  10) = ln   + C; 2 2 2a x + a  x − a   x 2 a  11) x2 + a dx = x + a + ln  x + x2 + a + C (a  0); 2 2    x a2 x 12) a2 − x2 dx = a2 − x2 + arcsin + C (a > 0). 2 2 a

Indefinite Integral

13

P ROPOSITION 2.2 (Properties of indefinite integrals).– Suppose that f : I → R and g: I → R. Then: 1) (linearity) for all α, β ∈ R,    (α f + βg)(x) dx = α f (x) dx + β g(x) dx, provided that both integrals on the right-hand side of the equality exist; 2) (change-of-variables formula)     f (x) dx = F(x) + C =⇒ f ϕ(t) ϕ (t) dt = F ϕ(t) + C, provided that R(ϕ) ⊂ I and ϕ is a differentiable function; 3) (integration-by-parts formula)    f (x)g (x) dx = f (x)g(x) − g(x) f  (x) dx, provided that f and g are differentiable functions and there exists at least one indefinite integral in the equality. R EMARK 2.2.– Since, strictly speaking, the indefinite integral is not a function, but rather a family of functions, the above equalities need some explanation. For example, the equality in item 1 means the following: if F and G are any primitives of f and g, respectively, then αF + βG is a primitive of the function α f + βg. P ROOF.– 1)   F∈ f (x) dx, G ∈ g(x) dx =⇒ F  = f, G = g =⇒ (αF + βG) = αF  + βG = α f + βg;  =⇒ αF + βG ∈ (α f + βg)(x) dx. 2) If F  (x) = f (x), x ∈ I, then

    F ϕ(t) t = F  ϕ(t) ϕ (t) ≡ f ϕ(t) ϕ (t).

14

Integral and Measure

3) From the equality ( f (x)g(x)) = f (x)g (x) + g(x) f  (x) we have 



f (x)g (x)dx& =









f (x)g(x) dx − g(x) f  (x) dx  & = f (x)g(x) − g(x) f  (x) dx.

 

R EMARK 2.3.– In practical integration, it is convenient to use the notion of differential. The differential of a differentiable function f : I → R at a point x ∈ I is the linear function d f (x) := f  (x) dx,

dx ∈ R.

By the definition of the derivative, we have: f (x + dx) − f (x) = f  (x) dx + o(dx),

dx → 0.

Therefore, the differential is often defined as the “principal linear part of the increment Δ f (x) = f (x + dx) − f (x) of a function f corresponding to the increment dx of the argument x” and is used for approximate calculations since, for “small” dx, Δ f (x) ≈ d f (x). However, in integration, the differentials are important because of another reason. Consider the differential of a composite function,     d f (ϕ(t)) = f ϕ(t)  dt = f  ϕ(t) ϕ (t) dt = f  ϕ(t) dϕ(t), or, for short, d f (ϕ) = f  (ϕ) dϕ. Thus, if in the equality d f (x) = f  (x) dx, instead of an “independent variable” x, we take a function ϕ, we get the true equality d f (ϕ) = f  (ϕ) dϕ. This property is called the invariance of the form of a differential. Using it, we can write the change-of-variables and integration-by-parts formulas as follows: 

 f (x) dx = F(x) + C =⇒

f (ϕ) dϕ = F(ϕ) + C;

Indefinite Integral



15

 f dg = f g −

g d f.

Simplicity and convenience of these formulas is one of the reasons for

using, for indefinite integrals, the notation f (x) dx, rather than f (x). E XAMPLES 2.1.–   sin4 x 1) sin3 x cos x dx = sin3 x d(sin x) = +C 4

4 (since ϕ3 dϕ = ϕ4 + C); 2) 

sin3 x dx = (1 − cos2 x) sin x dx = (cos2 x − 1) d(cos x)

= cos2 x d(cos x) − d(cos x) =

 3)

x dx 1 = 2 2 1+ x



cos3 x 3

d(x2 ) 1 = 1 + x2 2



− cos x + C;

d(1 + x2 ) 1 = ln(1 + x2 ) + C; 2 2 1+ x

4) 

x dx 1 = 4 2 1+ x



d(x2 ) 1 = arctg x2 + C; 2 2 2 1 + (x )

5) 

 ln x dx = ln x · x −

 x d(ln x) = x · ln x −

 = x · ln x −

dx = x · ln x − x + C;

1 x · dx x

16

Integral and Measure

6) 

 arctg x dx = x · arctg x −

x d(arctg x) 

= x · arctg x −

x dx 1 + x2

1 = x · arctg x − ln(1 + x2 ) + C; 2 7)  I :=

e x sin x dx = sin x d(e x ) = e x sin x − e x d(sin x)

= e x sin x − e x cos x dx

= e x sin x − cos x d(e x )  

= e x sin x − e x cos x − e x d(cos x)

= e x (sin x − cos x) − e x sin x dx = e x (sin x − cos x) − I =⇒ I = 12 e x (sin x − cos x) + C.

P.2. Problems P ROBLEM 2.1.– Find the indefinite integrals:



sin2 x cos x dx; sin2 x cos2 x dx; sin2 x cos3 x dx; sin2 x cos4 x dx.

P ROBLEM 2.2.– Find the indefinite integrals: 

x2 dx ; 1 + x2

Indefinite Integral



17

x3 dx ; 1 + x2



x4 dx . 1 + x2

P ROBLEM 2.3.– Find the conditions to be satisfied by the coefficients a, b, and c given that the integral: 

ax2 + bx + c dx x3 (x − 1)2

is a rational function.



P ROBLEM 2.4.– Find the indefinite integrals  x lnn x dx, n ∈ N.

xn ln x dx and

P ROBLEM 2.5.– Find the indefinite integrals:



xn e x dx,

n ∈ N;

xemx dx,

m ∈ N;

xn emx dx,

n, m ∈ N. 

P ROBLEM 2.6.– Find the indefinite integrals  x3 arctg x dx.

x2 arctg x dx and

 P ROBLEM 2.7.– Find the indefinite integral

eax sin bx dx, a, b ∈ R.

3 Definite Integral

3.1. Introduction The notion of definite integral is closely related to the problem of calculating the area of a geometric figure on the plane. Let us try to calculate the area of the figure that is under the graph of a continuous function over the interval [a, b] (such a figure is called a curvilinear trapezium). 1

Figure 3.1. Calculation of the area of a curvilinear trapezium

1 In fact, we do not know yet what the area of a figure is, except in the simplest cases (a rectangle, triangle, etc.); therefore, we will follow the intuitive notion of area.

20

Integral and Measure

Let us divide the interval [a, b] into small intervals bounded by the points a = x0 < x1 < x2 < · · · < xn = b (Figure 3.1). At the same time, the curvilinear trapezium will be divided into n bounded strips (that are also curvilinear trapeziums). Clearly, the area of the whole trapezium is the sum of small strips. Let us replace the area of each strip with base Δxk = xk − xk−1 by the area of the rectangle with the same base and height f (xk−1 ). If the strips are sufficiently narrow, we can expect that the area S of the trapezium is approximately equal to the sum of rectangles: S≈

n 

f (xk−1 )Δxk .

k=1

Instead of the values of f at the points xk−1 , we often take the values at arbitrary points ξk ∈ [xk−1 , xk ] (Figure 3.2): S≈

n 

f (ξk )Δxk .

[3.1]

k=1

Figure 3.2. Modified calculation of the area of a curvilinear trapezium

The narrower the strip, the more exact the value of the area S . We will get the exact value of S passing to the limit as all widths of strips tend to zero: S = lim

Δxk →0

n  k=1

f (ξk )Δxk .

Definite Integral

21

This limit (not yet defined rigorously) is called the (definite) Riemann integral of the function f in the interval [a, b] (or “from a to

b b”) and is denoted by a f (x) dx. Thus, the definite integral is equal to the area of the corresponding curvilinear trapezium. The “definition” above may be called “geometric” because it is based on the geometric problem of calculating an area and geometric arguments. We can use another “purely analytic” approach. Let us carefully look at our arguments. First, to define the integral as the area of a curvilinear trapezium, we replaced the curvilinear trapezium by a “close” figure, the union of “narrow” rectangles, the area of which we can calculate. Then, we passed to the limit by taking such figures increasingly close to the trapezium. Analytically, this means that we replaced the function f by a “close” step function ϕ defined by ϕ(x) = f (ξk ),

x ∈ [xk−1 , xk ),

k = 1, 2, . . . , n,

ϕ(b) = f (b).

Indeed, note that the sum in the approximate equality [3.1] is, in fact, the area of the figure under the graph of ϕ over the interval [a, b] and,  thus, the integral of the function ϕ in the interval [a, b]: nk=1 f (ξk )Δxk =

b ϕ(x) dx. Taking step functions increasingly “close” to the function f , a in the limit, we obtained the integral of the function f . Thus, the other way of defining the integral is as follows. First, we define the integrals of step functions. Then, we consider the functions that can be approximated by step functions, that is, the functions that can be expressed as the limit of a sequence of step functions. The integral of such a function is then defined as the limit of the corresponding integrals of step functions. In this context, a natural question arises: under which conditions does such a limit exist? Note that the pointwise convergence is not sufficient – for a pointwise converging sequence of step functions, the limit of the corresponding integrals may depend on the choice of the sequence (in similar cases, we speak about the incorrectness of a definition). Therefore, we need some additional conditions on convergence of sequences of step functions. The simplest (and, at the same time, the strongest) condition is the requirement of uniformity of convergence. Since the uniform

22

Integral and Measure

limits of sequences of step functions are the functions without discontinuities of the second kind (theorem 1.1), we will be able to define the integral namely for such functions. Later, in Chapter 8, replacing the uniform convergence condition by a weaker one, we will be able to extend the class of integrable functions. D EFINITION 3.1.– Suppose that a function ϕ ∈ S [a, b] takes the value  yk in the interval Ik , k = 1, 2, . . . , n ( nk=1 Ik = [a, b]). Then the definite integral of the function ϕ in the interval [a, b] is defined as the sum n 

yk |Ik |,

[3.2]

k=1

where |I| denotes the length of an interval I. It is denoted by 

b ϕ(x) dx, or, shortly, a ϕ.

b a

ϕ(x) dx,

[a,b]

P ROPOSITION 3.1 (The correctness of the definition of the integral).– The value of the sum in equation [3.2] (and thus the integral of a function ϕ ∈ S [a, b]) does not depend on the choice of a partition of the interval [a, b] by constancy intervals of ϕ. P ROOF.– Let {Ik , k = 1, 2, . . . , n}

and {Iˆj , j = 1, 2, . . . , m}

be two partitions of the interval [a, b] by the constancy intervals   ( nk=1 Ik = mj=1 Iˆj = [a, b]; Ik ∩ Ik = ∅ for k  k ; Iˆj ∩ Iˆj = ∅ for j  j ); suppose that ϕ(x) = yk for x ∈ Ik and ϕ(x) = yˆ j for x ∈ Iˆj . Then  yk = ϕ(x) = yˆ j if x ∈ Ik ∩ Iˆj  ∅. Moreover, |Ik | = mj=1 |Ik ∩ Iˆj | since   Ik = mj=1 (Ik ∩ Iˆj ), and, similarly, |Iˆj | = nk=1 |Ik ∩ Iˆj | since  Iˆj = nk=1 (Ik ∩ Iˆj ) (|∅| := 0). Therefore, n  k=1

yk |Ik | =

n  k=1

yk

m  n  m     I ∩ Iˆ  = yk Ik ∩ Iˆj  k j j=1

k=1 j=1

n  m m n  m         I ∩ Iˆ  = = yˆ j Ik ∩ Iˆj  = yˆ j yˆ j Iˆj . k j k=1 j=1

j=1

k=1

j=1

Definite Integral

23

The third equality follows from the fact that either yk = yˆ j or |Ik ∩ Iˆj | = 0; thus, we always have yk |Ik ∩ Iˆj | = yˆ j |Ik ∩ Iˆj |.   P ROPOSITION 3.2 (Elementary properties of the integral of step functions).– 1) Linearity: if f, g ∈ S [a, b] and α, β ∈ R, then  b  b  b (α f + βg) = α f +β g. a

a

a

2) Additivity: if f ∈ S [a, b] and a < c < b, then  b  c  b f= f+ f. a

a

a

a

c

3) Monotonicity: if f, g ∈ S [a, b] and f  g (i.e., ( f (x)  g(x) for all x ∈ [a, b]), then  b  b f g. 4) If f ∈ S [a, b], then   b   b      f . f     a a P ROOF.– We begin with an important remark. If f and g are two functions from S [a, b], then, without loss of generality, we may assume that they both have the same partition of the interval [a, b] by their constancy intervals. Indeed, if {Ik } is a system of constancy intervals of f , and {Iˆj } is a system of constancy intervals of g, then {Ik ∩ Iˆj : Ik ∩ Iˆj  ∅} is a system of intervals in which both functions f and g are constant. Thus, we further suppose that {Ik , k = 1, 2, . . . , n} is a system of constancy intervals of both step functions f and g and that, moreover, f (x) = yk and g(x) = zk for x ∈ Ik . 1) The value of the function α f + βg in the interval Ik is αyk + βzk . Therefore, by definition 3.1, 

b a

(α f + βg) =



k (αyk + βzk )|Ik |

=α

=α

b a



f +β

k yk |Ik | + β

b a

g.



k zk |Ik |

24

Integral and Measure

2) Let Ik0 be the interval containing the point c. Denote I 1 := Ik0 ∩ [a, c] and I 2 := Ik0 ∩ (c, b]. Then we have 

b

f =

a



 k

=

yk |Ik | + yk0 |Ik0 | +



yk |Ik |

    yk |Ik | + yk0 |I 1 | + yk0 |I 2 | + yk |Ik | k
k
 =



yk |Ik | =

k>k0



b

f+

a

k>k0

f. c

3) If f  g, then yk  zk for all k. Therefore, 

b

f=

a



yk |Ik | 

k





b

zk |Ik | =

g. a

k

4) Integrating the inequality −| f |  f  | f | and using properties 1 and 3, we have 

b

−



b

|f| 

a

a



b

f a

  | f | =⇒ 

b a

  b    f . f   a

 

D EFINITION 3.2.– The integral of a function f ∈ D[a, b] in the interval [a, b] is the limit 

b a





b

f=

b

f (x) dx := lim a

n→∞

ϕn (x) dx,

a

where {ϕn } is an arbitrary sequence of step functions converging to f uniformly in [a, b]. P ROPOSITION 3.3 (Existence of integral and correctness of definition).– The limit in the above definition always exists and does not depend on the choice of a sequence {ϕn } uniformly converging to f .

Definite Integral

25

P ROOF.– Existence. Let S [a, b]  ϕn ⇒ f in the interval [a, b]. Take arbitrary ε > 0. There exists N ∈ N such that   ϕn (x) − f (x) <

ε 2(b − a)

for all x ∈ [a, b] and n > N.

Therefore,       ϕn (x) − ϕm (x)  ϕn (x) − f (x) +  f (x) − ϕm (x) ε ε ε < + = 2(b − a) 2(b − a) b − a for all n, m > N and x ∈ [a, b]. Using the properties proved in proposition 3.3, we have:   b  b    b   b    ϕn − ϕm  =  (ϕn − ϕm )  |ϕn − ϕm |    a a a a  b  b ε ε  dx = dx b−a a a b−a ε = (b − a) = ε, n, m > N. b−a By the Cauchy criterion of convergence of a real sequence, the

b sequence of integrals { a ϕn } converges to a real number. Correctness. Let S [a, b]  ϕn ⇒ f and S [a, b]  ψn ⇒ f (in the interval [a, b]). Consider the new sequence {μn } = {ϕ1 , ψ1 , ϕ2 , ψ2 , . . . , ϕn , ψn , . . .}. It is clear that S [a, b]  μn ⇒ f in [a, b]. By the first part of the proof,

b there exists limn→∞ a μn ∈ R. Therefore, the limits of the subsequences

b

b { a ϕn } and { a ψn } of this sequence coincide.   E XAMPLE 3.1.– For any function f ∈ C[0, 1], define the step functions ϕn (x) := f

k , n

x∈

k k +1 , , k = 0, 1, 2, . . . . n n

26

Integral and Measure

We have already shown that ϕn ⇒ f (proposition 1.1, remark 1.2). Therefore, 

1

 f (x) dx = lim

n→∞

0

1

0 n−1 

k 1 f . n→∞ n n k=0

= lim

 n−1    k  k k + 1   ,  f n→∞ n n n  k=0

ϕn (x) dx = lim

Let us apply this equality to the function f (x) = x2 , x ∈ [0, 1]: 

1 0

n−1 2 n−1 k 1   k 2 x dx = lim = lim k=03 n→∞ n n→∞ n n k=0 n(n − 1)(2n − 1) 1 = lim = , n→∞ 3 6n3 2

where we used the well-known equality

n 

k2 =

k=1

n(n + 1)(2n + 1) , n ∈ N, 6

which can be easily checked by induction.

P ROPOSITION 3.4 (Properties of the integral).– 1) Linearity: if f, g ∈ D[a, b] and α, β ∈ R, then  b  b  b (α f + βg) = α f +β g. a

a

a

2) Additivity: if f ∈ D[a, b] and a < c < b, then  b  c  b f= f+ f. a

a

c



b

3) Monotonicity: if f, g ∈ D[a, b] and f  g, then 4) If f ∈ D[a, b], then   b   b      f . f     a a



g.

a



b

5) If D[a, b]  fn ⇒ f in the interval [a, b], then a

b

f a



b

fn →

f. a

Definite Integral

5 )

If fn ∈ D[a, b], n ∈ N, and the series f =

n=1

fn converges uniformly

n=1

in [a, b], then  b ∞   f= a

∞ 

27

b

fn .

a

P ROOF.– 1) Let S [a, b]  ϕn ⇒ f and S [a, b]  ψn ⇒ g in [a, b]. Then S [a, b]  αϕn + βψn ⇒ α f + βg. Therefore, by definition 3.2 and proposition 3.2, item 1, 

b

 (α f + βg) = lim

b

(αϕn + βψn )  a b  b  = lim α ϕn + β ψn n→∞ a a  b  b = α lim ϕn + β lim ψn n→∞ a n→∞ a  b  b =α f +β g. n→∞

a

a

a

2) Let S [a, b]  ϕn ⇒ f in [a, b]. Then, obviously, ϕn ⇒ f in both intervals [a, c] and [c, b]. Therefore, by definition 3.2 and proposition 3.2, item 2, 

b a



b

f = lim





ϕn +  c   = lim ϕn + lim ϕn = n→∞

n→∞

a

a

ϕn = lim

c

n→∞

n→∞

a b

c

c c a

b

 ϕn 

b

f+

f. c

3) First, suppose that f  0 and S [a, b]  ϕn ⇒ f . Since ϕn need not be non-negative, we slightly correct them by taking, instead, ϕ˜ n :=

28

Integral and Measure

max{ϕn , 0}  0. Then S [a, b]  ϕ˜ n ⇒ max{ f, 0} = f and n ∈ N. Therefore, 

b



b

f = lim

n→∞

a

b a

 ϕn  0 for all

 ϕn  0.

a

In the general case of f  g, we have that f − g  0 and, thus, by the preceding, 

b



b

( f − g)  0 =⇒

a



b

f−

a



b

g  0 =⇒

a



b

f

a

g. a

4) The proof is the same as that for step functions. 5) Since   εn := sup  fn (x) − f (x) → 0,

n → ∞,

x∈[a,b]

we have   b  b  fn −  a a

   b   b    f =  ( f − f )  | fn − f |   a n  a  b  εn dx = εn (b − a) → 0, n → ∞. a

5 ) It suffices to apply property 5 to the partial-sum sequence of the series: 

b

f=

 b ∞

a

 fn =

a n=1

b

lim

N 

a N→∞ n=1

fn = lim

N→∞

N   n=1

b a

fn =

∞   n=1

b

fn .

a

  P ROPOSITION 3.5.– Let f ∈ D[a, b]. Denote 

x

F(x) =

f (t) dt, a

x ∈ [a, b].

Definite Integral

29

Such a function F is called an integral with variable upper bound. Then F ∈ C[a, b]. If, moreover, f is continuous at a point x0 ∈ [a, b], then F is differentiable at x0 , and F  (x0 ) = f (x0 ). In particular, if f ∈ C[a, b], then F is a primitive function of f in the interval [a, b]. R EMARK 3.1.– A function F ∈ C[a, b] is called a generalized primitive of a function f : [a, b] → R if F  (x) = f (x) at all points x ∈ [a, b], except a countable set. Thus, in other words, proposition 3.5 states that an integral with variable upper bound of a function f ∈ D[a, b] is a generalized primitive of the latter. P ROOF .– If a  x0 < x  b, then, by proposition 3.2 (item 2),   x0    x     x    F(x) − F(x0 ) =  f− f  =   a   x0 a = M(x − x0 ) = M|x − x0 |,

  x  x  f  |f|  M  x0 x0

where M := sup x∈[a,b] | f (x)| < +∞. We get the same inequality for a  x < x0  b. From it we see that lim x→x0 F(x) = F(x0 ) for every x0 ∈ [a, b]. Thus, F ∈ C[a, b]. Suppose now that f is continuous at a point x0 ∈ [a, b]. Take an arbitrary ε > 0. Then there exists δ > 0 such that | f (t) − f (x0 )| < ε if |t − x0 | < δ, t ∈ [a, b]. For a  x0 < x  b such that |x − x0 | < δ, we have:   x      F(x) − F(x0 ) − f (x ) =  1 f (t) dt − f (x0 ) 0      x − x0 x0  x − x0 x   x  1  1 =  f (t) dt − f (x0 ) dt  x − x0 x0  x − x0 x0   x  x    1  1  f (t) − f (x ) dt = f (t) − f (x0 ) dt   0  |x − x0 | x0 |x − x0 |  x0 x 1  ε dt = ε. |x − x0 | x0

30

Integral and Measure

We get the same inequality for a  x < x0  b such that |x − x0 | < δ. Thus,    F(x) − F(x0 ) − f (x )  ε if |x − x | < δ, x ∈ [a, b]. 0  0  x − x0 This means that F  (x0 ) = lim x→x0

F(x)−F(x0 ) x−x0

= f (x0 ).

 

C OROLLARY 3.1 (Newton–Leibnitz formula2).– If F is a primitive of a function f ∈ C[a, b] in the interval [a, b], then  b f (x) dx = F(b) − F(a). a

x P ROOF.– By proposition 3.5, the function F1 (x) := a f (t) dt, x ∈ [a, b], also is a primitive of f . Therefore, by proposition 2.1, there exists a constant C such that F 1 (x) = F(x) + C, x ∈ [a, b]. Substituting x = a into this equality, we get 0 = F(a) +C, that is, C = −F(a). Then, substituting

b x = b, we get F 1 (b) = a f (t) dt = F(b) + C = F(b) − F(a).   R EMARK 3.2.– 1) It is common to denote b b F  = F(x) := F(b) − F(a). a

a

Therefore, the Newton–Leibnitz (NL) formula is often written as  b b f (x) dx = F(x) . a a

formula remains true for a > b if, in this case, we define

b 2) The NL a f := − b f . Indeed, we then have a 

b a



a

f := −

 f = − F(a) − F(b) = F(b) − F(a).

b

2 Also called the fundamental theorem of calculus.

Definite Integral

31

3) The formula also holds for f ∈ D[a, b] with any of its generalized primitives F. If f has a finite number of discontinuities, this can be checked very simply – it suffices to sum the NL formulas in the continuity intervals of f . In the general case (where f may have infinitely many discontinuities), the proof is somewhat more difficult (see problem 3.14). E XAMPLE 3.2.– The NL formula is very important in calculation of definite integrals. However, it is not always possible to find a primitive of a function. Moreover, many elementary functions have a non-elementary primitive. In such cases, integrals are often calculated by numerical methods. One of them is based on using the Taylor

1 formula. Suppose that we want to calculate the integral I := 0 sin x2 dx (primitives of the integrand function are not elementary). By the Taylor formula we have: sin t = t −

t3 t5 + + R6 (t) =: P(t) + R6 (t), 3! 5!

t ∈ [0, 1],

where   t7 1 R6 (t)  max (sin ξ)(7)  < 0,0002, 7! ξ∈[0,t] 7!

t ∈ [0, 1].

Therefore,   1   1  1     R (x2 ) dx 2 2 2   P(x ) dx  sin x − P(x ) dx = I − 6  0 0 0 1  0,0002 dx = 0,0002. 0

The integral I1 :=

1 0

P(x2 ) dx can easily be calculated:

1  x3 x6 x10  x7 x11  I1 = x − − dx = − −  6 120 3 6 · 7 120 · 11 0 0 1 1 1 = − + = 0, 3096 . . . . 3 42 13200  1

2

32

Integral and Measure

Thus, I = 0,3096 ± 0,0002. P ROPOSITION 3.6 (Change-of-variable formula).– Suppose that: f ∈ C[a, b], a function ϕ: [α, β] → [a, b] has a continuous derivative in the interval [α, β] (shortly, ϕ ∈ C 1 ([α, β], [a, b])), and ϕ(α) = a, ϕ(β) = b. Then  b  β  f (x) dx = f ϕ(t) ϕ (t) dt α

a

or, for short (see Remark 2.3) 

b

 f (x) dx =

β α

a

 f ϕ(t) dϕ(t).

R EMARK 3.3.– Recalling remark 3.2.2, we see that the formula also holds in the case a > b. P ROOF.– Applying the NL formula, we have 

b a

   β f = F(b) − F(a) = F ϕ(β) − F ϕ(α) = F ϕ(t)  α  β  β

   = F ϕ(t) t dt = F  ϕ(t) ϕ (t) dt α α β   = f ϕ(t) ϕ (t) dt. α

 

E XAMPLE 3.3.– Let us calculate  a I := a2 − x2 dx, a > 0. −a

We will use the change of variable x = ϕ(t) := a sin t, t ∈ [− π2 , π2 ]. By the change-of-variable formula,  I=

π 2

− π2



=a

2

 π 2

 a2 − a2 sin2 t d(a sin t)

a2 cos t dt = 2 − π2 2



π 2

− π2

=

π 2

− π2

a cos t a cos t dt

(1 + cos 2t) dt

Definite Integral

=

33

a2  sin 2t  π2 a2 π t+ = .  2 2 − π2 2

P ROPOSITION 3.7 (Integration-by-parts formula).– If f, g ∈ C 1 [a, b], then  b b  b  f (x)g (x) dx = f (x)g(x) − g(x) f  (x) dx a

a

a

or, for short, 

b a

b  b f dg = f g − g d f. a a

P ROOF.– Since f g is a primitive of the function ( f g) , by the NL formula we get b  b  f (x)g(x) = f (x)g(x)  dx a a b  b  = f (x)g (x) dx + g(x) f  (x) dx. a

a

 

E XAMPLE 3.4.– (Taylor’s formula with the integral remainder term) Let f ∈ C n+1 [x0 , x]. Applying the NL and integration-by-parts formulas, we get (in all integrals, the integration variable is t): 

= = = =





x

f (t) dt = f (x0 ) − f  (t) d(x − t) x x 0 0    x  x    f (x0 ) − f (t)(x − t) − (x − t) d f (t) x0  x0x f (x0 ) + f  (x0 )(x − x0 ) + f  (t)(x − t) dt x 0  1 x   f (x0 ) + f (x0 )(x − x0 ) − f (t) d (x − t)2 2 x0  f (x0 ) +  f (x0 )(x − x0 ) −  x  x 1 − f  (t)(x − t)2  − (x − t)2 f  (t) dt x0 2 x0  f (x0 ) + f (x0 )(x − x0 ) +

f (x) = f (x0 ) + =

x

34

Integral and Measure

1 1 + f  (x0 )(x − x0 )2 + 2 2



x

f  (t)(x − t)2 dt.

x0

Continuing, we finally get the Taylor formula with the integral remainder term: n  f (k) (x0 ) (x − x0 )k + Rn (x0 , x) k! k=0 x 1 Rn (x0 , x) = f (n+1) (t)(x − t)n dt. n! x0

f (x) =

with

P ROPOSITION 3.8 (Differentiation of functional sequences and series).– 1) Suppose that: a) fn ∈ C 1 [a, b], n ∈ N; b) ∃ limn→∞ fn (a) ∈ R; c) fn ⇒ g in [a, b]. Then there exists f := lim fn ∈ C 1 [a, b] (with uniform convergence), n→∞ and  f  (x) = lim fn (x) = g(x) , x ∈ [a, b]. n→∞

2) Suppose that: a) fn ∈ C 1 [a, b], n ∈ N;  b) the series ∞ n=1 fn (a) converges;   c) the series of functions ∞ n=1 fn converges uniformly in [a, b]. ∞ Then the series of functions n=1 fn also converges uniformly in [a, b], and    ∞ ∞ fn (x) = fn (x), x ∈ [a, b]. n=1

n=1

R EMARK 3.4.– 1) Instead of the interval endpoint a, we can take an arbitrary point of the interval [a, b].

Definite Integral

35

2) Here is a simple example showing that the uniform convergence of derivatives is an essential condition: 1n sin nx ⇒ 0, n → ∞, but ( 1n sin nx) = cos nx does not converge as n → ∞, except the point x = 0. P ROOF.– 1) We denote 

x

f (x) := lim fn (a) + n→∞

g(t) dt,

x ∈ [a, b].

a

Then f  (x) = g(x), x ∈ [a, b], and    fn (x) − f (x) = fn (x) − fn (a)− f (x) − f (a) + fn (a) − f (a) x x  NL === fn (t) dt − f  (t) dt + fn (a) − f (a) a a x     = fn (t) − f (t) dt + fn (a) − f (a) . a

Since fn (a) → f (a), n → ∞, it suffices to prove that that the integral with variable upper bound converges to 0 uniformly in [a, b]. Denote   εn := sup  fn (x) − f  (x) → 0, n → ∞. x∈[a,b]

Then   

x  x     sup x∈[a,b]  a fn (t) − f (t) dt  sup x∈[a,b] a  fn (t) − f  (t) dt  

x  sup x∈[a,b] a εn dt  εn (b − a) → 0, n → ∞. 2) It suffices to apply the first part of the proposition to the partialsum sequence of the functional series.   C OROLLARY 3.2 (Termwise differentiation and integration of power series).– A power series can be differentiated and integrated termwise  n in its convergence integral: if ∞ n=0 cn (x − a) is a power series with radius of convergence R, then    ∞ ∞ cn (x − a)n = ncn (x − a)n−1 , x ∈ (a − R, a + R), n=0

n=1

36

Integral and Measure

and  x  ∞



∞  cn cn (t − a) dt = (x − a)n+1 , n + 1 n=0 n=0

a

n

x ∈ (a − R, a + R).

P ROOF.– The radius of convergence of derivatives of a power series ∞ 

ncn (x − a)n−1

n=1

equals R =

1 1 = = R, √n √ lim supn n|cn | lim supn n |cn |

which coincides with the radius of convergence of the power series considered. Therefore, the series of derivatives converges uniformly in  a + R)  with 0 < R  < R. By proposition 3.8, every smaller interval (a − R, the power series can be termwise differentiated in every such interval and, thus, in the whole interval (a − R, a + R). Termwise integration is justified similarly by using property 5 of proposition 3.4.   E XAMPLE 3.5.–  xn 1) Consider the power series f (x) := ∞ n=1 n , x ∈ (−1, 1). In its convergence interval (−1, 1), it can be termwise differentiated: 

f (x) =

∞ 

xn−1 =

n=1

1 , 1− x

x ∈ (−1, 1).

Integrating the latter equality, we get 

x

f (x) = f (0) + 0

1 dt = − ln(1 − x), 1−t

x ∈ (−1, 1).

Definite Integral

37

Writing −x instead of x, we get a simple well-known formula, ln(1 + x) = − f (−x) =

∞  (−1)n+1 n=1

n

xn ,

x ∈ (−1, 1).

2) Integrating termwise the power series f (x) =

∞  (n + 1)xn ,

x ∈ (−1, 1),

n=0

we get 

x

f (t) dt =

0

  x   x ∞ ∞  (n + 1)tn dt = (n + 1) tn dt 0

=

∞ 

n=0

xn+1 =

n=0

n=0

x , 1− x

0

x ∈ (−1, 1).

Now, differentiating we get f (x) =

 x  1 = , 1− x (1 − x)2

x ∈ (−1, 1).

P ROPOSITION 3.9 (Mean-value theorem).– Suppose that f, g ∈ D[a, b], m  f (x)  M, x ∈ [a, b], and g  0. Then there exists μ ∈ [m, M] such that  b  b fg = μ g. a

a

If, moreover, f ∈ C[a, b], then there exists c ∈ [a, b] such that 

b a



b

f g = f (c)

g. a

In particular, when g(x) ≡ 1, there exists c ∈ [a, b] such that 

b a

f = f (c)(b − a).

38

Integral and Measure

P ROOF.– Integrating the inequality mg(x)  f (x)g(x)  Mg(x),

x ∈ [a, b],

we get 

b

m

 g

a

If then

b a

b

 fg  M

a

g = 0, then

b

μ := a b a

fg g

b

g. a

b a

f g = 0, and we can take μ ∈ [m, M]. If

b a

g > 0,

∈ [m, M].

If, moreover, f ∈ C[a, b], then by the Bolzano–Weierstrass theorem (on the intermediate values of a continuous function) the function f takes all values from the interval [m, M] := [ inf f (x), sup f (x)] x∈[a,b]

x∈[a,b]

and, in particular, the value μ.

 

P.3. Problems P ROBLEM 3.1.– Using the definition of the integral, find

π 0

sin x dx.

P ROBLEM 3.2.– Give an example of a sequence a) { fn } ⊂ S [a, b], b) { fn } ⊂ C[a, b] such that fn (x) → 0, x ∈ [a, b], but

b a

fn  0.

P ROBLEM 3.3.– Show the existence of the limit a = limn an and write it as an integral if a) an :=

1 1 1 + +··· + ; n+1 n+2 3n

Definite Integral

b) an := n

n 

1

k=1

n2 + k 2

39

;

1 k sin . n k=1 n n

c) an :=

P ROBLEM 3.4.– Find  π lim sinn x dx. n

0

P ROBLEM 3.5.– Find  1 lim f (xn ) dx, n

0

for f ∈ C[0, 1]. P ROBLEM 3.6.– Find the limits   n+1  x dx 4 lim n and n 1 + x5 n





2n

3

lim n n

n

 x dx . 1 + x5

P ROBLEM 3.7.– Show that if f ∈ C[0, ∞) and lim x→∞ f (x) = A ∈ R, then  1 x lim f (t) dt = A. x→∞ x 0

b P ROBLEM 3.8.– Show that if f ∈ D[a, b], f  0, and a f (x) dx = 0, then f ≡ 0. P ROBLEM 3.9.– Suppose that f ∈ C[a, b] is such that  b xn f (x) dx = 0 a

for all n ∈ N ∪ {0}. Show that f ≡ 0. P ROBLEM 3.10.– Suppose that f ∈ C[a, b] is such that  b xn f (x) dx = 0 a

40

Integral and Measure

for all n  2014. Prove that f ≡ 0. P ROBLEM 3.11.– A function f ∈ C[−1, 1] is such that 

1 −1

x2n f (x) dx = 0

for all n ∈ N ∪ {0}. Show that f is an odd function. P ROBLEM 3.12.– A function f ∈ C[a, b] is such that 

b

enx f (x) dx = 0

a

for all n ∈ N ∪ {0}. Show that f ≡ 0. P ROBLEM 3.13.– Find the generalized primitives of the functions f (x) := |x|, x ∈ R, and g(x) := sgn x, x ∈ R. P ROBLEM 3.14.– Prove that every regular function has a generalized primitive. Hint. Check the statement for the step functions and then, by passing to the limit, for the regular functions. P ROBLEM 3.15.– Prove that every f ∈ C 1 [a, b] can be expressed as the difference of two increasing functions in [a, b]. Express the function f (x) = sin x, x ∈ [−2π, 2π], by such a difference. P ROBLEM 3.16.– Let {xn , n ∈ N} ⊂ [0, 1]. Denote by νn (α, β) the number of indexes i  n for which xi ∈ (α, β). The sequence is said to be uniformly distributed in the interval [0, 1] if for all 0  α  β  1, lim n

νn (α, β) = β − α. n

Prove that, for any sequence {xn } that is uniformly distributed in [0, 1], 1 lim f (xi ) = n n i=1 n



1

f (x) dx. 0

Definite Integral

41

P ROBLEM 3.17.– For f ∈ C 1 [a, b], denote b−a   b − a r(n) := f a+k − n k=0 n n−1



b

f (x) dx. a

Prove that lim nr(n) = − n

b−a f (b) − f (a) . 2

P ROBLEM 3.18.– Let f ∈ C 1 [a, b] be such that f (a) = f (b) = 0. Prove that   b    (b − a)2   f (x) dx  max  f  (x) .   x∈[a,b] 4 a P ROBLEM 3.19.– Using the Taylor formula with the integral remainder term (example 3.4), derive the following expressions of the remainder term: 1) Lagrange form: f (n+1) (c) Rn (x0 , x) = (x − x0 )n+1 for some c ∈ (x0 , x). (n + 1)! 2) Cauchy form: f (n+1) (c) Rn (x0 , x) = (x − c)n (x − x0 ) for some c ∈ (x0 , x). n! 3) Peano form:  Rn (x0 , x) = o |x − x0 |n , x → x0 . P ROBLEM 3.20.– Find the sums of the series ∞  (−1)n 2n+1 x 2n + 1 n=0

and

∞ 

n(n + 1)xn .

n=1

P ROBLEM 3.21.– Let f ∈ D[a, b]. Denote 1 fh (x) = 2h



x+h

f (t) dt x−h

42

Integral and Measure

( f (t) := f (a), t  a; f (t) := f (b), t  b). Prove that fh (x) −→

1 f (x + 0) + f (x − 0) , 2

h ↓ 0, x ∈ [a, b].

In particular, if f ∈ C[a, b], then fh → f as h ↓ 0, . P ROBLEM 3.22.– Give an example showing that the mean-value theorem (theorem 3.9) does not hold if g is not positive (or negative) in the whole interval [a, b]. P ROBLEM 3.23.– Let f, g ∈ C[a, b], f be a monotone function, and g  0. Prove that there exists c ∈ [a, b] such that 

b



c

f g = f (a)

a



b

g + f (b)

a

g. c

P ROBLEM 3.24.– Let f, g ∈ C[a, b], f be a non-negative non-decreasing function, and g  0. Prove that there exists c ∈ [a, b] such that 

b a



c

f g = f (a)

g. a

x P ROBLEM 3.25.– Denote ln(x) := 1 1t dt, x > 0. Show that this function has the well-known properties of the logarithm: ln(xy) = ln(x) + ln(y) for x, y > 0; ln(xα ) = α ln(x) for x > 0 and α ∈ R.

4 Applications of the Integral

4.1. Area of a curvilinear trapezium Let us extend the notion of a curvilinear trapezium. By the latter, we now mean any set in the plane of the form A = {(x, y) : f1 (x)  y  f2 (x), x ∈ [a, b], where f1 , f2 ∈ D[a, b] are such that f1  f2 . We define its area as 

b

S (A) :=



f2 (x) − f1 (x) dx.

a

Such a definition can be motivated by the fact that this formula is obviously correct when f1 and f2 are step functions and, thus, A is a union of rectangles; see Figure 4.1. Therefore, it is natural to extend the formula to the class of functions that can be approximated by step functions. Let us emphasize that we did not derive or find the formula for the area of curvilinear trapezium but rather defined the area in terms of an integral. Such a way of introducing new notions is rather common in mathematics: if some property or notion (a formula, definition, integral, derivative, solution of an equation, etc.) is obvious or easily derived for a simple class of some objects (functions, sets, equations, etc.), then the same property or notion often can be and is taken as a definition

44

Integral and Measure

for a wider class of objects. Of course, usually, we need to justify such an extension by proving its existence, correctness, and other “good” properties.

Figure 4.1. Curvilinear trapezium bounded by step functions

Returning to the notion of area, note that if a plane set is not a curvilinear trapezium but can be expressed as a union of a finite  number of trapeziums, A = nj=1 K j , then its area is naturally defined as the sum S (A) :=

n 

S (K j ).

j=1

E XAMPLE 4.1.– Let us calculate the area of an ellipsis with semi-axes a and b. Put it on the coordinate plane so that its center coincides with the origin and the semi-axes would be on the coordinate axes. Then the ellipsis coincides with the set ! x2 y2 +  1 a2 b2 ! b 2 2 b 2 2 = (x, y) : − a −x y a − x , x ∈ [−a, a] . a a

AR = (x, y) :

So, its area is S (AR ) =

 a   b  b a2 − x2 − − a2 − x2 dx a −a a

Applications of the Integral

=



2b a

a

−a



3.3

a2 − x2 dx ===

45

2b πa2 · = πab. a 2

D EFINITION 4.1.– The section of a set (solid figure) A ⊂ R3 at a point x ∈ R is the set " # A x := (y, z) : (x, y, z) ∈ A ⊂ R2 . If A x = ∅ for x  [a, b], then the volume of A is defined as 

b

V(A) :=

S (A x ) dx a

(provided that S (A x ) is defined for all x ∈ [a, b] and the integral exists). We leave to the reader to justify this formula. P ROPOSITION 4.1 (Volume of a solid of rotation).– For a non-negative function f ∈ D[a, b], let " # A := (x, y, z) : y2 + z2  f 2 (x), x ∈ [a, b] ⊂ R3 (the solid figure obtained by rotating the curvilinear trapezium K = {(x, y) : 0  y  f (x), x ∈ [a, b]} in R3 about the x-axis). Then 

b

V(A) = π

f 2 (x) dx.

a

P ROOF .– " # A x = (y, z) : y2 + z2  f 2 (x) ,

x ∈ [a, b]

=⇒ S (A x ) = π f 2 (x), x ∈ [a, b]  b  b =⇒ V(A) = S (A x ) dx = π f 2 (x) dx. a

a

 

46

Integral and Measure

E XAMPLE 4.2.– 1) Let us calculate the volume of the ball Rr with radius r. Assuming that its center is at the origin, we have " # Rr = (x, y, z) : x2 + y2 + z2  r2 " # = (x, y, z) : y2 + z2  r2 − x2 , x ∈ [−r, r] , and, thus,  r  r 2  2 2 V(Rr ) = π r2 − x2 dx = π r − x dx −r −r   x3 r 4πr3  = = π r2 x − . 3 −r 3 2) Let A be a pyramid with height H and base area S . Suppose that

Figure 4.2. Calculation of the volume of a pyramid

it is positioned in the coordinate space so that its vertex is at the origin and the height is on the x-axis (Figure 4.2). Then S (A x ) x2 = 2 , x ∈ [0, H] S H S =⇒ S (A x ) = 2 x2 , x ∈ [0, H] H

Applications of the Integral

 =⇒ V =

H

S (A x ) dx =

0

47

S x3 H S H .  = 3 H2 3 0

D EFINITION 4.2.– Let functions f1 , f2 , . . . , fk : [a, b] → R. Define the function f : [a, b] → Rk as  f (x) := f1 (x), f2 (x), . . . , fk (x) , x ∈ [a, b]. Such a function is called a vectorial function of real variable (x); the functions fi are called the coordinate functions of f . We say that f is continuous (differentiable, regular, etc.) if all the functions fi , i = 1, 2, . . . , k, are continuous (differentiable, regular, etc.). Such functions are differentiated and integrated coordinate wise: 

b

 f  (x) := f1 (x), f2 (x), . . . , fk (x) , x ∈ [a, b],  b   b  b f (x) dx := f1 (x) dx, f2 (x) dx, . . . , fk (x) dx .

a

a

a

a

P ROPOSITION 4.2.– 1) If a function F: [a, b] → Rk is continuously differentiable in [a, b] (denoted F ∈ C 1 ([a, b], Rk )), then (the NL formula)  b F  (x) dx = F(b) − F(a). a

2) If a function f : [a, b] → Rk is regular in [a, b] (denoted f ∈ D([a, b], Rk )), then   b   b      f (x) dx f (x) dx   a a $ (where |y| := y21 + y22 + · · · + y2k for y = (y1 , y2 , . . . , yk ) ∈ Rk ). P ROOF.– 1) It suffices to apply the NL formula for each coordinate function of f . 2) Denote  b yk := fk (x) dx, a



b

y := (y1 , y2 , . . . , yk ) =

f (x) dx. a

48

Integral and Measure

Then |y| = 2

k 

y2i

 k  = yi

i=1

=

 b  k 

a

fi (x) dx a

i=1

  yi fi (x) dx 

b a

i=1

b

=

b

% & '

k 

y2i ·

% & ' k 

i=1

 b    f (x) dx, |y|  f (x) dx = |y|

a

fi2 (x) dx

i=1

a

where we used the Cauchy–Schwarz inequality ⎛ k ⎞2 k k   ⎜⎜⎜ ⎟ 2 ⎜⎜⎜ ai bi ⎟⎟⎟⎟⎟  a · b2i . i ⎝ ⎠ i=1

i=1

i=1

If |y|  0, we get the desired inequality by dividing the last inequality by |y|. If |y| = 0, the former is trivially obvious.   D EFINITION 4.3.– A continuous function γ: [a, b] → Rk (denoted as γ ∈ C([a, b], Rk )) is called a curve in the space Rk . It is called smooth if it is continuously differentiable, that is, γ ∈ C 1 ([a, b], Rk ).1 A curve γ ∈ C([a, b], Rk ) is called rectifiable if l(γ) = l(γ; a, b) := sup

 m

.   γ(xi ) − γ(xi−1 ) < +∞,

i=1

where the supremum is taken over all partitions a = x0 < x1 < · · · < xm = b, m ∈ N, of the interval [a, b]. The number l(γ) is called the length of a curve γ. R EMARK 4.1.– Geometrically, a curve is often understood not as a function γ itself, but as its image Γ := γ([a, b]). Then the function γ is called a parameterization of the curve Γ.

1 Often, it is additionally required that |γ (x)|  0, x ∈ [a, b].

Applications of the Integral

49

T HEOREM 4.1.– 1) (Formula of the length of a curve.) A smooth curve γ: [a, b] → Rk is rectifiable, and  b  b$    γ (x) dx = l(γ) = γ1 2 (x) + γ2 2 (x) + · · · + γk 2 (x) dx. a

a

2) If τ: [α, β] → [a, b] is a strictly increasing continuously differentiable function and γ˜ := γ ◦ τ: [α, β] → Rk (i.e., γ(t) ˜ := γ(τ(t))), t ∈ [α, β]), then l(γ) = l(˜γ), that is, the length of a curve does not depend on its parameterization. P ROOF .– 1) For any partition a = x0 < x1 < x2 < · · · < xm = b of [a, b], we have  m  m   xi     γ(x ) − γ(x ) =  γ (x) dx  i i−1  xi−1 i=1 i=1   b m x  i  γ (x) dx = γ (x) dx.  i=1

xi−1

a

Hence,  b γ (x) dx. l(γ)  a

Let us prove the opposite inequality. Take arbitrary ε > 0. Since γ is continuous and, thus, uniformly continuous in [a, b] (the Cantor theorem), there is δ > 0 such that |γ (x) − γ (y)| < ε if |x − y| < δ, x, y ∈ [a, b]. Take a partition a = x0 < x1 < x2 < · · · < xm = b of the interval [a, b] such that |xi − xi−1 | < δ, i = 1, 2, . . . , m. Then    γ (x) − γ (xi−1 ) < ε,

x ∈ [xi−1 , xi ],

and, thus,       γ (x)  γ (xi−1 ) + ε,

x ∈ [xi−1 , xi ].

50

Integral and Measure

From the last inequality we get  b m   γ (x) dx = a

xi

m      γ (x) dx 

xi−1

xi

   γ (xi−1 ) + ε dx

xi−1

i=1 i=1  m   xi m  xi    = γ (xi−1 ) dx + ε dx  x x i−1 i−1 i=1 i=1  m   xi          = γ (x) − γ (x) − γ (xi−1 ) dx + ε(b − a)  xi−1 xi

i=1

m       i=1

xi−1

  m   γ (x) dx + 

i=1

xi

   γ (x) − γ (xi−1 ) dx + ε(b − a)

xi−1

m  m   γ(x ) − γ(x ) +  ε(xi − xi−1 ) + ε(b − a) i i−1 i=1

 l(γ) + 2ε(b − a).

i=1

Since ε > 0 is arbitrary, we finally get  b γ (x) dx  l(γ). a

2) l(˜γ) = = = =



β

$

α

β

$

α β $ α b $ a

γ˜ 1 2 (t) + γ˜ 2 2 (t) + · · · + γ˜ k 2 (t) dt

   2

  γ1 τ(t) τ (t) + · · · + γk τ(t) τ (t) 2 dt   γ1 2 τ(t) + · · · + γk 2 τ(t) τ (t) dt

γ1 2 (τ) + · · · + γk 2 (τ) dτ.

 

R EMARK 4.2.– The formula of the length of a curve can be “derived” in another, easier way. First, we check that it is correct for polygonal lines; note that the derivatives of parameterizations of such curves are step functions. Then, believing (or proving) that smooth curves can be approximated by polygonal lines, we define the length of a curve by the same formula (section 4.1).

Applications of the Integral

51

E XAMPLE 4.3.– 1) The simplest parameterization of the graph of a function f (x), x ∈ [a, b], is γ(x) = (x, f (x)), x ∈ [a, b]. Therefore, its length is 

b

l=

$

1 + f  2 (x) dx.

a

2) Let us calculate the length of a circle with radius r. A possible, and the most natural, parameterization of the circle is γ(ϕ) := (r cos ϕ, r sin ϕ), ϕ ∈ [0, 2π]. Therefore, its length is  2π $  2π 2 2 l= (−r sin ϕ) + (r cos ϕ) dϕ = r dϕ = 2πr. 0

0

3) The length of a curve in polar coordinates: generalizing the previous example, we will derive a formula of the length of a curve Γ given by a polar equation r = r(ϕ), ϕ ∈ [ϕ1 , ϕ2 ]. Passing from polar to Cartesian coordinates, we get the following parameterization of the curve:   γ(ϕ) = x(ϕ), y(ϕ) = r(ϕ) cos ϕ, r(ϕ) sin ϕ ,

ϕ ∈ [ϕ1 , ϕ2 ].

Therefore, its length l(Γ) = l(γ) equals 

ϕ2

$

x 2 (ϕ) + y 2 (ϕ) dϕ ϕ1  ϕ2 $    = r (ϕ) cos ϕ − r(ϕ) sin ϕ 2 + r (ϕ) sin ϕ + r(ϕ) cos ϕ 2 dϕ ϕ1ϕ2 $ = r2 (ϕ) + r 2 (ϕ) dϕ. ϕ1

4.2. A general scheme for applying the integrals Suppose that we want to define (or “calculate”) the quantity Q = Q[a, b] that depends on the interval [a, b] and possesses the additivity property Q[a, b] = Q[a, c] + Q[c, b]

for a  c  b.

52

Integral and Measure

Also, suppose that, based on the properties of the quantity, we have a reason to believe that ΔQ(x) := Q[x, x + Δx] ≈ q(x) Δx for “small” Δx. Then we define 

b

Q[a, b] :=

q(x) dx.

[4.1]

a

Let us try to justify such an approach. Assume that (1) q ∈ C[a, b] and (2) the approximate equality means that Q[x, x + Δx] = q(x) Δx + o(Δx),

Δx → 0.

˜ ˜ + Δx) − Q(x) ˜ Denote Q(x) = Q[a, x]. Then Q[x, x + Δx] = Q(x by additivity, and, therefore,  Q[x, x + Δx] o(Δx)  ∃Q˜  (x) = lim = lim q(x) + = q(x) Δx→0 Δx Δx  Δx→0 b NL ˜ − Q(a) ˜ = Q[a, b]. =⇒ q(x) dx === Q(b) a

So, under clear additional assumptions, definition [4.1] becomes a rigorously proved equality. 4.3. Area of a surface of revolution Let, on the plane R2 , a curve with parameterization γ(t) = (x(t), y(t)), t ∈ [α, β], such that y = y(t)  0 be given. Rotating the curve around the x-axis, we get a surface. Let us try to derive a formula of its area. Intuitively, it is clear that surface areas have the additivity property. Consider the area S [t, t + Δt] of a “narrow” band of the surface corresponding to a “small” interval [t, t + Δt]. Such a band approximately is a cut cone with top and base radiuses r = y(t) and

Applications of the Integral

R = y(t + Δt), respectively, and slant height L = Therefore, for “small” Δt,  S [t, t + Δt] = π(r + R)L ≈ π y(t) + y(t + Δt)   ≈ 2πy(t)γ (t) Δt.



t+Δt t

53

|γ (s)| ds.

t+Δt 

γ (s) ds

t

Thus, we can define  S [α, β] := 2π

β α

 β $    y(t)γ (t) dt = 2π y(t) x 2 (t) + y 2 (t) dt. α

t Since the length of the part γ(s), s ∈ [α, t], of the curve γ is l(t) = |γ (s)| ds, we have dl(t) = l (t) dt = |γ (t)| dt. Therefore, the formula α obtained can be shortly written as follows:  S [α, β] = 2π

β α

y dl.

E XAMPLE 4.4.– Since the sphere of radius r can be obtained by rotating the semi-circle γ(ϕ) = (r cos ϕ, r sin ϕ), ϕ ∈ [0, π], around the x-axis, its area is  π  π $ S = 2π y dl = 2π r sin ϕ (−r sin ϕ)2 + (r cos ϕ)2 dϕ 0 0 π 2  = −2πr cos ϕ = 4πr2 . 0

4.4. Area of curvilinear sector Consider the curve on the plane defined by the polar equation r = r(ϕ)  0, ϕ ∈ [α, β]. The planar curve that is bounded by this curve and two rays ϕ = α and ϕ = β is called a curvilinear sector (Figure 4.3). Let us find its area denoted by S = S [α, β]. The area of the part of the sector that is between the rays at angles ϕ and ϕ + Δϕ to the x-axis when

54

Integral and Measure

Δϕ is “small” can be approximately considered as the circular sector with radius r(ϕ) and central angle Δϕ. Therefore, 1 S [ϕ, ϕ + Δϕ] ≈ r2 (ϕ) Δϕ 2

for “small” Δϕ,

and, thus, S = S [α, β] :=

1 2



β α

r2 (ϕ) dϕ.

Figure 4.3. A curvilinear sector

4.5. Applications in mechanics 1) Work of a variable force: suppose that a material point is moving along the x-axis from a point a to a point b under the action of a variable force. Suppose that, at a point x, this force equals F(x). When the point passes the “small” interval [x, x + Δx], the force may be assumed to be approximately equal to F(x) on the whole interval; thus, in this interval, the force does the work A[x, x+Δx] ≈ F(x) Δx. Therefore, the work done by the variable force on the whole interval [a, b] equals 

b

A = A[a, b] :=

F(x) dx. a

2) Static moment and mass center of a curve: suppose that, on the coordinate plane, we have a system of material points

Applications of the Integral

55

{(xi , yi ), i = 1, 2, . . . , n} with masses m1 , m2 , . . . , mn . Its static moment with respect to the x-axis is defined as M x :=

n 

mi y i .

i=1

Therefore, it is the sum of products of masses and distances of the points to the x-axis taking into account the signs of distances. This way, the statics moment of a mass system can be defined with respect to an arbitrary line. Now let us try to define the static moment of a “material” curve γ = (x, y): [a, b] → R2 . We will suppose that the mass of curve equals its length (that is, its density is 1). Since the static moment of system equals the sum of statics moments of all its points, the static moment possesses the additivity property. The static moment of a “small’ piece of the curve γ: [t, t + Δt] → R2 with respect to the x-axis is 

t+Δt 

γ (s) ds · y(t) ≈ γ (t) Δt · y(t) = y(t)γ (t) Δt.

M x [t, t + Δt] ≈ t

Therefore,  M x = M x [α, β] =

β α

 β    y(t)γ (t) dt = y dl. α

Similarly, the static moment of the curve with respect to the y-axis is equal to  My = My [α, β] =

β α

 x dl =

β α

  x(t)γ (t) dt.

56

Integral and Measure

A point (x0 , y0 ) is said to be the mass center of a material system if concentrating at this point the whole system mass M, the static moments with respect to any line remain the same. Applying this definition to the curve and coordinate axes, we get the equations My0 = M x , Mx0 = My . Therefore, My Mx , x0 = , M M

β where M = α |γ (t)| dt. y0 =

3) Static moment of a plane figure: by similar arguments, for static moments of the curvilinear trapezium " # A = (x, y) : f1 (x)  y  f2 (x), a  x  b , we can derive the following formulas, assuming that the mass of a plane figure equals its area:   1 b  2 2 1 b 2 Mx = y2 − y1 dx = f2 (x) − f12 (x) dx, 2 2  ba  b a  My = (y2 − y1 )x dx = f2 (x) − f1 (x) x dx. a

a

The mass center is defined by the same formulas as before with 

b

M=

( f2 (x) − f1 (x)) dx.

a

P.4. Problems P ROBLEM 4.1.– Find the areas of the figures bounded by the curves: 1) y = (x + 1)2 , y2 = x + 1; 2

2

2) y = 2x − x2 + e x , y = x2 − 4x + e x .

Applications of the Integral

57

P ROBLEM 4.2.– Find the areas of the figures bounded by curves given in polar coordinates: 1) r = 6 sin 3ϕ; 2) r = 6 sin ϕ, r = 4 sin ϕ; 3) r = sin ϕ + cos ϕ. P ROBLEM 4.3.– Find the lengths of the curves: 1) y = 1 − ln cos x, x ∈ [0, π/6]; √ 2) y = 1 − x2 + arcsin x, x ∈ [0, 7/9]; 3) y = ln(1 − x2 ), x ∈ [0, 1/4]. P ROBLEM 4.4.– Find the lengths of curves given in polar coordinates: 1) r = a(1 − cos ϕ); 2) r = a sin ϕ (a > 0). P ROBLEM 4.5.– Find the volumes of bodies bounded by the surfaces: 1) z = x2 + 4y2 , z = 2; 2) z = x2 + y2 /4 − z2 = 1, z = 0, z = 3; 3) z = x2 /16 + y2 /9 + z2 /4 = 1, z = 0, z = 1. P ROBLEM 4.6.– Find the volumes of bodies bounded by surfaces obtained by rotation of the following curves around the x-axis: 1) y = 2x − x2 , y = 2 − x; √ 2) y = x3 , y = x; 3) y = arcsin x, y = arccos x.

5 Other Definitions: Riemann and Stieltjes Integrals

5.1. Introduction Let us return to the “geometric” definition of the notion of integral (see Introduction 3.1 in Chapter 3). The so-called Riemann integral obtained this way is intuitively rather simple and is still widely preferred by practitioners, physicists, chemists, engineers, etc. Therefore, a mathematician should also know well the notion of the Riemann integral. However, its rigorous mathematical justification is more complicated. First, we emphasize that the Riemann integral is not some other integral. It is, in fact, the same integral, but it is defined in a slightly wider function class – the Riemann integral of a regular function coincides with its integral defined in Chapter 3 (theorem 5.1). How much precisely the class of Riemann-integrable functions is wider than the class D[a, b] will become clear later (theorem 8.3), when we will know the even more general Lebesgue integral. D EFINITION 5.1.– A partition of an interval [a, b] is a finite set P = {x0 , x1 , . . . , xn } such that a = x0 < x1 < x2 < · · · < xn = b. The intervals [xi−1 , xi ], i = 1, 2, . . . , n, are called the partition intervals of P, and the number |P| := maxi |xi − xi−1 | is called the mesh of P. A finite set of numbers ξ = (ξ1 , ξ2 , . . . , ξn ) is called a tag set of P if ξi ∈ [xi−1 , xi ],

60

Integral and Measure

i = 1, 2, . . . , n; the pair ((P, ξ) is called a tagged partition of the interval [a, b]. The Riemann integral sum of a function f : [a, b] → R corresponding to a tagged partition (P, ξ) is the sum S ( f, P, ξ) :=

n 

f (ξi )Δxi ,

Δxi := xi − xi−1 .

i=1

We write J = lim S ( f, P, ξ) |P|→0

if for every ε > 0, there exists δ > 0 such that for every tagged partition (P, ξ) of [a, b] with mesh |P| < ε, |S ( f, P, ξ) − J| < ε. In such a case, the function f is said to be Riemann-integrable in the interval [a, b], and the number J is called the Riemann integral of f in the interval [a, b]. We will temporarily denote it by  R

b

f (x) dx a

or  R

b

f. a

We denote by R[a, b] the class of all Riemann-integrable functions in the interval [a, b]. R EMARK 5.1.– 1) We can check that J = lim S ( f, P, ξ) ⇐⇒ J = lim S ( f, Pn , ξn ) |P|→0

n→∞

Other Definitions: Riemann and Stieltjes Integrals

61

for every sequence of tagged partitions {(Pn , ξn )} such that limn→∞ |Pn | = 0. The proof is similar to the proof of the equivalence of the definitions of the limit of a function in terms of ε − δ and in terms of sequences. 2) Let us try to compare this definition with the previous definition (see also section 3.1). We have  R

b

f = lim S ( f, Pn , ξn ) as |Pn | → 0 n→∞

a

(Pn = {xni , i = 0, 1, . . . , kn }). Notice that 

b

S ( f, P , ξ ) = n

n

ϕn ,

a

where ⎧ n n n ⎪ ⎪ ⎨ f (ξi ), x ∈ (xi−1 , xi ], ϕn (x) := ⎪ ⎪ ⎩ f (a), x = a. Therefore, the Riemann integral may also be interpreted as the integral obtained by passing to the limit from the integrals of step functions. The difference is in that the uniform convergence ϕn ⇒ f is replaced by the “convergence” of ϕn to f in another sense: we require the sequence of ϕn to coincide with f at an increasing number of points that are “uniformly distributed” in the interval [a, b]. P ROPOSITION 5.1 (Properties of the Riemann integral).– 1) Every f ∈ R[a, b] is a bounded function. 2) If f, g ∈ R[a, b] and α, β ∈ R, then α f + βg ∈ R[a, b], and  b  b  b R (α f + βg) = αR f + βR g. a a a  b  b 3) If f, g ∈ R[a, b] and f  g, then R f R g; in particular, a a   b   b     f . f  R R  a a

b

b 4) If R[a, b]  fn ⇒ f , then f ∈ R[a, b], and R a fn → R a f .

62

Integral and Measure

P ROOF.– 1) We prove by contradiction. Suppose that f is not bounded in the interval [a, b]. Let us take any partition P = {xi } of [a, b]. At least in one of the partition intervals, say [xi−1 , xi ], the function f is not bounded. Then, by choosing a tag ξi ∈ [xi−1 , xi ] we can make the summand f (ξi )Δxi of the integral sum S ( f, P, ξ), without changing the remaining summands, be arbitrarily large in absolute value. This means that, for any fixed partition P, the integral sum can be arbitrarily large in absolute value. Therefore, the integral sums cannot have a finite limit, a contradiction. 2) Let (P, ξ) be an arbitrary tagged partition of [a, b]. Then  S (α f + βg, P, ξ) = (α f + βg)(ξi )Δxi i

=α



f (ξi )Δxi + β

i



g(ξi )Δxi

i

= αS ( f, P, ξ) + βS (g, P, ξ). Taking the limit as |P| → 0, we get  b  ∃ R (α f + βg) = lim S (α f + βg, P, ξ) = αR a

|P|→0

b a

 b f + βR g. a

Let us justify this (we have not yet proved the linearity property of limits of Riemann integral sums, although we have just applied it). Take arbitrary ε > 0. By the definition of the Riemann integral, there exists δ > 0 such that if |P| < δ, then   b    ε f − S ( f, P, ξ) < R  |α| + |β| a and   b    ε g − S (g, P, ξ) < . R  |α| + |β| a

Other Definitions: Riemann and Stieltjes Integrals

63

b

b Then, denoting J = αR a f + βR a g, we get:    J − S (α f + βg, P, ξ)    b    b    = α R f − S ( f, P, ξ) + β R g − S (g, P, ξ)    a a   b    b       |α|R f − S ( f, P, ξ) + |β|R g − S (g, P, ξ) < ε,  a   a  which means that J = lim S (α f + βg, P, ξ), |P|→0

as claimed. 3) Taking the limit as |P| → 0 in the inequality   S ( f, P, ξ) = f (ξi )Δxi  g(ξi )Δxi = S (g, P, ξ), i

i

we get:  R

b a

 b f R g. a

Now try yourself to justify (as in the previous item) such a passing to the limit in the inequality. The inequality for the modulus of an integral is proved in the same way as for step functions (see proposition 3.2.4). 4) For an arbitrary ε > 0, there exists N ∈ N such that    fn (x) − f (x) <

ε , 2(b − a)

x ∈ [a, b], n > N.

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Integral and Measure

Then        fn (x) − fm (x)   fn (x) − f (x) +  f (x) − fm (x) <

ε ε ε + = , 2(b − a) 2(b − a) b − a

x ∈ [a, b], n, m > N.

Integrating this, we get:   b   b    b  b      f − f  R f − R f = R ( f − f )  R   n m n m  n m   a  a a a  b  b ε ε R dx = R 1 dx = ε, b−a a a b−a

n, m > N.

b  (R a 1 dx = b−a since S (1, P, ξ) = i 1 (xi − xi−1 ) = b−a). This means

b that the sequence of integrals {R a fn } is a Cauchy sequence and, thus,

b there exists a finite limit J := limn→∞ R a fn .

b It remains to check if J = R a f . Let (P, ξ) be an arbitrary tagged partition of [a, b]. Then   b      J − S ( f, P, ξ)   J − R  a

   b       fn  + R fn − S ( fn , P, ξ) + S ( fn − f, P, ξ).   a 

Let us again take arbitrary ε > 0. There exists N ∈ N such that   b  J − R  a

  ε fn  < ,  3

n > N,

and    fn (x) − f (x) <

ε , 3(b − a)

x ∈ [a, b], n > N.

Other Definitions: Riemann and Stieltjes Integrals

65

From the last inequality, we get that            S ( fn − f, P, ξ) =   fn (ξi ) − f (ξi )Δxi fn (ξi ) − f (ξi ) Δxi     i

<

ε 3(b − a)



i

Δxi =

i

ε 3

for n > N.

Therefore,     < 2ε   b    J − S ( f, P, ξ) + R fn − S ( fn , P, ξ) for n > N.  a  3 Let us fix any n0 > N. Then, there exists δ > 0 such that   b    ε fn0 − S ( fn0 , P, ξ) < R  3 a

if |P| < δ.

From this we get   2ε ε  J − S ( f, P, ξ) < + = ε if |P| < δ. 3 3 This means that f ∈ R[a, b]

and

 J=R

b

f. a

 

T HEOREM 5.1.– D[a, b] ⊂ R[a, b], and 

b a

 f =R

b

f a

for all f ∈ D[a, b]. R EMARK 5.2.– After the theorem is proved, the notation R before the Riemann integral becomes unnecessary.

66

Integral and Measure

P ROOF.– Step 1. First, consider simple functions f of the form (called the indicator of an interval I):  f (x) = 1II (x) :=

1, x ∈ I, 0, x  I.

Let I = [c, d], c < d (the closedness of the interval is not essential). Take arbitrary ε > 0 and let (P, ξ) be a tagged partition of [a, b] with mesh ε |P| < δ := min{d − c, }. 4 Denote k := min{i : xi  c},

l := min{i : xi  d}.

Then k < l, Δxk < δ, and Δxl < δ, and therefore,    b        f =  f (ξi )Δxi − (d − c) S ( f, P, ξ) −    a i            f (ξi )Δxi − (d − c) +  f (ξk )Δxk +  f (ξl )Δxl    k
b

f. a

Step 2. Now let f ∈ S [a, b]. Then f can be represented as a linear composition of the indicators of intervals. Indeed, if the function takes  the value yk in the constancy interval Ik ( k Ik = [a, b]), then  f (x) = yk 1IIk (x), x ∈ [a, b]. k

Other Definitions: Riemann and Stieltjes Integrals

67

Therefore, the coincidence of the integrals of f follows from their linearity and from their coincidence for the indicators of intervals (step 1):  R

b a

   b  b b  f =Ra y 1 I k k Ik = k yk R a 1IIk = k yk |Ik | = a f.

Step 1. Suppose that f ∈ D[a, b] and S [a, b]  ϕn ⇒ f . Then, by step 2,  b  b R ϕn = ϕn , a

n ∈ N.

a

Taking the limit as n → ∞, we get that  ∃R

b a

 b  b  f = lim R ϕn = lim ϕn = n→∞

n→∞

a

a

a

b

f.  

D EFINITION 5.2.– The (total) variation of a function f : [a, b] → R in the interval [a, b] is defined as   f (x ) − f (x ), V( f ) = V( f ; a, b) := sup i+1 i P

i

where the supremum is taken over all partitions P = {a = x0 < x1 < · · · < xm = b}, m ∈ N, of the interval [a, b]. If V( f ) < +∞, then f is said to be a function of finite variation (or of bounded variation). We denote by V[a, b] the set of all such functions. R EMARK 5.3.– Comparing definitions 5.2 and 4.3, we can see that, formally, a continuous finite-variation function f : [a, b] → R is a rectifiable curve in R (k = 1!), and its variation is equal to its length as of a curve. P ROPOSITION 5.2.– 1) If a function f : [a, b] → R has a bounded derivative f in the interval [a, b], except, possibly, a finite number of points of its

68

Integral and Measure

discontinuities of first kind, then f is a finite-variation function. In particular, every f ∈ C 1 [a, b] is of finite variation, which equals  b  f  (x) dx. V( f ) = [5.1] a

2) If a function f : [a, b] → R can be expressed as the difference of two increasing functions, then it is a finite-variation function. R EMARK 5.4.– The converse of the second statement is also true: every finite-variation function on the interval [a, b] can be represented as the difference of two increasing functions on the interval [a, b] (see problem 5.12). P ROOF.– 1) Denote M := sup xD | f  (x)| < +∞, where D is the set of discontinuities of f . If f ∈ C 1 [a, b] (D = ∅), then the statement, together with formula [5.1], is already proved! This is exactly theorem 4.1 in the case k = 1. In the general case, let us consider a partition P = {a = x0 < x1 < · · · < xm = b} in the interval [a, b]. Without loss of generality, we may, and do, assume that that the partition P is complemented by the discontinuity points of f (such a complement may only increase the sum in definition 5.2). For any point xi of the partition P, taking points xi < y1 < y2 < xi+1 , we have:          f (xi+1 ) − f (xi )   f (xi+1 ) − f (y2 ) +  f (y2 ) − f (y1 ) +  f (y1 ) − f (xi )       f (xi+1 ) − f (y2 ) + M(y2 − y1 ) +  f (y1 ) − f (xi ). Taking the limit as y1 ↓ xi and y2 ↑ xi+1 , we have:        f (xi+1 ) − f (xi )  M(xi+1 − xi ) +  f (xi+1 ) − f (xi+1 − 0) +  f (xi + 0) − f (xi ). Finally, by summation we obtain    f (x) − f (x − 0) V( f ; a, b)  M(b − a) + x∈T

  + f (x + 0) − f (x) < +∞.

Other Definitions: Riemann and Stieltjes Integrals

69

2) If f = f1 − f2 is the difference between two increasing functions, then for any partition P = {a = x0 < x1 < · · · < xm = b}, we have        f (xi+1 ) − f (xi )   f1 (xi+1 ) − f1 (xi ) +  f2 (xi+1 ) − f2 (xi )   = f1 (xi+1 ) − f1 (xi ) + f2 (xi+1 ) − f2 (xi ) . Therefore, by summation we get:   f (x ) − f (x ) V( f ; a, b) = sup i+1 i P i    f1 (b) − f1 (a) + f2 (b) − f2 (a) < +∞.

 

D EFINITION 5.3.– The Stieltjes integral of a function f ∈ S [a, b] with respect to a function g ∈ V[a, b] is the number 

b



b

f dg =

f (x) dg(x) :=

a

a

n−1   fk g(xk+1 ) − g(xk ) , k=0

where fk is the value of f in its constancy interval Ik with endpoints xk  xk+1 .1 The Stieltjes integral of a function f ∈ D[a, b] with respect to a function g ∈ V[a, b] is the number 

b a





b

f dg =

b

f (x) dg(x) := lim a

n→∞

ϕn (x) dg(x),

a

where (as before) {ϕn } is an arbitrary sequence of step functions converging to f uniformly in [a, b]. P ROPOSITION 5.3.– 1) (Existence of the Stieltjes integral and correctness of the definition.) The limit in the definition always exists and does not depend on the choice of the sequence {ϕn }; 1 That is, Ik = (xk , xk+1 ), or [xk , xk+1 ], or [xk , xk+1 ), or (xk , xk+1 ].

70

Integral and Measure

2) if f ∈ D[a, b] and g ∈ V[a, b], then   b    f dg  || f || V(g),   a

where || f || := sup x∈[a,b] | f (x)|; 3) if f ∈ C[a, b] and g ∈ C 1 [a, b], then  b  b f (x) dg(x) = f (x)g (x) dx. a

a

R EMARK 5.5.– From the proposition we see that the before used

b

b formal notation a f (x) dg(x) := a f (x)g (x) dx (see remark 2.3 and propositions 3.6 and 3.7) now has a precise meaning. P ROOF.– 1) If f ∈ S [a, b], then from the definition we see that, for every partition P = {a = x0 < x1 < · · · < xn = b} of [a, b] by the constancy intervals of a function f , we have   b       f dg  || f ||g(xi+1 ) − g(xi )  || f || V(g).   a i

Let S [a, b]  fn ⇒ f ∈ D[a, b]. Then   b    b  b       f dg − f dg = ( f − f ) dg    n m n m  a a a  || fn − fm || V(g) → 0, We see that { 

b a

b a

n, m → ∞.

fn dg} is a Cauchy sequence and, thus, there exists 

b

f dg = lim

n→∞

fn dg. a

The correctness of the definition can be proved as in proposition 3.3. 2) Suppose that S [a, b]  fn ⇒ f ∈ D[a, b]. Then   || fn || − || f ||  || fn − f || → 0.

Other Definitions: Riemann and Stieltjes Integrals

71

Therefore, the statement follows by passing to the limit in the inequality    

b a

 fn dg  || fn || V(g),

which holds, as we just checked, for step functions fn . 3) Denote ⎧  

⎪ n = a+k b−a , a+(k + 1) b−a , ⎪ ⎨ f (xnk ), x ∈ xnk , xk+1 n n fn (x) = ⎪ ⎪ ⎩ f (b), x = b.

[5.2]

Then (see proposition 1.1 and remark 1.2) S [a, b]  fn ⇒ f , and therefore, 

b

 fn dg →

a

b

f dg. a

However,   b     b   xnk+1       n  n n  fn dg − fg  =  f (xk ) g(xk+1 ) − g(xk ) − f (t)g (t) dt     n a a xk k  k  xn   n   xk+1 k+1   =  f (xnk )g (t) dt − f (t)g (t) dt   n n xk xk k k  n  xk+1   f (xn ) − f (t) g (t) dt  k =

k

xnk xnk+1 

k

xnk



 || fn − f ||

 f (t) − f (t) g (t) dt n



xnk+1  xn

g (t) dt

k b  k  g (t) dt → 0, = || fn − f || a

n → ∞.

 

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Integral and Measure

T HEOREM 5.2.– 1) Change-of-variable and Newton–Leibnitz formulas: if f ∈ C[a, b], ϕ ∈ C[α, β] ∩ V[α, β], and R(ϕ) ⊂ [a, b], then  β  ϕ(β)    f ϕ(t) dϕ(t) = f (x) dx = F ϕ(β) − F ϕ(α) , α

ϕ(α)

where F is any primitive function of f in [a, b]. 2) Integration-by-parts formula: if f, g ∈ C[a, b] ∩ V[a, b], then  b b  b f dg = f g − g d f. a

a

a

P ROOF.– 1) Denote tkn := α + k β−α n , k = 0, 1, . . . , n. By the Lagrange theorem, we get:    n  F ϕ(β) − F ϕ(α) = F ϕ(tk+1 ) − F ϕ(tkn ) 

n−1



k=0

=

n−1 

  n F  ξkn ϕ(tk+1 ) − ϕ(tkn )

k=0

n−1    n = f ξkn ϕ(tk+1 ) − ϕ(tkn ) k=0 n ). By the with points ξkn from the intervals with endpoints ϕ(tkn ) ir ϕ(tk+1 intermediate value theorem, we can write

    n  n F ϕ(β) − F ϕ(α) = f ϕ(sk ) ϕ(tk+1 ) − ϕ(tkn ) n−1 k=0 n ]. Denoting ψ(t) := f (ϕ(t)) and with snk ∈ [tkn , tk+1

 ψn (t) := f ϕ(snk )

n  for t ∈ tkn , tk+1 , ψn (β) := f ϕ(β) ,

Other Definitions: Riemann and Stieltjes Integrals

73

we have that ψn ⇒ ψ = f ◦ ϕ in the interval [α, β] (see again proposition 1.1 and remark 1.2). Therefore,   F ϕ(β) − F ϕ(α) =



β α

 ψn (t) dϕ(t) →

β α

 f ϕ(t) dϕ(t),

and thus,   F ϕ(β) − F ϕ(α) =



β α

 f ϕ(t) dϕ(t).

It remains to notice that the second equality,   F ϕ(β) − F ϕ(α) =



ϕ(β) ϕ(α)

f (x) dx,

is the usual Newton–Leibnitz formula in the interval [ϕ(α), ϕ(β)] ⊂ [a, b] (or [ϕ(β), ϕ(α)] ⊂ [a, b]). 2) Denoting xnk := a + b−a n k, we can write − f (a)g(a) =

f (b)g(b) =

n 

n 

n f (xnk )g(xnk ) − f (xnk−1 )g(xk−1

k=1

f (xnk ) −

 f (xnk−1 ) g(xnk ) +

k=1

k=1

Let 

n ), f (xkn ) for x ∈ [xnk , xk+1 f (b) for x = b,  n n ], g(xk+1 ) for x ∈ (xnk , xk+1 gn (x) := g(a) for x = a.

fn (x) :=

Then the last equality can be rewritten as  b  b  f g = gn d f + a a

n 

b

fn dg. a



 n f (xk−1 ) g(xnk ) − g(xnk−1 ) .

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Integral and Measure

Since fn ⇒ f and gn ⇒ g (proposition 1.1, remark 1.2), we have 

b

 fn dg →

a



b

f dg

b

and

a



b

gn d f →

a

g d f. a

Therefore, it remains to pass to the limit as n → ∞ in the previous equality.   R EMARK 5.6.– 1) Although the Stieltjes integral is also defined for discontinuous finite-variation functions, one can notice that the main statements are proved for continuous functions. A coherent theory of Stieltjes

b integrals a f dg in the general case is derivable when restricted to the left-continuous functions f ∈ D[a, b] and right-continuous functions g ∈ V[a, b]. As an example, we give an analog of the Newton–Leibnitz formula stated in theorem 5.2 for ϕ ∈ D[α, β] ∩ V[α, β]:   F ϕ(β) − F ϕ(α)  β       = f ϕ− (t) dϕ(t) + f ϕ(t) − f ϕ− (t) − f  ϕ− (t) Δϕ(t) , α

α
where ϕ− (t) := ϕ(t − 0) = lim s↑t ϕ(s), t ∈ (α, β], is a left-continuous function, and Δϕ(t) := ϕ(t) − ϕ− (t) is the jump size of the function ϕ at point t. Note that the function of t equals zero only in a finite or countable set (see theorem 1.1, part 2, and remark 1.3), and therefore the sum makes sense as the sum of (absolutely convergent) series. 2) A Stieltjes-type integral can be defined similarly to the Riemann integral as the limit of the integral sums S g ( f, P, ξ) :=

n 

 f (ξi ) g(xi ) − g(xi−1 )

i=1

corresponding to tagged partitions (P, ξ) as |P| → 0. The so-defined integral is called the Riemann–Stieltjes integral. However, as we will see in Part III, it is most convenient to consider such integrals in terms of the general measure theory.

Other Definitions: Riemann and Stieltjes Integrals

75

P.5. Problems P ROBLEM 5.1.– Let  1 if x is rational, f (x) = 0 otherwise. Prove that the function f is not Riemann-integrable in the interval [0, 1]. P ROBLEM 5.2.– Let ⎧ ⎪ 1 ⎪ ⎪ ⎨ sin , x ∈ (0, 1], f (x) := ⎪ x ⎪ ⎪ ⎩ 0, x = 0. Show that f ∈ R[a, b] but f  D[a, b]. P ROBLEM 5.3.– Let ⎧ ⎪ 1 ⎪ ⎪ ⎨ xα sin β , x ∈ (0, 1], f (x) := ⎪ x ⎪ ⎪ ⎩ 0, x = 0. Prove that f is a finite-variation function in the interval [0, 1] if and only if α > β. P ROBLEM 5.4.– Find the limit  xi−1  lim f (ξi ) g(x) dx |P|→0

i

xi

for f, g ∈ C[a, b], where ξ is a tag set of a partition P of the interval [a, b] (see definition 5.1). P ROBLEM 5.5.– Let f ∈ C[a, b] and P = {a = x0 < x1 < · · · < xk = b}. Prove that there exists a tag set ξ of P such that  b f (x) dx = S ( f ; P, ξ). a

P ROBLEM 5.6.– Give an example of a bounded convergent sequence { fn } ⊂ R[a, b] with limit f = limn→∞ fn that is not Riemann integrable.

76

Integral and Measure

P ROBLEM 5.7.– Find the total variations V( f ; −1, 1) and V(g; −1, 1) of the functions f (x) := |x| + |x + 1|, x ∈ [−1, 1], and g(x) = sgn x, x ∈ [−1, 1]. Find the integrals  1  1 (1 + x2 ) d f (x) and (1 + x2 ) dg(x). −1

−1

P ROBLEM 5.8.– Show that f ∈ V[a, b] is an increasing function if and only if V( f ; a, b) = f (b) − f (a). P ROBLEM 5.9.– Show the following properties of the variation of a function: V( f ± g; a, b)  V( f ; a, b) + V(g; a, b); V(α f ; a, b) = |α|V( f ; a, b);  V | f |; a, b  V( f ; a, b). P ROBLEM 5.10.– Prove that if f ∈ C[a, b], then V(| f |; a, b) = V( f ; a, b). P ROBLEM 5.11.– Let f, g ∈ V[a, b]. Show that f g ∈ V[a, b] and max{ f, g} ∈ V[a, b] and that if, moreover, g(x)  C > 0, x ∈ [a, b], then f g ∈ V[a, b]. Give an example which shows that from the condition g(x)  0, x ∈ [a, b], it does not follow that

f g

∈ V[a, b].

P ROBLEM 5.12.– Let f ∈ V[a, b]. Denote V(x) := V( f ; a, x) for x ∈ [a, b]. Prove that V and V − f are increasing (not strictly) functions. Therefore, every function f ∈ V[a, b] can be written as the sum of two increasing functions. P ROBLEM 5.13.– Express the function f (x) := sin x, x ∈ [0, 2π], as the sum of two increasing functions. P ROBLEM 5.14.– Show that V[a, b] ⊂ D[a, b]. P ROBLEM 5.15.– Check the linearity of the Stieltjes integral:  a

b



α1 f1 (x) + α2 f2 (x) dg(x) = α1



b a



b

f1 (x) dg(x) + α2

f2 (x) dg(x), a

provided that the integrals on the right-hand side of the equality exist.

Other Definitions: Riemann and Stieltjes Integrals

77

P ROBLEM 5.16.– Give an example where 

b

 f (x) dg(x) 

a



c

b

f (x) dg(x) +

a

f (x) dg(x) c

(i.e. the Stieltjes integral not always possesses the additivity property). Show that in such a case, both functions f and g have a discontinuity at the point c.

1 P ROBLEM 5.17.– Find 0 x d([3x] − x), where [x] denotes the integer part of a number x. P ROBLEM 5.18.– Let {gn } be an increasing sequence of increasing functions converging pointwise to a function g ∈ C[a, b]. Show that a) gn ⇒ g, but not necessarily V(gn − g; a, b) → 0;  b  b b) f (x) dgn (x) → f (x) dg(x) for every f ∈ C[a, b]. a

a

P ROBLEM 5.19.– An increasing function g is such that 

π

sin x dg(x) = g(π) − g(0).

0

Prove that g(x) = g(0) for x ∈ [0, π/2) and g(x) = g(π) for x ∈ (π/2, π]. P ROBLEM 5.20.– An increasing function g is such that 

2

f (x) dg(x) = f (1)

0

for every f ∈ C[0, 2]. Prove that g(x) = g(0) for x ∈ [0, 1) and g(x) = g(2) for x ∈ (1, 2]. P ROBLEM 5.21.– An increasing function g: [0, 1] → R is such that 

1 0

1  k f n k=1 n n

f (x) dg(x) =

for every f ∈ C[0, 1]. Find all such functions g.

78

Integral and Measure

P ROBLEM 5.22.– Find  1 lim xn dg(x) and n

0



1

lim n

(xn + e−nx ) dg(x),

0

provided that g is an increasing function on [0, 1].

6 Improper Integrals

D EFINITION 6.1.– Let f ∈ D[a,

c b) := D([a, b)), −∞ < a < b  +∞. If there exists the limit limc↑b a f (finite or infinite), it is called the improper integral of the function f in the interval [a, b) and is denoted

b

b

b−

b−0 f or f (x) dx (sometimes, f or f in order to emphasize a a a a b that the integral is improper). If a f ∈ R (i.e.,  ±∞), then one says

b

b that a f converges. Otherwise, one says that a f diverges. The improper integral of a function f ∈ D(a, b] (−∞  a < b < +∞) is defined analogously: 



b

b

f (x) dx := lim

f (x) dx.

c↓a

a

c

If f ∈ D(a, b) (−∞  a < b  +∞), then we define 



b

c

f := a

a



b

f+

f,

c ∈ (a, b).

c

(Check that the latter sum does not depend on the choice of c.) R EMARK 6.1.– 1) In this definition, instead of a function f ∈ D[a, b), we can take any function f : [a, b) → R that is integrable in the Riemann sense (or

80

Integral and Measure

any other sense) in every interval [a, c], a < c < b. In such a case, we speak of an improper Riemann integral.

c 2) If f ∈ D[a, b) and f  0, then the limit limc↑b a f always exists (finite or +∞). Therefore, for every such function f , we have: 

b

 f converges ⇐⇒

a

b

f < +∞.

a

3) Further, we formulate statements for functions f ∈ D[a, b) (all other cases are considered similarly). E XAMPLE 6.1.–

b dx 1) Let us consider the improper integral a (b−x) p (b < +∞). For a < c < b, we have  ⎧ ⎪ (b−x)−p+1 c  c  c ⎪ ⎪ − ⎪ dx −p+1 a , p  1, ⎨ −p c = (b − x) dx = ⎪ ⎪ p ⎪ ⎪ a (b − x) a ⎩ − ln(b − x) , p = 1, a  ⎧ 1  1 1 ⎪ ⎪ − , p  1, ⎨ p−1 (b−c) p−1 (b−a) p−1 =⎪ ⎪ ⎩ ln(b − a) − ln(b − c), p = 1. Therefore, 

b



c

f = lim c↑b

a

a

⎧ 1−p ⎪ ⎪ ⎨ (b−a) , p < 1, 1−p f =⎪ ⎪ ⎩ +∞, p  1.

Thus, this improper integral converges for p < 1 and diverges for p  1. Similarly, 

b

dx < +∞ ⇐⇒ p < 1. p a (x − a) ⎧ 1−p c ⎧ 1  +∞ ⎪ x ⎪ ⎪ ⎪ dx ⎨ 1−p 1 , p  1, ⎪ ⎨ p−1 , p > 1, 2) = lim ⎪ =⎪  ⎪ p c  ⎩ +∞, p  1. c→+∞ ⎪ x ⎩ ln x , p = 1. ⎪ 1 1

Improper Integrals

81

Therefore,  +∞ dx < +∞ ⇐⇒ p > 1. xp 1  +∞  c −x 3) e dx = lim e−x dx = lim (1 − e−c ) = 1. c→+∞

0

c→+∞

0

4) 

+∞ −∞

 +∞ dx dx + 2 1 + x2 −∞ 1+ x 0 0 c dx dx = lim + lim 2 c→−∞ c 1 + x c→+∞ 0 1 + x2 = lim (arctg 0 − arctg c) + lim (arctg c − arctg 0) = π.

dx = 1 + x2



0

c→−∞

c→+∞

P ROPOSITION 6.1 (Comparison of improper integrals).– Let f, g ∈ D[a, b). Then:  b  b 1) if 0  f  g, then f g; 1 a

a

f (x) 2) if f  0, g  0, and there exists μ := lim g(x) ∈ [0, +∞), then x↑b



b



b

g < +∞ =⇒

a

f < +∞.

a

In particular, if μ > 0, then  b  b g < +∞ ⇐⇒ f < +∞. a

a

P ROOF.– 1) It suffices to pass to the limit, as c ↑ b, in the inequality  c  c f g. a

a

1 An important obvious corollary: if 0  f  g and

b

b and if a f = +∞, then also a g = +∞.

b a

g < +∞, then also

b a

f < +∞;

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Integral and Measure

f (x) 2) Take c˜ < b such that 0  g(x) < μ + 1 for x ∈ (˜c, b), that is, 0  f (x)  (μ + 1)g(x) for x ∈ (˜c, b). Then, for c ∈ (˜c, b), we have:





c

c˜

f =





c˜

c

f f+ (μ + 1)g c˜ c˜  a c˜  c˜a c = f − (μ + 1) g + (μ + 1) g a a a c˜ c˜ b  f − (μ + 1) g + (μ + 1) g =: A < +∞.

a

f+



c

a

a

a

Therefore, 



b

c

f = lim c↑b

a

f  A < +∞.

 

a

E XAMPLE 6.2.– 1) 

+∞

e−x dx = 2

0



 +∞ 2 2 e−x dx + e−x dx 0  1+∞ 1  1 dx + e−x dx = 1 + e−1 < +∞. 1

0

1

2) Lets us consider the integral 

+∞

J :=

β

xα e−x dx,

α ∈ R, β > 0.

1

Let us check that it converges for all indicated values of the β parameters α and β. Taking f (x) := xα e−x , g(x) := x12 , x  1, we have: μ = lim

x→+∞

Since J < +∞.

f (x) β = lim xα+2 e−x = 0. x→+∞ g(x)

+∞ dx 1

x2

< +∞ (example 6.1), by proposition 6.1 we get that

Improper Integrals

83

b

f of a function f ∈ D[a, b)

b (−∞ < a < b  +∞) is said to converge absolutely if a | f | < +∞. If an

b

b improper integral a f converges while a | f | = +∞, then it is said to converge relatively.

b P ROPOSITION 6.2.– If f ∈ D[a, b) and a | f | < +∞, then the improper

b integral a f converges. D EFINITION 6.2.– The improper integral

a

P ROOF.– Denote f + := max{ f, 0} and f − := max{− f, 0}. Then f ± ∈ D[a, b), and we directly check that f = f + − f − and | f | = f + + f − .

b

b Since 0  f ±  | f | and a | f | < +∞, we have that 0  a f ± < +∞. Therefore, 

b

∃ a



c



c

( f + − f −) a c a c  b  + = lim f − lim f− = f+−

f = lim c↑b

c↑b

f = lim c↑b

a

c↑b

a

a

b a

f − ∈ R.

 

R EMARK 6.2.– For the improper integrals of functions f ∈ D[a, b), the Newton–Leibnitz, change-of-variable, and integration-by-parts formulas hold. This can be easily checked by passing to the limit, as c ↑ b, in the corresponding formulas in the interval [a, c]. However, we have additionally assumed that the limit exists at least in one side of the formula considered. 1) Let F  (x) = f (x), x ∈ [a, b). Then  c f (x) dx = F(c) − F(a), a  c < b.

[6.1]

a

If the limit F(b) := limc↑b F(c) exists, then passing to the limit in equation [6.1], we get: 

b a

b f (x) dx = F(b) − F(a) = F  . a

Formally, this formula does not differ from the previous Newton–Leibnitz formula. However, we have to realize that here F is

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Integral and Measure

not a primitive of f in the whole interval [a, b], but rather in the interval [a, b). The following cases are possible: 1) f (b) is not and cannot be defined because the limit limc↑b f (c) does not exist; 2) the derivative F  (b) does not exist or is not equal to f (b); and 3) b = +∞). To be more precise, we should write  a

b−

b− f (x) dx = F(b−) − F(a) = F  . a

However, if in calculations, we “forget” these nuances, the answers, as a rule, will be correct. . . . 2) Let us write the integration-by-parts formula in the interval [a, c] for functions f, g ∈ C 1 [a, b):  c  c

c  f (x)g (x) dx = f (x)g(x)  − g(x) f  (x) dx, a  c < b. a

a

a

If passing to the limit as c ↑ b, there exist limits of two terms in the formula (one of them, possibly, infinite), then there exists the limit of the third one. In such a case, we can write  b  b

b  f (x)g (x) dx = f (x)g(x)  − g(x) f  (x) dx, a

a

a

where, similarly as before,

b

c f (x)g(x)  := lim f (x)g(x)  . a a c↑b

3) The reader is encouraged to similarly consider the change-ofvariable formula. E XAMPLE 6.3.–  1 1 √ dx  = 2. 1) = −2 1 − x √ 0 0 1− x In view of remark 6.2, we could apply here √ the Newton–Leibnitz formula, although the function F(x) = −2 1 − x, x ∈ [0, 1], is not differentiable at the endpoint x = 1 of the interval.

Improper Integrals

85

2) 

+∞



−x

+∞

xe dx = −

0

−x

x de 0

+∞  = −[xe ] + 0

+∞

−x

e−x dx

0

c = − lim [xe−x ] + 1 = − lim ce−c + 1 = 1. c→+∞ c→+∞ 0

3) 

+∞

sin 1x x2

1



0

dx = −

sin t dt 1



1

=

(change of variable t = 1x )

sin t dt = 1 − cos 1.

0

Here we “did not notice” that the improper integral had become a “proper” one. To be precise, we should write as follows: 

+∞

sin 1x x2

1



c

dx = lim

c→+∞

sin 1x x2

1

  dx = lim − c→+∞

1 c

 sin t dt

1

 1 

1 c = lim cos t = lim cos − cos 1 = 1 − cos 1. c→+∞ c→+∞ 1 c

T HEOREM 6.1 (Abel–Dirichlet test).– Let functions f ∈ C[a, b) and g ∈ C 1 [a, b) satisfy the following conditions:   x  1)  f (t) dt  M < +∞, x ∈ [a, b); a

2) g is a decreasing function in the interval [a, b), and lim x↑b g(x) = 0. Then the improper integral   

b a

b a

f g converges, and

 f g  Mg(a).

x P ROOF .– Denote F(x) := a f (t) dt, x ∈ [a, b). By the integration-byparts formula, we have:  c c  c f g = Fg − Fg , a  c < b. a

a

a

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Integral and Measure

Since 

b



|Fg |

a

 M

b

|g |

(g  0)

a

  b  

   NL   = M g  === M  lim g(c) − g(a)  = Mg(a) < +∞, c↑b

a

b the improper integral a Fg (absolutely) converges, that is, there exists

c the finite limit limc↑b a Fg . Therefore, the improper integral 

b

 c c f g = lim Fg − lim Fg c↑b

a

a

c↑b

a

also converges since the first limit, because of the boundedness of the function F, is

 lim F(c)g(c) − F(a)g(a) = 0 − 0 · g(a) = 0. c↑b

 

+∞

E XAMPLE 6.4.– The improper integral 0 g(x) sin ax dx converges if g ∈ D[0, +∞) and g(x) ↓ 0 as x → +∞. For the proof, it suffices to check x that F(x) := 0 sin at dt, x ∈ [0, +∞), is a bounded function:  x  1    1 2 F(x) = − cos at  = |1 − cos ax|  , 0 a |a| |a|

x ∈ [0, +∞).

+∞ Similarly, we can check that the integral 0 g(x) cos ax dx also converges if g ∈ D[0, +∞) and g(x) ↓ 0 (a  0). In particular, the

+∞ integral 0 sinx x dx converges relatively.2 For this, it suffices to check 

+∞  that 0  sinx x  dx = +∞. We have 

+∞ 1

| sin x| dx  x

 1

+∞

sin2 x dx x

2 Meanwhile, we cannot calculate it. Later, we shall see that it equals π/2 (example 11.3).

Improper Integrals

1 = 2



+∞ 1

dx − x



+∞ 1

87

 cos 2x dx . x

The first integral diverges, while the second one converges. Therefore, their difference diverges. P ROPOSITION 6.3 (Integral test of series convergence).– If f : [1, +∞) the improper

+∞→ [0, +∞) is a decreasing function, then  integral 1 f (x) dx converges if and only if the series ∞ n=1 f (n) does. P ROOF.– By the monotonicity of f we have the inequalities f (n + 1)  f (x)  f (n),

x ∈ [n, n + 1], n ∈ N.

Integrating them, we get 

n+1

f (n + 1) 

f  f (n),

n ∈ N.

n

Summing the first N inequalities, we have: N 

f (n + 1) 

N  

n=1

n+1 n

n=1

f

N 

f (n) =: S N ,

N ∈ N,

n=1

that is, 

N+1

S N+1 − f (1) 

f  S N , N ∈ N.

1

Therefore, 

+∞



N+1

f = lim

1

if and only if

N→+∞

+∞

n=1

f < +∞

1

f (n) = limN→+∞ S N < +∞.

 

88

Integral and Measure

E XAMPLE 6.5.– 1) Consider the series the function f (x) :=

1 , xp

+∞

1 n=0 n p

(p > 0). Applying the proposition to

x ∈ [1, +∞),

we get  +∞ ∞  1 dx < +∞ ⇐⇒ < +∞ ⇐⇒ p > 1. p n xp 1 n=1 2)

∞ 

1 p < +∞ ⇐⇒ n ln n n=2 

+∞

⇐⇒ 2

d(ln x) = ln p x





+∞ 2

+∞ e2

dx < +∞ x ln p x

dt < +∞ ⇐⇒ p > 1. tp

P.6. Problems P ROBLEM 6.1.– Does the improper integral

∞ 0

sin x dx converge?

P ROBLEM 6.2.– Find the improper integrals:  ∞ (x + 1)3 dx 1) ; x5 1  2 dx 2) ; √3 |x − 1| 0  ∞ dx 3) . 2 2 x ln x P ROBLEM 6.3.– Prove the convergence of the following improper integrals:  ∞ 1) x100 e−0,01x dx; 1

Improper Integrals



∞

2) 

0 1

3) 0

89

x dx ; ex − 1 x ln 1x dx . 1+ x

6.4.– Let f : [0, ∞) → R be a decreasing function such that

P ROBLEM ∞ f (x) dx < +∞. Prove that: 0 1) f (x) → 0 as x → ∞;

2) x f (x) → 0 as x → ∞. P ROBLEM 6.5.– Let f : (0, 1] → R be a decreasing function such that

1 xα f (x) dx < +∞ for some α ∈ R. Prove that xα+1 f (x) → 0 as x ↓ 0. 0

+∞ P ROBLEM 6.6.– Let f ∈ C 1 (R) be such that −∞ | f  (x)| dx < +∞. Prove that there exist finite limits lim f (x) and lim f (x). x→+∞

x→−∞

P ROBLEM 6.7.– Let f ∈ C 1 (R) be such that 

+∞ −∞

 f (x) dx < +∞ and

+∞  −∞

 f  (x) dx < +∞.

Prove that f (x) → 0 as x → ±∞. P ROBLEM 6.8.– Let f ∈ C[0, ∞) and f  0. Denote   E= α∈R:

∞

. x f (x) dx < +∞ . α

0

Prove that E = ∅, R, (−∞, α), or (−∞, α]. Show by examples that all the cases are possible. P ROBLEM 6.9.– Give an example of a continuous function f : [1, ∞) → [0, ∞) such that 

∞ 1

f (x) dx < +∞ but

∞  n=1

f (n) = +∞.

90

Integral and Measure

P ROBLEM 6.10.– Give an example of a continuous function f : [1, ∞) → [0, ∞) such that 

∞

f (x) dx = ∞ but

1

∞ 

f (n) < +∞.

n=1

P ROBLEM 6.11.– For which α ∈ R the improper integral converges: 1) absolutely; 2) relatively?

1 0

xα sin 1x dx

Part 2

Integration of Several-variable Functions

7 Additional Properties of Step Functions

Our nearest aim is a generalization, in two directions, of the notion of an integral. First, we will consider the integrals of functions of several variables (including functions of a single variable). To this end, we will need to extend the notion of a step function in the multidimensional case. Second, instead of the uniform convergence of sequences of step functions, we will require the convergence in another, much weaker sense. Therefore, the class of limit (and, thus, of integrable) functions will be significantly wider. So we will get a more general Lebesgue integral which possesses very good properties. Although we will have to overcome some difficulties, the game is worth the candle. D EFINITION 7.1.– A k-dimensional rectangular box (or, shortly, k-box) is a set of the form k    I= Ii = I1 × I2 × · · · × Ik = x = (x1 , x2 , . . . , xk ) : xi ∈ Ii , i = 1, 2, . . . , k , i=1

where Ii ⊂ R are arbitrary intervals (including [a, a] = {a} and ∅). If all intervals Ii (called the sides of a k-box I) are bounded (closed, open, etc.), then I is also called a bounded (closed, open, etc.) k-box. The measure of such a k-box is defined as the number k  m(I) := |Ii | i=1

(where 0 · ∞ = ∞ · 0 := 0; c · ∞ := ∞ if c > 0).

94

Integral and Measure

D EFINITION 7.2.– A set N ⊂ Rk is called a zero-measure set (or a null set) if, for every ε > 0, there exists a sequence of k-boxes {I n , n ∈ N} such that N⊂

∞ 

I

n

and

n=1

∞ 

m(I n ) < ε.

n=1

In such a case, we also write m(N) = 0. We denote by N the class of all zero-measure sets. R EMARK 7.1.– 1) The notion of a zero-measure set depends on dimension k. For example, every interval I of non-zero length is not a zero-measure set in the real line R (k = 1), while when put in the plane R2 (k = 2), say, as the set I × {0}, already is a zero-measure set . 2) The class of zero-measure sets remains the same if, in the definition, we restrict ourselves to closed or open k-boxes. The case of closed k-boxes is obvious (m(I) = m(I)). Consider the case of open k-boxes. If N ∈ N, then by definition 7.2, for arbitrary ε > 0, there exists a countable system (sequence) of k-boxes {I n , n ∈ N} such that N⊂

∞ 

In

and

n=1

∞ 

ε m(I n ) < . 2 n=1

Taking, for each n ∈ N, an open k-box I˜n ⊃ I n such that m(I˜n )  ε m(I n ) + 2n+1 , we get N⊂

∞  n=1

I˜n

and

∞  n=1

m(I˜n ) 

∞ 

∞  ε ε ε m(I ) + < + = ε. n+1 2 2 2 n=1 n=1 n

P ROPOSITION 7.1 (Properties of zero-measure sets).– 1) if N ∈ N and A ⊂ N, then A ∈ N;  2) if Ni ∈ N, i = 1, 2, . . . , then ∞ i=1 Ni ∈ N.

Additional Properties of Step Functions

95

P ROOF.– 1) Obvious. 2) Take arbitrary ε > 0. For every i ∈ N, there exists a system of k-boxes {Iin , n ∈ N} such that Ni ⊂

∞

Iin

and

n=1

∞  n=1

m(Iin ) <

ε . 2i

Then ∞ i=1

Ni ⊂

∞ ∞ i=1 n=1

Iin

and

∞  ∞  i=1 n=1

m(Iin ) <

∞  ε = ε. 2i i=1

It remains to note that the system {Iin , i ∈ N, n ∈ N} is countable as a countable union of countable systems (and their sum does not depend on the summation order).   E XAMPLE 7.1.– 1) An arbitrary finite or countable set is a zero-measure set. Indeed, any one-point set is a zero-measure set, and every countable set is a countable union of one-point sets. It remains to apply item 2 of proposition 7.1. 2) The boundary of a bounded k-box is a zero-measure set. For example, the boundary of a k-box [a1 , b1 ] × [a2 , b2 ] × · · · × [ak , bk ] consists of 2k (k − 1)-boxes {a1 } × [a2 , b2 ] × · · · × [ak , bk ], {b1 } × [a2 , b2 ] × · · · × [ak , bk ], . . ., [a1 , b1 ] × [a2 , b2 ] × · · · × {bk }, which are all zero-measure sets (since each has a zero-length side). 3) Let us check that any smooth curve Γ in Rk (k  2) is a zeromeasure set. Suppose that  Γ = γ [a, b] ,

 γ ∈ C 1 [a, b], Rk .

96

Integral and Measure

Take arbitrary δ > 0 and a partition a = t0 < t1 < · · · < tn = b of [a, b] such that maxi |ti+1 − ti | < δ. Denote M := maxt∈[a,b] |γ (t)|. By the Lagrange theorem we have that   γ(t) − γ(ti )  M|t − ti | < Mδ

for t ∈ [ti , ti+1 ],

or  γ(t) ∈ Bi := B γ(ti ), M|ti+1 − ti | for t ∈ [ti , ti+1 ]. Let Ii denote the closed k-box with center γ(ti ) and edges of length 2M|ti+1 − ti |. Then Bi ⊂ Ii , and, therefore, γ(t) ∈ Ii

for t ∈ [ti , ti+1 ].

This means that Γ⊂

n−1

Ii .

i=0

Let us estimate the sum of the measures of all Ii : n−1 

m(Ii ) =

i=0

n−1  

 2M|ti+1 − ti | k = (2M)k |ti+1 − ti |k−1 |ti+1 − ti | n−1

i=0

i=0

< (2M)k δk−1

n−1 

|ti+1 − ti | = (2M)k (b − a)δk−1

i=0

= Cδ

k−1

.

Since the constant C does not depend on a partition of [a, b], and ε > 0 was arbitrary, taking δ := (ε/C)1/k−1 , we have Γ⊂

n−1 i=0

Ii

and

n−1  i=0

m(Ii ) < ε.

Additional Properties of Step Functions

97

7.1. The notion “almost everywhere” A property of points of Rk is said to hold almost everywhere (a.e.) in a set A ⊂ Rk if the property holds for all points of A except a zeromeasure set. E XAMPLE 7.2.– 1) fn → f a.s. in A :⇐⇒ {x ∈ A : fn (x) → f (x)} ∈ N. 2) A function f : A → R = [−∞, ∞] is finite a.e. if {x ∈ A : f (x) = ±∞} ∈ N. 3) A function f : A → R is continuous a.e. in A ⊂ Rk if it is discontinuous in a zero-measure set. For example, every function f ∈ D[a, b] is continuous a.e. (in the interval [a, b]) since the set of its discontinuity points is countable ((1) of theorem 1.1) and therefore is a zero-measure set ((1) of example 7.1). 4) A function f : A \ N → B is said to be defined a.e. in the set A ⊂ Rk if N is a zero-measure set. D EFINITION 7.3.– A function ϕ: I → R (where I ⊂ Rk is a k-box) is called a step function if there exists a finite system of k-boxes I1 , I2 , . . . , In ⊂ I (Ii ∩ I j = ∅ for i  j) such that ϕ is constant in each of  these k-boxes and equals zero in the set I \ nj=1 I j . The integral of such a function in I is defined as the number  ϕ := I

n 

y j m(I j ),

j=1

where y j is the value of ϕ in I j . The integral is also denoted by 

 ϕ(x) dx,

ϕ(x1 , x2 , . . . , xk ) dx1 dx2 · · · dxk ,

I

I

and 

bk ak

 ...

b2  b1 a2

a1

ϕ(x1 , x2 , . . . , xk ) dx1 dx2 · · · dxk

98

Integral and Measure

if I = [a1 , b1 ] × [a2 , b2 ] × · · · × [ak , bk ]. Often, especially in the two- and three-dimensional cases, the integration dimension k is shown by the number of symbols : 

 f (x1 , x2 ) dx1 dx2 ,

f (x1 , x2 , x3 ) dx1 dx2 dx3 , etc.

I

I

We denote by S (I) the class of all step functions in I. R EMARK 7.2.– It is often convenient to express step functions in terms of the indicators of k-boxes. The indicator of a set A is the function  1 for x ∈ A, 1IA (x) := 0 for x  A. If a step function ϕ takes the values yi in (non-intersecting) k-boxes Ii , i = 1, 2, . . . , n, then ϕ=

n 

y j 1II j .

j=1

On the contrary, every function ϕ of such a form is a step function (even if Ii ∩ I j  ∅ for some i, j). P ROPOSITION 7.2 (Properties of the integrals of step functions).– 1) Linearity: if f, g ∈ S (I), α, β ∈ R, then    (α f + βg) = α f + β g. I

I

I

if f, g ∈ S (I), f  g (i.e., f (x)  g(x), x ∈ I), then

2) Monotonicity:

f  I g; in particular, I      f   | f |. I

I

Additional Properties of Step Functions

99

P ROOF.– The proof is the same as in the case of an interval (proposition 3.2).   L EMMA 7.1 (First main lemma).– If {ϕn } ⊂ S (Rk ) is a decreasing sequence of step functions, converging almost everywhere to 0, then limn→∞ Rk ϕn = 0, or shortly,  ϕn ↓ 0 a.e. =⇒ ϕn ↓ 0. Rk

Note that here a sequence {ϕn } is supposed to be decreasing and nonnegative everywhere (i.e. in the whole Rk ), while converging to 0 almost everywhere. P ROOF .– Let I be any closed k-box such that ϕ1 = 0 outside it: ϕ1 (x) = 0 for x ∈ Rk \ I. Then also ϕn (x) = 0, x ∈ Rk \ I, n ∈ N. Denote " # N0 := x ∈ Rk : lim ϕn (x)  0 ∈ N. n→∞

 

Also let N1 be the union of the boundaries of constancy k-boxes of all functions ϕn , n ∈ N. It is clear that N1 ∈ N (see (2) of example 7.1 and (2) of proposition 7.1). Therefore, N := N0 ∪ N1 ∈ N. Let us take an arbitrary ε > 0 and denote  M := max ϕ1 (x), ε˜ := ε/ M + m(I) > 0. x∈I

By the definition of a zero-measure set (definition 7.2), there exists a sequence of open k-boxes {I j } such that ∞ j=1

Ij ⊃ N

and

∞ 

m(I j ) < ε. ˜

j=1

Now let x ∈ I \ N. Since limn→∞ ϕn (x) = 0 (as x  N0 ), there exists n x ∈ N such that 0  ϕnx (x) < ε. ˜ Since x is an inner point of a constancy k-box of ϕnx (as x  N1 ), there exists a neighborhood U(x) of x such that 0  ϕnx (y) = ϕnx (x) < ε˜

for y ∈ U(x).

100

Integral and Measure

Obviously, ∞



Ij ∪

j=1

U(x) ⊃ N ∪ (I \ N) = I.

x∈I\N

So we have a covering of I by open sets. Since I is a compact set, we can choose a finite subcovering I j1 , I j2 , . . . , I j p , U(x1 ), U(x2 ), . . . , U(x s ) such that p

I jr ∪

r=1

s

U(xt ) ⊃ I.

t=1

Denote K = max{n x1 , n x2 , . . . , n x s }. If a point y ∈ I belongs to some neighborhood U(xt ), t = 1, 2, . . . , s, then 0  ϕn (y) < ε˜ for n  K. p Otherwise (i.e. when y ∈ r=1 I jr ), we only have 0  ϕn (y)  M. Therefore, we can write: " # 0  ϕn (y)  max ε1 ˜ II (y), M1II j 1 (y), M1II j 2 (y), . . . , M1II j p (y)  ε1 ˜ II (y) + M

p 

1II jr (y),

y ∈ Rk , n  K.

r=1

By integrating this inequality we have:  0

 Rk

ϕn  ε˜

= εm(I) ˜ +M

Rk

1II + M

p   r=1

Rk

1II j r

p   m 1II jr < εm(I) ˜ + M ε˜ = ε,

n  K.

r=1

This means that  lim ϕn = 0. n→∞

Rk

C OROLLARY 7.1 (Continuity of the integral with respect to monotonic sequences).– If S (Rk )  ϕn ↑ ϕ ∈ S (Rk ) a.e.

(ϕn ↓ ϕ a.e.), n → ∞,

Additional Properties of Step Functions

101

then 





Rk

ϕn ↑

Rk

ϕ

 Rk

ϕn ↓

Rk

 ϕ,

n → ∞.

P ROOF.– (Case “↑”) Denote ϕ˜ n := ϕ − ϕn ↓ 0 a.e., n → ∞. Then, using Lemma 7.1, we have:  Rk

 ϕ− =⇒



Rk 

ϕn =

Rk

Rk

ϕn ↑

(ϕ − ϕn ) ↓ 0,

Rk

ϕ,

n→∞

n → ∞.

 

L EMMA 7.2 (Second main lemma).– Let {ϕn } ⊂ S (Rk ) be an increasing sequence such that1  A := lim ↑ ϕn < +∞. n→∞

Rk

Then lim ↑ ϕn < +∞ a.e., n→∞

that is, " # N := x ∈ Rk : lim ↑ ϕn (x) = +∞ ∈ N. n→∞

P ROOF .– First, consider the case ϕn  0, n ∈ N. Denote 2A ! Nε := x ∈ Rk : lim ϕn (x) > , ε > 0, n→∞ ε ! 2A Nn,ε := x ∈ Rk : ϕn (x) > , ε > 0, n ∈ N. ε

1 To emphasize that a sequence is increasing or decreasing, we will use the notation lim ↑ or lim ↓, respectively, instead of lim . n→∞

n→∞

n→∞

102

Integral and Measure

Note that Nε =

∞

Nn,ε ,

ε > 0.

n=1

Indeed, since the sequence {ϕn } increases, we have 2A 2A ⇐⇒ ∃ n ∈ N : ϕn (x) > ε ε ∞ ⇐⇒ ∃ n ∈ N : x ∈ Nn,ε ⇐⇒ x ∈ Nn,ε .

x ∈ Nε ⇐⇒ lim ϕn (x) > n→∞

n=1

The set N1,ε consists of the constancy k-boxes of the function ϕ1 in which its values are greater than 2A ε . Let us denote them by I1 , I2 , . . . , In1 and their sum by S 1 . Since ϕ1  2A ε 1IN1,ε , we have:  A



Rk

n1  2A 2A  1IN1,ε = 1II ε j=1 Rk j Rk ε n1 2A  2A ε = m(I j ) = S 1 =⇒ S 1  . ε j=1 ε 2

ϕ1 

The set N2,ε consists of the k-boxes I1 , I2 , . . . , In1 (since ϕ2  ϕ1 ) and the constancy k-boxes In1 +1 , . . . , In2 of ϕ2 in which ϕ2 > 2A ε  ϕ1 (n2 = n1 if there are no such k-boxes). Denoting S 2 :=

n2 

m(I j ),

j=1

we get that S 2  2ε . Continuing this way, we can write: Nr,ε =

nr j=1

Ij

and S r :=

nr 

ε m(I j )  , 2 j=1

r ∈ N.

Additional Properties of Step Functions

103

Note that N ⊂ Nε =

∞

Nr,ε =

r=1

nr ∞

Ij =

r=1 j=1

n∞

Ij

j=1

(here n∞ := sup{nr , r ∈ N}) and n∞ 

m(I j ) = lim

r→∞

j=1

nr 

m(I j ) 

j=1

ε < ε. 2

This means that N ∈ N. In the general case, denote ϕ˜ n := ϕn − ϕ1  0, n ∈ N. Then     lim ϕ˜ n = lim ϕn − ϕ1 = A − ϕ1 < +∞. n→∞

n→∞

Rk

Rk

Rk

Rk

By the preceding, we have that limn→∞ ϕ˜ n < +∞ a.e.; therefore, lim ϕn = lim ϕ˜ n + ϕ1 < +∞ a.e.

n→∞

 

n→∞

C OROLLARY 7.2.– If {ϕn } ⊂ S (Rk ) is a sequence of non-negative functions such that ∞   ϕn < +∞, Rk

n=1

then ∞ 

ϕn < +∞ a.e.

n=1

P ROOF.– It suffices to apply the second main lemma to the (increasing)  sequence of partial sums of the series n ϕn : lim ↑ N→∞

  N Rk n=1

ϕn = lim ↑

N  

N→∞ n=1

=⇒

∞  n=1

Rk

ϕn =

∞   n=1

ϕn = lim ↑

N 

N→∞ n=1

Rk

ϕn < +∞

ϕn < +∞ a.e.

 

104

Integral and Measure

P.7. Problems P ROBLEM 7.1.– Which of the following notions make sense: 1) an almost everywhere continuous function; 2) an almost everywhere discontinuous function; 3) an almost everywhere uniformly continuous function; 4) an almost everywhere bounded function; 5) an almost everywhere positive function; 6) a function converging to zero almost everywhere? P ROBLEM 7.2.– Can the union of a family of zero-measure sets be a non-zero-measure set? P ROBLEM 7.3.– Give an example

of a sequence of step functions {ϕn } such that ϕn (x) → 0, x ∈ R, but R ϕn → 0. P ROBLEM 7.4.– Give an example of a sequence of step functions {ϕn } such that  lim ↑ ϕn (x) dx < ∞ but sup ϕ(x) = +∞ a.e. n

R

x∈R

P ROBLEM 7.5.– Prove that N ⊂ Rk is a zero-measure set if and only if, for all ε > 0, there exists an increasing sequence of step functions {ϕn } such that  lim ↑ ϕn  ε and sup ϕn (x)  1, x ∈ N. n

Rk

n∈N

This property is often used to generalize the notion of a zero-measure set.

8 Lebesgue Integral

D EFINITION 8.1.– A sequence {ϕn } ⊂ S (I) (where I ⊂ Rk is a k-box) is called a Cauchy sequence, or a fundamental sequence, if  ∀ε > 0, ∃ N ∈ N : |ϕn − ϕm | < ε for n, m > N. [8.1] I

To distinguish such Cauchy sequences of step functions from Cauchy sequences of numbers, the former are called Cauchy sequences in the L1 sense or Cauchy sequences in the mean. D EFINITION 8.2.– A function f : I → R = [−∞, +∞] is said to be Lebesgue-integrable in a k-box I if there exists a Cauchy sequence of step functions {ϕn } ⊂ S (I) converging to f almost everywhere (a.e.) in I. The Lebesgue integral of such a function f in I is the number   f := lim ϕn . I

n→∞

I

The other notations are, as before,   f (x) dx, f (x1 , x2 , . . . , xk ) dx1 dx2 · · · dxk , etc. I

I

(see definition 7.3). We denote by Lebesgue-integrable functions in I.

L(I) the class of all

106

Integral and Measure

R EMARK 8.1.–

1) Note that the limit of the sequence { I ϕn } in the definition always exists. Indeed, for such a sequence, we have:       8.1 =⇒ ∀ε > 0, ∃ N ∈ N :  ϕn − ϕm  < ε for n, m > N.  I  I Therefore, by

the Cauchy criterion of convergence, there exists a finite limit lim I fn . n→∞

2) Let us compare definition 8.2 with definition 3.2 of the integral in the class D[a, b]. For functions f ∈ D[a, b], we constructed a sequence {ϕn } ⊂ S [a, b] such that – fn ⇒ f in [a, b] and then, having checked that – { fn } is a Cauchy sequence,

proved the existence of the integral I f := limn→∞ I fn and the correctness of its definition (i.e. the independence of the latter limit on the choice of a sequence {ϕn }). Now we have weakened condition 1) by replacing the uniform convergence by the convergence a.e.; however, at the same time, we require condition 2), which is not automatically fulfilled now. While the existence of the integral is, as before, easily obtained, the proof of the correctness of its definition is more difficult. 8.1. Proof of the correctness of the definition of integral Step 1. Let {ϕn } ⊂ S (I) be a Cauchy sequence such that 0  ϕn → 0 a.e. in I. Let us prove that λ := limn→∞ I ϕn = 0. Take an arbitrary ε > 0. By [8.1] there exists a sequence {ni } ⊂ N such that  ε |ϕn − ϕni | < i forall n > ni , i ∈ N. 2 I

Lebesgue Integral

107

Denote ψm := min{ϕn1 , ϕn2 , . . . , ϕnm }, m ∈ N. Clearly, 0  ψm ↓ 0 a.e., m → ∞ since 0  ψm  ϕnm → 0 a.e. Therefore, by the first principal lemma (lemma 7.1),  ψm ↓ 0, m → ∞. I

Integrating the inequality 0  ϕnm − ψm = max{ϕnm − ϕn1 , ϕnm − ϕn2 , . . . , ϕnm − ϕnm−1 , 0} 

m−1 

|ϕnm − ϕni |,

m ∈ N,

i=1

we get  0 I

(ϕnm − ψm ) 

m−1  i=1

I

|ϕnm − ϕni | <

m−1  i=1

ε < ε, 2i

m ∈ N.

Therefore, 



0  λ = lim

m→∞

I

ϕnm = lim

m→∞

 I

(ϕnm − ψm ) + lim

m→∞

ψm  ε + 0 = ε. I

The arbitrariness of ε > 0 yields λ = 0. Step 2. Now let{ϕn } ⊂ S (I) be a Cauchy sequence such that ϕn → 0 a.e.

in I (i.e., now we do not require that ϕn  0 ). Let us prove that ϕ → 0, n → ∞. I n Consider the sequences of non-negative functions {ϕ+n } and {ϕ−n }.1 Since 0  ϕ±n  |ϕn | → 0 a.e. and   |ϕ±n − ϕ±m |  |ϕn − ϕm |, n, m ∈ N I

I

1 Recall that x± := max{±x, 0}, x = x+ − x− , |x| = x+ + x− , x± = 12 (|x| ± x), |x± − y± |  |x − y|.

108

Integral and Measure

(i.e. {ϕ±n } also are Cauchy sequences), by step 1 we have that I ϕ±n → 0, n → ∞. Therefore,     lim fn = lim ( fn+ − fn− ) = lim fn+ − lim fn− = 0. n→∞

n→∞

I

n→∞

I

I

n→∞

I

Step 3. Let {ϕn } and {ϕ˜ n } be two Cauchy sequences of step functions such that ϕn → f and ϕ˜ n → f a.e. in I. Denote ψn := ϕn − ϕ˜ n . Then ψn → 0 a.e. in I (indeed, if ϕn (x) → f (x), x ∈ I \ N1 , N1 ∈ N, and ϕ˜ n (x) → f (x), x ∈ I \ N2 , N2 ∈ N, then ψn (x) → 0, x ∈ I \ (N1 ∪ N2 ), and N1 ∪ N2 ∈ N).2 Since    |ψn − ψm |  |ϕn − ϕm | + |ϕ˜ n − ϕ˜ m |, n, m ∈ N, I

I

I

{ψn } is also a Cauchy sequence. By step 2 we have       lim ψn = 0 =⇒ lim ϕn − ϕ˜ n = 0 =⇒ lim ϕn = lim ϕ˜ n . n→∞

n→∞

I

I

I

n→∞

I

n→∞

I

Thus, the limit limn→∞ I ϕn does not depend on the choice of the Cauchy sequence {ϕn } ⊂ S (I) converging a.e. to f . P ROPOSITION 8.1 (Elementary properties of the integral).– 1) Linearity: if f, g ∈ L(I), | f |, |g| < +∞ a.e.,3 α, β ∈ R, then α f + βg ∈ L(I), and    (α f + βg) = α f + β g; I

I

I

2) if f, g ∈ L(I), then max{ f, g}, min{ f, g}, | f | ∈ L(I);

3) (Monotonicity) If f, g ∈ L(I), f  g, then I f  I g; in particular,      f   | f |;   I

I

2 Further, we will omit similar arguments related to the notion almost everywhere. 3 Further (lemma 8.1, item 3), we will see that the latter condition is not needed since every integrable function is finite almost everywhere.

Lebesgue Integral

4) If f ∈ L(I) and g = f a.e. in I, then g ∈ L(I), and

g= I

I

109

f.

P ROOF .– 1) If {ϕn } ⊂ S (I) and {ψn } ⊂ S (I) are Cauchy sequences, then so is the sequence {αϕn + βψn } since     (αϕ + βψ ) − (αϕ + βψ )  |α| |ϕ − ϕ | + |β| |ψ − ψ |, n n m m n m n m I

I

I

n, m ∈ N.

If, moreover, ϕn → f and ψn → g a.e. in I, then αϕn + βψn → α f + βg a.e. in I. Therefore, by the definition of the integral (definition 8.2), α f + βg ∈ L(I), and 



    (αϕn + βψn ) = lim α ϕn + β ψn

(α f + βg) = lim

n→∞

I

n→∞

I





= α lim

n→∞

ϕn + β lim I

n→∞

I





ψn = α I

I

f +β I

g. I

2) If {ϕn } ⊂ S (I) is a Cauchy sequence, so is {| fn |} since    |ϕ | − |ϕ |  |ϕ − ϕ |, n, m ∈ N. n m n m I

I

If, moreover, ϕn → f a.e. in I, then |ϕn | → | f | a.e. in I, and therefore, | f | ∈ L(I). The integrability of max{ f, g} and min{ f, g} now follows from property 1 and from the equalities max{ f, g} =

1 f + g + | f − g| , 2

min{ f, g} =

1 f + g − | f − g| . 2

3) Let f ∈ L(I) and f  0. Take a Cauchy sequence {ϕn } ⊂ S (I) such that ϕn → f a.e. in I. Then S (I)  ϕ+n → f + = f a.e. in I. Since {ϕ+n } is also a Cauchy sequence (see step 2 of the proof of proposition 7.1) and fn+  0, we have   f = lim ϕ+n  0. I

n→∞

I

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Integral and Measure

In the general case, we have    f  g =⇒ f − g  0 =⇒ f − g = ( f − g)  0. I

I

I

4) If {ϕn } ⊂ S (I) is a Cauchy sequence and ϕn → f a.e. in I, then also ϕn → g a.e. in I. Therefore, g ∈ L(I), and    g = lim fn = f.   n→∞

I

I

I

Now, our aim is proving highly important Beppo Levi and Fatou– Lebesgue theorems that we state side-by-side and split their proofs into several parts. T HEOREM 8.1 (Beppo Levi theorem).–

1) If { fn } ⊂ L(I), fn ↑ f ( fn ↓ f ) a.e. in I, and limn→∞ I fn < +∞

(respectively, limn→∞ I fn > −∞), then   f ∈ L(I) and f = lim fn . n→∞ I I  2) If { fn } ⊂ L(I), fn  0, n ∈ N, and ∞ n=1 I fn < +∞, then  ∞ ∞   f := fn ∈ L(I) and f= fn . I

n=1

n=1

I

T HEOREM 8.2 (Fatou–Lebesgue theorem).– 1a) If { fn } ⊂ L(I), g ∈ L(I), fn  g, n ∈ N, and lim supn→∞ then   lim sup fn ∈ L(I) and lim sup fn  lim sup fn ; n→∞

I

n→∞

n→∞

I

1b) If { fn } ⊂ L(I), g ∈ L(I), fn  g, n ∈ N, and lim inf n→∞ then   lim inf fn ∈ L(I) and lim inf fn  lim inf fn ; n→∞

I n→∞

n→∞



f I n

> −∞,

f I n

< +∞,

I

2) If { fn } ⊂ L(I), g ∈ L(I), | fn |  g, n ∈ N, and fn → f , then   f ∈ L(I) and fn → f. I

I

Lebesgue Integral

111

D EFINITION 8.3.– A function f : I → R (where I ⊂ Rk is a k-box) is said to belong to class L+ (I) if there exists a sequence {ϕn } ⊂ S (I) such that 1) ϕn ↑ f a.e. in I; 2) {ϕn } is a Cauchy sequence. R EMARK 8.2.– Comparing this definition with definition 8.2, we see that the class L+ (I) consists of functions f ∈ L(I) for which there exists an increasing Cauchy sequence {ϕn } ⊂ S (I) that converges to f a.e. Note that the role of the class L+ (I) is auxiliary. After we prove the main theorems of integral calculus (Bepo Levi, Fatou–Lebesgue, Fubini, change-of-variables formula), we will be able to “forget” it. L EMMA 8.1 (Properties of function classes L(I) and L+ (I)).– 1) Under condition (a) of definition

8.3, condition (b) is equivalent to the following condition:(b) lim ↑ I ϕn < +∞; n→∞

2) Every function f ∈ L(I) can be expressed as the difference of two functions from L+ (I), that is, ∀ f ∈ L(I),

∃ g, h ∈ L+ (I) : f = g − h

a.e.;

2ε ) ∀ f ∈ L(I), ∀ ε > 0, ∃ g, h ∈ L+ (I) : f = g−h a.e., h  0, and

3) If f ∈ L(I), then | f | < +∞ a.e. in I.

I

h < ε;

P ROOF .– 1) Let condition (a) be satisfied. It is obvious that (b) implies (b) .

Let us prove the converse. Denote A := lim ↑ I ϕn < +∞. Take arbitrary n→∞

ε > 0. There exists N ∈ N such that  A − ε < ϕn  A for n > N I

and







|ϕn − ϕm | = I



(ϕn − ϕm ) = I

ϕn − I

ϕm < A − (A − ε) = ε I

(the same is true for interchanged n and m).

for n  m > N

112

Integral and Measure

2) Let {ϕn } ⊂ S (I) be a Cauchy sequence such that ϕn → f a.e. (in I). Without loss of generality, we may suppose that ∞  

A :=

n=1

 |ϕn+1 − ϕn | + I

|ϕ1 | < +∞. I

Indeed, otherwise, instead of {ϕn }, we could take a subsequence {ϕnk } such that  1 |ϕnk+1 − ϕnk | < k , k ∈ N. 2 I Denote ϕ˜ m := ϕˆ m :=

m 

|ϕn+1 − ϕn | + |ϕ1 | ∈ S (I),

n=1 m 

( fn+1 − fn )+ + f1+ ∈ S (I),

n=1

m  ϕ¯ m := ( fn+1 − fn )− + f1− ∈ S (I),

m ∈ N.

n=1

Using the inequalities ϕˆ m  ϕ˜ m and ϕˆ m  ϕ˜ m , we get that   lim ↑ ϕˆ m  lim ↑ ϕ˜ m = A < +∞, m→∞



lim ↑ m→∞

m→∞

I



I

ϕ¯ m  lim ↑ m→∞

I

ϕ˜ m = A < +∞. I

By the first statement of the lemma, g := lim ↑ ϕˆ m ∈ L+ (I), m→∞

h := lim ↑ ϕ¯ m ∈ L+ (I), m→∞

and therefore, ϕˆ − ϕ¯ ∈ L(I). It remains to notice that   m m  + + − − g − h = lim (ϕn+1 − ϕn ) + ϕ1 − (ϕn+1 − ϕn ) − ϕ1 m→∞

n=1

n=1

Lebesgue Integral m  1 = lim (ϕn+1 − ϕn ) + ϕ1 = lim ϕm+1 = f m→∞

m→∞

n=1

113

a.e.

2ε ) By the property just proved, we can write f = g−h

with g ∈ L+ (I)

and h ∈ L+ (I).

Take sequences {gn }, {hn } ⊂ S (I) such that g = lim ↑ gn ,

h = lim ↑ hn

n→∞

a.e.

n→∞

and 





g = lim ↑ n→∞

I

gn < +∞,

h = lim ↑ n→∞

I

hn < +∞. I

Take arbitrary ε > 0. There exists n ∈ N such that    0  (h − hN ) = h − hN < ε. I

I

I

Denote g˜ := g − hN , h˜ := h − hN . Then g˜ = g − hN = lim (gn − hN ) ∈ L+ (I) n→∞

and, similarly, h˜ = h − hN ∈ L+ (I). It remains to notice that h˜  0,

I

˜ h˜ < ε, and f = g˜ − h.

3) Every function f ∈ L+ (I) is a.e. finite by Lemma 7.2:  S (I)  fn ↑ f a.e., lim fn < +∞ =⇒ f < +∞ a.e., n→∞

I

and thus, every f ∈ L(I) is a.e. finite as the difference of two functions from L+ (I).  

114

Integral and Measure

8.2. Proof of the Beppo Levi theorem Step 1 (1, L+ ).4 Suppose that +



+

L = L (I)  fn ↑ f

and

fn < +∞

lim

n→∞

I

(here and below, the convergence is a.e. in the k-box I). Let us prove

+ that f ∈ L and I f = lim ↑ I fn . By the definition of class L+ , n→∞

" # ∃ gm,n ∈ S (I), m ∈ N : gm,n ↑ fn ,

∀n ∈ N,

m → ∞.

Denote hm := max gm,n = max{gm,1 , gm,2 , . . . , gm,n } ∈ S (I), nm

m ∈ N.

Note that {hm } is an increasing sequence: hm = max gm,n  max gm+1,n  max gm+1,n = hm+1 , nm

nm

nm+1

m ∈ N.

Moreover, hm = maxnm gm,n  maxnm fn = fm , m ∈ N. Integrating the inequality gm,n  hm  fm , we get 

n  m,





gm,n  I

[8.2]

hm  I

fm ,

n  m.

[8.3]

I

Taking the limit as m → ∞ in [8.2], we have fn = lim ↑ gm,n  lim ↑ hm  f, m→∞

m→∞

n ∈ N.

[8.4]

4 The shorthand (1, L+ ) means that in step 1, we prove item 1 of the B. Levy theorem for functions from class L+ = L+ (I).

Lebesgue Integral

Now, taking the limit as m → ∞ in [8.3], we have     fn = lim ↑ gm,n  lim ↑ hm  lim ↑ fm < +∞. m→∞

I

m→∞

I

m→∞

I

115

[8.5]

I

Taking the limit in [8.4], we have f = lim ↑ fn  lim ↑ hm  f =⇒ f = lim ↑ hm .

[8.6]



f = lim ↑ I hm . Now,

n→∞

m→∞

m→∞

Therefore, by lemma 8.1, item 1, f ∈ L+ and

I

m→∞

let us take the limit in inequality [8.5]: 



lim ↑



fn  lim ↑

n→∞

m→∞

I

hm  lim I

m→∞



=⇒ lim ↑ m→∞

fm I



hm = lim ↑ m→∞

I

fm < +∞.

[8.7]

I

From [8.6] and [8.7] we finally obtain    f = lim ↑ hm = lim ↑ fm . m→∞

I

I

m→∞

I

Step 2 (2, L+ ). Suppose that fn ∈ L+ , fn  0, n ∈ N, and ∞   n=1

fn < +∞. I

Let us apply step 1 to the sequence of partial sums of the series Note that S N :=

N 

fn ↑ f,



N → ∞,

n=1

and  lim ↑ N→∞

S N = lim ↑ I

N→∞

  N I n=1

fn = lim ↑

N  

N→∞ n=1

fn = I

∞   n=1

fn < +∞. I

fn .

116

Integral and Measure

In view of step 1, we get that 

f = lim ↑ S N ∈ L+

 f = lim ↑

and

N→∞

SN =

N→∞

I

I

∞  

fn . I

n=1

Step 3 (2, L). Suppose that fn ∈ L = L(I), fn  0, n ∈ N, and ∞ ε n=1 I fn < +∞. By Lemma 8.1, item 2 , 

+

∀n ∈ N,

∃ gn , hn ∈ L : fn = gn − hn ,

hn  0,

hn < I

1 . 2n

From step 2 and from the inequality ∞  

hn < I

n=1

∞  1 = 1 < +∞ 2n n=1

it follows that h :=

∞ 

hn ∈ L



+

h=

and I

n=1

∞   n=1

hn . I

Since L+  gn = fn + hn  0 and ∞   n=1

gn = I

∞   n=1

fn + I

∞  

hn  I

n=1

∞   n=1

fn + 1 < +∞, I

again applying step 2, we get g :=

∞ 

gn ∈ L

+

 g=

and I

n=1

∞   n=1

gn < +∞. I

Since g are h finite a.e. (8.1, item 3), the following equalities hold a.e.: g−h =

∞  n=1

gn −

∞  n=1

hn =

∞ ∞   (gn − hn ) = fn , n=1

n=1

Lebesgue Integral

117

and hence f ∈ L. Moreover, ∞  

fn = I

n=1

=

∞  

(gn − hn ) =

∞  

n=1

I

n=1

I

gn −





hn I



hn = I



gn − I

n=1

∞   n=1

∞  

g− I





h= I

(g − h) = I

f. I

Step 4 (1, L). Suppose that  fn ∈ L,

fn ↑ f,

lim ↑ n→∞

fn < +∞. I

Since (denoting f0 := 0) n ∞   f = lim ↑ fn = lim ↑ ( fk − fk−1 ) = ( fk − fk−1 ) n→∞

n→∞

k=1

k=1

and ∞   k=1

( fk − fk−1 ) = lim ↑ n→∞

I

n   k=1



  fk−1 = lim ↑ fn < +∞,

fk − I

n→∞

I

I

from step 3 we have  f ∈L

f=

ir I

∞   k=1

 ( fk − fk−1 ) = lim ↑ I

n→∞

fn . I

C OROLLARY 8.1.–   1) If fn ∈ L(I) and ∞ | fn | < +∞, then the series ∞ n=1 I  n=1 fn converges absolutely a.e., its sum f := ∞ f ∈ L, and n=1 n  f= I

∞   n=1

fn . I

2) Let f ∈ L(I). Then  | f | = 0 ⇐⇒ f = 0 I

a.e.

118

Integral and Measure

P ROOF.– 1) Since 0  fn±  | fn |

∞  

and

I

n=1

fn± 

∞  

| fn | < +∞, I

n=1

by the Levi theorem we have that   ∞ ∞ ∞    ± ± fn ∈ L(I) and fn = fn± . I n=1

n=1

n=1

I

Therefore (8.1, item 3), ∞ 

fn± < +∞ a.e.,

n=1

and thus ∞ 

| fn | =

n=1

∞ 

fn+ +

n=1

∞ 

fn− < +∞ a.e.

n=1

Moreover, ∞   n=1

fn = I

∞  

( fn+ − fn− )

n=1 I ∞ 

fn+ −

8.1

===

I n=1

=

∞  

fn+ −

∞  

fn−

n=1 I  ∞ n=1 I     ∞ ∞  − + − fn = ( fn − fn ) = fn . I n=1

I n=1

I n=1

2) “⇒” Denote fn = n| f |, n ∈ N. Then   lim ↑ fn = lim n | f | = lim (n · 0) = 0 < +∞. n→∞

I

n→∞

I

n→∞

By the Levi theorem, g := lim ↑ fn ∈ L(I), and therefore g < +∞ a.e. Since "

n→∞

  # " # " # x ∈ I : g(x) = +∞ = x ∈ I :  f (x) > 0 = x ∈ I : f (x)  0 ,

we have that f = 0 a.e. “⇐”. If f = 0, then | f | = 0 a.e., and therefore, by proposition 8.1, item 4, I | f | = I 0 = 0.  

Lebesgue Integral

119

8.3. Proof of the Fatou–Lebesgue theorem 1a) Denote h1 := supn1 fn . Since h1 = lim ↑ maxnN fn , maxnN fn ∈ N→∞

L(I) for all n ∈ N (by proposition 8.1, item 2), and lim ↑ I maxnN fn  N→∞

g < +∞, by the Levi theorem we get that h ∈ L(I). Similarly, hN := 1 I supnN fn ∈ L(I), N ∈ N. By integrating the inequality fn  hN , n  N, we get 

 fn 

hN , n  N,

I

I

and therefore,   sup fn  hN , nN

I

N ∈ N.

I

Taking the limit in the latter inequality as N → ∞, we get 



−∞ < lim sup n→∞



fn = lim ↓ sup N→∞ nN

I



8.1

===

fn  lim ↓ I

lim ↓ hN =

I N→∞

N→∞



hN I



lim ↓ sup fn =

I N→∞ nN

lim sup fn . I

n→∞

1b) The proof is similar. 2) Note that f = lim fn = lim sup fn = lim inf fn n→∞

n→∞

n→∞

and the conditions of both items 1a and 1b are satisfied. Therefore,       f = lim inf fn  lim inf fn  lim sup fn  lim sup fn = f, I

I n→∞

n→∞

I

n→∞

I

I

n→∞

I

120

Integral and Measure

and thus, 





fn = lim inf

lim sup n→∞

n→∞

I

fn = I

f. I

So we have   lim fn = f. n→∞

I

I

P ROPOSITION 8.2.– If f : I → R (where I ⊂ Rk is a compact k-box) is a bounded and a.e. continuous function, then f ∈ L(I); in particular, C(I) ⊂ L(I). P ROOF.– To simplify the notation, we restrict ourselves to the case I = [a, b] (k = 1). For every n ∈ N, let us partition the interval I into 2n equal subintervals {Ink , k = 1, 2, . . . , 2n }. Denote " # fn (x) = lkn := inf f (t) : t ∈ Ink for x ∈ Ink , n ∈ N. The sequence { fn } is increasing and bounded. Therefore, we can define A(x) := limn ↑ fn (x), x ∈ [a, b]. Clearly, A(x)  f (x). Let us show that A(x) = f (x) if x ∈ [a, b] is a continuity point of the function f , and thus, fn ↑ f a.e. So, let x0 be an arbitrary continuity point of f . Take arbitrary ε > 0. Then   ε ∃ δ > 0 :  f (x) − f (x0 ) < 2

for x ∈ Uδ (x0 ) ∩ [a, b].

Taking N such that b−a < δ, we will have m(Ink ) < δ for n  N. 2N Therefore, if x0 ∈ Ink , then Ink ⊂ Uδ (x0 ) for n > N. Hence, " # ε fn (x0 ) = lkn  inf f (t) : t ∈ Uδ (x0 )  f (x0 ) − > f (x0 ) − ε 2 This means that limn→∞ fn (x0 ) = f (x0 ). Thus, fn ↑ f a.e. in [a, b]. Moreover,  lim ↑ n→∞

a

b



b

fn  a

C = C(b − a) < +∞,

for n  N.

Lebesgue Integral

121

where C := sup{ f (x) : x ∈ [a, b]} < +∞ (recall that f is bounded). By the Levi theorem, f ∈ L[a, b] (and I fn ↑ I f ).   R EMARK 8.3.– Actually, we have proved that every bounded and a.e. continuous function f : I → R belongs to class L+ (I). By slightly modifying the proof we can prove even more: T HEOREM 8.3 (Riemann-integrability criterion).– A function f : [a, b] → R is Riemann-integrable if and only if f is bounded and a.e. continuous in the interval [a, b]. Thus, R[a, b] ⊂ L[a, b]. P ROOF .– Sufficiency. Modifying the proof of proposition 8.2, take an arbitrary sequence {Ink , k = 1, 2, . . . , kn }, n ∈ N, of I with meshes δn := maxk |Ink | → 0 as n → ∞ (so, we do not require the sequence to be increasingly refined). Now we define the sequence of functions { fn } as follows:  fn (x) := f ξnk for x ∈ Ink , n ∈ N, where ξnk are arbitrary points in Ink . We similarly get that fn (x) → f (x) if x is a continuity point of f (although we already do not have the monotonicity of the sequence { fn }). Since | fn (x)|  C, by the Fatou–

b

b Lebesgue theorem we get a fn → a f . However, 

b a

fn =

  f ξnk m(Ink ),

n ∈ N,

k

are Riemann integral sums of f . By the definition of the Riemann integral,  b  b f ∈ R[a, b] and R f= f. a

a

Necessity. The boundedness of a Riemann-integrable function is proved in proposition 5.1. Take arbitrary ε > 0. Using the notation of the proof of proposition 8.2, in each interval Ink , take a point ξnk such that f (ξnk ) < lkn + ε. Then  b     f ξnk m(Ink )  lnk + ε m Ink = fn + ε(b − a). k

k

a

122

Integral and Measure

Taking the limit as n → ∞, we get  R

b



b

f  lim ↑ n

a



b

fn + ε(b − a) =

a

A(x) dx. a

Since ε > 0 is arbitrary, we have  R

b

 f

b

A(x) dx.

a

a

Similarly, denoting " # gn (x) = ukn := sup f (t) : t ∈ Ink for x ∈ Ink , n ∈ N, V(x) := lim ↓ gn (x)  f (x), n

and

x ∈ [a, b],

we check that V(x) = f (x) if x ∈ [a, b] is a continuity point of f and that  R

b

 f

a

b

V(x) dx. a

Therefore, 

b

 V(x) − A(x) dx = 0.

a

Since V − A  0, we have V − A = 0 a.e. Therefore, V(x) = A(x) = f (x) for almost all x ∈ [a, b], that is, f is continuous at almost all x ∈ [a, b].   D EFINITION 8.4.– A function f : Rk → R = [−∞, +∞] is called measurable if there exists a sequence of step functions {ϕn } converging a.e. to f .

Lebesgue Integral

123

A function f : A → R (A ⊂ Rk ) is said to be measurable (in a set A) if its extension  fˆ(x) :=

f (x), x ∈ A, 0, x  A,

is a measurable function. In view of this, below we state and prove all the propositions for functions defined on the whole space Rk . We denote by M(A) the class of all measurable function on a set A. Also, for short, we denote M := M(Rk ). P ROPOSITION 8.3.– 1) L := L(Rk ) ⊂ M; 2) if f, g ∈ M, | f | < +∞ a.e., |g| < +∞ a.e., and α, β ∈ R, then α f +βg ∈ M, f g ∈ M, max{ f, g} ∈ M, min{ f, g} ∈ M, and | f | ∈ M; if, moreover, g  0 a.e., then gf ∈ M. R EMARK 8.4.– If at some point, an arithmetic operation is not defined (e.g. we get an expression of type (+∞) − (+∞) or 0 · (+∞)), then the result of such a operation can be defined arbitrarily, say, equal to zero. No problems arise since such points are zero-measure sets and the choice of function values in these sets have no influence on measurability or integrability. P ROOF.– 1) This is a direct consequence of the definition of class L. 2) If {ϕn } ⊂ S (Rk ), {ψn } ⊂ S (Rk ), ϕn → f a.e., and ψn → g a.e., then S (Rk )  αϕn + βψn → α f + βg a.e., and thus α f + βg ∈ M. The remaining operations are considered similarly.

 

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Integral and Measure

P ROPOSITION 8.4.– 1) If f ∈ M, g ∈ L, | f |  g a.e., then f ∈ L; 2) if fn ∈ M, n ∈ N, and fn → f a.e., then f ∈ M; 3) if f : Rk → R is an a.e. continuous function, then f ∈ M. P ROOF.– 1) Let {ϕn } ⊂ S (Rk ) be a sequence such that ϕn → f a.e. Denote fn := max{−g, min{ϕn , g}}, n ∈ N. Since S (Rk ) ⊂ L and g ∈ L, we have that all fn ∈ L. Since fn → max{−g, min{ f, g}} = f a.e. and | f˜n |  g, by the Fatou–Lebesgue theorem we get that f ∈ L. 2) Consider an arbitrary integrable function g > 0.5 Denote gn :=

g fn g + | fn |

and note that |gn | =

g| fn | < g. g + | fn |

Moreover, gn → h :=

gf g+|f|

a.e.

By the Fatou–Lebesgue theorem we get that h ∈ L. We can express the function f in terms of h: g f = gh + | f |h = gh + f |h| =⇒ f =

gh . g − |h|

By proposition 8.3, item 2, it follows that f ∈ M.

5 An example of such a function is g(x) = min{1, 1/|x|k+1 }, although at the moment we have no possibility to check this, except in the case of k = 1.

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3) We start with an obvious remark: if g ∈ L(I), where I ⊂ Rk is a compact k-box and gˆ is its (trivial) extension in Rk , defined by  g(x), x ∈ I, gˆ (x) := 0, x  I,

then gˆ ∈ L(Rk ) and Rk gˆ = I g. Indeed, from the definition of a step function and its integral we see that this property holds for all ϕ ∈ S (I). Therefore, if {ϕn } ⊂ S (I) is a Cauchy sequence converging a.e. in I to a function g, then the corresponding sequence of extensions {ϕˆ n } is also a Cauchy sequence in the whole Rk to the function gˆ . Hence,     gˆ = lim ϕˆ n = lim ϕn = g. Rk

n→∞

n→∞

Rk

I

I

Denote " # fn (x) = min  n, max{−n, f (x)} , f (x), x ∈ In , fˆn (x) = n 0, x  In .

x ∈ In := [−n, n]k ,

Every function fn is almost everywhere continuous and bounded on In (| fn |  n). Therefore (proposition 8.2), fn ∈ L(In ), and in view of the remark above, fˆn ∈ L = L(Rk ) ⊂ M. Since fˆn (x) → f (x), x ∈ Rk , by part 2 of the proposition we get that f ∈ M.   D EFINITION 8.5.– If f  0 and f ∈ M \ L, then we assume that the integral of the function f equals +∞:  f = +∞. Rk

If f ∈ M and at least one of the functions f + , f − is integrable, then f is said to be quasi-integrable, and its integral is the number    + f := f − f −. Rk

Rk

Rk

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R EMARK 8.5.– The latter equality is true if f ∈ L or f + , f − ∈ L. Therefore, this definition actually is a “legalization” of infinite values (+∞ and −∞) of the integral. Thus, from now on, remains

the integral

undefined only of those functions f for which Rk f + = Rk f − = +∞. We will further see that many properties of integrable functions also hold for the class of quasi-integrable functions. This is sometimes convenient since we do not have a priori to wonder about the integrability of the functions considered. P ROPOSITION 8.5.–

1) If f and g are quasi-integrable functions and f  g a.e., then f  Rk g. Rk

2) If 0  fn ↑ f a.e., fn ∈ M, n ∈ N, then Rk fn ↑ Rk f . 2 . If 0  fn ∈ M, then   ∞ ∞   fn = fn . Rk n=1

Rk

n=1

P ROOF.–

1) If Rk g = +∞ or Rk f = −∞, the inequality is obvious. Therefore, consider the case where   −∞ < f and g < +∞. Rk

Rk

Then Rk g+ < +∞ and Rk f − < +∞. Since 0  f +  g+ and 0  g−  f − , it follows that   + 0 f  g+ < +∞ Rk

Rk

and  0

Rk

g− 

 Rk

f − < +∞.

Therefore, f, g ∈ L(I), for which the inequality is already proved in proposition 8.1, item 3.

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127

2) The case where limn→∞ Rk fn < +∞ is the B. Levi theorem. Therefore, it remains to check the proposition when

limn→∞ Rk fn = +∞. In this case, it suffices to pass to the limit in the inequality   fn  f Rk

to get that

Rk

Rk

f = +∞.

2 ) It suffices to apply the property just proved to the partial-sum  sequence of the series ∞   n=1 fn . D EFINITION 8.6.– A set A ⊂ Rk is said to be measurable (in the Lebesgue sense) if 1IA ∈ M. Its Lebesgue measure is the number  m(A) := 1IA . Rk

R EMARK 8.6.– In the cases of bounded k-box and zero-measure set, this definition is compatible with the previous definitions (definitions 7.2 and 7.3):

– for a bounded k-box I, 1II ∈ S (Rk ) and Rk 1II = 1 · m(I) = m(I);

– for N ∈ N, 1IN = 0 a.e., and thus, Rk 1IN = 0. P ROPOSITION 8.6.– 1) Let A and B be measurable sets. Then A ∩ B, A ∪ B, A \ B, Ac = Rk ⊂ A are also measurable sets. If, in addition, A ∩ B = ∅, then m(A ∪ B) = m(A) + m(B) (additivity of measure); if A ⊂ B, then m(A)  m(B) (monotonicity of measure). 2) If {An , n ∈ N} is a sequence of measurable sets, then the union ∞ ∞ n=1 An and intersection n=1 An are measurable sets. If, in addition, Ai ∩ A j = ∅ for all i  j, then ∞ ∞    m An = m(An ) n=1

n=1

(σ-additivity of measure). 3) Let {An , n ∈ N} be a sequence of measurable sets.

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Integral and Measure

a) If An ⊂ An+1 , n ∈ N, then m

∞ 

 An = lim ↑ m(An ); n→∞

n=1

b) if An ⊃ An+1 , n ∈ N, and m(A1 ) < +∞, then m

∞ 2



An = lim ↓ m(An ). n→∞

n=1

R EMARKS 8.1.– 1) The following notation is often used: An ↑ A : ⇐⇒ An ⊂ An+1 , n ∈ N,

and

A=

∞

An ;

n=1

An ↓ A : ⇐⇒ An ⊃ An+1 , n ∈ N,

and

A=

∞ 2

An .

n=1

Therefore, properties 3a,b can be written in short as follows: An ↑ A =⇒ m(An ) ↑ m(A); An ↓ A, m(A1 ) < +∞ =⇒ m(An ) ↓ m(A). From this it is clear why the property is called the continuity of measure with respect to monotone sequences. 2) The condition m(A1 ) < +∞ (clearly, it suffices to require that m(An0 ) < +∞ for some n0 ∈ N) is essential. Consider the example An = [n, +∞), n ∈ N (k = 1). Then m(An ) = +∞, n ∈ N, but m

∞ 2 n=1

 An = m(∅) = 0.

Lebesgue Integral

129

P ROOF.– 1) If A and B are measurable sets, then by proposition 8.3, item 2, 1IA , 1I B ∈ M =⇒ 1IA∪B = max{1IA , 1IB } ∈ M, 1IA∩B = min{1IA , 1IB } ∈ M, 1IAc = 1 − 1IA ∈ M, and, thus, A ∪ B, A ∩ B, and Ac are measurable sets. From this we also have that A \ B = A ∩ Bc is a measurable set. Suppose, in addition, that A ∩ B = ∅. Then 1IA∪B = 1IA + 1IB , and thus,  m(A ∪ B) =

 Rk

1IA∪B =

 Rk

1IA +

Rk

1I B = m(A) + m(B).

2) If An , n ∈ N, are measurable sets, then, by the previous statement N n=1 An , N ∈ N, also are measurable sets. Therefore (proposition 8.5, item 2), 1I∞n=1 An = lim ↑ 1IN N→∞

n=1 An

∈M

(check the equality), and thus the set ∪∞ n=1 An is measurable. The measurability of an intersection of measurable sets is checked similarly. If, in addition, Ai ∩ A j = ∅ for i  j, then 1I∞

n=1 An

=

∞ 

1IAn

n=1

(check again!), and, therefore, by proposition 8.5, item 2 , ∞ ∞  ∞         m An = 1I ∞n=1 An = 1IAn = m(An ). n=1

Rk

n=1

Rk

n=1

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Integral and Measure

3a) If A = lim ↑ An , then 1IA = lim ↑ 1IAn , and, therefore, by n→∞

n→∞

proposition 8.5, item 2,   m(A) = 1IA = lim 1IAn = lim ↑ m(An ). Rk

n→∞

Rk

n→∞

3b) The proof is similar by applying the B. Levi theorem (theorem 8.1) to the decreasing sequence 1IAn ↓ 1IA and noticing that

1I = m(A1 ) < +∞.   Rk A1 P ROPOSITION 8.7.– All open and all closed sets are measurable. P ROOF.– Since closed sets are complements of open sets, it suffices to prove the proposition for open sets. To simplify the notation, we only consider the case k = 2. So, let G ⊂ R2 be an arbitrary open set. For any x ∈ G, take a neighborhood U(x) = Uε x (x) ⊂ G and then a rectangle A(x) ⊂ U(x) (for example, A(x) = (x1 − √ε , x1 + √ε ) × (x2 − √ε , x2 + √ε ), 2 2 2 2 x = (x1 , x2 )). For every rectangle A(x), we can choose another rectangle ˆ = (a, b) × (c, d) ⊂ A(x) such that a, b, c, d are rational numbers and A(x) ˆ x ∈ A(x). Clearly, ˆ = G. A(x) x∈G

ˆ Since the family of different rectangles A(x), x ∈ G, is at most countable (finite or countable), the set G, as the union of at most countable family system of rectangles, which are measurable sets, is a measurable set (proposition 8.6, item 2).   D EFINITION 8.7.– The integral of a measurable function f : Rk → R in a measurable set A ⊂ Rk is the number   f := f 1IA , A

Rk

provided that the latter integral is defined. In case the integral A f is finite, we shall write f ∈ L(A) and say that the function f is integrable in a set A.

Lebesgue Integral

131

P ROPOSITION 8.8.– 1) If f ∈ L(A1 ) ∩ L(A2 ), A1 ∩ A2 = ∅, then f ∈ L(A1 ∪ A2 ), and    f= f+ f A1 ∪A2

A1

A2

(additivity of the integral).  If f ∈ L(A), A = ∞ n=1 An , Ai ∩ A j = ∅ for all i  j, where An all are measurable sets, then  ∞   f= f A

An

n=1

(σ-additivity of the integral). 2) If An , n ∈ N, are measurable sets, An ↑ A, and f ∈ M, then   f→ f, An

A

provided that at least one of the following conditions is satisfied: a) f ∈ L(A);

b) lim ↑ A | f | < +∞; n→∞

n

c) f  0. 3) If An , n ∈ N, are measurable sets, An ↓ A, and f ∈ L(A1 ), then   f→ f. An

A

P ROOF.– 1) If A1 ∩ A2 = ∅, then 1IA1 ∪A2 = 1IA1 + 1IA2 , and therefore,       f= f 1IA1 ∪A2 = f 1IA1 + f 1IA2 = f+ A1 ∪A2

Rk

Rk

Rk

A1

If f ∈ L(A), then also | f | ∈ L(A). For all N ∈ N, we have  N  N     f 1 I = | f | 1IAn = | f |1IN An  | f |1IA ∈ L.  An  n=1  n=1

n=1

f. A2

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Integral and Measure

However, N 

f 1IAn → f 1IA ,

N → ∞.

n=1

Therefore, we can apply the Fatou–Lebesgue theorem to the sequence N fN := f n=1 1IAn , N ∈ N: 

 f = lim

N→∞

A

=

Rk

∞   n=1

fN = lim

N→∞

  N Rk n=1

f 1IAn = lim

N→∞

N   n=1

Rk

f 1IAn

f. An

2a) Since 

   f 1IAn   | f |1IA ,

n ∈ N,

 Rk

| f |1IA =

| f | < +∞, A

and f 1IAn → f 1IA ,

n → ∞,

then, again by applying the Fatou–Lebesgue theorem, we get     f= f 1IAn → f 1IA = f. Rk

An

Rk

A

2b) It suffices to notice that condition (b) is equivalent to condition (a). Indeed, by proposition 8.5, item 2,   lim ↑ | f | = | f | < +∞ =⇒ f ∈ L(A). n→∞

2c)

An

A

f  0 =⇒ f 1IAn ↑ f 1IA

Lebesgue Integral





=⇒

f= An

 Rk

f 1IAn ↑

Rk

133

 f 1IA =

f. A

 

3) Similar by applying the Levi theorem.

C OROLLARY 8.1.– Let f ∈ D[0, ∞). Then f ∈ L[0, ∞) if and only if the ∞ improper integral 0 f (x) dx converges absolutely, and in such a case, the latter coincides with the Lebesgue integral.6 P.8. Problems P ROBLEM 8.1.– Is the function f (x) = 1, x ∈ R, a step function? measurable? integrable? quasi-integrable? P ROBLEM 8.2.– Is the function f (x) = x, x ∈ R, measurable? integrable? quasi-integrable? P ROBLEM 8.3.– Is the function f (x) = x2 , x ∈ R, measurable? integrable? quasi-integrable? P ROBLEM 8.4.– Find the Lebesgue integrals 

∞

e

−[x]

 dx;

0

0

∞

1 dx; [x + 1][x + 2]



∞ 0

1 dx. [x]!

Here [x] is the integer part of a number x. P ROBLEM 8.5.– Prove that a measurable function f : A → [0, +∞) is integrable in a measurable set A ⊂ Rk if and only if 1)

∞

2)

∞

n=1 m{x

∈ A : f (x)  n} < +∞;

n=1 n m{x

∈ A : n  f (x) < n + 1} < +∞.

6 In particular, this means that there are no relatively converging Lebesgue integrals.

134

Integral and Measure

k P ROBLEM 8.6.– Let

f ∈ L(A), A ⊂ R , m(A) < +∞. Prove that the Lebesgue integral A f (x) dx can be expressed as the limit

 f (x) dx = lim

n→∞

A

k k +1! m x ∈ A :  f (x)  . n n n

k k

P ROBLEM 8.7.– Let { fn } be a sequence of non-negative functions and suppose that fn (x) → f (x), x ∈ I. Show that     f  lim inf fn , but not necessarily f = lim inf fn . n

I

I

I

n

I

P ROBLEM 8.8.– Give an example of a sequence { fn } ⊂ L(I) converging to a function f ∈ L(I) such that   lim fn = f, n

I

I

but g := supn | fn |  L(I), that is, the sequence { fn } has no dominating integrable function. P ROBLEM 8.9.– Suppose that { fn } ⊂ C[0, 1] (or { fn } ⊂ S [0, 1]), 0  fn (x)  1, x ∈ [0, 1], n ∈ N, and fn → 0. Then, by the

1 Fatou–Lebesgue theorem we immediately obtain that 0 fn (x) dx → 0. Prove this without using the Lebesgue integration theory. Remark. Undoubtedly, the reader will meet serious challenges while solving this problem. We included this problem for comparison of possibilities of the Riemann and Lebesgue integration theories. P ROBLEM 8.10.– Suppose that 0  fn → f , n → ∞, in [a, b] and

b

b f → c. Show that a f ∈ [0, c] and that all the values in the interval a n [0, c] are possible.

P ROBLEM 8.11.– Suppose that fn ∈ L(A), 0  fn → f , and A fn → A f ,

n → ∞. Prove that A | fn − f | → 0. Hint. Use the inequality ( fn − f )−  f and apply the Fatou–Lebesgue theorem.

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135

P ROBLEM 8.12.– Suppose that

| fn |  gn , fn , gn ∈ L(A), fn → f , gn → g, and A gn → A g. Prove that A fn → A f . P ROBLEM 8.13.– Let f : R → R be a measurable function. Prove that there exists a strictly positive function g such that the product f g ∈ L(R). P ROBLEM 8.14.– Let f : A → R be a bounded function on a measurable set A ⊂ Rk such that there exist constants C > 0 and α < 1 such that   " # C m x ∈ A :  f (x) > ε  α , ε

ε > 0.

Prove that f ∈ L(A). P ROBLEM 8.15.– Let f : A → R be a function on a bounded measurable set A ⊂ Rk such that there exist constants C > 0 and α > 1 such that " m x∈A:

  # C  f (x) > λ  α , λ

λ > 0.

Prove that f ∈ L(A). P ROBLEM 8.16.– Let a function f : A → R be bounded and integrable on A. Must the following functions be integrable: | f |, | f |2 , | f |−1 , 1+|f f | ? P ROBLEM 8.17.– Explain why functions f ∈ D[a, b] are sometimes called uniformly measurable functions. P ROBLEM 8.18.– The norm (or, more precisely, the L1 -norm) of a function f ∈ L(I) is defined as    || f || :=  f (x) dx. I

Check the following properties of the norm: 1)  f   0;  f  = 0 ⇐⇒ f = 0 a.e.; 2) α f  = |α| f , α ∈ R, f ∈ L(I); 3)  f + g   f  + g, f, g ∈ L(I). P ROBLEM 8.19.– A sequence { fn } ⊂ L(I) is said to converge to a function f in the space L(I) (or in the L1 sense) if  fn − f  → 0, n → ∞.

136

Integral and Measure

A sequence { fn } ⊂ L(I) is said to be a Cauchy (or fundamental) sequence in the space L(I) (or in the L1 sense) if  fn − fm  → 0, n, m → ∞, that is, if ∀ε > 0, ∃ N ∈ N :  fn − fm  < ε for all n, m > N. Prove that the space L(I) is complete, that is, that every Cauchy sequence { fn } ⊂ L(I) converges in L(I) to some function from L(I). Hint. See the proof of correctness of the definition of the Lebesgue integral (page 106). P ROBLEM 8.20.– Prove that for every function f ∈ L[a, b], there exists a sequence { fn } ⊂ C[a, b] that converges to f in L[a, b]. Hint. First, construct a sequence {ϕn } ⊂ S [a, b] converging to f in L[a, b]. Then approximate each function ϕn by a continuous function in the L1 sense. k P ROBLEM 8.21.– A function f : I → R (where

I ⊂ R is a measurable set) is called a square-integrable function if I f 2 (x) dx < +∞. The set of all such functions is denoted L2 (I). The norm (or, more precisely, the L2 -norm) of a function f ∈ L2 (I) is defined as

 f 2 :=



1/2 f 2 (x) dx . I

Check the following properties of the L2 -norm: 1)  f 2  0;  f 2 = 0 ⇔ f = 0 b.v.; 2) α f 2 = α f 2 , α ∈ R, f ∈ L2 (I); 3) (Cauchy–Schwarz inequality)     f (x)g(x) dx   f  g , f, g ∈ L2 (I); 2 2   I

4)  f + g2   f 2 + g2 , f, g ∈ L2 (I). P ROBLEM 8.22.– Prove that if m(I) < +∞, then L(I) ⊂ L2 (I). P ROBLEM 8.23.– Give an example of a function f ∈ L2 [0, 1] \ L[0, 1].

Lebesgue Integral

137

P ROBLEM 8.24.– Give an example of a function f ∈ L(R) \ L2 (R). P ROBLEM 8.25.– Let f : R → R. Prove that if the sets {x ∈ R : f (x) < a} are measurable for all a ∈ R, then the function f is measurable. P ROBLEM 8.26.– Prove that for every function f ∈ C[a, b],  b  1/p    f (x) p dx lim = max  f (x).

p→∞

x∈[a,b]

a

Hint. First, check this relation for step functions. P ROBLEM 8.27.– Find the minimal value of the product 

b a



b

f (x) dx · a

1 dx, f (x)

where f is a positive function. For which functions is the minimum achieved? Hint. Apply the Cauchy–Schwarz inequality. P ROBLEM 8.28.– For a measurable set A ⊂ [0, 1], denote f (x) := m(A ∩ [0, x]), x ∈ [0, 1]. Prove that f ∈ C[0, 1].

9 Fubini and Change-of-Variables Theorems

In the multidimensional case, there is no analogue of the Newton–Leibnitz theorem. Happily, calculation of multidimensional integrals can be reduced to that of one-dimensional integrals. T HEOREM 9.1 (Fubini theorem).– If f ∈ L(I1 × I2 ) (where I1 , I2 ⊂ R are intervals), then      1 1 f= f (x, y) dy dx = f (x, y) dx dy. I1

I1 ×I2

I2

I2

I1

Here the first (say) equality is understood as follows. The integral  Φ(x) := f (x, y) dy I2

exists for almost all x ∈ I1 , that is for all x ∈ I1 \ N, m(N) = 0. Defining Φ arbitrarily on N (e.g. Φ(x) = 0, x ∈ N), we have 

 f=

I1 ×I2

Φ(x) dx. I1

R EMARK 9.1.– 1) Naturally, a more general statement is true when I1 and I2 are boxes of higher dimensions:

140

Integral and Measure

If f ∈ L(I1 × I2 ), where I1 ⊂ Rk and I2 ⊂ Rm are k- and m-boxes, then  f =

  I1

I1 ×I2

=

1 f (x1 , . . . , xk , y1 , . . . , ym ) dy1 · · · dym dx1 · · · dxk I2

  I2

1 f (x1 , . . . , xk , y1 , . . . , ym ) dx1 · · · dxk dy1 · · · dym . I1

2) If D = {(x, y) : y1 (x)  y  y2 (x), a  x  b} ⊂ R2 is a measurable curvilinear trapezium, then, for all f ∈ L(D),  f (x, y) dx dy =

 b 

D

a

y2 (x)

1 f (x, y) dy dx.

y1 (x)

Indeed, 

 f = D

= = =

R2

f 1ID =

  R

R

  R

a

R

R

1 f (x, y)1ID (x, y) dy dx

1 f (x, y)1I[a,b] (x)1I[y1 (x),y2 (x)] (y) dy dx 

1I[a,b] (x)

 b 

 

y2 (x)

R

1 1I[y1 (x),y2 (x)] (y) f (x, y) dy dx

1 f (x, y) dy dx.

y1 (x)

In particular, taking f ≡ 1, we get m(D) =

 b  a

y2 (x) y1 (x)

 b 1  1 dy dx = y2 (x) − y1 (x) dx. a

This is the same formula we obtained while defining the area of a curvilinear trapezium. Similarly, if a curvilinear cylinder " # V = (x, y, z) ∈ R3 : z1 (x, y)  z  z2 (x, y), (x, y) ∈ D ⊂ R3

Fubini and Change-of-Variables Theorems

is a measurable set, then     f= V

D

141

1 f (x, y, z) dz dx dy.

z2 (x,y) z1 (x,y)

P ROOF.– [of the Fubini theorem] F denotes the set of functions f ∈ L(I1 × I2 ) for which the first equality of the Fubini theorem holds. We have to show that F coincides with L(I1 × I2 ). Step 1. We first check that the indicators of rectangles belong to F . Let f = 1IA×B , where A ⊂ I1 and B ⊂ I2 are intervals. Then the left-hand side of the equality is  f = 1 · m2 (A × B) = m(A) · m(B), I1 ×I2

where m2 and m are the Lebesgue measures on the plane R2 and the line R, respectively. The right-hand side of the equality is   I1

  1 1 f (x, y) dy dx = 1IA×B (x, y) dy dx I2

= =

I1

I2

I1

I2

   

1 1IA (x)1IB (y) dy dx 

 1 1I B (y) dy dx = 1IA (x)m(B) dx

1IA (x) I1

I2



= m(B)

I1

1IA (x) dx = m(B) · m(A). I1

Actually, in this simple case, the Fubini theorem reflects the fact that the area of a rectangle equals the product of the lengths of its sides. Step 2. Using the linearity of the integrals (one- and two-dimensional), we can easily check that F is a linear space, that is, from f, g ∈ F and α, β ∈ R it follows that α f + βg ∈ F :    (α f + βg) = α f +β g I1 ×I2

I1 ×I2

I1 ×I2

142

Integral and Measure

 

=α

I1

  1 1 f dy dx + β g dy dx I2

I1

I2

    1 = α f dy + β f dy dx I1

=

I2

  I1

I2

1 (α f + βg) dy dx.

I2

Since every step function is a linear combination of the indicators of rectangles, we get, in particular, that S (I1 × I2 ) ⊂ F . Step 3. Now let us show the following property of the set F : If F  fn ↑ f (everywhere) and lim ↑

fn n→∞ I1 ×I2

< +∞, then f ∈ F .

Denote  F n (x) :=

fn (x, y) dy,

x ∈ I1 , and

I2

F(x) := lim ↑ Fn (x), n→∞

x ∈ I1 .

Then, applying the B. Levi theorem to the sequence { fn }, we get that    F n (x) dx = fn ↑ f < +∞. I1

I1 ×I2

I1 ×I2

Applying the same theorem to the sequence {Fn }, we get that F ∈ L(I1 ) and   F n (x) dx ↑ F(x) dx. I1

I1

Thus, we have   f= F(x) dx. I1 ×I2

I1

Fubini and Change-of-Variables Theorems

143

Therefore, it remains to check that F(x) = I f (x, y) dy for almost all 2 x ∈ I1 . Since fn (x, y) ↑ f (x, y) for all (x, y) ∈ I1 × I2 , by applying the B. Levi theorem once more (to the sequence { fn (x, ·)} with any fixed

x ∈ I1 ), we get that Fn (x) = I fn (x, y) dy ↑ I f (x, y) dy if 2

2 F(x) = limn→∞ Fn (x) < +∞. So, we have that F(x) = I f (x, y) dy if 2 F(x) < +∞. It remains to recall that F ∈ L(I1 ) and, therefore, F < +∞ almost everywhere (lemma 8.1, item 3) Thus, f ∈ F . Step 4. Let us show that f = 0 a.e. =⇒ f ∈ F .

If f = 0 a.e., then I ×I f = 0. So, we have to verify that the iterated 1 2 integral also equals 0.1 We consider several particular cases. 1) Suppose that 0  f  1 (and f = 0 a.e.). Denote " # N := (x, y) ∈ R2 : f (x, y)  0 ∈ N. By the definition of a zero-measure set, for every m ∈ N, there exists a sequence of rectangles {Ikm , k ∈ N} such that ∞

Ikm

⊃N

and

k=1

∞ 

m2 (Ikm ) <

k=1

1 . m

Denote ϕm n

=

n 

1IIkm ∈ S (I1 × I2 ).

k=1

The functions ϕm n possess the following properties: ∞  m i) lim ϕn (x, y) = 1IInm (x, y)  1IN (x, y)  f (x, y), (x, y) ∈ I1 × I2 , n→∞

m ∈ N;

n=1

1 This is not as obvious as it may seem at first sight.

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Integral and Measure

 ϕm n

ii)



∞  

1I

Ikm

k=1I ×I 1 2

I1 ×I2

=

∞ 

m2 (Ikm ) <

k=1

1 , m ∈ N, n ∈ N; m

iii) ϕm := lim ↑ ϕm n ∈ F (by step 3). n→∞

Without loss of generality, we may assume that for every n ∈ N, the m+1 m sequence {ϕm n , m ∈ N} is decreasing: ϕn (x, y)  ϕn (x, y). Otherwise, we could sequentially (for m = 1, 2, . . .) replace the functions ϕm+1 by the n m }, preserving, at the same time, properties (i)– functions min{ϕm+1 , ϕ n n (iii). In such a case, for all (x, y) ∈ I1 × I2 , there exists the limit h(x, y) = limm→∞ ϕm (x, y). From (i)–(iii) we get that h possesses the following corresponding properties: i ) h(x, y)  f (x, y), (x, y) ∈ I1 × I2 ;

ii ) I ×I h = lim ↓ ϕm = 0; 1

2

m→∞ I ×I 1 2

iii ) h ∈ F .2 Denote H(x) :=  H(x) dx = I1

I2

h(x, y) dy, x ∈ I1 . Since

  I1

 1 h(x, y) dy dx = h = 0, I2

I1 ×I2

we have H(x) = 0 for almost all x ∈ I1 ; for all those x, h(x, y) = 0 for almost all y ∈ I2 , and thus, for all those x ∈ I1 also f (x, y) = 0 for almost all y ∈ I2 . Therefore, I f (x, y) dy = 0 for almost all x ∈ I1 ; thus, 2

  I1

1 f (x, y) dy dx = 0. I2

2) Now suppose that f  0 (and f = 0 a.e.). Then from case (1) we have that min{1, n−1 f } ∈ F .

2 By step 3, with increasing sequences replaced by decreasing ones.

Fubini and Change-of-Variables Theorems

145

Therefore, by step 2, fn := min{n, f } = n min{1, n−1 f } ∈ F . Since fn ↑ f and limn→∞

f I1 ×I2 n

= 0 < +∞, by step 6 we have f ∈ F .

3) General case: since f + , f − ∈ F (case (2)), by step 2 we get that f = f+− f− ∈ F. Step 5. Suppose that f ∈ L+ = L+ (I1 × I2 ). By the definition of class L , there exists {ϕn } ⊂ S (I1 × I2 ) ⊂ F such that ϕn ↑ f a.e.

a sequence

and limn→∞ I ×I ϕn = I ×I f < +∞. Denote +

1

2

1

2

fˆ := lim ↑ ϕn ∈ F . n→∞

Then fˆ = f a.e. Denoting f˜ := f − fˆ, we have that f˜ = 0 a.e. By step 4, ϕ ∈ F . Therefore, by step 2, f = fˆ + f˜ ∈ F . Step 6. If f ∈ L(I1 × I2 ), then f can be expressed by the difference f = h − g with h, g ∈ L+ ⊂ F . Therefore, by step 2, f ∈ F .   T HEOREM 9.2 (Tonelli theorem).– If f ∈ M(I1 × I2 ) (with I1 , I2 intervals) and f  0, then    1 f= f (x, y) dy dx ( +∞). I1 ×I2

I1

I2

P ROOF .– If f ∈ L(I1 × I2 ), then this is the Fubini theorem. Therefore, let f  L(I1 × I2 ), that is,  f = +∞. I1 ×I2

Denote fn = min{n, f }1I[−n,n]2 .

146

Integral and Measure

Then fn ↑ f , and therefore, by proposition 8.5, item 2 (p. 126),   fn ↑ f = +∞. [9.1] I1 ×I2

I1 ×I2

Since 

 fn 

I1 ×I2

n1I[−n,n]2  n(2n)2 < +∞,

n ∈ N,

I1 ×I2

we have that fn ∈ L(I1 × I2 ) and, by the Fubini theorem,      1 1 fn = fn (x, y) dy dx  f (x, y) dy dx. I1

I1 ×I2

I2

I1

I2

The last integral in the inequality is well defined since, again by proposition 8.5, item 2,   Φ(x) := f (x, y) dy = lim ↑ fn (x, y) dy, x ∈ I1 , I2

n→∞

I2

and therefore, the integrand function Φ is a non-negative measurable function as the limit of a sequence of non-negative integrable (and, thus, measurable) functions. Taking the limit in the inequality, in view of [9.1], we get that   1 f (x, y) dy dx = +∞.   I1

I2

C OROLLARY 9.1.– If f ∈ M(I1 × I2 ) and    1  f (x, y) dy dx < +∞, I1

I2

then f ∈ L(I1 × I2 ) (and, thus, for f , the Fubini theorem is applicable). P ROOF.– By the Tonelli theorem,  | f | < +∞ =⇒ | f | ∈ L(I1 × I2 ) =⇒ f ∈ L(I1 × I2 ). I1 ×I2

 

Fubini and Change-of-Variables Theorems

147

R EMARK 9.2.– 1) If f (x, y) = g(x)h(y), g ∈ L(I1 ), h ∈ L(I2 ),3 then         f = g(x)h(y) dy dx = g(x) h(y) dy dx I1 ×I2   I1  I2 I1 I2 = h(y) dy · g(x) dx = g· h. I2

For example,  R2

I1

 dx dy = (1 + x2 )(1 + y2 )

I1

I2



dx 2 = π2 . 2 1 + x R

2) Let f (x, y) :=

x2 − y2 , (x2 + y2 )2

(x, y) ∈ [0, 1]2 .

Then  1  0

1

0

Similarly,  1  1 0

0

 1  1 2 2   x −y f (x, y) dy dx = dy dx (x2 + y2 )2 0 0  1  1  y   = dy dx x2 + y2 y 0 0  1  1 y 1 dx π = dx = = . 2 + y2 y=0 2 +1 4 x x 0 0

 π f (x, y) dx dy = − . 4

We see that iterated integrals with a different integration order may be different. From the Fubini theorem we see that this is possible only in the case where the function f is not integrable (in this case, f  L([0, 1]2 ).

3 The function f is called the tensor product of g and h and is denoted by g ⊗ h.

148

Integral and Measure

D EFINITION 9.1.– The Jacobian of a differentiable function f : D → Rk (where D ⊂ Rk is an open set) at a point x ∈ D is the number      J f (x) := det f (x) =    

∂ f1 ∂x1 (x) ∂ f2 ∂x1 (x)

.. . ∂ fk ∂x1 (x)

∂ f1 ∂x2 (x) ∂ f2 ∂x2 (x)

... ... .. .

.. . ∂ fk ∂x2 (x) . . .

∂ f1 ∂xk (x) ∂ f2 ∂xk (x)

.. . ∂ fk ∂xk (x)

     .  

In view of the coordinate expression of f ,  f (x) = f1 (x1 , x2 , . . . , xk ), . . . , fk (x1 , x2 , . . . , xk ) , the Jacobian is denoted as J f (x) =

D( f1 , f2 , . . . , fk ) . D(x1 , x2 , . . . , xk )

P ROPOSITION 9.1.– If f : D → D˜ and g: D˜ → Rk (where D, D˜ ⊂ Rk are open sets) are differentiable functions, then  Jg◦ f (x) = Jg f (x) · J f (x),

x ∈ D.

P ROOF.– It follows from the equality  (g ◦ f ) (x) = g f (x) ◦ f  (x) and the property of determinants det(AB) = det A · det B.

 

T HEOREM 9.3 (Change of variables in multivariate integrals).– If  (where D, D  ⊂ Rk are open sets) is a bijection (on-to-one g ∈ C 1 (D, D)  correspondence) and Jg (x)  0, x ∈ D, then, for all f ∈ L(D), 

  D

   f g(x)  Jg (x) dx

f (y) dy = D

Fubini and Change-of-Variables Theorems

149

or  f (y1 , . . . , yk ) dy1 · · · dyk      D(g1 , g2 , . . . , gk )   dx1 · · · dxk . = f g1 (x), . . . , gk (x)  D(x1 , x2 , . . . , xk )  D

 D

P ROOF.– To simplify the notation, we restrict ourselves to the  is a two-dimensional case (k = 2). So, we will prove that if g: D → D 2  ⊂ R are open sets), continuously differentiable bijection (where D, D  = L(g(D)), Jg (x)  0, x ∈ D, then, for all f ∈ L(D)  f (y1 , y2 ) dy1 dy2 = g(D)      D(g1 , g2 )   dx1 dx2 , f g1 (x1 , x2 ), g2 (x1 , x2 )  D(x1 , x2 )  D or, for short, 

 f=

g(D)

f ◦ g|Jg |. D

We divide the proof into several steps. Step 1. First, note that if the change-of-variables theorem (CVT) holds for a set D, then it also holds for every measurable subset D1 ⊂ D:     f = f · 1Ig(D1 ) = f · 1Ig(D1 ) ◦ g |Jg | g(D1 ) D  g(D)  = f ◦ g · 1Ig(D1 ) ◦ g · |Jg | = f ◦ g · 1ID1 |Jg | D  D = f ◦ g|Jg |. D1

Step 2. The CVT holds when D is an open rectangle (a, b)× (c, d) and the function g does not change one coordinate:   g(x1 , x2 ) = x1 , g2 (x1 , x2 ) or g(x1 , x2 ) = g1 (x1 , x2 ), x2 .

150

Integral and Measure

Consider, for example, the first case. Thus, suppose that  g(x1 , x2 ) = x1 , g2 (x1 , x2 ) ,

x = (x1 , x2 ) ∈ D = (a, b) × (c, d).

Then   1 Jg (x) =  ∂g2  ∂x (x) 1

 0  ∂g2 (x)  0, ∂g2  = ∂x2 ∂x2 (x)

From the continuity of g it is clear that

x ∈ D. ∂g2 ∂x2

has a constant sign in the

2 rectangle D. Let, for example, ∂g ∂x2 (x) > 0, x ∈ D, that increasing function with respect to x2 . Therefore,

is, g2 (x1 , x2 ) be an

" # g(D) = (y1 , y2 ) : a < y1 < b, g2 (y1 , c) < y2 < g2 (y1 , d) , and  f (y1 , y2 ) dy1 dy2  b  g2 (y1 ,d) FT === dy1 f (y1 , y2 ) dy2  ab  gd2 (y1 ,c)  ∂g2 = dy1 f y1 , g2 (y1 , x2 ) (y1 , x2 ) dx2 ∂x2 a b c d    = dx1 f x1 , g2 (x1 , x2 )  Jg (x1 , x2 ) dx2 c  a = f ◦ g|Jg |. g(D)

D

Step 3. The CVT holds for the function ϕ = g◦h if it holds for g and h (with appropriate domain of definition and range). Applying proposition 9.1, we have:  ϕ(D)





f =

f= f ◦ g |Jg | h(D)  g(h(D))  = f ◦ g ◦ h |Jg ◦ h||Jh | = f ◦ ϕ |Jϕ |. D

D

Fubini and Change-of-Variables Theorems

151

Step 4. We show that for every point x0 ∈ D, there is its rectangular neighborhood (an open rectangle such that the point x0 belongs to it) where the CVT holds. Since  ∂g  1 0 ) ∂g1 (x0 )   (x ∂x ∂x    0, 2 Jg (x0 ) =  ∂g12 ∂g2 0  0  ∂x (x ) ∂x (x )  1

2

∂g1 0 ∂g2 0 1 0 ∂g2 0 at least one of the products ∂g ∂x1 (x ) · ∂x2 (x ) and ∂x2 (x ) · ∂x1 (x ) is not zero. Consider, for example, the first product. Denote

 h(x) := g1 (x), x2 ,

x = (x1 , x2 ) ∈ D.

Then

  ∂g1 (x0 ) 0 Jh (x ) =  ∂x1  0



∂g1 0  ∂x2 (x ) 

1

∂g1 0  = ∂x1 (x )  0.

By the inverse function theorem,4 there exists a neighborhood V of x such that the function h: V → W = h(V) has a continuously differentiable inverse function h−1 : W → V. Denote 0

 k(t) = k(t1 , t2 ) := t1 , g2 ◦ h−1 (t) ,

t = (t1 , t2 ) ∈ W.

Then    k ◦ h(x) = h1 (x), g2 ◦ h−1 h(x) = g1 (x), g2 (x) = g(x),

x ∈ V.

Diminishing, if necessary, the sets V and W, we may assume that W is a subset of a rectangle where the CVT holds. It remains to apply steps 2 and 3. Step 5. We are now able to prove the theorem in the general case. For every point x ∈ D, let us take an open rectangle V(x)  x in which

4 Theorem. If the Jacobian of a function h ∈ C 1 (D, Rk ) (where D ⊂ Rk is an open set) does not vanish at a point x0 , then there exist two open sets V  x0 and W in Rk such that f : V → W is a bijection, the inverse function g := f −1 ∈ C 1 (W, V), and g (y) = [ f  (g(y))]−1 , y ∈ W.

152

Integral and Measure

the CVT holds. In view of step 1, we may assume that all vertices of all the rectangles have rational coordinates. The set of all such (different)  rectangles is countable. Enumerate them V1 , V2 , . . . . Clearly, ∞ V =  n−1 ∞ n=1 n V(x) = D. Denote U := V \ V , n ∈ N. Then U =D n n i n x∈D i=1 n=1 and Ui ∩ U j = ∅ for i  j. From step 1, the CVT holds in every Un . It remains to use the σ-additivity of the integral (theorem 8.6, item 1):  f= g(D)

∞   n=1

f=

g(Un )

∞   n=1

 f ◦ g|Jg | =

Un

f ◦ g|Jg |. D

 

E XAMPLES 9.1.– 1) Polar change of variables: recall that a pair of numbers (r, ϕ) is called the polar coordinates of a point (x, y) ∈ R2 if x = r cos ϕ, y = r sin ϕ.  ⊂ R2 so that D˜ would fill the plane R2 Let us try to choose open sets D, D as much as possible and the function g

 D  (r, ϕ) → (x, y) ∈ D would be a bijection with Jacobian non-vanishing in D. A possible version is " # D = (r, ϕ) : r > 0, ϕ ∈ (0, 2π) ,

 = R2 ⊂ [0, ∞) × {0} . D

Let us calculate the Jacobian of the function g:    ∂x ∂x   D(x, y)  ∂r ∂ϕ   cos ϕ −r sin ϕ  Jg (r, ϕ) = = ∂y  =  sin ϕ r cos ϕ  D(r, ϕ)  ∂y ∂r ∂ϕ  = r(cos2 ϕ + sin2 ϕ) = r. So, by the CVT, we have   f (x, y) dx dy = f (r cos ϕ, r sin ϕ) r dr dϕ.  D

D

Since   = 0 m R2 \ D

and

 m [0, +∞) × [0, 2π] \ D = 0,

Fubini and Change-of-Variables Theorems

we can write   f (x, y) dx dy = R2

2π  +∞ 0

153

f (r cos ϕ, r sin ϕ) r dr dϕ.

0

Emphasize that the CVT is often applied taking no care that the bijectivity of the function g is violated on a zero-measure set. The latter example shows that such an application can be easily justified. By applying the last formula we will calculate the following integral:  I= 1 2

=



R2

e

−(x2 +y2 )



2π  +∞

dx dy =

e 0

2π

−r2



2π 

r dr dϕ =

0

0

1 2 +∞ − e−r  dϕ r=0 2

dϕ = π.

0

However, by the Fubini theorem, 

+∞   +∞

 +∞    +∞ 2  −x2 I= e e dy dx = e e−y dy dx −∞ −∞ −∞ −∞  +∞  +∞   +∞ 2 2 2 2 = e−y dy · e−x dx = e−x dx . −x2 −y2

−∞

−∞

−∞

From this we find the important integral 

+∞ −∞

e−x dx = 2

√ π.

Similarly, the polar change of variables in the circle CR with center at the origin and radius R yields the equality 



2π  R

f (x, y) dx dy = CR

0

f (r cos ϕ, r sin ϕ)r dr dϕ.

0

For example, 



2π  R

(x + y ) dx dy = 2

CR

2 p

0

0



2π  R

(r ) r dr dϕ = 2 p

0

0

r2p+1 dr dϕ

154

Integral and Measure

=

R2p+2 2p + 2

 0

2π

dϕ =

R2p+2 π p+1

(p > −1).

2) Spherical change of variables: let us change the variables (x, y, z) in the triple integral  f (x, y, z) dx dy dz, BR

where BR is the ball with center at the origin (0, 0, 0) and radius R, by the variables (r, ϕ, θ) as follows: ⎧ ⎪ x = r cos ϕ sin θ, ⎪ ⎪ ⎨ y = r sin ϕ sin θ, ⎪ ⎪ ⎪ ⎩ z = r cos θ. The triple of numbers (r, ϕ, θ) that satisfies these equalities is called the spherical coordinates of a point (x, y, z). Their geometrical meaning is shown in Figure 9.1.

Figure 9.1. Spherical change of variables

So, consider the corresponding function g: g

(r, ϕ, θ) → (x, y, z) = (r cos ϕ sin θ, r sin ϕ sin θ, r cos θ).

Fubini and Change-of-Variables Theorems

155

Its Jacobian is    cos ϕ sin θ −r sin ϕ sin θ r cos ϕ cos θ  D(x, y, z)  Jg (r, ϕ, θ) = =  sin ϕ sin θ r cos ϕ sin θ r sin ϕ cos θ  D(r, ϕ, θ)   cos θ 0 −r sin θ  = −r2 sin θ. In order to satisfy the conditions of CVT (bijectivity of g and Jg  0), instead of the ball BR , we take a slightly smaller domain  = g(D), D

where

" # D := (r, ϕ, θ) : 0 < r < R, 0 < ϕ < 2π, 0 < θ < π .

 is a zero-measure set, we have Since BR \ D    f= f= f ◦ g|Jg |,  D

BR

D

or completely, 

=

2 2 +z2
dϕ 0

0

f (x, y, z) dx dy dz  R sin θ dθ f (r cos ϕ sin θ, r sin ϕ sin θ, r cos θ) r2 dr. 0

We see that the spherical change of variables is convenient to use when integrating in the ball. Sometimes, it is useful when the integrand is of the form f (x, y, z) = h(x2 + y2 + z2 ) (then, after the spherical change of variables, f (x, y, z) = h(r2 )). When the integration domain is the cylinder " # CR,h = (x, y, z) : x2 + y2 < R2 , 0 < z < h or the integrand is of the form f (x, y, z) = h(x2 + y2 , z)), it is convenient to use the cylindrical change of variables x = r cos ϕ, y = r sin ϕ, z = h (i.e. g(r, ϕ, h) = (r cos ϕ, r sin ϕ, h)), the Jacobian of which is    cos ϕ −r sin ϕ 0  D(x, y, z)  =  sin ϕ r cos ϕ 0  = r, D(r, ϕ, h)  0 0 1 

156

Integral and Measure

and therefore, 

 f=

CR,h



h



2π

dz

dϕ

0

0

R

f (r cos ϕ, r sin ϕ, z) r dr.

0

3) Let us find the area of the figure bounded by the curves y = 2x, y2 = x, x2 = 2y, x2 = y (see Figure 9.2). 2

Figure 9.2.

Of course, the area can be calculated directly; however, this is rather inconvenient. The figure can be described by inequalities as follows: 4 " # 3 y2 x2 D = (x, y) : x < y2 < 2x, y < x2 < 2y = (x, y) : 1 < < 2, 1 < <2 . x y Therefore, we may expect that after change of variables x, y by new 2 2 variables u = yx , v = xy , the problem will be simplified since, in terms of the new variables, the domain D is described by the simpler inequalities 1 < u < 2, 1 < v < 2. We easily express x and y in terms of u and v, respectively: 1

2

x = u3 v3 ,

2

1

y = u3 v3 .

The Jacobian of this change of variables is   2 2 1 1  D(x, y)  13 u− 3 v 3 23 u 3 v− 3  1 = 1 1 2 2  =− . D(u, v)  23 u− 3 v 3 13 u 3 v− 3  3

Fubini and Change-of-Variables Theorems

157

Therefore,  S (D) =

 2 2   − 1  du dv = 1 . dx dy =  3 3 D 1 1

P.9. Problems P ROBLEM 9.1.– The iterated integrals of the double integral 

∞ ∞ 0

e−xy sin x sin y dx dy

0

are finite and coincide. Does the double integral exist? P ROBLEM 9.2.– The iterated integrals of the double integral  1

1

−1 −1

xy (x2 + y2 )2

dx dy

are finite and coincide. Does the double integral exist? P ROBLEM 9.3.– Check the equality 





b

f (x)g(y) dx dy =

d

f (x) dx

D

a

g(y) dy, c

when D = [a, b] × [c, d], f ∈ L[a, b], g ∈ L[c, d]. P ROBLEM 9.4.– Check the equality 



a

y

dy 0

0

 f (x) dx =

a

(a − x) f (x) dx

0

for f ∈ L[0, a]. P ROBLEM 9.5.– Check the formula for the area of a rectangle when the edges of the latter are not parallel to the coordinate axes. Based on this, prove the invariance of the integral with respect to rotation of the coordinates.

158

Integral and Measure

P ROBLEM 9.6.– Change the integration order in the following integrals:  −1  0  0  0 1) √ dx √ f dy + dx f dy; 

− 2 π/4

2)

− 2−x2 sin x



f dy +

dx 

0



1

3)

0 x

dx 0

0

−1 π/2

 

π/4



e

f dy +



f dy;

dx 0 1

f dy.

dx 1

x cos x

ln x

P ROBLEM 9.7.– Find the following integrals where the domain D ⊂ R2 is bounded by the given curves:  √ 1) (54x2 y2 + 150x4 y4 ) dx dy, D : x = 1, y = x3 , y = − x;  D 2) y2 e−xy/8 dx dy, D : x = 0, y = 2, y = x/2; D  3) x4 dx dy, D : x2 /4 + y2 /25 = 1. D

P ROBLEM 9.8.– Find the following integrals where the domain D ⊂ R3 is bounded by the given surfaces:  xyz 1) y2 z cos dx dy dz, D : x = 0, x = 3 y = 0, y = 1, z = 3 D 0, z = 2π;  dx dy dz y z x 2)  6 , D : 8 + 3 + 5 = 1, x = 0, y = 0, z = 0; y D 1+ x + + z 8 3 5  3) |z| dx dy dz, D : x2 + y2  4, x2 + y2 + z2 = 16. D

P ROBLEM 9.9.– Perform the spherical change of variables in the integral  f (x, y, z) dx dy dz D

where D = {(x, y, z) : 1  x2 + y2 + z2  2, x  0, y  0, z  0}. P ROBLEM 9.10.– Find the volumes of the bodies bounded by surfaces:

Fubini and Change-of-Variables Theorems

1) z = −2x2 − 2y2 − 1, z = 4y − 1; 2) x = 5y2 − 1, x = −3y2 + 1, z = 2 −  3) z = 1 − x2 − y2 , 3z/2 = x2 + y2 .



x2 + 6y2 , z = −1 −



159

x2 + 6y2 ;

P ROBLEM 9.11.– Find the volume of the body V := (x, y, z) : 25  x2 + y2 + z2  100, 5 

√ ! x2 + y2 √ y , 3 − 3 . 99 x

P ROBLEM 9.12.– Let Bk (r) denote the ball in Rk with radius r > 0, and let Vk = m(Bk (1)). 1) Show that m(Bk (r)) = Vk rk , r > 0. 2) Find Vk . Hint. Beginning from the known values (k = 1, 2), derive a recurrent formula that expresses Vk+2 in terms of Vk , using in the space Rk+2 = Rk × R2 the polar change of variables for the variables xk+1 and xk+2 . P ROBLEM 9.13.– Find the Lebesgue measure of the set D := {x ∈ Rk : xi  0, x1 + · · · + xk  1}. P ROBLEM 9.14.– Find the integrals 

∞ ∞ 0

0



dx dy x2 + y2 (x2 + y2 + 1)

 1

1

; 0

0



dx dy |x2 − y2 |

.

P ROBLEM 9.15.– Prove that  ∞ ∞ dx dy < +∞. 2 2 0 0 (x + y)(x + y + 1) P ROBLEM 9.16.– Find the values of the parameter α ∈ R such that 

dx dy D

if

(x2 + y2 )α

< +∞

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Integral and Measure

1) D = B(0, 1) = {(x, y) ∈ R2 : x2 + y2  1}; 2) D = B(0, 1)c ; 3) D = R2+ . P ROBLEM 9.17.– Let f ∈ L(R), f  0. 1) Is the set A( f ) := {(x, y) ∈ R2 : 0 < y < f (x)} measurable?

If yes, is it true that R f = m2 (A( f ))? 2) Is the graph of f , G( f ) := {(x, y) ∈ R2 : y = f (x)}, a measurable set in R2 ? If yes, is it true that m2 (G( f )) = 0? P ROBLEM 9.18.– Let f ∈ L[0, ∞), f  0, and r(x) := that  ∞  ∞ r(x) dx = x f (x) dx. 0

∞ x

f (t) dt. Show

0

P ROBLEM 9.19.– Let D = (0, 1) × (0, 1). Give an example of a positive

1 function f ∈ C(D) ∩ L(D) such that for some x ∈ (0, 1), 0 f (x, y) dy = +∞.

10 Applications of Multiple Integrals

10.1. Calculation of the area of a plane figure We define the area of a set (plane figure) A ⊂ R2 as its Lebesgue measure   s(A) := m(A) = dx dy = 1IA (x, y) dx dy. R2

A

If the set A is of the form " # A = (x, y) : f1 (x)  y  f2 (x, y), x ∈ [a, b] (curvilinear trapezium), then, by the Fubini theorem,  S (A) =

R2

 = 

R

  1I[a,b] (x) 1I[ f1 (x), f2 (x)] (y) dy dx

b

=

1I[ f1 (x), f2 (x)] (y)1I[a,b] (x) dy dx

R



f2 (x) − f1 (x) dx.

a

So, we have justified the definition of the area of a curvilinear trapezium given in section 4.1.

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Integral and Measure

10.2. Calculation of the volume of a solid We define the area of a measurable set (solid) A ⊂ R3 as its Lebesgue measure   V(A) := m(A) = dx dy dz = 1IA (x, y, z) dx dy dz. R3

A

If a set A is of the form " # A = (x, y, z) : (x, y) ∈ D, z1 (x, y)  z  z2 (x, y) (cylindrical solid), then similarly to the above, we get   V(A) = z2 (x, y) − z1 (x, y) dx dy. D

Recall that (definition 4.1) the section of a set A ⊂ R3 at a point x ∈ R is the set " # A x := (y, z) ∈ R2 : (x, y, z) ∈ A . By the Fubini theorem,  V(A) = =

R3

   

=

1IA (x, y, z) dx dy dz =

R

R

R2

   R

R2

    1IA x (y, z) dy dz dx = R

 1IA (x, y, z) dy dz dx  dy dz dx

Ax

S (A x ) dx.

Thus, definition 4.1 is also justified. 10.3. Calculation of the area of a surface Without going into details of the theory of surfaces, we will derive a formula for the area of a surface in one particular case. We will use

Applications of Multiple Integrals

163

the fact that the area S (Φ) of a plane figure Φ inclined by the angle α against some plane is expressed by the formula S (Φ) =

S (D) , | cos α|

where D is the projection of the figure Φ onto the plane. Now consider a surface of the form " # Φ = (x, y, z) : z = f (x, y), (x, y) ∈ D , where f is a continuously differentiable function in a domain D ⊂ R2 . Let us partition the surface Φ by “small” pieces Φk : Φ=

n

Φk ,

" # Φk := (x, y, z) : z = f (x, y), (x, y) ∈ Dk .

k=1

In each subdomain Dk , let us take an arbitrary point Pk = (xk , yk ) and at the corresponding point Mk = (xk , yk , zk ) = (xk , yk , f (xk , yk )) on the subsurface Φk draw a tangent plane. Its equation can be written by applying the Taylor formula as follows: z − zk =

∂f ∂f (xk , yk )(x − xk ) + (xk , yk )(y − yk ) ∂x ∂y $  +o (x − xk )2 + (y − yk )2 , (x, y) → (xk , yk ).

Discarding the last “small” term, we get the equation of a plane, z − zk =

∂f ∂f (xk , yk )(x − xk ) + (xk , yk )(y − yk ), ∂x ∂y

which is called the tangent plane of the surface Φ at the point (xk , yk )  k , the set of points (x, y, z) (actually, at the point (xk , yk , zk )). Denoted by Φ  k ) ≈ S (Φk ). Recalling such that (x, y) ∈ Dk . Intuitively, it is clear that S (Φ the formula for the angle α between two planes a1 x + b1 y + c1 z + d1 = 0

and a2 x + b2 y + c2 z + d2 = 0,

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Integral and Measure

cos α = $

|a1 a2 + b1 b2 + c1 c2 | , $ a21 + b21 + c21 · a22 + b22 + c22

we get 6  k ) = S (Dk ) = S (Φ cos α

1+

∂ f

2  ∂ f 2 (xk , yk ) + (xk , yk ) · S (Dk ). ∂x ∂y

Therefore,   k ) S (Φ) = S (Φk ) ≈ S (Φ k

=

k

6



1+

∂ f

k

2  ∂ f 2 (xk , yk ) + (xk , yk ) · S (Dk ). ∂x ∂y

[10.1]

On the right-hand side, we see the integral on the domain D of the (“step-like”) function ϕ that takes the values 6 1+

∂ f

2  ∂ f 2 (xk , yk ) + (xk , yk ) ∂x ∂y

on the subdomains Dk . By refining the partitions of D, the latter function approaches the function 6 ψ(x, y) =

1+

∂ f

2  ∂ f 2 (x, y) + (x, y) , ∂x ∂y

(x, y) ∈ D.

Therefore, the sum on the right-hand side of equation [10.1] will approach the integral  6 1+ D

∂ f

2  ∂ f 2 (x, y) + (x, y) dx dy, ∂x ∂y

Applications of Multiple Integrals

165

which can be naturally called the area S (Φ) of the surface Φ. We rewrite the latter expression in short as  $ S (Φ) = 1 + f x 2 + fy 2 dx dy. D

The reader may try, by using this formula, to rederive the formula for the area of the surface of revolution that is obtained by rotating the graph of a function y = f (x), x ∈ [a, b], around the x-axis (see section 4.3): 

b

S = 2π

$

f (x) 1 + f  2 (x) dx.

a

10.4. Calculation of the mass of a body Suppose that we know the density p(x, y, z)  0 of a material solid D ⊂ R3 at each point (x, y, z) ∈ D. If p is a measurable function taking finitely many values p1 , p2 , . . . , pn in the corresponding (measurable)  sets D1 , D2 , . . . , Dk ( nk=1 Dk = D), then the mass M(D) of the solid D is M(D) =

n 

pk m(Dk ) =

  n D k=1

k=1

 pk 1IDk =

p. D

Therefore, it is natural to extend this formula to the general case: if p ∈ L(D), then  M(D) := p(x, y, z) dx dy dz. D

The “derivation” of this formula can be motivated in more detail. Noticing that the formula holds for step-like functions, consider the Cauchy sequence of step } that converges to p a.e. in D.

densities {pn

Since the limit limn→∞ D pn equals p, it can be considered the D mass of the solid D (provided that p ∈ L(D), p  0). However, the following question may arise: what is the density of a solid at a point? In the case of a homogeneous solid, the density is constant and equals

166

Integral and Measure

the ratio of the mass and volume. In the general case, we have to pass to the limit: M(Uε (a)) . V(Uε (a))

p(a) = lim ε↓0

So, we have arrived at the closed circle that reminds the question of priority of the egg or hen: we define the mass by using the notion of density, while to define the density, we must know the mass. However, in concrete cases, such “problems” usually do not arise since the density is found, for example, as a solution of some equation derived by using the laws of physics. 10.5. The static moment and mass center of a body Suppose that, in the space R3 , we are given a system of material points {(xi , yi , zi ), i = 1, 2, . . . , n} of masses m1 , m2 , . . . , mn . Its static moment with respect to the xy plane is defined as the number M xy :=

n 

mi z i

i=1

(the sum of the products of masses of the points and their distances from the plane, taking the signs into account; see section 4.5, item 2). Based on this definition, let us try to derive a formula for the static moment of a material body D ⊂ R3 with density p(x, y, z)  0, (x, y, z) ∈ D. Let us divide the body D into “small” pieces Di , i = 1, 2, . . . , n. Intuitively, it is clear that we will not make a large error by assuming that the mass of each piece Di is concentrated at some of its points (xi , yi , zi ). Therefore, the static moment of the body D with respect to the xy plane is M xy ≈

n 

 n  zi M(Di ) = zi p(x, y, z) dx dy dz

i=1

=

i=1

n   i=1

Di

zi p(x, y, z) dx dy dz Di

Applications of Multiple Integrals

≈

n  

 zp(x, y, z) dx dy dz =

zp(x, y, z) dx dy dz.

Di

i=1

167

D

Refining the partitions of D, in the limit, instead of an approximate formula, we get an exact one:  M xy = zp(x, y, z) dx dy dz. D

Similarly, we get the formulas for the static moments with respect to the other coordinate planes:   M xz = yp(x, y, z) dx dy dz, Myz = xp(x, y, z) dx dy dz. D

D

The mass center of the body is defined as the point (x0 , y0 , z0 ) such that when concentrating at this point the total its mass M, its static moments with respect to all coordinate planes remain the same (cf. section 4.5). Therefore, Mx0 = Myz ,

My0 = M xz ,

Mz0 = M xy ,

and thus, x0 =

Myz , M

y0 =

M xz , M

z0 =

Mxy . M

E XAMPLE 10.1.– Let us find the mass center (x0 , y0 , z0 ) of the half-ball " # B+R = (x, y, z) : x2 + y2 + z2  R2 , z  0 with density p ≡ 1. By symmetry it is clear that x0 = y0 = 0 (of course, this can also be calculated directly). Let us find z0 . Applying the spherical change of variables, we get  M xy = =

 B+R

πR4 , 4

z dx dy dz =

M = V(B+R ) =



2π

π 2

dϕ 0

2πR3 3

 dθ

0

R

r cos θ r2 sin θ dr

0

=⇒

z0 =

M xy 3R = . M 8

11 Parameter-dependent Integrals

11.1. Introduction Suppose that we are given a function f : X × Y → R (X ⊂ Rm , Y ⊂ Rk ). If for (almost) every fixed x ∈ X, the function f (x, y), y ∈ Y (or, for short, f (x, ·)), is integrable on the set Y, then the function  J(x) := f (x, y) dy, x ∈ X, Y

is called the parameter-dependent integral (depending on the parameter x ∈ X) or shortly PDI. In this chapter, we are interested in the following questions: 1) The continuity of the function J or, more generally, the possibility to pass to the limit under the sign of integral, that is, the question when  lim J(x) = lim f (x, y) dy. x→x0

Y x→x0

2) The differentiability of J and conditions allowing one to differentiate under the sign of integral, that is, the question when ∂J(x) = ∂xi

 Y

∂ f (x, y) dy. ∂xi

170

Integral and Measure

R EMARK 11.1.– Conditions of integrability of a PDI are given in the Fubini theorem: If f ∈ L(X × Y), then J ∈ L(X), and        J(x) dx = f (x, y) dy dx = f (x, y) dx dy. X

X

Y

Y

X

D EFINITION 11.1.– A PDI  J(x) = f (x, y) dy, x ∈ X, Y

converges normally with respect to x ∈ X (or in X) if there exists a function g ∈ L(Y) such that    f (x, y)  g(y),

x ∈ X, y ∈ Y.

In such a case, one also says that the function system { f (x, ·), x ∈ X} is normally integrable or that an (integrable) function g dominates the system { f x }. P ROPOSITION 11.1.– Let a function f : X × Y → R satisfy the following conditions: 1) For all y ∈ Y, there exists the limit f0 (y) := lim f (x, y) ∈ R. x→x0

2) There exists a deleted neighborhood U˙ = Uε (x0 ) \ {x0 } of the point x such that  the PDI 0

J(x) =

f (x, y) dy Y

˙ that is, there exists a function g ∈ L(Y) such normally converges in U, that    f (x, y)  g(y), x ∈ U˙ ∩ X, y ∈ Y. Then there exists the limit  lim J(x) = J0 := f0 (y) dy, x→x0

Y

Parameter-dependent Integrals

171

that is, one can pass to the limit under the integral sign:   lim f (x, y) dy = lim f (x, y) dy. x→x0

Y x→x0

Y

In particular, if, in addition, for every y ∈ Y, the function f (·, y) is

continuous at a point x0 ∈ X,1 then the PDI J(x) = Y f (x, y) dy is continuous at the point x0 . P ROOF.– Let X \ {x0 }  xn → x0 , n → ∞. We have to check that J(xn ) → J0 . From condition 1 we have that f (xn , y) → f0 (y),

n → ∞, y ∈ Y.

Therefore, there exists N ∈ N such that xn ∈ U˙ for all n > N, and from condition 2 we have  n   f (x , y)  g(y), y ∈ Y, n > N. By the Fatou–Lebesgue theorem we get   n J(xn ) = f (x , y) dy → f0 (y) dy = J0 , Y

Y

n → ∞.

 

E XAMPLE 11.1.– 1) For any f ∈ L[0, +∞), denote  +∞ F(p) := e−px f (x) dx, p  0. 0

The function F is called the Laplace transform of f . Note the following facts: i) for every fixed x ∈ [0, +∞), the integrand e−px f (x) is continuous as a function of p; ii) |e−px f (x)|  | f (x)|, x ∈ [0, +∞), p ∈ [0, +∞), and thus the PDI F(p) normally converges with respect to p ∈ [0, +∞).

1 That is, condition 1 is satisfied with f0 (y) = f (x0 , y).

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Integral and Measure

By proposition 11.1 the function F is continuous at all points p ∈ [0, +∞), that is, F ∈ C[0, +∞). 2) For any f ∈ L(R), denote  fˆ(t) :=

+∞

−∞ +∞

=

−∞

eitx f (x) dx



cos tx f (x) dx + i

+∞

−∞

sin tx f (x) dx,

t ∈ R.

The complex function fˆ is called the Fourier transform of f . As in the first case, we can easily check that fˆ ∈ C(R, C). Note that, in both cases, we do not require the continuity of the function f . 3) Now, we give an example that shows that the continuity of the integrand function with respect to parameter is not sufficient for the continuity of the PDI. Consider the function 

∞

J(x) :=

xe−xy dy,

x ∈ [0, +∞).

0

We directly calculate that  J(x) =

0 for x = 0, 1 for x > 0.

So, despite the fact that the integrand function f (x, y) = xe−xy is continuous with respect to x (even with respect to both variables (x, y)), the function J is discontinuous at the point 0. 4) If f ∈ C(X × Y) where X and Y are compact sets, then the PDI J(x) := Y f (x, y) dy, x ∈ X, is continuous in X. Indeed, in this case, the function f is bounded, that is, M := sup(x,y)∈X×Y | f (x, y)| < +∞, and therefore, in proposition 11.1, we can take g(y) = M for all y ∈ Y.

T HEOREM 11.1.– Let a PDI J(x) = Y f (x, y) dy, x ∈ X, satisfy the following conditions:

Parameter-dependent Integrals

1) For every fixed y ∈ Y ⊂ Rk , there exists

173

∂ f (x0 ,y) ∂xi . 0

2) There exists a neighborhood U of a point x ∈ X in which the PDI  ∂f (x, y) dy ∂x i Y normally converges. Then there exists   ∂J 0 ∂ ∂f 0 (x ) = f (x0 , y) dy = (x , y) dy. ∂xi ∂xi Y ∂x i Y P ROOF.– Without loss of generality, we may assume that X ⊂ R (m = 1). From condition 2, there exists a function g ∈ L(Y) such that  ∂ f (x, y)     g(y), ∂x

x ∈ U, y ∈ Y.

From the Lagrange mean value theorem2 we have    f (x, y) − f (x0 , y)  sup  ∂ f (x,y)  · |x − x0 | x∈U ∂x    x − x0 |x − x0 |  g(y), x ∈ U˙ = U \ {x0 }, y ∈ Y. Since for all y ∈ Y, lim

x→x0

f (x, y) − f (x0 , y) ∂ = f (x0 , y), ∂x x − x0

we can apply proposition 11.1 to the PDI  Y

f (x, y) − f (x0 , y) dy, x − x0

˙ x ∈ U,

2 Theorem. If a function f ∈ C[a, b] is differentiable in the interval (a, b), then there exists a point c ∈ (a, b) such that f (b) − f (a) = f  (c)(b − a).

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Integral and Measure

and obtain ∂ ∂x

 f (x , y) dy = lim 0

x→x0

Y

Y

f (x, y) dy −



Y x − x0 0

f (x0 , y) dy

f (x, y) − f (x , y) = lim dy x − x0 x→x0 Y  f (x, y) − f (x0 , y) = lim dy 0 x − x0 Y x→x 0 ∂ f (x , y) = dy. ∂x Y

 

E XAMPLE 11.2.– 1) Let us consider the PDI 

+∞

J(a) := 0

dx π = √ , a + x2 2 a

a > 0.

Formal differentiation of the integral with respect to a yields: 



J (a) = −

+∞

dx π = − 3/2 , a > 0, 2 2 4a 0 (a + x ) +∞ dx π =⇒ = 3/2 , a > 0. 2 )2 (a + x 4a 0

To justify the differentiation under the integral sign, fix any a0 > 0 and ε ∈ (0, a0 ). Then 1 1 < , (a + x2 )2 (a0 − ε + x2 )2

x ∈ [0, +∞), a ∈ U := Uε (a0 ).

The on the right-hand side is integrable. Therefore, the

+∞function dx PDI 0 (a+x2 )2 , a ∈ U, converges normally (in U), and we can apply theorem 11.1. Iterating the differentiation, we similarly get 

+∞ 0

dx (2n − 1)!! π = · n+1/2 , 2 n+1 2(2n)!! a (a + x )

n ∈ N.

Parameter-dependent Integrals

175

2) Differentiating the equality 

1 0

1 xa−1 dx = , a

a > 0,

with respect to a, we get 

1

xa−1 ln x dx = −

0

1 , a2

a > 0.

The differentiation under the integral sign can be justified similarly. Iterating the differentiation, we get 

1

xa−1 lnn x dx = (−1)n

0

n! , an+1

a > 0, n ∈ N.

b(x) 3) If the bounds of a one-dimensional PDI J(x) = a(x) f (x, y) dy are also functions of the parameter x, then such an integral can be reduced to a PDI in the interval [0, 1]. Namely, by the change of variable y = a(x)(1 − z) + b(x)z, z ∈ [0, 1], we have 

1

J(x) =

f˜(x, z) dz,

0

where f˜(x, z) = f (x, a(x)(1 − z) + b(x)z)(b(x) − a(x)), z ∈ [0, 1]. Formal differentiation under the integral sign gives a formula for the derivative of J: 

∂ f˜(x, z) dz ∂x 0  1   1 ∂f ∂f = (x, . . .) + (x, . . .) · a (x)(1 − z) + b (x)z × ∂x ∂y 0   ! × b(x) − a(x) + f (x, . . .) · b (x) − a (x) dz  1 ∂f = (x, . . .)(b(x) − a(x) dz + ∂x

0    1 ∂ f (x, . . .) a (x)(1 − z) + b (x)z + dz ∂z 0

J  (x) =

1

176

Integral and Measure

 = 

∂ f (x, y) dy ∂x a(x)

  1 + f x, a(x)(1 − z) + b(x)z a (x)(1 − z) + b (x)z  b(x)

b(x)

=

a(x)

z=0

  ∂f (x, y) dy + b (x) f x, b(x) − a (x) f x, a(x) . ∂x

The differentiation is “legal” (and, thus, the formula is correct) if, for example, the functions f, a, b are continuously differentiable. 4) When the Laplace transform of a function f ∈ L[0, +∞) (example 11.1) is differentiable? Formal differentiation of  +∞ F(p) = e−px f (x) dx 0

with respect to p gives  +∞  F (p) = − e−px x f (x) dx. 0

The integrand can be simply estimated:  −px    e x f (x)   x f (x), x, p ∈ [0, +∞). Thus, together with the integrability of f , we need the integrability of the function f1 (x) := x f (x). These two integrability conditions can be unified as follows:  +∞ (1 + x)| f (x)| dx < +∞. 0

Iterating the differentiation, we see that the Laplace transform F(p) is k times (continuously) differentiable if  +∞ (1 + xk )| f (x)| dx < +∞. 0

It is interesting to note that the smoothness of the Laplace transform depends not on the smoothness of the function, but rather on the rate of its convergence to zero at infinity. Similar properties are possessed by the Fourier transform.

Parameter-dependent Integrals

177

11.2. Improper PDIs The notion of normal convergence cannot be applied to the study of convergence, continuity, and differentiability of improper relatively convergent parameter-dependent integrals since from the requirement of normal convergence there obviously follows the absolute convergence of the integral (for all parameter values). The way out of this problem is a slight generalization of the notion of convergence of PDIs. D EFINITION 11.2.– An improper PDI 



b

c

f (x, y) dy = lim

J(x) :=

c↑b

a

f (x, y) dy,

x ∈ I,

a

is said to converge uniformly (in I) if   lim sup  c↑b x∈I

b c

 f (x, y) dy = 0.

[11.1]

R EMARK 11.2.– Let us compare the notions of normal and uniform convergences of PDIs. If the PDI in the definition were convergent normally, then we would have  b  f (x, y) dy = 0. lim sup c↑b x∈I

[11.2]

c

So, relation [11.1] is obtained from [11.2] by taking the modulus outside the integral sign. P ROPOSITION 11.2.– If f ∈ C(I × [a, b)) (where I ⊂ R is any interval) and an improper PDI 

b

J(x) =

f (x, y) dy,

x ∈ I,

a

converges uniformly in a neighborhood of a point x0 ∈ I, then J(x) is continuous at x0 .

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Integral and Measure

P ROOF.– Consider the difference  b      b  J(x) − J(x0 ) =  f (x, y) dy − f (x0 , y) dy a  ac     f (x, y) − f (x0 , y) dy  a    b +  f (x, y) dy  c b  +  f (x0 , y) dy, x ∈ I, a  c < b. c

Take arbitrary ε > 0. From the above definition,   ∃ c0 ∈ (a, b) : 

b c0

   f (x, y) dy + 

b c0

 ε f (x0 , y) dy < 2

for x ∈ U,

where U is the neighborhood of x0 from the statement of proposition. Without loss of generality, we may assume that its closure U ⊂ I. Applying proposition 11.1 to the (proper) PDI  c0   f (x, y) − f (x , y) dy, J1 (x) := 0 a

we get that J1 (x) → 0 as x → x0 , that is, there exists a neighborhood U1 ⊂ U of x0 such that J1 (x) < ε2 for x ∈ U1 . Therefore, finally,    J(x) − J(x0 ) < ε for x ∈ U1 . This means that lim J(x) = J(x0 ).

x→x0

 

T HEOREM 11.2 (Abel–Dirichlet test of uniform convergence of improper PDI).– Suppose that functions f ∈ C(I ×[a, b)) and g ∈ C 1 [a, b) satisfy the following conditions:

t 1) The function F(x, t) = a f (x, y) dy, x ∈ I, t ∈ [a, b), is uniformly bounded:   F(x, t)  M, x ∈ I, t ∈ [a, b);

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179

2) g is a decreasing function, and lim x↑b g(x) = 0. Then the improper PDI 

b

J(x) =

f (x, y)g(y) dy,

x ∈ I,

a

converges uniformly in I. P ROOF.– By the Abel–Dirichlet test of convergence of an improper integral (theorem 6.1) we have    

b c

 f (x, y)g(y) dy  Mg(c),

c ∈ [a, b).

Therefore,   sup  x∈I

b c

 f (x, y)g(y) dy  Mg(c) → 0,

c ↑ b.

E XAMPLE 11.3.– We will apply the theorem to the improper PDI 

+∞

J(α) = 0

e−αx

sin x dx, x

α  0.

Taking f (α, x) = e−αx sin x,

1 g(x) = , x

x  1,

we have   −αx      t −αx e (α sin x − cos x) t  F(α, t) =  e sin x dx =  1  1 + α2 1 |e−αt (α sin t − cos t) − e−α (α sin 1 − cos 1)| = 1 + α2 2(α + 1)  < 4, α  0, t ∈ [1, +∞). 1 + α2

 

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Integral and Measure

We see that the conditions of theorem 11.2 are satisfied for the improper PDI  +∞ sin x J1 (α) := e−αx dx, α  0. x 1 Therefore, the latter improper PDI converges uniformly; thus, the improper PDI 

1

J(α) = J1 (α) +

e−αx

0

sin x dx, x

α  0,

also converges uniformly. Since the integrand function is continuous with respect to α, by the theorem the function J(α), α  0, is also a continuous function. In particular,  +∞ sin x lim J(α) = dx. α↓0 x 0 We will apply this equality for calculation of the latter integral. Formally differentiating J(α), we get  +∞  J (α) = − e−αx sin x dx, α > 0. 0

It is easy to see that this PDI (proper) converges uniformly in the neighborhood U(α0 ) = (α0 /2, 3α/2) of any α0 > 0: α0

|e−αx sin x|  e− 2 x ,

x ∈ [0, +∞), α ∈ U(α0 ).

Therefore, differentiation under the integral sign is justified. However, a direct calculation gives J  (α) = −

1 =⇒ J(α) = − arctg α + C, 1 + α2

α > 0.

Since limα→+∞ J(α) = 0, we have C = π2 . Hence, J(α) =

π − arctg α, 2

α > 0,

Parameter-dependent Integrals

and, thus,  +∞ 0

181

sin x π dx = lim J(α) = . α↓0 x 2

By the change of a variable x = at, we get a more general formula  +∞ sin at π dt = sgn a, a ∈ R. t 2 0 Now we will generalize the notion of normal integrability of a system of functions. D EFINITION 11.3.– A system of integrable functions { fa , a ∈ A} on a set X ⊂ Rk is said to be uniformly integrable if the following conditions are satisfied: 1)

 | fa | < ∞.

sup a∈A

[11.3]

X

2) For every decreasing sequence of measurable sets X ⊃ An ↓ ∅ (i.e.  An ⊃ An+1 , n ∈ N, and ∞ n=1 An = ∅), we have  sup | fa | ↓ 0, n → ∞. [11.4] a∈A

An

In such a case, we also say that the PDI J(a) := uniformly in A (or with respect to a ∈ A).

f (x) dx converges X a

R EMARK 11.3.– From the properties of the integral we know that  ∀a ∈ A, | fa | ↓ 0, n → ∞, An

provided that An ↓ ∅. Therefore, condition 1 means that the latter relation holds uniformly for all functions fa . P ROPOSITION 11.3.– Let { fa , a ∈ A} be a system of integrable functions on a set X ⊂ Rk . This system is uniformly integrable if it satisfies at least one of the following conditions:

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Integral and Measure

1) { fa } is a normally integrable system. 2) There exists an increasing sequence of measurable sets Kn ⊂ X (Kn ⊂ Kn+1 ) such that i) { fi } is normally integrable on each set Kn (not necessarily on X);

 | fa | ↓ 0, n → ∞;

ii) sup a∈A

X\Kn

3) m(X) < +∞, and  sup | fa | ↓ 0, n → ∞. a∈A {| fa |>n}

P ROOF.– Since condition 1 follows from condition 2 (by taking Kn = X, n ∈ N), we begin with a proof of sufficiency of condition 2. From condition 2 it follows that  sup | fa | < ∞, n ∈ N. a∈A

Kn

Taking any n such that  sup | fa | < ∞, a∈A X\Kn

we obtain relation [11.3]:    sup | fa |  sup fa + sup | fa | < ∞. a∈A

X

a∈A

Kn

a∈A X\Kn

Now let us take an arbitrary sequence of measurable sets An ↓ ∅. In view of condition 2(i), let gn ∈ L(Kn ) be functions such that    fa (x)  gn (x),

x ∈ Kn , a ∈ A.

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183

Then, for all n, m ∈ N,  

 | fa | = sup

sup a∈A

a∈A

An

| fa | + An ∩K m

 sup a∈A





 | fa |

An ∩Kmc

gm + An ∩Km

=

| fa |

X\K  m

gm + sup

a∈A X\Km

An ∩Km





| fa |.

Taking the limit as n → ∞, in view of An ∩ Km ↓ ∅, n → ∞, we get   lim ↓ sup | fa |  sup | fa |. n→∞ a∈A

a∈A X\Km

An

Now taking the limit as m → ∞, from condition 2(ii) we get  lim ↓ sup | fa |  0, n→∞ a∈A

An

so that the latter limit equals 0, that is condition [11.4] is satisfied. The sufficiency of condition 3 is proved similarly. First, take any n such that  sup | fa | < ∞. a∈A {| fa |>n}

Then



 | fa |  nm(X) + sup

sup a∈A

a∈A {| fa |>n}

X

| fa | < ∞,

that is, condition [11.3] is satisfied. Let us prove [11.4]. We have 

 | fa | = sup

sup a∈A



An

a∈A

| fa | + An ∩{| fa |>K}



 An ∩{| fa |K}

| fa |

184

Integral and Measure

   sup a∈A

| fa | + {| fa |>K}

 sup

a∈A {| fa |>K}



 K

An ∩{| fa |K}

| fa | + K · m(An ).

Let us take the limit as n → ∞:   lim ↓ sup | fa |  sup | fa |, n→∞ a∈A

An

a∈A {| fa |>K}

where we used the fact that from An ↓ ∅, m(A1 ) < +∞ it follows that m(An ) ↓ 0. Now it suffices to take the limit as K → ∞ and to apply condition 3.   R EMARK 11.4.– Condition 2 is usually applied by choosing the sets Kn such that, for each n ∈ N, m(Kn ) < +∞ and { fa } is uniformly bounded (with respect to a ∈ A) in the set Kn :   Cn := sup sup  fa (x) < +∞, n ∈ N, a∈A x∈Kn

or    fa (x)  Cn < +∞,

x ∈ Kn , a ∈ A, n ∈ N.

In this case, condition 2(i) is satisfied with gn (x) := Cn , x ∈ Kn . In the one-dimensional case (X = (c, d)), we typically take Kn = [cn , dn ] ↑ X = (c, d) (cn ↓ c  −∞, dn ↑ d  +∞). T HEOREM 11.3 (Generalization of the Fatou–Lebesgue theorem).– If { fn } ⊂ L(X) is a uniformly integrable sequence3 and fn → f a.e. in X, then X fn → X f .

3 That is, the normal integrability condition is replaced by a weaker condition of uniform integrability.

Parameter-dependent Integrals

185

P ROOF.– Without loss of generality, we may assume that fn → f on the whole set X. By the Fatou–Lebesgue theorem we have    | f |  lim inf | fn |  sup | fn | < ∞, n

X

X

X

that is, f ∈ L(X). Let g > 0 be any integrable function on X . For arbitrary ε > 0, denote 3 4   An := x ∈ X : sup  fn (x) − f (x)  εg(x) . kn

Since fn → f , we have supkn | fn (x) − f (x)| ↓ 0, n → ∞, for all x ∈ X. Therefore, An ↓ ∅, n → ∞. Let us estimate the integral of the difference: 





| fn − f |  X

| fn − f | + | fn − f |εg

ε  ε

| fn − f |>εg



g+ X

X

ε

| fn | +

| fn − f |>εg

g+

m

and thus, limn

X

f X n

X

=

X

f.

|f|

| fn − f |>εg An

|f| 

| fm | + An

Therefore,   lim sup | fn − f |  ε g, n



| fn | +

An 

g + sup X

| fn − f |

 |f| ↓ ε

An

g. X

 ε > 0 =⇒ lim n

| fn − f | = 0, X

 

C OROLLARY 11.1.– Proposition 11.1 remains true if the word “normally” is replaced by the word “uniformly.” P ROOF.– In the proof of proposition 11.1, instead of the Fatou–Lebesgue theorem, it suffices to apply its generalization, theorem 11.3.  

186

Integral and Measure

E XAMPLE 11.4.– We give an example of a uniformly integrable, but not normally integrable, function system. Let 1 fk (x) = 1I[k,k+1) (x), x

x ∈ [1, +∞), k ∈ N.

The function system { fk , k ∈ N} is not normally integrable since 1 sup fk (x) = , x k∈N

x ∈ [1, +∞),

and 

+∞ 1

1 dx = +∞, x

and therefore, for any dominating function g, we have 1 g(x)  , x

 x ∈ [1, +∞) =⇒

+∞

g = +∞.

1

Now let us check that the family { fk } is uniformly integrable. In proposition 11.3, take Kn = [1, n), n ∈ N. It is clear that 1) sup sup | fk (x)| = 1 < +∞, and k∈N x∈Kn





k∈N

+∞

fk = sup

2) sup [1,+∞)\Kn

k∈N

n

fk 

1 →0 n

since, for all k, n ∈ N, 

+∞ n

⎧ ⎧ ⎪ ⎪ ⎪ n > k, ⎪ n > k, ⎨ 0, ⎨ 0,

k+1 dx

k+1 dx fk = ⎪ ⎪ ⎪ ⎪ ⎩ ⎩ x ,nk k ,nk k  k 1 0, n > k, = 1  . n k, n  k

Parameter-dependent Integrals

187

P.11. Problems P ROBLEM 11.1.– The Euler gamma function (or the second-type Euler integral) is the PDI  ∞ Γ(x) := e−t t x−1 dt, x > 0. 0

Prove the following properties of the gamma function: 1) Γ ∈ C ∞ (0, +∞); 2) Γ is a convex function; 3) Γ(x + 1) = xΓ(x), x > 0; Γ(n + 1) = n!, n ∈ N; 4) lim x→+0 Γ(x) = lim x→+∞ Γ(x) = +∞. P ROBLEM 11.2.– The beta function (or the first-type Euler integral) is the PDI  1 B(x, y) := t x−1 (1 − t)y−1 dt, x, y > 0. 0

Prove that B ∈ C ∞ ((0, +∞)2 ). P ROBLEM 11.3.– Let f ∈ C 1 [0, +∞) and f  ∈ L[0, +∞), and let F be the Laplace transform of f (example 11.1). Prove that the Laplace transform of the function f1 = f  is F1 (p) = pF(p) − f (0), p  0. P ROBLEM 11.4.– Express the derivatives of the elliptic functions  π/2 $ E(x) := 1 − x2 sin2 ϕ dϕ, 0 < x < 1, 0

and



π/2

F(x) := 0

$

dϕ

,

0 < x < 1,

1 − x2 sin2 ϕ

in terms of E and F. Show that the function E satisfies the differential equation 1 1 E  (x) + E  (x) + E(x) = 0. x 1 − x2

188

Integral and Measure

P ROBLEM 11.5.– Show that the Bessel function 

1 Jn (x) = π

π

cos(nϕ − x sin ϕ) dϕ

0

satisfies the Bessel equation x2 Jn (x) + xJn (x) + (x2 − n2 )Jn (x) = 0. P ROBLEM 11.6.– Let, g ∈ C[0, 1]. Denote  K(x, y) := 

x(1 − y), x  y, y(1 − x), x > y, 1

f (x) :=

K(x, y)g(y) dy,

x ∈ [0, 1].

0

Show that f (x), x ∈ [0, 1], satisfies the following differential equation with boundary conditions: 

f  (x) = −g(x), f (0) = f (1) = 0.

P ROBLEM 11.7.– Find the limits ∞ dx 1) lim ; n→∞ 0 1 + xn  π/2 sinn x 2) lim dx; n→∞ 0 1 + x2  π/2 3) lim e−n sin x dx. n→∞

0

P ROBLEM 11.8.– Let f ∈ C[0, ∞) be a bounded function. Find 

∞

lim

y→0

0

y f (x) dx. x2 + y2

P ROBLEM 11.9.– Let f ∈ C(R). Prove that 1 lim t→0 t

 a

b



f (x + t) − f (x) dx = f (b) − f (a).

Parameter-dependent Integrals

189

P ROBLEM 11.10.– Is it true that  1  1 2x −x2 /ε2 2x 2 2 lim e dx = lim 2 e−x /ε dx? ε→0 0 ε2 ε→0 ε 0 P ROBLEM 11.11.– We say that fn → f in measure in a set X ⊂ Rk if, for all ε > 0,   " # m x ∈ X :  fn (x) − f (x) > ε → 0,

n → ∞.

Prove that if m(X) < +∞, then in theorem 11.3, the convergence fn → f almost everywhere may be replaced by the convergence fn → f in measure. P ROBLEM 11.12.– Prove the convergence of the improper Fresnel integrals 



∞

2

sin x dx

∞

and

0

cos x2 dx.

0

Hint: Use example 11.3. P ROBLEM 11.13.– For x  0, denote 

x

f (x) :=

e

−t2



2

dt ,

e

g(x) :=

0

0

1) Find f  (x) and g (x). 2) Check that f  (x) + g (x) = 0, x  0. 3) Show that f (x) + g(x) = π4 , x  0. 4) Find the integral  ∞  x 2 −t2 e dt = lim e−t dt. 0

1 −x2 (t2 +1)

x→∞

0

t2 + 1

dt.

Part 3

Measure and Integration in a Measure Space

12 Families of Sets

12.1. Introduction In the previous chapters, we first defined the integrals for step functions and then for the limits of their sequences in one sense or another. The bases of the “steps” of step functions were intervals in the line, rectangles in the plane, and, in the general case, k-boxes in the space Rk . In the definitions of the integral, it was important that we could measure those steps in some way. The intervals were measured by their lengths, the rectangles by their area, and k-boxes by their (Lebesgue) measures. However, in many areas of mathematics and physics, a notion of the integral in more general domains is needed, with other measures, and with “steps” more general than k-boxes. It is rather natural to try to imagine the most general situation where, in some abstract set, we can measure its every subset. Unfortunately, as a rule, such a “maximalist” approach faces principal difficulties. For example, even in the real line, we cannot define the Lebesgue measure of its every subset, and, therefore, we are forced to settle for the fact that we can only measure the sets that we called measurable in part II. So, to create a coherent general integration theory, we have to first postulate that we can measure only some sets. Then, if such sets are not sufficiently numerous, we can try to extend the measure as much as possible to a wider family of sets, preserving, at the same time, all “good” properties of the measure.

194

Integral and Measure

So, by a family of sets we mean a set of subsets of some fixed set E (sometimes called the universal set). In the measure theory, the most important set families are algebras and σ-algebras. D EFINITION 12.1.– A family A of subsets of a set E is called an algebra if it contains the whole set E and is closed under set operations of complement and union, that is, satisfies the following conditions: 1) E ∈ A; 2) If A ∈ A, then Ac ∈ A; 3) If A, B ∈ A, then A ∪ B ∈ A . The class A is called a σ-algebra if, in addition, it is closed under the operation of countable union, that is, satisfies the following condition:  4) If An ∈ A, n ∈ N, then ∞ n=1 An ∈ A. The pair (E, A) consisting of a set E and a σ-algebra A of subsets of E is called a measurable space. The sets from A are called measurable sets (or, more precisely, A-measurable sets). E XAMPLE 12.1.– 1) The simplest and minimal (σ-)algebra of subsets of a set E consists of only two sets, ∅ and E. It is called the trivial (σ-)algebra. 2) The other extremal is the maximal family of sets consisting of all subsets of E, which we denote by P(E). Because of the above-mentioned reasons, this σ-algebra has a practical meaning only if E is a finite or countable set. 3) Let A be the family of sets consisting of all finite unions of intervals of E = R. Then A is an algebra but not a σ-algebra. 4) Let E be an arbitrary set, and let A consist of all finite or countable sets A ⊂ E and complements of such sets. The practical importance of such a σ-algebra is not high. It is mostly used for illustration. 5) E = Rk , and A is the family of all Lebesgue-measurable sets A ⊂ E. Then A is a σ-algebra (see proposition 8.6).

Families of Sets

195

P ROPOSITION 12.1.– Let A be an algebra of subsets of a set E. Then: 1) ∅ ∈ A; 2) If A, B ∈ A, then A ∩ B ∈ A, A\B ∈ A, AB ∈ A;  3) If A is a σ-algebra and An ∈ A, n ∈ N, then ∞ n=1 An ∈ A. P ROOF .– It follows from the definition of a σ-algebra and from the following equalities: 1) ∅ = E c ; 2) A ∩ B = (Ac ∪ Bc )c ; A\B = A ∩ Bc ; AB = (A\B) ∪ (B\A);  ∞ c c 3) ∞ n=1 An = ( n=1 An ) .

 

Let {Aα , α ∈ I} be a collection of algebras (σ-algebras) of subsets  of a set E. Then α∈I Aα also is an algebra (σ-algebra). Therefore, for an arbitrary family B of subsets of E, there exists the minimal algebra (σ-algebra) A containing B. Indeed, this algebra (σ-algebra) is the intersection of all algebras (σ-algebras) that contain B. This intersection is not empty since there is at least one (σ-)algebra, P(E), that contains B. D EFINITION 12.2.– The minimal algebra (σ-algebra) that contains a family C of sets is called the algebra (σ-algebra) generated by C. The σ-algebra of sets generated by a family of sets C is denoted by σ(C). In the case of a metric space E (in particular, R or Rk ), the σ-algebra generated by all open (for example) balls, denoted by B = B(E), is called the Borel σ-algebra, and the sets B ∈ B are called Borel sets of E. R EMARK 12.1.– The argument that shows the existence of the generated σ-algebra works in the general situation: if, in a set E, some set operations are defined, then, for any family of subsets of E, there exists the minimal family of subsets of E containing C and closed under those operations. Recall that a sequence of sets {An , n ∈ N} is said to be increasing (decreasing) if An ⊂ An+1 (An ⊃ An+1 ) for all n ∈ N. For any increasing  (decreasing) sequence of sets {An , n ∈ N}, we denote lim ↑ An = ∞ n=1 An n ∞ (lim ↓ An = n=1 An ). n

196

Integral and Measure

If A = lim ↑ An (A = lim ↓ An ), then we will also write An ↑ A (An ↓ A). n

n

D EFINITION 12.3.– The family M of subsets of E is called a monotone family if it is closed with respect to operations lim ↑ and lim ↓. The minimal monotone family of sets containing a family of sets C is called the monotone family of sets generated by C. We denote it by m(C). P ROPOSITION 12.2.– An algebra A of subsets of a set E is a σ-algebra if and only if it is a monotone family. R EMARK 12.2.– This statement is important because verification that a set family is monotone is usually simpler than that it is a σ-algebra. Even more important is theorem 12.1 proved below. P ROOF.– It is clear that every σ-algebra is a monotone family (from definition 12.1 and proposition 12.1). However, let A be both an algebra and a monotone family, and let An ∈ A, n ∈ N. Then, for all m ∈ N, m n=1 An ∈ A since A is an algebra, and thus, ∞ n=1

An = lim ↑ m

 m

 An ∈ A

n=1

since A is a monotone family.

 

T HEOREM 12.1.– If A is an algebra, then σ(A) = m(A). P ROOF.– Denote B = σ(A) and M = m(A). By proposition 12.2, B is a monotone family. Since M is the minimal monotone family containing A, it follows that M ⊂ B. Let us show that M is an algebra. Then from proposition 12.2 it will follow that M is a σ-algebra and, therefore, B ⊂ M. First, we check that the family M is closed under complements. To this end, it suffices to show that the family M := {B ∈ M : Bc ∈ M} ⊂ M

Families of Sets

197

coincides with M. Note that M contains A and is a monotone family: if B = limn ↑ Bn or B = limn ↓ Bn with Bn ∈ M , n ∈ N, then, respectively, Bc = lim ↓ Bcn ∈ M n

or

Bc = lim ↑ Bcn ∈ M, n

that is, B ∈ M . Therefore, from A ⊂ M ⊂ M it follows that M = M (since M is the minimal monotone family containing A). Now, let us show that M is closed under intersections. To this end, it suffices to show that, for every A ∈ M, the family MA := {B ∈ M : A ∩ B ∈ M} ⊂ M coincides with M. Note that since limn (A ∩ Bn ) = A ∩ limn Bn for every monotone (increasing or decreasing) sequence {Bn , n ∈ N}, MA is a monotone family. First, let A ∈ A. Then, for all B ∈ A, A ∩ B ∈ A ⊂ M, which means that A ⊂ MA ⊂ M, and, thus, MA = M. Now let A ∈ M. Take arbitrary B ∈ A. Since, as just proved, M = MB , we have that B ∈ M (since A ⊂ M) and A ∩ B ∈ M, that is, B ∈ MA . Therefore, we again have A ⊂ MA ⊂ M, and thus MA = M for all A ∈ M.   P.12. Problems In problems 12.1–12.3, given a family D of subsets of a set E, find the algebra A and σ-algebra A˜ = σ(D) generated by D. P ROBLEM 12.1.– D consists of a single set D ⊂ E. P ROBLEM 12.2.– Fix a set D ⊂ E. D consists of all subsets of E containing D, that is D = {F : D ⊂ F ⊂ E}. P ROBLEM 12.3.– D is the family of all sets consisting of: i) a single element; ii) two elements. In problems 12.4–12.6, given a family D of subsets of a set E, find the σ-algebra A = σ(D) generated by D.

198

Integral and Measure

P ROBLEM 12.4.– Let f : E → E be a bijection (one-to-one map). A set A ⊂ E is called f -invariant if f (A) ⊂ A and f −1 (A) ⊂ A. D is the family set of all f -invariant sets. P ROBLEM 12.5.– E := R3 . A set A ⊂ E us called a cylinder if from (x, y, z) ∈ A it follows that {(x, y, z ) : z ∈ R} ⊂ A. D is the family of all cylinders. P ROBLEM 12.6.– E := R2 , D is the family of all subsets of E that can be covered by a finite number of horizontal straight lines. P ROBLEM 12.7.– Let f : E 1 → E2 , and let A be an algebra (σ-algebra) of subsets of E2 . Show that " # f −1 (A) = A ⊂ E1 : A = f −1 (B), B ∈ A is an algebra (σ-algebra) of subsets of E1 . P ROBLEM 12.8.– Recall that the Borel σ-algebra B = B(R) on R is the σ-algebra generated by the family of open intervals. i) Show that B is generated by the intervals (−∞, r) with rational r. ii) Show that all intervals I ⊂ R are Borel sets, that is, belong to B. iii) Show that the set of irrational numbers is a Borel set. P ROBLEM 12.9.– Prove that, in E = Rk , all open and all closed sets are Borel sets. Is this true for an arbitrary metric space E? P ROBLEM 12.10.– Give an example of a set E and a family M of its subsets that contains ∅ and E and is a monotone family but not a σ-algebra.

13 Measure Spaces

D EFINITION 13.1.– A measure on a family A of subsets of a set E is a map μ : A → R+ = [0, +∞] such that: 1) μ(∅) = 0; 2) (σ-additivity) if {Ai , i ∈ N} is a sequence of pairwise disjoint sets  from A and ∞ i=1 Ai ∈ A, then ∞ ∞    μ Ai = μ(Ai ). i=1

i=1

If μ(E) < ∞, then the measure μ is called finite. If E can be written  as a countable union of sets of finite measure (i.e. E = ∞ n=1 En , E n ∈ A, μ(En ) < +∞, n ∈ N), then the measure μ is called σ-finite. R EMARK 13.1.– If A = B ∪ C and B ∩ C = ∅, we will write A = B + C.   Similarly, if A = i∈I Ai and Ai ∩ A j = ∅, i  j, we will write A = i∈I Ai . Then the σ-additivity of a measure may be written as follows: Ai ∈ A,

i ∈ N,

∞ 

∞ ∞    Ai ∈ A =⇒ μ Ai = μ(Ai ).

i=1

i=1

i=1

T HEOREM 13.1.– Every measure μ in an algebra A possesses the following properties: 1) (monotonicity) if A1 , A2 ∈ A, A1 ⊂ A2 , then μ(A1 )  μ(A2 );

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Integral and Measure

2) (strong additivity) for all A1 , A2 ∈ A, μ(A1 ) + μ(A2 ) = μ(A1 ∪ A2 ) + μ(A1 ∩ A2 ); 3) (σ-semiadditivity) if Ai ∈ A, i ∈ N, and

∞

i=1 Ai

∈ A, then

∞ ∞    μ Ai  μ(Ai ); i=1

i=1

4a) (continuity from below) if An ↑ A in A, then μ(An ) ↑ μ(A); 4b) (continuity from above) if An ↓ A in A and μ(An ) < ∞ for some n, then μ(An ) ↓ μ(A). P ROOF.– 1) From the equality A2 = A1 + (A2 \A1 ), which holds if A1 ⊂ A2 , we have μ(A2 ) = μ(A1 ) + μ(A2 \A1 )  μ(A1 ). 2) Using the equalities A1 = (A1 ∩ A2 ) + (A1 ∩ Ac2 ), A2 = (A1 ∩ A2 ) + (Ac1 ∩ A2 ), A1 ∪ A2 = (A1 ∩ A2 ) + (Ac1 ∩ A2 ) + (A1 ∩ Ac2 ), we get the corresponding equalities for measures: μ(A1 ) = μ(A1 ∩ A2 ) + μ(A1 ∩ Ac2 ); μ(A2 ) = μ(A1 ∩ A2 ) + μ(Ac1 ∩ A2 ); μ(A1 ∪ A2 ) = μ(A1 ∩ A2 ) + μ(Ac1 ∩ A2 ) + μ(A1 ∩ Ac2 ). Adding the first two equalities, in view of the third one, we get μ(A1 ) + μ(A2 ) = μ(A1 ∪ A2 ) + μ(A1 ∩ A2 ).

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3) Since ∞

∞  

Ai =

i=1

Ai \

i=1

i−1

 Aj ,

j=1

using the σ-additivity and monotonicity properties of a measure, we get ∞ ∞  i−1 ∞      μ Ai = μ Ai \ Aj  μ(Ai ). i=1

i=1

j=1

i=1

4a) Let An ↑ A. Since A = lim ↑ An = n

∞

An =

n=1

∞ 

(An \An−1 ) + A1 ,

n=2

we have   μ(A) = μ lim ↑ An = μ(An \An−1 ) + μ(A1 ) ∞

n

= lim ↑

k 

k→∞ n=2

= lim ↑ μ k→∞

n=2

μ(An \An−1 ) + μ(A1 )

k 

 An \An−1 + A1 = lim ↑ μ(Ak ). k→∞

n=2

4b) Now suppose that An ↓ A and μ(An0 ) < ∞ for some n0 . By applying property 4a to the sequence {An0 \An , n ∈ N} we get    μ(An0 ) − μ lim ↓ An = μ An0 \ lim ↓ An = μ lim ↑ (An0 \An ) n n n = lim ↑ μ(An0 \An ) = lim ↑ μ(An0 ) − μ(An ) n n = μ(An0 ) − lim ↓ μ(An ). n

Therefore,  μ lim ↓ An = lim ↓ μ(An ). n

n

 

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Integral and Measure

P ROPOSITION 13.1.– Suppose that a mapping μ : A → R+ on an algebra A satisfies the following conditions: 1) μ(∅) = 0; 2) (additivity) if {Ai , i ∈ I} is a finite system of pairwise disjoint sets from A, then    μ Ai = μ(Ai ). i∈I

i∈I

Then μ is a measure on A if and only if at least one of the following two conditions is satisfied: 3) μ is continuous from below, that is, A  An ↑ A ∈ A =⇒ μ(An ) ↑ μ(A); 4) μ is finite (μ(E) < +∞) and continuous from above, that is, A  An ↓ ∅ =⇒ μ(An ) ↓ 0. P ROOF.– First, let μ be continuous from below. Then μ

∞ 





An = μ lim ↑ n

n=1

= lim ↑ n

n 

n   Ak = lim ↑ μ Ak



n

k=1

n 

μ(Ak ) =

k=1

∞ 

k=1

μ(Ak ).

k=1

When μ is finite and continuous from above, we use the equality μ

∞ 

n ∞ n ∞           Ak = μ Ak + μ Ak = μ(Ak ) + μ Ak .

k=1

k=1

k=n+1

Since ∞ 

Ak ↓ ∅,

n → ∞,

k=n+1

by assumption 4 we have μ

∞   k=n+1

 Ak ↓ 0,

n → ∞.

k=1

k=n+1

Measure Spaces

203

Therefore, ∞ n ∞     μ Ak = lim ↑ μ(Ak ) = μ(Ak ). k=1

n

k=1

 

k=1

D EFINITION 13.2.– A measure space is a triple (E, A, μ), where E is a nonempty set, A is a σ-algebra of its subsets, and μ is a measure on the σ-algebra A. Now we introduce the notions of upper and lower limits of sequences of sets similar to those for real number sequences. D EFINITION 13.3.– The upper and lower limits of a sequence of sets {An , n ∈ N} are defined, respectively, as  2 lim sup An = lim ↓ sup Am = Am , ∞

n

n

mn



n=1 mn



lim inf An = lim ↑ inf Am = n

n

mn

∞ 2

Am .

n=1 mn

It is clear that every σ-algebra is closed under the operations lim supn and lim inf n . We can easily check that lim supn An (lim inf n An ) consists of all points x ∈ E that belong to infinitely many sets An (respectively, to all sets An , except a finite number of them). Clearly, lim inf n An ⊂ lim supn An . When these sets coincide, they are denoted as limn An . Note that if a sequence of sets {An , n ∈ N} is increasing (decreasing), then lim ↑ An = limn An (lim ↓ An = limn An ). n

n

P ROPOSITION 13.2 (Continuity of measure).– Let (E, A, μ) be a  measure space. Suppose that An ∈ A, n ∈ N, and μ ∞ n=1 An < ∞ (or at ∞ least μ( n=n0 An ) < ∞ for some n0 ). Then   μ lim inf An  lim inf μ(An )  lim sup μ(An )  lim sup An . n

n

n

n

In particular, if limn An exists, then limn μ(An ) = μ(limn An ).

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Integral and Measure

R EMARK 13.2.– The first two inequalities hold without the assumption  μ( n An ) < ∞. P ROOF.– Since μ(inf mn Am )  μ(An ) and μ(supmn Am )  μ(An ), n ∈ N, using the continuity of measure with respect monotone sequences (theorem 13.1, 4a–b), we get    μ lim inf An = μ lim ↑ inf Am = lim ↑ μ inf Am  lim inf μ(An ) n

and

n

mn

n

n

mn

 μ lim sup An = μ(lim ↓ sup Am ) = lim ↓ μ(sup Am )  lim sup μ(An ). n

n

mn

n

mn

n

  P ROPOSITION 13.3 (Borel–Cantelli lemma).– If {An , n ∈ N} is a  sequence of measurable sets from (E, A, μ) such that ∞ n=1 μ(An ) < ∞, then μ(lim sup An ) = 0. n

P ROOF.– Letting n → ∞ in the inequality   μ sup Am  μ(Am ), mn

mn

we get      μ lim sup An ) = μ lim ↓ sup Am = lim ↓ μ sup Am = 0. n

n

mn

n

mn

 

D EFINITION 13.4.– Let (E, A, μ) be a measure space. A set N ⊂ E (not necessarily from A) is a null set (with respect to measure μ) or a zeromeasure set if there is a set A ∈ A such that N ⊂ A and μ(A) = 0. A measure space (E, A, μ) is called complete if A contains all null subsets of E. It is clear that any subset of a null set also is a null set. Moreover, from theorem 13.1 (property 3) it follows that a countable union of null sets also is a null set.

Measure Spaces

205

T HEOREM 13.2.– Let (E, A, μ) be a measure space, and let N denote all its null sets. 1) A := {A ∪ N : N ∈ N} = σ(A ∪ N), that is, the family A of all sets of the form A∪ N, A ∈ A, N ∈ N, coincides with the σ-algebra generated by the families A and N; 2) The formula μ(A ∪ N) = μ(A) correctly defines a unique measure μ extending the measure μ to the σ-algebra A. 3) The measure space (E, A, μ) is complete (and is called the completion of the space (E, A, μ)). P ROOF.– 1) Clearly, A ∪ N ⊂ A ⊂ σ(A ∪ N). Therefore, it suffices to check that A is a σ-algebra. First, the family A is closed under countable  ∞ unions. Indeed, if Ai ∈ A, Ni ∈ N, i ∈ N, then ∞ i=1 Ai ∈ A and i=1 Ni ∈ N, and, therefore, ∞ i=1

∞ ∞     (Ai ∪ Ni ) = Ai ∪ Ni ∈ A. i=1

i=1

The family A is also closed with respect to complement. Indeed, let A ∈ A, N ∈ N, N ⊂ B ∈ A, μ(B) = 0. Then (A ∪ N)c = (A ∪ B)c + B ∩ (A ∪ N)c ∈ A since (A ∪ B)c ∈ A, B ∩ (A ∪ N)c ⊂ B, and thus B ∩ (A ∪ N)c ∈ N. 2) Correctness. Let us show the correctness of the formula μ(A ∪ N) = μ(A), that is, that from A1 ∪ N1 = A2 ∪ N2 ∈ A (with Ai ∈ A and Ni ∈ N, i = 1, 2) it follows that μ(A1 ) = μ(A2 ). So, let A1 ∪ N1 = A2 ∪ N2 . Then A1 \A2 ⊂ (A1 ∪ N1 )\A2 = (A2 ∪ N2 )\A2 ⊂ N2 , and, therefore, μ(A1 \A2 ) = 0. Similarly, μ(A2 \A1 ) = 0. From this we have μ(A1 )  μ(A1 \A2 ) + μ(A2 ) = μ(A2 )  μ(A2 \A1 ) + μ(A1 ) = μ(A1 ),

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Integral and Measure

and thus μ(A1 ) = μ(A2 ). σ-additivity. Checking the σ-additivity of μ is simple: ∞ ∞ ∞ ∞ ∞         μ (Ai ∪ Ni ) = μ Ai ∪ Ni = μ Ai = μ(Ai ) i=1

i=1

=

∞ 

i=1

i=1

i=1

μ(Ai ∪ Ni ).

i=1

Uniqueness. Now let μ1 be a measure on the σ-algebra A that coincides with μ on the σ-algebra A. Take an arbitrary set A ∪ N ∈ A (A ∈ A, N ∈ N). If N ⊂ B ∈ A and μ(B) = 0, then μ(A) = μ1 (A)  μ1 (A ∪ N)  μ1 (A ∪ B) = μ(A ∪ B)  μ(A) + μ(B) = μ(A), and thus μ1 (A ∪ N) = μ(A) = μ(A ∪ N). So, μ1 coincides with μ, which means that the measure μ can be extended to a measure on A in a unique way. Completeness. It remains to check the completeness of (E, A, μ). Let A be a null set with respect to μ. Then there is B ∈ A such that A ⊂ B and μ(B) = 0. By part 1, the set B is of the form B = A ∪ N, where A ∈ A, N ∈ N, and μ(A ∪ N) = μ(A) = 0. If N ⊂ B ∈ A, where μ(B) = 0, then A ⊂ B ⊂ A ∪ B ∈ A and μ(A ∪ B)  μ(A) + μ(B) = 0. Therefore, A ∈ N ⊂ A.   P.13. Problems P ROBLEM 13.1.– Find lim supn An and lim inf n An if An = B for odd n and An = C for even n. P ROBLEM 13.2.– Find lim supn An and lim inf n An if E = R and An = (−∞, an ), n ∈ N.

Measure Spaces

207

P ROBLEM 13.3.– Check and interpret in terms of sets the following relations: 

lim inf An n

c

= lim sup(Acn ); n

lim sup(An ∪ Bn ) = lim sup An ∪ lim sup Bn ; n

n

n

lim inf An ∩ lim sup Bn ⊂ lim sup(An ∩ Bn ) ⊂ lim sup An ∩ lim sup Bn . n

n

n

n

n

Show by examples that, in the last relation, the symbol ⊂ cannot be replaced by the equality signs. P ROBLEM 13.4.– For a sequence of sets {An , n ∈ N}, denote D1 = A1 , Dn+1 = Dn An+1 , n ∈ N. Prove that the sequence {Dn } has a limit if and only if limn An = ∅.1 P ROBLEM 13.5.– Let μ be a σ-finite measure on a σ-algebra A of subsets of a set E, and let D = {Di , i ∈ I} ⊂ A be a family of pairwise disjoint sets. Prove that the family of sets D ∈ D such that μ(D) = ∞ is at most countable. P ROBLEM 13.6.– Let μ be a σ-finite measure on a σ-algebra A of sets of E such that μ(E) = +∞. Prove that for every M > 0, there is a set A ∈ A such that M < μ(A) < +∞. P ROBLEM 13.7.– Let E be an infinite set, and let  μ(A) =

0 if A is a finite set, +∞ otherwise.

Show that μ: P(E) → R is an additive function but not a measure (σ-additive function). P ROBLEM 13.8.– Let E be an infinite set, and let the family A of its subsets consist of sets A that either are finite or their complements Ac 1 If we call the operation  “addition,” then this statement can be interpreted as follows: a series of sets converges if and only if its generic term converges to “zero,” the empty set.

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Integral and Measure

are finite; in the former case, let μ(A) = 0, otherwise, let μ(A) = 1. Show that: i) A is an algebra but not a σ-algebra; ii) μ is an additive function on A. For which sets E the function μ is a measure on A? P ROBLEM 13.9.– Give an example where An ↓ ∅ but limn μ(An )  0. P ROBLEM 13.10.– Let B be the Borel σ-algebra of Rk , A be the σ-algebra of Lebesgue-measurable sets of Rk , and m be the Lebesgue measure on Rk . Show that: i) (Rk , A, m) is a complete measure space; ii) (Rk , A, m) is the completion of (Rk , B, m). P ROBLEM 13.11.– Let E = R, and let Q be the set of all rational numbers. Suppose that Q is enumerated in a sequence {rn }. Define the μ on the Borel σ-algebra B on E by  μ(A) := 2−n , A ∈ B. rn ∈A

Find the complement of the measure space (R, B, μ). P ROBLEM 13.12.– Let (E, A, μ) be the complement of a measure space (E, A, μ). Prove that C ∈ A ⇐⇒ ∃ A, B ∈ A such that A ⊂ C ⊂ B and μ(B\A) = 0. P ROBLEM 13.13.– Prove that if A, B,C are sets from σ-algebra A of subsets of a set E and μ is a measure on A, then μ(A) + μ(B) + μ(C) + μ(A ∩ B ∩ C) = μ(A ∪ B ∪ C) + μ(A ∩ B) + μ(A ∩ C) + μ(B ∩ C).

14 Extension of Measure

The aim of this chapter is a proof of the important theorem about an extension of a measure on an algebra to a measure on the σ-algebra generated by that algebra. Extension of a measure is not a new situation for the reader. At the beginning of Part II, we could measure k-boxes (and their finite unions as well), then zero-measure sets, and, finally, we defined the Lebesgue measures of (Lebesgue)-measurable sets. In the general case, typically, we first try to define a measure on a relatively simple and scanty algebra (or even on a simpler family, semi-algebra). Usually, the additivity of a “candidate” measure is almost obvious, and the main difficulty is checking its σ-additivity. Of course, we would like to extend the measure as much as possible to the wider family of sets, the σ-algebra generated by the initial algebra. Happily, no principal difficulties arise – the possibility of extension is guaranteed by the main theorem of this chapter. T HEOREM 14.1.– Let μ be a measure defined on an algebra A of subsets of a set E. Then, on the σ-algebra σ(A), there exists a measure  ∗ ∗  μ (called an extension of the measure μ) such μ that μ A = μ, that is, μ∗ (A) = μ(A) for all A ∈ A. If μ is σ-finite, then such an extension is unique and also σ-finite.

210

Integral and Measure

P ROOF.– Step 1. Define the new family D of subsets of E consisting of all sets of the form lim ↑ An , where An ∈ A, n ∈ N. Let us check that, for D ∈ D, n the limit μ◦ (D) := lim ↑ μ(An ) n

does not depend on the choice of an increasing sequence {An } ⊂ A with limit lim ↑ An = D. So, let {Am , m ∈ N} and {An , n ∈ N} be two increasing sequences of sets from A such that ∞

Am =

m=1

∞

An .

n=1

Since for any m ∈ N, the sequence {Am ∩ An , n ∈ N} increases and ∞ n=1

An ⊃

∞

(Am ∩ An ) = Am ∩

n=1

∞

An = Am ,

n=1

that is, lim ↑ An ⊃ lim ↑(Am ∩ An ) = Am , n

n

by the monotonicity and continuity from below μ we have lim ↑ μ(An )  lim ↑ μ(Am ∩ An ) = μ(Am ) n

n

for all m ∈ N.

Now, taking the limit as m → ∞, we get lim ↑ μ(An )  lim ↑ μ(Am ). n

m

By exchanging the sequences we get the opposite inequality lim ↑ μ(Am )  lim ↑ μ(An ). m

n

Extension of Measure

211

Therefore, lim ↑ μ(An ) = lim ↑ μ(An ). n

n

Step 2. Let us check the following properties of the set family D and function μ◦ : D → R+ : i) A ⊂ D, and μ◦ (A) = μ(A) for all A ∈ A; ii) if D1 , D2 ∈ D, then D1 ∪ D2 , D1 ∩ D2 ∈ D, and μ◦ (D1 ∪ D2 ) + μ◦ (D1 ∩ D2 ) = μ◦ (D1 ) + μ◦ (D2 ); iii) if D1 ⊂ D2 , D1 , D2 ∈ D, then μ◦ (D1 )  μ◦ (D2 ); iv) if Dn ↑ D, Dn ∈ D, then D ∈ D and μ◦ (D) = limn μ◦ (Dn ). Property (i) is obvious. Now, consider two increasing sequences of sets from A such that An,1 ↑ D1 and An,2 ↑ D2 . Note that An,1 ∪ An,2 ↑ D1 ∪ D2 and An,1 ∩ An,2 ↑ D1 ∩ D2 . Therefore, taking the limit, as n → ∞, in the equality μ(An,1 ) + μ(An,2 ) = μ(An,1 ∩ An,2 ) + μ(An,1 ∪ An,2 ), we get property (ii). To check property (iii), taking, as before two sequences An,1 ↑ D1 and An,2 ↑ D2 , we get, as in step 1, μ(D1 ) = lim μ(An,1 )  lim μ(An,2 ) = μ(D2 ). n

n

The proof of (iv) is not so obvious. Suppose that Dn = lim ↑ Am,n , n ∈ N, where Am,n ∈ A, m, n ∈ N. m

Let Bm := Bm =



nm

nm Am,n

Am,n ⊂

∈ A. The sequence {Bm } is increasing since

nm

Am+1,n ⊂

nm+1

Am+1,n = Bm+1 .

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Integral and Measure

Since Am,n ⊂ Bm ⊂ Dm ,

n  m,

we have μ(Am,n )  μ(Bm )  μ◦ (Dm ),

n  m.

Taking the limit in the above inequalities as m → ∞, we get lim ↑ Am,n = Dn ⊂ lim ↑ Bm ⊂ lim ↑ Dm = D m

m

m

and lim ↑ μ(Am,n ) = μ◦ (Dn )  lim ↑ μ(Bm )  lim ↑ μ◦ (Dm ). m

m

m

Now taking the limits as n → ∞, we finally get D = lim ↑ Dn ⊂ lim ↑ Bm ⊂ D, n

m

that is, D = lim ↑ Bm ∈ D, m

and lim ↑ μ◦ (Dn )  lim ↑ μ(Bm )  lim ↑ μ◦ (Dm ), n

m

m

that is, lim ↑ μ◦ (Dn ) = lim ↑ μ(Bm ) = μ◦ (D). n

n

Also, note that from (ii) and (iv) it follows that D is the smallest family containing A and closed under operations of intersection and countable union.

Extension of Measure

213

Step 3. In the class P(E) of all subsets of E, define the function μ∗ by " # μ∗ (B) = inf μ◦ (D) : B ⊂ D ∈ D . We will check the following properties of μ∗ : i) μ∗ (D) = μ◦ (D) for D ∈ D; μ∗ (A) = μ(A) for A ∈ A; ii) μ∗ (B1 ∪ B2 ) + μ∗ (B1 ∩ B2 )  μ∗ (B1 ) + μ∗ (B2 ); in particular, iii) μ∗ (B ∩ C) + μ∗ (Bc ∩ C)  μ∗ (C); iv) B1 ⊂ B2 =⇒ μ∗ (B1 )  μ∗ (B2 ); v) Bn ↑ B =⇒ μ∗ (Bn ) ↑ μ∗ (B). Property (i) is obvious. ii) If at least one of the numbers μ∗ (B1 ) and μ∗ (B2 ) equals ∞, the inequality is obvious. Suppose that μ∗ (B1 ) < ∞ and μ∗ (B2 ) < ∞. By the definition of μ∗ , for arbitrary ε > 0, there exist D1 , D2 ∈ D such that Bi ⊂ Di ,

μ∗ (Bi ) +

ε  μ◦ (Bi ), 2

i = 1, 2.

Then μ∗ (B1 ) + μ∗ (B2 ) + ε  μ◦ (D1 ) + μ◦ (D2 ) = μ◦ (D1 ∪ D2 ) + μ◦ (D1 ∩ D2 )  μ∗ (B1 ∪ B2 ) + μ∗ (B1 ∩ B2 ). It remains to take the limit as ε → 0. iii) follows from (ii) by taking B1 = B∩C, B2 = Bc ∩C: then B1 ∪ B2 = C, B1 ∩ B2 = ∅. iv) It is a consequence of property (iii) in step 2. v) If μ(Bn0 ) = ∞ for some n0 , then μ(B)  μ(Bn )  μ(Bn0 ) = ∞ for all n  n0 , and the property is obvious. Therefore, suppose that μ(Bn ) < ∞ for all n. Take arbitrary ε > 0. Then there exists a sequence {Dn , n ∈ N} of sets from D such that Bn ⊂ Dn ,

μ∗ (Bn ) +

ε  μ◦ (Dn ), 2n

n ∈ N.

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Integral and Measure

 Denote Dn = mn Dm . Clearly, {Dn } is an increasing sequence of sets from D and Bn ⊂ Dn . Let us show by induction on n that  1 μ∗ (Bn ) + ε 1 − n  μ◦ (Dn ), 2

n ∈ N.

For n = 1, the inequality holds because of the choice of D1 = D1 . Suppose that it holds for some n. Since Bn ⊂ Dn ∩ Dn+1 ∈ D (because Dn ⊃ Bn and Dn+1 ⊃ Bn+1 ⊃ Bn ), we have μ◦ (Dn+1 ) = μ◦ (Dn ∪ Dn+1 ) = μ◦ (Dn ) + μ◦ (Dn+1 ) − μ◦ (Dn ∩ Dn+1 )   1   ε   μ∗ (Bn ) + ε 1 − n + μ∗ (Bn+1 ) + n+1 − μ∗ (Bn ) 2 2  1  ∗ = μ (Bn+1 ) + ε 1 − n+1 . 2 Thus, the inequality is proved. Taking the limit as n → ∞, we get  lim ↑ μ∗ (Bn ) + ε  lim ↑ μ◦ (Dn ) = μ◦ lim ↑ Dn  μ∗ (B) n

n

n

since lim ↑ Dn = n

∞ n=1

Dn =

∞ n=1

Dn ⊃

∞

Bn = B.

n=1

Since ε > 0 is arbitrary, we finally have lim ↑ μ∗ (Bn )  μ∗ (B). n

The opposite inequality follows from μ∗ (Bn )  μ∗ (B), n ∈ N (property (iv)), and so the equality lim ↑ μ∗ (Bn ) = μ∗ (B) is proved. n

Step 4. Define the new family G of all subsets G of E that satisfy the inequality μ∗ (G ∩ D) + μ∗ (Gc ∩ D)  μ∗ (D) for all D ∈ D. Let us show that G is a σ-algebra and that the restriction of μ∗ to the σ-algebra G is a measure.

Extension of Measure

215

The inequality is obvious when μ∗ (D) = ∞; therefore, it suffices to consider the sets D ∈ D such that μ(D) < ∞. Since the opposite inequality always holds, G ∈ G is equivalent to the equality μ∗ (G ∩ D) + μ∗ (Gc ∩ D) = μ∗ (D) for all D ∈ D. By the symmetry of the definition of the family G it follows that G ∈ G =⇒ Gc ∈ G. From the equality μ∗ (∅) + μ∗ (C) = μ∗ (C) we have that ∅, E ∈ G. Let G1 ,G2 ∈ G and D ∈ D. Applying property (ii) of step 3 to the sets Bi = Gi ∩ D, i = 1, 2, and then to Bi = Gci ∩ D, i = 1, 2, we get the following inequality:   μ∗ (G1 ∪ G2 ) ∩ D + μ∗ (G1 ∩ G2 ) ∩ D  μ∗ (G1 ∩ D) + μ∗ (G2 ∩ D). [14.1] Similarly,   μ∗ (Gc1 ∪ Gc2 ) ∩ D + μ∗ (Gc1 ∩ Gc2 ) ∩ D  μ∗ (Gc1 ∩ D) + μ∗ (Gc2 ∩ D), that is,   μ∗ (G1 ∪ G2 )c ∩ D + μ∗ (G1 ∩ G2 )c ∩ D  μ∗ (Gc1 ∩ D) + μ∗ (Gc2 ∩ D). [14.2] The sum of their left-hand sides equals 2μ∗ (D) by assumption. However, by property (ii) of step 3, we have   μ∗ (G1 ∪ G2 ) ∩ D + μ∗ (G1 ∪ G2 )c ∩ D  μ∗ (D),   μ∗ (G1 ∩ G2 ) ∩ D + μ∗ (G1 ∩ G2 )c ∩ D  μ∗ (D).

[14.3] [14.4]

Four inequalities [14.1–14.4] all hold only in case they all are in fact equalities. This means that G1 ∪G2 ,G1 ∩G2 ∈ G and G is an algebra. Let us check that μ∗ is additive in the family G. Indeed, take G1 ,G2 ∈ G such that G1 ∩ G2 = ∅. If at least one of the numbers μ∗ (G1 ) and μ∗ (G2 ) (say, the first one) equals ∞, then μ∗ (G1 + G2 )  μ∗ (G1 ) = ∞, and therefore

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Integral and Measure

μ∗ (G1 + G2 ) = μ∗ (G1 ) + μ∗ (G2 ) = ∞. If μ∗ (G1 ) < ∞ and μ∗ (G2 ) < ∞, then take sets D  D1 ⊃ G1 and D  D2 ⊃ G2 such that μ◦ (D1 ) < ∞ and μ◦ (D2 ) < ∞. Then μ◦ (D1 ∪ D2 ) = μ◦ (D1 ) + μ◦ (D1 ) − μ◦ (D1 ∩ D2 ) < ∞. Now, taking D = D1 ∪ D2 , from the equality   μ∗ (G1 + G2 ) ∩ D + μ∗ (G1 ∩ G2 ) ∩ D = μ∗ (G1 ∩ D) + μ∗ (G2 ∩ D) we have μ∗ (G1 + G2 ) = μ∗ (G1 ) + μ∗ (G2 ). Let {Gn } be any increasing sequence of sets from G. Then, by properties (v) and (iv) of step 3, for all D ∈ D, we respectively have    μ∗ Gn ∩ D = lim ↑ μ∗ (Gn ∩ D), n  n c  μ∗ Gn ∩ D  μ∗ (Gcm ∩ D), m ∈ N. n

Taking the limit in the last inequality as m → ∞ and adding the inequality obtained to the preceding equality, we get μ∗



   c  Gn ∩ D + μ∗ Gn ∩ D n n   lim μ∗ (Gn ∩ D) + μ∗ (Gcn ∩ D) = μ∗ (D). n→∞

By the definition of the family G it follows that ∪nGn ∈ G. This means that G is a σ-algebra. Moreover, since, on the σ-algebra G, the function μ∗ is continuous from below (that is, ∗ ∗ μ (lim ↑ Gn ) = lim ↑ μ (Gn )), it is a measure (proposition 13.1). n

n

Thus, given a measure μ on an algebra A of subsets of E, we have constructed the measure μ∗ on the σ-algebra G. Clearly, A ⊂ G and μ∗ (A) = μ(A) for all A ∈ A (property (i) of step 3), that is, μ∗ is an

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217

extension of μ to the σ-algebra G and thus to its sub-σ-algebra σ(A) ⊂ G. Step 5. It remains to prove the uniqueness of extension when the measure μ is σ-finite. First, consider the case where μ is finite. Consider two of its extensions μ1 and μ2 to the σ-algebra σ(A). Denote by M the family of all sets A such that μ1 (A) = μ2 (A). Clearly, A ⊂ M. Since for any monotone (increasing or decreasing) sequence {An , n ∈ N} of sets from M,   μ1 lim An = lim μ1 (An ) = lim μ2 (An ) = μ2 lim An , n

n

n

n

it follows that M is a monotone family. By theorem 12.1 we have that σ(A) = m(A) ⊂ M. Thus, μ1 and μ2 coincide on the σ-algebra σ(A). If μ is a σ-finite measure, there exists a sequence of finite-measure  (nonintersecting) sets {En } that cover E, that is, ∞ n=1 En = E. Let again μ1 and μ2 be two extensions of μ to the σ-algebra σ(A). Applying the just proved uniqueness of extension of the measure on each set E n , for every A ∈ A, we have ∞ ∞      μ1 (A) = μ1 A ∩ En = μ1 (A ∩ En ) n+1

=

∞ 

μ1 (A ∩ En ) =

n+1

= μ2

n+1 ∞ 

μ2 (A ∩ En )

n+1

∞ ∞      (A ∩ En ) = μ2 A ∩ E n = μ2 (A). n+1

n+1

 

We now introduce the auxiliary notion of a semi-algebra, which is very useful when considering measures on the real line and products of spaces. D EFINITION 14.1.– A family S of subsets of a set E is called a semialgebra if it satisfies the following conditions: 1) ∅, E ∈ S;

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2) S is closed with respect to intersections, that is, S 1 , S 2 ∈ S ⇒ S 1 ∩ S 2 ∈ S. 3) The complement S c of every S ∈ S can be expressed by a finite union of pairwise disjoint sets from S. E XAMPLE 14.1.– Let E = R, and let S consist of the sets ∅, R and all the intervals of the forms (−∞, a), [a, ∞) and [a, b) (that is, all rightopen intervals). Then S is a semi-algebra. Indeed, conditions 1 and 2 are clearly satisfied, and condition 3 follows from the equalities (−∞, a)c = [a, ∞),

[a, ∞)c = (−∞, a),

and [a, b)c = (−∞, a) ∪ [b, ∞). Recall that a function μ: S → R+ is said to be additive (σ-additive) if for any finite (countable) family {S i , i ∈ I} of sets from S such that its  union i∈I S i ∈ S, we have    μ Si = μ(S i ). i∈I

i∈I

T HEOREM 14.2.– Let S be a semi-algebra of subsets of a set E. Then  the algebra A generated by S consists of all sums A = i∈I S i of finite families {S i , i ∈ I} of disjoint sets from S. If μ: S → R+ is an additive function such that μ(∅) = 0, then the formula    μ Si = μ(S i ) i∈I

i∈I

correctly defines the unique additive extension μ of μ to the algebra A. If μ is σ-additive, then μ also is σ-additive. In this case, there exists a measure extending the measure μ to the σ-algebra σ(S) (which coincides with σ(A)). If μ is σ-finite, then such an extension is unique.  P ROOF.– Let A denote the family of all finite unions i∈I S i (S i ∈ S). Let us check that A is an algebra. Indeed,

Extension of Measure

219

1) ∅, E ∈ A; 2) A is closed with respect to intersections since      Si ∩ S j = (S i ∩ S j ); i∈I

j∈J

i∈I, j∈J

 3) A is closed with respect to complement. Indeed, let i∈I S i ∈ A (S i ∈ S, i ∈ I). Then S ic ∈ A by the definition of a semi-algebra. Since the family A is closed with respect to intersections, we have   c 2 Si = S ic ∈ A. i∈I

i∈I

 Thus, A is an algebra. However, all sums i∈I S i belong to the algebra generated by S. Therefore, the family A coincides with that algebra. To prove the correctness of the definition of μ , we have to check that the value of μ (A) depends only on the set A ∈ A itself and not on its  expression in the form A = i∈I S i . Suppose that   A= Si = S j . i∈I

j∈J

We have to check that Si = Si∩





i∈I μ(S i ) =



 j∈J μ(S j ).

Note that

    Si = Si∩ S j = (S i ∩ S j )

i∈I

Similarly,  S j = (S i ∩ S j )

j∈J

for all i ∈ I.

j∈J

for all j ∈ J.

i∈I

By the additivity of μ in the semi-algebra S, we have  i∈I

    μ (S i ∩ S j ) = μ(S i ∩ S j ) i∈I j∈J  i∈I  j∈J     = μ(S i ∩ S j ) = μ (S i ∩ S j ) = μ(S j ).

μ(S i ) =

j∈J i∈I

j∈J

i∈I

j∈J

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Integral and Measure

Now let us check the additivity (σ-additivity) of μ . Let  A= A j , A j ∈ A. j∈J

Moreover, suppose that the sets A and A j , j ∈ J, are expressed as follows:   j j Aj = S i , S i ∈ S; A= S k , S k ∈ S. i∈I j

k∈K

We will use the equalities      j Sk = A∩Sk = Aj ∩S k = Si ∩Sk j∈J  j = (S i ∩ S k ),

j∈J i∈I j

k ∈ K,

j∈J i∈I j

and j

j

Si = A∩Si =

 k∈K

  j j Sk ∩Si = (S i ∩ S k ),

i ∈ I j , j ∈ J.

k∈K

Since μ is additive (σ-additive) on the semi-algebra S, we obtain   j μ (A) = μ(S k ) = μ(S i ∩ S k ) =

k∈K   j∈J i∈I j k∈K

k∈K j∈J i∈I j j μ(S i ∩ S k ) =

 j∈J i∈I j

j

μ(S i ) =



μ (A j ).

j∈J

The uniqueness of extension is obvious. The last statement of the theorem follows from the theorem on the extension of a measure.   P.14. Problems P ROBLEM 14.1.– Let E be the set of all rational numbers of the interval [0, 1], and let A be the algebra consisting of all finite unions of “semiclosed intervals” E ∩ [a, b) (where a and b are rational numbers).

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221

Show that σ(A) = P(E). For a set A ∈ P(E), denote by μ1 (A) the number of elements of A (μ1 (A) := ∞ for infinite A). Define another measure by μ2 (A) = 2μ1 (A). Check that μ1 and μ2 coincide on the algebra A but do not coincide on the σ-algebra σ(A). Thus, the uniqueness does not hold without the assumption that the measure μ is σ-finite. P ROBLEM 14.2.– Let S1 and S2 be semi-algebras of subsets of a set E. Prove that the family S := {S 1 ∩ S 2 : S 1 ∈ S1 , S 2 ∈ S2 } also is a semialgebra and that the algebra (σ-algebra) generated by S coincides with the algebra (σ-algebra) generated by the family S1 ∪ S2 . P ROBLEM 14.3.– Let E be the set of all rational numbers of the interval [0, 1], and let S be the semi-algebra consisting of all “semiclosed intervals” E ∩ [a, b) (where a and b are rational numbers). Define on S the function μ by  μ E ∩ [a, b) = b − a. Prove that 1) μ is an additive function continuous from above and below; 2) μ is not a σ-additive function. So proposition 13.1 does not hold with an algebra replaced by a semi-algebra.

15 Lebesgue–Stieltjes Measures on the Real Line and Distribution Functions

We now apply the above theory to the case of real line. S denotes the semi-algebra of subsets of E = R constructed in example 14.1 and consists of ∅, R, and all the intervals of the forms (−∞, a), [a, ∞), [a, b). From theorem 14.2, the algebra A generated by this semi-algebra consists of all finite unions of such intervals. As already mentioned, the σ-algebra B = B(R) generated by S (and by the algebra A as well) is called the Borel σ-algebra, and the sets belonging to B are called Borel sets (of the real line). The one-point sets are Borel sets  1 since {a} = ∞ n=1 [a, a + n ) is a countable intersection of the intervals [a, a + 1n ) ∈ S ⊂ B). From this it follows, in particular, that all countable sets are Borel sets as countable unions of one-point sets. For example, the set of rational numbers and its complement, the set of irrational numbers, are Borel sets. All intervals also are Borel sets. For example, [a, b] = [a, b) ∪ {b}, etc. Among measures on the Borel σ-algebra on the real line, particularly important are the Lebesgue–Stieltjes measures that assign finite values to finite intervals. Since measures are functions of sets, it is difficult to consider them in terms of classic calculus. Therefore, it is important to establish relations between notions of measure theory and notions of classic calculus. We will show that there is a certain class of functions on the real line that is in one-to-one correspondence (up to an additive constant) with the class of Lebesgue–Stieltjes measures.

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Integral and Measure

D EFINITION 15.1.– Finite, non-decreasing, and left-continuous1 function F: R → R is called a distribution function. For any distribution function F, in the semi-algebra S, define the function μ by  μ [a, b) = F(b) − F(a), −∞ < a  b  ∞,  μ (−∞, b) = F(b) − F(−∞), −∞ < b  ∞, where F(±∞) := lim x→±∞ F(x). T HEOREM 15.1.– The above-defined function μ: S → R+ is σ-additive, that is, In ∈ S,

n ∈ N,

∞ 

In ∈ S

=⇒

n=1

∞ ∞    μ In = μ(In ). n=1

n=1

From this theorem it follows that the function μ can be extended to a measure (which we also denote by μ) in the Borel σ-algebra B generated by the semi-algebra S. Moreover, by theorem 13.2, the measure μ can be extended to a complete measure μ in the σ-algebra B. The measures μ and μ are called Lebesgue–Stieltjes measures (corresponding to the distribution function F). We also say that the function F generates the measure μ. The sets from the σ-algebra B are called Lebesgue–Stieltjes sets (or sets measurable in the Lebesgue–Stieltjes sense). In particular, the function F(x) ≡ x generates the Lebesgue measure, which we already know, and the σ-algebra obtained by completion of the Borel σ-algebra in measure coincides with the σ-algebra of Lebesgue-measurable sets (this is not obvious). P ROOF.– [of theorem 15.1] Step 1. We first show that if I=

∞ 

In ,

I = [a, b), In = [an , bn ),

n=1

1 An equivalent theory can be given in terms of right-continuous functions.

Lebesgue–Stieltjes Measures on the Real Line and Distribution Functions

225

then μ(I) 

∞ 

μ(In ).

n=1

Take any k ∈ N. By changing, if necessary, the index order we may suppose that a  a1  b1  a2  · · ·  ak  bk  b. Then k 

μ(In ) =

n=1

k 

n=1



k 

F(bn ) − F(an )



 

 F(bn ) − F(an ) + F(an+1 ) − F(bn ) k−1

n=1

=

k−1 

n=1

 F(an+1 ) − F(an ) = F(bk ) − F(a1 )

n=1

 F(b) − F(a) = μ(I).  Thus, we have that kn=1 μ(In )  μ(I) for all k ∈ N. Taking the limit ∞ as k → ∞, we get n=1 μ(In )  μ(I). Step 2. Let us show the opposite inequality, that μ(I)   I= ∞ n=1 In , I = [a, b), In = [an , bn ).

∞

n=1 μ(In )

if

Take an arbitrary ε ∈ (0, b − a). Denote I ε = [a, b − ε]. From the leftcontinuity of F it follows that ∀n ∈ N,

 ε ∃δn > 0 : μ [an − δn , an ) = F(an ) − F(an − δn ) < n . 2

Denote Inε = (an − δn , bn ). Note that Iε ⊂ I =

∞  n=1

In ⊂

∞ n=1

Inε .

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Integral and Measure

Since the interval I ε is a compact and Inε are open intervals, there exists n0 ∈ N such that Iε ⊂

n0

Ikε .

k=1

Take k1  n0 such that a ∈ Ikε1 . If b − ε  Ikε1 , then take k2  n0 such that bk1 ∈ Ikε2 , etc. Continue the process until we get b − ε ∈ Ikεm for some m. Then ε

I ⊂

m

Ikεi .

i=1

Without loss of generality (omitting, if necessary, the unchosen intervals and changing the index numbering), we may suppose that ε

I ⊂

m

Ikε

k=1

and a 1 − δ1 < a < b 1 , ak+1 − δk+1 < bk < bk+1 , k = 1, 2, . . . , m − 1, am − δm < b − ε < bm . Then, we get  μ [a, b − ε)  F(bm ) − F(a1 − δ1 ) = F(b1 ) − F(a1 − δ1 ) + 

m 

k=1

 F(bk+1 ) − F(bk )

k=1

F(bk ) − F(ak − δk )



  ε F(bk ) − F(ak ) + 2k k=1 k=1 ∞ ∞  

  F(bk ) − F(ak ) + ε = μ(Ik ) + ε. 

m 

m 

k=1

m

k=1

Lebesgue–Stieltjes Measures on the Real Line and Distribution Functions

Taking the limit as ε ↓ 0, we get μ(I) 

227

∞

n=1 μ(In ).

So, from steps 1 and 2 it follows that if I =  [an , bn ), then μ(I) = ∞ n=1 μ(In ).

∞

n=1 In ,

I = [a, b), In =

Step 3. Now it remains to prove that for any infinite interval I ∈ S, I=

∞ 

In ,

In ∈ S, n ∈ N

=⇒

μ(I) =

n=1

∞ 

μ(In ).

n=1

From the definition of μ it easily follows that μ(I) = lim ↑ μ(I ∩ [−k, k)) for all I ∈ S. For example, if I = (−∞, b), then k

I ∩ [−k, k) = [−k, b) for k  b, and therefore,    lim ↑ μ I ∩ [−k, k) = lim ↑ μ [−k, b) = lim ↑ F(b) − F(−k) k

k

k

= F(b) − F(−∞) = μ(I).

Continuing this equality, we have    ∞  μ(I) = lim ↑ μ In ∩ [−k, k) k

n=1

∞ ∞      = lim ↑ μ In ∩ [−k, k) = lim ↑ μ In ∩ [−k, k) k

= =

∞  n=1 ∞ 

k

n=1



lim ↑ μ In ∩ [−k, k)

n=1



k

∞     μ lim ↑ In ∩ [−k, k) = μ(In ). k

n=1

n=1

where we could interchange the symbols lim ↑ and k

[−k, k)) ↑ μ(In ), k → ∞.2

∞

n=1

since 0  μ(In ∩  

2 Check that lim ↑ lim ↑ xnk = lim ↑ lim ↑ xnk if all the sequences {xnk , k ∈ N} (n ∈ N) and n

k

k

{xnk , n ∈ N} (k ∈ N) are increasing.

n

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Integral and Measure

Thus, for any distribution function, we can construct a Lebesgue–Stieltjes measure on the Borel σ-algebra B of the real line or on the completed σ-algebra B. A natural question arises whether for every Lebesgue–Stieltjes measure there exists a distribution function that generates the measure? The answer is positive: P ROPOSITION 15.1.– Every Lebesgue–Stieltjes measure on the Borel σ-algebra B of the real line has a unique (up to an additive constant) distribution function that generates this measure. P ROOF.– Let μ be an arbitrary Lebesgue–Stieltjes measure on B, i.e. a measure that takes finite values on finite intervals. Take any fixed point a0 ∈ R and define the function  μ[a0 , x), x > a0 , F(x) = −μ[x, a0 ), x  a0 . Let us check that F is a distribution function. Its monotonicity obviously follows from the monotonicity of the measure. The left-continuity follows from the continuity of measure: xn ↑ x > a0 =⇒ =⇒ xn ↑ x  a0 =⇒ =⇒

[a0 , xn ) ↑ [a0 , x) F(xn ) = μ[a0 , xn ) ↑ μ[a0 , x) = F(x); [xn , a0 ) ↓ [x, a0 ) F(xn ) = −μ[xn , a0 ) ↑ −μ[x, a0 ) = F(x).

We easily check that the distribution function F generates the measure μ: ⎧ ⎪ μ[a0 , b) − μ[a0 , a) = F(b) − F(a) if a0 < a, ⎪ ⎪ ⎨ μ[a, a0 ) + μ[a0 , b) = F(b) − F(a) if a < a0  b, μ[a, b) = ⎪ ⎪ ⎪ ⎩ μ[a, a ) − μ[b, a ) = F(b) − F(a) if b < a . 0

0

0

Now, suppose that two distribution functions F1 and F2 generate the same Lebesgue–Stieltjes measure μ. Fix any point a0 ∈ R. Then  μ[a0 , x), x > a, F 1 (x) − F 1 (a0 ) = F 2 (x) − F2 (a0 ) = −μ[x, a0 ), x  a.

Lebesgue–Stieltjes Measures on the Real Line and Distribution Functions

229

Therefore, F 1 (x) − F2 (x) = F1 (a0 ) − F 2 (a0 ) = const. Thus, any two distribution functions that generate the same Lebesgue–Stieltjes measure differ by a constant.   P.15 Problems In problems 15.1–15.5, μ is a Lebesgue–Stieltjes measure on the Borel σ-algebra B of the real line, and F is a distribution function that generates μ. P ROBLEM 15.1.– Define the measure μ by  μ(B) :=

1, 0 ∈ B, 0, 0  B.

It is called the Dirac measure. Write F. Which sets belong to the completed σ-algebra B? P ROBLEM 15.2.– Let μ(B) be equal to the number of natural numbers belonging to a set A. Write F. Which sets belong to the completed σ-algebra B? P ROBLEM 15.3.– Let all the rational numbers be written in a sequence {rn , n ∈ N}. Define  μ(B) := 2−n , B ∈ B. rn ∈B

Prove that F is continuous at all irrational points and discontinuous at all rational points. Which sets belong to the completed σ-algebra B? P ROBLEM 15.4.– Prove that the measure μ({x}) (of a single-point set {x}) equals zero if and only if F is continuous at the point x.

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Integral and Measure

P ROBLEM 15.5.– Let f : R → R be a non-negative continuous function, and let  b  μ [a, b) := f (x) dx, −∞ < a  b < +∞. a

Prove that this formula defines a Lebesgue–Stieltjes measure on the real line. ( f is called the density of such a measure.) Write F. P ROBLEM 15.6.– Give an example of a measure on the σ-algebra B that is not a Lebesgue–Stieltjes measure. Further, by the measure we mean the Lebesgue measure m on the real line (F(x) ≡ x). P ROBLEM 15.7.– Prove that the measure of any countable set equals 0. P ROBLEM 15.8.– Let A be the set of all rational numbers of the interval [0, 1] enumerated in some way: A = {r1 , r2 , . . .}. For ε > 0 and i ∈ N, let denote Bi (ε) the closed interval of length ε · 2−i with center ri . Also, denote B(ε) =

∞ i=1

Bi (ε),

∞   2 1 B= B . n i=1

Prove that: i) m(B) = 0; ii) B is not a countable set. P ROBLEM 15.9.– Let An ⊃ An+1 , m(An ) = ∞, n ∈ N. May the intersection ∞ n=1 An have a) an infinite measure, b) finite positive measure, c) zero measure? P ROBLEM 15.10.– May a measurable unbounded set have a finite positive measure? P ROBLEM 15.11.– Let A be a measurable set of positive measure. Prove that there exist two points in A such that the distance between them is an irrational number.

Lebesgue–Stieltjes Measures on the Real Line and Distribution Functions

231

P ROBLEM 15.12.– Let A be the set of points in the interval [0, 1] such that, in their binary fraction representations, there are zeros at all even places. Prove that m(A) = 0. P ROBLEM 15.13.– Let A1 ⊂ [0, 1], A2 ⊂ [0, 1] and m(A1 ) + m(A2 ) > 1. Prove that m(A1 ∩ A2 ) > 0. P ROBLEM 15.14.– Let Ai ⊂ [0, 1], i = 1, 2, . . . , n, and m(A1 ) + m(A2 ) + · · · + m(An ) > n − 1. Prove that n 2  m Ak > 0. k=1

16 Measurable Mappings and Real Measurable Functions

For any mapping f : E1 → E 2 , we can define the “inverse” mapping f −1 : P(E2 ) → P(E1 ) by " # f −1 (B) = x ∈ E 1 : f (x) ∈ B ,

B ⊂ E2 ,

This mapping preserves the set operations of union, intersection, and taking the complement, i.e. f −1 (∅) = ∅, f

−1

  Bi = f −1 (Bi ), i∈I

 f −1 (Bc ) = f −1 (B) c ,

f −1 (E2 ) = E1 , f

−1

2  2 Bi = f −1 (Bi )

i∈I

i∈I

i∈I

for any family {Bi , i ∈ I} of subsets of E 2 . For any family C of subsets of a set E 2 , we denote f −1 (C) := { f −1 (B), B ∈ C}. From the above formulas it follows that if A is a σ-algebra of subsets of E 2 , then f −1 (A) is a σ-algebra of subsets of E 1 .

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Integral and Measure

L EMMA 16.1.– Let f : E 1 → E2 . Then, for any family C of subsets of E2 ,   σ f −1 (C) = f −1 σ(C) . P ROOF.– Denote A1 = σ( f −1 (C)) and A2 = σ(C). Since f −1 (A2 ) is a σ-algebra that contains f −1 (C), we have f −1 (A2 ) ⊃ A1 . However, the family B2 = {B ⊂ E 2 : f −1 (B) ∈ A1 } is a σ-algebra of subsets of E2 that contains C. Therefore, B2 ⊃ A2 , and f −1 (A2 ) ⊂ f −1 (B2 ) ⊂ A1 . Thus, A1 = f −1 (A2 ), and the lemma is proved.

 

D EFINITION 16.1.– Let (E 1 , A1 ) and (E2 , A2 ) be measurable spaces. The mapping f : E 1 → E 2 is called measurable (to be precise, A1 /A2 measurable) if f −1 (A2 ) ⊂ A1 or, in other words, f −1 (B) ∈ A1 for all B ∈ A2 . In such a case, we will write f : (E1 , A1 ) → (E 2 , A2 ). P ROPOSITION 16.1.– Let (E1 , A1 ) and (E 2 , A2 ) be measurable spaces. Suppose that a family C ⊂ A2 generates the σ-algebra A2 and, for a mapping f : E 1 → E2 , f −1 (C) ⊂ A1 . Then f is measurable. P ROOF.– From lemma 16.1 we can write   f −1 (A2 ) = f −1 σ(C) = σ f −1 (C) ⊂ σ(A1 ) = A1 .

 

The Borel σ-algebra B in the extended real line R is defined as the σ-algebra generated by the intervals [−∞, a), a ∈ R. It is easy to see that B consists of all the sets of the form B, B ∪ {−∞}, B ∪ {−∞, +∞}, and B ∪ {+∞}, where B are Borel sets on the real line. D EFINITION 16.2.– A real measurable function in a measurable space (E, A) is a mapping f : (E, A) → (R, B). A simple real function in (E, A) is a finite real measurable function taking finitely many different values. In the particular (and typical) case where E is a metric space and A is the σ-algebra of its Borel subsets, real measurable functions in (E, A) are called Borel functions.

Measurable Mappings and Real Measurable Functions

235

When speaking about real measurable functions, we will usually omit the word “real” for brevity. In this chapter, we further consider a fixed measurable space (E, A). Recall that the indicator of a set A ⊂ E is the function  1 if x ∈ A, 1IA (x) = 0 if x  A. If A ∈ A, then 1IA (another frequent notation is χA ) is a simple function. The following properties of indicators are obvious: 1IAc = 1I − 1IA ; 1IA+B = 1IA + 1IB ; 1IA∩B = 1IA · 1IB ; 1Isup{A,B} = sup{1IA , 1IB }; 1Iinf{A,B} = inf{1IA , 1IB }. Here we denoted sup{A, B} := A ∪ B and inf{A, B} := A ∩ B. Such a notation is motivated by analogy with real numbers: for example, A ∪ B is the minimal set containing both A and B (i.e. “greater than or equal to” both). Note that every simple function can be written as a linear combination of indicators of measurable sets. Indeed, suppose that different values of a simple function f are y1 , y2 , . . . , yn . Then, denoting Ak = f −1 (yk ) = {x : f (x) = yk } ∈ A, we have f (x) =

n 

yk 1IAk (x),

x ∈ E.

k=1

However, every function of this form (with Ak ∈ A) is a simple function. P ROPOSITION 16.2 (Properties of simple functions).– Let f, g ∈ S and α, β ∈ R. Then α f + βg ∈ S, f g ∈ S, max{ f, g} ∈ S, min{ f, g} ∈ S.

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Integral and Measure

P ROOF.– For simple functions   f= yk 1IAk and g = z j 1I B j , j∈J

k∈K

we have α f + βg =



(αyk + βz j )1IAk ∩B j ∈ S,

k∈K j∈J

fg =



(yk z j )1IAk ∩B j ∈ S,

k∈K j∈J

max{ f, g} =



max{yk , z j }1IAk ∩B j ∈ S,

k∈K j∈J

min{ f, g} =



min{yk , z j }1IAk ∩B j ∈ S

k∈K j∈J

since Ak ∩ B j ∈ A for all k ∈ K, j ∈ J.

 

Recall the notation " # f + (x) = max f (x), 0 ,

" # f − (x) = max − f (x), 0

for a function f : E → R. The functions f + and f − are called the positive and negative parts of f . It is obvious that f ±  0, f = f + − f − and | f | = f + + f − . Note that f + and f − are measurable functions if f is. Indeed, for all B ⊂ R, we have ⎧ −1  ⎪ ⎪ if 0  B, ⎨ f B ∩ (0, ∞] + −1 ( f ) (B) = ⎪   ⎪ −1 −1 ⎩ f B ∩ (0, ∞] ∪ f [−∞, 0] if 0 ∈ B. Therefore, if f is measurable and B ∈ B, then ( f + )−1 (B) ∈ A. The measurability of f − can be checked similarly. Further, for short, the sets of the form "

# x : f (x) ∈ B ,

"

# x : f (x)  g(x) ,

"

# x : lim sup fn (x)  a , n

Measurable Mappings and Real Measurable Functions

237

and so on, are denoted as { f ∈ B},

"

{ f  g},

# lim sup fn  a ,

and so on.

n

P ROPOSITION 16.3.– A function f : E → R is measurable if and only if there exists a sequence { fn , n ∈ N} of simple functions that converges to f at all points x ∈ E. If f  0, then the sequence { fn } can be taken positive and increasing. P ROOF.– Necessity. Let f be a measurable function. Since f is a difference of two non-negative measurable functions, f + − f − , it suffices to consider the case of non-negative f . Denote fn =

n·2n  i−1 i=1

2n

1I{ i−1n  f < 2

i } 2n

+ n1I{ f n}

or, in other words, ⎧ ⎪ i−1 i−1 i ⎪ ⎪ ⎨ n if  f (x) < n , n fn (x) = ⎪ 2 2 2 ⎪ ⎪ ⎩n if f (x)  n.

i = 1, 2, . . . , n 2n ,

Note that the sequence { fn , n ∈ N} is increasing and f (x) − 2−n  fn (x)  f (x) if f (x) < n; fn (x) = n  f (x) if f (x)  n. From this it follows that lim ↑ fn (x) = f (x) for all x ∈ E. n

Sufficiency. Let { fn , n ∈ N} be a sequence of simple functions such that limn fn (x) = f (x) for all x ∈ E. Then, for all a ∈ R, "

# " 1# x : f (x) < a − x : f (x) < a = k k=1 ∞

=

∞

3 14 lim inf x : fn (x) < a − n→∞ k k=1

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Integral and Measure

=

∞ ∞ 2 ∞ 3

x : fn (x) < a −

k=1 m=1 n=m

14 ∈ A. k

Indeed, these equalities of sets follow from the relations f (x) < a ⇐⇒ ∃ k ∈ N : f (x) < a −

1 k

 1 ⇐⇒ ∃ k ∈ N : ∃m ∈ N : ∀n  m, fn (x) < a − . k Since the family of intervals [−∞, a), a ∈ R, generates the σ-algebra B, from proposition 16.1 it follows that f is a measurable function.   T HEOREM 16.1.– Let f and g be measurable functions. Then 1) c f is a measurable function for all c ∈ R; 2) f + g is a measurable function, provided that f (x) + g(x)  (±∞) + (∓∞) (for all x ∈ E); 3) f g is a measurable function, provided that f (x)g(x)  0 · ∞ or ∞ · 0; 4) f /g is a measurable function, provided that f (x)/g(x) 

0 0

or

∞ ∞;

5) If { fn } is a sequence of measurable functions, then supn fn , inf n fn , lim supn fn , and lim inf n fn are measurable functions. P ROOF.– Properties 1–4 can be simply checked by writing f and g as limits of simple functions and applying propositions 16.2 and 16.3. 5) Since "

# " # x: inf fn (x) < a = x: fn (x) < a ∈ A, n

a ∈ R,

n

and the family of intervals [−∞, a), a ∈ R, generates the σ-algebra B, inf n fn is a measurable function by proposition 16.1; the function supn fn = − inf n (− fn ) is also measurable. From this we immediately obtain that the functions lim supn fn = inf n supmn fm and lim inf n fn = supn inf mn fm are also measurable.  

Measurable Mappings and Real Measurable Functions

239

P.16. Problems P ROBLEM 16.1.– Show that the composition f  ◦ f of two measurable mappings f : (E, A) → (E  , A ) and f  : (E  , A ) → (E  , A ) (defined by ( f  ◦ f )(x) = f  ( f (x))) is a measurable function. P ROBLEM 16.2.– Which functions f : E → R are measurable in two “extreme” cases, A = P(E) and A = {∅, E}? P ROBLEM 16.3.– Show that every monotone function f : R → R is a Borel function. P ROBLEM 16.4.– Prove that every continuous function f : R → R is a Borel function. P ROBLEM 16.5.– If f is a measurable function, then, for all a ∈ R, {x: f (x) = a} is a measurable set. Is the converse true? P ROBLEM 16.6.– Let f be a function such that f 2 is a measurable function. Show that the function f itself need not be measurable. P ROBLEM 16.7.– Prove that f is a measurable function if f 3 is. P ROBLEM 16.8.– Let f : R → R be a differentiable function. Prove that its derivative f  is a Borel function. P ROBLEM 16.9.– Let f : R → R, and let Q be the set of rational numbers. Prove that 1IQ f is a Borel function. P ROBLEM 16.10.– Let f : (E, A) → (R, B), and let μ be a finite measure on (E, A). Prove that for every ε > 0, there exists a set A ∈ A such that f is bounded on A and μ(Ac ) < ε. Is this true if the measure μ is not finite?

17 Convergence Almost Everywhere and Convergence in Measure

Let (E, A, μ) be a measure space. We say that a property P(x) of points x from a set E holds almost everywhere (a.e.) with respect to the measure μ (or μ-almost everywhere or, shortly, μ-a.e.) if " # μ x ∈ E : element x does not possess the property P(x) = 0. For example, a measurable function f : E → R is said to be i) finite a.e. if μ{x ∈ E : | f (x)| = +∞} = 0; ii) zero a.e. if μ{x ∈ E : f (x)  0} = 0; iii) continuous a.e. if the measure of its discontinuity points T is zero: μ(T ) = 0. If E 0 ∈ A and μ(E0c ) = 0, then a function f : E0 → R is said to be defined a.e. It is often convenient to suppose that such a function is extended in some way to a function on the whole space E; for example, we set f (x) := 0 for x ∈ E 0c . We will say that two measurable functions f and g are equal a.e. or equivalent if their difference f − g equals zero a.e. In such a case, we a.e. will write f = g a.e. or f === g. It is easy to see that if a.e.

f === g and

a.e.

f1 === g1 ,

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Integral and Measure

then a.e.

a.e.

c f === cg,

f + f1 === g + g1 ,

a.e.

f f1 === gg1 ,

etc.,

where we suppose that all expressions considered are defined a.e., that is, at all points of E except a set of zero measure (where, as mentioned a.e. before, we may suppose that they are equal to zero). Similarly, if fi === gi , i ∈ N, then a.e.

a.e.

sup fi === sup gi , i∈N

inf fi === inf gi . i∈N

i∈N

i∈N

D EFINITION 17.1.– A sequence of real measurable functions { fn , n ∈ N} converges almost everywhere if a.e.

lim sup fn === lim inf fn . n→∞

n→∞

Clearly, the limit of such a sequence limn→∞ fn is defined almost everywhere. P ROPOSITION 17.1 (Cauchy criterion).– A sequence of a.e. finite measurable functions { fn , n ∈ N} converges almost everywhere to an a.e. finite measurable function if and only if it is a Cauchy sequence a.e., that is, a.e.

lim ( fn − fm ) === 0.

n,m→∞

P ROOF.– This criterion simply follows from the Cauchy convergence criterion for real sequences since, for any x ∈ E, the limit lim fn (x) exists if and only if lim ( fn (x) − fm (x)) = 0,

n,m→∞

and thus, 3

4 3 4  x : lim sup fn (x) = lim inf fn (x) ∈ R = x : lim fn (x) − fm (x) = 0 . n

n

n,m→∞

 

Convergence Almost Everywhere and Convergence in Measure

243

P ROPOSITION 17.2.– For the convergence a.e. of a sequence { fn , n ∈ N} of measurable functions, it suffices that there exists a series of positive  numbers ∞ n=1 εn < ∞ for which ∞   "  # μ x :  fn+1 (x) − fn (x) > εn < ∞. n=1

P ROOF.– For n ∈ N, denote  "  # An = x :  fn+1 (x) − fn (x) > εn . By the Borel–Cantelli lemma (proposition 13.3) we have that μ(lim supn An ) = 0. So, if x  lim supn An , the inequality | fn+1 (x) − fn (x)| > εn holds for finitely many n, that is, | fn+1 (x) − fn (x)|  εn starting with some n = n0 (depending on x). Therefore, n ∞  ∞ 0 −1     f (x) − f (x)   f (x) − f (x) + εn < ∞. n+1 n n+1 n n=1

n=n0

n=1

From this it follows that for all x ∈ (lim supn An )c , there exists a finite limit  f (x) = lim fN (x) = lim f1 (x) + N

= f1 (x) +

N

∞  

N  

fn+1 (x) − fn (x)



n=1

fn+1 (x) − fn (x) .

 

n=1

D EFINITION 17.2.– A sequence of real measurable functions { fn , n ∈ N} converges in measure to an a.e. finite measurable function f if, for all ε > 0,  "  # μ x :  fn (x) − f (x) > ε → 0,

n → ∞. μ

In such a case, we write f = μ-limn fn or fn → f .

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Integral and Measure

T HEOREM 17.1 (Cauchy criterion of convergence in measure).– A sequence of a.e. measurable functions { fn , n ∈ N} converges in measure if and only if it is a Cauchy sequence in the sense of convergence in measure, that is, for all ε > 0,  "  # μ x :  fn (x) − fm (x) > ε → 0, m, n → ∞. P ROOF.– Sufficiency. Since "

     # "  # x :  fn (x) − fm (x) > ε ⊂ x :  fn (x) − f (x) +  f (x) − fm (x) > ε 3   ε4 3   ε4 ⊂ x :  fn (x) − f (x) > ∪ x :  fm (x) − f (x) > , 2 2

we have  "  # μ x :  fn (x) − fm (x) > ε 3   ε4  μ x :  fn (x) − f (x) > 3   ε24 +μ x :  fm (x) − f (x) > → 0, 2

n, m → ∞. μ

Necessity. Suppose that ( fn − fm ) → 0, n, m → ∞. First, we will show that there exists a subsequence { fn j , j ∈ N} that converges everywhere. Let n1 = 1. If n j−1 is already chosen, let n j be the minimal natural number N > n j−1 for which 3  4  μ x :  fr (x) − f s (x) > 2− j < 3− j

for r, s  N.1

Note that ∞ 3 ∞    1 4  1 μ x :  fn j+1 (x) − fn j (x) > j < < ∞. 2 3j j=1 j=1

By proposition 17.2, the sequence { fn j , j ∈ N} converges a.e. to a measurable function f .

1 Instead of 2− j and 3− j , we can take any positive numbers ε j and εj such that  +∞ and j εj < +∞.



j εj

<

Convergence Almost Everywhere and Convergence in Measure

245

Denote Bj =

∞ 3

4 x : | fnk+1 (x) − fnk (x)| > 2−k ,

j ∈ N.

k= j

Take an arbitrary ε > 0 and take j such that have, for all x ∈ Bcj and i > j,    fni (x) − fn j (x) =

∞

−k k= j 2

< ε. Then we

 i−1      fnk+1 (x) − fnk (x)    k= j

∞  ∞      fnk+1 (x) − fnk (x)   2−k < ε. k= j

k= j

Taking the limit as i → ∞, we get    f (x) − fn j (x)  ε for all x ∈ Bcj , except a zero-measure set. Therefore, ∞   "  #   μ x : f (x) − fn j (x) > ε  μ(B j )  3−k → 0,

j → ∞.

k= j

From this, passing to the limit in the inequality  "  # μ x :  fn (x) − f (x) > ε 3   ε4 3   ε4  μ x :  fn (x) − fn j (x) > + μ x :  fn j (x) − f (x) > 2 2 as n, j → ∞, we get that μ{x : | fn (x) − f (x)| > ε} → 0, n → ∞, that is, μ fn → f , n → ∞.   P ROPOSITION 17.3.– If a sequence of a.e. finite measurable functions { fn , n ∈ N} converges in measure to an a.e. finite measurable function f , then there exists a subsequence { fn j , j ∈ N} converging a.e. If the measure μ is finite, then every converging a.e. sequence of measurable functions converges in measure.

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Integral and Measure

P ROOF.– The first statement was checked in the proof of theorem 17.1. Now suppose that fn → f a.e. Then, for every ε > 0,  "  # lim sup μ x :  fn (x) − f (x) > ε n   " ##  μ lim sup{x :  fn (x) − f (x) > ε " n #  μ x : −ε + lim sup fn (x) < f (x) < ε + lim inf fn (x) c = 0, n

n

μ

and, thus, fn → f , n → ∞.

 

P.17. Problems μ

μ

P ROBLEM 17.1.– Suppose that fn → f , gn → g, n → ∞, and that the measure μ is finite. Prove that μ

1) a fn + bgn → a f + bg, a, b ∈ R; μ

2) | fn | → | f |; μ

3) fn2 → fn ; μ

4) fn gn → f g. μ

Is it true that 1/ fn → 1/ f , provided that f (x)  0 for all x? P ROBLEM 17.2.– Give examples of sequences of a.e. finite measurable functions { fn , n ∈ N} such that: i) { fn } converges a.e. but does not converge in measure; ii) { fn } converges in measure but does not converge a.e.2 P ROBLEM 17.3.– Let A = P(N), and let the measure μ(A) of a set A ⊂ N be defined as the number of its elements. Prove that the convergence in measure μ of a sequence of functions defined on N is equivalent to the uniform convergence.

2 Note that does not converge a.e. is stronger than is not convergent a.e.

Convergence Almost Everywhere and Convergence in Measure

247

P ROBLEM 17.4.– Let f and fn , n ∈ N, be a.e. finite measurable functions, εn ↓ 0, n → ∞, and ∞   "  # μ x :  fn (x) − f (x) > εn < ∞. n=1

Prove that fn → f μ-a.e. P ROBLEM 17.5.– Let (E, A, μ) be a measure space, and (E, A, μ) be its complement. Prove that for every real measurable function f in (E, A), there is a real measurable function f  in (E, A) such that f  = f μ-a.e.

18 Integral

In this chapter, we assume that (E, A, μ) is a fixed measure space. D EFINITION 18.1.– The integral of a non-negative simple function f =  k yk 1IAk (i.e., f (x) = yk for x ∈ Ak ) with respect to measure μ is the sum 

yk μ(Ak ).

[18.1]

k

We denote it by    f dμ, f (x) μ(dx), or, shortly, by f. E

E

R EMARK 18.1.– We first define the integral for non-negative simple functions because we want to avoid infinities of opposite signs in the sum [18.1]. We first must

check that the definition is correct, that is, the value of the integral E f dμ depends only on the function f and not on its  representation in the form f = yk 1IAk . Indeed, let z1 , z2 , . . . , zn be all different values of f . Then  k

yk μ(Ak ) =

n   l=1 k:yk =zl

yk μ(Ak ) =

n   l=1 k:yk =zl

zl μ(Ak )

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Integral and Measure

=

n  l=1

zl



μ(Ak ) =

k:yk =zl

n 

zl μ{ f = zl }.

l=1

Thus, the sum [18.1] does not depend on the representation of f . We now state several properties of the integral in the class S+ that follow directly from the definition. P ROPOSITION 18.1 (Properties of the integral of non-negative simple functions).–   1) 1IA = μ(A), 1 = μ(E);   2) f  0; f = 0 if and only if f = 0 a.e.;   3) (c f ) = c f for all c  0;    4) ( f + g) = f + g;   5) If f  g, then f  g;  6) If f  c = const, then f  cμ(E). P ROOF.–

1) 1IA = 1 · μ(A) + 0 · μ(Ac ) = μ(A).  2) If f = k yk 1IAk  0 (= 0), then all yk  0 (respectively, all yk = 0),

 and therefore f = k yk μ(Ak )  0 (respectively, f = 0. Conversely, if the sum [18.1] equals zero, then, for all k, either yk = 0 or μ(Ak ) = 0. Therefore, 

⎛ ⎞ ⎜⎜⎜ ⎟⎟⎟  f = 0 =⇒ μ{x : f (x)  0} = μ ⎜⎜⎜⎜⎝ Ak ⎟⎟⎟⎟⎠ = μ(Ak ) = 0 k:yk 0

=⇒ f = 0 a.e.

k:yk 0

Integral

3) If f =



k yk 1IAk ,

 (c f ) =



251

then

cyk μ(Ak ) = c



k

 yk μ(Ak ) = c

f.

k

4) First, note that any two simple functions can be represented by linear combinations of the indicators of the same measurable sets.   Indeed, let f = i yi 1IAi and g = j z j 1I B j . Then f=



yi 1IAi ∩B j

and g =

i, j



z j 1IAi ∩B j .

i, j

Thus, without loss of generality, we may suppose that simple functions f and g have the same constancy sets Ai , that is,   f= yi 1IAi and g = zi 1IAi . i



i

Then      f+ g= yi μ(Ai ) + zi μ(Ai ) = (yi + zi )μ(Ai ) = ( f + g). i

5)

i

i



f = (g + ( f − g)) = g + ( f − g)  g.

6) The property follows from properties 5, 3, and 1:    f  c = c 1 = cμ(E).

 

T HEOREM 18.1.– The integral in the class S+ is continuous with respect to monotone sequences:

1) If S+  fn ↓ f ∈ S+ and f1 < ∞, then fn ↓ f ;

2) If S+  fn ↑ f ∈ S+ , then fn ↑ f .

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Integral and Measure

P ROOF.–

1) First, suppose that fn ↓ 0. Clearly, f1 < ∞ ⇔ μ{ f1 > 0} < ∞. Denote by A the maximal value of the function f1 . Then fn  A for all n ∈ N. For all ε > 0, we have 0  fn  A · 1I{ fn >ε} + ε · 1I{0< fn ε} ,

n ∈ N.

Integrating these inequalities, we get  0

fn  Aμ{ fn > ε} + εμ{0 < fn  ε}  Aμ{ fn > ε} + εμ{ fn > 0}  Aμ{ fn > ε} + εμ{ f1 > 0}.

Since μ{ f1 > ε}  μ{ f1 > 0} < ∞ and { fn > ε} ↓ ∅, taking the limit as n → ∞, from the continuity of measure from above (theorem 13.1, property 4b) we get  0  lim ↓ fn  εμ{ f1 > 0}. n

Now taking the limit as ε → 0, we get  lim ↓ fn = 0. n

Now suppose that f ↓ f and f < ∞. Then ( f − f ) ↓ 0 and ( f1 − n 1 n



f )  f1 < ∞. By the above, fn − f = ( fn − f ) ↓ 0, that is, fn ↓ f .

f < ∞. Then f − fn ↓ 0 and

2) Now let fn ↑ f and ( f − f1 )  f < ∞. Therefore, applying the property to

just proved

the sequence { f − f }, we get f − f = n

n ( f − fn ) ↓ 0, that is, fn ↑ f .

Ir remains to prove property 2 in the case f = ∞. The latter equality means that there is a value y > 0 of the simple function f such that μ{ f = y} = ∞. Then μ{ f > ε} = ∞ for all 0 < ε < y. Since "

# " # x : fn (x) > ε ↑ x : f (x) > ε

Integral

253

and fn  ε1I{ fn >ε} , we get 

 fn  εμ{ fn > ε} ↑ εμ{ f > ε} = ∞ =

f,

where we used the continuity of measure from below (theorem 13.1, property 4a).   Our nearest aim is an extension of the integral to a large class of measurable functions. We begin with non-negative functions. First, we need the following lemma. L EMMA 18.1.– Let { fn , n ∈ N} and {gn , n ∈ N} be two increasing sequences of non-negative simple functions such that lim ↑ fn  lim ↑ gn . n

n

Then  lim ↑

 fn  lim ↑

n

gn .

n

P ROOF.– For every fixed n ∈ N, we have " # lim ↑ gm  lim ↑ min{ fn , gm } = min fn , lim ↑ gm = fn . m

m

m

By the previous theorem,  lim ↑

 gm  lim ↑

m

 min{ fn , gm } =

m

fn ,

n ∈ N.

Now, taking the limit as n → ∞, we get  lim ↑ m

as claimed.

 gm  lim ↑ n

fn ,  

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Integral and Measure

C OROLLARY 18.1.– Let { fn , n ∈ N} and {gn , n ∈ N} be two increasing sequences of non-negative simple functions such that lim ↑ fn = lim ↑ gn . n

n

Then  lim ↑ n

 fn = lim ↑ n

gn .

D EFINITION 18.2.– Let f be a non-negative measurable function, and let { fn , n ∈ N} be as sequence of simple non-negative functions 1 converging

to f . The integral of f with respect to measure μ,

denoted as f = E f (x) μ(dx) (or, shortly, f ), is the limit lim ↑ fn . This n definition is correct since by corollary 18.1 the latter limit does not depend on the choice of the sequence { fn } ⊂ S+ . Thus,   f := lim ↑ fn if f = lim ↑ fn ( fn ∈ S+ ). n

n

We further denote by M + the class of all non-negative measurable functions. T HEOREM 18.2 (Properties of the integral of non-negative measurable functions).–

1) 0  f  ∞, f ∈ M + ;

2) (α1 f1 + α2 f2 ) = α1 f1 + α2 f2 for all f1 , f2 ∈ M + and α1 , α2  0;

3) If f1 , f2 ∈ M + and f1  f2 , then f1  f2 ;

4) If fn ↑ f , fn ∈ M + , then fn ↑ f ;

5) If f ∈ M + , then f = 0 if and only if f = 0 μ-a.e.;

6) If f < +∞, then f < +∞ μ-a.e.

1 Such a sequence always exists by proposition 16.3.

Integral

255

P ROOF.– 1) Obvious. 2) Let fi = lim ↑ fn,i , fn,i ∈ S+ , i = 1, 2. Since n

α1 f1 + α2 f2 = lim ↑ (α1 fn,1 + α2 fn,2 ), n

we have 

 (α1 f1 + α2 f2 ) = lim ↑

(α1 fn,1 + α2 fn,2 )   = α1 lim ↑ fn,1 + α2 lim ↑ fn,2 n n  = α1 f1 + α2 f2 . n

If, moreover, f1  f2 , then from lemma 18.1 we get property 3. 4) (compare the proof of the B. Levi theorem, p. 114.) Let fn = lim ↑ m

gm,n , n ∈ N, gm,n ∈ S+ ). Denote hm := max gm,n = max{gm,1 , gm,2 , . . . , gm,m } ∈ S+ , nm

m ∈ N.

Note that {hm } is an increasing sequence: hm = max gm,n  max gm+1,n  max gm+1,n = hm+1 , nm

nm

nm+1

m ∈ N.

Moreover, hm = max gm,n  max fn = fm , nm

nm

m ∈ N.

Integrating the inequalities gm,n  hm  fm ,

n  m,

[18.2]

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Integral and Measure

we get 

 gm,n 

 hm 

fm ,

n  m.

[18.3]

Taking the limit, as m → ∞, in inequalities [18.2], we have fn = lim ↑ gm,n  lim ↑ hm  f, m→∞

m→∞

n ∈ N.

[18.4]

Taking the limits, as m → ∞, in inequalities [18.3], we have     fn = lim ↑ gm,n  lim ↑ hm  lim ↑ fm . [18.5] m→∞

m→∞

m→∞

Now, take the limits in inequalities [18.4]: f = lim ↑ fn  lim ↑ hm  f =⇒ f = lim ↑ hm . n→∞

m→∞

[18.6]

m→∞

Since

hm ∈ S+ , from the definition of the integral we get that lim ↑ hm . Finally, take the limits in inequalities [18.5]:

f=

m→∞

 lim ↑ n→∞

 fn 

5) Suppose that

 f = lim ↑ m→∞

 hm  lim ↑ m→∞

 fm =⇒

f = 0. Since

f = f · 1I{ f n−1 } + f · 1I{ f 0 ,

 f = lim ↑ n→∞

fn .

Integral

257

we have that 3 4 " # μ x : f (x) > 0 = lim ↑ μ x : f (x)  n−1 = 0, n

that is, f = 0 a.e. If S+  fn ↑ f = 0 a.e., then, also, f n = 0 a.e., n ∈ N, since 0  fn  f .

Therefore, fn = 0; thus, f = lim ↑ fn = 0. n

5) Since   f  (+∞)1I{ f =+∞} = (+∞) · μ{ f = +∞}, it follows that μ{ f = +∞} = 0.

 

R EMARK 18.2.– Let us compare the proof of property 4 with Step 1 of the proof of the B. Levi theorem (see section 8.2, p. 114), where a similar property is proved in the class L+ . This is rather natural since the definitions of function classes M + and L+ are very similar. The functions from M + are limits of increasing sequences of simple functions, while the functions from L+ are those of step functions. However, there are also subtler differences. In the case of the Lebesgue integral, at first sight it may seem that restricting on non-negative functions of class L+ , we will get the functions from M + whose integrals are finite. However, this is not so because the class of simple functions is much wider than that of step functions as the latter are constant on intervals, while simple functions are constant on arbitrary measurable sets. Therefore, the limits of increasing sequences of step functions do not exhaust the class of all non-negative Lebesgue-measurable functions. All the more surprising is the fact that the differences of functions from L+ already do exhaust all integrable functions, as well as the differences of functions from the wider class M + (compare lemma 8.1 and definition 18.3 below). Slightly simplifying the situation, we can say that this shows the equivalence of two integration schemes, Daniel’s in Chapter 8 and Lebesgue’s in this chapter. One of them starts with the definition and properties of the integral and ends with the definition and properties of the measure,

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Integral and Measure

while the other proceeds in the reverse order. As we could see, the principle of “conservation of difficulties” is at work, and the choice between two approaches is basically a matter of taste. D EFINITION 18.3.– A measurable function f is called quasi-integrable

(with respect to measure μ) if at least one of the integrals f + and f − is finite. The integral of such a function f is defined as    + f= f − f −. A measurable function is called integrable if

f + < ∞ and

f − < ∞.2

T HEOREM 18.3.– The integral in the class of quasi-integrable functions possesses the following properties:

1) f ∈ R; f ∈ R if and only if f is an integrable function, and in this case, f is a.e. finite;

2) If f  0, then f  0;

3) (c f ) = c f for all c ∈ R;

4) ( f + g) = f + g if f − and g− (or f + and g+ ) are integrable functions;

5) If f  g, then f  g;

6) If fn ↑ f and f1− is integrable, then fn ↑ f ;

if fn ↓ f and f1+ is integrable, then fn ↓ f ;

∞ ∞ 6 . If fn  0, n ∈ N, then n=1 fn = n=1 fn ;

7) If f = 0 a.e., then f = 0. P ROOF.– Properties 1, 2, and 3 directly follow from the definition. The a.e. finiteness of an integrable function follows from property 6 in theorem 18.2.

2 In other words, a quasi-integrable function is integrable if its integral is finite.

Integral

259

4) Let h1 and h2 be any non-negative measurable functions, at least one integrable, such that

h(x)

:= h1 (x) − h2 (x)  ∞ − ∞, x ∈ E. Let us check that then h = h1 − h2 .3 Since h+ = max{0, h}  max{0, h 1 } = h 1 and, similarly, h−  h2 , we have that at least one of the integrals h+ , h− is finite, and, therefore, h is quasi-integrable. Since h = h+ − h− = h1 − h2 , we have h+ + h2 = h− + h1 , and, thus,          + − + − h + h2 = h + h1 =⇒ h = h − h = h1 − h2 . Now consider functions f and g satisfying the conditions of property 4. Using the property just proved, we get 





 + + (f − f +g −g ) = ( f + g ) − ( f − + g− )       = ( f + + g+ ) − ( f − + g− ) = f + + g+ − f − − g−   = f + g.

( f + g) =

+

−

+

−



5) If f = −∞, then the inequality is trivial. If f > −∞, then g=

(g − f ) + f  f .

6) Let fn ↑ f and f1− < ∞. Then fn−  f1− , n ∈ N, and f −  f1− . Therefore, the functions fn and f are quasi-integrable. Since 0  fn + f1− ↑ f + f1− , applying property 4 of theorem 18.2, we get 

 fn +

f1−

 ↑

 f+

f1−

 =⇒

 fn ↑

f.

The case of a decreasing sequence is considered similarly.

3 That is, in the definition of the integral h := h+ − h− , instead of h± , we can take arbitrary non-negative measurable functions such that their difference equals h.

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Integral and Measure

6 ) Applying property 6 to the sequence of partial sums of the series, we have:    ∞ n  fn = lim ↑ fk n

n=1

= lim ↑ n

fk

property 6

k=1

= lim ↑ n

k=1

  n n  

fk =

∞  

fn .

n=1

k=1

7) f = 0 a.e. ⇒ f ± = 0 a.e. ⇒

f=

f+−

f − = 0.

 

C OROLLARY 18.2 (Fatou–Lebesgue theorem).– For any sequence of measurable functions { fn , n ∈ N} and integrable functions g and h,   fn  g, n ∈ N =⇒ lim sup fn  lim sup fn ; n n   fn  h, n ∈ N =⇒ lim inf fn  lim inf fn . n

n

In particular, if { fn , n ∈ N} converges and there exists an integrable function g such that | fn |  g, n ∈ N, then   lim fn = lim fn . n

n

P ROOF.– 4 Since supn fn  g and g is integrable, the function (supm fm )+ also is integrable. Therefore, taking the limit as n → ∞ in the inequality   sup fm  sup fm , n ∈ N, mn

mn

and using property 6 in theorem 18.3, we get    lim sup fn = lim ↓ sup fm  lim ↓ sup fm n

n

mn

4 Compare with the proof in section 8.3, p. 119.

n

mn

Integral

 =

261

 lim ↓ sup fm = mn

lim sup fn . n

The second statement is proved similarly. If | fn |  g or, equivalently, −g  fn  g, n ∈ N, where g is an integrable function, from the properties just proved we get     lim inf fn  lim inf fn  lim sup fn  lim sup fn . n

n

n

n

Therefore, if there exists limn fn , that is,

lim supn fn = lim inf n fn , then there also exists limn fn , which equals limn fn .   R EMARK 18.3.– If f is a quasi-integrable and f  = f a.e.,



function

   then f = f . Indeed, f = ( f − f ) + f = f . Therefore, all the statements of theorem 18.3 and corollary 18.2 remain valid if we only require all the conditions to be satisfied almost everywhere (since all the functions involved can be replaced by a.e. equivalent functions so that the conditions would be satisfied everywhere). For example, if f  g a.e., then f   g with f  = f · 1I{ f g} === f, a.e.

g = g · 1I{ f g} === g. a.e.

D EFINITION 18.4.– The indefinite integral of a non-negative measurable function f is the function A f defined on the σ-algebra A by   f = ( f 1IA ). A

The value

A

f is called the integral of a function f over a set A.

From the definition we immediately get the following properties of the indefinite integral, where all the functions and sets are assumed to be measurable. P ROPOSITION 18.2.–

1) 0  A f  f ; A f = 0 ⇔ μ{A ∩ { f > 0}} = 0;

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Integral and Measure

2) If {Ai , i ∈ I} is a countable system of pairwise non-intersecting and  A := i∈I Ai , then   f= f; A

Ai

i∈I



3) A1 ⊂ A2 ⇒ A f  A f ;

1 2 4) An ↑ A ⇒ A f ↑ A f ;

n

An ↓ A, A f < ∞ ⇒ A f ↓ A f . n

1

Note that from properties 1 and 2 it follows that λ(A) := measure on the σ-algebra A.

A

f is a

Finally, a few remarks about a particular important case. Let E = R, and let μ be a Lebesgue–Stieltjes measure (on the corresponding σ-algebra L of sets measurable in the Lebesgue–Stieltjes sense) generated by a distribution function F. Then the integral of a measurable (in the Lebesgue–Stieltjes sense) function f : (R, L) → (R, B) with respect to measure μ is called the Lebesgue–Stieltjes integral and is denoted +∞

+∞ f (x) dF(x),

−∞

+∞ f dF,

f (x) F(dx).

or

−∞

−∞

Certainly, we can also consider the Lebesgue–Stieltjes integral of a function f over a set A ∈ L:      f (x) dF(x) = f dF = f (x) F(dx) . A

A

A

In particular, if A = [a, b], then the integral 

 f (x) dF(x) [a,b]

b

f (x) dF(x).

is also denoted by a

In the particular case F(x) ≡ x, we have the Lebesgue integral already known from Chapter 17.

Integral

263

P.18. Problems P ROBLEM 18.1.– Let μ be a finite measure. Prove that a measurable function f is integrable if and only if: i)

∞   "  # μ x :  f (x)  n < ∞; n=1

ii)

∞ 

  " # nμ x : n   f (x) < n + 1 < ∞ and μ{x : | f (x)| = +∞} = 0.

n=1

P ROBLEM 18.2.– Let a measurable function f be such that the product f g is integrable for every integrable function g. Prove that there then exists a constant C < ∞ such that |ϕ|  C a.e. (In such a case, f is said to be essentially bounded.) P ROBLEM 18.3.– Let f be an integrable function. Prove that the set A = {x : | f (x)| > 0} has a σ-finite measure, that is, it can be represented  as the union A = ∞ n=1 En where μ(E n ) < ∞, n ∈ N. Is this true if f is a quasi-integrable function? P ROBLEM 18.4.– For an integrable function f , denote 

1

( f ) = 0

| f (x)| μ(dx). 1 + | f (x)|

Prove the following properties of the functional : 1) | f |  |g| =⇒ ( f )  (g); 2) ( f + g)  ( f ) + (g); μ

3) fn → 0 ⇐⇒ ( fn ) → 0.

P ROBLEM 18.5.– Prove that if f is an integrable function and A f = 0 for all measurable sets A, then f = 0 a.e.

P ROBLEM 18.6.– Let g be a measurable function such that f g  0 for every bounded non-negative measurable function f . Prove that g  0 a.e.

1 P ROBLEM 18.7.– Find 0 f (x) dx if

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Integral and Measure

⎧ 2 ⎪ x for irrational x > 13 , ⎪ ⎪ ⎪ ⎪ ⎨ 3 f (x) = ⎪ x for irrational x < 13 , ⎪ ⎪ ⎪ ⎪ ⎩ 0 otherwise; ⎧ 1 ⎪ ⎪ ⎨ √ x for irrational x, f (x) = ⎪ ⎪ ⎩ x3 for rational x.

i)

ii)

P ROBLEM 18.8.– Let μ be the counting measure on N, that is, μ(A) equals the number of elements of a set A ⊂ N (A = P(N)). Prove that a  function f : N → R is μ-integrable if and only if ∞ n=1 | f (n)| < +∞, and, in this case,  N

f dμ =

∞ 

f (n).

n=1

P ROBLEM 18.9.– Let {nk , k ∈ N} be a strictly increasing sequence of natural numbers, and let A := {x ∈ (−π, π) : ∃ limk (sin nk x)}. Prove that m(A) = 0 (m is the Lebesgue measure). P ROBLEM 18.10.– Let { fn } be an increasing sequence of integrable functions such that, for some constant C > 0, fn  C for all n ∈ N. Prove that f := limn fn is an integrable function and   f = lim fn . n

P ROBLEM 18.11.– Let { fn } be a sequence of non-negative functions

such that, for some constant C > 0, f  C for all n ∈ N. Prove that if n

fn → f , n → ∞, then also f  C. Is it true that 

 f = lim n

fn ?

P ROBLEM 18.12.– Let f1 and f2 be measurable functions in measure spaces (E1 , A1 , μ1 ) and (E2 , A2 , μ2 ) with finite measures. They are said to be identically measurable if for all a ∈ R, " # " # μ1 x : f1 (x) < a = μ2 x : f2 (x) < a .

Integral

Prove that, in such a case,   f1 (x) μ1 (dx) = E1

265

f2 (x) μ2 (dx).

E2

P ROBLEM 18.13.– Let ϕ be a mapping from a measure space (E, A, μ) into a measurable space (E 0 , A0 ). On the latter, define the measure μϕ−1 by μϕ−1 (A) := μ(ϕ−1 (A)), A ∈ A0 . Prove that for any real measurable function f on (E 0 , A0 , μϕ−1 ), the following equality holds (change-ofvariable theorem):    −1 f (y) μϕ (dy) = f ϕ(x) μ(dx), E0

E

provided that at least one of the integrals is defined. Show that from this theorem the statement of the previous problem follows. P ROBLEM 18.14.– Let f ∈ L(E, A, μ). Prove that for all ε > 0, there exists δ > 0 such that A | f | dμ < ε whenever μ(A) < δ (the absolute continuity of the integral). P ROBLEM 18.15.– Prove the properties of the indefinite integral (proposition 18.2).

19 Product of Two Measure Spaces

In this chapter, we suppose that we are given two fixed measure spaces (E1 , A1 , μ1 ) and (E2 , A2 , μ2 ) with σ-finite measures μ1 and μ2 . Recall that the direct (or Cartesian) product of two sets E 1 and E 2 is defined as the set E 1 × E2 := {(x1 , x2 ) : x1 ∈ E1 , x ∈ E2 }. The section of a set A ⊂ E1 × E 2 at a point x1 ∈ E1 is the set A x1 := {x2 ∈ E2 : (x1 , x2 ) ∈ A}. For a fixed x1 , the mapping A → A x1 maps the set family P(E1 × E 2 ) into the set family P(E2 ) and preserves the set operations of union, intersection, and complement, that is, for arbitrary family of sets {Ai , i ∈ I} ⊂ E1 × E 2 , we have the following equalities: 2  2 Ai = Aix1 , i∈I

x1

i∈I

  Ai = Aix1 , i∈I

x1

(Ac ) x1 = (A x1 )c .

i∈I

The section of a mapping f : E1 × E2 → E (where E is an arbitrary set) at a point x1 ∈ E 1 is the mapping f x1 : E 2 → E defined by f x1 (x2 ) = f (x1 , x2 ). For any fixed x1 ∈ E 1 , the mapping f → f x1 preserves all usual function operations, including the pointwise convergence.

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Integral and Measure

We call a rectangle in E1 × E2 a set of the form A1 × A2 , where A1 ⊂ E1 and A2 ⊂ E2 . If A1 ∈ A1 and A2 ∈ A2 , then the rectangle A1 × A2 is said to be measurable. A rectangle A1 × A2 is an empty set if and only if at least one of its “sides” A1 and A2 is empty. The section (A1 × A2 ) x1 of any rectangle A1 × A2 at a point x1 is either A2 or ∅ if, respectively, x1 ∈ A1 or x1  A1 . P ROPOSITION 19.1.– The family of all measurable rectangles A1 × A2 is a semi-algebra of subsets of E 1 × E2 . P ROOF.– Clearly, ∅ = ∅ × ∅ and E1 × E 2 are measurable rectangles. The intersection of two measurable rectangles is a measurable rectangle, since (A1 × A2 ) ∩ (A1 × A2 ) = (A1 ∩ A1 ) × (A2 ∩ A2 ). The third axiom of semi-algebras is satisfied since (A1 × A2 )c = Ac1 × A2 + E1 × Ac2 .

 

D EFINITION 19.1.– The σ-algebra generated by all measurable rectangles A1 × A2 is called the product of σ-algebras A1 and A2 and is denoted by A1 ⊗ A2 (or, sometimes, A1 × A2 ). The measurable space (E 1 × E2 , A1 ⊗ A2 ) is called the product of measurable spaces (E 1 , A1 ) or (E2 , A2 ). P ROPOSITION 19.2.– 1) For any set A ∈ A1 ⊗ A2 , its section A x1 at any point x1 ∈ E 1 is a measurable set, that is, belongs to the σ-algebra A2 . 2) For any measurable function in the space (E 1 × E 2 , A1 ⊗ A2 ), its section f x1 at any point x1 ∈ E1 is a measurable function in the space (E 2 , A2 ). P ROOF.– Denote by C x1 the family of all subsets A of E1 × E 2 such that the section A x1 ∈ A2 . We easily verify that all measurable rectangles belong to C x1 and that C x1 is closed under operations of complement

Product of Two Measure Spaces

269

and countable intersection. Therefore, C x1 ⊃ A1 ⊗ A2 , and the first part of the theorem is proved. The second part follows from the equality  ( f x1 )−1 (B) = f −1 (B) x1 .

 

C OROLLARY 19.1.– A non-empty rectangle B1 × B2 ⊂ E1 × E 2 is measurable in the space (E1 × E 2 , A1 ⊗ A2 ), that is, belongs to A1 ⊗ A2 , if and only if B1 ∈ A1 and B2 ∈ A2 . A real function f (x) = f1 (x1 ) · f2 (x2 ), x = (x1 , x2 ) ∈ E1 × E2 , which is not identically zero, is measurable (with respect to A1 ⊗ A2 ) if and only if f1 and f2 are measurable with respect to A1 and A2 , respectively. P ROOF .– If B1 × B2  ∅, then also B1  ∅. Therefore, there exists x1 ∈ B1 , and by proposition 19.2 we get that B2 = (B1 × B2 ) x1 ∈ A2 . Similarly, B1 ∈ A1 . The proof for functions is similar.   T HEOREM 19.1 (Fubini theorem).– 1) In the measurable space (E1 × E 2 , A1 ⊗ A2 ), there exists a unique measure μ satisfying the condition μ(A1 × A2 ) = μ1 (A1 ) · μ2 (A2 )

for all

A1 ∈ A1 , A2 ∈ A2 . [19.1]

It is called the product of measures μ1 and μ2 and is denoted by μ1 × μ2 . The measure space (E 1 × E2 , A1 ⊗ A2 , μ1 × μ2 ) is called the product of measure spaces (E1 , A1 , μ1 ) and (E2 , A2 , μ2 ). 2) Let f be an arbitrary quasi-integrable real function in a product measure space (E1 × E2 , A1 ⊗ A2 , μ1 × μ2 ). Then,    f dμ = μ1 (dx1 ) μ2 (dx2 ) f x1 (x2 ) E1 ×E2



E1

=



E2

μ2 (dx2 ) E2

E1

μ1 (dx1 ) f x2 (x1 ).

P ROOF .– 1) By the theorems on extension of a measure (theorems 14.1 and 14.2), to prove the existence and uniqueness of such an extension of μ, it suffices to check that μ defined by equation [19.1] is a σ-additive

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Integral and Measure

function on the semi-algebra of measurable rectangles A1 × A2 in E1 ×  i i E2 . Suppose that A1 × A2 = ∞ i=1 A1 × A2 or, in other words, 1IA1 ×A2 (x1 , x2 ) = 1IA1 (x1 )1IA2 (x2 ) =

∞  i=1

1IAi (x1 )1IAi (x2 ). 1

2

Integrating this equality in the set E2 with respect to measure μ2 , for every x1 ∈ E1 , we get 1IA1 (x1 )μ2 (A2 ) =

∞  i=1

1IAi (x1 )μ2 (Ai2 ). 1

Now, let us integrate the equality obtained in the set E 1 with respect to μ1 : μ(A1 × A2 ) = μ1 (A1 )μ2 (A2 ) =

∞ 

μ1 (Ai1 )μ2 (Ai2 )r

i=1

=

∞ 

μ(Ai1 × Ai2 ).

i=1

The σ-additivity of μ is proved, as required. 2) Let F denote the class of all non-negative A1 ⊗ A2 -measurable functions f for which the Fubini theorem holds. More precisely, F consists of all non-negative A1 ⊗ A2 -measurable functions f for which the function  g(x1 ) := f x1 (x2 ) μ2 (dx2 ) E2

is A1 -measurable and the following equality holds:   f dμ = g(x1 ) μ1 (dx1 ). E1 ×E2

E1

The function class F contains all indicators of measurable rectangles 1IA1 ×A2 . Indeed,   μ2 (dx2 )1IA1 ×A2 (x1 , x2 ) = 1IA1 (x1 ) μ2 (dx2 )1IA2 (x2 ) = 1IA1 (x1 )μ2 (A2 ) E2

E2

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271

is an A1 -measurable function of x1 ∈ E 1 , and  1IA1 ×A2 dμ = μ(A1 × A2 ) = μ1 (A1 )μ2 (A2 ) E1 ×E2



=



E1

= E1

 1IA1 (x1 )μ1 (dx1 ) · μ2 (A2 )

(1IA1 (x1 )μ2 (A2 )) μ1 (dx1 ).

From the linearity of the integral, now it follows that the class F also contains all linear combinations of the indicators of measurable rectangles, that is, simple functions in (E1 × E2 , A1 ⊗ A2 ). Now, let f be any non-negative A1 ⊗A2 -measurable function. By proposition 16.3, there exists an increasing sequence of simple functions fn ↑ f . From the definition of the integral of non-negative functions (definition 18.2), we get that, for all x1 ∈ E1 ,   gn (x1 ) := μ2 (dx2 ) f xn1 (x2 ) ↑ μ2 (dx2 ) f x1 (x2 ) = g(x1 ) E2

E2

and 

 f dμ = lim ↑ n

E1 ×E2

 f dμ = lim ↑

gn (x1 ) μ1 (dx1 )

n

E1 ×E2

n

E1

 =

g(x1 ) μ1 (dx1 ). E1

In other words, f ∈ F , that is, the Fubini theorem holds for non-negative functions (the second equality is obtained similarly by interchanging the measures μ1 and μ2 ). Note that if f is integrable with respect to μ, then g is integrable with respect to μ1 and thus, is finite a.e. in E1 . Therefore, the section f x1 is integrable with respect to μ2 for μ1 -almost all x1 ∈ E 1 . In the general case where f takes values of opposite signs and is quasi-integrable (say, f + dμ < ∞), we get that the function ( f + ) x1 =

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Integral and Measure

( f x1 )+ is μ2 -integrable for μ1 -almost all x1 ∈ E1 and thus, f x1 is quasiintegrable for μ1 -almost all x1 ∈ E 1 . Therefore, the function g(x1 ) :=

f (x )μ (dx2 ) is defined as μ1 -a.e. Moreover, from the inequality E 2 x1 2 2 + g (x1 )  E f x+1 (x2 )μ2 (dx2 ) it follows that 2



g+ (x1 )μ1 (dx1 )  E1



 μ1 (dx1 ) E1

 =

E2

f x+1 (x2 )μ2 (dx2 )

f + dμ < ∞,

E1 ×E2

that is, g is quasi-integrable with respect to μ1 . Now, applying the equality obtained for non-negative functions f + and f − and using the linearity of the integral, we get the equality for the function f = f + − f − as well.   C OROLLARY 19.2.– 1) A measurable function f on (E1 × E 2 , A1 ⊗ A2 , μ1 × μ2 ) equals zero μ1 × μ2 -almost everywhere if and only if μ1 -almost all its sections f x1 equal zero μ2 -almost everywhere. 2) If a measurable function f on (E 1 × E2 , A1 ⊗ A2 , μ1 × μ2 ) is integrable, then μ1 -almost all its sections f x1 are integrable (with respect to μ2 ). P ROOF.– 1) If f  0, then from the equality    f dμ = μ1 (dx1 ) μ2 (dx2 ) f x1 (x2 ) E1

E1 ×E2

E2

we get that f = 0 μ-a.e. if and only if  g(x1 ) = μ2 (dx2 ) f x1 (x2 ) = 0 E1

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273

for μ1 -a.a. x1 , which, in turn, holds if and only if for μ1 -a.a. x1 , the section f x1 = 0 μ2 -a.e. In the general case, note that f = 0 a.e. if and only if f + = 0 a.e. and f − = 0 a.e. 2) The proof is similar to that of part 1.

 

C OROLLARY 19.3.– Let a function f : E1 × E2 → R be measurable with respect to σ-algebra A1 ⊗ A2 (the completion of σ-algebra A1 ⊗ A2 with respect to measure μ1 ×μ2 ). Then μ1 -almost all sections f x1 of f are measurable with respect to σ-algebra A2 (the completion of σ-algebra A2 with respect to measure μ2 ). If f is quasi-integrable (in particular, non-negative), then the equalities of theorem 19.1, part 2, remain valid. P ROOF.– Let fˆ be an A1 ⊗ A2 -measurable function such that {x ∈ E1 × E2 : f (x)  fˆ(x)} is a null set.1 Note that from A ∈ A1 ⊗ A2 and μ(A) = 0, it follows that μ2 (A x1 ) = 0 for μ1 -almost all x1 . We can see this from the equalities    μ(A) = 1IA dμ = μ1 (dx1 ) 1IA x1 (x2 )μ2 (dx2 ) E 1 ×E 2

= E1

E1

E2

μ2 (A x1 )μ1 (dx1 ).

Therefore, for μ1 -almost all x1 , the set " # " # x : f (x1 , x2 )  fˆ(x1 , x2 ) x1 = x2 : f x1 (x2 )  fˆx1 (x2 ) is a null set on (E2 , A2 , μ2 ). Since, for all x1 , the function fˆx1 is A2 measurable (proposition 19.2), it follows that f x1 is A2 -measurable for μ1 -almost all x1 . If f is quasi-integrable, then so is fˆ. Therefore, it is clear that   f dμ = fˆ dμ, E1 ×E2

E1 ×E2

1 The existence of such a function is obvious in the case of a simple function and, in the general case, follows by passing to the limit.

274

and

Integral and Measure



 E2

f x1 (x2 )μ2 (dx2 ) =

E2

fˆx1 (x2 )μ2 (dx2 ) μ1 -a.e.

Since the equalities of the Fubini theorem hold for the function fˆ, they also hold for f .   Up to now, we only considered the products of two measure spaces. In the general case of the product of any (finite) number of measure spaces, we state, without proof, the Fubini theorem as follows. Let (E i , Ai , μi ), i = 1, 2, . . . , n, be measure spaces. Recall that the Cartesian product of sets E1 , E 2 , . . . , En is defined as the set n ą

" # E i = E 1 × E2 × · · · × E n := x = (x1 , x2 , . . . , xn ) : xi ∈ Ei , i = 1, . . . , n

i=1

Ś 7 (sometimes, notation is used, ni=1 E i ). Subsets of ni=1 E i of Śanother n the form i=1 Ai , where Ai ⊂ ŚEn i , are called k-boxes. If Ai ∈ Ai , i = 1, 2, . . . , n, then the k-box i=1 Ai is called a measurable k-box. Again, we can easily check that the class of all measurable k-boxes is a semi-algebra. It is generated that σ-algebra, denoted by 8n A = A ⊗ A2 ⊗ · · · ⊗ An , is called the product of σ-algebras i 1 i=1 A1 , A2 , . . . , An . Repeating almost literally the proof of theorem 19.1, we arrive at the Fubini theorem for a finite product of measure spaces. T HEOREM 19.2.–

Śn n 1) In the measurable space ( i=1 Ei , ⊗i=1 Ai ), there exists a unique Śn measure μ (denoted i=1 μi or μ1 × μ2 × · · · × μn ) such that n Ś  0 n μ Ai = μ(Ai ), Ai ∈ Ai , i = 1, 2, . . . , n. i=1

i=1

2) For every quasi-integrable real function f on the measure space 8n Ś  Śn n (E, A, μ) := i=1 Ei , i=1 μi , we have the equality i=1 Ai ,     f dμ = μ1 (dx1 ) μ2 (dx2 ) · · · μn (dxn ) f (x1 , x2 , . . . , xn ) E

E1

E2

En

and yet (n! − 1) similar equalities obtained by changing the integrations order.

Product of Two Measure Spaces

275

The left-hand side of the latter formulas is understood as follows. For any fixed x1 , x2 , . . . , xn−1 , f (x1 , x2 , . . . , xn−1 , xn ) is an An -measurable function of xn ,; integrating it with respect to the measure μn , we get the function  g(x1 , x2 , . . . , xn−1 ) = f (x1 , x2 , . . . , xn−1 , xn ) μn (dxn ), En

which is defined for μ1 × · · · × μn−1 -almost all (x1 , x2 , . . . , xn−1 ). This function is integrated, as a function of xn−1 , with respect to the measure μn−1 , and so on. Corollaries 19.2 and 19.3 are also easily generalized. At the end of the section, we give some terminology remarks for readers who skipped Part 2. Let all Ei = R, Ai = B be the Borel σ-algebra in R, and8 μi be the LebesgueŚmeasure on B, i = 1, 2, . . . , n. n n n n Let, further, Bn := i=1 μi . The sets from B are i=1 Ai and μ := called Borel sets of the space Rn (or sets measurable in the Borel sense), the σ-algebra Ln = Bn obtained by completion of Bn with respect to μn is called the Lebesgue σ-algebra of Rn , and the sets from Ln are called Lebesgue-measurable sets of Rn (or sets measurable in the Lebesgue sense). The measure μn on the σ-algebra Bn or Ln is called the Lebesgue measure in the space Rn . In the expressions containing integrals, the symbols dμn or μn (dx) are often replaced by the symbols dx or dx1 dx2 · · · dxn . P.19. Problems P ROBLEM 19.1.– Let E be an infinite non-countable set, and A the σ-algebra of its subsets generated by all single-point sets. Check that the set Δ = {(x, x) : x ∈ E} (the “diagonal” of E × E) does not belong to the σ-algebra A ⊗ A, although all its sections belong to the σ-algebra A. Therefore, the inverse of proposition 19.2 is not true in general. P ROBLEM 19.2.– Let A1 and A2 be two families of subsets of two sets E 1 and E 2 , respectively. Is the family {A1 × A2 : A1 ∈ A1 , A2 ∈ A2 } of subsets E 1 × E2 semi-algebra if A1 and A2 are algebras? or semialgebras?

276

Integral and Measure

P ROBLEM 19.3.– Show by a counter example that the product of two complete measure spaces may be an incomplete measure space. P ROBLEM 19.4.– Let E 1 = E2 = [0, 1], and A1 = A2 = B[0, 1] be the Borel σ-algebra of subsets of [0, 1]. Let μ1 be the Lebesgue measure, and μ2 be the counting measure, that is, μ2 (A) is equal to the number of points of a set A (μ(A) = ∞ if A is an infinite set). 1) Prove that the “diagonal” Δ = {(x, y) : x = y ∈ [0, 1]} ∈ A1 ⊗ A2 . 2) Find the integrals   μ1 (dx) 1IΔ (x, y) μ2 (dy) and 

 μ2 (dy)

1IΔ (x, y) μ1 (dx),

showing that they are not equal. Thus, the Fubini theorem does not hold if the measures μ1 and μ2 are not both σ-finite. P ROBLEM 19.5.– Prove that the Borel σ-algebra B(R2 ) coincides with the product of σ-algebras B(R) ⊗ B(R).

Bibliography

[BOU 04] B OURBAKI N., Functions of a Real Variable: Elementary Theory, Springer-Verlag, Berlin/Heidelberg, 2004. [DIE 69] D IEUDONNÉ J., Foundations of Modern Analysis, Academic Press, New York/London, 1969. [DOR 87] D OROGOVTSEV A.Y., Mathematical Analysis. A Problem Book [in Russian], Vishcha Shkola, Kiev, 1987. [HAL 74] H ALMOS P.R., Measure Theory, Springer-Verlag, New York, NY, 1974. [KAB 83] K ABAILA V., Mathematical Analysis, part 1 [in Lithuanian], Mokslas, Vilnius, 1983. [KAB 86] K ABAILA V., Mathematical Analysis, part 2 [in Lithuanian], Mokslas, Vilnius, 1986. [KUB 70] K UBILIUS J., Theory of Real-Variable Functions [in Lithuanian], Mintis, Vilnius, 1970. [LOÈ] L OÈVE M., Probability Berlin/Heidelberg, 1977.

Theory,

4th

ed.,

Springer-Verlag,

ˇ [MAC 76] M ACKEVI CIUS V., Elements of Measure Theory [in Lithuanian], Vilnius University Press, Vilnius, 1976. ˇ [MAC 91] M ACKEVI CIUS V., Integral [in Lithuanian], Vilnius University Press, Vilnius, 1991. ˇ [MAC 98] M ACKEVI CIUS V., Integral and Measure [in Lithuanian], TEV, Vilnius, 1998.

278

Integral and Measure

ˇ [MIS 98] M ISEVI CIUS E., Mathematical Analysis, part 1 [in Lithuanian], TEV, Vilnius, 1998.

[NEV 65] N EVEU J., Mathematical Foundations of the Calculus of Probability, Holden-Day, San Francisco, CA, 1965. [OCH 81] O CHAN Y.S., Collection of Problems and Theorems in the Theory of Functions of a Real Variable [in Russian], Prosveshchenie, Moscow, 1981. [RIE 90] R IESZ F., Szökefalvi-Nagy B., Functional Analysis, Dover, New York, NY, 1990. [RUD 87] RUDIN W., Principles of Mathematical Analysis, McGraw-Hill, Singapore, 1987. [ZAM 74] Z AMANSKII M., Introduction to Modern Algebra and Analysis [in Russian], Nauka, Moscow, 1974.

Index

A additivity of Lebesgue measure, 127 antiderivative, 11 area of a curvilinear sector, 53 of a sphere, 53 of a surface, 162 surface, of a solid of revolution, 52

criterion of Riemann-integrability, 121 D, E differentiation of functional sequences and series, 34 termwise of power series, 35 under the integral sign, 172 extension of a measure, 209

C

F

center mass, of a material body, 166 change of variables cylindrical, 155 class set, 194 completion of a measure space, 205 continuity of a measure, 203 convergence absolute, of an improper integral, 83 relative, of an improper integral, 83 uniform of PDI, 181

family of sets, monotone, generated by, 196 of sets, monotonic, 196 formula change-of-variable, 32, 83 integration-by-parts, 72, 83 Newton–Leibnitz, 30, 83 Taylor’s, with the integral remainder term, 33 function Bessel, 188 continuous almost everywhere, 97

280

Integral and Measure

defined almost everywhere, 97 distribution, 224 elliptic, 187 finite almost everywhere, 97, 241 Lebesgue-integrable, 105 primitive, 11 quasi-integrable, 125 Riemann-integrable, 60, 121 step, 4, 97 I, L indicator of a set, 235 of an interval, 66 integral improper, 177 indefinite, 12 Lebesgue, 105, 262 Lebesgue–Stieltjes, 262 of a non-negative measurable function, 254 of a quasi-integrable function, 125 of a step function, 22, 97 parameter-dependent, 169 Riemann, 21, 60 Stieltjes, 69 Stieltjes, of a step function, 69 integration termwise of power series, 35 lemma Borel–Cantelli, 204

finite, 199 Lebesgue, of a set, 127 measure space complete, 204 moment static, of a material solid, 166 monotonicity of Lebesgue measure, 127 N, R plane tangent, 163 product of σ-algebras, 268 of measurable spaces, 268 of two measure spaces, 269 of two measures, 269 rectangle measurable, 268 S section of a set, 267 of mapping, 267 set Lebesgue-measurable, 127 measurable, 194 null, 204 universal, 194 zero-measure, 204 space measurable, 194 static moment of a curve, 54 of a plane figure, 56 T, V

M mapping measurable, 234 mass of a solid, 165 mass center of a curve, 54 of a plane figure, 56 measure σ-finite, 199

test Abel–Dirichlet, 85 integral, of series convergence, 87 theorem change-of-variables, 148 Fatou–Lebesgue, 260 Fubini, 269

Index

fundamental, of calculus, 30 mean-value, 37 Tonelli, 145 trapezium curvilinear, 140

volume of a ball, 46 of a body, 162 of a body of rotation, 45

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