Lebesgue Integral (Compact Textbooks in Mathematics) [1st ed. 2021] 3030601625, 9783030601621

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Compact Textbooks in Mathematics

Liviu C. Florescu

Lebesgue Integral

Compact Textbooks in Mathematics

Compact Textbooks in Mathematics This textbook series presents concise introductions to current topics in mathematics and mainly addresses advanced undergraduates and master students. The concept is to offer small books covering subject matter equivalent to 2- or 3-hour lectures or seminars which are also suitable for self-study. The books provide students and teachers with new perspectives and novel approaches. They may feature examples and exercises to illustrate key concepts and applications of the theoretical contents. The series also includes textbooks specifically speaking to the needs of students from other disciplines such as physics, computer science, engineering, life sciences, finance. • compact: small books presenting the relevant knowledge • learning made easy: examples and exercises illustrate the application of the contents • useful for lecturers: each title can serve as basis and guideline for a semester course/lecture/seminar of 2–3 hours per week.

More information about this series at http://www.springer.com/series/ 11225

Liviu C. Florescu

Lebesgue Integral

Liviu C. Florescu Faculty of Mathematics ”Al. I. Cuza” University Ia¸si, Romania

ISSN 2296-4568 ISSN 2296-455X (electronic) Compact Textbooks in Mathematics ISBN 978-3-030-60162-1 ISBN 978-3-030-60163-8 (eBook) https://doi.org/10.1007/978-3-030-60163-8 Mathematics Subject Classification: 26-01, 26Axx © The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Switzerland AG 2021 This work is subject to copyright. All rights are solely and exclusively licensed by the Publisher, whether the whole or part of the material is concerned, specifically the rights of translation, reprinting, reuse of illustrations, recitation, broadcasting, reproduction on microfilms or in any other physical way, and transmission or information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed. The use of general descriptive names, registered names, trademarks, service marks, etc. in this publication does not imply, even in the absence of a specific statement, that such names are exempt from the relevant protective laws and regulations and therefore free for general use. The publisher, the authors, and the editors are safe to assume that the advice and information in this book are believed to be true and accurate at the date of publication. Neither the publisher nor the authors or the editors give a warranty, expressed or implied, with respect to the material contained herein or for any errors or omissions that may have been made. The publisher remains neutral with regard to jurisdictional claims in published maps and institutional affiliations. This book is published under the imprint Birkhäuser, www.birkhauser-science.com, by the registered company Springer Nature Switzerland AG. The registered company address is: Gewerbestrasse 11, 6330 Cham, Switzerland

v

Introduction Until the end of the nineteenth century, mathematical analysis focused on the study of continuous functions and on the Riemann integral as the main measuring instrument. The need to model increasingly complex phenomena justified the efforts of mathematicians to widen the class of integrable functions. Inspired by the work of E. Borel and C. Jordan, H. Lebesgue built in around 1900 a measure theory, which is used later to define a more general integral than the Riemann integral, an integral which bears his name. If f : [a, b] → [m, M] is a bounded function, then G. Darboux n n introduced the sums sδ = k=1 mk · (xk − xk−1 ) and Sδ = k=1 Mk · (xk − xk−1 ), where δ = {x0 , x1 , . . . , xn } is a partition of the interval [a, b], mk = infx∈[xk−1 ,xk ] f (x), and Mk = supx∈[xk−1 ,xk ] f (x), for any k = 1, · · · n. The function f can be integrated in the Riemann sense if the distance between the two sums can be as small as we want. Henry Lebesgue had the idea to reverse things: let  = {y0 , y1 , . . . , yn } be a partition of the interval [m, M] and, for any k = 1, · · · n, let Ek = {x ∈ [a, b] : yk−1 ≤ f (x) < yk }. If we can give a meaning to the “length” (“measure”) of the sets Ek , λ(Ek ), then we can consider the sums   σ = nk=1 yk−1 · λ(Ek ) and  = nk=1 yk · λ(Ek ). The function f will be “integrable” if the difference  − σ can be small enough for fairly fine partitions . The development of this simple idea leads to a significant extension of the Riemann integral. In the figure below, we have represented the Darboux sums attached to a function f and to the partition δ = {x0 = 0, x1 , x2 , x3 , x4 , x5 } and next the Lebesgue sum relating to the partition  = {y0 = 0, y1 , y2 , y3 , y4 } for the same function (⊡ Fig. 1). In the left, Sδ is the area of the polygon upper bounded by the thickened line and sδ is the area of the polygon upper bounded by the dashed line. In the right, the area of the polygon upper bounded by the thickened line is equal to the Lebesgue sum  . To give us a first idea of the difference in approach between the two constructions, let us evaluate the sums of Darboux and the sums of Lebesgue for the Dirichlet function: f : [0, 1] → [0, 1],  f (x) =

1, x ∈ Q ∩ [0, 1], 0 , x ∈ (R \ Q) ∩ [0, 1],

and for the partitions δ = {x0 , x1 , · · · , xn } and  = {y0 , y1 , · · · , yp } of [0, 1]. It is obvious that Sδ = 1, sδ = 0; therefore, f is not Riemann

vi

Introduction

6

6

Darboux sums

Lebesgue sum

M = y4 y3 y2 y1 E4 x3

x4 x5

m = y0

? Q  o Q  MB  x5 o S o Z   Q  B  S ZZ Q  B  SE  ZE Q 1 2 Q E3

o o

x1 x2

o o

x0

-

⊡ Fig. 1 Riemann and Lebesgue sums

integrable. However, E1 = (R \ Q) ∩ [0, 1] and Ek = ∅, for any k = 2, · · · , p. By anticipating certain results of measure theory, we can evaluate λ(E1 ) = 1 and λ(Ek ) = 0, for k = 2, · · · , p and then  = y1 and σ = y0 = 0. Thus,  − σ = y1 is less than the mesh of the partition . So, f is integrable in the sense of Lebesgue. The theory of measure and integration is mainly the work of H. Lebesgue. Many other mathematicians, among whom we cite E. Borel, G. Vitali, F. Riesz, D. Egoroff, N. Lusin, J. Radon, M. Fréchet, H. Hahn, O. Nikodym, and C. Carathéodory, contributed to the development of this theory. The introduction into this theory is as necessary (because of its multiple applications) as difficult for the uninitiated. Most of the measure theory treaties involve a large accumulation of knowledge and some important theoretical difficulties. Hence, the idea of providing a less knowledgeable reader with a material allowing him or her to easily access the definition and main properties of the Lebesgue integral. The results presented in this book refer to the measure and integral on R and R2 . In order to take a broader look at the measure theory, we have introduced, at the end of each chapter, a section entitled “Abstract Setting,” where the properties of the measure on abstract spaces are presented (without demonstrations). As we have noted, for the introduction of the Lebesgue integral, it is necessary to measure certain subsets of R that are more complicated than the intervals. First of all, we must establish the conditions under which a definition of “measurability” can be considered acceptable. Can we measure all the subsets of R? The answer is no! What are the properties of the sets to which we can assign a measure? These questions are answered in the first chapter of the book. The measure of open sets is introduced as a natural consequence of the structure of open sets in R. The outer measure is then defined, and finally the class of the Lebesgue measurable sets L. L is strictly included in the family of all subsets of R but has the same cardinal number as the latter. In  Chap. 2, we deal with the so-called measurable functions. For a measurable function f , it is possible to assign a measure to sets of form E = {x : c ≤ f (x)
0, [a + ε, b − ε] ⊆ ∞ ∗ p=1 ]ap , bp [, and therefore there is p0 ∈ N such that [a + ε, b − ε] ⊆

p0  k=1

Ik .

(*)

3 1.1 · Measure of Open Sets

Indeed, if we claim that [a + ε, b − ε] has no finite subcover, then, for all p ∈ N∗ , p there exists xp ∈ [a + ε, b − ε] \ k=1 ]ak , bk [. The sequence (xp )p is bounded. Then let (xkp )p be a subsequence convergent to x ∈ [a + ε, b − ε] (BolzanoWeierstrass theorem). Let p1 ∈ N∗ such that x ∈ Ip1 ; as Ip1 is a neighborhood of x, there is p2 ∈ N∗ , p2 > p1 such that xkp ∈ Ip1 , for all p ≥ p2 . But xkp2 ∈ kp2 [a + ε, b − ε] \ k=1 Ik which, because of p1 < p2 ≤ kp2 , contradicts xkp2 ∈ Ip1 (see also Lemma 7.1.6). The relation (∗) allows us to renumber the finite family of intervals {Ik : k = 1, · · · p0 } so that a1 < a + ε < b − ε < bp0 . Then b − a − 2ε < bp0 − a1 ≤ ∞ p0 k=1 |Ik |. Because ε is arbitrarily positive, k=1 |Ik | ≤ b − a = |I0 | ≤

∞ 

|Ip |.

p=1

 (b) If I0 =]a, +∞[, then, for any n ∈ N, ]a, n[⊆ ∞ p=1 Ip . Using the previous point, ∞ ∞ n − a ≤ k=1 |Ik | from where k=1 |Ik | = +∞ = |I0 |. The same reasoning exists in other possible cases for the unbounded interval I0 .  (2) If I0 is unbounded, then |I0 | = +∞ ≥ ∞ p=1 |Ip |. Let us assume that I0 =]a, b[ is bounded; then the intervals Ip =]ap , bp [ are all bounded. n  For every n ∈ N∗ , Ip ⊆ I0 . Since I1 , . . . , In are pairwise disjoint, we can renumber p=1

them so that a ≤ a1 < b1 ≤ a2 < b2 ≤ · · · ≤ an < bn ≤ b. Then |I0 | = b − a ≥ (b1 − a1 ) + (b2 − a2 ) + · · · + (bn − an ) =  from where ∞ p=1 |Ip | ≤ |I0 |.

n p=1

|Ip |, for any n ∈ N∗ , 

Definition 1.1.2 A subset A ⊆ R is said to be open if each time that it contains a point x, it contains also an open interval which itself contains x (for all x ∈ A, there is I ∈ I such as x ∈ I ⊆ A). The family τu of all open subsets of R is a topology on R, that is to say it satisfies the following properties: (T1 ) D ∩ G ∈ τu , for every D, G ∈ τu ; (T2 ) ∪γ ∈ Dγ ∈ τu , for every {Dγ : γ ∈ } ⊆ τu ; (T3 ) ∅, R ∈ τu . τu is called the usual topology of R.

1

4

1

Chapter 1 • Lebesgue Measure on R

We note that the open intervals are open sets, and therefore, according to (T2 ), the countable unions (even uncountable ones) of open intervals are open sets. We can show that each open set is a countable union of open intervals. Theorem 1.1.3 (structure of open sets in R) Any open set D ⊆ R is a countable union of pairwise disjoint open intervals. This representation of D is unique, up to the order of the family intervals.

Proof If D = ∅, then it is a countable union of empty open intervals of form ]a, a[. If D is not empty, for every x ∈ D, there exist a0 , b0 ∈ R such that x ∈]a0 , b0 [⊆ D. We denote Ax = {a ∈ R : there exists b ∈ R such that x ∈]a, b[⊆ D}, Bx = {b ∈ R : there exists a ∈ R such that x ∈]a, b[⊆ D}. We note that a0 ∈ Ax , b0 ∈ Bx , and then Ax = ∅ = Bx . Let ax = inf Ax ∈ [−∞, +∞[, bx = sup Bx ∈] − ∞, +∞], and let Ix =]ax , bx [. Let’s show that x ∈ Ix ⊆ D. Indeed, ax ≤ a0 < x < b0 ≤ bx , and then x ∈ Ix . For every y ∈ Ix , ax < y < bx , and then there is a ∈ Ax and b ∈ Bx such that ax ≤ a < y < b ≤ bx . Given the definitions of the sets Ax and Bx , there exist a1 , b1 ∈ R such that x ∈]a, b1 [⊆ D and x ∈]a1 , b[⊆ D. ]a, b1 [ and ]a1 , b[ are non-disjoint intervals, and then ]a, b1 [∪]a1 , b[=]a2 , b2 [, where a2 = min{a, a1 }, b2 = max{b, b1 }. It’s obvious that ]a2 , b2 [⊆ D and a2 ≤ a < y < b ≤ b2 from where y ∈ D. It follows that Ix ⊆ D. Then Ix is the largest open interval that contains the point x and is included in D. This maximility character of Ix allows us to show that, for every x, y ∈ D, Ix = Iy , or Ix ∩ Iy = ∅. Indeed, if we assume that Ix ∩ Iy = ∅, then I = Ix ∪ Iy is an interval and x ∈ I ⊆ D, y ∈ I ⊆ D. Using the maximality of intervals Ix and Iy , we obtain I ⊆ Ix and I ⊆ Iy , which leads us to Iy ⊆ Ix and respectively to Ix ⊆ Iy ; therefore Ix = Iy . The family of these maximal intervals ID = {Ix : x ∈ D} is countable. Indeed, for every x ∈ D, let a rational number qx ∈ Ix and let the mapping ϕ : ID → Q, defined by ϕ(Ix ) = qx (here Q denotes the set of rational numbers). It should be noted that if Ix = Iy , then Ix appears once as an element of the family ID , and then the same rational point qx is chosen in Ix = Iy ; therefore ϕ is well-defined. If Ix = Iy , then Ix ∩ Iy = ∅, and then ϕ(Ix ) = qx = qy = ϕ(Iy ). Therefore ϕ is injective and then ID is countable. Let {In : n ∈ N} be an enumeration of family intervals ID , where In ∩ Im = ∅, for all m, n ∈ N, n = m. Then D = ∪x∈D Ix = ∪{I : I ∈ ID } = ∪∞ n=1 In ; hence D is a countable union of pairwise disjoint open intervals.

1

5 1.1 · Measure of Open Sets

Let I = {In : n ∈ N} be another representation of D with open and disjoint intervals: D = ∪n∈N In . Then, for any n ∈ N and every x ∈ In , x ∈ In ⊆ D. By the maximal character of the interval Ix , In ⊆ Ix . But Ix ∈ ID , and then there exists mn ∈ N such that In ⊆ Ix = Imn . Let In =]an , bn [; if we assume that In = Imn , then an ∈ Imn ⊆ D or bn ∈ Imn ⊆ D. Let now p ∈ N, p = n such that an ∈ Ip or bn ∈ Ip , but this is nonsense because Ip and In are disjoint. Therefore In = Imn , and then I ⊆ ID . On the other hand, for any n ∈ N and for every x ∈ In ⊆ D, there is Im n ∈ I such that x ∈ Im n ⊆ D; it also follows that Im n ⊆ In , and with a reasoning similar to the one above, it follows that In = Im n ∈ I . Therefore ID = I , which ensures the uniqueness of decomposition of D. 

Definition 1.1.4 With the notations of the above theorem, we will say that D =

∞ 

In is the

n=1

representation of the set D or that {In : n ∈ N} are the intervals of the representation of D.  ¯ + , defined by λ(D) = ∞ |In |, where {In : n ∈ N} are The mapping λ : τu → R n=1 the intervals of the representation of D, is said to be the measure of open sets. We note that, because of the uniqueness of representation, the above definition is consistent (in a series with positive terms, we can sum the terms in an indifferent order and always get the same sum).

In the following theorem, we mention some important properties of the measure of open sets.

Theorem 1.1.5 The measure of open sets has the following properties: (1) λ(I ) = |I |, for every I ∈ I , (2) λ(∅) = 0, λ(R) = +∞, (3) x + D ∈ τu and λ(x + D) = λ(D), for every x ∈ R and D ∈ τu , (4) xD ∈ τu and λ(xD) = |x|λ(D), for every x ∈ R \ {0} and D ∈ τu , (5) λ(D) ≤ λ(G), for every D, G ∈ τu , with D ⊆ G, ∞ (6) λ(∪∞ n=1 Dn ) = n=1 λ(Dn ), ∀(Dn ) ⊆ τu , Dn ∩ Dm = ∅ ∀n = m, ∞ (7) λ(∪∞ n=1 λ(Dn ), for every (Dn ) ⊆ τu . n=1 Dn ) ≤

Proof Properties (1) and (2) are obvious.  To show (3), it suffices to note that if D = ∞ n=1 In is the representation of open set ∞ D, then x + D = n=1 (x + In ) ∈ τu and that {x + In : n ∈ N∗ } are the intervals of the representation of x + D.

6

1

Chapter 1 • Lebesgue Measure on R

 ∞ Then λ(x + D) = ∞ n=1 |x + In | = n=1 |In | = λ(D).  (4) For every x ∈ R \ {0} and every D ∈ τu , x · D = {xy : y ∈ D}. If D = ∞ n=1 In is the ∞ representation of set D, then x · D = n=1 (x · In ) ∈ τu is the representation of x · D,   and so |λ(x · D)| = ∞ |x · In | = |x| · ∞ n=1  n=1 |In | = |x| · λ(D). ∞ (5) Let D = n=1 In ⊆ G = ∞ J be the representations of the two open sets. m m=1 For every m ∈ N∗ , let Nm = {n ∈ N∗ : In ⊆ Jm } ⊆ N∗ . We note that some of the Nm can be empty (it could be that some Jm does not contain any interval In ); let M = {m ∈ N∗ : Nm = ∅} ⊆ N∗ . ∗ Then {N m : m ∈ M} is a partition of N , which means that: ∗ (a) N = Nm and m∈M

(b) Nm ∩ Np = ∅, for all m, p ∈ M, m = p. ∞ 

Indeed, for any n ∈ N∗ , In ⊆ D ⊆ G =

Jm , and then there exists m ∈ N∗ such

m=1

that In ∩ Jm = ∅. By the maximal character of Jm , we have In ⊆ Jm and then n ∈ Nm and m ∈ M. If we assume that it exists n ∈ Nm ∩ Np , then In ⊆ Jm and In ⊆ Jp , which contradicts that Jm and Jp are disjoint for m = p.  In ⊆ Jm . Since the intervals In are pairwise disjoint, For every m ∈ M, n∈Nm



according to (2) of Lemma 1.1.1, we obtain

|In | ≤ |Jm |, for any m ∈ M. Then

n∈Nm

λ(D) =

∞ 

 

|In | =

n=1

|In | ≤

m∈M n∈Nm

 m∈M

|Jm | ≤



|Jm | = λ(G).

m∈N∗

∞ n (6) For every n ∈ N∗ , let Dn = k=1 Ik be the representation of Dn ; then D = ∞ ∞ n ∞ n=1 Dn = n=1 k=1 Ik is the representation of D, and therefore λ(D) =

∞ ∞  

|Ikn | =

n=1 k=1

∞ 

λ(Dn ).

n=1

  (7) Let D = ∞ ∈ τ . Let D = ∞ n=1 D p=1 Ip be the representation of D, and for any n∞ un n ∈ N, let Dn = k=1 Ik be the representation of Dn . Then, for any p ∈ N, Ip = Ip ∩ D =

∞  n=1

(Ip ∩ Dn ) =

∞ ∞   n=1 k=1

where Ip ∩ Ikn ∈ I , for every p, k, n ∈ N.

(Ip ∩ Ikn ),

1

7 1.1 · Measure of Open Sets

According to (1) of Lemma 1.1.1, |Ip | ≤ λ(D) =

∞  p=1

=

∞  ∞  n=1 k=1

|Ip | ≤

∞  ∞  ∞ 

∞ ∞

|Ip ∩ Ikn | =

p=1 n=1 k=1

λ(D ∩ Ikn ) =

∞  ∞  n=1 k=1

n=1

k=1

|Ip ∩ Ikn |, and then

∞  ∞  ∞ 

|Ip ∩ Ikn | =

n=1 k=1 p=1

|Ikn | =

∞ 

λ(Dn ).

n=1

In the above relations, we have used that for all n, k ∈ N, D ∩ Ikn = the representation of open set D ∩ Ikn .

∞

p=1 (Ip

∩ Ikn ) is 

Definition 1.1.6 Property (3) of the previous theorem is called the property of translation invariance of the Lebesgue measure λ. Property (5) tells us that λ is monotonic. Property (6) is said to be the property of σ -additivity and (7) the property of σ -subadditivity of the measure λ.

In the spaces Rn , n ≥ 2, a result like the Theorem 1.1.3 does not work. For example, in R2 , we cannot represent open sets as a countable union of a pairwise disjoint family of open two-dimensional intervals (rectangles). However, it will work a result of the representation of an open set as a countable union of the almost disjoint closed rectangles (the interiors of the rectangles are pairwise disjoint)—see Theorem 5.1.12. We have a similar result in the case of R. Definition 1.1.7 An interval J ∈ J is said to be closed if: (a) J is bounded and of the form J = [a, b], with a, b ∈ R or (b) J is unbounded and of the form ] − ∞, b] or [a, +∞[, with a, b ∈ R. A point x is interior for the interval J = [a, b] if a < x < b (if J =] − ∞, b], then x < b, and if J = [a, +∞[, then a < x).

Theorem 1.1.8 Any non-empty open set of R is a countable union of nonoverlapping closed intervals (without common interior points).  If D = ∞ {Jn : n ∈ N} is a countable family of almost disjoint closed n=1 Jn , where intervals, then λ(D) = ∞ n=1 |Jn |.

8

1

Chapter 1 • Lebesgue Measure on R

Proof  ∗ Let D = ∞ n=1 In , where, for any n ∈ N , In =]an , bn [ is an open interval (bounded or unbounded) and In ∩ Im = ∅, for all n = m (the representation of D). It is enough to represent each open interval ]an , bn [ as a countable union of closed nonoverlapping intervals. Let αpn ↓ an and βpn ↑ bn such that an < αpn < βqn < bn , for all n, p, q ∈ N. Then ]an , bn [=

∞ 

n [αp+1 , αpn ] ∪ [α0n , β0n ] ∪

p=0

∞ 

n [βpn , βp+1 ].

p=0

Now let D = ∪∞ n=1 Jn , where {Jn : n ∈ N} is a family of almost disjoint closed intervals. (a) If we assume that the intervals Jn are all bounded, then, for all n ∈ N∗ , Jn = [an , bn ], where an , bn ∈ R. For every ε > 0 and every n ∈ N∗ , there exist the open intervals In , Kn such that In ⊆ Jn ⊆ Kn and ε ε , |Kn | ≤ |Jn | + n 2n 2



ε ε ε ε (we choose In = an + 2n+1 , bn − 2n+1 , bn + 2n+1 and Kn = an − 2n+1 ). ∞ ∞ Let I = n=1 In and K = n=1 Kn ; then I ⊆ D ⊆ K and, since the intervals In are pairwise disjoint, |Jn | ≤ |In | +

λ(D) ≥ λ(I ) =

∞ 

|In | ≥

n=1

∞ 

|Jn | − ε.

(1.1)

n=1

On the other hand, using the monotonicity and the σ -subadditivity of the measure λ, we get λ(D) ≤ λ(K) ≤

∞  n=1

|Kn | ≤

∞ 

|Jn | + ε.

(1.2)

n=1

Since ε is arbitrarily positive, according to (1) and (2), it follows that λ(D) = ∞ n=1 |Jn |. (b) If, among the Jn , there is an interval unbounded, Jn0 = [an0 , +∞[, then, for all ε > 0, let In0 =]an0 + ε, +∞[⊆ Jn0 ⊆ D; then +∞ = |In0 | = λ(D). Therefore λ(D) =  +∞ = |Jn0 | = ∞  n=1 |Jn |. Similarly, if Jn0 =] − ∞, bn0 ] Remark 1.1.9 We note that an open set D can have several representations like countable union of almost disjoint closed intervals; for each of these representations, the sum of the lengths of the intervals is the same—the measure of D.

9 1.2 · Lebesgue Outer Measure

1.2

Lebesgue Outer Measure

We will now extend the measure of open sets to an application defined on P (R) (the family of all subsets of R). Definition 1.2.1 ¯ + , defined by The mapping λ∗ : P (R) → R λ∗ (A) = inf{λ(D) : D ∈ τu , A ⊆ D}, for all A ⊆ R, is called Lebesgue outer measure on R. By definition, we immediately observe that 

 ∞ ∞   λ (A) = inf (bn − an ) : A ⊆ ]an , bn [ , for every A ⊆ R. ∗

n=0

n=0

Remark 1.2.2 We note that for every open set D, λ∗ (D) = λ(D). Indeed, from the definition of outer measure, λ∗ (D) ≤ λ(D) (among the open sets containing D appears the set itself). On the other hand, for every open set G with D ⊆ G, λ(D) ≤ λ(G) (see Property (5) of Theorem 1.1.5), and so λ(D) ≤ λ∗ (D). It follows that the Lebesgue outer measure is an extension of measure of open sets to the family of all subsets of R. It is not a measure because, as we will see in 1.2.7, it is not σ -additive.

The Lebesgue outer measure has the following properties:

Theorem 1.2.3 (1) λ∗ (∅) = 0, (2) λ∗ (A) ≤ λ∗ (B), for every A, B ∈ P (R), A ⊆ B,  ∞ ∗ (3) λ∗ ( ∞ n=1 An ) ≤ n=1 λ (An ), for every (An )n ⊆ P (R).

Proof (1) Due to the previous remark, λ∗ (∅) = λ(∅) = 0, because ∅ ∈ τu and λ(∅) = 0. (2) Since A ⊆ B, {λ(D) : D ∈ τu , B ⊆ D} ⊆ {λ(G) : G ∈ τu , A ⊆ G}; hence, passing to infimum, λ∗ (A) ≤ λ∗ (B). (3) If there is n ∈ N∗ such that λ∗ (An ) = +∞, then the inequality is obvious. Now assume that for any n ∈ N∗ , λ∗ (An ) < +∞. For every ε > 0 and every n ∈ N∗ ,  there exists Dn ∈ τu such that An ⊆ Dn and λ(Dn ) < λ∗ (An ) + 2εn . Then D = ∞ n=1 Dn ∈ ∞    ∞ ∞ ∗ τu and n=1 An ⊆ D. Therefore λ∗ ( ∞ n=1 An ) ≤ λ(D) ≤ n=1 λ(Dn ) ≤ n=1 λ (An ) + ∗ ε. Because ε is positive and arbitrarily small, we get the σ -subadditivity of λ . 

1

10

1

Chapter 1 • Lebesgue Measure on R

Remarks 1.2.4 (i) λ∗ ({x}) = 0, for every x ∈ R. Indeed, for every x ∈ R and for any n ∈ N∗ , {x} ⊆]x − n1 , x + n1 [, from where ∗ λ ({x}) ≤ n2 , for any n ∈ N∗ , and then λ∗ ({x}) = 0. (ii) λ∗ (A ∪ B) ≤ λ∗ (A) + λ∗ (B), for every A, B ⊆ R. Let {An : n ∈ N∗ }, so that A1 = A, A2 = B and, for any n ≥ 3, An = ∅; then due to (1) and (3) of the previous theorem, ∗

∗

λ (A ∪ B) = λ

∞  n=1

An

≤

∞ 

λ∗ (An ) = λ∗ (A) + λ∗ (B).

n=1

This property is called finite subadditivity; it can be extended by complete induction to a finite number of subsets of R. (iii) For any interval J ∈ J , λ∗ (J ) = |J |. Indeed, if J is an open interval, then the property results from Remark 1.2.2. If J is not open, then it differs from an open interval by two points, at most. The property results from (i).

The outer measure has a property similar to that of Theorem 1.1.8.  Proposition 1.2.5 Let A = ∞ n=1 Jn , where Jn are nonoverlapping closed intervals (the interiors of the intervals are pairwise disjoint); then λ∗ (A) =

∞ 

|Jn |.

n=1

Proof If there is n0 ∈ N∗ so that the interval Jn0 is unbounded, then λ∗ (A) ≥ λ∗ (Jn0 ) = |Jn0 | = +∞, and therefore equality is checked. We now assume that |Jn | < +∞, for all n ∈ N∗ . For every ε > 0 and every n ∈ N∗ ,  there exists an open interval In ⊆ Jn such that |Jn | < |In | + 2εn . Let D = ∞ n=1 In ∈ τu . Then D ⊆ A, and, since the intervals In are pairwise disjoint, λ∗ (A) ≥ λ∗ (D) = λ(D) = ∞ ∞ ∞ ∗ ∗ n=1 |In | ≥ n=1 |Jn | − ε, from where λ (A) ≥ n=1 |Jn |. The σ -subadditivity of λ assures us the conversely inequality. 

The following theorem shows that the outer measure is invariant by translations.

Theorem 1.2.6 λ∗ (x + A) = λ∗ (A), for every x ∈ R and every A ⊆ R. λ∗ (x · A) = |x| · λ∗ (A), for every x ∈ R \ {0} and every A ⊆ R.

1

11 1.2 · Lebesgue Outer Measure

Proof For every D ∈ τu with A ⊆ D, x + A ⊆ x + D; from (3) of Theorem 1.1.5, it results that λ∗ (x + A) ≤ λ(D) and then λ∗ (x + A) ≤ λ∗ (A). Since this last inequality takes place for every x ∈ R and every A ⊆ R, λ∗ (A) = λ∗ (−x + (x + A)) ≤ λ∗ (x + A). For every D ∈ τu with A ⊆ D, x · D ∈ τu and x · A ⊆ x · D. According to (4) of Theorem 1.1.5, λ∗ (x · A) ≤ λ(x · D) = |x| · λ(D), and then λ∗ (x · A) ≤ |x| · λ∗ (A).

(*)

The inequality (∗) occurs for every x = 0 and every A ⊆ R and then λ∗ (A) = λ∗



 1 1 · (x · A) ≤ · λ∗ (x · A) x |x| 

where we get the inverse inequality of (∗).

From the above, λ∗ checks properties (b) and (c) presented at the beginning of this chapter; this function is an invariant σ -subadditive extension of the interval length. The following example shows that λ∗ is not σ -additive (does not check property a) formulated in the introduction to this chapter). Vitali Set 1.2.7 On the set I = [0, 1], we define a relation by xy if and only if x − y ∈ Q. We can easily see that  is reflexive, symmetric, and transitive, therefore an equivalence relation on I . For every x ∈ I , the equivalence class of representative x is [x] = {y ∈ I : xy} = {y ∈ I : y − x ∈ Q} = {y ∈ I : y ∈ x + Q} = I ∩ (x + Q). It follows that [x] is a countable set, for every x ∈ I . Two distinct equivalence classes are disjoint, and the union of these classes is I . Because each equivalence class is not empty, the axiom of choice assures us that there exists a set V ⊆ I containing a single element of each equivalence class. So for any equivalence class [x], card(V ∩ [x]) = 1, where card(A) denotes the cardinal number of A. Let us denote by Q1 = Q ∩ [−1, 1] the set of rational numbers in the interval [−1, 1]; then [0, 1] ⊆



(r + V ) ⊆ [−1, 2].

(*)

r∈Q1

Indeed, for every x ∈ [0, 1], there exists x1 ∈ V ∩ [x]; then xx1 , and so x − x1 = r ∈ Q. It follows that x = r + x1 ∈ r + V and r = x − x1 ∈ Q1 . The second inclusion is obvious if we note that for all r ∈ Q1 , r + V ⊆ r + [0, 1] ⊆ [−1, 2]. Using successive the results (iii) of Remarks 1.2.4 and (2) and (3) of Theorems 1.2.3 and 1.2.6, we will get from the first inclusion of (∗) that ⎛ ∗

∗⎝

1 = λ ([0, 1]) ≤ λ

 r∈Q1

from where λ∗ (V ) > 0.

⎞ (r + V )⎠ ≤

 r∈Q1

λ∗ (r + V ) =

 r∈Q1

λ∗ (V ),

Chapter 1 • Lebesgue Measure on R

12

1

The sets {r + V : r ∈ Q1 } are pairwise disjoint. Indeed, if we assume that, for r, s ∈ Q1 , r = s there exists x ∈ (r + V ) ∩ (s + V ), then x − r, x − s ∈ V , but (x − r)(x − s). Then x − r and x − s are two different elements of V belonging to the same equivalence class; this is a contradiction because V contains only one element in each equivalence class. If we assume that λ∗ is σ -additive, ⎛ ∗⎝

λ

 r∈Q1

⎞ (r + V )⎠ =



λ∗ (r + V ) =

r∈Q1



λ∗ (V ) = +∞.

(**)

r∈Q1

⎛ According to the second inclusion of (∗), λ∗ ⎝



⎞ (r + V )⎠ ≤ λ∗ ([−1, 2]) = 3, which

r∈Q1

contradicts (∗∗). Therefore λ∗ is not σ -additive. It follows that the extension made in Definition 1.2.1 is too wide, λ∗ not meeting the requirements specified at the beginning of this chapter.

In the following proposition, we give other calculation formulas for the outer measure of a set. Proposition 1.2.8 For every set A ⊆ R,  λ∗ (A) = inf{ ∞ (bn − an ) : A ⊆ ∪∞ n=1 ]an , bn ]} n=1 ∞ = inf{ n=1 (bn − an ) : A ⊆ ∪∞ n=1 [an , bn ]}. Proof  ∞ Let λ∗1 (A) = inf{ ∞ open intervals n=1 (bn − an ) : A⊆ ∪n=1 ]an , bn ]}. For every cover with  ∞ ∗ ]an , bn [, n ∈ N of A, we have A ⊆ n=1 ]an , bn ], and therefore λ∗1 (A) ≤ ∞ n=1 (bn − an ), from where λ∗1 (A) ≤ λ∗ (A). If λ∗1 (A) = +∞, the equality is demonstrated. We assume that λ∗1 (A) < +∞. For every ε > 0, there is a sequence of semi-closed  ∞ ∗ intervals (]an , bn ])n≥1 such that A ⊆ ∞ n=1 ]an , bn ] and λ1 (A) + ε > n=1 (bn − an ). Then ∞   ε

∗ A⊆ an , bn + n , and therefore λ∗ (A) ≤ ∞ n=1 (bn − an ) + ε < λ1 (A) + 2ε. Since ε 2 n=1 is arbitrarily positive, we obtain the inverse inequality λ∗ (A) ≤ λ∗1 (A). The proof of the second formula is similar. 

Although the outer measure is not, in general, σ -additive, on certain sequences of sets, it verifies the property of σ -additivity. For two non-empty sets B, C ⊆ R, let us denote with d(B, C) = = inf{|x − y| : x ∈ B, y ∈ C} the distance between B and C. Obviously, if B and C have one point in common, then d(B, C) = 0. The converse is not true. Indeed, if B = { n1 : n ∈ N∗ } and C = {0}, then B ∩ C = ∅, and d(B, C) = 0. In general, we can show that d(B, C) = 0 if and only if there exist two sequences (xn )n ⊆ B, (yn )n ⊆ C such that xn − yn → 0.

13 1.2 · Lebesgue Outer Measure

If B = ∅ or C = ∅, we agree that d(B, C) = +∞. Theorem 1.2.9 (1) λ∗ (B ∪ C) = λ∗ (B) + λ∗ (C), for every B, C, with d(B, C) > 0. p p (2) λ∗ (∪n=1 An ) = n=1 λ∗ (An ), for every A1 , . . . , An ⊆ R with d(An , Am ) > 0, for all n, m ∈ {1, · · · , p}, n = m. ∞ ∗ (3) λ∗ (∪∞ n=1 λ (An ), for every sequence (An )n≥1 of subsets of R with n=1 An ) = d(An , Am ) > 0, for all n = m.

Proof (1) From the property of finite subadditivity of λ∗ (see (ii) of Remark 1.2.4), λ∗ (B ∪ C) ≤ λ∗ (B) + λ∗ (C). If λ∗ (B ∪ C) = +∞, equality is evident. Now let us assume that λ∗ (B ∪ C) < +∞; using Proposition 1.2.8, for every ε > 0,  ∗ there exists {]an , bn ] : n ∈ N} such that B ∪ C ⊆ ∞ n=0 ]an , bn ] and λ (B ∪ C) + ε > ∞ n=0 (bn − an ). Without limiting the generality, it can be assumed that, for all n ∈ N, bn − an < δ ≡ d(B, C) (otherwise the intervals ]an , bn ] will be divided into a sufficient number of subintervals of the same kind, the lengths of which make it possible to verify the above requirement). Then, for all n ∈ N, the interval ]an , bn ] intersects only one of sets B and C. Let N1 = {n ∈ N :]an , bn ] ∩ B = ∅} and N2 = {n ∈ N :]an , bn ] ∩ C = ∅}. Note that N1 ∩ N2 = ∅ and B⊆



]an , bn ], C ⊆

n∈N1

Therefore λ∗ (B) ≤



]an , bn ].

n∈N2



λ∗ (B) + λ∗ (C) ≤

n∈N1 (bn



− an ), and λ∗ (C) ≤

(bn − an ) +

n∈N1

≤



 n∈N2



n∈N2 (bn

(bn − an ) =

− an ), and then



(bn − an ) ≤

n∈N1 ∪N2

(bn − an ) < λ∗ (B ∪ C) + ε.

n∈N

Since ε is arbitrary positive, we obtain the inverse inequality and therefore the required equality. (2) The reasoning is done by induction; for p = 2, the demonstration was made at the previous point. Suppose the property is checked for p − 1, and let {A1 , · · · , Ap } be a family of p subsets of R for which the distance between any two is strictly positive.

1

Chapter 1 • Lebesgue Measure on R

14

1

Let us denote B =

p

n=1 An

and C =

p−1 n=1

An . Then

d(C, Ap ) = min{d(An , Ap ) : n = 1, · · · , p − 1} > 0. Using (1) and the recurrence hypothesis, λ∗ (B) = λ∗ (C ∪ Ap ) = λ∗ (C)+ p−1 p +λ∗ (Ap ) = n=1 λ∗ (An ) + λ∗ (Ap ) = n=1 λ∗ (An ).  p p ∗ (3) For every p ∈ N∗ , ∪∞ , and then λ∗ (∪∞ n=1 An n=1 An ⊇ n=1 An ) ≥ λ ( n=1 An ) = p ∞ ∗ ∗ ∞ ∗ n=1 λ (An ). The inverse inequality comes n=1 λ (An ). Therefore λ (∪n=1 An ) ≥ ∗ from the property of σ -subadditivity of λ . 

At the end of this paragraph, we will present a very important concept in the theory of measure and integration—that of a null set or negligible set. Definition 1.2.10 A subset A ⊆ R is a Lebesgue null set or a Lebesgue negligible set if λ∗ (A) = 0. According to the definition, A is a null set if and only if, for every ε > 0, there exists a  ∞ sequence of open intervals (In )n∈N ⊆ I such that A ⊆ ∞ n=0 In and n=0 |In | < ε. Note that in the case of Jordan null sets, the cover by a family of open intervals is finite, so that any Jordan null set is a Lebesgue null set. Because we will not be working with Jordan null sets, in what follows, we will use the term null set for a Lebesgue null set.

Examples 1.2.11 (i) For every x ∈ R, {x} is a null set (see (i) of Remark 1.2.4). (ii) Any subset of a null set is a null set (see (2) of Theorem 1.2.3).  (iii) Any countable union of null sets is a null set. Indeed, let A = ∞ n=1 An where, for any  ∗ (A ) = 0 and so A is a null set. n ∈ N∗ , λ∗ (An ) = 0; then λ∗ (A) ≤ ∞ λ n n=1 (iv) From (i) and (iii), every countable set is a null set. Particularly, N, Z and Q are null sets (here Z denotes the set of the integers). But there are also uncountable null sets; such is the triadic Cantor set that we will build in 1.3.16. Another example is presented in Exercise (4) of  Sect. 1.5.

1.3

Lebesgue Measurable Sets

In this section, we will specify the subsets of R to whom a measure can be assigned and which properties have that measure. For every A ⊆ R, we defined λ∗ (A) = inf{λ(D) : D ∈ τu , A ⊆ D}. If we assume that λ∗ (A) < +∞, then, for every ε > 0, there exists D ∈ τu with A ⊆ D such that λ(D) < λ∗ (A) + ε or λ(D) − λ∗ (A) < ε. On the other hand, D = A∪(D\A), from where λ(D) = λ∗ (D) ≤ λ∗ (A)+λ∗(D\A) and then λ(D) − λ∗ (A) ≤ λ∗ (D \ A).

15 1.3 · Lebesgue Measurable Sets

Definition 1.3.1 A set A ⊆ R is measurable (in the sense of Lebesgue) if, for all ε > 0, there exists D ∈ τu such that A ⊆ D and λ∗ (D \ A) < ε. We will denote with L(R) or L the class of measurable sets of R. The restriction λ = λ∗ |L is called the Lebesgue measure on R. For every A ∈ L, we will denote with L(A) = {B ⊆ A : B ∈ L} the family of all measurable subsets of A.

Remark 1.3.2 τu ⊆ L; indeed, if G ∈ τu , then, for every ε > 0, there exists D = G ⊇ G such that λ∗ (D \ G) = λ∗ (∅) = 0 < ε. From this follows that λ is an extension of the measure of open sets and then the notation performed is therefore not confusing.

Theorem 1.3.3 (1) Every null set is a measurable set. ∞ (2) n=1 An ∈ L, for every (An )n ⊆ L.

Proof (1) Let A ⊆ R be a null set (λ∗ (A) = 0); for every ε > 0, there exists D ∈ τu such that A ⊆ D and λ(D) < ε. Then λ∗ (D \ A) ≤ λ∗ (D) = λ(D) < ε, and therefore A ∈ L.  ∗ (2) Let A = ∞ n ∈ N∗ , there n=1 An , where {An : n ∈ N } ⊆ L; for every ε > 0 and every ∞ ε ∗ exists Dn ∈ τu such that An ⊆ Dn and λ (Dn \ An ) < 2n . Let D = n=1 Dn ∈ τu ;  ∞ ∗ then A ⊆ D and D \ A = ∞ n=1 (Dn \ A) ⊆ n=1 (Dn \ An ), from where λ (D \ A) ≤ ∞ ∗  n=1 λ (Dn \ An ) ≤ ε. Remarks 1.3.4 (i) ∅ ∈ L. (ii) For every A ∈ L with λ(A) = 0 and every B ⊆ A, λ∗ (B) = 0; therefore B ∈ L. We will say that λ is complete on L or that L is complete with respect to λ. (iii) A ∪ B ∈ L, for every A, B ∈ L (A ∪ B = A ∪ B ∪ ∅ ∪ · · · ∪ ∅ ∪ · · · ). (iv) Each interval is a measurable set. Indeed, open intervals are open sets and therefore measurable; each other interval differs up to two points from an open interval.

We will show that the closed sets are also measurable. We recall that A ⊆ R is a closed set if its complement is an open set (R \ A ∈ τu ) or equivalent if, for any convergent sequence of A, the limit belongs to set A. In particular, any closed interval (see Definition 1.1.7) is a closed set. By passing to the complementary in the properties of open sets (see Definition 1.1.2), we obtain that the union of a finite family of closed sets is a closed set and that any intersection (finite or infinite) of closed sets is a closed set.

1

16

1

Chapter 1 • Lebesgue Measure on R

A set A ⊆ R is a compact set if it is bounded and closed, or equivalent, if any sequence of A has a convergent subsequence to a point of A. Any bounded closed interval (Definition 1.1.7) is a compact set. First, we will present a lemma. Lemma 1.3.5 If F is a closed set and K is a compact set such that F ∩ K = ∅, then d(F, K) = inf{|x − y| : x ∈ F, y ∈ K} > 0. Proof If we assume that d(F, K) = inf{|x − y| : x ∈ F, y ∈ K} = 0, then there exist two sequences, (xn )n ⊆ F and (yn )n ⊆ K, such that xn − yn → 0. K being compact, (yn )n has a subsequence (ykn )n converging to y ∈ K. It follows that xkn → y and, since F is closed, y ∈ F . So y ∈ F ∩ K, which contradicts the hypothesis that F and K are disjoint. 

Theorem 1.3.6 Any closed set is a measurable set.

Proof (a) First we assume that F is a bounded closed set; therefore F is a compact, and λ∗ (F ) < +∞ (see Exercise (5) in  Sect. 1.5). According to the outer measure definition, for any ε > 0, there exists D ∈ τu such that F ⊆ D and λ(D) < λ∗ (F ) + ε. Then D \ F ∈ τu , and according to Theorem 1.1.8,  D\F = ∞ {J : n ≥ 1} is a family of nonoverlapping closed intervals; n=1 Jn , where  n moreover, λ(D \ F ) = ∞ n=1 |Jn |.  For all m ∈ N∗ , m n=1 Jn = J is a closed subset of D \F and then F ∩J = ∅. From the previous lemma, d(F, J ) > 0, and then Theorem 1.2.9 assures us that λ∗ (F ∪ J ) = λ∗ (F ) + λ∗ (J ). It follows that λ(D) ≥ λ∗ (F ∪ J ) = λ∗ (F ) + λ∗ (J ). Any two intervals of the family {Jn : n = 1, · · · , m} have at most one point in common (in which case their union is a closed interval) or are disjoint (the case where the distance between them is strictly positive); applying again Theorem 1.2.9, we get   ∗ ∗ λ∗ (J ) = m |J |. Then m n=1 n=1 |Jn | ≤ λ(D) − λ (F ) < ε, for any m ∈ N . Then ∞ n λ(D \ F ) = n=1 |Jn | < ε, from where F ∈ L.  (b) If F is an unbounded closed set, then F = ∞ n=1 (F ∩ [−n, n]) is a countable union of bounded closed sets, hence of measurable sets; from (2) of Theorem 1.3.3, F ∈ L. 

For every A ⊆ R, we denote by Ac = R \ A the complement of A. Theorem 1.3.7 The complement of a measurable set is a measurable set.

1

17 1.3 · Lebesgue Measurable Sets

Proof Let A ∈ L; for any n ∈ N∗ , there exists Dn ∈ τu such that A ⊆ Dn and λ∗ (Dn \ A) <  ∞ c c Then A ⊆ ∞ n=1 Dn , and so B = n=1 Dn ⊆ A . We can then write Ac = B ∪ (Ac \ B).

1 n.

(*)

The sets Dnc = R \ Dn are closed, for all n ∈ N∗ ; according to Theorem 1.3.6, Dnc ∈ L;  c therefore B = ∞ Theorem 1.3.3). n=1 Dn ∈ L (see point (2) of c On the other hand, Ac \ B = Ac ∩ B c = ∞ n=1 Dn \ A, and then A \ B ⊆ Dn \ A, for 1 ∗ ∗ c ∗ any n ∈ N . It follows that λ (A \ B) < n , for any n ∈ N from where λ∗ (Ac \ B) = 0. Since Ac \ B is a null set, it is measurable, and then, according to (∗), Ac is a union of the two measurable sets, so it is measurable.  Corollary 1.3.8 (1) A ∩ B ∈ L, for every A, B ∈ L. (2) A \ B ∈ L, for every A, B ∈ L. ∞ (3) n=1 An ∈ L, for every (An )n ⊆ L. Proof The proof is an immediate consequence of the previous theorem and of the relations: (A ∩  ∞ c c B)c = Ac ∪ B c , A \ B = A ∩ B c , and ( ∞  n=1 An ) = n=1 An .

The following corollary gives a characterization of the measurable sets using closed sets. Corollary 1.3.9 A ∈ L if and only if, for every ε > 0, there exists a closed set F ⊆ A, such that λ∗ (A \ F ) < ε. Proof A ∈ L if and only if Ac ∈ L and so, equivalent, if, for every ε > 0, there exists D ∈ τu such that Ac ⊆ D and λ∗ (D \ Ac ) < ε which is equivalent to the existence of the closed set F = D c ⊆ A with λ∗ (A \ F ) = λ∗ (A ∩ F c ) = λ∗ (A ∩ D) = λ∗ (D \ Ac ) < ε. 

Theorem 1.3.10 The set function λ : L → [0, +∞], λ(A) = λ∗ (A), for every A ∈ L, is σ -additive, which means that, for every pairwise disjoint sequence (An )n ⊆ L, λ

∞  n=1

An

=

∞  n=1

λ(An ).

18

1

Chapter 1 • Lebesgue Measure on R

Proof  Let (An )n ⊆ L be a pairwise disjoint sequence, and let A = ∞ n=1 An . (a) First we assume that all the sets An are bounded. According to the previous corollary, for all n ∈ N∗ and all ε > 0, there exists a closed set Fn ⊆ An such that λ∗ (An \ Fn ) < 2εn . Then λ∗ (An ) ≤ λ∗ (An \ Fn ) + λ∗ (Fn ) < λ∗ (Fn ) + 2εn . For every n = m, An is disjoint from Am , and so Fn ∩ Fm = ∅; the sets Fn being bounded and closed (therefore compact), we can apply the result of Lemma ∞ 1.3.5

to  obtain d(Fn , Fm ) > 0, for all n = m. According to Theorem 1.2.9, λ∗ Fn = ∞ 

n=1

λ∗ (Fn ). We deduce that

n=1

∗

∗

λ (A) = λ

∞ 

An

∗

≥λ

n=1

∞ 

Fn

=

n=1

∞ 

λ∗ (Fn ) >

n=1

∞ 

λ∗ (An ) − ε.

n=1

∞

Since ε is arbitrarily positive, λ∗ (A) ≥ n=1 λ∗ (An ). The inverse inequality is always verified (see (3) of Theorem 1.2.3), and then we obtain the required equality. (b) Suppose now that An are not all bounded. For every p ∈ N, we denote Ip = [−p, p];  then ∞ \ Ip , J0 = {0}; p=0 Ip = R, and Ip ⊆ Ip+1 . For every p ∈ N, let Jp+1 = Ip+1  Jp are measurables (unions of two intervals), pairwise disjoint, and ∞ p=0 Jp = R. For  p p p every p ∈ N and every n ∈ N∗ , let An = An ∩ Jp ; then A = n,p An and the sets An are bounded and pairwise disjoint. Using case (a), we obtain λ∗ (A) =

∞  ∞  n=1 p=0

p λ∗ (An )

=

∞  n=1

⎛ λ∗ ⎝

∞  p=0

⎞ p An ⎠

=

∞ 

λ∗ (An ).



n=1

In the following theorem, we list some consequences of σ -additivity of Lebesgue measure on L. We recall that, for a sequence (xn )n ⊆ R, lim infn xn = supn∈N infk≥n xk , and lim supn xn = infn∈N supk≥n xk . For every sequence of sets (An )n ⊆ P (R),  ∞ ∞ ∞ lim infn An = ∞ n=0 k=n Ak , and lim supn An = n=0 k=n Ak . Theorem 1.3.11 ¯ has the following properties: The measure λ : L → R n  n+ n (1) λ k=1 Ak = k=1 μ(Ak ), for every (Ak )k=1 ⊆ L, and Ak ∩ Al = ∅, for any k = l. (2) λ(A) ≤ λ(B), for every A, B ∈ L , A ⊆ B. (3) λ(B \ A) = λ(B) − λ(A), for every A, B ∈ L, A ⊆ B, λ(A) < +∞. (4) λ(A ∪ B) + λ(A ∩ B) = λ(A) + λ(B), for every A, B ∈ L. (Continued )

19 1.3 · Lebesgue Measurable Sets

Theorem 1.3.11 (continued)   ∞ An ≤ ∞ (5) λ n=1 λ(An ), for every (An ) ⊆ L. n=1  ∞ (6) λ An = limn λ(An ), ∀(An ) ⊆ L, An ⊆ An+1 , for any n ∈ N∗ . n=1 ∞  ∗ (7) λ n=1 An = limn λ(An ), if (An ) ⊆ L, An+1 ⊆ An , for any n ∈ N and λ(A1 ) < +∞. (8) λ(lim infn An ) ≤ lim infn λ(An ), for every (An ) ⊆ L. ∞  (9) lim supn λ(An ) ≤ λ(lim supn An ), ∀(An ) ⊆ L, λ n=1 An < +∞.

Proof   (1) λ( nk=1 Ak ) = λ( ∞ k > n, Ak = ∅. By applying the property k=1 Ak ), where, for all   n of σ -additivity of the measure, we obtain λ( nk=1 Ak ) = ∞ k=1 λ(Ak ) = k=1 λ(Ak ) (for any k > n, λ(Ak ) = 0). (2) The monotonicity is the consequence of point (2) of Theorem 1.2.3 and of the fact that λ = λ∗ |L . (3) The property results from λ(A) < +∞ and from λ(B) = λ(A ∪ (B \ A)) = λ(A) + λ(B \ A). (4) If λ(A ∩ B) = +∞, then the relation is obvious. Suppose that λ(A ∩ B) < +∞, and apply the finite additivity of measure λ in the relation: A ∪ B = [A \ (A ∩ B)] ∪ (A ∩ B) ∪ [B \ (A ∩ B)]. From property (3), we obtain λ(A ∪ B) = [λ(A) − λ(A ∩ B)] + λ(A ∩ B) + [λ(B) − λ(A ∩ B)] which immediately leads us to the desired relation. (5) The property results from (3) of Theorem 1.2.3.  (6) Let (An )n ⊆ L be an increasing sequence of sets, and let A = ∞ n=1 An . If there is n0 ∈ N∗ such that λ(An0 ) = +∞, then, for any n ≥ n0 , λ(An ) = +∞, and hence λ(A) = +∞ = limn λ(An ). Suppose that λ(An ) < +∞, for any n ∈ N∗ ; then the associated disjoint sequence (Bn )n is defined by B1 = A1 , Bn = An \ An−1 , for any n ≥ 2. We can easily show that  A= ∞ n=1 Bn and that (Bn ) are pairwise disjoint. Then, using (3), λ(A) = λ

∞  n=1

Bn

=

∞  n=1

λ(Bn ) = lim n

n 

λ(Bk ) =

k=1

= lim[λ(A1 ) + λ(A2 \ A1 ) + . . . + λ(An \ An−1 )] = n

= lim[λ(A1 ) + λ(A2 ) − λ(A1 ) + · · · + λ(An ) − λ(An−1 )] = lim λ(An ). n

n

1

Chapter 1 • Lebesgue Measure on R

20

1

(7) Let (An ) ⊆ L be a decreasing sequence of sets with λ(A1 ) < +∞, and let A = ∞ A1 \ An , for any n ∈ N∗ , is n=1 An ; then the sequence (Bn )n defined by Bn =  ∞ ∗ an increasing one (Bn ⊆ Bn+1 , for any n ∈ N ), and ∞ n=1 Bn = A1 \ ( n=1 An ). ∞ By applying property (6), we obtain λ( n=1 Bn ) = limn λ(Bn ). Since λ(A1 ) < +∞, λ(An ) < +∞, for any n ∈ N∗ , and then we can use (3). It follows that   λ(A1 ) − λ( ∞ ) − λ(An )], from where λ( ∞ n=1 An ) = limn [λ(A n=1 An ) = limn λ(An ). ∞1 (8) Let (Bn ), defined by Bn = k=n Ak , for any n ≥ 1. It’s easy to note that Bn ⊆  Bn+1 , for any n ∈ N∗ and ∞ n=1 Bn = lim infn An . From property (6), it follows that ∞ λ(lim infn An ) = λ( n=1 Bn ) = limn λ(Bn ). But, for any n ∈ N∗ , Bn ⊆ An , from where λ(Bn ) ≤ λ(An ), for any n ∈ N∗ , and so, passing to the lower limit limn λ(Bn ) = lim infn λ(Bn ) ≤ lim infn λ(An ), which leads to announced inequality. (9) Let (An ) be a sequence with the properties required in the statement, and let A = ∞ (Bn ) by Bn = A \ An . Then lim infn Bn = A \ n=1 An . We define the sequence  ∞ lim supn An ; for any n ∈ N∗ , ∞ k=n Bk = A \ k=n Ak , and so, from (8), it follows that λ(lim infn Bn ) ≤ lim infn λ(Bn ), or λ(A \ lim supn An ) ≤ lim infn λ(A \ An ). Since λ(A) < +∞, we can use (3) λ(A) − λ(lim sup An ) ≤ lim inf[λ(A) − λ(An )] = λ(A) − lim sup λ(An ). n

n

n



Definition 1.3.12 Property (1) is said to be the property of finite additivity the Lebesgue measure λ; property (2) is the property of monotonicity, (5) is the σ -subadditivity, and (6) and (7) are the properties of continuity of the measure λ from below, respectively from above.

An immediate consequence of Theorem 1.3.11 is the property of continuity of Lebesgue outer measure from below. ¯ + be the Lebesgue outer measure, and let Proposition 1.3.13 Let λ∗ : P (R) → R (An )n∈N∗ ⊆ P (R) (not necessarily measurable) such that An ⊆ An+1 , for any n ∈ N∗ . Then ∞

 ∗ λ An = lim λ∗ (An ). n=1

Proof Let A =

∞  n=1

n

An ⊆ R. Using the monotonicity property of the outer measure (see (2) of

Theorem 1.2.3), there exists limn λ∗ (An ) ≤ λ∗ (A). If there were n ∈ N∗ such that λ∗ (An ) = +∞, then limn λ∗ (An ) = +∞ = λ∗ (A). Let us assume that λ∗ (An ) < +∞, for any n ∈ N∗ . For every n ∈ N∗ , there exists Dn ∈ τu ⊆ L such that An ⊆ Dn and λ(Dn ) < λ∗ (An ) + n1 . For every n ∈ N∗ , let

1

21 1.3 · Lebesgue Measurable Sets

 Gn = ∞ ∈ N∗ and any k ≥ n, An ⊆ Ak ⊆ Dk , we obtain k=n Dk ∈ L. Since, for any n ∞ An ⊆ Gn ⊆ Dn , and therefore A = ∞ n=1 An ⊆ n=1 Gn = lim infn Dn . From (8) of Theorem 1.3.11, λ∗ (A) ≤ λ∗ (lim inf Dn ) = λ(lim inf Dn ) ≤ lim inf λ(Dn ) ≤ n

n

n

≤ lim λ∗ (An ) ≤ λ∗ (A). n



Remark 1.3.14 The outer measure λ∗ is not continuous from above (does not verify a property similar to property (7) in Theorem 1.3.11). Indeed, let V ⊆ [0, 1] be Vitali set (see 1.2.7), let Q1 = Q ∩ [−1, 1] = {q1 , · · · , qn , · · · }, and let Vn = qn + V ⊆ [−1, 2]; then,  for any n ∈ N∗ , λ∗ (Vn ) = λ∗ (V ) > 0. The sequence (An )n , defined by An = ∞ k=n Vk ,  is decreasing (An ⊇ An+1 ), and λ∗ (A1 ) ≤ 3. Furthermore, ∞ A = ∅. Indeed, if we n n=1  assume that there is x ∈ ∞ A , then x ∈ A , and so there is k ≥ 1 such that x ∈ V ; but n 1 k n=1 x ∈ Ak+1 , and then there is l ≥ k + 1 such that x ∈ Vl . So Vk ∩ Vl = ∅ which contradicts the fact that sets Vn are pairwise disjoint (see 1.2.7).  ∗ ∗ ∗ Therefore λ∗ ( ∞ n=1 An ) = 0 < λ (V ) = limn λ (Vn ) ≤ limn λ (An ).

In the following theorem, we underline the invariance of the Lebsegue measure by translations as well as the behavior with respect to homotheties.

Theorem 1.3.15 (1) x + A ∈ L and λ(x + A) = λ(A), for every A ∈ L and every x ∈ R. (2) x · A ∈ L and λ(x · A) = |x| · λ(A), for every A ∈ L and every x ∈ R \ {0}.

Proof (1) Since A is measurable, for every ε > 0, there exists D ∈ τu such that A ⊆ D and λ∗ (D \ A) < ε. Then, according to (3) of Theorem 1.1.5, x + D ∈ τu , x + A ⊆ x + D, and λ∗ ((x + D) \ (x + A)) = λ∗ (x + (D \ A)) = λ∗ (D \ A) < ε (see Theorem 1.2.6); we deduce that x + A ∈ L. Equality is also the consequence of Theorem 1.2.6. (2) Let x ∈ R \ {0}. For every ε > 0, there exists D ∈ τu with A ⊆ D and λ∗ (D \ A) < ε |x| . From (4) of Theorem 1.1.5, x · D ∈ τu , and x · A ⊆ x · D. Then, according to Theorem 1.2.6, λ∗ ((x · D) \ (x · A)) = λ∗ (x · (D \ A)) = |x| · λ∗ (D \ A) < ε. Therefore, x · A ∈ L. 

We have mentioned (see example (iii) of 1.2.11) that any countable set is a null set. There are also examples of uncountable sets which are null sets. One such example is the Cantor ternary set. Cantor Ternary Set 1.3.16 Cantor set is a subset ofI = [0, 1] obtained by an iterative 1 2 construction. We remove the central third of I , J1 = , ; then we repeat the operation 3 3

22

1

Chapter 1 • Lebesgue Measure on R

⊡ Fig. 1.1 Cantor’s set construction

[0, 1] \ ∪3n=1 Jn [0, 1] \ (J1 ∪ J2 ) [0, 1] \ J1 [0, 1]

1 2 33 33

0

1 32

7 8 33 33 2 32

1 3

19 20 33 33

2 3

7 32

25 26 33 33 8 32

-

1



     1 2 1 2 on the two remaining intervals 0, ∪ , 1 from where we remove J2 = 2 , 2 ∪ 3 3 3 3   7 8 . , 32 32         2 7 1 2 1 8 ∪ , 2 ∪ 2 , 1 , we remove the set At the third stage of the set 0, 2 ∪ 2 , 3     3 3 3 3 3 1 2 7 8 19 20 25 26 J3 = 3 , 3 ∪ 3 , 3 ∪ 3 , 3 ∪ 3 , 3 and so on. 3 3 3 3 3 3 3 3 In the above figure, we have marked with a bold line what remains of the interval I after we have removed J1 , J2 and J3 (⊡ Fig. 1.1). ∞  1 2 Let J = Jn and D = I \J . It is obvious that λ(J1 ) = |J1 | = , λ(J2 ) = 2 , λ(J3 ) = 3 3 n=1

22 2n−1 . We can easily see that λ(J ) = . Since the sets Jn are pairwise disjoint, λ(J ) = n 33 3n ∞ ∞  n   1 2 λ(Jn ) = = 1, and then λ(D) = 0 (see (3) of Theorem 1.3.11). To establish 2 3 n=1 n=1 the cardinal of set D, we will use the ternary number system; all x ∈ I is written x = ∞  xn = 03 , x1 · · · xn · · · , where xn ∈ {0, 1, 2}. For the uniqueness of the writing, we agree 3n n=1 that between two expressions of the same number, choose the one containing an infinite number of non-zero digits ( 31 = 03 , 1 will be written 03 , 022 · · · , 23 = 03 , 2 will be written 03 , 122 · · · , etc.). We can see that x ∈ J1 if and only if x = 03 , 1x2 x3 · · · and x ∈ J2 if and only if x = 03 , x1 1x3 · · · , where x1 ∈ {0, 2}. In general, x ∈ Jn if and only if ∞  x = 03 , x1 x2 · · · xn−1 1xn+1 · · · , where x1 , x2 , · · · xn−1 ∈ {0, 2}. Then x ∈ J = Jn if and n=1

only if x = 03 , x1 x2 x3 · · · , and there is k ∈ N∗ such that xk = 1. Therefore x ∈ D = I \ J if and only if x = 03 , x1 x2 x3 · · · xn · · · and xn ∈ {0, 2} for all n ∈ N∗ .  xn Let f : D → I be the function defined by f (x) = ∞ n=1 2n+1 = 02 , z1 z2 · · · zn · · · (numbers written in base 2), where x = 03 , x1 x2 · · · xn · · · and, for any n ∈ N∗ , zn = x2n . Then f is a bijection:

1

23 1.3 · Lebesgue Measurable Sets

 ∞ yn xn f is strictly increasing: Let x = ∞ n=1 3n , y = n=1 3n ∈ D with x < y; there exists n0 ∈ N∗ such that xn = yn , for any n < n0 and xn0 < yn0 . Since x, y ∈ D, xn0 = 0, and n0 −1 xn  n0 −1 yn xn 1 yn0 = 2. Then f (x) = n=1 + ∞ n=n0 +1 2n+1 ≤ n=1 2n+1 + 2n0 < f (y) (y has an 2n+1 infinity of non-zero digits). f is surjective: For every z ∈ [0, 1], let z = 02 , z1 z2 · · · zn · · · be the writing of z in base 2. If x = 03 , x1 x2 x3 · · · xn · · · , where xn = 2zn , for all n ∈ N∗ , then x ∈ D and f (x) = z. Therefore card(D) = card([0, 1]) = card(R) = c > ℵ0 = card(N), and then D is not countable. The set D is a not countable null set. The set of Cantor  will be defined by adding to Dthe 2 2 8 2 8 20 26 points located at the right end of the intervals Jn : E = , , , , , , ,··· . 3 32 32 33 33 33 33 So the Cantor set is C = D ∪ E. D and E are null sets, and then C is a null set. Moreover card(C) = card(D) = c (see Exercise (1) of  Sect. 1.5). Noting that J \ E is an open set (countable union of open intervals) and that C = I \ (J \ E), we note that C is closed and bounded and therefore is a compact set. Exercise (4) of  Sect. 1.5 gives another example of a null set which has the power of the continuum. Remarks 1.3.17 (i) Because Lebesgue measure is complete (see (ii) of Remarks 1.3.4), the subsets of C are measurable; therefore P (C) ⊆ L ⊆ P (R). Then card(P (C)) = 2c ≤ card(L) ≤ card(P (R)) = 2c , and so card(L) = 2c . (ii) Although having the same cardinality, however, L is a proper subset of P (R). Indeed, we note that λ∗ is σ -additive on L (Theorem 1.3.10), but it is not σ -additive on P (R) (see 1.2.7). The Vitali set (see 1.2.7) is not measurable Lebesgue; indeed, if we assume that V ∈ L, then r + V ∈ L, for ⎛ any r ∈ Q1 ,⎞and therefore, from Theorem 1.3.10 and    (r + V )⎠ = λ(r + V ) = λ(V ) = +∞ from (1) of Theorem 1.3.15, λ ⎝ r∈Q1

r∈Q1

r∈Q1

which contradicts (∗) of 1.2.7. We have noticed in 1.2.7 that the existence of V depends essentially on the axiom of choice. There exist models of the theory of sets without axiom of choice, where all the sets of real numbers are measurable in the sense of Lebesgue.

Next we will define an important class of measurable sets, the class of Borel sets. Definition 1.3.18 Let A ⊆ P (R); A is said to be a σ -algebra on R if: ∞ (1) n=1 An ∈ A, for every (An )n ⊆ A; (2) A \ B ∈ A, for any A, B ∈ A; (3) R ∈ A. (Continued )

24

1

Chapter 1 • Lebesgue Measure on R

Definition 1.3.18 (continued) Let U ⊆ P (R); then there is the smallest σ -algebra on R, A(U ), which contains the class U (see Proposition 1.4.3). A(U ) is said to be the σ -algebra generated by U .. The σ -algebra generated by the usual topology τu , A(τu ) = Bu , is said to be the family of Borel sets of (R, τu ); every B ∈ Bu is called a Borel set. The open and closed subsets of (R, τu ) are Borel sets. The countable sets are Borel sets (as countable union of closed sets).

Remark 1.3.19 Because τu ⊆ L (see Remark 1.3.2), L contains the σ -algebra generated by τu , Bu (Definition 1.3.18). We can show that card(Bu ) = c (see Theorem 3.3.18 of [8]). Then card(Bu ) = c < 2c = card(L). A subset of a Borel null set is not necessarily a Borel set; indeed the Cantor set C is a Borel set, but P (C)  Bu (card(Bu ) = c < 2c = card(P (C))). Therefore the restriction of Lebesgue measure on Bu is not complete.

Although L contains “many more” elements than Bu , the sets of L differ from those of Bu by null sets. Theorem 1.3.20 A ∈ L if and only if A = B ∪ N, where B ∈ Bu and λ(N) = 0.

Proof Let A ∈ L; then Ac ∈ L, and so, for any n ∈ N∗ , there exists Dn ∈ τu such that Ac ⊆ Dn and λ(Dn \ Ac ) = λ(Dn ∩ A) < n1 . The set Fn = Dnc ⊆ A is closed, and therefore  B= ∞ n=1 Fn ∈ Bu . Moreover λ(A \ B) = λ

∞  n=1

(A ∩ Dn )
0; there exists D ∈ τu such that T ⊆ D and λ(D) < λ∗ (T ) + ε. Because D, A ∈ L and λ is a measure on L, λ(D) = λ(D ∩ A) + λ(D \ A) ≥ λ∗ (T ∩ A) + λ∗ (T \ A), from where λ∗ (T ) + ε > λ∗ (T ∩ A) + λ∗ (T \ A), for any ε > 0. (C) is sufficient. If λ∗ (A) < +∞, then, for every ε > 0, there exists D ∈ τu such that A ⊆ D and λ(D) < λ∗ (A) + ε. We use the condition (C) with T = D: λ(D) = λ∗ (D) = λ∗ (D ∩ A) + λ∗ (D \ A) = λ∗ (A) + λ∗ (D \ A), from where λ∗ (D \ A) < ε. If λ∗ (A) = +∞, we denote, for any n ∈ N∗ , An = A∩]−n, n[. Since ]−n, n[∈ τu ⊆ L, according to the necessity of (C), λ∗ (T ) = λ∗ (T ∩] − n, n[) + λ∗ (T \] − n, n[), for every T ⊆ R.

(1.3)

We write (C) for A with Tn = T ∩] − n, n[: λ∗ (Tn ) = λ∗ (Tn ∩ A) + λ∗ (Tn \ A).

(1.4)

From 1.3 and 1.4, we obtain λ∗ (T ) = λ∗ (Tn ∩ A) + λ∗ (Tn \ A) + λ∗ (T \] − n, n[), for every T ⊆ R.

(1.5)

In 1.5, we substitute T with T \ An : λ∗ (T \ An ) = λ∗ (Tn \ A) + λ∗ (T \] − n, n[).

(1.6)

Finally, from 1.5 and 1.6, λ∗ (T ) = λ∗ (T ∩ An ) + λ∗ (T \ An ), for every T ⊆ R. Therefore, λ∗ (An ) < +∞ and An satisfy the condition (C), for all n ∈ N∗ . After the first  part of the sufficiency, An ∈ L, for any n ∈ N∗ . Then A = ∞  n=1 An ∈ L.

26

Chapter 1 • Lebesgue Measure on R

1.4

Abstract Setting

1

So far we have built a measure on R—the Lebesgue measure. In this section, we will approach a more abstract point: we will consider a general measure defined on a class of the subsets of an abstract space. Since the construction that we present generalizes that of the Lebesgue measure, only a part of the properties will be found in the general case. Definition 1.4.1 Let X be an abstract set, let P (X) be the family of all subsets of X, and let A ⊆ P (X); A is said to be a σ -algebra on X if: ∞ (1) n=1 An ∈ A, for every (An )n ⊆ A; (2) A \ B ∈ A, for any A, B ∈ A; (3) X ∈ A. If A is a σ -algebra on X, then (X, A) is a measurable space.

Remarks 1.4.2 (i) Let A be a σ -algebra on X; then (a) ∅ = X \ X ∈ A. (b) A ∈ A if and only if Ac ∈ A. (c) A ∪ B = A ∪ B ∪ ∅ ∪ · · · ∪ ∅ ∪ · · · ∈ A, for any A, B ∈ A. (d) A ∩ B = (Ac ∪ B c )c ∈ A, for any A, B ∈ A. ∞ ∞ c )c ∈ A, for every (A ) ⊆ A. (e) n n n n=1 An = ( n=1 A  ∞ ∞ ∞ (f) lim infn An = n=1 k=n Ak ∈ A, lim sup n An = ∞ n=1 k=n Ak ∈ A, for every (An )n ⊆ A. (ii) From Theorems 1.3.3 and 1.3.6 and Corollary 1.3.8, it follows that the family of measurable sets in the sense of Lebesgue is a σ -algebra on R and that (R, L) is a measurable space. Proposition 1.4.3 Let U ⊆ P (X); then there is a smallest σ -algebra on X, A(U ), which contains the class U . Proof It is easy to prove that any intersection (finite or infinite) of σ -algebras is a σ -algebra. Then A(U ) is the intersection of all σ -algebras which contains the class U (at least P (X) is such σ -algebra). A(U ) is the smallest σ -algebra containing U . 

1

27 1.4 · Abstract Setting

Definition 1.4.4 The σ -algebra A(U ) is called the σ -algebra generated by U . If τ is a topology on X, then the σ -algebra generated by τ , A(τ ) = B, is said to be the family of Borel sets of (X, τ ); every B ∈ B is called a Borel set. The open sets and the closed sets of (X, τ ) are Borel sets.

Definition 1.4.5 ¯ + be a set function; γ is called a Let A be a σ -algebra on X, and let γ : A → R measure on X and the triple (X, A, γ ) a measure space if: (1) γ (∅) = 0.  ∞ (2) γ ( ∞ n=1 An ) = n=1 γ (An ), for every (An )n ⊆ A, An ∩ Am = ∅ and every n = m. If γ (X) < +∞, then γ is a finite measure on X; if γ (X) = 1, γ is said to be a probability on X.  If X = ∞ n=1 An and, for any n ∈ N, An ∈ A and γ (An ) < +∞, then γ is called σ -finite. The measure γ is complete on A (or the σ -algebra A is complete with respect to γ ) if, for every A ∈ A with γ (A) = 0 and every B ⊆ A, it follows that B ∈ A (obviously, from (2) of Theorem 1.4.7 γ (B) = 0). If τ is a topology on the measure space (X, A, γ ) such that τ ⊆ A (and so A(τ ) = B ⊆ A), then γ is called regular when, for every A ∈ A and every ε > 0, there exist an open set D and a closed set F such that F ⊆ A ⊆ D and γ (D \ F ) < ε.

Examples 1.4.6



1, x ∈ A, δx is a probability complete 0, x ∈ / A. on X; it is called the Dirac measure (the unitmass concentrated in x). ¯ + , defined by γ (A) = card(A) , A = finite, γ is a complete (ii) Let γ : P (N) → R +∞ , A = infinite. σ -finite measure on N; it is called the counting measure. (i) Let x ∈ X and δx : P (X) → R+ , δx (A) =

The proof of the following theorem is similar to that of Theorem 1.3.11. Theorem 1.4.7 ¯ be a measure on (X, A); then: Let γ : A → R n +  n n (1) γ k=1 Ak = k=1 γ (Ak ), for every (Ak )k=1 ⊆ A, Ak ∩ Al = ∅ and every k = l. (Continued )

Chapter 1 • Lebesgue Measure on R

28

1

Theorem 1.4.7 (continued) (2) γ (A) ≤ γ (B), for every A, B ∈ A with A ⊆ B. (3) γ (B \ A) = γ (B) − γ (A), for every A, B ∈ A, A ⊆ B, γ (A) < +∞. (4) γ (A ∪ B) + γ (A ∩ B) = γ (A) + γ (B), for every A, B ∈ A. ∞   (5) γ An ≤ ∞ n=1 γ (An ), for every (An ) ⊆ A. n=1  ∞ (6) γ A = lim n γ (An ), for every (An ) ⊆ A with An ⊆ An+1 for any n=1 n n ∈ N. ∞  (7) γ n=1 An = limn γ (An ), for every (An ) ⊆ A with An+1 ⊆ An , for any n ∈ N, and γ (A1 ) < +∞. (8) γ (lim infn An ) ≤ lim infn γ (An ), for every (An ) ⊆ A. ∞  (9) lim supn γ (An ) ≤ γ (lim supn An ), ∀(An ) ⊆ A, γ n=1 An < +∞.

Definition 1.4.8 Property (1) is said to be the property of finite additivity of the measure γ ; property (2) is the property of monotonicity, (5) is the σ -subadditivity, and (6) and (7) are the properties of continuity of the measure γ from below, respectively from above.

Remark 1.4.9 The family of Lebesgue measurable subsets of R, L, is a σ -algebra and Lebesgue measure, λ, is a measure σ -finite, complete and regular on L. If A ∈ L, then L(A) is a σ -algebra on A, and the restriction of λ on L(A) is a measure on A.

1.5

Exercises

(1) Let A = {a0 , a1 , · · · , an , · · · } be a countable set, and let B be an infinite set. (a) Show that there is a countable set C = {c0 , c1 , · · · , cn , · · · } ⊆ B such that definition B \ C is infinite; deduce from here that ℵ0 ===== = card(A) ≤ card(B) (ℵ0 is the smallest transfinite cardinal). (b) Suppose⎧ that A ∩ B = ∅; show that the function f : A ∪ B → B, defined by ⎪ x, x ∈ B \ C, ⎨ f (x) = c2k−1 , x = ck ∈ C, is a bijection. ⎪ ⎩ c , x = a ∈ A, 2k k definition

Deduce from here that card(B) = card(A ∪ B) ===== = card(A) + card(B). (2) Show that the following functions are bijections: ad − bc c−d ·x+ . (a) f :]a, b[→]c, d[, f (x) = a−b a−b x−a . (b) g :]a, b[→]0, +∞[, g(x) = b−x

29 1.5 · Exercises

(c) h :]0, +∞[→ R, h(x) = ln x. From the above, it will be concluded that card(]a, b[) = card(]c, d[) = = card(]0, +∞[) = card(R) = c. (3) Show that: (a) d(B, C) ≤ d(A, C), for every A, B, C ⊆ R with A ⊆ B. (b) d(A ∪ B, C) = min{d(A, C), d(B, C)}, for every A, B, C ⊆ R. (4) Let A ⊆]0, 1[ be the set of numbers which, written in base 10, use only the digits 0 and 1. Show that card(A) = c (= card(R)) and that λ∗ (A) = 0. Indication: A =

∞

A for which, in decimal writing, the digit n=1 An , where An is the set of elements of   ! 1 ∪ 1 + ∪∞ A  and that, for all n=2 n 10 10

1 appears for the first time in the place n. It is shown that A1 = 8 n, p ≥ 1, d(An , An+p ) > n+p > 0 and An = 10 · An+1 . 10 ∗

(5) Show that, for every bounded set A ⊆ R, λ (A) < +∞. Does the converse hold true? (6) Let a continuous function f : R → R such that A = {x ∈ R : f (x) = 0} is a null set (λ∗ (A) = 0). Show that f (x) = 0, for every x ∈ R. (7) Let C ⊆ R be a bounded closed set (a compact). Show that λ∗ (C) = 0 if and  only if, for every ε > 0, there is {I1 , · · · , In } ⊆ I such that C ⊆ nk=1 Ik and n k=1 |Ik | < ε (a compact set is a Lebesgue null set if and only if it is a Jordan null set). Indication: see Lemma 7.1.6

¯ + be a set function defined by λ∗ (A) = sup{λ(F ) : F ⊆ (8) Let λ∗ : P (R) → R A, F closed set}, for every A ⊆ R; λ∗ is called the inner Lebesgue measure on R. Show that, for every A ⊆ R, λ∗ (A) ≤ λ∗ (A); if A ∈ L, then λ∗ (A) = λ∗ (A). Conversely, if λ∗ (A) = λ∗ (A) < +∞, then A ∈ L. Calculate λ∗ (R \ V ), where V is a Vitali set. (9) Let A ⊆ R; for any n ∈ N∗ , we denote Dn = {x ∈ R : d(x, A) < n1 }. Show that (Dn )n ⊆ τu and that if A is compact, then λ(A) = limn λ(Dn ). Show that the compactness hypothesis cannot be removed. (10) Show that A ∈ L if and only if, for every ε > 0, there exist a closed set F and an open set D such that F ⊆ A ⊆ D and λ(D \ F ) < ε. (11) Let A ⊆ B ⊆ C, A, C ∈ L, and λ(A) = λ(C) < +∞; show that B ∈ L. (12) Show that Lebesgue measure has the Darboux property: (a) For every A ∈ L and every b ∈ R with 0 < b < λ(A), there is B ∈ L, B ⊆ A such that λ(B) = b. (b) For every bounded sets A, B ∈ L, A ⊆ B and for any c ∈ R with λ(A) < c < λ(B), there is C ∈ L such that A ⊆ C ⊆ B and λ(C) = c. Indication. a). Let A be bounded from below, let t0 = inf A, and let f : [t0 , +∞) → R be a function defined by f (t) = λ(A ∩ [t0 , t]). Then f is Lipschitz continuous, f (t0 ) = 0 and limt→+∞ f (t) = λ(A). Since f has the Darboux property and b ∈ f ((t0 , +∞)), there exists t such that f (t) = b; we consider B = A∩[t0 , t]. If A is not bounded from below, we reason the same for the set An = A ∩ [−n, n]. b). If λ(A) < c < λ(B), then 0 < c − λ(A) < λ(B \ A), and the question is reduced to case (a).

1

30

1

Chapter 1 • Lebesgue Measure on R

(13) Let A, B ∈ L with λ(A) < +∞ and λ(B) < +∞; show that |λ(A) − λ(B)| ≤ λ(AB), where AB = (A \ B) ∪ (B \ A) is the symmetric difference of A and B. (14) Let n ∈ N∗ , and let A1 , A2 , · · · , An ⊆ [0, 1] be Lebesgue measurable sets with n n k=1 λ(Ak ) > n − 1; show that λ(∩k=1 Ak ) > 0. Indication. Show that λ(∩nk=1 Ak ) = 1 − λ(∪nk=1 ([0, 1] \ Ak )).

(15) Let (An )n ⊆ L with λ(An ∩ Am ) = 0, for all n = m. Show that λ

∞  n=1

An

=

∞ 

λ(An ).

n=1

(16) Let A ⊆ R be a null set; show that R \ A is dense in R (for every x, y ∈ R with x < y, there exists z ∈ R \ A such that x < z < y). Does the converse hold true? (17) Let A ⊆ R, let τA = {D ∩ A : D ∈ τu }—the usual topology on A—and let BA be the σ -algebra generated by τA on A (Definition 1.4.4). Show that BA = {B ∩ A : B ∈ Bu }. (τu is the usual topology on R (Definition 1.1.2), and Bu is the family of Borel sets of (R, τu ) (Definition 1.3.18).) Indication. Let A = {B ∩ A : B ∈ Bu }; it’s easy to verify that BA ⊆ A. For reverse inclusion, show that Bu = {B ∈ Bu : B ∩ A ∈ BA }.

(18) Let γ be a measure on a measurable space (X, A), and let A¯ = {AN : A ∈ A, N ⊆ B, B ∈ A and γ (B) = 0}. Show that A¯ is a σ -algebra on X and γ¯ : A¯ → R+ , γ¯ (AN) = γ (A) is a measure complete on A¯ . γ¯ is called the completion of γ . Show that A ⊆ A¯ and that the restriction of γ¯ to A coincides with γ . The measure γ¯ is the smallest complete extension of γ . (19) Let γ be a σ -finite measure on the measurable space (X, A), and let {Ai : i ∈ I } ⊆ A be a family of pairwise disjoint sets. Show that N = {i ∈ I : γ (Ai ) > 0} is a countable set.   Indication. Suppose that γ (X) < +∞, and let C= {C ⊆ I : C be countable}. For every C∗ ∈ C , i∈C γ (Ai ) = γ ( i∈C Ai ) ≤ γ (X) < +∞. Hence s = supC∈C i∈C γ (Ai ) < +∞. Therefore, ∀n ∈ N , ∃Cn ∈ C such that   s − n1 < i∈Cn γ (Ai ). Show that C0 = ∞ n=1 Cn ∈ C and that γ (Ai ) = 0, for every i ∈ I \ C0 . ∞ Let now X = n=1 Xn with γ (Xn ) < +∞, for any n ∈ N∗ ; we can still assume that Xn ⊆ Xn+1 , ∀n ∈ N∗ . Let  Cn = {i ∈ I : γ (Ai ∩ Xn ) > 0}, ∀n ∈ N∗ ; show that C0 = ∞ n=1 Cn = {i ∈ I : γ (Ai ) > 0}.

31

Measurable Functions

In this chapter, we will present and study a large class of functions—that of measurable functions. In the following chapter, we will have identified the integrable functions among the measurable functions. The class of measurable functions contains most of the known functions (continuous, monotonic, integrable Riemann functions); in addition, this class benefits from a number of remarkable properties for the passage to the limit.

2.1

Definitions. Properties

Recall that a function f : A ⊆ R → R is continuous at a point x ∈ A if, for every open interval I ∈ I with f (x) ∈ I , there exists J ∈ I such that x ∈ A ∩ J ⊆ f −1 (I ) or, equivalent, for every sequence (xn )n ⊆ A, xn → x, it follows that f (xn ) → f (x). The function f is continuous on A if it is continuous at every point of set A. A simple characterization of the global continuity (which we will demonstrate below) asserts that a function is continuous on A if and only if the inverse image of any open set of R is an open set of A (a set of form A ∩ G, with G ∈ τu ). Proposition 2.1.1 A function f : A ⊆ R → R is continuous on A if and only if, for every D ∈ τu , there exists G ∈ τu such that f −1 (D) = A ∩ G. Proof Let us assume that f is continuous on A, and let D ∈ τu ; if f −1 (D) = ∅, then G = ∅ is the desired set. If f −1 (D) = ∅, then, for every x ∈ f −1 (D), f (x) ∈ D, and then there exists I ∈ I such that f (x) ∈ I ⊆ D; f is continuous at x, and then there exists an open interval  Jx ∈ I such that x ∈ A ∩ Jx ⊆ f −1 (I ); the set G = x∈f −1 (D) Jx meets the conditions of the proposition. Conversely, we assume that the inverse image by f of any open set of R is an open set of A. Let x ∈ A be an arbitrary point, and let I ∈ I such that f (x) ∈ I ; since I ∈ τu , there exists G ∈ τu with x ∈ f −1 (I ) = A ∩ G. Let J ∈ I such that x ∈ J ⊆ G; then x ∈ A ∩ J ⊆ A ∩ G = f −1 (I ). Therefore f is continuous at x, for every x ∈ A. 

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2021 L. C. Florescu, Lebesgue Integral, Compact Textbooks in Mathematics, https://doi.org/10.1007/978-3-030-60163-8_2

2

32

2

Chapter 2 • Measurable Functions

In the previous proposition, the open set D can be replaced by open intervals; so the function f : A ⊆ R → R is continuous on A if and only if, for all I ∈ I , there is G ∈ τu such that f −1 (I ) = A ∩ G. Definition 2.1.2 Let A ∈ L and let f : A → R; f is said to be a measurable function on A in the Lebesgue sense if f −1 (] − ∞, a[) ∈ L(A), for any a ∈ R. The set of all the functions measurable on A will be denoted with L(A); let L+ (A) be the measurable and positive functions on A.

Remarks 2.1.3 (i) If A is a null set, then f ∈ L(A), for every f : A → R (see (ii) of Examples 1.2.11 and (1) of Theorem 1.3.3). (ii) For every B ∈ L(A) (B ∈ L, B ⊆ A) and for every f ∈ L(A), the restriction of f on B, f |B ∈ L(B); indeed, (f |B )−1 (] − ∞, a[) = f −1 (] − ∞, a[) ∩ B ∈ L(B), for any a ∈ R. (iii) Let A, B, C ∈ L, A = B ∪ C. f ∈ L(A) if and only if f |B ∈ L(B) and f |C ∈ L(C) (f −1 (] − ∞, a[) = (f |B )−1 (] − ∞, a[) ∪ (f |C )−1 (] − ∞, a[), for any a ∈ R). (iv) The above definition can be extended to the case where the function f can also take the value +∞ (f : A →]−∞, +∞]). As in Definition 2.1.2, the function f will be called measurable on A ∈ L if f −1 (] − ∞, a[) ∈ L(A), for any a ∈ R. We remark that f is measurable on A if and only if Z = f −1 (+∞) ∈ L(A) and f |A\Z is measurable on  A\Z ∈ L. Indeed, this results immediately if we notice that Z = A\ ∞ n=1 (f < n) and −1 −1 −1 that, for any a ∈ R, (f |A\Z ) (]∞, a[) = f (]−∞, a[) = f (]−∞, a[)∩(A\Z).

Definition 2.1.4 Let A ∈ L, and let BA = {B ∩A : B ∈ Bu } ⊆ L(A) be the family of Borel sets on A (see Exercise (17) of  Sect. 1.5); it is obvious that BR = Bu . A function f : A → R is said to be a Borel function on A if, for any a ∈ R, f −1 (] − ∞, a[) ∈ BA . If A = R, we say that f is a Borel function. It is obvious that every Borel function on A is measurable on A.

Proposition 2.1.5 Let A ∈ L. (1) Every continuous function on A is a Borel function on A. (2) Every monotonic function on A is a Borel function on A. Proof (1) Let f be a continuous function on A. According to Proposition 2.1.1, for any a ∈ R, there exists G ∈ τu ⊆ Bu such that f −1 (] − ∞, a[) = A ∩ G ∈ BA . (2) Let us assume that f : A → R is an increasing function on A, and let a ∈ R. If card(f −1 (] − ∞, a[)) ≤ 1, then f −1 (] − ∞, a[) ∈ BA (see Definition 1.3.18). If

33 2.1 · Definitions. Properties

card(f −1 (] − ∞, a[)) ≥ 2, then, A∩] − ∞, x0 [⊆ f −1 (] − ∞, a[) ⊆ A∩] − ∞, x0 ] where x0 = sup f −1 (] − ∞, a[) ∈] − ∞, +∞]. Indeed, for every x ∈ A∩] − ∞, x0 [, there exists y ∈ f −1 (] − ∞, a[) ⊆ A such that x < y. It follows that f (x) ≤ f (y) < a and then x ∈ f −1 (] − ∞, a[). The second inclusion is obvious. From the two inclusions, it follows that f −1 (] − ∞, a[) coincides with the left set or with the right set of the above inclusions (the two sets do not differ only with one point); but the two sets are Borel sets of A.  Remark 2.1.6 Because there are continuous functions which are not monotonic and monotonic functions which are not continuous, it follows that the two classes (the class of continuous functions and that of monotonic functions) are strictly included in the class of Borel functions.

Since all Borel functions are measurable, we can formulate the following corollary. Corollary 2.1.7 Let A ∈ L. (1) Every continuous function on A is measurable on A. (2) Every monotonic function on A is measurable on A. 

1, x ∈ A , the 0, x ∈ /A characteristic function of A. Then χA ∈ L(R) if and only if A ∈ L; χA is a Borel function if and only if A ∈ Bu .

Proposition 2.1.8 For every A ⊆ R, let χA : R → R, χA (x) =

Proof If χA ∈ L(R) (χA is a Borel function), then Ac = R\A = χA−1 (]−∞, 12 [) ∈ L(∈ Bu ), from ⎧ ⎪ ⎨∅ ,a ≤ 0 where A ∈ L(∈ Bu ). Conversely, if A ∈ L, then χA−1 (] − ∞, a[) = Ac , 0 < a ≤ 1 , and ⎪ ⎩ R ,1 < a therefore χA−1 (] − ∞, a[) ∈ L(∈ Bu ), for any a ∈ R, which means that χA ∈ L(R) (χA is a Borel function).  Remark 2.1.9 Let Q ⊆ R be the subset of rational numbers; since Q ∈ Bu , it follows that χQ (the Dirichlet function) is a Borel function, and so it is measurable.

In the following theorem, several characterizations of the measurability of a function on a set are presented.

2

34

2

Chapter 2 • Measurable Functions

Theorem 2.1.10 Let A ∈ L and let f : A → R; the following are equivalent: (1) f ∈ L(A). (2) f −1 (] − ∞, a]) ∈ L, for any a ∈ R. (3) f −1 (]a, +∞[) ∈ L, for any a ∈ R. (4) f −1 ([a, +∞[) ∈ L, for any a ∈ R. (5) f −1 (I ) ∈ L, for any I ∈ I . (6) f −1 (D) ∈ L, for every D ∈ τu . (7) f −1 (B) ∈ L, for every B ∈ Bu .

Proof (1) ⇒ (2): f −1 (] − ∞, a]) =

∞  n=1

f −1

 −∞, a +

1 n

 , for any a ∈ R.

(2) ⇒ (3): f −1 (]a, +∞[) = A \ f −1 (] − ∞, a]), for any a ∈ R.   ∞  1 (3) ⇒ (4): f −1 ([a, +∞[) = f −1 a − , +∞ , for any a ∈ R. n n=1 (4) ⇒ (5): Every open interval I ∈ I is of the form I =] − ∞, b[, I =]a, +∞[ or I =]a, b[ with a < b. f −1 (] − ∞, b[) = A \ f −1 ([b, +∞[), for any b ∈ R,   ∞  1 −1 −1 f (]a, +∞[) = f a + , +∞ , for any a ∈ R and n n=1

f −1 (]a, b[) = f −1 (] − ∞, b[) ∩ f −1 (]a, +∞[), for all a, b ∈ R with a < b. In all cases −1 f (I ) ∈ L. (5) ⇒ (6): From the open set structure theorem (Theorem 1.1.3), for every D ∈  ∞ −1 (D) = −1 (I ) ∈ L. τu , D = ∞ n n=1 In where {In : n ≥ 1} ⊆ I . Then f n=1 f −1 (6) ⇒ (7): Let C = {C ⊆ R : f (C) ∈ L}; it is easy to show that C is a σ -algebra on R, and according to condition (6), τu ⊆ C . Since Bu is the smallest σ -algebra that contains τu , Bu ⊆ C (see Definition 1.3.18). (7) ⇒ (1): Every open interval ] − ∞, a[ is an open set and therefore is a Borel set.  Remarks 2.1.11 (i) Equivalences (1)–(7) remain valid if we replace “f ∈ L(A)” by “f a Borel function on A” and L by Bu . In particular, f is a Borel function on A if and only if f −1 (B) ∈ BA , for every B ∈ Bu . (ii) In general, the inverse image of a measurable set by a measurable function is not measurable. Let f : D → I = [0, 1] be the function defined in  Sect. 1.3.16; let us remember that f is strictly increasing and surjective and therefore bijective. Then the inverse function of f , g = f −1 : [0, 1] → D, is also strictly increasing, so it is measurable (see (2) of Corollary 2.1.7). In  Sect. 1.2.7, we denoted with V ⊆ [0, 1] the Vitali set; V is not measurable ((ii) of Remark 1.3.17), but N = g(V ) ⊆ D is measurable ((ii) of Remark 1.3.4). Then N ∈ L and g −1 (N) = V ∈ / L.

35 2.1 · Definitions. Properties

(iii) If f ∈ L(A), then f −1 (J ) ∈ L, for every J ∈ J . Indeed, it suffices to note that J ⊆ Bu . Notations Let f, g, fn : A → R, for any n ∈ N; for the simplification of the writing, we will use in the current way the following abbreviations: (f =A g) ≡ {x ∈ A : f (x) = g(x)} (f =A g) ≡ {x ∈ A : f (x) = g(x)} (fn →A f ) ≡ {x ∈ A : fn (x) → f (x)} (fn A f ) ≡ {x ∈ A : fn (x)  f (x)}

In the same way, we can clearly see the meaning of the notations (f >A 0), (f −∞, for every x ∈ A; (3) f = lim supn fn ∈ L(A), if lim supn fn (x) ∈ R, for every x ∈ A; (4) f = lim infn fn ∈ L(A), if lim infn fn (x) ∈ R, for every x ∈ A; (5) If fn (x) → f (x) ∈ R, for every x ∈ A, then f ∈ L(A). · (6) If fn − → f , then f ∈ L(A). A

Proof  (1) For every a ∈ R, f −1 (]a, +∞[) = n∈N fn−1 (]a, +∞[) ∈ L.  (2) For every a ∈ R, f −1 (] − ∞, a[) = n∈N fn−1 (] − ∞, a[) ∈ L. (3) lim supn fn = infn supk≥n fk ∈ L(A) (see (1) and (2)). (4) lim infn fn = supn infk≥n fk ∈ L(A) (see (1) and (2)). (5) If fn (x) → f (x), for every x ∈ A, then f = lim infn fn ∈ L(A). (6) Let N = (fn A f ); then λ(N) = 0, and fn (x) → f (x), for every x ∈ A \ N. The previous point assures us that f |A\N ∈ L(A \ N), and (i) of Remarks 2.1.3 assures us that f |N ∈ L(N). From (iii) of Remarks 2.1.3, f ∈ L(A).  Remark 2.1.19 Properties (1)–(5) remain valid for Borel functions but not property (6)! In particular, if f and g are Borel functions on A, then sup{f, g} is a Borel function on A.

Theorem 2.1.20 Let f, g ∈ L(A) and let α ∈ R; then f + g ∈ L(A), α · f ∈ L(A), and f · g ∈ L(A).

Proof For every a ∈ R, (f + g < a) =



[(f < r) ∩ (g < a − r)] ∈ L;

r∈Q

hence f + g ∈ L(A). a! If α > 0, then, for any a ∈ R, (α · f > a) = f > ∈ L; if α < 0, then (α · f > α a! a) = f < ∈ L. α

2

38

2

1 4

Chapter 2 • Measurable Functions

Finally, if f, g ∈ L(A), then f 2 ∈ L(A) (see Corollary 2.1.17), and therefore f · g = " # (f + g)2 − (f − g)2 ∈ L(A). 

Definition 2.1.21 Let f : A → R; we will define f + , f − : A → R by f + = sup{f, 0}, f − = sup{−f, 0}. f + is called the positive part and f − the negative part of f . It is obvious that f = f + − f − and |f | = f + + f − .

Proposition 2.1.22 Let A ∈ L. (1) f ∈ L(A) if and only if f + ∈ L(A) and f − ∈ L(A). (2) f ∈ L(A) implies |f | ∈ L(A). Proof (1) If f ∈ L(A), then, according to Theorem 2.1.18, f + ∈ L(A), and f − ∈ L(A); the converse theorem is provided by the preceding theorem. (2) is a consequence of Theorem 2.1.15.  Remark 2.1.23 f is a Borel function on A if and only if f + and f − are Borel functions on A.

2.2

Different Types of Convergence

Let fn , f : A → R; the sequence (fn )n∈N converges pointwise on A to f if, for every p → f . In the previous paragraph, we x ∈ A, fn (x) → f (x). We denote this by fn − A

introduced convergence almost everywhere; we recall that a sequence (fn ) converges almost everywhere to the function f on A ⊆ R if λ∗ ((fn A f )) = 0. We will denote · · → f . In other words, fn − → f if and only if there exists a null set N such this by fn − A

A

p

that fn −−→ f . A\N

·

We have shown in (6) of Theorem 2.1.18 that if A ∈ L, (fn ) ⊆ L(A) and fn − → f, A

then f ∈ L(A). In this paragraph, we will introduce two other types of convergence for the sequences of measurable functions, and we will analyze the links between these convergences. u → f denotes the uniform convergence on A of (fn )n If fn , f : A → R, then fn − A

to f : for every ε > 0, there exists n0 ∈ N such that, for any n ≥ n0 and every x ∈ A, |fn (x) − f (x)| < ε.

2

39 2.2 · Different Types of Convergence

Definition 2.2.1 Let A ∈ L, (fn ) ⊆ L(A), and let f ∈ L(A): 1. (fn ) converges almost uniformly on A to f if, for every ε > 0, there exists u a.u. Aε ∈ L such that λ(Aε ) < ε and fn −−−→ f . We will denote this with fn −→ A\Aε

A

f. 2. (fn ) converges in measure on A to f if limn λ((|fn − f | ≥ ε)) = 0, for every λ ε > 0; we will denote this with fn − → f. A

The sequence (fn )n is convergent in measure on A if there exists f ∈ L(A) λ → f. such that fn − A

3. (fn ) is Cauchy in measure on A if, for every ε > 0, lim λ((|fm − fn | ≥ ε)) = 0. m,n→∞

Theorem 2.2.2 Let (fn ) ⊆ L(A) and let f ∈ L(A): a.u. λ (1) If fn −→ f , then fn − → f; a.u.

A

A ·

A

A

(2) If fn −→ f , then fn − → f. (3) Every sequence convergent in measure on A is Cauchy in measure.

Proof a.u. u (1) Since fn −→ f , for every ε > 0, there exists Aε ∈ L with λ(Aε ) < ε and fn −−−→ f . A

A\Aε

It follows that for every η > 0, there exists n0 ∈ N such that, for any n ≥ n0 and every x ∈ A \ Aε , |fn (x) − f (x)| < η or, in other words, A \ Aε ⊆ (|fn − f | < η). If we go to the complement with respect to A in the last inclusion, we get (|fn − f | ≥ η) ⊆ Aε from where λ(|fn − f | ≥ η) < ε, and so limn λ(|fn − f | ≥ η) = 0, for every η > 0, λ which implies fn − → f. A

a.u.

u

A

A\Aε

(2) Since fn −→ f , for every ε > 0, there exists Aε ∈ L with λ(Aε ) < ε and fn −−−→ f p

→ f ). If we from where (fn )n is pointwise convergent to f on A \ Aε , or A \ Aε ⊆ (fn − A

go to the complement with respect to A in the last inclusion, we get (fn  f ) ⊆ Aε , and therefore λ∗ (fn  f ) ≤ ε, for every ε > 0. It follows that λ∗ (fn  f ) = 0, and · then fn − → f. A

(3) Let (fn )n ⊆ L(A) be a sequence convergent in measure on A; then there exists f ∈ λ → f. L(A) such that fn − A

For every ε > 0, limn λ((|fn − f | ≥ 2ε )) = 0; therefore, for every η > 0, there exists n0 ∈ N, such that, for any n ≥ n0 , λ((|fn − f | ≥ 2ε )) < η2 . Now let m, n ≥ n0 ; because

40

2

Chapter 2 • Measurable Functions

|fm − fn | ≤ |fm − f | + |f − fn |, |fm − f |
0, λ(|gp | >

np (np −1) 2

np (np −1) ∈ {1, 2, . . . , np }. 2

! kp −1 kp ε) ≤ λ = n1p np , np

2

< p ≤

np (np +1) , 2

and then

λ

→ 0; hence gp − → 0. R



On the other hand, for every x ∈]0,

1[ and

every n ∈ N, n ≥ 3, there exist k , k ∈ n(n−1) k −1 k k −1 k

{1, . . . , n} such that x ∈ + n , n \ n , n , and therefore there exists pn = 2

k , pn

= n(n−1) + k

such that gpn (x) = 1 and gpn

(x) = 0, which shows that (gp (x))p∈N is 2 divergent. So (gp ) is not convergent a.e. on ]0, 1[ to 0 from where it turns out that (gp ) is not almost uniformly convergent on ]0, 1[ to 0.

Theorem 2.2.4 Let A ∈ L, (fn ) ⊆ L(A), f, g ∈ L(A). λ λ (1) Let fn − → f ; then fn − → g if and only if f = g a.e. A ·

A ·

A

A

(2) Let fn − → f ; then fn − → g if and only if f = g a.e. a.u.

a.u.

A

A

(3) Let fn −→ f ; then fn −→ g if and only if f = g a.e.

2

41 2.2 · Different Types of Convergence

Proof λ λ (1) (⇒): We assume that fn − → f, fn − → g and let ε > 0; from the inequality |f − g| ≤ A

A

|f − fn | + |fn − g|, it follows the inclusion (|fn − f | < 2ε ) ∩ (|fn − g| < 2ε ) ⊆ (|f − g| < ε). We go to the complement in the last inclusion, and we get (|f − g| ≥ ε) ⊆ (|fn − f | ≥ ε2 ) ∪ (|fn − g| ≥ 2ε ) from where, using the monotonicity and the finite subadditivity of measure, λ(|f −g| ≥ ε) ≤ λ(|fn −f | ≥ ε2 )+λ(|fn −g| ≥ 2ε ). Passing to the limit in the previous inequality, we get that, for every ε) = 0. ⎛ ε > 0, λ(|f − g| ≥ ⎞  ∞   1 ⎠ On the other hand, λ(f = g) = λ(|f − g| > 0) = λ ⎝ |f − g| ≥ p p=1   ∞  1 = 0, and then f = g a.e. ≤ λ |f − g| ≥ p p=1

λ

(⇐): We assume that fn − → f and f = g a.e. For every ε > 0, (|fn − g| ≥ A

ε) ⊆ (|fn − f | ≥ ε) ∪ (f = g); by applying the properties of monotonicity and finite subadditivity of the measure, we obtain λ(|fn −g| ≥ ε) ≤ λ(|fn −f | ≥ ε) and, passing to the limit, limn λ(|fn − g| ≥ ε) = 0. (2) (⇒): From inclusion (f = g) ⊆ (fn  f ) ∪ (fn  g) and from the properties of the measure λ, it follows that λ(f = g) = 0. (⇐): The inclusion (fn  g) ⊆ (fn  f ) ∪ (f = g) leads us to λ(fn  g) = 0. · a.u. a.u. (3) (⇒): If fn −→ f and fn −→ g, then, from point (2) of Theorem 2.2.2, fn − → f , and A

·

A

A

fn − → g. According to the previous point, f = g a.e. A

a.u.

(⇐): Let us assume that fn −→ f and that f = g a.e.; for every ε > 0, there A

u

u

A\Aε

A\(Aε ∪(f =g))

exists Aε ∈ L such that λ(Aε ) < ε and fn −−−→ f . Then fn −−−−−−−−−→ f or u

a.u.

fn −−−−−−−−−→ g, and, since λ(Aε ) = λ(Aε ∪ (f = g)) < ε, it follows that fn −→ g. A A\(Aε ∪(f =g)) 

Theorem 2.2.5 (Riesz) (1) Any Cauchy in measure sequence on A ∈ L has an almost uniform convergent subsequence on A. a.u. λ (2) If fn − → f , then there exists kn ↑ +∞ such that fkn −→ f. A

A

(3) Any Cauchy in measure sequence on A ∈ L is convergent in measure on A.

Proof (1) Let (fn )n ⊆ L(A) be a Cauchy in measure sequence on A; for every ε > 0, limm,n→+∞ λ(|fn − fm | ≥ ε) = 0. Hence, for every ε > 0, there exists kε ∈ N such that, for any k ≥ kε , λ(|fk − fkε | ≥ ε) < ε. We will give values for ε in the set { 21k : k ∈ N}. (0) ε = 1, there exists k0 , such that λ(|fk − fk0 | ≥ 1) < 1, ∀k > k0 , (1) ε = 12 , there exists k1 > k0 , such that λ(|fk − fk1 | ≥ 12 ) < 12 , ∀k > k1 ,

42

2

Chapter 2 • Measurable Functions

··· (n) ε = 21n , there exists kn > kn−1 , s.t. λ(|fk − fkn | ≥ 21n ) < 21n , ∀k > kn , ··· If we replace k = kn+1 > kn in the relation (n), then we get: λ(|fkn+1 − fkn | ≥ 21n ) < 21n , for any n ∈ N.  1 For every n ∈ N, we denote Bn = ∞ i=n (|fki+1 − fki | ≥ 2i ), and we note that ∞ ∞ 1 1 1 λ(Bn ) ≤ i=n λ(|fki+1 − fki | ≥ 2i ) < i=n 2i = 2n−1 .  1 Let B = ∞ n=1 Bn ; then, for any n ∈ N, λ(B) ≤ λ(Bn ) < 2n−1 , from where it follows that λ(B) = 0.  For every x ∈ A \ B = ∞ n=1 (A \ Bn ), there exists n0 ∈ N such that x ∈ A \ Bn0 ; hence, for any n ≥ n0 , |fkn+1 (x) − fkn (x)| < 21n . Therefore, for any n > m ≥ 1 n0 , |fkn (x) − fkm (x)| ≤ |fkn (x) − fkn−1 (x)| + · · · + |fkm+1 (x) − fkm (x)| < 2n−1 + 1 1 1 + · · · + < . It follows that the sequence (f (x)) is a Cauchy on R, and m k n n−2 m−1 n 2 2 2 then there exists limn fkn (x) ∈ R, for everyx ∈ A \ B. limn→∞ fkn (x), x ∈ A \ B We will define f : A → R by f (x) = 0, x ∈ B. ·

Then fkn − → f and, according to (6) of Theorem 2.1.18, f ∈ L(A). A

a.u.

We will show that fkn −→ f . A

For every ε > 0, there exists n0 ∈ N such that u

1 2n0 −1

We show that fkn −−−→ f . A\Bn0  For every x ∈ A \ Bn0 = ∞ n=n0 (|fkn+1 − fkn | < for any n ≥ n0 . So, like above, |fkn (x) − fkm (x)|
0, |fn − fkn | < ε2 ∩  n  |fkn − f | < 2ε ⊆ (|fn − f | < ε) from where, passing to the complement, (|fn − f | ≥ ε) ⊆ |fn − fkn | ≥

ε! ε! ∪ |fkn − f | ≥ 2 2

hence λ(|fn − f | ≥ ε) ≤ λ |fn − fkn | ≥

ε! ε! + λ |fkn − f | ≥ . 2 2 λ

→ f. Then limn λ((|fn − f | ≥ ε)) = 0, for every ε > 0; therefore, fn −



A

λ

·

A

A

Corollary 2.2.6 If fn − → f , then there exists kn ↑ +∞ such that fkn − → f. Remark 2.2.7 From (3) of the previous theorem, the space L(A) is complete with respect to the convergence in measure. In fact we can show that this convergence is pseudo-metrizable. Let d : L(A) × L(A) → R+ , d(f, g) = min{1, infα>0 {α + λ(|f − g| ≥ α}}; d is a pseudoλ metric on L(A) and d(fn , f ) → 0 if and only if fn − → f ; therefore, the pseudo-metric space A

(L(A), d) is complete.

The convergence almost everywhere does not imply the convergence almost uniform (see (i) of Examples 2.2.3). On the sets of finite measure this implication works.

Theorem 2.2.8 (Egoroff) · a.u. Let A ∈ L with λ(A) < +∞, and let (fn )n ⊆ L(A) such that fn − → f ; then fn −→ f. A

A

Proof · According to (6) of Theorem 2.1.18, f ∈ L(A). Since fn − → f , the set B = (fn →A f ) is A

a null set which is equivalent to B ∈ L and λ(B) = 0. For all k, m ∈ N∗ , we denote   1 Ek,m = x ∈ A \ B : |fn (x) − f (x)| < , for any n ≥ k . m  We remark that Ek,m = ∞ n=k |fn − f | < Ek,m ∈ L, for all k, m ∈ N∗ . Moreover,

1 m

!

\ B. Since |fn − f | ∈ L(A), it follows that

Ek,m ⊆ Ek+1,m ⊆ A \ B, for all k, m ∈ N∗ . For all x ∈ A \ B, fn (x) → f (x), and then there exists k0 ∈ N∗ such that, for any n ≥ k0 , |fn (x) − f (x)| < m1 or x ∈ Ek0 ,m . We have shown that, for any m ∈ N∗ , the sequence

44

2

Chapter 2 • Measurable Functions

 (Ek,m )k∈N∗ is increasing and that ∞ k=1 Ek,m = A\B. We now use the propriety of continuity from below of measure (see (6) of Theorem 1.3.11), and to obtain λ(A \ B) = lim λ(Ek,m ), for any m ∈ N∗ . k→+∞

But λ(A \ B) = λ(A) < +∞, and, according to (3) of Theorem 1.3.11, for every ε > 0, and any m ∈ N∗ , there exists km ∈ N∗ such that λ(A \ Ekm ,m ) = λ(A) − λ(Ekm ,m )
0, let Aε = ∞ ,m ). m=1 (A \ Ekm ∞ ε Then λ(Aε ) ≤ m=1 λ(A\Ekm ,m ) < ∞ m=1 2m = ε. Let us show that (fn )n is uniformly convergent on A \ Aε to f .  ∗ For every x ∈ A \ Aε = ∞ m=1 Ekm ,m , for any m ∈ N and for any n ≥ km , |fn (x) − 1 f (x)| < m . Let η be arbitrary positive, and let m0 ∈ N∗ such that m10 < η. There exists nε = km0 ∈ N

such that, for any n ≥ nε , and every x ∈ A \ Aε , |fn (x) − f (x)| < u

a.u.

A\Aε

A

1 m0

< η, from where

fn −−−→ f . Therefore fn −→ f .



·

λ

A

A

Corollary 2.2.9 If fn − → f and λ(A) < +∞, then fn − → f. Remark 2.2.10 The condition λ(A) < +∞ of Egoroff’s theorem is essential. Indeed, if fn = χ[n, n + 1] , then (fn ) is pointwise convergent on R to the function 0, but it does not converge in measure, and even less it does not converge almost uniformly.

In the following diagram, we summarize the relationships between the different types of convergences: the uniform convergence (U), the pointwise convergence (P), the almost uniform convergence (AU), the almost everywhere convergence (AE), and the convergence in measure (M). The dotted arrows in ⊡ Fig. 2.1 represent the convergence of a subsequence.

U

AU

2.2.2

M

2.2.5

2.2.8 λ(A) 0, there exists Aε ∈ L with λ(Aε ) < ε such that fkn −−−→ f . The functions fkn are A\Aε

bounded (they are simple functions), and since uniform convergence preserves the property of being bounded, f is bounded on A \ Aε . Therefore there exists k > 0 such that, for every x ∈ A \ Aε , |f (x)| < k or, equivalent, A \ Aε ⊆ (|f | < k). Passing to the complement set in the last relation, (|f | ≥ k) ⊆ Aε , and then λ(|f | ≥ k) < ε. Now let f ∈ L(A) such that, for every ε > 0, there exists k > 0 so that λ(|f | ≥ k) < ε. Then, for all n ∈ N∗ , there exists kn > 0 such that λ(|f | ≥ kn ) < n1 . (kn )n can be chosen a strictly increasing sequence of natural numbers, and then kn ≥ n, for all n ∈ N∗ . We denote An = (|f | ≥ kn ). Then −kn < f (x) < kn , for every x ∈ A \ An . Therefore A \ An ⊆ f −1 ([−kn , kn [) =

nk n −1

f −1



k=−nkn

Since f ∈ L(A), Ak,n = f −1

fn =

nk n −1  −nkn



k k+1 , n n

k k+1 , n n

 .

 ∈ L(A), and then

k ·χ ∈ E (A). n Ak,n

Let us show that (fn )n converges in measure on A to f . We can easily see that, for all n ∈ N∗ , (|fn − f | > n1 ) ⊆ An .

49 2.3 · The Structure of Measurable Functions

For every ε > 0, there exists n0 ∈ N∗ such that

1 n

< ε, for all n ≥ n0 . Then

  1 1 λ((|fn − f | ≥ ε)) ≤ λ ≤ λ(An ) < , for any n ≥ n0 , |fn − f | > n n λ

from where limn λ((|fn − f | ≥ ε)) = 0, for every ε > 0; hence fn − → f , and so f ∈ Lt (A). A



Remarks 2.3.7 (i) The previous theorem states that a function is totally measurable on A if and only if it is measurable and asymptotically bounded on A: For every ε > 0, there exists Aε ∈ L with λ(Aε ) < ε, and f is bounded on A \ Aε (Aε = (|f | > k)). Obviously, a bounded function on a measurable set is totally measurable if and only if it is measurable.  (ii) If λ(A) < +∞, then Lt (A) = L(A). Indeed, let f ∈ L(A); then ∞ n=0 (|f | ≥ n) = ∅, and since λ(A) < +∞, we can use the property of continuity from above of λ ((7) of Theorem 1.3.11). So limn λ(|f | ≥ n) = 0. It follows that, for every ε > 0, there exists n0 ∈ N such that λ(|f | ≥ n0 ) < ε. The function f is then asymptotically bounded on A, and so f ∈ Lt (A). (iii) Let f ∈ L(A) and B ∈ L(A) with λ(B) < +∞; then f · χB ∈ Lt (A). Indeed, f · χB ∈ L(A) (see Exercise (4) of  Sect. 2.5). From (ii), f |B ∈ Lt (B). to $ According $  Theorem 2.3.6, for every ε > 0, there exists k > 0 such that λ $f |B $ ≥ k = λ({x ∈ B : |f (x)| ≥ k}) = λ({x ∈ A : |f · χB | ≥ k}) < ε, from where f · χB ∈ Lt (A). (iv) The function f : R → R, f (x) = x is continuous everywhere on R, and then it is measurable on R; however f is not asymptotically bounded on R (for any k > 0, λ(|f | > k) = +∞), and therefore it is not totally measurable. The suite of simple n.2n  k functions (fn ), fn = · χ k k+1 , is pointwise convergent to f on R. 2n , n 2n 2n k=−n.2

We have noticed that the almost everywhere continuous functions are measurable (see (2) of Theorem 2.1.13). The Dirichlet function (see Remark 2.1.9) is an example of a measurable function which is discontinuous at each point of R. The following theorem shows us that, although the measurable functions can be discontinuous everywhere, they are continuous on sets whose complement has the arbitrarily small measure (continuity related to the topology induced on these sets).

Theorem 2.3.8 (Lusin) Let A ∈ L; then f ∈ L(A) if and only if, for every ε > 0, there exists a closed set Fε ⊆ A, such that λ(A \ Fε ) < ε and f |Fε is continuous (for the relative topology of τu on Fε ).

2

50

2

Chapter 2 • Measurable Functions

Proof (1) The condition is necessary. We will prove this part in three steps.  (I) Let f = nk=1 ak χA ∈ E (A). Then, for any k = l, ak = al , Ak ∩ Al = ∅, Ak ∈ k  L(A), and nk=1 Ak = A. Corollary 1.3.9 assures us that, for every ε > 0 and any k = 1, . . . , n, there exists a closed set Fk ⊆ Ak such that λ(Ak \ Fk ) < 2εk .  Then Fε = nk=1 Fk ⊆ A is also closed, and λ(A \ Fε ) = λ

n 

Ak \ Fε

=

k=1

≤

n 

λ(Ak \ Fk )
0, there exists Aε ∈ L(A) with λ(Aε ) < ε4 and fn −−−→ f . A \ Aε ∈ A\Aε

L; according to Corollary 1.3.9, there exists a closed set Gε ⊆ A \ Aε such that λ((A \ Aε ) \ Gε ) < ε4 . From the first step of the demonstration, for any n ∈ N, there exists a closed set Fnε ⊆ A, such that fn |Fnε is continuous and ε λ(A \ Fnε ) < 2n+1 .  ε Let Fε = Gε ∩ ∞ The set Fε is closed, and λ(A \ Fε ) ≤ n=1 Fn ⊆ A.  ∞ ε ε ε ε λ(A \ Gε ) + n=1 λ(A \ Fn ) < ε2 + ∞ n=1 2n+1 = 2 + 2 = ε. u

According to the definition, Fε ⊆ Gε ⊆ A \ Aε , and then fn −→ f , and for Fε

any n ∈ N, fn |Fε is continuous for the relative topology on Fε . The preserving continuity by uniform convergence then assures us that f |Fε is continuous.  (III) f ∈ L(A). For all n ∈ N, let An = A∩] − n, n[∈ L(A); then A = ∞ n=1 An . Since λ(An ) < +∞, f · χAn ∈ Lt (A) (see (iii) of Remark 2.3.7). From step II, for every ε > 0 and any n ∈ N∗ , there exists a closed set Fnε ⊆ A such that (f · χAn )|F ε is continuous and λ(A \ Fnε ) < 2εn . n

Let Fε =

∞

n=1

Fnε ; it follows that Fε ⊆ A is a closed set and

λ(A \ Fε ) = λ

∞  n=1

(A \

Fnε )

≤

∞  n=1

λ(A \ Fnε ) < ε.

51 2.3 · The Structure of Measurable Functions

Let us show that f |Fε is continuous. For every x ∈ Fε ⊆ A, there exists n0 ∈ N∗ such that x ∈ An0 . Let (xp )p ⊆ Fε , xp → x. There exists p0 ∈ N so that xp ∈ A∩] − n0 , n0 [= An0 , for any p ≥ p0 . Since (xp )p ⊆ Fnε0 and x ∈ Fnε0 , for any p ≥ p0 , f (xp ) = (f · χAn )(xp ) = (f · χAn )|F ε (xp ) → (f · χAn )|F ε (x) = f (x). 0

0

n0

0

n0

Consequently f |Fε is continuous. (2) The condition is sufficient. According to the hypothesis, for all n ∈ N∗ , there exists a closed set Fn ⊆ A such that λ(A \ Fn ) < n1 and f |Fn is continuous. Let F = ∞ ∞ 1 ∗ n=1 Fn ∈ L(A); then λ(A\F ) = λ( n=1 (A\Fn )) ≤ n , for any n ∈ N , from where ∗ λ(A \ F ) = 0. Therefore N = A \ F is a null set. For any n ∈ N , f |Fn is continuous; it follows that f |Fn ∈ L(Fn ) (see (i) of Corollary 2.1.7). Then, for any a ∈ R and any n ∈ N∗ , An = {x ∈ Fn : f (x) < a} ∈ L. The set A0 = {x ∈ N : f (x) < a} ⊆ N is a  null set so that A0 ∈ L. Thus {x ∈ A : f (x) < a} = A0 ∪ ∞ n=1 An ∈ L, from where f ∈ L(A).  Corollary 2.3.9 Let A ∈ L such that λ(A) < +∞; then f ∈ L(A) if and only if, for every ε > 0, there exists a compact set Kε ⊆ A such that λ(A \ Kε ) < ε and f |Kε is continuous. Proof We need only to prove that the condition is necessary. For any n ∈ N, let An = A∩] − n, n[∈  L(A); then A = ∞ n=1 An ; hence λ(A) = limn λ(An ). For every ε > 0, there exists n0 ∈ N such that λ(A \ An0 ) = λ(A) − λ(An0 ) < 2ε (see (3) of Theorem 1.3.11). From the previous theorem, there is a closed set Fε ⊆ An0 such that λ(An0 \ Fε ) < 2ε and f |Fε is continuous. Then λ(A \ Fε ) < ε and Fε is a bounded closed set and hence compact. 

We will present in the following another theorem to approximate measurable functions with continuous functions. Theorem 2.3.10 (Borel) Let A ∈ L; for every f ∈ L(A) and every ε > 0, there exists a continuous function fε : A → R, such that λ(f = fε ) < ε. Moreover, we can choose f so that supx∈A |fε (x)| ≤ supx∈A |f (x)|.

Proof Let f ∈ L(A) and let α = supx∈A |f (x)| ∈ [0, +∞]; then f (A) ⊆ I = [−α, α]. According to the Lusin theorem, for every ε > 0, there exists a closed set Fε ⊆ A such that f |Fε is continuous and λ(A \ Fε ) < ε; f (Fε ) ⊆ [−α, α]. We will now use a classic theorem of topology, the Tietze extension theorem: For every closed set F ⊆ R and every continuous mapping f : F → R, there exists a continuous extension f¯ : R → R of f . Moreover, if

2

52

2

Chapter 2 • Measurable Functions

f (F ) ⊆ [−α, +α], then f¯ can be chosen so that f¯(R) ⊆ [−α, +α] (for a demonstration on normal spaces, see Theorem 5.1, page 149 in [4]). If α < +∞, let f¯ : R → [−α, +α] be a continuous function on R such that f¯|Fε = f , and let fε = f¯|A : A → R. It’s obvious that λ(f = fε ) ≤ λ(A \ Fε ) < ε and supx∈R |fε (x)| ≤ α = supx∈R |f (x)|. If α = +∞, then supx∈A |fε (x)| ≤ +∞ = supx∈F |f (x)|.  Example 2.3.11 Let Q = {q1 , q2 , . . . , qn , . . .} ⊆ R be the set of rational numbers; then Q ∈ Bu , and so χQ ∈ L(R) (see Remark 2.1.9). For every ε > 0 and any n ∈ N∗ , let Inε = qn −

ε 2n+1

, qn +

ε

2n+1

.

ε Then Q ⊆ ∪∞ n=1 In = Dε ∈ τu . Fε = R \ Dε ⊆ R \ Q is a closed set. λ(R \ Fε ) = λ(Dε ) ≤ ∞ ε n=1 2n = ε. χQ (x) = 0, for every x ∈ Fε , then χQ is continuous on this set. We note that the function χQ is not continuous in any point of R. The function fε = 0 is continuous on R and λ∗ (χQ = fε ) = 0 < ε.

2.4

Abstract Setting

Just like at the end of  Chap. 1, we will present the measurable functions and their properties on an abstract measure space. Definition 2.4.1 Let (X, A, γ ) be a measure space where γ is a measure σ -finite and complete; the function f : X → R is A-measurable if, for any a ∈ R, f −1 (] − ∞, a[) ∈ A. We denote M(X) (or simply with M when there is no danger of confusion) the set of all measurable functions on X.

For every A ⊆ X, χA ∈ M if and only if A ∈ A. A function A-simple (or simple, if there is no risk of confusion) is a function f : X → R for which f (X) = {a1 , . . . , ap } ⊆ R and Ai = f −1 ({ai }) ∈ A, for any i = 1, . . . , p. We denote with E (X) the set of all simple functions; for every f ∈ E (X), f = p i=1 ai ·χAi , where {A1 , . . . , Ap } is a partition A-measurable of X. Obviously, E (X) ⊆ M(X). If X is provided with a topology τ and if τ ⊆ A, then the real continuous functions on (X, τ ) are measurable (C(X) ⊆ M(X)). If we replace correctly A by X, L by A, and L(A) by M(X), then the results from Theorems 2.1.10, 2.1.13, 2.1.18, 2.1.20, and Proposition 2.1.22 are preserved. A set A ⊆ X is a null set if A ∈ A and γ (A) = 0. Since γ is a complete measure, every subset of a null set is a null set. A property P is satisfied γ -almost everywhere (almost everywhere or a.e.) on X if {x ∈ X : x does not have the property P } is a null set.

53 2.4 · Abstract Setting

Theorem 2.4.2 Let f : X → R; the following are equivalent: (1) f ∈ M(X). (2) f −1 (] − ∞, a]) ∈ A, for any a ∈ R. (3) f −1 (]a, +∞[) ∈ A, for any a ∈ R. (4) f −1 ([a, +∞[) ∈ A, for any a ∈ R. (5) f −1 (I ) ∈ A, for every I ∈ I . (6) f −1 (D) ∈ A, for every D ∈ τu . (7) f −1 (B) ∈ A, for every B ∈ Bu .

Theorem 2.4.3 Let (fn ) ⊆ M(X); then: (1) f = supn fn ∈ M(X), if supn fn (x) < +∞, for every x ∈ X. (2) f = infn fn ∈ M(X), if infn fn (x) > −∞, for every x ∈ X. (3) f = lim supn fn ∈ M(X), if lim supn fn (x) ∈ R, for every x ∈ X. (4) f = lim infn fn ∈ M(X), if lim infn fn (x) ∈ R, for every x ∈ X. (5) If fn (x) → f (x) ∈ R, for every x ∈ X, then f ∈ M(X). · (6) If fn − → f , then f ∈ M(X). X

Theorem 2.4.4 Let f, g : X → R. (1) If f ∈ M(X) and f = g a.e., then g ∈ M(X). (2) If f is continuous a.e. on X, then f ∈ M(X).

Theorem 2.4.5 Let f, g ∈ M(X) and let α ∈ R; then f + g ∈ M(X), α · f ∈ M(X) and f · g ∈ M(X).

Proposition 2.4.6 (1) f ∈ M(X) if and only if f + ∈ M(X) and f − ∈ M(X). (2) If f ∈ M(X), then |f | ∈ M(X).

We can define, as in Definition 2.2.1, the convergence in measure and almost uniform convergence, and we find the results Theorems 2.2.2, 2.2.4, 2.2.5, 2.2.8, Corollary 2.2.6, Remark 2.2.7 and Corollary 2.2.9.

2

54

2

Chapter 2 • Measurable Functions

Definition 2.4.7 Let (fn ) ⊆ M(X) and f ∈ M(X). a.u. 1. (fn ) converges almost uniformly to f on X (fn −→ f ) if, for every ε > 0, X

u

there exists Aε ∈ A such that γ (Aε ) < ε and fn −−−→ f . X\Aε γ

2. (fn ) converges in measure to f on X (fn − → f ) if, for every ε > X

0, limn γ ((|fn − f | ≥ ε)) = 0. The sequence (fn )n is convergent in measure on X if there is f ∈ M(X) such γ → f. that fn − X

3. (fn ) is Cauchy in measure if, for every ε > 0, limm,n→∞ γ ((|fm −fn | ≥ ε)) = 0.

Theorem 2.4.8 Let (fn ) ⊆ M(X) and f ∈ M(X); then: γ a.u. (1) If fn −→ f , then fn − → f. a.u.

X

X ·

X

X

→ f. (2) If fn −→ f , then fn − (3) Every sequence convergent in measure is Cauchy in measure on X.

Theorem 2.4.9 Let (fn ) ⊆ M(X), f, g ∈ M(X). γ γ (1) Let fn − → f ; then fn − → g if and only if f = g a.e. X ·

X ·

X

X

→ f ; then fn − → g if and only if f = g a.e. (2) Let fn − a.u.

a.u.

X

X

(3) Let fn −→ f ; then fn −→ g if and only if f = g a.e.

Theorem 2.4.10 (Riesz) (1) Every sequence Cauchy in measure on X has a subsequence almost uniformly convergent on X. γ a.u. (2) If fn − → f , then there exists kn ↑ +∞ such that fkn −→ f. X

X

(3) Every sequence Cauchy in measure on X is convergent in measure.

γ

·

X

X

Corollary 2.4.11 If fn − → f , then there exists kn ↑ +∞ such that fkn − → f.

55 2.5 · Exercises

Theorem 2.4.12 (Egoroff) · a.u. If γ (X) < +∞, (fn )n ⊆ M(X), and fn − → f , then fn −→ f. X

X

·

γ

X

X

Corollary 2.4.13 If fn − → f and γ (X) < +∞, then fn − → f.

In this abstract framework, we can also prove the theorem of approximation of measurable functions with simple functions (see Theorem 2.3.3).

Theorem 2.4.14 (1) If f : X → R+ , f ∈ M+ (X), there exists (fn ) ⊆ E+ (X), fn ↑ f . p → f. (2) If f ∈ M(X), there exists (fn ) ⊆ E (X), fn − X

u

(3) If f ∈ M(X) is bounded, there exists (fn ) ⊆ E (X), fn − → f. X

Let (X, τ ) be a normal topological space (see [4, page 144]; particularly, X can be a metric space or even a normed space), let A be a σ -algebra on X such that τ ⊆ A, and let γ be a σ -finite complete regular measure on (X, A); in this context we can formulate the generalizations of the theorems of Lusin and Borel.

Theorem 2.4.15 (Lusin) f ∈ M(X) if and only if, for every ε > 0, there exists a closed set Fε ⊆ X such that γ (X \ Fε ) < ε and f |Fε is continuous.

Theorem 2.4.16 (Borel) For every f ∈ M(X) and every ε > 0, there exists a continuous function fε : X → R such that γ (f = fε ) < ε. Moreover, we can choose f so that supx∈X |fε (x)| ≤ supx∈X |f (x)|.

2.5

Exercises

(1) Let (An )n ⊆ L(R) be a sequence of pairwise disjoint measurable sets (An ∩Am = ∅, for all n = m), and let (an )n ⊆ R. Show that the function f : R → R, defined ∞  by f (x) = an · χAn (x), for every x ∈ R, is measurable. n=0

2

56

2

Chapter 2 • Measurable Functions

(2) Let A ⊆ R be a non-measurable Lebesgue set (A ∈ / L(R)), and let f : R → R,  1, x ∈ A . Show that |f | ∈ L(R) but f ∈ / L(R). defined by f (x) = −1, x ∈ /A (3) Check ifthe Riemann function: f : [0, 1] → R, 0, x ∈ (R \ Q) ∩ [0, 1] ∪ {0} f (x) = 1 , , x = pq ∈ [0, 1], p, q ∈ N∗ , p, q are relatively prime q is Lebesgue measurable. Indication. Show that f is continuous at all irrational points and in 0 and discontinuous at rational points of ]0, 1].

(4) Let f : A → R, f  ∈ L(A) and let B ∈ L(A); show that the function g : A → R, 0, x ∈ A \ B , is Lebesgue measurable. defined by g(x) = f (x), x ∈ B (5) Let A ∈ L, a, b ∈ R, a = 0, f : aA+b → R and g : A → R, g(x) = f (ax+b), for every x ∈ A. Show that f ∈ L(aA + b) if and only if g ∈ L(A). Indication. Show that, for any α ∈ R, g−1 (]−∞, α[) = − ab + a1 f −1 (]−∞, α[), and use Theorem 1.3.15. (6) Let I ∈ I , and let f : I → R be a differentiable function on open interval I ; show that the derivative of f , f , is Lebesgue measurable.     1 Indication. Show that, for every x ∈ R, f (x) = limn→∞ n · f x + − f (x) . n

(7) Let A ∈ L, f ∈ L(A) and let a ∈ R; show that (f = a) = {x ∈ A : f (x) = a} ∈ L. (8) Let A ∈ L and let f, g ∈ L(A); show that (f < g) = {x ∈ A : f (x) < g(x)} ∈ L. (9) Let A ∈ L, f ∈ L(A), and let f  = inf{α + λ(|f | > α) : α > 0}. Show that: (a) f  = 0 if and only if f = 0 a.e., (b)  − f  = f , for every f ∈ L(A), (c) f + g ≤ f  + g, for every f, g ∈ L(A), λ (d) fn − → f if and only if fn − f  → 0. A Indication. See Remark 2.2.7

(10) Check if the following sequences are convergent a.e., almost uniform, or in measure on their definition sets: (a) fn : R → R, fn = n1 · χ[0, n] . (b) fn : R → R, fn = n · χ 1 . [0, n ]  sin nx (11) Show that the series ∞ n=1 n converges on R to a measurable function. (12) Show that the sequence (fn )n ⊆ R, fn (x) = sin nx, is not convergent in measure to 0. (13) Let A ∈ L with λ(A) < +∞. (a) f ∈ L(A) ⇒ ∀ε > 0, ∃k > 0 such that λ(|f | ≥ k) < ε. λ (b) fn − → f ⇒ ∀ε > 0, ∃k > 0 such that λ(|fn | ≥ k) < ε, ∀n ∈ N. A

57 2.5 · Exercises

λ

λ

A

A

(c) fn − → f ⇒ fn2 − → f 2. λ

λ

λ

A

A

A

(d) fn − → f, gn − → g ⇒ fn gn − → fg.

(a) (b) (c) (d)

Indications. See (ii) of Remarks 2.3.7. We will be using inclusion (|fn | ≥ k + 1) ⊆ (|fn − f | ≥ 1) ∪ (|f | ≥ k) and point (a). ε ) ∪ (|f | ≥ k) ∪ (|f | ≥ k) and point (b). We will be using inclusion (|fn2 − f 2 | ≥ ε) ⊆ (|fn − f | ≥ 2k n We will take into account the relation fn gn = 14 [(fn + gn )2 − (fn − gn )2 ] and point (c).

(14) Show that the sequence (fn )n ⊆ L(R), defined by fn (x) = x + n1 , is convergent in measure to the function f ∈ L(R), f (x) = x, but (fn2 )n is not convergent in measure to f 2 . Explain the result. (15) Show that, for every A ∈ L and every f ∈ L+ (A), f = sup{ϕ : ϕ ∈ E+ (A), ϕ ≤ f }.

2

3

59

Lebesgue Integral

In this chapter, we will build the Lebesgue integral, first for the measurable and positive functions and then for the measurable functions in general. We will show that the family of Lebesgue integrable functions on a set A ∈ L is organized as a vector subspace of the space L(A) and that the integral is a linear operator on this space. We will present the main properties of the class of integrable functions and of the integral; among these stand out the properties of passing to the limit under the integral. At the end of the chapter, we will carry out a comparative study of the two integrals, Riemann and Lebesgue.

3.1

Integrals of Nonnegative Measurable Functions

Let A ∈ L and let E+ (A) be the set of all simple and positive functions on A; if f ∈  E+ (A), then f = pi=1 ai χA , where {ai : i = 1, . . . , p} ⊆ R+ , ai = aj , for any i = j i and Ai = f −1 ({ai }) ∈ L(A), for every i = 1, . . . , p. Suppose that, among the values ai , there is also the value 0; then {Ai : i = 1, . . . , p} is a L-measurable partition of the set A. Definition 3.1.1 Let f ∈ E+ (A), f =

p

% f dλ =

p 

ai λ(Ai ) ∈ [0, +∞] is & the integral of f over the set A. The function f is integrable over A if A f dλ < +∞. We denote with E+1 (A) the set of all simple positive integrable functions over A. p We remark that i=1 ai λ(Ai ) < +∞ if and only if, for those i for whom λ(Ai ) = +∞, we have ai = 0 (we will agree that 0 · (+∞) = 0). i=1 ai χAi ;

we say that

A

i=1

(Continued )

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2021 L. C. Florescu, Lebesgue Integral, Compact Textbooks in Mathematics, https://doi.org/10.1007/978-3-030-60163-8_3

60

Chapter 3 • Lebesgue Integral

Definition 3.1.1 (continued)

3

p If B ∈ L, B ⊆ A, then the restriction of f on the set B is f |B = i=1 ai χA ∩ B ∈ i E+ (B) (we recall that we identify χA ∩ B with the restrictions of these characteristic i functions to the set B). & & & p It is obvious that B f |B dλ = i=1 ai λ(Ai ∩ B) = A f · χB dλ ≤ A f dλ; we will & denote this integral with B f dλ. If f ∈ E+1 (A), then f |B ∈ E+1 (B).

If B ∈ L(A), then χB = χB +0·χA \ B ∈ E+ (A) and if and only if λ(B) < +∞.

& A

χB dλ = λ(B); χB ∈ E+1 (A)

Remarks 3.1.2 p q (i) If f = i=1 ai χAi = j =1 bj χBj , where {Ai : i = 1, . . . , p} and {Bj : j = 1, . . . , q} are measurable partitions of A, then p 

ai λ(Ai ) =

q p  

ai λ(Ai ∩ Bj ) =

i=1 j =1

i=1

q p  

bj λ(Ai ∩ Bj ) =

i=1 j =1

q 

bj λ(Bj ).

j =1

The equality in the middle occurs because, for all (i, j ) for which Ai ∩Bj = ∅, ai = bj . Thus, the integral of the function f is well defined. & p q (ii) If f = i=1 ai χA , g = j =1 bj χB : A → R and f = g a.e., then A f dλ = i j & gdλ. Indeed, since f = g a.e., λ(Ai ∩ Bj ) = 0, for all i, j for which ai = bj . Then &A    f dλ = ai =bj ai λ(Ai ∩ Bj ) + ai =bj ai λ(Ai ∩ Bj ) = ai =bj bj λ(Ai ∩ Bj ) = A & B gdλ. Proposition 3.1.3 Let A ∈ L, c ≥ 0, and f, g ∈ E+ (A); then & & (1) cf ∈ E+ (A) and A cf dλ = c A f dλ. & & & (2) f + g ∈ E+ (A) and A (f + g)dλ = A f dλ + A gdλ. & & (3) A f dλ ≤ A gdλ if f ≤ g a.e. & & (4) B f dλ ≤ C f dλ, for every B, C ∈ L, B ⊆ C ⊆ A. & (5) For every ε > 0, there exists δ > 0 such that B f dλ < ε, for any B ∈ L(A) with λ(B) < δ. Proof p p If f = i=1 ai χA , then cf = i=1 cai χA , and therefore (1) is obvious. i i p q (2) Let f = i=1 ai χA , g = j =1 bj χB , where {Ai : i = 1, . . . , p} and {Bj : j = i j 1, . . . , q} are measurable partitions of A; then f +g =

q p   i=1 j =1

ai χA

i ∩ Bj

+

q p   i=1 j =1

bj χA

i ∩ Bj

=

q p   (ai + bj )χA i=1 j =1

i ∩ Bj

;

3

61 3.1 · Integrals of Nonnegative Measurable Functions

therefore f + g ∈ E+ (A) and % (f + g)dλ =

p  q  (ai + bj )λ(Ai ∩ Bj ) =

A

=

i=1 j =1

p 

⎛ ai ⎝

p 

⎞ λ(Ai ∩ Bj )⎠ +

j =1

i=1

=

q 

q 

q 

p 

%

% gdλ. A

q

, where {Ai : i = 1, . . . , p} and {Bj : p q j = 1, . . . , q} are measurable partitions of A. Since f ≤ g a.e., i=1 j =1 ai χA ∩ B ≤ i j p q i=1 j =1 bj χAi ∩ Bj a.e.; we notice that, for any pair (i, j ) for which ai > bj , λ(Ai ∩ Bj ) = 0. Therefore % f dλ = A

≤

i=1 ai χAi , g

q p  

=

λ(Ai ∩ Bj ) =

f dλ + A

p

i=1

bj λ(Bj ) =

j =1

(3) Let f =

bj

j =1

ai λ(Ai ) +

i=1



ai λ(Ai ∩ Bj ) =

i=1 j =1



j =1 bj χBj



ai λ(Ai ∩ Bj ) ≤

(ai ≤bj )

bj λ(Ai ∩ Bj ) =

(ai ≤bj )

q p  

% bj λ(Ai ∩ Bj ) =

gdλ. A

i=1 j =1

& & & & (4) If we remark that B f dλ = A f · χB dλ, C f dλ = A f · χC dλ and that f χB ≤ f χC , then (4) results from (3). p 5). Let f = i=1 ai χA ; if f = 0, then the condition is obviously verified. i If f = 0, then let M = max{ai : i = 1, . . . , p} > 0. For every ε > 0, there exists & p ε δ = M > 0 such that, for every B ∈ L(A) with λ(B) < δ, B f dλ = i=1 ai λ(Ai ∩ B) ≤ p M · i=1 λ(Ai ∩ B) = M · λ(B) < M · δ = ε. 

For every A ∈ L and every f ∈ L+ (A), f = sup{ϕ : ϕ ∈ E+ (A), ϕ ≤ f } (see Exercise 15) of 2.5). Therefore, the following definition of the integral for the measurable and positive functions is as natural as possible: Definition 3.1.4 For every f ∈ L+ (A) we say that %

%



f dλ = sup A

A

ϕdλ : ϕ ∈ E+ (A), ϕ ≤ f

∈ [0, +∞] (Continued )

62

3

Chapter 3 • Lebesgue Integral

Definition 3.1.4 (continued) is the integral of f over the measurable set A. The function f is integrable over A & if A f dλ < +∞. L1+ (A) denotes the set of all positive and integrable functions over A. & & If B ∈ L(A), then f χB ∈ L+ (A) and A f χB dλ = B f dλ. The restriction of f to B, f |B , is positive and measurable on B (see Definition 2.1.2). If f ∈ L1+ (A), then & & f |B ∈ L1+ (B) and B f |B dλ = B f dλ.

Remarks 3.1.5 (i) The definition of the integral for the measurable and positive functions does not contradict that of the integral for simple and positive functions. Indeed, if f ∈ E+ (A), & '& ( then A f dλ is the biggest element of the set A ϕdλ : ϕ ∈ E+ (A), ϕ ≤ f and then 1 1 the upper bound% for this set. One can notice that E + (A) ⊆ L+ (A). % (ii) f dλ = sup ϕdλ : ϕ ∈ E+ (A), ϕ ≤ f a.e. . Indeed, if ϕ ∈ E+ (A) and ϕ ≤ f A

A

a.e., then λ((ϕ > f )) = 0. Therefore ϕ1 = ϕ · χ(ϕ ≤ f ) ∈ E+ (A), ϕ1 ≤ f and & & A ϕ1 dλ = A ϕdλ.

In the following proposition, we highlight certain immediate properties of the integral of the measurable and positive functions. Proposition 3.1.6 Let f, g ∈ L+ (A) and c ≥ 0; then & & (1) cf ∈ L+ (A) and A cf dλ = c A f dλ. & & (2) A f dλ ≤ A gdλ, if f ≤ g a.e.. & & (3) B f dλ ≤ C f dλ, for every B, C ∈ L, B ⊆ C ⊆ A. Proof (1) If c = 0, then cf = 0 and so If c > 0, then

& A

cf dλ = c

& A

f dλ.

%

% cf dλ = sup A

A

 ϕdλ : ϕ ∈ E+ (A), ϕ ≤ cf

=

  % 1 1 ϕdλ : ϕ ∈ E+ (A), ϕ ≤ f = = sup c c A c  %  % = sup c ψdλ : ψ ∈ E+ (A), ψ ≤ f = c f dλ. A

A

(2) We can see that if ϕ ≤ f a.e., then ϕ ≤ g a.e., which leads to % A

 %  ϕdλ : ϕ ∈ E+ (A), ϕ ≤ f a.e. ⊆ ϕdλ : ϕ ∈ E+ (A), ϕ ≤ g a.e. . A

The inequality results from (ii) of Remark 3.1.5.

3

63 3.1 · Integrals of Nonnegative Measurable Functions

(3) We remark that f χB ≤ f χC and then we use (2).



The following theorem plays an extremely important role in Lebesgue integral theory.

Theorem 3.1.7 (Monotone Convergence Theorem) Let A ∈ L, f : A → R+ , and let (fn ) ⊆ L+ (A) be an increasing sequence such that & & fn ↑ f ; then f ∈ L+ (A) and A fn dλ ↑ A f dλ.

Proof & It is obvious that f ∈ L+ (A) (see 5) of 2.1.18) and, since fn ≤ f , for any n ∈ N, A fn dλ ≤ %  & f dλ. In addition, from (2) of the previous proposition, the sequence f dλ is n A A n∈N & increasing in [0, +∞], and therefore there exists limn A fn dλ ∈ [0, +∞]; then %

% fn dλ ≤

lim n

A

(1)

f dλ A

p Let t ∈]0, 1[ be an arbitrary number, and let ϕ = i=1 ai χA ∈ E+ (A), ϕ ≤ f , an arbitrary i simple function but, for the time being, fixed. For any n ∈ N, let Bn = {x ∈ A : fn (x) ≥ tϕ(x)} ∈ L. Then Bn ⊆ Bn+1 , for any n ∈ N and

∞ 

Bn = A.

(2)

n=1

The inclusion Bn ⊆ Bn+1 is a consequence of the fact that the sequence (fn )n is increasing. The equality is shown by double inclusion; the inclusion ⊆ is true because Bn ⊆ A, for any n ∈ N. Let’s show the inclusion ⊇. Let x ∈ A; if ϕ(x) = 0, then x ∈ Bn , for any n ∈ N, because the functions fn are positive. If ϕ(x) > 0, then tϕ(x) < ϕ(x) ≤ f (x); since fn (x) ↑ f (x), there exists n ∈ N such that tϕ(x) < fn (x) and so x ∈ Bn . Now, using the measure property of continuity from below (see the property 6) of Theorem 1.3.11) and the relations (2), it follows that % tϕdλ = A

p  i=1

= lim n

p  i=1

tai λ(Ai ) =

p 

tai λ

i=1

∞ 

(Ai ∩ Bn ) =

% tai λ(Ai ∩ Bn ) = lim n

(3)

n=1

%

%

tϕdλ ≤ lim Bn

n

fn dλ ≤ lim Bn

n

fn dλ. A

64

Chapter 3 • Lebesgue Integral

& & & From (3), A ϕdλ ≤ 1t · limn A fn dλ, and, since the function ϕ ≤ f is arbitrary, A f dλ ≤ & 1 t · limn A fn dλ. If in the previous relations, t → 1, then

3

%

% f dλ ≤ lim

fn dλ.

(4)

The inequalities (1) and (4) end the proof.



n

A

A

The above theorem can be extended to the case where f : A → [0, +∞] (the function f can also take the value +∞). Corollary 3.1.8 Let A ∈ L and let (fn )n ⊆ L+ (A) be an increasing sequence such that fn ↑ f , where f : A → [0, +∞]. Then Z = f −1 (+∞) ∈ L(A), f is measurable on A, and if we denote by &

% f dλ =

A\Z

+∞, if λ(Z) > 0

A

%

, then

% fn dλ =

lim n

f dλ, if λ(Z) = 0

f dλ.

A

A

Proof p For every p, n ∈ N, we denote Zn = {x ∈ A : fn (x) ≥ p}; since fn are measurable  ∞ p p functions, Zn ∈ L(A). If we remark that Z = ∞ p=1 n=1 Zn , then Z ∈ L(A). Let gn = fn |A\Z and g = f |A\Z . Since gn ↑ g it follows from the previous theorem that g ∈ L+ (A \ Z) & & and A\Z gn dλ ↑ A\Z gdλ. According to (iv) of Remark 2.1.3, f is measurable on A. If λ(Z) = 0, then %

%

%

fn dλ =

%

fn dλ =

A

A\Z

%

gn dλ ↑ A\Z

gdλ = A\Z

f dλ. A\Z

If λ(Z) > 0, then, for all n, p ∈ N, %

%

%

fn dλ ≥

p

fn dλ ≥

A

Z

p

fn dλ ≥ p · λ(Z ∩ Zn ).

Z∩Zn

Then, using the property of continuity from below of λ,

%

fn dλ ≥ p · λ Z ∩

lim n

A

Therefore limn Corollary 3.1.9

∞ 

p Zn

= p · λ(Z), for any p ∈ N.

n=1

&

fn dλ = +∞.

A

&

A (f

+ g)dλ =

 & A

f dλ +

&

A gdλ,

for every f, g ∈ L+ (A).

3

65 3.1 · Integrals of Nonnegative Measurable Functions

Proof Let (fn )n , (gn )n ⊆ E+ (A) such that fn ↑ f and gn ↑ g (see Theorem 2.3.3); then fn + gn ↑ f + g, and, according to the previous theorem and Proposition 3.1.3, %

% (f + g)dλ = lim n

A

%

%

(fn + gn )dλ = lim n

A

fn dλ + lim A

n

gn dλ = A

%

% f dλ +

= A



gdλ. A

Corollary 3.1.10 (Beppo Levi) Let (fn )n ⊆ L+ (A) be a sequence such that the function  ∞ series ∞ n=1 fn is pointwise convergent on A, and let f = n=1 fn ; then f ∈ L+ (A) and % f dλ = A

∞ % 

fn dλ.

n=1 A

Proof n The sequence of partial sums of the series (sn )n , defined by sn = k=1&fk , consists of & measurable functions and sn ↑ f . It follows that f ∈ L+ (A) and A sn dλ ↑ A f dλ. On the other hand, from the previous corollary, % sn dλ = A

n % 

fk dλ −−−−→ n→+∞

k=1 A

and therefore

&

A f dλ

=

∞ % 

fk dλ,

k=1 A

∞ &



n=1 A fn dλ.

Corollary 3.1.11 Let f ∈ L+ (A) and let (An )n ⊆ L(A) be a sequence of pairwise disjoint sets; then % f dλ =

∞

n=1 An

∞ % 

f dλ.

n=1 An

Proof  For any n ∈ N∗ , let fn = f · χAn ⊆ L+ (A); since ∞ n=1 fn = f · χ∪∞ An ≤ f , the series n=1 ∞ n=1 fn converges punctually on A. We are in the hypotheses of the previous corollary and we therefore obtain %

% f dλ =

∞

n=1 An

A

f · χ∪∞

n=1 An

dλ =

%  ∞ A n=1

f · χAn dλ =

∞ %  n=1 An

f dλ.



Corollary 3.1.12 (Fatou’s Lemma) Let (fn ) ⊆ L+ (A) such that f (x) = lim infn fn (x) < +∞, for every x ∈ A; then %

% lim inf fn dλ ≤ lim inf A

n

n

fn dλ. A

66

3

Chapter 3 • Lebesgue Integral

Proof We recall that lim infn fn = supn∈N infk≥n fk . If, for any n ∈ N, gn = infk≥n fk , then (gn )n ⊆ L+ (A) is an increasing sequence, and f = lim infn fn = supn∈N gn = limn gn . Therefore f ∈ L+ (A), and, according to the monotone convergence theorem & & (Theorem 3.1.7), A gn dλ ↑ A f dλ. & & On the other hand, since gn ≤ fn , A gn dλ ≤ A fn dλ, for any n ∈ N, from where & & & & & limn A gn dλ = lim infn A gn dλ ≤ lim infn A fn dλ. So A f dλ ≤ lim infn A fn dλ.  Proposition 3.1.13 Let A ∈ L and let f ∈ L+ (A); then % f dλ = 0 if and only if f = 0 a.e. A

Proof & (⇒): We suppose that A f dλ = 0 and let, for any n ∈ N∗ , An = (f ≥ n1 ) ∈ L. Then  An ⊆ An+1 , for any n ∈ N∗ , and ∞ n=1 An = (f > 0), from where λ(f = 0) = λ(f > 0) = limn λ(An ). & & & On the other hand, for any n ∈ N∗ , An f dλ ≤ A f dλ = 0, from where An f dλ = 0. & Since An f dλ ≥ n1 λ(An ), for any n ∈ N∗ , λ(An ) = 0. It follows that λ(f = 0) = 0, and therefore f = 0 a.e. (⇐): If f = 0 a.e., then, for every ϕ ∈ E+ (A) with ϕ ≤ f, ϕ = 0 a.e. If ϕ = & p any i for which λ(Ai ) = 0, ai = 0, and so A ϕdλ = 0. Since the i=1 ai χAi , then, for &  function ϕ is arbitrary, A f dλ = 0. Corollary 3.1.14

% (1) Let A ∈ L, B ∈ L(A) with λ(B) = 0 and let f ∈ L+ (A); then f dλ = 0. B % % · (2) Let A ∈ L and let f, g ∈ L+ (A) with f = g on A; then f dλ = gdλ. A

A

Proof %

% f dλ =

(1) We remark that B

A

f χB dλ and that f χB = 0 a.e.

(2) According to Corollary 3.1.11 and the previous point 1), %

% f dλ = A

% (f =g)

f dλ +

% (f =g)

f dλ =

% (f =g)

f dλ =

(f =g)

% gdλ =

gdλ.  A

Remark 3.1.15 The Dirichlet function, χQ , is zero a.e. on R. It follows from Proposition 3.1.13 that this function is integrable and that its integral is 0. From (3) of Proposition 3.1.6, this function is integrable over every measurable set, and its integral is 0. Note that the Dirichlet function is everywhere discontinuous, so it is not Riemann integrable over closed intervals (see Lebesgue Criterion 7.1.7).

3

67 3.1 · Integrals of Nonnegative Measurable Functions

Theorem 3.1.16 Let f ∈ L1+ (A), let L(A) be the σ -algebra of all measurable subsets of A (see & Definition 1.3.1), and let γ : L(A) → R+ , γ (B) = B f dλ, for every B ∈ L(A). Then γ is a finite measure on L(A) which checks the following two conditions: & (1) For every ε > 0, there exists δ > 0 such that γ (B) = B f dλ < ε, for every B ∈ L(A) with λ(B) < δ. (2) For every ε > 0, there exists A0 ∈ L(A) with λ(A0 ) < +∞ such that γ (A \ & A0 ) = A\A0 f dλ < ε.

Proof L(A) = {B ∈ L : B ⊆ A} is a σ -algebra on A. We will show that γ verifies the conditions of Definition 1.4.5: & γ (∅) = ∅ f dλ = 0 because λ(∅) = 0.  Let (Bn )n ⊆ L(A), Bn ∩ Bm = ∅, for all n = m ,and let B = ∞ n=1 Bn ∈ L(A); by applying the Beppo Levi theorem (see Corollary 3.1.11), % γ (B) =

f dλ = B

%  ∞ B

f χBn

∞ % 

dλ =

f dλ =

n=1 Bn

n=1

∞ 

γ (Bn ).

n=1

& Therefore γ is a measure on A, and, since γ (A) = A f dλ < +∞, γ is finite. (1) For any n ∈ N, let An = (f > n) ∈ L(A); then An ⊇ An+1 , for any n ∈ N and ∞ we remark that f χA \ An ↑ f and, according to n=0 An = (f = +∞) = ∅. Then & & monotone convergence theorem, A\An f dλ ↑ A f dλ. Since f = f χAn + f χA \ An , & it follows that An f dλ ↓ 0. Then, for every ε > 0, there exists n0 ∈ N such that & ε An0 f dλ < 2 . Let δ = 2nε 0 > 0 and let B ∈ L(A) be an arbitrary set with λ(B) < δ; then %

%

γ (B) = B


0 such that |f (x)| ≤ k, for every x ∈ A. The integral of the constant function & k ∈ E+ (A) is A kdλ = k · λ(A) < +∞. Therefore k ∈ E+1 (A) ⊆ L1+ (A). Theorem 3.2.6 assures us that f ∈ L1 (A). Corollary 3.2.9 Any function Riemann integrable over a bounded and closed interval [a, b] is Lebesgue integrable over [a, b]: R[a,b]  L1 ([a, b]). Proof Any Riemann integrable function over [a, b] is bounded and measurable (see Corollary 2.1.14); therefore, according to the previous corollary, it is Lebesgue integrable.

We noted in Remark 3.1.15 that the Dirichlet function is Lebesgue integrable on any closed interval [a, b] but it is not Riemann integrable on [a, b]. Therefore, the inclusion in the previous corollary is strict. As we will show later in Theorem 3.3.11, R[a,b] is a dense subspace of L1 ([a, b]). Theorem 3.2.10 For every A ∈ L, L1 (A) is a real vector space, and the mapping I : L1 (A) → & R, I (f ) = A f dλ, is a linear operator: & & & (1) A (f + g)dλ = A f dλ + A gdλ, for every f, g ∈ L1 (A); (2)

&

A cf dλ

=c

&

A f dλ,

for every f ∈ L1 (A), and every c ∈ R.

Proof To show that L1 (A) is a vector space, it was enough to show that it is closed for sum and multiplication with scalars (L1 (A) is a subset of the vector space of L(A)). Let then f, g ∈ L1 (A); from Theorem 3.2.2 it follows that |f |, |g| ∈ L1+ (A), and so & & & Corollary 3.1.9 tells us that h = |f |+|g| ∈ L1+ (A) ( A hdλ = A |f |dλ+ A |g|dλ < +∞). On the other hand, |f + g| ≤ |f | + |g| = h, and then Theorem 3.2.6 assures us that f + g is integrable. From the relation (f + − f − ) + (g + − g − ) = f + g = (f + g)+ − (f + g)−

3

72

Chapter 3 • Lebesgue Integral

we obtain that

3

(f + g)+ + f − + g − = (f + g)− + f + + g + . By integrating the previous equality (we observe that all the intervening functions are integrable and positive) and by using again Corollary 3.1.9, we obtain %

(f + g)+ dλ +

%

A

f − dλ +

%

A

g − dλ = A

%

(f + g)− dλ +

A

%

f + dλ + A

%

g + dλ,

A

hence, all the terms being finite, %

%

%

(f + g)dλ =

f dλ +

A

A

gdλ. A

Let now f ∈ L1 (A) and let c ∈ R; then |c · f | = |c| · |f | ∈ L1+ (A), from where c · f ∈ L1 (A) and %

%

(c · f )+ dλ −

(c · f )dλ = A

%

A

(c · f )− dλ. A

If c > 0, then (c · f )+ = c · f + and (c · f )− = c · f − , from where, using (1) of Proposition 3.1.6, %

% (c · f )dλ = c A

A

f + dλ − c

%

f − dλ = c A

% f dλ. A

If c < 0, then the proof is similar by noting that (c · f )+ = −c · f − and (c · f )− = −c · f + . 

Theorem 3.2.11 Let A ∈ L, λ(A) > 0 and let f, g ∈ L(A). & (1) If f ≥ 0 a.e., then f has an integral over A and A f dλ ≥ 0. & & (2) If f ≤ g a.e. and f, g have integrals, then A f dλ ≤ A gdλ. (3) If f = g, a.e. and if f has an integral over A, then g has also an integral over A & & and A f dλ = A gdλ; f ∈ L1 (A) if and only if g ∈ L1 (A).

Proof & (1) If f ≥ 0 a.e., then f − = 0 a.e., and so A f − dλ = 0 (see 1) of Corollary 3.2.3). & & Therefore f has an integral, and A f dλ = A f + dλ ≥ 0. & & (2) If f ≤ g a.e., then f + ≤ g + and f − ≥ g − a.e. on A. Then A f dλ = A f + dλ − & − & + & − & A f dλ ≤ A g dλ − A g dλ = gdλ.

73 3.3 · The Space of Integrable Functions

& & (3) Since f has an integral over A, A f + dλ < +∞ or A f − dλ < +∞. Let us suppose & + that A f dλ < +∞; then, since f + − g + = 0 a.e., f + − g + ∈ L1 (A) and & & + & + + + A (f −g )dλ = 0 (see Corollary 3.2.3). It follows that A g dλ = A f dλ < +∞; therefore g has an integral. & & & Similarly, if A f − dλ < +∞, then A g − dλ = A f − dλ. & & According to (2), it is obvious that A f dλ = A gdλ.  Remark 3.2.12 Let f ∈ L1 (A) and let B ∈ L(A) such that λ(A \ B) = 0. Then f = f · χB & & a.e. and so A f dλ = B f dλ.

3.3

The Space of Integrable Functions

In this section, we present some important properties of the integral (absolute continuity, countable additivity with respect to the integration domain, etc.). We also show that the space of integrable functions is organized as a complete seminormed space.

Theorem 3.3.1 Let A ∈ L and f, g ∈ L1 (A); then & & (1) A f dλ ≤ A gdλ, if f ≤ g a.e. & (2) For every ε > 0, there exists δ > 0, such that B |f |dλ < ε, for every B ∈ L(A) with λ(B) < δ. (3) For every ε > 0, there exists A0 ∈ L(A) with λ(A0 ) < +∞, such that & A\A0 |f |dλ < ε; % ∞ %  f dλ = f dλ, for every (An )n ⊆ L(A), An ∩ Am = ∅, for (4)  ∞ n=1 An

n = m.

n=1 An

Proof (1) If f ≤ g a.e., then from (1) of Theorem 3.2.5 and from Theorem 3.2.10 it follows that & & & g − f ∈ L1+ (A) and A gdλ − A f = A (g − f )dλ ≥ 0, hence the inequality required. (2) and (3) are immediate consequences of Theorem 3.1.16.  (4) Apply Corollary 3.1.11 to positive functions f + and f − . Remarks 3.3.2 (i) Property (2) is the property of absolute continuity of the Lebesgue integral. In fact, this property says that the signed measure γ : L(A) → R, defined by γ (B) = & B f dλ, is absolutely continuous with respect to the Lebesgue measure λ, γ  λ (see Definition 6.2.1 and Proposition 6.2.3). (ii) Property (3) shows that, for an integrable function, the integral depends on the behavior of this function on the sets of finite measures.

3

74

3

Chapter 3 • Lebesgue Integral

Theorem 3.3.3 Let A ∈ L and  · 1 : L1 (A) → R+ , defined by & f 1 = A |f |dλ, for every f ∈ L1 (A); then (1) f 1 = 0 if and only if f = 0 a.e., (2) cf 1 = |c| · f 1 , for every f ∈ L1 (A) and any c ∈ R, (3) f + g1 ≤ f 1 + g1 , for every f, g ∈ L1 (A).

Proof & (1) f 1 = A |f |dλ = 0 if and only if |f | = 0 a.e. (see Proposition 3.1.13), and this happens if and only if f = 0 a.e. (2) is the consequence of (1) of Proposition 3.1.6. (3) Using (2) of Proposition 3.1.6 and Corollary 3.1.9, we obtain: %

%

f + g1 =

%

|f + g|dλ ≤ A

|f |dλ + A

|g|dλ = f 1 + g1 .



A

Remark 3.3.4 It follows from the previous theorem that  · 1 is a seminorm on the vector space L1 (A). ·1 is not a norm on L1 (A) because there are positive functions whose integral is zero and which are not zero a.e. (see Remark 3.1.15). · The relation = is an equivalence relation on L1 (A) (it is reflexive, symmetric, and transitive). The quotient space, L1 (A)|=· , is L1 (A) = {[f ] : f ∈ L1 (A)}, where [f ] = ·

{g ∈ L1 (A) : f = g} is the equivalence class to which belongs f . Since the Lebesgue integral is the same for two functions equal almost everywhere (see Theorem 3.2.11), we can define coherently [f ]1 = f 1 , for every [f ] ∈ L1 (A). The application thus defined is a norm on L1 (A).

Definition 3.3.5 A sequence (fn ) ⊆ L1 (A) is L1 -convergent to f ∈ L1 (A) if fn − f 1 → 0 (for every ε > 0, there exists n0 ∈ N such that, for any n ≥ n0 , fn − f 1 = & ·1 → f . The L1 -seminorm  · 1 A |fn − f |dλ < ε). We denote this situation by fn −− A

is said to be the seminorm of L1 -convergence on L1 (A). A sequence (fn ) ⊆ L1 (A) is L1 -Cauchy if, for every ε > 0, there is n0 ∈ N, such that, for any m, n ≥ n0 , fm − fn 1 < ε. If F ⊆ L1 (A), then f ∈ L1 (A) is a L1 -adherent point for F if there exists (fn ) ⊆ F ·1

1

such that fn −−→ f ; we denote f ∈ F . A

3

75 3.3 · The Space of Integrable Functions

Theorem 3.3.6 Let (fn ) ⊆ L1 (A) and f ∈ L1 (A). ·1

λ

A

A

1). If fn −−→ f , then fn − → f. 2). If (fn )n is L1 -Cauchy, then (fn )n is Cauchy in measure.

Proof (1) If (fn )n ⊆ L1 (A) is L1 -convergent to f ∈ L1 (A), then, for every ε > 0, fn − & & f 1 = A |fn − f |dλ ≥ (|fn −f |≥ε) |fn − f |dλ ≥ ε · λ(|fn − f | ≥ ε), from where λ(|fn − f | ≥ ε) ≤

1 ε

· fn − f 1 and then limn λ(|fn − f | ≥ ε) = 0. ε being positive

λ

arbitrary, fn − → f. A

(2) Similarly, we notice that, for all ε > 0, λ(|fn − fm | ≥ ε) ≤ 1ε · fn − fm 1 . If (fn )n is L1 -Cauchy, then fn − fm 1 −−−−−→ 0, and so limn,m λ(|fn − fm | ≥ ε) = 0, for m,n→∞

every ε > 0.

 Remark 3.3.7 The converses of the previous theorem are not true. For example, the sequence (fn )n , fn = n · χ 1 ∈ L1 ([0, 1]), is convergent in measure to 0 (why?) but is not L1 [0, n ]

convergent (why?).

Theorem 3.3.6 allows us to show that the seminormed space (L1 (A),  · 1 ) is complete. Theorem 3.3.8 The seminormed space (L1 (A),  · 1 ) is complete (every L1 -Cauchy sequence is L1 convergent).

Proof From (2) of Theorem 3.3.6, it follows that any sequence L1 -Cauchy, (fn )n ⊆ L1 (A), is Cauchy in measure. From (1) of Riesz theorem (Theorem 2.2.5), there exists a subsequence (fkn )n of (fn )n almost uniform convergent to a function f : A → R. Point (2) of · Theorem 2.2.2 tells us that fkn − → f and then f ∈ L(A) (see 3) of Theorem 2.1.13). A

Let’s fix a m ∈ N; for any n ∈ N, let gn = |fm −fkn | ∈ L+ (A). We apply to the sequence (gn )n Fatou’s lemma (Corollary 3.1.12): %

% lim inf gn dλ ≤ lim inf A

n

n

gn dλ. A

76

Chapter 3 • Lebesgue Integral

Since lim infn gn = |fm − f | a.e., it follows that %

3

A

|fm − f |dλ ≤ lim inf fm − fkn 1 n

and, since limm,n→+∞ fm − fkn 1 = 0, it results, on the one hand, that fm − f ∈ L1 (A) and then f = fm − (fm − f ) ∈ L1 (A) and, on the other hand, that fm − f 1 → 0.

The following result shows a very practical condition to pass to the limit under the Lebesgue integral.

Theorem 3.3.9 (Dominated Convergence Theorem) Let (fn ) ⊆ L(A) and g ∈ L1 (A) such that · (1) fn − → f and A

(2) |fn | ≤ g, a.e., for any n ∈ N. ·1

%

Then (fn ) ⊆ L1 (A), f ∈ L1 (A), fn −−→ f , and A

% fn dλ →

A

f dλ. A

Proof From Theorem 3.2.6, it follows that (fn )n ⊆ L1 (A); (3) of Theorem 2.1.13 assures us that f ∈ L(A). If we go to the limit in inequality (2), we get |f | ≤ g a.e.; again Theorem 3.2.6 leads us to f ∈ L1 (A). Now we remark that |fn − f | ≤ |fn | + |f | ≤ 2g. If hn = 2g − |fn − f |, then (hn )n ⊆ L+ (A). Then, we apply Fatou’s lemma (Corollary 3.1.12) to the sequence (hn )n : %

% lim inf hn dλ ≤ lim inf n

A

n

hn dλ. A

Because lim infn hn = 2g a.e., we obtain %

 %  gdλ + lim inf − |fn − f |dλ =

% gdλ ≤ 2

2 A

n

A

A

%

% gdλ − lim sup

=2 A

n

|fn − f |dλ, A

from where lim supn fn − f 1 ≤ 0. It follows that lim supn fn − f 1 = 0; therefore, there is limn fn$%− f 1 = 0.% $ % $ $ & $ Since $ fn dλ − f dλ$$ ≤ |fn − f |dλ = fn − f 1 , it results that A fn dλ → A A A &  A f dλ.

77 3.3 · The Space of Integrable Functions

Corollary 3.3.10 (Bounded Convergence Theorem) Let A ∈ L with λ(A) < +∞, let (fn ) ⊆ L(A) and let c ∈ R+ such that · → f and (1) fn − A

(2) |fn | ≤ c, for any n ∈ N. ·1

Then (fn ) ⊆ L1 (A), f ∈ L1 (A), fn −−→ f and

&

A

fn dλ →

A

&

A f dλ.

Proof The theorem results from the dominated convergence theorem if we notice that, on sets of finite measure, the constant functions are integrable (see Corollary 3.2.8); then we can take g = c. 

In what follows, we highlight two dense subsets in L1 (A). Theorem 3.3.11 Let E 1 (A) = E (A) ∩ L1 (A) be the set of all integrable simple functions on A, and let C1 (A) = C(A) ∩ L1 (A) be the set of all integrable continuous functions on A; then 1

(1) E 1 (A) = L1 (A) and 1

(2) C1 (A) = L1 (A).

Proof 1 1 By definition E 1 (A) ⊆ L1 (A) and C1 (A) ⊆ L1 (A). We have to show the reverse inclusions. (1) For every f ∈ L1 (A), f = f + − f − and f + , f − ∈ L1+ (A). Taking into account (1) of approximation theorem of the measurable functions (see Theorem 2.3.3), there exist two sequences (gn )n , (hn )n ⊆ E+ (A) such that gn ↑ f + and hn ↑ f − . Then fn = gn − hn → f and (fn )n ⊆ E (A). For any n ∈ N, |fn | ≤ gn + hn ≤ f + + f − = |f | ∈ L1+ (A). From the dominated convergence theorem (Theorem 3.3.9), (fn )n ⊆ L1 (A); then (fn )n ⊆ E (A) ∩ L1 (A) = 1

E 1 (A), and, more, fn − f 1 → 0. It follows that f ∈ E 1 (A) . 1 (2) We will first show that E 1 (A) ⊆ C1 (A) .

p For every f = i=1 ai · χA ∈ E 1 (A), let M = supx∈A |f (x)| = i = max{|a1 |, · · · , |ap |}. From Borel theorem (Theorem 2.3.10), for every ε > 0, there exists ε fε ∈ C(A) such that λ(f = fε ) < 2M and supx∈A |fε (x)| ≤ M. Let B = (f = fε ) ∈ L(A). Then fε = fε · χB + fε · χA \ B = fε · χB + f · χA \ B , and so |fε | ≤ M · χB + |f | ∈ 1 (A) ⊆ L1 (A). From Theorem 3.2.6, f ∈ L1 (A) and then f ∈ C (A). Moreover E+ ε ε 1

%

%

f − fε 1 =

|f − fε |dλ = A

% |f − fε |dλ ≤

B

(|f | + |fε |)dλ ≤ 2M · λ(B) < ε. B

3

78

Chapter 3 • Lebesgue Integral

If ε takes the values n1 , n ∈ N∗ , one by one, we can find the sequence (fn )n ⊆ C1 (A) such that f − fn 1
0 such that g is L-Lipschitz, that is to say: |g(x) − g(y)| ≤ L · |x − y|, for every x, y ∈ A. g is locally Lipschitz if for each x ∈ R, there exists an open set D ⊆ R such that x ∈ D and g is Lipschitz on D (for a certain constant L which can depend on x).

Remark 3.5.2 (i) Every locally Lipschitz function g : R → R is continuous on R. Indeed, for every x ∈ R and every (xn )n ⊆ R, xn → x, there exists an open set D ⊆ R and L > 0, such that x ∈ D and g is L-Lipschitz on D. Let n0 ∈ N such that xn ∈ D, for any n ≥ n0 . Then |g(xn ) − g(x)| ≤ L · |xn − x|, for any n ≥ n0 , from where g(xn ) → g(x). (ii) Let g : R → R be a function differentiable on an interval I ⊆ R with the bounded derivative on I ; then g is Lipschitz on I . Indeed, let L = supx∈I |g (x)|; then g is L-Lipschitz. Proposition 3.5.3 The function g : R → R is locally Lipschitz on R if and only if g is Lipschitz on every compact subset of R. Proof The sufficiency of the condition is obvious. The necessity. Suppose, reducing to the absurd, that there is a compact K ⊆ R, so that g is not Lipschitz on K. Then, for any n ∈ N, there exist xn , yn ∈ K such that |g(xn )−g(yn )| > n · |xn − yn |. Since K is compact, possibly passing to a subsequence, we can assume that xn → x, yn → y, where x, y ∈ K. If x = y, then there exist an open set D ⊆ R and L > 0 such that x ∈ D and g is L-Lipschitz on D; since xn , yn → x, there exists n0 ∈ N such that xn , yn ∈ D, for any n ≥ n0 . Then L · |xn − yn | ≥ |g(xn ) − g(yn )| > n · |xn − yn |, for any n ≥ n0 , which is absurd (xn = yn , for any n ∈ N). If x = y, |x − y| > 0. Because |xn − yn | → |x − y|, there exists n0 ∈ N such that |xn −yn | > 12 ·|x−y|, for any n ≥ n0 . Since g is continuous on R (see Remark 3.5.2) and since K is compact, g(K) is also compact, and so it is bounded. Let d = supu,v∈K |g(u) − g(v)| < +∞. Then n · |x − y| < n · |xn − yn | < |g(xn ) − g(yn )| ≤ d, for any n ≥ n0 , 2 which is also absurd.



83 3.5 · Properties of the Lebesgue Integral

Remarks 3.5.4 (i) The function g : R → R is said to be of class C 1 on R if g is differentiable and the derivative g is continuous on R; it is denoted by g ∈ C 1 (R). If g is of class C 1 on R, then g is locally Lipschitz on R. Indeed, for every compact K ⊆ R, there exists an interval compact I such that K ⊆ I . The derivative of g is continuous on I , hence it is bounded. According to (ii) of Remark 3.5.2, g is Lipschitz on I , and then it is Lipschitz on K. (ii) Every locally Lipschitz function on R is Lipschitz on every bounded subset of R.

Theorem 3.5.5 Let g : R → R be a locally Lipschitz function. (a) For every null set N ⊆ R, g(N) is a null set. (b) For every A ∈ L, g(A) ∈ L.

Proof (a) First, let us suppose that N is a bounded null set, and let a ∈ R∗+ such that N ⊆]−a, a[⊆ [−a, a]. The function g is Lipschitz on [−a, a]; let L > 0 such that |g(x) − g(y)| ≤ L · |x − y|, for every x, y ∈ [−a, a]. Since N is a null set, for every ε > 0, there exists a sequence of open intervals  ∞ ε (]ap , bp [)p such that N ⊆ ∞ p=1 ]ap , bp [⊆ [−a, a] and p=1 (bp − ap ) < 2L . For any p ≥ 1 and for every x, y ∈]ap , bp [, |g(x) − g(y)| ≤ L · |x − y| ≤ L · (bp − ap ); therefore g(]ap , bp [) is bounded. Then, for any p ∈ N∗ , there exists an open interval ]cp , dp [ such that g(]ap , bp [) ⊆]cp , dp [ and dp − cp ≤ 2L · (bp − ap ).  ∞ Consequently, g(N) ⊆ ∞ p=1 ]cp , dp [ and p=1 (dp − cp ) < ε. Therefore g(N) is a null set. If N is a unbounded null set, then we can represent it as a countable union of ∞ bounded null sets; for example, N = n=1 Nn , where Nn = N ∩ [−n, n]. Then ∞ g(N) = n=1 g(Nn ), which shows that g(N) is a countable union of null sets, so it is a null set. (b) As in the previous point, we first assume that A ∈ L is bounded. Using Corollary 1.3.9, for all n ∈ N∗ , we can find a closed set Fn ⊆ A such that λ(A \ Fn ) < n1 . Let B = ∞ ∞ 1 ∗ n=1 Fn ∈ L. Then λ(A \ B) = λ( n=1 (A \ Fn )) ≤ λ(A \ Fn ) < n , for any n ∈ N . Therefore N = A \ B is a null set and, it is obvious that A = B ∪ N. According to Remark 3.5.2, g is continuous on R, and, since every Fn is compact,  g(Fn ) is also compact, for any n ∈ N∗ . Therefore ∞ n=1 g(Fn ) ∈ Bu ⊆ L (see Definition 1.3.18). As stated by a), g(N) is a null set, so that g(N) ∈ L. Then  g(A) = g(B) ∪ g(N) = ∞ n=1 g(Fn ) ∪ g(N) ∈ L. If A ∈ L is unbounded, we can represent it as a countable union of bounded  ∗ measurable sets: A = ∞ n=1 An , where, for any n ∈ N , An = A ∩ [−n, n]. Then ∞ g(A) = n=1 g(An ); every An is measurable and bounded so that g(An ) ∈ L,

3

84

3

Chapter 3 • Lebesgue Integral

for any n ∈ N∗ . Therefore g(A) is a countable union of mesurable sets, so it is measurable. 

We will now present a change of variable theorem for the Lebesgue integral.

Theorem 3.5.6 Let g : R → R be an injective function of class C 1 (g is differentiable with continuous derivative on R); then, for every A ∈ L, g(A) ∈ L and %

|g |dλ.

λ(g(A)) =

(*)

A

For every Borel function f : R → R and for every set A ∈ L, f ∈ L1 (g(A)) if and only if (f ◦ g) · |g | ∈ L1 (A) and then %

%

(f ◦ g) · |g |dλ.

f dλ = g(A)

(**)

A

Proof Let g : R → R be an injection of class C 1 . We will first note that any continuous injection on R is strictly monotonic. Let us suppose that g is strictly increasing (the demonstration is similar if we assume that g is strictly decreasing). Then g (x) ≥ 0, for every x ∈ R. According to (i) of Remark 3.5.4, g is locally Lipschitz. Let A ∈ L; point b) of the previous theorem assures us that g(A) ∈ L. Let’s calculate λ(g(A)). Because we assumed that g is an increasing function, for every open interval I =]a, b[⊆ R, g(I ) =]g(a), g(b)[ is still an open interval and %

b

λ(g(I )) = g(b) − g(a) =

g (x)dx =

%

g dλ = I

a

%

|g |dλ.

I

 Let now D ⊆ R be an open set, and let D = ∞ n=1 In be the representation of D (see Theorem 1.1.3); then, since the open intervals g(In ) are pairwise disjoint (g is injective), λ(g(D)) = λ

∞  n=1

g(In ) =

∞  n=1

λ(g(In )) =

∞ %  n=1 In

|g |dλ =

%

|g |dλ D

(at the last above equality, we used Corollary 3.1.11). Let A ∈ L be a bounded measurable set. For any n ∈ N∗ , let Dn ∈ τu such that A ⊆ Dn and λ(Dn \ A) < n1 . It is obvious that we can assume that all the open sets Dn are bounded and that Dn+1 ⊆ Dn .

3

85 3.5 · Properties of the Lebesgue Integral

 Let B = ∞ n=1 Dn ; then N = B \ A is a null set and A = B \ N. Then g(A) = g(B) \ g(N), and, since N is a null set, λ(g(A)) = λ(g(B)) (see a) of previous theorem). The function g carries bounded sets into bounded sets (g is strictly increasing). Then g(B) = ∞ n=1 g(Dn ), g(Dn+1 ) ⊆ g(Dn ) and λ(g(Dn )) < +∞. The measure λ is continuous from above (see 7) of Theorem 1.3.11); therefore λ(g(B)) = limn λ(g(Dn )). Hence, %

%

λ(g(A)) = λ(g(B)) = lim λ(g(Dn )) = lim n

n

%

|g |dλ = Dn

|g |dλ.

|g |dλ = B

A

In the penultimate equality, we took into account the fact that the mapping A → γ (A) = &

A |g |dλ is a finite measure on L(R) (see Theorem 3.1.16) and this measure is continuous from above (see 7) of Theorem 1.4.7); the last equality is the consequence of Remark 3.2.12. If A ∈ L is unbounded, then it is expressed as an union of an increasing sequence of  measurable bounded sets: A = ∞ n=1 An , where An = A ∩ [−n, n], for any n ∈ N. Then ∞ g(A) = n=1 g(An ), and, using the continuity property from below of measure (the property 6) of Theorem 1.3.11) and the monotone convergence theorem (Theorem 3.1.7), %

|g |dλ = lim

λ(g(A)) = lim λ(g(An )) = lim n

n

%

n

An

A

χAn · |g |dλ =

%

|g |dλ. A

Now consider a Borel function f : R → R and a set A ∈ L; as demonstrated above, g(A) ∈ L and the relation (∗) is verified. Moreover, according to Theorem 2.1.15, f ◦ g ∈ L(A) and then (f ◦ g) · |g | ∈ L(A) (see Theorem 2.1.20). We will prove the theorem for different situations in which the function f can be found. (1) Suppose that f = χB , where B ∈ Bu . Then, using the relation (∗), %

(f ◦ g) · |g |dλ =

% χ

A

g−1 (B)

A

= λ(g(A ∩ g

−1

· |g |dλ =

% A∩g −1 (B)

|g |dλ =

% (B)) = λ(g(A) ∩ B) =

f dλ. g(A)

We can remark that f ∈ L1 (g(A)) if and only if λ(g(A) ∩ B) < +∞ and that this last inequality occurs if and only if (f ◦ g) · |g | ∈ L1 (A). The above equality is precisely the relation (∗∗) in this case. p (2) Let now f = i=1 ai · χB be a positive Borel simple function; then ai ≥ 0 and i

Bi ∈ Bu , for any i = 1, · · · , p. For any i = 1, · · · , p, χB ∈ L1 (g(A)) if and only if i

(χB ◦ g) · |g | ∈ L1 (A) and i

% f dλ = g(A)

=

p  i=1

p 

% ai ·

A

i

i=1

g(A)

χ

· |g |dλ =

% ai ·

χB dλ =

g−1 (Bi )

p 

% ai · A

i=1

%  p A i=1

ai · χ

(χB ◦ g) · |g |dλ = i

g−1 (Bi )

· |g |dλ =

% A

(f ◦ g) · |g |dλ.

86

3

Chapter 3 • Lebesgue Integral

(3) Let f be a positive Borel function; according to Remark 2.3.4, there exists an increasing sequence of positive Borel simple functions (fn )n such that fn ↑ f . Using the monotone convergence theorem (Theorem 3.1.7), we obtain %

%

%

f dλ = lim g(A)

n

fn dλ = lim g(A)

n

(fn ◦ g) · |g |dλ =

A

%

(f ◦ g) · |g |dλ. A

(4) Finally, if f is any Borel function, then f = f + − f − , where f + and f − are positive Borel functions. The proof results from the simple observation that (f ◦ g)+ = (f + ◦g) and (f ◦ g)− = (f − ◦ g). 

3.5.2 Integrals Depending on a Parameter When we integrate a function which depends on a parameter, the problems which arise concern the study of the integral depending on the mentioned parameter. Let A ∈ L(R), let T ⊆ R and let f : T × A → &R such that f (t, ·) ∈ L1 (A), for every t ∈ T . Then, the function F : T → R, F (t) = A f (t, x)dλ(x), is well defined; F is said to be an integral depending on a parameter. We want to specify the conditions under which the function F is continuous or differentiable on T . Theorem 3.5.7 With the above notations, suppose that there is a positive function g ∈ L1 (A) such that |f (t, x)| ≤ g(x), for every t ∈ T and for almost every x ∈ A. Let t0 be a limit point of T (there exists a sequence (tn )n ⊆ T , tn = t0 , tn → t0 ). (1) If it exists limt→t0 f (t, x) = ϕ(x), for almost every x ∈ A, then ϕ ∈ L1 (A) and & & limt→t0 A f (t, x)dλ(x) = A ϕ(x)dλ(x). (2) If f (·, x) is continuous at t0 ∈ T , for almost every x ∈ A, then F = & A f (·, x)dλ(x) is continuous at t0 . (3) If f (·, x) is continuous on T , for almost every x ∈ A, then F is continuous on T .

Proof · (1) Let (tn )n ⊆ T , tn → t0 , tn = t0 , ∀n ∈ N; then f (tn , ·) − → ϕ. According to (7) A

of Theorem 2.1.18, ϕ ∈ L(A). Since |f (tn , ·)| ≤ g almost everywhere, then |ϕ| ≤ g almost everywhere. Theorem 3.2.6 assures us that ϕ ∈ L1 (A). In conformity with the & & dominated convergence theorem (3.3.9), A f (tn , x)dλ(x) → A ϕdλ. The sequence (tn )n & & being arbitrary, there exists limt→t0 A f (t, x)dλ(x) = A ϕ(x)dλ(x). (2) and (3) are consequences of (1). 

3

87 3.5 · Properties of the Lebesgue Integral

Theorem 3.5.8 Let A ∈ L, let T ⊆ R be an open interval, and let f : T × A → R, such that f (t, ·) ∈ L1 (A), for every t ∈ T and g ∈ L1 (A). We suppose that ∂f (a) There exists (t, x), for every t ∈ T and almost for every x ∈ A, and $ $ ∂t $ ∂f $ (b) $$ (t, x)$$ ≤ g(x), for every t ∈ T and for almost every x ∈ A. ∂t Then ∂f (1) (t, ·) ∈ L1 (A), for every t ∈ T , ∂t & (2) F : T → R, F (t) = A f (t, x)dλ(x), is differentiable on T and F (t) =

% A

∂f (t, x)dλ(x), for every t ∈ T . ∂t

Proof Let t ∈ T be an arbitrary but fixed point, and let a > 0 such that ]t − a, t + a[⊆ T . The function r : (] − a, a[\{0}) × A → R, r(h, x) = h1 · [f (t + h, x) − f (t, x)], verifies the hypotheses (1) of the previous theorem. Indeed, for every h ∈] − a, a[ and every x ∈ A, there ∂f exists s between t and t + h such that f (t + h, x) − f (t, x) = (s, x) · h and then, for ∂t every h ∈] − a, a[\{0} and for almost every x ∈ A, $ $ $ ∂f $ 1 $ · |f (t + h, x) − f (t, x)| ≤ $ (s, x)$$ ≤ g(x). |r(h, x)| = |h| ∂t Moreover, for almost every x ∈ A, there exists lim r(h, x) = h→0

Then lim r(h, ·) = h→0

∂f (t, ·) ∈ L1 (A) and ∂t

% lim

h→0 A

% r(h, x)dλ(x) = A

∂f (t, x). ∂t

∂f (t, x)dλ(x). ∂t

The proof ends if we notice that % r(h, x)dλ(x) = A

1 [F (t + h) − F (t)]. h



3.5.3 Jensen’s Inequality Let J ⊆ R be an interval and let f : J → R; f is called convex on J if, for every x, y ∈ J and every t ∈]0, 1[ we have f (tx + (1 − t)y) ≤ tf (x) + (1 − t)f (y). We will need some useful lemmas.

88

Chapter 3 • Lebesgue Integral

Lemma 3.5.9 f : J → R is convex on J if and only if, for every x ∈ J , the function gx : J \ {x} → R, defined by

3

gx (y) =

f (y) − f (x) , for every y ∈ J, y −x

is increasing on J \ {x}. Proof Let us suppose that f is convex on J and let an arbitrary x ∈ J . Every two points y, z ∈ J \ {x} with y < z can be found in one of the situations: x−z x−z If y < z < x, then t = x−y ∈]0, 1[ and z = ty + (1 − t)x. Hence f (z) ≤ x−y · f (y) + z−y x−y

· f (x). In the two members of the inequality, we multiply by x − y and add xf (x); after some calculations, we obtain gx (y) ≤ gx (z). In the cases y < x < z and x < y < z, the procedure is similar, getting every time gx (y) ≤ gx (z). Conversely, for every x, y ∈ J and t ∈ (0, 1), let z = tx + (1 − t)y; if x < y, then x < f (x) − f (z) f (y) − f (z) z < y. Since gz is increasing, gz (x) ≤ gz (y), from where ≤ . x −z y −z Then we obtain f (tx + (1 − t)y) = f (z) ≤

y −z z−x · f (x) + · f (u) = tf (x) + (1 − t)f (y). y−x y −x



Lemma 3.5.10 f : J → R is convex on J if and only if, for every interior point x0 of the interval J , there exist a, b ∈ R such that f (x) ≥ ax+b, for every x ∈ J and f (x0 ) = ax0 +b. Proof Let the function f : J → R be convex on J , and let x0 be an interior point of J ; according (x0 ) to the previous lemma, the function x → f (x)−f is increasing on J \ {x0 }, and then there x−x0 exists   f (x) − f (x0 ) f (x) − f (x0 ) a = lim = sup : x ∈ J, x < x0 . x↑x0 x − x0 x − x0 (x0 ) ≥ a, It follows that, for every x < x0 , f (x) ≥ ax +(f (x0 )−ax0 ). For all x > x0 , f (x)−f x−x0 and then f (x) ≥ ax + (f (x0 ) − ax0 ). Necessity results if we take b = f (x0 ) − ax0 . Conversely, for every x, y ∈ J and every t ∈ (0, 1), x0 = tx +(1−t)y is an interior point of J . Then there exist a, b ∈ R such that f (z) ≥ az + b, for any z ∈ J and f (x0 ) = ax0 + b. Therefore tf (x) + (1 − t)f (y) ≥ t (ax0 + b) + (1 − t)(ax0 + b) = ax0 + b = f (x0 ) = f (tx + (1 − t)y). 

Lemma 3.5.11 Let a convex function f : J → R; then f is continuous on the interval I formed by the interior points of the interval J .

89 3.5 · Properties of the Lebesgue Integral

Proof If J ∈ I , then I = J ; generally card(J \ I ) ≤ 2. Let x ∈ I and let y, z ∈ I , such that y < x < z. For every (xn )n ⊆ J , xn ↑ x, we can consider without loss of generality that y < xn < x < z, for all n ∈ N. According to (x) Lemma 3.5.9, gx , defined by gx (u) = f (u)−f , is increasing on J \ {x}, and then gx (y) ≤ u−x gx (xn ) ≤ gx (z). If we multiply the above inequalities by xn − x < 0, we obtain xn − x xn − x ·[f (y)−f (x)] ≥ f (xn )−f (x) ≥ ·[f (z)−f (x)], for any n ∈ N. Passing y −x z−x to the limit in the above inequalities, we obtain f (xn ) → f (x). Therefore limu↑x f (u) = f (x). Similarly, it is shown that limu↓x f (u) = f (x); therefore f is continuous at x, for every x ∈ I.  Remark 3.5.12 Let J ∈ J and let f : J → R be a convex function; then f is a Borel function on J , and so it is measurable on J . Indeed, if I is the interval of all interior points of J and if a ∈ R, then f −1 (] − ∞, a[) = [f −1 (] − ∞, a[) ∩ I ] ∪ [f −1 (] − ∞, a[) ∩ (J \ I )]. According to the previous lemma, f is continuous on I ; hence f −1 (]−∞, a[)∩I ∈ τu ⊆ Bu ; since card(f −1 (] − ∞, a[) ∩ (J \ I )) ≤ 2, f −1 (] − ∞, a[) ∩ (J \ I ) ∈ Bu , and then f −1 (] − ∞, a[) ∈ Bu .

Theorem 3.5.13 (Jensen’s Inequality) Let J ⊆ R be an interval, let F : J → R be a convex function, let A ∈ L with 0 < λ(A) < +∞, and let f : A → J , f ∈ L1 (A). Then F ◦ f has an integral on A and  F

1 · λ(A)

%

 f dλ ≤

A

1 · λ(A)

% (F ◦ f )dλ. A

Proof According to Lemma 3.5.10, for every y0 ∈ J there exist a, b ∈ R such that F (y) ≥ ay + b, & 1 for every y ∈ J and F (y0 ) = ay0 + b. We will take y0 = λ(A) · A f dλ ∈ I (see Exercise 14) of 3.7). For every x ∈ A, y = f (x) ∈ J and then F (f (x)) ≥ af (x) + b. Since F is a Borel function (see Remark 3.5.12), Theorem 2.1.15 assures us that F ◦ f ∈ L(A). Since λ(A) < +∞, the function g : A → R, g(x) = af (x) + b, is integrable on A, & and, since F ◦ f ≥ g, (F ◦ f )− ≤ g − . Then A (F ◦ f )− dλ < +∞. It follows that F ◦ f

3

90

3

Chapter 3 • Lebesgue Integral

has an integral on A (see Definition 3.2.1). According to (2) of Theorem 3.2.11, % % (F ◦ f )dλ ≥ a f dλ + bλ(A) = ay0 λ(A) + bλ(A) = F (y0 ) · λ(A), A

A



which is exactly the inequality to be demonstrated. Corollary 3.5.14 Under the conditions of the previous theorem, if λ(A) = 1, then  % f dλ ≤ (F ◦ f )dλ.

% F A

3.6

A

Abstract Setting

Let (X, A) be a measurable space, and let γ be a σ -finite measure on A. We denote with E (X) the set of all A-simple functions, with E+ (X) the subset of positive simple functions and with M(X) the set of all measurable functions on X. For every f = p i=1 ai · χA ∈ E+ (X) and every A ∈ A, let’s define i

% f dγ = A

p 

ai · γ (Ai ∩ A) ∈ [0, +∞].

i=1

& It is said that f is integrable on A if A f dγ < +∞. Let E+1 (X) be the set of all positive simple functions integrable on X. If we replace L with A and E+ (A) with E+ (X) we find, in this abstract framework, the properties of Proposition 3.1.3: Proposition 3.6.1 Let f, g ∈ E+ (X) and c ∈ R+ ; then & & (1) cf ∈ E+ (X) and X cf dγ = c X f dγ . & & & (2) f + g ∈ E+ (X) and X (f + g)dγ = X f dγ + X gdγ . & & (3) X f dγ ≤ X gdγ , if f ≤ g, γ -a.e. & & (4) B f dγ ≤ C f dγ , for every B, C ∈ A with B ⊆ C. & (5) For every ε > 0, there exists δ > 0 such that B f dγ < ε, for every B ∈ A with γ (B) < δ.

Definition 3.6.2 Let f ∈ M+ (X) be a positive measurable function on X and let A ∈ A; we define %

% f dγ = sup A

f is integrable if functions on X.

A

& X

 ϕdγ : ϕ ∈ E+ (X), ϕ ≤ f

∈ [0, +∞].

f dγ < +∞. Let L1+ (X) be the set of all positive integrable

91 3.6 · Abstract Setting

We find the results of Proposition 3.1.6: Proposition 3.6.3 Let A ∈ A, f, g ∈ M+ (X) and c ≥ 0; then: & & (1) cf ∈ M+ (X) and A cf dγ = c A f dγ . & & (2) A f dγ ≤ A gdγ , if f ≤ g. & & (3) B f dγ ≤ C f dγ , for every B, C ∈ A, B ⊆ C.

We can also prove the monotone convergence theorem, Beppo Levi’s theorem, and Fatou’s lemma. Theorem 3.6.4 (Convergence Monotone Theorem) Let f : X → R+ and (fn ) ⊆ M+ (X) such that fn ≤ fn+1 , for any n ∈ N and & & fn ↑ f ; then f ∈ M+ (X) and X fn dγ ↑ X f dγ .

Theorem 3.6.5 (Beppo Levi)  Let (fn )n ⊆ M+ (X) be a sequence for which the corresponding series ∞ n=1 fn is ∞ pointwise convergent on X and let f = n=1 fn ; then f ∈ M+ (X) and % f dγ = X

∞ % 

fn dγ .

n=1 X

Theorem 3.6.6 (Fatou’s Lemma) Let (fn ) ⊆ M+ (X) such that f = lim infn fn < +∞; then %

% lim inf fn dγ ≤ lim inf X

n

n

fn dγ . X

Let f ∈ M(X); then f + = sup{f, 0} ∈ M+ (X), f − = sup{−f, 0} ∈ M+ (X), and f = f + − f − , |f | = f + + f − . If one of the functions f + or f − is integrable, then we can define %

%

f + dγ −

f dγ = X

X

%

f − dγ . X

If the two functions f + and f − are integrable, then it is said that f is integrable. We denote L1 (X) the set of all integrable functions & on X. & For every A ∈ A and every f ∈ L(X), A f dγ = X χA · f dγ . The function f is integrable on A if χA · f ∈ L1 (X); L1 (A) is the set of functions integrable on A.

3

92

3

Chapter 3 • Lebesgue Integral

 Let f = ni=1 ai χA be a A-simple function: {a1 , . . . , ap } ⊆ R and {A1 , . . . , Ap } i is a partition A-measurable of X. We can remark that f ∈ L1 (X) if and only if ai = 0 for any i ∈ {1, · · · , p} for which γ (Ai ) = +∞. Let E 1 (X) ⊆ L1 (X) be the set of all A-simple integrable functions. With the appropriate adaptations, the results 3.2.2–3.2.8, 3.2.10, 3.3.1, 3.3.3, and 3.3.6–3.3.10 also work in an abstract framework. Theorem 3.6.7 Let f ∈ M(X); then f ∈ L1 (X) if and only if |f | ∈ L1+ (X), and, in this case, $% $ % $ $ $ f dγ $ ≤ |f |dγ . $ $ X

X

Theorem 3.6.8 Let f ∈ M(X). & (1) If f = 0 γ -a.e. on X, then f ∈ L1 (X) and X f dγ = 0. & (2) If γ (A) = 0, then f ∈ L1 (A) and A f dγ = 0.

Theorem 3.6.9 Let f ∈ L1 (X). & (a) If A f dγ ≥ 0, for every A ∈ A, then f ≥ 0 γ -a.e. on X. & (b) If A f dγ ≤ 0, for every A ∈ A, then f ≤ 0 γ -a.e. on X. & (c) If A f dγ = 0, for every A ∈ A, then f = 0 γ -a.e. on X.

Theorem 3.6.10 Let f ∈ M(X) and g ∈ L1+ (X) such that |f | ≤ g γ -a.e.; then f ∈ L1 (X).

Theorem 3.6.11 If f ∈ L1 (X) and A ∈ A, then f ∈ L1 (A).

Theorem 3.6.12 A bounded measurable function is integrable on a finite measure set.

93 3.6 · Abstract Setting

Theorem 3.6.13 & L1 (X) is a real vector space, and the mapping I : L1 (X) → R, I (f ) = X f dγ , is a linear%operator: % % (1)

%X

(2)

(f + g)dγ = f dγ + gdγ , for every f, g ∈ L1 (X); X % X cf dγ = c f dγ , for every f ∈ L1 (X) and any c ∈ R.

X

X

Theorem 3.6.14 Let f,%g ∈ L1 (X);% then f dγ ≤

(1) X

gdγ , if f ≤ g. X

% |f |dγ < ε, for every A ∈ A

(2) For every ε > 0, there exists δ > 0 such that A

with γ (A) < δ. (3) For % every ε > 0, there exists A0 ∈ A with γ (A0 ) < +∞ such that |f |dγ < ε. X\A0

% (4)

∪∞ n=1 An n = m.

f dγ =

∞ % 

f dγ , for every (An )n ⊆ A with An ∩ Am = ∅, for all

n=1 An

Theorem 3.6.15 & Let  · 1 : L1 (X) → R+ , defined by f 1 = X |f |dγ , for every f ∈ L1 (X); then (1) f 1 = 0 if and only if f = 0 γ -a.e. (2) cf 1 = |c| · f 1 , for every f ∈ L1 (X) and any c ∈ R. (3) f + g1 ≤ f 1 + g1 , for every f, g ∈ L1 (X).

Theorem 3.6.16 Let (fn ) ⊆ L1 (X) and let f ∈ L1 (X). ·1

γ

(1) If fn −−→ f , then fn − → f. (2) If

X (fn )n is

X

 · 1 -Cauchy, then (fn )n is Cauchy in measure.

3

94

3

Chapter 3 • Lebesgue Integral

Theorem 3.6.17 (Dominated Convergence Theorem) Let (fn ) ⊆ M(X) and let g ∈ L1 (X) such that · (1) fn − → f and X

(2) |fn | ≤ g, for any n ∈ N. ·1

Then (fn ) ⊆ L1 (X), f ∈ L1 (X), fn −−→ f and X

& X

fn dγ →

& X

f dγ .

Theorem 3.6.18 (Bounded Convergence Theorem) Let us suppose that γ (X) < +∞, (fn ) ⊆ M(X) and c ∈ R+ such that · (1) fn − → f. X

(2) |fn | ≤ c, for any n ∈ N. ·1

Then (fn ) ⊆ L1 (X), f ∈ L1 (X), fn −−→ f and X

& X

fn dγ →

& X

f dγ .

The seminormed space (L1 (X),  · 1 ) is complete, and the set of integrable simple functions, E 1 (X) = E (X) ∩ L1 (X), is dense in L1 (X) with respect to the topology of  · 1 - convergence.

3.7

Exercises

& (1) If A ∈ L and f ∈ L&+ (A) with 0 ≤ f (x) ≤ a, then 0 ≤ A f dλ ≤ aλ(A). (2) If f ∈ L+ (A), then A f dλ ≥ aλ((f ≥ a)), for any a > 0. If f ∈&L1+ (A), then lima→+∞ aλ((f ≥ a)) = 0. (3) Compute [0,+∞[ e−[x] dλ(x) ([·] is the ceiling function—[x] is the least integer greater than or equal to x). (4) Let f : [a, b] → R defined by f (x) = αx + β, with α > 0 and f (a) > 0. For , k = 0, 1, . . . , n and le fn : [a, b] → any n ∈ N∗ , we denote ckn = a + k · b−a n n−1 n n n R, fn (x) = k=0 f (ck ) · (ck+1 & − ck ). Show that (fn )n ⊆ E+ ([a, b]) and that fn ↑ f . Compute the integral [a,b] f dλ, and compare with the Riemann integral &b a f (x)dx. (5) Let f : [a, b] → [m, M] be a bounded measurable function & (then f is a Lebesgue integrable function; see Corollary 3.2.8); let I = [a,b] f dλ ∈ R. For every partition  = {y0 , y1 , · · · , yn } of the interval [m, M], let σ = n n k=1 yk−1 · λ(Ek ),  = k=1 yk · λ(Ek ), where, for any k = 1, · · · n, Ek = {x ∈ [a, b] : yk−1 ≤ f (x) < yk } ∈ L([a, b]). Show that sup∈D([a,b]) σ = inf∈D([a,b])  = I . Show that, for every ε > 0, there exists δ > 0 such that, for every  ∈ D([a, b]) with  < δ,  − σ < ε.

95 3.7 · Exercises



1 , x ∈ [c, d], 0 , x ∈ [a, b] \ [c, d]. Show that, for every ε > 0, there exists a continuous function g : [a, b] → R such that % |f − g|dλ < ε.

(6) Let [c, d] ⊆ [a, b] and f : [a, b] → R, f (x) =

[a,b]

Indication: We define g(x) = f (x) if x ∈ [a, c − ε] ∪ [c, d] ∪ [d + ε, b] and piecewise linear on [c − ε, c]

(7)

and [d, d + ε]. Let f ∈ L1 (A)

and B, C ∈ L(A); show that %

%

%

f dλ = B∪C

%

f dλ + B

f dλ − C

f dλ. B∩C

(8) Let f, g ∈ L1 (A) two bounded functions on A; show that fg, f 2 , g 2 ∈ L1 (A) and that %  % % 1 2 2 |fg|dλ ≤ f dλ + g dλ . 2 A A A (9) For everyfunction f ∈ L1+ (A) and for any p ∈ N, we define fp : A → R+ by & & f (x) , f (x) ≤ p . Show that (fp ) ⊆ L1+ (A) and A fp dλ ↑ A f dλ. fp (x) = p , f (x) > p & 1 Calculate this way ]0,1] f dλ, where f (x) = √ , for every x ∈]0, 1]. 3 x +∞ (−1)n (10) We consider the function f : R → R, f = n=1 n χ[n, n + 1[ ; show that it is measurable and bounded, but not integrable on R (see Corollary 3.2.8).  (−1)n (11) Show that the function f = +∞ n=1 n2 χ[n, n + 1[ is integrable on R. p (12) Let f = i=1 ai · χA ∈ E (A); show that f ∈ L1 (A) if and only if, for any i i ∈ {1, · · · , p} for which ai = 0 it follows that λ(Ai ) < +∞. Deduce by that E (A) ⊆ L1 (A), if λ(A) < +∞. (13) Let A ∈ L, a, b ∈ R, a = 0, f : aA+b → R and g : A → R, g(x) = f (ax+b), for every x ∈ A. Show that f ∈ L1 (aA + b) if and only if g ∈ L1 (A) and that %

%

%

f (y)dλ(y) = |a| · aA+b

g(x)dλ(x) = |a| · A

f (ax + b)dλ(x). A

Note that in this case, unlike that of Theorem 3.5.6 (where f is assumed to be Borel function), the function f is an arbitrary measurable function. Indication. According to Exercise (5) of 2.5, f ∈ L(aA + b) if and only if g ∈ L(A). The demonstration p k=1 ai · χE ∈ E (aA + b), (3)

is made in the following cases: (1) f = χE ∈ L(aA + b), (2) f = f ∈ L+ (aA + b) and (4) f ∈ L(aA + b).

i

(14) Let A ∈ L with 0 < λ(A) n)

f (x)dλ(x) = 0.

(21) Let (X, A, γ ) be an abstract measure space where γ is a positive σ -finite and ¯ + by γA (B) = complete measure and let A ∈ A; let’s define γA : A → R γ (A ∩ B). Show that every γA -integrable function is γ -integrable. (22) Let A ∈ L with λ(A) = 1. Show that %

&

e

A

f (x)dλ(x)

≤

ef (x) dλ(x), for every f ∈ L1 (A), A

%

%

 g(x)dλ(x) , for every g ∈ L1 (A), g > 0.

ln(g(x))dλ(x) ≤ ln A

A

(23) Let A ∈ L with 0 < λ(A) < +∞; show that, for every f ∈ L1 (A) and any p ≥ 1, $ $% %  p1 $ $ p−1 p $ f dλ$ ≤ (λ(A)) p · |f | dλ . $ $ A

A

Indication. Jensen’s inequality will be applied to the function F (y) = |y|p .

97

The Lp Spaces

This chapter is devoted to a class of Banach spaces constructed using the notion of an integrable function—Lebesgue spaces or classical Banach spaces. In the first paragraph, we will present the algebraic and topological structure of these spaces. Paragraph 2 is reserved for the study of density properties in Lp . A limit case of Lebesgue spaces, the space L∞ , is studied in the third paragraph, and in the last paragraph, an introduction in the theory of Fourier series on L2 is made.

4.1

Algebraic and Topological Structure

Definition 4.1.1 Let A ∈ L, p ∈ R, p ≥ 1; a function f : A → R is said to be p-integrable on A if f ∈ L(A) and |f |p ∈ L1 (A). We denote with Lp (A) the set of all p-integrable functions on A. In the particular case p = 1, we find the space L1 (A) studied in the previous chapter; indeed, f ∈ L1 (A) if and only if |f | ∈ L1 (A) (see Theorem 3.2.2). We define the application  · p : Lp (A) → R+ by %

1 p |f | dλ , for every f ∈ Lp (A).

f p =

p

A

Proposition 4.1.2 Lp (A) is a real vector space. Proof The set F (A, R) of all real functions defined on A is a vector space with respect to the usual operations of addition and multiplication with scalars (operations punctually defined). To show that Lp (A) ⊆ F (A, R) is a vector subspace, it is enough to show that the sum of two functions of Lp (A) remains in Lp (A) and that the product of a real scalar with a function of Lp (A) remains in Lp (A). © The Author(s), under exclusive license to Springer Nature Switzerland AG 2021 L. C. Florescu, Lebesgue Integral, Compact Textbooks in Mathematics, https://doi.org/10.1007/978-3-030-60163-8_4

4

Chapter 4 • The Lp Spaces

98

Let f, g ∈ Lp (A); then f, g ∈ L(A), and so f + g ∈ L(A). On the other hand, the function h : R → R, h(y) = |y|p , is continuous, and therefore h◦(f +g) = |f +g|p ∈ L(A) (see Theorem 2.1.15).

4

|f + g|p ≤ (|f | + |g|)p ≤ 2p · max{|f |p , |g|p } ≤ 2p · (|f |p + |g|p ).

(*)

The function 2p · (|f |p + |g|p ) is integrable, and, from (∗), it dominates the measurable function |f + g|p ; Theorem 3.2.6 assures us that |f + g|p ∈ L1 (A), and so f + g ∈ Lp (A). For any c ∈ R and every f ∈ Lp (A), c · f ∈ L(A) and |c · f |p = |c|p · |f |p ∈ L1 (A); hence c · f ∈ Lp (A).  Lemma 4.1.3 Let p, q > 1 such that

a·b ≤

1 1 + = 1; for all a, b ≥ 0, p q

ap bq + . p q

Proof The function defined by x → f (x) = ax −

ap xq − is differentiable over [0, +∞[; its p q 1

derivative, f (x) = a −x q−1 , takes the value zero for x0 = a q−1 . We immediately notice that 1

f is increasing on [0, x0 ] and decreasing on [x0 , +∞[. Hence f (x) ≤ f (x0 ) = f (a q−1 ) = 1

a · a q−1 −

ap p

−

q a q−1

q

= a p · (1− p1 − q1 ) = 0, for every x ≥ 0, which proves the lemma. 

Theorem 4.1.4 (Hölder Inequality) Let p, q > 1 such that p1 + q1 = 1. For every f ∈ Lp (A) and every g ∈ Lq (A), f · g ∈ L1 (A) and %

%

 1 % 1 p q |f |p dλ · |g|q dλ .

|f · g|dλ ≤ f p · gq = A

A

A

Proof & If f p = 0, then A |f |p dλ = 0, from where f = 0 a.e. (Theorem 3.3.3). Then f · g = 0 & a.e., and, applying Theorem 3.3.3 again, f · g1 = A |fg|dλ = 0 ≤ f p · gq . We reason in the same way if gq = 0. Let us suppose that f p = 0 = gq . In the inequality of the previous lemma, let us |f (x)| |g(x)| replace a = and b = ; we obtain then f p gq |f (x)|p |g(x)|q |f (x)g(x)| ≤ p + q , for every x ∈ A. f p · gq pf p qgq

4

99 4.1 · Algebraic and Topological Structure

The same inequality can be written: |fg| ≤ ||f ||p · ||g||q

1 1 p q . p · |f | + q · |g| p||f ||p q||g||q

(*)

As f ∈ Lp (A) and g ∈ Lq (A), it follows that fg ∈ L(A) and |f |p , |g|q ∈ L1 (A). Then h = f p · gq

1 1 p q p |f | + q |g| pf p qgq

∈ L1 (A).

From (∗), |fg| ≤ h; Theorem 3.2.6 leads to fg ∈ L1 (A). The integral being monotonic (see 1) of Theorem 3.3.1) and linear (Theorem 3.2.10), we can now integrate the inequality (∗) and then

% |fg|dλ ≤ f p · gq A

= f p · gq

%

1 |f | dλ + q qgq A

1 p pf p

%

|g| dλ =

p

1 1 p q p · ||f ||p + q · gq p||f ||p qgq

q

A

= f p · gq .



Theorem 4.1.5 (Minkowski Inequality) For any p ≥ 1 and every f, g ∈ Lp (A), f + gp ≤ f p + gp .

Proof If p = 1, then the inequality is obvious (see 3) of Theorem 3.3.3). Let us suppose that p > 1 and f, g ∈ Lp (A). From Proposition 4.1.2, f + g ∈ Lp (A). If f + gp = 0, then the Minkowski inequality is obvious. p Let’s then suppose, more, that f + gp > 0. Let q = p−1 > 1; then p1 + q1 = p−1 q 1, and the function h = |f + g| ∈ L (A). Indeed, h is the composition between the measurable function f + g and the continuous function l : R → R, l(y) = |y|p−1 ; from Proposition 2.1.15, h = l ◦ (f + g) ∈ L(A). Moreover, |h|q = |f + g|p ∈ L1 (A) (f + g ∈ Lp (A)), and so h ∈ Lq (A). From the inequality of Hölder, it follows that |f + g|p−1 · |f | = h|f | ∈ L1 (A) and |f + g|p−1 · |g| = h|g| ∈ L1 (A). Then, using the properties of monotonicity and linearity of the integral,  p f + gp =

%

% |f + g|p dλ = A

|f + g|p−1 · |f + g|dλ ≤ A

Chapter 4 • The Lp Spaces

100

%

%

≤

|f + g|p−1 (|f | + |g|)dλ =

% |f + g|p−1 |f |dλ +

A

A

|f + g|p−1 |g|dλ. A

But, according to the Hölder inequality, %

4

|f + g|p−1 · |f |dλ = f h1 ≤ f p · hq , A

% |f + g|p−1 · |g|dλ = gh1 ≤ gp · hq . A

Then  p   ||f + g||p ≤ ||f ||p + ||g||p · ||h||q =   = ||f ||p + ||g||p ·

%

1 q |f + g|(p−1)q dλ =

A

  = ||f ||p + ||g||p ·

%

 p−1 p    p−1 |f + g| dλ = ||f ||p + ||g||p · ||f + g||p . p

A

p−1  , we obtain Simplifying the previous inequality with ||f + g||p ||f + g||p ≤ ||f ||p + ||g||p .



Theorem 4.1.6 (Lp (A),  · p ) is a seminormed space.

Proof f p = 0 if and only if f = 0, a.e. (Theorem 3.3.3).

%

1 p |c · f |p dλ = |c| · f p .

For any c ∈ R and for every f ∈ Lp (A), c · f p = Triangle inequality is actually Minkowski inequality.

A



Definition 4.1.7 If fn − f p −−−−→ 0, then we say that the sequence (fn ) ⊆ Lp (A) is Lp n→+∞

·p

convergent to f ∈ Lp (A) and we write fn −−→ f . The mapping  · p is called the seminorm of Lp -convergence.

A

(Continued )

101 4.1 · Algebraic and Topological Structure

Definition 4.1.7 (continued) The sequence (fn ) is Lp -Cauchy if limm,n→+∞ fm − fn p = 0. f ∈ Lp (A) is Lp -adherent point for F ⊆ Lp (A) if there exists a sequence (gn ) ⊆ ·p

p

F such that gn −−→ f . We denote with F the set of all Lp -adherent points of F . It A

p

can be easily shown that f ∈ F if and only if, for every ε > 0, there exists g ∈ F p such that f − gp < ε. The set F ⊆ Lp (A) is dense if F = Lp (A).

Definition 4.1.8 Let p ≥ 1; the la relation f ∼ g if and only if f = g a.e. is an equivalence relation on Lp (A) (∼ is reflexive, symmetric, and transitive). We remark that if f ∈ Lp (A) and f = g a.e., then g ∈ Lp (A) and f p = gp . Indeed, if f ∈ Lp , then & & g ∈ L(A) (see 1) of Theorem 2.1.13); since |f |p = |g|p a.e., A |f |p dλ = A |g|p dλ (Theorem 3.2.11), and so g ∈ Lp (A) and f p = gp . Let Lp (A)|∼ = Lp (A) be the quotient set. The elements of this set are of the form: [f ] = {g ∈ Lp (A) : f ∼ g}.

We remark that Lp (A) is a real vector space: [f ] + [g] = [f + g], c · [f ] = [c · f ] ∈& Lp (A), for &every f, g ∈ Lp (A) and for any c ∈ R. Moreover, for every g ∈ [f ], A |f |p dλ = A |g|p dλ. Therefore, we can correctly define the application:  · p : Lp (A) → R+ , [f ]p = f p , for every [f ] ∈ Lp (A).

Theorem 4.1.9 The space (Lp (A),  · p ) is a real normed space.

Theorem 4.1.10 Let p ≥ 1, (fn )n ⊆ Lp (A) and f ∈ Lp (A). ·p

λ

→ f. (1) If fn −−→ f , then fn − (2) If

A (fn )n is

A

a Lp -Cauchy sequence, then (fn )n is Cauchy in measure.

Proof (1) For every ε > 0 and any n ∈ N, let An (ε) = (|fn − f | > ε) ∈ L. Then |fn − f |p ≥ |fn − f |p χAn (ε) ≥ εp χAn (ε) .

4

Chapter 4 • The Lp Spaces

102

Now using the monotonicity of the integral, we get %

p

fn − f p =

|fn − f |p dλ ≥ εp λ(An (ε)), for every ε > 0, and any n ∈ N. A

4

It follows that, for every ε > 0 and any n ∈ N, λ(An (ε)) ≤ ·p

1 εp

p

· fn − f p .

Since fn −−→ f , it follows that limn λ(An (ε)) = 0, for every ε > 0, that implies A

λ

→ f . We can show (2) similarly, by replacing An (ε) with Am,n (ε)= (|fm −fn | > ε).  fn − A

Using Theorem 4.1.10, we will show that the space (Lp (A),  · p ) is a complete seminormed space; the proof is an adapted version of the proof of Theorem 3.3.8.

Theorem 4.1.11 The seminormed space (Lp (A),  · p ) is complete (every Lp -Cauchy sequence is Lp convergent). The normed space (Lp (A),  · p ) is a Banach space (complete normed space).

Proof From point (2) of Theorem 4.1.10, it follows that any Lp -Cauchy sequence, (fn )n ⊆ Lp (A), is Cauchy in measure. From (1) of Riesz theorem (Theorem 2.2.5), there is a subsequence (fkn )n of (fn )n almost uniform convergent to a function f : A → R. Point (2) of · Theorem 2.2.2 tells us that fkn − → f and then f ∈ L(A) (see 3) of Theorem 2.1.13). A

Let’s fix m ∈ N; for any n ∈ N, let gn = |fm −fkn |p ∈ L+ (A). We apply to the sequence (gn )n the Fatou’s lemma (Corollary 3.1.12): %

% lim inf gn dλ ≤ lim inf A

n

n

gn dλ. A

As lim infn gn = |fm − f |p a.e., it follows that %

p

A

|fm − f |p dλ ≤ lim inf fm − fkn p . n

From limm,n→+∞ fm − fkn 1 = 0, it follows, on one hand, that fm − f ∈ Lp (A) and then f = fm − (fm − f ) ∈ Lp (A) and, on the other hand, that fm − f p → 0. 

If λ(A) < +∞, then we can compare the spaces Lp (A) and the topologies generated by the seminorms  · p on these spaces for p ≥ 1.

4

103 4.1 · Algebraic and Topological Structure

Theorem 4.1.12

·r

·p

A

A

If λ(A) < +∞ and 1 ≤ p < r, then Lr (A) ⊆ Lp (A). If fn −−→ f , then fn −−→ f, for every (fn ) ⊆ Lr (A), f ∈ Lr (A).

Proof Let p1 =

r p

> 1 and q1 =

r r−p

> 1; then r p

1 p1

+

1 q1

= 1. For every f ∈ Lr (A), f ∈ L(A)

and |f |r ∈ L1 (A). It follows that |f |p ∈ L (A) = Lp1 (A). Since λ(A) < +∞, 1 ∈ Lq1 (A). We can now apply the Hölder inequality (see Theorem 4.1.4) to the functions |f |p and 1; therefore |f |p · 1 = |f |p ∈ L1 (A). It follows, on the one hand, that f ∈ Lp (A), and then Lr (A) ⊆ Lp (A). On the other hand, the inequality of Hölder leads to %

% |f |p dλ ≤ A

 1 % 1 p1 q1 |f |p·p1 dλ · 1dλ ,

A

A

or equivalent to %

p r r−p r−p p |f |r dλ · λ(A) r = f r · λ(A) r .

p

f p ≤ A

The last inequality implies f p ≤ [λ(A)]

r−p pr

· f r , for every f ∈ Lr (A).

Hence, for every (fn )n ⊆ Lr (A) and f ∈ Lr (A), fn − f p ≤ [λ(A)]

r−p pr

· fn − f r ,

which shows that if the sequence (fn )n Lr -converges to f , then it Lp -converges to f .



Remarks 4.1.13 (i) In the previous theorem, the condition λ(A) < +∞ is essential. Indeed, let the function 1 f : [1, +∞[→ R, f (x) = . Then f ∈ C([1, +∞[) ⊆ L([1, +∞[) and, for any x p > 1, %

% [1,+∞[

|f |p dλ = 1

+∞

$+∞ 1 1 1 1−p $ dx = · x = . $ 1 xp 1−p p−1

It follows that f ∈ Lp ([1, +∞[), for any p > 1, but f ∈ / L1 ([1, +∞[).

Chapter 4 • The Lp Spaces

104

(ii) If A is a set of finite measure and if p < r, the trace of the topology induced by the seminorm  · p on the subset Lr (A) ⊆ Lp (A) is less fine than the topology induced by the seminorm  · r on the same subset.

4

4.2

Density Properties in Lp

In this paragraph we will highlight various dense sets in Lp (A) (see Definition 4.1.7). Theorem 4.2.1 Let A ∈ L and p ≥ 1; then (1) E 1 (A) = E (A) ∩ Lp (A). p (2) E 1 (A) = Lp (A).

Proof  (1) Let f ∈ E (A); then f = ni=1 ai · χA , where Ai ∈ L(A) are pairwise disjoints. Then i  f ∈ Lp (A) if and only if ni=1 |ai |p λ(Ai ) < +∞ which says that, for any i for which λ(Ai ) = +∞, ai = 0 and this one is equivalent to f ∈ E 1 (A). p (2) The inclusion E 1 (A) ⊆ Lp (A) is obvious. Let’s prove reverse inclusion. For every f ∈ Lp (A), f = f + − f − , where f + , f − ∈ L+ (A). There exist two sequences (un )n , (vn )n ⊆ E+ (A) such that un ↑ f + and vn ↑ f − (see 1) of Theorem 2.3.3). The sequence (fn )n ⊆ E (A), fn = un − vn , is pointwise convergent on A to f ; moreover, for any n ∈ N, |fn |p ≤ (un + vn )p ≤ (f + + f − )p = |f |p ∈ L1 (A), from where, using Theorem 3.2.6, (fn )n ⊆ E (A) ∩ Lp (A) = E 1 (A). On the other hand, |fn − f |p ≤ (|fn | + |f |)p ≤ 2p (|fn |p + |f |p ) ≤ 2p+1 · |f |p , and, since fn → f , it follows from the dominated convergence theorem (Theorem 3.3.9), p

%

fn − f p =

|fn − f |p dλ → 0, A ·p

p

which is equivalent to fn −−→ f . Therefore f ∈ E 1 (A) . A



105 4.2 · Density Properties in Lp

Remark 4.2.2 Lp (A) is the completion of (E 1 (A),  · p ).

Theorem 4.2.3 Let p ≥ 1 and let Cp (A) = C(A) ∩ Lp (A) be the set of all continuous p-integrable functions; then p

Cp (A) = Lp (A).

The proof is an immediate adaptation of the case p = 1 (see Theorem 3.3.11), and we leave it to the reader. Remark 4.2.4 If A is a compact set, then Cp (A) = C(A), and so, in this case, C(A) is dense in Lp (A).

We conclude this section with an important property of spaces Lp ([a, b])—the separability.

Theorem 4.2.5 For every compact interval [a, b] ⊆ R and any p ≥ 1, the space (Lp ([a, b]),  · p ) is separable (contains a countable, dense subset).

Proof Let P be the set of restrictions of polynomial functions with rational coefficients at the interval [a, b]; P is a countable set. p Let us show that P is a dense subset of Lp (A), therefore that P = Lp ([a, b]). On the one hand, P ⊆ C([a, b]), and, from the previous remark, C([a, b]) = Cp ([a, b]) ⊆ Lp ([a, b]); hence P ⊆ Lp ([a, b]). On the other hand, according to Theorem 4.2.3, for every f ∈ Lp ([a, b]), and every ε > 0, there exists g ∈ C([a, b]) such that f − gp < ε3 . Now let us remember Weierstrass theorem of uniform approximation of continuous functions with polynomials. From this theorem, there is a polynomial h on [a, b] such that g − h∞ = sup |g(x) − h(x)| < x∈[a,b]

ε 1

3(b − a) p

.

Obviously, due to the density of rationals in R, we can uniformly approximate h with polynomials of P . Therefore there exists l ∈ P such that h − l∞
0. Since |f |p ∈ L1+ (R), we can use point (2) of Theorem 3.1.16; so there is k ∈ N such that % R\[−k,k]

|f |p dλ
α) > 0, then the only element α ∈ R+ for which |f | ≤ α a.e. is α = +∞ and therefore f ∞ = +∞. If it exists α ∈ R+ such that λ(|f | > α) = 0, then f ∞ < +∞. (ii) In general, the essential supremum of a function |f | is smaller than its supremum: f ∞ ≤ sup |f (x)|. x∈A

Indeed, if f = χQ , then supx∈R |f (x)| = 1, and f ∞ = 0 (λ(|f | > 0) = λ(Q) = 0; hence |f | = 0 a.e.). (iii) Let J ⊆ R be an interval, and let f : J → R be a continuous function on J ; then f ∞ = supx∈J |f (x)|. Indeed, as we saw in the previous point, f ∞ ≤ supx∈J |f (x)|. For any c ∈ R with c < supx∈J |f (x)|, there exists x0 ∈ J such that c < |f (x0 )|. From the continuity of |f | at x0 , there exists δ > 0 such that (x0 − δ, x0 + δ) ∩ J ⊆ (|f | > c) from where 0 < λ((x0 − δ, x0 + δ) ∩ J ) ≤ λ(|f | > c). Then c ≤ f ∞ , which implies that supx∈J |f (x)| ≤ f ∞ . Lemma 4.3.3 Let f ∈ L(A); then (1) f ∞ = inf{supx∈A\N |f (x)| : N ∈ L, λ(N) = 0}. (2) |f | ≤ f ∞ almost everywhere.

Chapter 4 • The Lp Spaces

108

4

Proof (1) Let α0 = inf{supx∈A\N |f (x)| : N ∈ L, λ(N) = 0} ∈ R+ . For any α ∈ R with |f | ≤ α a.e., we denote N = (|f | > α) ∈ L; then λ(N) = 0. By definition, α0 ≤ supx∈A\N |f (x)| ≤ α. It follows that α0 ≤ f ∞ . If α0 = +∞, then the equality is obvious. Let us suppose that α0 < +∞; by the definition of α0 , for every ε > 0, there exists N ∈ L with λ(N) = 0 such that α0 + ε > supx∈A\N |f (x)|. It follows that |f (x)| ≤ α0 + ε, for every x ∈ A \ N or (|f | > α0 + ε) ⊆ N, and then |f | ≤ α0 + ε a.e. Considering the meaning of the essential supremum of f , f ∞ ≤ α0 + ε, for every ε > 0, and so f ∞ ≤ α0 . The two inequalities show that f ∞ = α0 . (2) If f ∞ = +∞, the inequality is obvious. Let us suppose that f ∞ < +∞; from the previous point, for any n ∈ N∗ , there exists Nn ∈ L with λ(Nn ) = 0 such that sup |f (x)| < f ∞ +

x∈A\Nn

Let N =

∞

n=1

1 . n

Nn ∈ L; then λ(N) = 0 and

|f (x)| < f ∞ +

∞  1 , for every x ∈ A \ N = (A \ Nn ), for any n ∈ N∗ . n n=1

Passing to the limit in the last inequality, we obtain |f (x)| ≤ f ∞ , for every x ∈ A \ N or |f | ≤ f ∞ a.e.  Remark 4.3.4 From point (2) of the previous lemma, it follows that f ∞ is the smallest element of the set {α ∈ R+ : |f | ≤ αa.e.}; because this last set is an interval, {α ∈ R+ : |f | ≤ α a.e.} = [f ∞ , +∞].

Definition 4.3.5 For every A ∈ L, L∞ (A) = {f ∈ L(A) : f ∞ < +∞}.

Theorem 4.3.6 Equipped with the usual operations of addition and multiplication with scalars, L∞ (A) is a real vector space and  · ∞ is a seminorm on this space.

4

109 4.3 · The L∞ Space

Proof Let f, g ∈ L∞ (A); then |f + g| ≤ |f | + |g| ≤ f ∞ + g∞ a.e. Therefore f + g∞ ≤ f ∞ + g∞ < +∞; it follows that f + g ∈ L∞ (A) and that  · ∞ satisfies the triangle inequality. For every f ∈ L∞ (A) and for any c ∈ R, |c · f | = |c| · |f | ≤ |c| · f ∞ a.e. from where c · f ∞ ≤ |c| · f ∞ < +∞.

(*)

Therefore c · f ∈ L∞ (A). If c = 0, then it is obvious that c · f ∞ = |c| · f ∞ . If c = 0, then we apply the inequality of (∗) for 1c ∈ R and c · f ∈ L∞ (A): ) ) )1 ) ) · (c · f )) ≤ 1 · c · f ∞ . )c ) |c| ∞ Therefore |c| · f ∞ ≤ c · f ∞ which, with the inequality (∗), leads us to equality c · f ∞ = |c| · f ∞ . It follows that  · ∞ is a seminorm on L∞ (A).  Remark 4.3.7 The relation ∼, defined by f ∼ g if and only if f = g a.e., is an equivalence relation on L∞ (A). We denote the quotient set L∞ (A)|∼ with L∞ (A); for every [f ] ∈ L∞ (A), [f ]∞ = f ∞ . According to Lemma 4.3.3, f ∞ = 0 if and only if f = 0 a.e.; it follows that the previous definition does not depend on the representative f of the equivalence class [f ]. We can easily deduce that (L∞ (A),  · ∞ ) is a normed space. Proposition 4.3.8 Let (fn )n ⊆ L∞ (A) and f ∈ L∞ (A). ·∞

u

A

A\B

(1) fn −−−→ f ⇔ there exists B ∈ L(A) with λ(B) = 0 such that fn −−→ f. (2) (fn )n is Cauchy sequence in (L∞ (A),  · ∞ ) if and only if there exists B ∈ L(A) with λ(B) = 0 such that (fn )n is uniformly Cauchy on A \ B. Proof ·∞ (1) We suppose that fn −−−→ f ; for any p ∈ N∗ , there exists np ∈ N such that fn − f ∞ < A

for any n ≥ np . According to Lemma 4.3.3, it follows that |fn − f | < p1 a.e., or that   1 are null sets, for any p ∈ N∗ and for any n ≥ np . Then the set An,p = |fn − f | ≥ p ∞ ∞ B = p=1 n=np An,p is also a null set.  ∞ 1 For every x ∈ A \ B = ∞ p=1 n=np (A \ An,p ), we have that |fn (x) − f (x)| < p , for 1 p,

any p ∈ N∗ and for any n ≥ np from where fn −−→ f . u

A\B

u

Conversely, if we assume that there is B ∈ L(A) with λ(B) = 0 such that fn −−→ f , A\B

then, for every ε > 0, there exists nε ∈ N such that |fn (x) − f (x)| < ε, for any n ≥ nε and every x ∈ A \ B. It follows that A \ B ⊆ (|fn − f | < ε), which leads, by passing to the

110

Chapter 4 • The Lp Spaces

complementary, to (|fn − f | ≥ ε) ⊆ B. Therefore |fn −f | < ε a.e. and then fn −f ∞ ≤ ε, ·∞

→ f. for any n ≥ nε or fn −−− A



(2) is demonstrated in the same way.

4 Theorem 4.3.9 (L∞ (A),  · ∞ ) is a complete seminormed space, and then (L∞ (A),  · ∞ ) is a Banach space.

Proof Let (fn )n ⊆ (L∞ (A),  · ∞ ) be a Cauchy sequence. According to (2) of previous proposition, there exists B ∈ L(A) with λ(B) = 0 such that (fn )n is uniformly Cauchy on A \ B; it means that, for every ε > 0, there exists nε ∈ N such that, for all m, n ≥ nε , |fm (x) − fn (x)| < ε, for every x ∈ A \ B.

(*)

From (∗), for every x ∈ A \ B, (fn (x))n is a Cauchy sequence in R, and then there exists limn fn (x) ∈ R.  limn fn (x), x ∈ A \ B, Let f : A → R defined by f (x) = 0, x ∈ B. The sequence (fn )n converges a.e. to f and then, from (3) of Theorem 2.1.13, f ∈ L(A). From (∗), with n = nε and m → ∞, we obtain |f (x)−fnε (x)| ≤ ε, for every x ∈ A\B; it follows that |f (x)| ≤ |f (x) − fnε (x)| + |fnε (x)| < ε + |fnε (x)|, for every x ∈ A \ B, from where |f | ≤ ε + |fnε | ≤ ε + fnε ∞ a.e. We deduce that f ∞ < +∞ and then f ∈ L∞ (A). We again use the relation (∗), in which n → ∞, to obtain |fm (x) − f (x)| ≤ ε, for any m ≥ nε and every x ∈ A \ B. Since λ(B) = 0, |fm − f | ≤ ε a.e. or fm − f ∞ ≤ ε, for ·∞

→ f. any m ≥ nε ; then fn −−− A



The following proposition is an extension of the Hölder inequality (Theorem 4.1.4). Proposition 4.3.10 For every f ∈ L1 (A) and every g ∈ L∞ (A), the function f · g is integrable (f · g ∈ L1 (A)) and % f · g1 = A

|f · g|dλ ≤ f 1 · g∞ .

Proof From Lemma 4.3.3, |g| ≤ g∞ almost everywhere, and then |f · g| ≤ g∞ · |f |, a.e. It follows from Theorem 3.2.6 that f · g ∈ L1 (A), and, by integrating the inequality above, we obtain the extension of the inequality of Hölder presented in the statement. 

4

111 4.3 · The L∞ Space

If A has a finite measure, we can compare the spaces Lp , 1 ≤ p < +∞ with L∞ (see Theorem 4.1.12). Theorem 4.3.11 (Riesz)  p ∞ Let λ(A) < +∞; then L∞ (A)  ∞ p=1 L (A) and, for every f ∈ L (A), f ∞ = lim f p . p→+∞

Proof Let an arbitrary 1 ≤ p < +∞; if f ∈ L∞ (A), then f ∈ L(A), and, from Lemma 4.3.3, |f | ≤ f ∞ a.e., from where p

|f |p ≤ ||f ||∞ a.e.

(1) p

But, in the finite measure spaces, the constant functions are integrable. Hence f ∞ ∈ (f ∞ ∈ R+ ), and then, according to Theorem 3.2.6, |f |p ∈ L1 (A). It follows that  p p f ∈ L (A), which proofs the inclusion L∞ (A) ⊂ ∞ p=1 L (A). Now, by integrating the inequalities, (1), we get % p |f |p dλ ≤ f ∞ · λ(A),

L1 (A)

A

form where 1

f p ≤ [λ(A)] p · f ∞ , for every f ∈ L∞ (A).

(2)

From (2) it follows that the trace of the topology generated by the seminorm  · p on L∞ (A) is less fine than the topology generated by  · ∞ . We have observed that inequality (2) occurs for all f ∈ L∞ (A) and for any p ≥ 1; passing to the upper limit in this inequality, we get lim sup f p ≤ f ∞ , for every f ∈ L∞ (A).

(3)

p→+∞

In the above we have assumed that λ(A) > 0; in the particular case where λ(A) = 0, it follows that, for any p ≥ 1, L1 (A) = Lp (A) = L∞ (A) = L(A) (λ is a complete measure) and f p = 0 = f ∞ , for any p ≥ 1. In this situation it is obvious that limp f p = f ∞ . Returning to inequality (3), if f ∞ = 0, then, from (3), it follows that there exists limp f p = 0 = f ∞ . We suppose that f ∞ > 0. For any 0 < α < f ∞ , from the definition of f ∞ , it follows that λ(|f | > α) > 0; let Aα = (|f | > α) ∈ L(A). For any p ≥ 1, we obtain %

1 p 1 " #1 |f |p dλ ≥ α p · λ(Aα ) p = α · [λ(Aα )] p .

f p ≥ Aα

(4)

Chapter 4 • The Lp Spaces

112

1

Since λ(Aα ) ∈]0, +∞[, there exists limp→+∞ [λ(Aα )] p = 1. If we go to the lower limit for p → ∞ in (4), we obtain lim inf f p ≥ α, for any α < f ∞ .

p→+∞

4

(5)

But (5) implies that f ∞ ≤ lim inf f p .

(6)

p→+∞

From (3) and (6) we deduce that there exists limp f p = f ∞ .



Remark 4.3.12 In the statement of Riesz theorem, we specified that the inclusion ∞ p L∞ (A)  p=1 L (A) is strict. Indeed, if the function f :]0, 1] → R is defined by f (x) = ln x, then, according to (iii) of Remark 4.3.2, f ∞ = supx∈]0,1] |f (x)| = +∞. & p Therefore f ∈ / L∞ (]0, 1]). On the other hand, for any p ≥ 1, f p = ]0,1] |f |p dλ = &1 1 p β p 0+0 | ln x| dx < +∞ (for β = 2 there exists limx↓0 x · | ln x| = 0 < +∞; then the p generalized Riemann integral is convergent). It follows that f ∈ L (]0, 1]), for any p ≥ 1.

4.4

Fourier Series in L2 ([−π, π ])

From the start, we observe that a good number of the results of this section remain true if we replace the closed interval [−π, π] by some other measurable set. Let us remember that the space L2 ([−π, π]) is a Banach space with respect to the norm  · 2 : %  12 2 2 L [−π, π] → R+ , f 2 = f dλ , for every f ∈ L2 ([−π, π]) (we will [−π,π ]

use the current, instead of equivalence classes L2 ([−π, π]) = L2 ([−π, π])|∼ , their representatives). Moreover, Theorem 4.2.5 assures us that the space (L2 ([−π, π]),  · 2 ) is a separable Banach space. Note that if p = q = 12 , then we can write the inequality of Hölder: for every f, g ∈ L2 ([−π, π]), f ·g ∈ L1 ([−π, π]) and fg1 ≤ f 2 ·g2 (see Theorem 4.1.4). So we can define the application (·, ·) : L2 ([−π, π]) × L2 ([−π, π]) → R by % (f, g) =

fgdλ. [−π,π ]

The demonstration of the following proposition is a simple application of the above definition. Proposition 4.4.1 The application (·, ·) defined above is an inner product on L2 ([−π, π]), that is to say it checks the conditions: (1) (f, f ) ≥ 0, ∀f ∈ L2 ([−π, π]) and (f, f ) = 0 if and only if f = 0 a.e. (2) (f, g) = (g, f ), for every f, g ∈ L2 ([−π, π]).

113 4.4 · Fourier Series in L2 ([−π, π])

(3) (f + g, h) = (f, h) + (g, h), for every f, g, h ∈ L2 ([−π, π]). (4) (c · f, g) = c · (f, g), for every f, g ∈ L2 ([−π, π]) and any c ∈ R.

The norm associated with the inner product is defined in a standard way through √ f  = (f, f ); we immediately note that f  = f 2 , for every f ∈ L2 ([−π, π]). The space L2 ([−π, π]) is then a separable Hilbert space (a separable Banach space whose norm is induced by an inner product). In these spaces we can define the notion of orthogonality: two vectors f, g ∈ L2 ([−π, π]) are orthogonal when (f, g) = 0, and this is denoted by f ⊥ g. A sequence (fn )n ⊆ L2 ([−π, π]) is said to be orthogonal if fn ⊥ fm , for any m = n; moreover, if fn 2 = 1, for any n ∈ N, then the sequence is orthonormal. A general result of the functional analysis assures us that in any separable Hilbert space, there are orthonormal sequences and that each element of the space is expressed by the sum of a series constructed with the elements of a such sequence (see, e.g., Theorems V.9 and V.10 of [3]). In what follows, we will emphasize an important orthonormal sequence in L2 ([−π, π])—the trigonometric system. Definition 4.4.2 For any n ∈ N, let fn : [−π, π] → R defined by: =1 , for every x ∈ [−π, π] , f0 (x) f2n−1 (x) = cos nx , for every x ∈ [−π, π] , and any n ≥ 1, f2n (x) = sin nx , for every x ∈ [−π, π] , and any n ≥ 1. The sequence (fn )n ⊆ C([−π, π]) ⊆ R([−π, π]) ⊆ L2 ([−π, π]) is the trigonometric system. We remark that, for every x ∈ [−π, π], (fn (x))n∈N = (1, cos x, sin x, cos 2x, sin 2x, . . . , cos nx, sin nx, . . .) .

Proposition 4.4.3 The trigonometric system is an orthogonal sequence in L2 ([−π, π]): fn ⊥ fm , for all n = m and f0 2 =

√ √ 2π , fn 2 = π, for any n ≥ 1.

Proof Let’s show that (fn , fm ) = 0, for all n = m. Since (fn )n ⊆ R([−π, π]), it follows that, for all n, m ∈ N, % (fn , fm ) =

%

[−π,π ]

% (f0 , f2m−1 ) =

π

−π

fn fm dλ =

cos mxdx =

π

−π

fn (x)fm (x)dx.

1 sin mx|π−π = 0, for any m ≥ 1. m

4

Chapter 4 • The Lp Spaces

114

% (f0 , f2m ) =

π −π

sin mxdx = − %

(f2n−1 , f2m−1 ) =

4 =

1 · 2

%

π

−π

% %

π −π

%

1 2

%

π

cos nx cos mxdx =

π

−π

cos nx sin mxdx =

[sin(n + m)x − sin(n − m)x] dx = 0, for all n, m ≥ 1.

(f2n , f2m ) =

=

−π

[cos(n + m)x + cos(n − m)x]dx = 0, for all n, m ≥ 1, n = m.

(f2n−1 , f2m ) = 1 = 2

π

1 cos mx|π−π = 0, for any m ≥ 1. m

π

−π

sin nx sin mxdx =

[− cos(n + m)x + cos(n − m)x] dx = 0, for all n, m ≥ 1, n = m.

π

It follows * that fn ⊥ fm , for all n, m ∈ N, n = m. &π 2 ||fn ||2 = −π fn (x)dx, for any n ∈ N and then ||f0 ||2 =

√

2π , +%

||f2n−1 ||2 =

−π

+% ||f2n ||2 =

π

+%

π

cos2 nxdx

= +%

sin nxdx =

−π π

2

−π

π

−π

√ 1 + cos 2nx dx = π , 2

√ 1 − cos 2nx dx = π. 2



Definition 4.4.4 The sequence (en )n , where, for any n ∈ N, en = sequence. For every x ∈ [−π, π],  (en (x))n∈N =

1 fn 2

· fn , is an orthonormal

 1 1 1 1 1 √ , √ cos x, √ sin x, . . . , √ cos nx, √ sin nx, . . . . π π π π 2π

For every f ∈ L2 ([−π, π]) the Fourier coefficients associated with f are defined by cn = (f, en ) = f1n 2 · (f, fn ), for any n ∈ N, and the Fourier series associated with f and with the trigonometric system is (Continued )

115 4.4 · Fourier Series in L2 ([−π, π])

Definition 4.4.4 (continued) x →

∞ 

∞

cn ·en (x) =

n=0

1 1  ·(f, f0 )+ · [(f, f2n−1 ) · cos nx + (f, f2n ) · sin nx] . 2π π n=1

To simplify the writing, let us denote, for any n ∈ N, an =

1 · π

% [−π,π ]

f (x) cos nxdλ(x), bn =

1 · π

% [−π,π ]

f (x) sin nxdλ(x).

Then the Fourier series associated with f is ∞

a0  + (an cos nx + bn sin nx). 2 n=1

Theorem 4.4.5 Let f ∈ L2 ([−π, π]), and let (an )n , (bn )n the sequences defined above by an = 1 1 π (f, f2n−1 ), bn = π (f, f2n ). We denote a0  + (ak cos kx + bk sin kx), ∀n ∈ N∗ , ∀x ∈ [−π, π]. 2 n

Sn (x) =

k=1

(Sn )n is the sequence of partial sums of the Fourier series associated with f . For every sequences (αn )n , (βn )n ⊆ R, let α0  + (αk cos kx + βk sin kx), ∀n ∈ N∗ , ∀x ∈ [−π, π] 2 n

Tn (x) =

k=1

be a trigonometric polynomial. Then (1) f − Sn 2 ≤ f − Tn 2 , for any n ∈ N, &  a2 1 1 2 2 2 2 (2) 20 + ∞ n=1 (an + bn ) ≤ π · f 2 = π · [−π,π ] f dλ, (3) limn→+∞ an = 0 = limn→+∞ bn .

Proof Tn being an arbitrary trigonometric polynomial of indicated form,  α02 · (1, 1) + [αk2 · (cos k·, cos k·) + βk2 · (sin k·, sin k·)] = 4 n

(Tn , Tn ) =

k=1

4

Chapter 4 • The Lp Spaces

116

 α2 = 0 · 2π + (αk2 + βk2 ) · π = 4 n



k=1

n α02  2 2 + (αk + βk ) · π. 2 k=1

Then, for any n ∈ N,

4

f − Tn 22 = (f − Tn , f − Tn ) = (f, f ) − 2(f, Tn ) + (Tn , Tn ) = = f 22 − 2

n 

[αk (f, f2k−1 ) + βk (f, f2k )] − α0 (f, f0 ) + (Tn , Tn ) =

k=1

= f 22 − πα0 a0 − 2π

n n   π (αk ak + βk bk ) + α02 + π (αk2 + βk2 ) = 2 k=1

k=1

 π (α0 − a0 )2 + π [(αk − ak )2 + (βk − bk )2 ]− 2 n

= f 22 +

k=1

−π

a02  2 + (ak + bk2 ) . 2 n

k=1

In particular, if instead of Tn we put Sn , we obtain f

− Sn 22

=

f 22

−π

n a02  2 2 + (ak + bk ) . 2

(*)

k=1

By comparing the two members of the relation (∗), it immediately follows the inequality of point (1) From the last equality, it turns out that, for any n ∈ N, a02  2 1 1 1 + (ak + bk2 ) = · f 22 − · f − Sn 22 ≤ · f 22 . 2 π π π n

k=1

Passing to the limit for n → +∞ in the above inequality, we get the inequality of (2).  2 2 Finally, from the inequality of point (2), it turns out that the series ∞ n=1 (an + bn ) is convergent; so the its general term tends to 0, and this leads to the equality (3).  Remarks 4.4.6 (i) The inequality of (1) shows us that the sequence of partial sums for the Fourier series associated with f gives the best approximation in norm of f among the other trigonometric polynomials. We will demonstrate later that this sequence L2 -converges to f . (ii) The inequality (2) is called the Bessel inequality. As we will show below, this will actually become an equality.

4

117 4.4 · Fourier Series in L2 ([−π, π])

(iii) The condition (3) can still be written: % lim n

%

[−π,π ]

f (x) cos nxdλ(x) = 0 = lim n

[−π,π ]

f (x) sin nxdλ(x).

In the literature, this result is often encountered as the Riemann-Lebesgue lemma.

To prove that the Fourier series associated with a function L2 -converges to that function, we need some helpful results. Lemma 4.4.7 Let (dn )n∈N be the sequence of real numbers defined by % dn =

π

−π

cos2n tdt, for any n ∈ N.

For any n ∈ N, and for every x, y ∈ R, let Dn (x, y) = d1n · cos2n x−y 2 . Then, for every r ∈]0, π[: & y+r (1) limn y−r Dn (x, y)dx = 1, uniformly with respect to y ∈ [−π + r, π − r]. &π (2) −π Dn (x, y)dx = 1, for any n ∈ N and for every y ∈ R. Proof We remark that, for any n ≥ 1, % dn =

π

−π

cos2n−1 t (sin t) dt = cos2n−1 t sin t|π−π + %

+(2n − 1)

π −π

cos2n−2 t sin2 tdt = (2n − 1)dn−1 − (2n − 1)dn .

From here results the recurrence relation: dn =

2n − 1 dn−1 , for any n ≥ 1. 2n

In the previous relation, we give values to n, from 1 to a number m ∈ N∗ , and, if we multiply the relations, we found: dm =

(2m − 1)!! (2m − 1)!! d0 = · 2π, (2m)!! (2m)!!

(1)

where (2m − 1)!! = 1 · 3 · 5 · · · (2m − 1) and (2m)!! = 2 · 4 · 6 · · · (2m). We remark that, for any m ∈ N∗ , 2m − 1 2m − 2 < . 2m − 1 2m

(2)

Chapter 4 • The Lp Spaces

118

In inequality (2), we give values to m, from 2 to a number n ≥ 2, and by multiplying the relations, we found (2n − 2)!! (2n − 1)!! 0 such that |f (x) − f (y)|
0, there exists g ∈ C([−π, π]) such that f − g2 < ε. The function g being continuous on [−π, π], there exists M > 0 such that |g(x)| ≤ M, for every x ∈ [−π, π]. For any n ∈ N, let Tn : [−π, π] → R, % π % π 1 x −y g(x)dx. Tn (y) = Dn (x, y)g(x)dx = cos2n d 2 n −π −π We notice that, for every y ∈ [−π, π] and for any n ∈ N, % π 1 Tn (y) = n (1 + cos x cos y + sin x sin y)n g(x)dx. 2 dn −π We can easily demonstrate by induction that, for any n ∈ N, and any k = 0, 1, . . . , n, there exist two continuous functions ckn , dkn : R → R, such that (1 + cos x cos y + sin x sin y)n =

n  k=0

(ckn (x) cos ky + dkn (x) sin ky), ∀x, y ∈ R.

4

121 4.4 · Fourier Series in L2 ([−π, π])

Then, for any n ∈ N and for every y ∈ R, Tn (y) =

 n % π 1  n c (x)g(x)dx cos ky+ k 2n dn −π k=0

+

 n % π 1  n d (x)g(x)dx sin ky. k 2n dn −π k=0

For any n ∈ N, and any k = 0, 1, . . . , n, we denote αkn =

%

1 2n dn

π −π

ckn (x)g(x)dx and βkn =

1 2n dn

%

π −π

dkn (x)g(x)dx.

Then Tn = α0n f0 +

n  (αkn f2k−1 + βkn f2k ). k=1

From (1) of Theorem 4.4.5, for any n ∈ N, f − Sn 2 ≤ f − Tn 2 ≤ f − g2 + g − Tn 2 .

(*) u

On the other hand, from Lemma 4.4.8, for every [a, b] ⊆ (−π, π), Tn −−→ g. [a,b]

·

It follows that |g − Tn |2 −−−−→ 0. [−π,π ]

Then, for any n ∈ N and for every y ∈ R, % |Tn (y)| ≤

π

−π

% Dn (x, y)|g(x)|dx ≤ M

π −π

Dn (x, y)dx = M and so,

|Tn − g|2 ≤ (|Tn | + |g|)2 ≤ (M + M)2 = 4M 2 .   So we can apply to |Tn − g|2 n∈N the bounded convergence theorem (Corollary 3.3.10). &π ·2 Then limn −π |Tn − g|2 dλ = 0; in other words Tn −→ g. Using the relation (∗), it follows that lim supn f − Sn 2 ≤ ε, for every ε > 0, from ·2

where Sn −→ f , which concludes the demonstration.



Corollary 4.4.10 (Parseval Equality) For every f ∈ L2 ([−π, π]), let (an )n∈N , (bn )n∈N be the Fourier coefficients associated with f ; then ∞

a02  2 1 + (an + bn2 ) = 2 π n=1

%

π −π

f 2 (x)dx.

Chapter 4 • The Lp Spaces

122

Proof According to the previous theorem, if f ∈ L2 ([−π, π]), then the Fourier series associated with f L2 -converges to f . The result follows by passing to the limit in the relation (∗) of the proof of Theorem 4.4.5. 

4

Corollary 4.4.11 (1) Every function f ∈ L2 ([−π, π]), for which all the Fourier coefficients are zero, is zero almost everywhere. (2) If two functions f, g ∈ L2 ([−π, π]) have the same Fourier coefficients, then f = g is almost everywhere. Proof (1) If an = bn = 0, for any n ∈ N, then the Fourier series associated has partial sums Sn nules, for any n ∈ N. & ·2 As Sn −→ f, f 2 = Sn − f 2 → 0. Then A |f |2 dλ = 0, and so f = 0 a.e. (see 1) of Theorem 3.3.3). (2) If f, g ∈ L2 ([−π, π]) have the same Fourier coefficients, then f − g has all the Fourier coefficients zero, and so, according to point (1), f − g = 0 a.e.  Examples 4.4.12 ⎧ ⎪ ⎨ −1, x < 0, (i) Let f : [−π, π] → R, f (x) = signx = 0, x = 0, ⎪ ⎩ 1, x > 0. It is obvious that f ∈ L2 ([−π, π]). Since f is an odd function, an = 0, for any n ∈ N: bn =

1 · π

%

π

−π

f (x) · sin nxdx =

# −2 " (−1)n − 1 . nπ

Parseval equality is ∞ 

bn2 =

n=1

1 · π

%

π −π

f 2 (x)dx = 2,

from where ∞  n=1

As

∞

1 π2 = . (2n − 1)2 8

1 n=1 n2

=

∞

1 n=1 (2n−1)2

+

1 4

·

∞

1 n=1 n2 ,

it follows that

123 4.5 · Abstract Setting

∞  π2 1 . = 2 n 6 n=1

(ii) Let f : [−π, π] → R, f (x) = |x|. Since f is an even function, for all n ∈ N∗ , &π &π bn = 0. a0 = π1 · −π |x|dx = π, and, for any n ≥ 1, an = π1 · −π |x| cos nxdx = 2 [(−1)n − 1]. n2 π Parseval equality is ∞

a02  2 1 + an = · 2 π n=1

%

π −π

x 2 dx =

2π 2 , 3

from where ∞ 

2 a2n−1 =

n=1

∞ 16  π2 π2 1 or 2 . = 6 π (2n − 1)4 6 n=1

Considering that

∞

1 n=1 n4

=

∞

1 n=1 (2n−1)4

+

1 16

∞

1 n=1 n4 ,

it follows that

∞  1 π4 = . 4 n 90 n=1

4.5

Abstract Setting

¯ + be a complete and σ -finite Let (X, A) be a measurable space, and let γ : A → R measure on the σ -algebra A. As above, we will limit to describe the essential points and to state the main theorems (without demonstration) for the theory of spaces Lp in this abstract framework. Definition 4.5.1 Let p ≥ 1; a function A-measurable f : X → R is said to be p-integrable if |f |p ∈ L1+ (X). Let Lp (X) be the set of all p-integrable functions. We notice that, according to Theorem 3.6.7, L1 (X) is accurate to the set of all integrable functions on X.  Let f = ni=1 ai χA be a simple function: {a1 , . . . , ap } ⊆ R and {A1 , . . . , Ap } is a i  A-measurable partition of X. Since |f |p = ni=1 |ai |p χA , f ∈ Lp (X) if and only i if ai = 0, for any i for which γ (Ai ) = +∞. Therefore f ∈ Lp (X) if and only if f ∈ E 1 (X) ⊆ L1 (X) (E 1 (X) is the set of all integrable A-simple functions). & 1 Let  · p : Lp (X) → R+ , defined by f p = X |f |p dγ p .

4

124

Chapter 4 • The Lp Spaces

The inequalities of Hölder and Minkowski (see Theorems 4.1.4 and 4.1.5) can also be demonstrated in this context:

4

Theorem 4.5.2 (Hölder Inequality) Let p, q > 1 such that p1 + q1 = 1. For every f ∈ Lp (X) and every g ∈ Lq (X), f · g ∈ L1 (X) and %

% f · g1 =

 1 % 1 p q q |f | dγ · |g| dγ = f p · gq .

|f · g|dγ ≤

p

X

X

X

Theorem 4.5.3 (Minkowski Inequality) For any p ≥ 1 and for every f, g ∈ Lp (X), f + g ∈ Lp (X) and f + gp = % =

1 |f + g|p dγ

X

p

%

1

≤

|f |p dγ

p

%

1

+

X

|g|p dγ

p

= f p + gp .

X

We can also show that space (Lp (X),  · p ) is a complete seminormed space. Theorem 4.1.12 is still valid: Theorem 4.5.4 If γ (X) < +∞ and if 1 ≤ p < r, then Lr (X) ⊆ Lp (X) and f p ≤ [γ (X)]

r−p rp

· f r , for every f ∈ Lr (X).

The relation defined by f ∼ g if and only if f = g γ -almost everywhere is an equivalence relation on Lp (X); then we denote by Lp (X)|∼ = Lp (X) the quotient set. We can define correctly the mapping:  · p : Lp (X) → R+ , [f ]p = f p , for every equivalence class [f ] ∈ Lp (X). The space (Lp (X),  · p ) is a Banach space. The spaces (Lp (X),  · p ), p ≥ 1 are called classic Banach spaces. Remark 4.5.5 If X = N is the set of natural numbers and if γ is the counting measure (see (ii) of 1.4.6), then Lp (N) = Lp (N) = p is the space of all real p-summable sequences ∞ 1 p p < +∞). (x = (xn )n∈N , xp = n=0 |xn |

We can also find some density results in Lp (X).

125 4.6 · Exercises

Theorem 4.5.6 (1) E 1 (X) is dense in (Lp (X),  · p ). (2) If (X, τ ) is a normal topological space, A is a σ -algebra on X such that τ ⊆ A, and if γ is a complete σ -finite regular measure on (X, A), then Cp (X) = C(X)∩ Lp (X) is dense in (Lp (X),  · p ). Moreover, if (X, τ ) is compact and if γ is a finite measure, then the set of all continuous functions on X, C(X), is dense in (Lp (X),  · p ).

We can also define L∞ (X) as the set of all measurable functions f : X → R for which f ∞ = inf{α ∈ R+ : |f | ≤ α γ − a.e.} < +∞. The space (L∞ (X),  · ∞ ) is a complete seminormed space, and the quotient space (L∞ (X),  · ∞ ) is a Banach space. If γ (X) < +∞, then we can prove the Riesz theorem (Theorem 4.3.11).

4.6

Exercises

(1) Let fn :]0, 1] → R, fn (x) =

n √ . Show that 1+n x

(a) (fn ) ⊆ L2 (]0, 1]). (b) There exists f :]0, 1] → R such that fn (x) → f (x), for every x ∈]0, 1]. (c) (fn ) don’t converge in (L2 (]0, 1]),  · 2 ) to f . (2) Let f, g be two Riemann integrable functions on [a, b]; show that %

2

b

f (x)g(x)dx

%

b

≤

a

a

%

b

f 2 (x)dx ·

g 2 (x)dx. a

(3) Let f ∈ L1 (R), f (x) > 0, for every x ∈ R; show that −1

(4)

1 ∈ / L1 (R). f 1

Indication: One use Hölder inequality of with the functions f 2 , f 2 and p = q = 12 . √ Let f : [0, 1] → R+ such that f ∈ L1 ([0, 1]); show that

%

. [0,1]

+% f dλ ≤

f dλ. [0,1]

Compare with Jensen’s inequality. 1 (5) Let f :]0, +∞[→ R, f (x) = √ . x · ex Show that f ∈ L1 (]0, +∞[) \ L2 (]0, +∞[). · → f. (6) Let (fn ) ⊆ Lp (A) and f ∈ Lp (A) such that fn − ·p

A

Show that fn −−→ f if and only if fn p → f p . A

4

126

Chapter 4 • The Lp Spaces

Indication: For the implication “⇐” it will first be shown that, for all a, b ∈ R, and for any p ≥ 1,

(7)

4

|a + b|p ≤ 2p−1 (|a|p + |b|p ); then Fatou’s lemma will be applied to the sequence (gn )n where gn =   2p−1 |f |p + |fn |p − |f − fn |p . ·p p , (fn )n ⊆ Lp (A) and f ∈ Lp (A) such that fn −−→ f . Let p > 1, q = A p−1 q

Show that, for every g ∈ L (A), %

% (fn · g)dλ → A

(f · g)dλ. A

(8) Prove the following properties in L2 (A): (a) |(f, g)| ≤ f 2 · g2 , for every f, g ∈ L2 (A). ·2

(b) If fn −−→ f , then (fn , g) → (f, g), where (fn )n ⊆ L2 (A), f, g ∈ L2 (A). A

(c) f + g22 + f − g22 = 2 · (f 22 + g22 ), for every f, g ∈ L2 (A). ·2

(d) If f ⊥ gn , for any n ∈ N and gn −−→ g, then f ⊥ g. A

(9) Write Parseval equality for functions: (a) f : [−π, π] → R, f (x) = x 2 . (b) f : [−π, π] → R, f (x) = χ[0, α] , α ∈]0, π]; see the case α = (10) Show that the series

π . 2

∞  sin nx √ is pointwise convergent on [−π, π], but it cannot n n=1

be the Fourier series for any function f ∈ L2 ([−π, π]).

127

Lebesgue Integral on R2

Let R × R = R2 = {(x1 , x2 ) : x1 , x2 ∈ R} and R × R × R = R3 = {(x1 , x2 , x3 ) : x1 , x2 , x3 ∈ R}; by the usual operations of addition and multiplication with real scalars (operations defined on the coordinates), these sets are organized * like real vector spaces. The mappings defined by (x1 , x2 ) → (x1 , x2 ) = x12 + x22 and respectively * (x1 , x2 , x3 ) → (x1 , x2 , x3 ) = x12 + x22 + x32 , are norms on these spaces, that is to say that the following properties are satisfied: (1) x = 0 if and only if x = 0 (0 = (0, 0) or 0 = (0, 0, 0)). (2) c · x = |c| · x, for any c ∈ R and for every x ∈ R2 (or x ∈ R3 ). (3) x + y ≤ x + y, for every x, y ∈ R2 (or x, y ∈ R3 ).

Any norm on a vector space defines a metric on this space; the mapping (x, y) → d(x, y) = x − y is a metric on R2 (R3 ); it has the following properties: (1) d(x, y) = 0 if and only if x = y. (2) d(x, y) = d(y, x), for every x, y ∈ R2 (or x, y ∈ R3 ). (3) d(x, y) ≤ d(x, z) + d(y, z), for every x, y, z ∈ R2 (or x, y, z ∈ R3 ). Let x ∈ R2 (R3 ) and let r > 0; we call open ball centered at x and of radius r the set S(x, r) = {y ∈ R2 (R3 ) : d(x, y) < r}. The set T (x, r) = {y ∈ R2 (R3 ) : d(x, y) ≤ r} is the closed ball centered at x and of radius r.

5.1

Lebesgue Measure on R2

First, we will define some topological notions in R2 . With slight adaptations, they can be easily rewritten for R3 .

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2021 L. C. Florescu, Lebesgue Integral, Compact Textbooks in Mathematics, https://doi.org/10.1007/978-3-030-60163-8_5

5

128

Chapter 5 • Lebesgue Integral on R2

Definition 5.1.1 A set D ⊆ R2 is open set if D = ∅ or, in the case where D = ∅, for every x ∈ D, there exists r > 0 such that S(x, r) ⊆ D. A set F ⊆ R2 is closed if this complement, F c , is open. It is said that the family of open sets of R2 is the usual topology of R2 ; it is denoted by τu2 . The family of closed sets is denoted by F 2 .

5

Using the above definition, it is easy to prove the following proposition: Proposition 5.1.2 The usual topology has the following properties: (1) For every D, G ∈ τu2 , D ∩ G ∈ τu2 .  (2) For every {Di : i ∈ I } ⊆ τu2 , i∈I Di ∈ τu2 . 2 2 (3) R , ∅ ∈ τu . The family of closed sets has the dual properties: (1 ) For every F, H ∈ F 2 , F ∪ H ∈ F 2 .  (2 ) For every {Fi : i ∈ I } ⊆ F 2 , i∈I Fi ∈ F 2 .

2 2 (3 ) R , ∅ ∈ F .

Definition 5.1.3 A neighborhood of x ∈ R2 is a set V ⊆ R2 which contains an open ball: S(x, r) ⊆ V . Let V (x) be the family of all neighborhoods of x. A point x ∈ R2 is said to be adherent to A ⊆ R2 when every neighborhood V of x meets A (V ∩ A = ∅), in other words, every open ball S(x, r) meets A. Let A¯ be the set of all adherent points of A; set A¯ is said to be the closure of A. A point x ∈ R2 is called interior point of A ⊆ R2 when A is a neighborhood of x. The interior of A, A◦ , is the set of all interior points of A. A set A ⊆ R2 is bounded if there is r > 0 such that A ⊆ S(0, r). A sequence (xn )n ⊆ R2 converges to x ∈ R2 if xn − x → 0. A set K ⊆ R2 is compact if it is bounded and closed.

Remarks 5.1.4 (i) x ∈ A¯ if and only if there exists a sequence (xn )n ⊆ A, xn → x. (ii) F is closed set if and only if F = F¯ . (iii) Let (xn )n ⊆ R2 , xn = (x1n , x2n ), for any n ∈ N and let x = (x1 , x2 ) ∈ R2 . Then xn → x if and only if x1n → x1 and x2n → x2 . (iv) A set K ⊆ R2 is compact if and only if every sequence in K has a subsequence convergent to a point of K. (v) A set K ⊆ R2 is compact if and only if from any open cover of K, we can extract a finite subcover (the proof is similar to that of Lemma 7.1.6).

5

129 5.1 · Lebesgue Measure on R2

6

y

y2

I

x2 x (0, 0)

x1

y1

⊡ Fig. 5.1 A two-dimensional open interval

Definition 5.1.5 Let x = (x1 , x2 ), y = (y1 , y2 ) ∈ R2 ; then x ≤ y if x1 ≤ y1 and x2 ≤ y2 . This is a partial order on R2 (it is not possible to compare any two elements of R2 ). If x = (x1 , x2 ) ≤ y = (y1 , y2 ), then we can define the two-dimensional open interval: I =]x, y[=]x1 , y1 [×]x2 , y2 [= {(z1 , z2 ) ∈ R2 : x1 < z1 < y1 , x2 < z2 < y2 } The above ⊡ Fig. 5.1 shows us that the interval I is the interior of a rectangle whose sides are parallel to the coordinate axes. We will denote with I (R2 ), or with I if there is no danger of confusion, the family of open intervals; it’s obvious that I ⊆ τu2 . We can also define the two-dimensional closed intervals: [x, y] = [x1 , y1 ] × [x2 , y2 ]; every closed interval is a closed set. We can also define the other types of intervals, generically marked by J = |x, y| = |x1 , y1 | × |x2 , y2 | (the vertical bar can be a closed bracket or an open bracket). We will denote with J (R2 ), or with J if there is no risk of confusion, the family of all the intervals in R2 . In the plan, we will consider only bounded intervals! For every interval J = |x, y| = |x1 , y1 | × |x2 , y2 | ∈ J , the measure of J is |J | = (y1 − x1 ) · (y2 − x2 ); |J | is the area of the rectangle J . For every J = |x1 , y1 | × |x2 , y2 | ∈ J and every z = (z1 , z2 ) ∈ R2 , z + J = |x1 + z1 , y1 + z1 | × |x2 + z2 , y2 + z2 | ∈ J is the image of J by the translation with z; it is obvious that |z + J | = |J |. (Continued )

Chapter 5 • Lebesgue Integral on R2

130

Definition 5.1.5 (continued) A partial order relation is defined similarly in R3 . You can also define threedimensional intervals (these will be parallelepipeds with edges parallel to the coordinate axes). The measure of such an interval will be the volume of the parallelepiped. We also observe that the translation of an interval is an interval of the same type and that the measure is invariant to translations.

5

We will present below the construction of Lebesgue’s measure on R2 ; the concepts and results can be easily adapted to R3 . The method of construction of the measure on R2 follows the same steps as in the case of R: the measure of open sets will be defined, then the outer measure will be constructed in the plane, and the Lebesgue measurable sets will be defined in R2 . Recall that, in the definition of the open sets measure of R, the theorem of the open sets structure plays a major role: any open set of R is uniquely written as a countable union of pairwise disjoint open intervals (Theorem 1.1.3). In R2 , we no longer have a representation of this type; however, we can give a theorem of representation of open sets as a countable union of the two-dimensional nonoverlapping (without common interior points) closed intervals (we have given a similar theorem in the case of R, Theorem 1.1.8). To be able to demonstrate such a result, we need some preparatory lemmas.  Lemma 5.1.6 Let J, J1 , J2 , . . . , Jn ∈ J such that J ⊇ nk=1 Jk and, for any k = l, Jk and n Jl do not have common interior points; then |J | ≥ k=1 |Jk |. Proof Let J1 , J2 , . . . , Jn ∈ J be arbitrary two-dimensional nonoverlapping intervals so that their union is included in the interval J . For any k ∈ {1, . . . , n}, we extend the sides of the rectangle Jk to the border of J . We thus obtain a partition of J into nonoverlapping rectangles: K1 , . . . , Km and a partition N0 , N1 , . . . , Nn of the set {1, . . . , m} such that, for any k ∈ {1, · · · , n}, Jk = ∪i∈Nk Ki and ∪i∈N0 Ki = J \ ∪nk=1 Jk . We suppose that J = |a, b| × |c, d|; the intersection points of the sides extensions of the intervals Jk with the sides of J determine two partitions a = a0 < a1 < . . . < ap = b and c = c0 < c1 < . . . < cq = d of intervals |a, b|, respectively |c, d| (m = p · q). Then |J | = (b − a) · (d − c) = (b − a) ·

q−1 

(cj +1 − cj ) =

j =0

=

p−1  q−1 

m 

i=0 j =0

l=1

(ai+1 − ai ) · (cj +1 − cj ) =

|Kl | =

 i∈N0

|Ki | +

n   k=1 i∈Nk

|Ki |.

5

131 5.1 · Lebesgue Measure on R2

Similarly, it is shown that, for any k ∈ {1, . . . , n}, |Jk | = |J | ≥

n  

|Ki | =

k=1 i∈Nk

n 

 i∈Nk

|Ki |. It follows that

|Jk |.



k=1

 From the above demonstration, we note that if N0 = ∅, then J = nk=1 Jk and |J | = n k=1 |Jk |. In ⊡ Fig. 5.2, we have imagined a possible situation to illustrate the previous lemma. The intervals delimited by continuous lines are the intervals Jk (the large rectangle is J ). The extensions of the sides of the rectangles Jk were marked by dotted lines. Then, J ⊃ ∪4k=1 Jk , N0 = {1, 6, 7, 9}, N1 = {4}, N2 = {2, 5}, N3 = {3}, andN4 = {8}. According to this partition, J1 = K4 , J2 = K2 ∪ K5 , J3 = K3 , andJ4 = K8 . c3 = d

J4 K7

K8

K9

c2 J1

J2

K4

K5

K6

c1

J3 K2

K1

K3

c = c0 a = a0

a2

a1

a3 = b

⊡ Fig. 5.2 A possible situation for the interval J = [a, b] × [c, d] and the intervals J1 , · · · , Jn

In the following lemma, we will no longer consider the intervals Jk with disjoint interiors. Lemma 5.1.7 Let J1 , . . . , Jn ∈ J such that J = |J | ≤

n  k=1

|Jk |

n k=1

Jk ∈ J ; then

Chapter 5 • Lebesgue Integral on R2

132

Proof We proceed as in the previous lemma demonstration by extending the sides of the intervals  Jk and obtaining a new partition of J : J = m , Nn } a cover (but not l=1 Kl and {N1 , . . . a partition!) of {1, . . . , m} such that, for any k = 1, · · · , n, Jk = i∈Nk Ki (in this case N0 = ∅). Then

5

|J | =

m 

|Kl | ≤

l=1

n  

|Ki | =

k=1 i∈Nk

n 

|Jk |.

Lemma 5.1.8 Let J1 , . . . , Jn , K1 , . . . , Kp ∈ J s.t. Ji and Jl don’t have common interior points, then n 

|Ji | ≤

p 



k=1

n

i=1 Ji

⊆

p

j =1 Kj .

If, for any i = l,

|Kj |.

j =1

i=1

Proof  Let Lij = Ji Kj ∈ J ; then, for any i = l, Lij and Llj do not have common interior  points and are included in Kj . According to Lemma 5.1.6, ni=1 |Lij | ≤ |Kj |, for any j = 1, . . . , p, from where p n  

|Lij | =

i=1 j =1

But Ji = n 

p  n 

|Lij | ≤

j =1 i=1

p

j =1 Lij

|Ji | ≤

|Kj |.

j =1

and then, according to Lemma 5.1.7, |Ji | ≤

p n  

|Lij ≤

i=1 j =1

i=1

p 

p  j =1

|Kj |.

p

j =1 |Lij |

from where



Lemma 5.1.9 Let (Jn )n and (Km )m be two sequences of closed intervals such that ∞ ∞ n=1 Jn ⊆ m=1 Km . If, for any l = n, Jl and Jn don’t have common interior points, then ∞  n=1

|Jn | ≤

∞ 

|Km |.

m=1

Proof  ∞ Let us suppose that ∞ n=1 |Jn | > m=1 |Km |. Then there exist N ∈ N and ε > 0 such that ∞ N |J | > |K |+ε. For any m ∈ N∗ , let Im be an open interval such that Km ⊆ Im m n=1 n m=1 ε and |Im | < |Km | + m . ∞ 2 The set C = N n=1 Jn is compact (bounded and closed) and C ⊆ m=1 Im . Therefore, {Im : m ∈ N∗ } is an open cover of the compact set C; from this open cover, we can extract a  finite subcover (see (v) of 5.1.4). Let M ∈ N∗ such that C ⊆ M m=1 Im .

5

133 5.1 · Lebesgue Measure on R2

As

N

N 

n=1 Jn

|Jn | ≤

n=1

⊆

M 

M

m=1 Im ,

|Im | ≤

m=1

we can use Lemma 5.1.8:

∞ 

|Im |
0 such that the open square P =]x1 − ε, x1 + ε[×]x2 − ε, x2 + ε[⊆ D. Let n ∈ N such that 21n < ε and let k = [2n x1 ], l = [2n x2  ] ∈ Z, where [a]is the leastinteger greater than or equal to a. Then, the closed square k k+1 l l+1 K = n, n × n, n (obtained in the network 2kn ) contains x and is contained 2 2 2 2 in P . K is either contained in one of the sets Jm , m < n, or is one of the squares constituting  ∞ the set Jn . It follows that x ∈ ∞ n=1 Jn and then D = n=1 Jn .

135 5.1 · Lebesgue Measure on R2

 ∞ If D admits two of these representations D = ∞ K of which, for each, n=1 Jn = ∞m=1 m ∞ the interiors are disjoint, then according to Lemma 5.1.9, n=1 |Jn | ≤ m=1 |Km | and ∞ ∞ |K | ≤ |J |, from where the equality requested.  m n m=1 n=1

From the proof of Theorem 5.1.12, it follows that any non-empty open set can be represented as a countable union of nonoverlapping dyadic intervals. We can now define the measure of open sets in R2 . Definition 5.1.13 Let D ∈ τu2 ; if D = ∅, then μ(D) = 0. If D = ∅, then, according to the  previous theorem, D = ∞ n=1 Jn , where Jn are two-dimensional nonoverlapping closed intervals; we say that this writing of D is a representation of the open set D.  We will then define the measure of D by μ(D) = ∞ n=1 |Jn |. The definition is consistent because, using Theorem 5.1.12 again, the above sum does not depend on the representation of the set D.

¯ + ; in the following theorem, we will give So we defined an application μ : τu2 → R some of the properties of open sets measure.

Theorem 5.1.14 (1) (2) (3) (4) (5) (6)

μ(∅) = 0, μ(R2 ) = +∞. μ(x + D) = μ(D), for every x ∈ R2 and every D ∈ τu2 . μ(D) ≤ μ(G), for every D, G ∈ τu2 with D ⊆ G.   2 μ( ∞ Dn ) = ∞ n=1 μ(Dn ), for every (Dn )n ⊆ τu pairwise disjoint. n=1 ∞ ∞ μ( n=1 Dn ) ≤ n=1 μ(Dn ), for every (Dn )n ⊆ τu2 . μ(I ) = |I |, for every I ∈ I

Proof (1) R2 admits the following representation as a counting union of nonoverlapping closed squares: R2 =



([k, k + 1] × [l, l + 1]) .

k,l∈Z

It follows that the measure of R2 is an infinite sum of 1 (the area of these squares), and then it is +∞.  ∞ (2) If D = ∞ n=1 Jn is a representation of D, then x+D = n=1 (x+Jn ) is a representation of x + D (since Jn are closed intervals without common interior points, so are the  ∞ intervals x + Jn . It is obvious that μ(x + D) = ∞ n=1 |x + Jn | = n=1 |Jn | = μ(D).

5

136

5

Chapter 5 • Lebesgue Integral on R2

 ∞ (3) Let D = ∞ of the open sets D and n=1 Jn and let G = m=1 Km be the representations  ∞ G. Since D ⊆ G, we can use Lemma 5.1.9; then μ(D) = ∞ n=1 |Jn | ≤ m=1 |Km | = μ(G).  n (4) For any n ∈ N∗ , let Dn = ∞ be a representation of open set Dn . The set D = k=1 Jk  ∞ ∞ ∞ n n D is open and D = n n=1 n=1 k=1 Jk is a representation of it. Indeed, Jk are n m closed intervals. Two distinct intervals Jk and Jl have no common interior points because, if m = n, then Jkn ∩ Jlm = ∅ (are included in the disjointed sets Dn and Dm ), and, if m = n, then k = l and so Jkn and Jln have no common interior points. It follows that μ(D) =

∞ ∞  

|Jkn | =

∞ 

n=1 k=1

μ(Dn ).

n=1

  (5) Let D = ∞ Jk be a representation of open set D = ∞ D and, for any n ∈ N∗ , k=1 ∞ n=1 n∞ ∞ n ∞ n let Dn = k=1 Jk a representation of Dn . Since k=1 Jj ⊆ n=1 k=1 Jk , we can use Lemma 5.1.9 and therefore μ(D) =

∞  k=1

|Jk | ≤

∞  ∞ 

|Jkn | =

n=1 k=1

∞ 

μ(Dn ).

n=1

(6) Let I =]x, y[∈ I ⊆ τu2 , where x = (x1 , x2 ) and y = (y1 , y2 ); then I = ]x1 , y1 [×]x2 , y2 [. Consider the sequences: x1n ↓ x1 , y1m ↑ y1 with x10 < y10 and the p q sequences x2 ↓ x2 , y2 ↑ y2 with x20 < y20 . Then, the open intervals ]x1 , y1 [ and ]x2 , y2 [ can be written as countable unions of closed intervals: ]x1 , y1 [= [x10 , y10 ] ∪

∞  n=0

]x2 , y2 [= [x20 , y20 ] ∪

∞ 

[x1n+1 , x1n ] ∪

[y1m , y1m+1 ] and

m=0

∞ 

p+1

[x2

∞ 

p

, x2 ] ∪

p=0

q

q+1

[y2 , y2

],

q=0

so that their Cartesian product will be a countable union of two-dimensional nonoverlapping closed intervals: ! I = [x10 , y10 ] × [x20 , y20 ] ∪ ∪

∪

∞ 

p+1

[x10 , y10 ] × [x2

∞ !  ! p q q+1 [x10 , y10 ] × [y2 , y2 ] ∪ , x2 ] ∪

p=0

q=0

∞ 

∞ !  [x1n+1 , x1n ] × [x20 , y20 ] ∪

n=0

n,p=0

p+1

[x1n+1 , x1n ] × [x2

! p , x2 ] ∪

5

137 5.1 · Lebesgue Measure on R2

∪

∞ 

q

q+1

[x1n+1 , x1n ] × [y2 , y2

∞ !  ! ] ∪ [y1m , y1m+1 ] × [x20 , y20 ] ∪

n,q=0

∪

∞ 

m=0

p+1

[y1m , y1m+1 ] × [x2

∞ !  p , x2 ] ∪

m,p=0

q

q+1

[y1m , y1m+1 ] × [y2 , y2

! ] .

m,q=0

This being a representation of the open set I , μ(I ) will be the sum of the surfaces of the closed rectangles on the components. If we consider that ∞ ∞   (x1n − x1n+1 ) = x10 − x1 , (y1m+1 − y1m ) = y1 − y10 , n=0 ∞ 

m=0

p

p+1

(x2 − x2

) = x20 − x2 and

p=0

∞  q+1 q (y2 − y2 ) = y2 − y20 , q=0

a simple calculation show that μ(I ) = (y1 − x1 ) · (y2 − x2 ) = |I |.



Once the measure is defined for the open sets, we will proceed as in the case of R: we will define the Lebesgue outer measure in the plan, and then we will define the measurable sets and the Lebesgue measure in the plan. The demonstrations being identical, we will limit ourselves to presenting the main definitions and results concerning the Lebesgue measure on R2 . Definition 5.1.15 ¯ + , defined by The mapping μ∗ : P (R2 ) → R μ∗ (E) = inf{μ(D) : D ∈ τu2 , E ⊆ D}, for every E ⊆ R2 , is the Lebesgue outer measure on R2 .

The outer measure has the following properties:

Theorem 5.1.16 (1) μ∗ (∅) = 0, (2) μ∗ (E) ≤ μ∗ (F ) and E ⊆ F ,  ∞ ∗ 2 (3) μ∗ ( ∞ n=1 En ) ≤ n=1 μ (En ), for every (En )n ⊆ P (R ).

138

Chapter 5 • Lebesgue Integral on R2

Remarks 5.1.17 (i) μ∗ (D) = μ(D), for every D ∈ τu2 . (ii) μ∗ ({x}) = 0, for every x ∈ R2 . (iii) μ∗ (J ) = |J |, for every J ∈ J .   (iv) μ∗ ( nk=1 Ek ) ≤ nk=1 μ∗ (Ek ), for all n ∈ N∗ and every E1 , · · · , En ∈ P (R2 ). (v) μ∗ (x + E) = μ∗ (E), for every x ∈ R2 and every E ⊆ R2 .

5 Definition 5.1.18 A set E ⊆ R2 is a null set or Lebesgue-negligible if μ∗ (E) = 0. According to the definition, E is a null set if and only if, for all ε > 0, there exists  a sequence of nonoverlapping closed intervals (Jn )n such that E ⊆ ∞ n=0 Jn and ∞ |J | < ε. n=0 n A property P is satisfied almost everywhere on the set E ⊆ R2 (P is μalmost everywhere, or a.e. accomplished) if the set EP = {x ∈ E : x does not have the property P } is a null set, which means that μ∗ (AP ) = 0.

Definition 5.1.19 A set E ⊆ R2 is measurable (in the sense of Lebesgue) if, for every ε > 0, there exists D ∈ τu2 such that E ⊆ D and μ∗ (D \ E) < ε. Let L(R2 ) be the family of all measurable sets of R2 ; the mapping μ = μ∗ |L(R2 ) is σ -additive. μ is the Lebesgue measure on R2 . If E ∈ L(R2 ), then L(E) = {F ⊆ E : F ∈ L(R2 )} is the family of all measurable subsets of E. We can easily prove that E ∈ L(R2 ) if and only if, for every ε > 0, there exists a closed set F and an open one D such that F ⊆ E ⊆ D and μ(D \ F ) < ε.

Remarks 5.1.20 (i) τu2 ⊆ L(R2 ). It follows that μ is the extension of the measure of open sets, and then the notation performed does not give rise to confusion. (ii) E ∈ L(R2 ), for every E ⊆ R2 with μ∗ (E) = 0. ∞ 2 2 (iii) n=1 En ∈ L(R ), for every (En )n ⊆ L(R ). 2 (iv) For every E ∈ L(R ) with μ(E) = 0 and every F ⊆ E, F ∈ L(R2 ). (v) Every interval J ∈ J is measurable and μ(J ) = |J |.

It follows that L(R2 ) is a σ -algebra on R2 and that μ is a complete measure on L(R2 ). Then μ will have all the properties of Theorem 1.3.11. At the end of this paragraph, we will give two results showing how the measure of a set of L(R2 ) can be calculated using the Lebesgue measure on R.

139 5.1 · Lebesgue Measure on R2

Theorem 5.1.21 Let A, B ∈ L(R); then A × B ∈ L(R2 ) and μ(A × B) = λ(A) · λ(B).

Proof (1) At the beginning, we assume that λ(A) < +∞ and λ(B) < +∞. (a) If A and B are intervals, then, according to (v) of 5.1.20, A × B ∈ J (R2 ) and μ(A × B) = |A × B| = |A| · |B| = λ(A) · λ(B). (b) Let A, B ∈ τu ; then, according to the theorem of structure of open sets of R ∞ ∞ (Theorem 1.1.3), A = n=1 In and B = m=1 Jm , where (In )n , (Jm )m are sequences of open interval, pairwise disjoint. A × B = ∪n ∪m (In × Jm ) and, since In × Jm are open two-dimensional intervals disjoint, μ(A × B) =



|In × Jm | =

n,m



|In | · |Jm | = λ(A) · λ(B).

n,m

(c) Let now the case where A, B ∈ L(R). Using the characterization given in Exercise 10) of 1.5, for every ε > 0, there exist the closed sets F1 , F2 and the open sets D1 , D2 of R such that F1 ⊆ A ⊆ D1 , F2 ⊆ B⊆ D2 , and λ(D1 \ F1 ) < ε1 , ε ε λ(D2 \ F2 ) < ε2 , where ε1 = min 1, 2(λ(B)+1) and ε2 = min 1, 2(λ(A)+1) . Then F1 × F2 is closed, D1 × D2 is open in R2 , and F1 × F2 ⊆ A × B ⊆ D1 × D2 .  Since (D1 × D2 ) \ (F1 × F2 ) ⊆ [(D1 \ F1 ) × D2 ] [D1 × (D2 \ F2 )], we obtain que μ[(D1 × D2 ) \ (F1 × F2 )] ≤ μ[(D1 \ F1 ) × D2 ] + μ[D1 × (D2 \ F2 )] = = λ(D1 \ F1 ) · λ(D2 ) + λ(D1 ) · λ(D2 \ F2 ) < ε1 · λ(D2 ) + λ(D1 ) · ε2 < < ε1 · (λ(B) + ε2 ) + ε2 · (λ(A) + ε1 ) ≤

ε ε + = ε. 2 2

It follows that A × B ∈ L(R2 ). On the one hand, μ(A × B) ≤ μ(D1 × D2 ) = λ(D1 ) · λ(D2 ) < (λ(A) + ε1 ) · (λ(B) + ε2 ) = λ(A) · λ(B) + ε1 · (λ(B) + 1) + ε2 · (λ(A) + 1) < λ(A) · λ(B) + ε. On the other hand, μ(A × B) ≥ μ(F1 × F2 ) = μ(D1 × D2 ) − μ((D1 × D2 ) \ (F1 × F2 )) > λ(D1 ) · λ(D2 ) − ε ≥ λ(A) · λ(B) − ε. Since ε is arbitrarily positive, it follows that μ(A × B) = λ(A) · λ(B). (2) Now we consider the case where A or B can have the measure +∞. For any n ∈ N, let  ∞ An = A ∩ [−n, n] and Bn = B ∩ [−n, n]. Then A = ∞ n=0 An , B = n=0 Bn , and,

5

Chapter 5 • Lebesgue Integral on R2

140

 since (An )n and (Bn )n are increasing sequences, A × B = ∞ n=0 (An × Bn ). Since An and Bn have a finite measure, it appears from (1) that An × Bn ∈ L(R2 ), for any n ∈ N; therefore A × B ∈ L(R2 ). The sequences (An )n and (Bn )n are increasing and then the sequence (An × Bn )n is also increasing; using the property of continuity form below of the measure (property 6) of Theorem 1.4.7) and point (1) of this proof, it follows that μ(A × B) = lim μ(An × Bn ) = lim λ(An ) · λ(Bn ) = λ(A) · λ(B).

5

n

n



For every E ⊆ R2 and every x, y ∈ R, let Ex = {y ∈ R : (x, y) ∈ E} be the section of E at x and E y = {x ∈ R : (x, y) ∈ E} be the section of E at the second variable y.

Theorem 5.1.22 Let E ∈ L(R2 ). Then, for almost every x ∈ R, Ex ∈ L(R). Let us define f : R → [0, +∞] letting f (x) = λ(Ex ) (f (x) = 0 at the points where Ex ∈ / L(R)), and let Z = f −1 (+∞) = {x ∈ R : λ(Ex ) = +∞}; then Z ∈ L(R), f is measurable and positive on R, and &

% μ(E) =

R

λ(Ex )dλ(x) =

R\Z

λ(Ex )dλ(x), λ(Z) = 0, +∞, λ(Z) > 0.

.

Proof (1) We will first deal with the case where E is bounded; therefore Ex is bounded, for every x ∈ R and so Z = ∅. 2 (a) Let  E = [a, b] × [c, d] be a closed interval of R . For every x ∈ R, Ex = ∅ ,x ∈ / [a, b] ∈ L(R) and λ(Ex ) = (d − c) · χ[a, b] (x). It follows that [c, d] , x ∈ [a, b] x → λ(Ex ) is a measurable function and % R

% λ(Ex )dλ(x) =

[a,b]

(d − c)dλ = (b − a) · (d − c) = μ(E).

(b) Let E ∈ τu2 ; according to the theorem of structure of open sets of R2 (Theo n n rem 5.1.12), E = ∞ n=1 J , where J are two-dimensional nonoverlapping closed

141 5.1 · Lebesgue Measure on R2

 n n intervals. For every x ∈ R, Ex = ∞ n=1 Jx . Since Jx are closed intervals in R, it follows that Ex is a Borel set of R and therefore Ex ∈ L(R).  n The intervals Jxn have no common interior points; then λ(Ex ) = ∞ n=1 λ(Jx ) (see Theorem 1.1.8). Then the function x → λ(Ex ) is the pointwise limit n k of the sequence of functions x → k=1 λ(Jx ); according to point a), this sequence is a sequence of measurable functions, and so its pointwise limit is measurable also (see 5) of Theorem 2.1.18). Now we use the theorem of Beppo Levi (Corollary 3.1.10) and point a) to obtain % R

λ(Ex )dλ(x) =

∞ %  n=1 R

λ(Jxn )dλ(x) =

∞ 

μ(Jn ) = μ(E).

n=1

(c) Now let E be a closed set of R2 . Since E is bounded, there exists an open interval I =]a, b[×]c, d[⊆ R2 such that E ⊆ I . The set D = I \ E is open and μ(E) = & & μ(I ) − μ(D). From b), μ(D) = R λ(Dx )dλ(x) = R [λ(Ix ) − λ(Ex )]dλ(x) = & (b − a) · (d − c) − R λ(Ex )dλ(x). It follows that μ(E) = μ(I ) − μ(D) = & R λ(Ex )dλ(x). (d) Let E ∈ L(R2 ) be any measurable bounded set and let 0 < δ < 1 be an arbitrary number. ▬ From the characterization mentioned by Definition 5.1.19, there exists an open set D 1 and a closed one F 1 such that F 1 ⊆ E ⊆ D 1 and μ(D 1 \ F 1 ) < 14 · δ 2 . The set U 1 = D 1 \ F 1 is open and then, using point b), we obtain % μ(U 1 ) =

R

λ(Ux1 )dλ(x)
0, there is i such that · < ε; since x ∈ / Ei , λ(Uxi ) < 21i · n1 < ε. We have shown that, for every x ∈ R \ N and every ε > 0, there exist a closed set Fxi and an open one Dxi in R such that Fxi ⊆ Ex ⊆ Dxi and λ(Dxi \ Fxi ) < ε. It means that Ex ∈ L(R), for every x ∈ R \ N. Therefore, Ex ∈ L(R) for almost every x ∈ R. The function x → λ(Ex ) is therefore well defined on R (in the points of N, we agree to give of this function the value 0). Moreover, λ(Ex ) = limi→∞ λ(Dxi ); then the above function is the sum of a series of measurable functions (see point b)), and so it is measurable and positive. Since λ(Fxi ) ≤ λ(Ex ) ≤ λ(Dxi ), for every x ∈ R and any i, we can use points b) and c) to obtain 1 2i

% μ(F i ) =

1 n

% λ(Fxi )dλ(x) ≤

R

< μ(F i ) +

% R

λ(Ex )dλ(x) ≤

R

λ(Dxi )dλ(x) = μ(D i )
n.

Therefore, (fn )n is increasing and fn ↑ f , where f : R → [0, +∞], f (x) = λ(Ex ). We remark that the function f can also take the value +∞ and then let Z = f −1 (+∞). According to Corollary 3.1.8, Z ∈ L(R), f is measurable on R, and % lim n

&

% R

fn dλ =

R

f dλ =

R\Z

λ(Ex )dλ(x), λ(Z) = 0,

(3)

+∞, λ(Z) > 0.



By (1), (2), and (3), we obtain the conclusion of the theorem.

In ⊡ Fig. 5.3, we have represented a set E ⊆ R2 and two of it sections, one at x and one at y; the integrals on R of their measures give the measure of the set (the “area” of E).

6 6

E

y μ(E) =

Ex

μ(E) = ?

 R

 R

λ(Ex )dλ(x) λ(E y )dλ(y)



x

Ey

-

⊡ Fig. 5.3 The sections Ex and E y of a set E ∈ L(R2 )

Remarks 5.1.23 (i) In general, the sections of a measurable set in R2 are not all measurable in R. For example, we consider a set N ∈ P (R) \ L(R) (in (ii) of Remark 1.3.17, we have justified the existence of such sets), and let E = {0} × N ⊆ R2 ; E ⊆ {0} × R and μ({0} × R) = 0 (see Exercise 1) of 5.5). Since the measure μ is complete, E ∈ L(R2 ). In the section of this set at 0, E0 = N and so it is not measurable in R.

144

Chapter 5 • Lebesgue Integral on R2

(ii) Similarly, we can show that &

% μ(E) =

R\Z

λ(E )dλ(y) = y

R

λ(E y )dλ(y), λ(Z) = 0, +∞, λ(Z) > 0.

, where

Z = {y ∈ R : λ(E y ) = +∞}. With this extended form of the integral in mind, we will write

5

% μ(E) =

%

R

λ(Ex )dλ(x) =

R

λ(E y )dλ(y).

(iii) If A, B ∈ L(R) and if E ∈ L(A × B), then %

% λ(Ex )dλ(x) =

μ(E) = A

λ(E y )dλ(y). B

Indeed, in this case, for every x ∈ R \ A and every y ∈ R \ B, Ex = E y = ∅. (iv) The formula given in the preceding theorem also allows a quick demonstration of the principle of Cavalieri in plane: If two plane figures are bounded by two parallel lines and if each other line parallel to these two lines cuts the two figures into segments of the same length, then the two figures have the same area. Indeed, let E, F ⊆ R2 be two measurable sets so that any straight line parallel to Oy intersects them after two linear sets of the same length, so λ(Ex ) = λ(Fx ), for all x ∈ R (“length” of a set of R is here the Lebesgue measure of this set). The previous theorem assures us that μ(E) = μ(F ), and, if the “surface” of a set of planes is its Lebesgue measure, we obtain the conclusion of the principle. (v) The process of building the Lebesgue measure on R2 can be easily adapted for R3 (and more generally for Rn , n ≥ 1). Thus, the σ -algebra of Lebesgue measurable subsets L(R3 ) can be constructed in R3 , and then a complete measure, θ, can be constructed on this σ -algebra; this measure has all the properties of the Lebesgue measure on R2 . For every M ∈ L(R3 ) and for every x ∈ R, we can define the section of M at x: Mx = {(y, z) ∈ R2 : (x, y, z) ∈ M} ∈ L(R2 ). We can show that, λ-almost for every x ∈ R, Mx ∈ L(R2 ). The function x → μ(Mx ) is measurable and positive and % θ(M) =

R

μ(Mx )dλ(x).

(vi) The formula given in the previous point provides a quick demonstration of the principle of Cavalieri in space: If two solids are bounded by two parallel planes and if all the intersections of these solids with a plane parallel to the first two have the same area, then the solids have the same volume.

5

145 5.2 · Lebesgue Multiple Integrals

Indeed, let M, N ⊆ R3 be measurable, so that any plane parallel to yOz intersects them after sets with the same area, so μ(Mx ) = μ(Nx ), for all x ∈ R (“area” of a set of R2 is here Lebesgue’s measure of this set). According to point (v), θ(M) = θ(N), and, as “volume” of a set of R3 is the Lebesgue measure of this set, we obtain the conclusion of the principle.

The following theorem is an immediate consequence of Theorem 5.1.22. Theorem 5.1.24 Let A ∈ L(R) and let f ∈ L1 (A). Then the function t → λ(|f | > t) is integrable on [0, +∞[ and %

% |f |dλ = A

[0,+∞[

λ(|f | > t)dλ(t).

Proof Let us denote E = {(x, t) ∈ A × [0, +∞[: 0 ≤ t < |f (x)|}. Let’s show that E ∈ L(A × [0, +∞[).  If f is a positive simple function, then f = nk=1 ak χA , where {a1 , · · · , an } k ⊆ R and {A1 , · · · , An } is a L-partition of A. It is immediately noticed that E = n + k=1 (Ak × [0, ak [) and, according to Theorem 5.1.21, E ∈ L(A × [0, +∞[). Let now f ∈ L1 (A); according to Theorem 2.3.3, there exists a strictly increasing sequence of positive simple functions (fn )n ⊆ E+ (A) such that fn ↑ |f | (why ?). Based on the theorem proven above, En = {(x, t) ∈ A×[0, +∞[: 0 ≤ t < fn (x)} ∈ L(A×[0, +∞[).  ∞ Then E = lim infn En = ∞ n=1 k=n Ek ∈ L(A × [0, +∞[). Therefore, according to Theorem 5.1.22 and to (iii) of Remark 5.1.23, we obtain % μ(E) =

% λ(Ex )dλ(x) =

A

[0,+∞[

λ(E t )dλ(t).

The demonstration ends if we notice that Ex = [0, |f (x)|[ and that E t = (|f | > t).

5.2



Lebesgue Multiple Integrals

¯ +. In the previous paragraph, we built a complete measure μ : L(R2 ) → R Definition 5.2.1 Let E ∈ L(R2 ) and let f : E → R; the function of two variables f is said to be μmeasurable on E, or simply measurable, if, for every a ∈ R, f −1 (] − ∞, a[) = {(x, y) ∈ E : f (x, y) < a} ∈ L(R2 ) (Continued )

146

Chapter 5 • Lebesgue Integral on R2

Definition 5.2.1 (continued) We denote with L(E) the set of all measurable functions on E and with L+ (E) the subset of measurable and positive functions.

5

As in the one-dimensional case, we can show that L(E) is, concerning the usual operations of addition and multiplication with scalars, a real vector space, containing all the real continuous functions on E, C(E). Definition 5.2.2 A function f : E → R is said to be simple on E ∈ L(R2 ) if it takes a finite number p of values on measurable subsets of E, so it is of form f = ak χE , where k  k=1 1, (x, y) ∈ Ek , {E1 , · · · , Ep } is a measurable partition of E and χE (x, y) = k 0, (x, y) ∈ E \ Ek . Let E (E) be the set of all simple functions on E; we will notice that E (E) ⊆ L(E).

We can define the μ-almost everywhere (a.e.) limit of a sequence of measurable functions, and we can prove that it is a measurable function. In addition, the composition of a continuous function with a measurable function is measurable. As in the case of R, we can define the almost uniform convergence and the convergence in measure; the relationships between them and the convergence a.e. are the same as in the one-dimensional case. As in the one-dimensional case, the Riesz theorem (any convergent in measure sequence admits an almost uniform convergent subsequence) and the theorem of Egorov (convergence a.e. on sets of finite measures leads to almost uniform convergence) are true. We can also prove a theorem of simple functions approximation of measurable functions: Theorem 5.2.3 Let E ∈ L(R2 ) and f : E → R. (1) f ∈ L+ (E) if and only if there is a sequence (fn )n ⊆ E+ (E) such that fn ↑ f . (2) f ∈ L(E) if and only if there is a sequence (fn )n ⊆ E (E) pointwise convergent to f on E. (3) If the function f ∈ L(E) is bounded, then there exists a sequence (fn )n ⊆ E (E) uniform convergent on E to f .

The integral of the functions of two variables is introduced by following the same steps as in the case of the functions of a variable. Let E ∈ L(R2 ) and f : E → R.

147 5.2 · Lebesgue Multiple Integrals

(1) If f =

p k=1

ak χE ∈ E+ (E), then k

%%

%% f dμ =

f (x, y)dμ(x, y) =

E

E

p 

ak μ(Ek ).

k=1

(2) If f ∈ L+ (E), then % %

%%

 ϕdμ : ϕ ∈ E+ (E), ϕ ≤ f .

f dμ = sup E

E

(3) If f ∈ L(E) and if at least one of the integrals of its positive part f + = sup{f, 0} or of its negative part f − = sup{−f, 0} is finite, then %%

%%

%%

+

f dμ =

f − dμ ∈ [−∞. + ∞].

f dμ −

E

E

E

A function f ∈ L(E) is μ-integrable if its integral is finite (the integrals of f + and of f − are finite); the integral of f is %%

%%

%%

+

f dμ =

f − dμ ∈ R.

f dμ −

E

E

E

Let L1 (E) be the set of all integrable functions on E and let L1+ (E) be the subset of positive and integrable functions. L1 (E) is a vector subspace of L(E) and the integral is a linear operator on this subspace. As in the one-dimensional case, we can show that f ∈ L1 (E) if and only if |f | ∈ 1 L+ (E) and that $% % $ %% $ $ $ $≤ f dμ |f |dμ. $ $ E

E

We can also prove here the monotone convergence theorem: %%

%% fn dμ ↑

E

f dμ, if (fn )n ⊆ L+ (E) and fn ↑ f, E

and the Fatou’s lemma: if (fn )n ⊆ L+ (E) and if lim infn fn : E → R, then %%

%% lim inf fn dμ ≤ lim inf

E

n

n

fn dμ. E

We can also formulate the Beppo Levi theorem. The integral is σ -additive with respect to the domain of integration:

5

Chapter 5 • Lebesgue Integral on R2

148

Let (En )n ⊆ L(R2 ) with En ∩ Em = ∅, for n = m; if f ∈ L1 (∪∞ n=1 En ), then f ∈ L1 (En ), for any n ∈ N and %% ∪∞ n=1 En

5

f dμ =

∞ %%  n=1

f dμ. En

&& The mapping  · 1 : L1 (E) → R+ , defined by f 1 = E |f |dμ, for every f ∈ L1 (E), is a seminorm and the space (L1 (E),  · 1 ) is complete. As in the one-dimensional case, the quotient space L1 (E) = L1 (E)|=· is a Banach · space with respect to the norm  · 1 , defined by [f ]1 = f 1 (here = is the equality μ-a.e.). We can formulate and prove the dominated convergence theorem.

Theorem 5.2.4 Let E ∈ L(R2 ), (fn )n ⊆ L(E) and g ∈ L1 (E) such that a.e. (1) fn −−→ f and E

(2) |fn | ≤ g, a.e. on E, for any n ∈ N. Then (fn )n ⊆ L1 (E), f ∈ L1 (E) and %%

%% fn dμ →

f dμ.

E

E

We will mention a variable change formula similar to that of the functions of one variable (see Theorem 3.5.6). Theorem 5.2.5 Let g = (g1 , g2 ) : R2 → R2 be an injective function of class C 1 (there exist all partial derivatives of g1 and g2 and they are continuous on R2 ); then, for every E ∈ L(R2 ), g(E) ∈ L(R2 ) and %% μ(g(E)) =

$ $ $detJg (x, y)$ dμ(x, y).

E

For every Borel function f $: R2 → R and for every set E ∈ L(R2 ), f ∈ L1 (g(E)) if $ and only if (f ◦ g) · $detJg $ ∈ L1 (E) and then %

%

$ $ (f (g(x, y)) · $detJg (x, y)$ dμ(x, y).

f (x, y)dμ(x, y) = g(E)

E

149 5.3 · Fubini’s Theorem

∂gWe∂ghave

denoted with detJg the determinant of the Jacobian matrix of g: Jg = 1

1

∂x ∂y ∂g2 ∂g2 ∂x ∂y

. The proof in the one-dimensional case was essentially based on the

monotonicity of g. In the two-dimensional case, in which it is unable to use the monotonicity, the demonstration turns out to be much more difficult; we will indicate, for those interested, the Theorem 3.7.1 of the monograph [1]. Finally, the spaces Lp , 1 ≤ p ≤ +∞ can be defined in the same way in R2 , and properties similar to those of the one-dimensional case can be demonstrated. Remark 5.2.6 All this construction of the integral can be easily adapted for R3 . Using the integral of R2 , we can still give a formula to calculate the measure of the sets of R3 (further, from the formula presented in point (v) de 5.1.23). For every x, y ∈ R, the section of M ∈ L(R3 ) at (x, y) is M(x,y) = {z ∈ R : (x, y, z) ∈ M}. Then M(x,y) ∈ L(R), for μ-almost every (x, y) ∈ R2 . The function (x, y) → λ(M(x,y) ) is measurable and positive and %% θ(M) =

5.3

R2

λ(M(x,y) )dμ(x, y).

Fubini’s Theorem

In this section, we will examine conditions under which the double integral can be reduced to two iterated integrals. Definition 5.3.1 Let A, B ∈ L(R) and let f ∈ L(A × B). The following integrals, if they exist, will be called iterated integrals of the function f : % % A

  % % f (x, y)dλ(y) dλ(x), f (x, y)dλ(x) dλ(y). B

B

A

In Theorem 5.1.22, we have shown that, for all E ∈ L(R2 ), Ex ∈ L(R), for almost every x ∈ R, the function x → λ(Ex ) is measurable and positive on R and μ(E) = & R λ(Ex )dλ(x). By replacing E with χE and noting that, for all x, y ∈ R, χEx (y) = χE (x, y), we can rewrite the formula of measure calculation of E: % %

%% R2

χE dμ =

R

R

 χE (x, y)dλ(y) dλ(x).

5

Chapter 5 • Lebesgue Integral on R2

150

If A, B ∈ L(R) and E ⊆ A × B, then the above formula becomes % %

%% A×B

χE dμ =

A

B

 χE (x, y)dλ(y) dλ(x).

We will show that the above relation works for any measurable and positive application.

5 Theorem 5.3.2 (Tonelli) If f ∈ L+ (A × B), then, λ-almost for every x ∈ A, the function vx = f (x, ·) : B → R+ is measurable and positive on B; the function u : A → [0, +∞], defined by & u(x) =

B

vx dλ =

& B

f (x, y)dλ(y), vx ∈ L1 (B) , +∞, vx ∈ / L1 (B)

is measurable and positive on A; and %%

% f dμ =

A×B

A

¯ + or udλ ∈ R % %

%% f (x, y)dμ(x, y) = A×B

A

B

 ¯ +. f (x, y)dλ(y) dλ(x) ∈ R

Proof (1) We suppose first that f = χE ∈ L+ (A × B). Then E ∈ L(A × B), vx = χEx , u(x) = λ(Ex ), almost for every x ∈ R, and so the proof is a consequence of (iii) of Remark 5.1.23. p (2) Let f = i=1 ai χE ∈ E+ (A × B) ⊆ L+ (A × B). i p vx = i=1 ai χEi x is then measurable for almost every x ∈ A, and u(x) = p i=1 ai λ(Ei x ) is also measurable and %% f dμ(E) = A×B

=

ai λ(Ei x ) dλ(x) =

% % B

p 

% λ(Ei x )dλ(x) =

ai

i=1

A

% , %  p

i=1

= A

ai μ(Ei ) =

i=1

%  p A

p 

 f (x, y)dλ(y) dλ(x).

A

B

i=1

-

ai χE (y) dλ(y) dλ(x) = ix

5

151 5.3 · Fubini’s Theorem

(3) Let now f ∈ L+ (A × B); then there exists an increasing sequence of positive simple functions, (fn )n ⊆ E+ (A × B), which converges to f (see 5.2.3). From (2), for any n ∈ N, almost for every x ∈ A, vxn = fn (x, ·) is measurable. It follows that, almost for every x ∈ A, the sequence (vxn )n = (fn (x, ·))n is a sequence of measurable functions (a countable union of null sets is a null set), and therefore its limit, vx = f (x, ·), is measurable on B. Since vxn = fn (x, ·) ↑ f (x, ·) = vx , we can apply the monotone convergence theorem (Theorem 3.1.7), and so we obtain % un (x) = B

% vxn dλ =

%

%

fn (x, ·)dλ ↑

f (x.·)dλ =

B

B

vx dλ = u(x). B

Because (un )n is a sequence of measurable functions on A, it follows that u is measurable and positive on A but which can take the value +∞ (see (iv) of Remark 2.1.3). Applying again the monotone convergence theorem, the result of point (2), and Corollary 3.1.8, we obtain %%

% %

%% f dμ = lim n

A×B

% = lim n

fn dμ = lim n

A×B

% %

% un dλ =

A

udλ = A

A

A

 fn (x, y)dλ(y) dλ(x) =

B

 f (x, y)dλ(y) dλ(x).



B

Remark 5.3.3 With a similar proof, we obtain, in the hypotheses of Tonelli’s theorem, % %

%% f (x, y)dμ(x, y) = A×B

B

 f (x, y)dλ(x) dλ(y).

A

The following theorem shows that, if the function f is integrable on A × B, the iterated integrals exist and are equal to the integral of the function.

Theorem 5.3.4 (Fubini) Let A, B ∈ L(R) and f ∈ L1 (A × B); then, λ-almost for every x ∈ A, the function vx = f (x, ·) : B → R is integrable on B; the function u : A → R, defined by & u(x) = B f (x, y)dλ(y), is integrable on A (at the points x ∈ A where vx is not integrable, we consider u(x) = 0); and %%

% f dμ =

A×B

udλ or A

% %

%%

 f (x, y)dλ(y) dλ(x).

f (x, y)dμ(x, y) = A×B

A

B

Chapter 5 • Lebesgue Integral on R2

152

Proof The positive part of f is f + = sup{f, 0} and the negative part f − = sup{−f, 0}. Since f ∈ L1 (A × B), f + , f − ∈ L1+ (A × B) ⊆ L+ (A × B). From Tonelli’s theorem, it follows that %%

f + (x, y)dμ(x, y) =

% %

A×B

5

%%

A

f − (x, y)dμ(x, y) =

 f + (x, y)dλ(y) dλ(x) < +∞ and B

% %

A×B

A

 f − (x, y)dλ(y) dλ(x) < +∞. B

It follows that, almost for every x ∈ A, vx+ = f + (x, ·), vx− = f − (x, ·) ∈ L1+ (B); therefore vx ∈ L1 (B). and also u+ , u− ∈ L1+ (A) and then u ∈ L1 (A) and %%

%%

f + dμ −

f dμ = A×B

A×B

% % = A

%%

f − dμ = A×B

 (f + (x, y) − f − (x, y))dλ(y) dλ(x) =

B

% % = A

 f (x, y)dλ(y) dλ(x).



B

Remark 5.3.5 With a similar demonstration, we obtain in the hypotheses of Fubini’s theorem  %% % % f (x, y)dμ(x, y) = f (x, y)dλ(x) dλ(y). A×B

B

A

We conclude this section with another iteration result, an immediate consequence of the theorems of Tonelli and Fubini. Theorem 5.3.6 Let f ∈ L(A × B); if one of the integrals % %

%%

  % % |f (x, y)|dλ(y) dλ(x), |f (x, y)|dλ(x) dλ(y)

|f |dμ, A×B

A

B

B

A

is finite, then % %

%%



f dμ = A×B

% %

f (x, y)dλ(y) dλ(x) = A

B

 f (x, y)dλ(x) dλ(y).

B

A

5

153 5.4 · Abstract Setting

Proof %% Tonelli’s theorem applied to the function |f | gives

|f |dμ = A×B

% % = A

  % % |f (x, y)|dλ(y) dλ(x) = |f (x, y)|dλ(x) dλ(y).

B

B

A

From the hypothesis, one of the above integrals is finite, so all of them are finite. It follows that f is integrable on A × B; using Fubini’s theorem, we get the desired result. 

5.4

Abstract Setting

Let (X, A), (Y, B ) be two measurable spaces; the set of measurable rectangles

R = {A × B : A ∈ A, B ∈ B } is not a σ -algebra on the Cartesian product X × Y . Let then C = A ⊗ B be the σ -algebra generated by R; we recall that C is the smalest σ -algebra on X × Y which contains R. C is the product of σ -algebras A and B . For every function f : X × Y → R, the sections by the two variables x ∈ X and y ∈ Y are fx : Y → R, fx (v) = f (x, v) and f y : X → R, f y (u) = f (u, y). We can show that, for any function C -measurable, f : X × Y → R, every x ∈ X and every y ∈ Y , fx is B -measurable, and f y is A-measurable. If C ⊆ X × Y, x ∈ X, and y ∈ Y , the sections of C are Cx = {v ∈ Y : (x, v) ∈ C}, respectively C y = {u ∈ X : (u, y) ∈ C}; it is obvious that (χC )x = χCx and (χC )y = χC y . If C ∈ C , then Cx ∈ B and C y ∈ A. We suppose that (X, A) and (Y, B ) are provided with two measures σ -finite, γ and respectively ν.

Theorem 5.4.1 For every C ∈ C let fC : X → R+ , fC (x) = ν(Cx ) and gC : Y → R+ , gC (y) = & & γ (C y ). Then, fC is A-measurable, gC is B-measurable, and X fC dγ = Y gC dν. The set function η : C → R+ , defined by % η(C) =

% fC dγ =

X

gC dν, for every C ∈ C , Y

is a measure σ -finite on (X × Y, C ). For every A × B ∈ R ⊆ C , η(A × B) = γ (A) · ν(B). η is the only measure on the product space (X × Y, C ) verifying (∗).

(*)

Chapter 5 • Lebesgue Integral on R2

154

Definition 5.4.2 η is called the product of measures γ and ν, and we denote η = γ ⊗ ν.

5

Remarks 5.4.3 Let λ be the Lebesgue measure on (R, L) and let μ be the Lebesgue measure on (R2 , L(R2 )). (i) We have shown in Theorem 5.1.21 that, for every A, B ∈ L, A × B ∈ L(R2 ) and that μ(A × B) = λ(A) · λ(B). It follows that L ⊗ L ⊆ L(R2 ) and that μ|L⊗L = λ ⊗ λ. (ii) In general, the product measure is not complete even if the two factors are complete measures. Indeed, let V ⊆ R be Vitali’s set (1.2.7 and 1.3.17). (λ ⊗ λ)(R × {0}) = 0, V × {0} ⊆ R × {0} but C = V × {0} ∈ / L ⊗ L (C 0 = V ∈ / L). Since μ is complete ∞ 2 on L(R ), and μ(R × {0}) = μ(∪n=1 ([−n, n] × {0})) = limn μ([−n, n] × {0}) = limn (2n · 0) = 0, V × {0} ∈ L(R2 ). We deduce from here that L ⊗ L  L(R2 ). In fact the Lebesgue measure on R2 , μ, is the completion of λ ⊗ λ (see Exercise 18) of 1.5).

Let L1 (X × Y ) be the space of all integrable functions on (X × Y, A ⊗ B , γ ⊗ ν) (see  Sect. 3.6). If f ∈ L1 (X × Y ), then the integral of f is %

% f (x, y)dη(x, y) = X×Y

f (x, y)d(γ ⊗ ν)(x, y). X×Y

We find in this abstract framework the theorems of Tonelli and Fubini. Theorem 5.4.4 (Tonelli) & If f : X × Y → R+ is C -measurable and positive, then the functions x → Y fx dν & y and y → Y f dγ are measurable and % %

%



f dη =

% %



fx dν dγ =

X×Y

X

Y

¯ +. f dγ dν ∈ R y

Y

X

Theorem 5.4.5 (Fubini) Let f ∈ L1 (X × Y ); then, γ -almost for every x ∈ X, fx ∈ L1 (Y ) and ν-almost for every y ∈ Y , f y ∈ L1 (X). & & Let the functions u : X → R, u(x) = Y fx dν, and v : Y → R, v(y) = X f y dγ . Then u ∈ L1 (X), v ∈ L1 (Y ), and %

%

%

f dν = X×Y

udγ = X

% %

%

  % % f (x, y)dν(y) dγ (x) = f (x.y)dγ (x) dν(y).

f dη = X×Y

X

vdν, or Y

Y

Y

X

5

155 5.5 · Exercises

5.5

Exercises

(1) Let A, B ⊆ R; if λ∗ (A) = 0, then μ∗ (A × B) = 0 and so A × B ∈ L(R2 ). Indication: First of all it will be considered the case where B = [a, b] is a closed bounded interval; it will

(2)

(3)

then be taken into account that R can be represented as a countable union of such intervals. Show that any straight line in the plane is a null set in R2 .

Indication: A straight line parallel to the axis Oy or to the Ox has the following form: {a}×R or respectively R × {b}, and the result follows from the previous point. Let now be a line of equation y = ax + b; we will assume that / a > 0. We will first show that the set 6ε E = {(x, ax + b) : x ≥ 0} is a null set. For every ε > 0, let ε1 = and let the sequence (xnε )n where aπ 2   1 1 x0ε = 0 and xnε = ε1 · 1 + + · · · + , for any n ∈ N∗ . We are going to build closed intervals Jn = 2 n ∞ ∞ ε ] × [ax ε + b, ax ε [xnε , xn+1 n n+1 + b], for any n ∈ N. We show that E ⊆ n=0 Jn and that n=0 |Jn | = ε . 2 Let E = [0, 1] × {0}, F = {0} × [0, 1] ⊆ R ; show that μ(E) = μ(F ) = 0 and

μ(E + F ) = 1 (E + F = {x + y : x ∈ E, y ∈ F }). (4) Using the formula given in Theorem 5.1.22, calculate μ(E), where E = {(x, y) ∈ R2 : x 2 + y 2 = r 2 }. (5) For every E ∈ L(R2 ), x = (x1 , x2 ) ∈ R2 and every a ∈ R let x + E = {x + y = (x1 +y1 , x2 +y2 ) : y = (y1 , y1 ) ∈ E} and a·E = {ay = (ay1, ay2 ) : y = (y1 , y2 ) ∈ E}. Show that x + E, a · E ∈ L(R2 ) and that μ(x + E) = μ(E), μ(a · E) = a 2 · μ(E).

Indication: It will be used the first part of Theorem 5.2.5 for the following variable changes g, h : R2 → R2 , g(y) = x + y, h(y) = a · y . A direct demonstration can also be given: the proof of the sets measurability is

similar to that of Theorem 1.3.15 and for the measure calculation is used Theorem 5.1.22.

(6) Calculate θ (M), where M = {(x, y, z) ∈ R3 : x 2 + y 2 + z2 ≤ r 2 }. (7) Show that if A, B ∈ L(R), f ∈ L1 (A), and g ∈ L1 (B), then the function h : A × B → R, h(x, y) = f (x) · g(y), for every (x, y) ∈ A × B, is integrable on A × B and %

%%

 %  f (x)dλ(x) · g(y)dλ(y) .

f (x) · g(y)dμ(x, y) = A×B

A

B

(8) Let f : [−1, 1] × [−1, 1] → R be defined by  f (x, y) =

, (x, y) = (0, 0) . 0 , (x, y) = (0, 0)

xy (x 2 +y 2 )2

Show that f ∈ / L1 ([−1, 1] × [−1, 1]) but %

% [−1,1]

 [−1,1]

f (x, y)dλ(y) dλ(x) =

%

%

 f (x, y)dλ(x) dλ(y).

[−1,1]

[−1,1]

157

Signed Measures

In Theorem 3.3.1, we have seen that the integral of a function can be considered as a set function σ -additive with respect to the set on which we integrate. This justifies the interest in σ -additives set functions with real values. We will discuss in this chapter some interesting properties of these set functions. We will show that such a function is a difference of two positive measures, and we will present the conditions under which such set functions can be represented as integrals of certain measurable functions. The last paragraph of the chapter presents some connections between the integral and the derivative.

6.1

Decomposition Theorems

In this chapter, we will use the concepts and results of paragraphs 1.4, 2.4, and 3.6 generically titled “Abstract Setting.” Let (X, A) be a measurable space where X is an abstract set and A is a σ -algebra on X. For every B ∈ A, we denote by A(B) the family of sets C ⊆ B, C ∈ A. A(B) is a ¯ + be a complete, σ -finite positive measure σ -algebra on B. Let γ : A → [0, +∞] = R on X. Let us denote by M(X), the vector space of all real measurable functions on X; by M+ (X), the subset of measurable and positive functions; and by E+ (X), the subset of positive simple functions. In paragraph 3.6, we defined the integral of every f ∈ M+ (X) by %

%

 ϕdγ : ϕ ∈ E+ (X), ϕ ≤ f

f dγ = sup X

∈ [0, +∞].

X

L1+ (X) denotes the set of measurable and positive functions which have a finite integral. Finally, for all f ∈ M(X) for which f − = sup{−f, 0} ∈ L1+ (X), let νf : A → ] − ∞, +∞] defined by %

%

νf (B) =

f + dγ −

f dγ = B

B

%

f − dγ , for every B ∈ A. B

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2021 L. C. Florescu, Lebesgue Integral, Compact Textbooks in Mathematics, https://doi.org/10.1007/978-3-030-60163-8_6

6

158

Chapter 6 • Signed Measures

If f + ∈ L1+ (X), then νf takes the values in R; in this case the function f is integrable on X. Let L1 (X) be the vector space of integrable functions on X. According to (4) of Theorem 3.6.14, the integral is σ -additive with respect to the set on which we integrate. It follows that νf is a σ -additive set function. This example justifies the following extension of the concept of measure.

6

Definition 6.1.1 Let ν : A → [−∞, +∞]; ν is called a signed measure if it satisfies: (1) ν(∅) ∞= 0, ∞   (2) ν Bn = ν(Bn ), for every sequence of pairwise disjoint sets (Bn )n ⊆ n=1

n=1

A. (3) ν(A) ⊆] − ∞, +∞] or ν(A) ⊆ [−∞, +∞[ (ν assumes at most one of the values −∞, +∞).

In the following, we will consider that the signed measures take values in ] − ∞, +∞]. The results of this chapter also apply to measures taking values in [−∞, +∞[. A measure ν taking values in R =] − ∞, +∞[ is said to be a finite signed measure. ν is said to be σ -finite if there is a countable partition of X, {Xn : n ∈ N} ⊆ A, such that ν(Xn ) ∈ R, for every n ∈ N. As we noted above, if γ is a positive, complete, and σ -finite measure on X, for every & f ∈ M(X) with f − ∈ L1+ (X), νf : A →]−∞, +∞], defined by νf (B) = B f dγ , is a signed measure; we say that νf is the measure generated by f ; νf is also noted as f · γ . If f ∈ L1 (X), then νf is a finite signed measure.

A signed measure retains certain properties of a positive measure (see Theorem 1.4.7). Proposition 6.1.2 Let ν : A →] − ∞, +∞] be a signed measure; then: (1) ν(B ∪ C) = ν(B) + ν(C), for every B, C ∈ A with B ∩ C = ∅ (ν is finitely additive). (2) ν(B \ C) = ν(B) − ν(C), ∀B, C ∈ A with C ⊆ B and ν(C) < +∞.  (3) ν( ∞ n=1 Bn ) = limn ν(Bn ), for every increasing sequence (Bn )n ⊆ A (ν is continuous from below).  (4) ν( ∞ n=1 Bn ) = limn ν(Bn ), for every decreasing sequence (Bn )n ⊆ A, with ν(B1 ) < +∞ (ν is continuous from above). Proof  (1) Let B1 = B, B2 = C, and Bn = ∅, for any n ≥ 3; then ν(B ∪ C) = ν( ∞ n=1 Bn ) = ∞ n=1 ν(Bn ) = ν(B) + ν(C) + ν(∅) + · · · + ν(∅) + · · · = ν(B) + ν(C). (2) B = (B \C)∪C; using the finite additivity of ν, it follows that ν(B) = ν(B \C)+ν(C) and, since ν(C) < +∞, ν(B \ C) = ν(B) − ν(C).

6

159 6.1 · Decomposition Theorems

 (3) Let B = ∞ n=1 Bn . If there is n0 ∈ N such that ν(Bn0 ) = +∞, then, for any n > n0 , ν(Bn ) = ν(Bn \ Bn0 ) + ν(Bn0 ) = +∞ and so ν(B) = ν(B \ Bn0 ) + ν(Bn0 ) = +∞ = limn ν(Bn ). Let us suppose that ν(Bn ) < +∞, for any n ∈ N∗ . Then, since B = B1 ∪ (B2 \ B1 ) ∪ · · · ∪ (Bn \ Bn−1 ) ∪ · · · , it follows that ν(B) = ν(B1 ) + ν(B2 \ B1 ) + · · · + ν(Bn \  Bn−1 ) + · · · = ν(B1 ) + ∞ n=2 [ν(Bn ) − ν(Bn−1 )] = n = limn [ν(B1 ) + k=2 (ν(Bk ) − ν(Bk−1 ))] = limn ν(Bn ).  ∗ (4) Let B = ∞ n=1 Bn ; since ν(B1 ) = ν(B1 \ Bn ) + ν(Bn ), for any n ∈ N and ν(B1 ) = ν(B1 \ B) + ν(B), it follows that ν(B) < +∞ and ν(Bn ) < +∞, for any n ∈ N∗ .  The sequence (B1 \Bn )n is increasing and ∞ n=1 (B1 \Bn ) = B1 \B. Using properties (2) and (3), ν(B1 ) − ν(B) = ν(B1 \ B) = limn ν(B1 \ Bn ) = ν(B1 ) − limn ν(Bn ), from where it follows the property of continuity from above.  Remark 6.1.3 Signed measures do not verify all the properties of the positive measures of Theorem 1.4.7. In general, a signedmeasure is not monotonic. Indeed, let f : [0, 2] → R 1 , x ∈ [0, 1] ; f ∈ L1 ([0, 2]). The signed measure be the function defined by f (x) = −1 , x ∈]1, 2] generated by f is defined by νf (B) = λ(B ∩ [0, 1]) − λ(B∩]1, 2]), for every B ∈ L([0, 2]) (see Definition 6.1.1). The set [0, 1] ∈ L([0, 2]), νf ([0, 1]) = 1, and νf ([0, 2]) = 0. The property of finite subadditivity (and therefore that of σ -subadditivity) is not verified either by the signed measures. For the above function f , we consider the measurable sets B =] 23 , 2] and C = [0, 32 [; then νf (B ∪ C) = νf ([0, 2]) = 0 > − 16 = νf (B) + νf (C).

Definition 6.1.4 Let ν : A → (−∞, +∞] be a signed measure. A set B ∈ A is said to be ν-positive (ν-negative) if, for every C ∈ A(B) (C ∈ A and C ⊆ B), ν(C) ≥ 0 (respectively ν(C) ≤ 0). If B is ν-positive (ν-negative), then ν(B) ≥ 0 (ν(B) ≤ 0). The converse is not satisfied; indeed, in the example given in the above remark, νf ([0, 2]) = 0, but the set [0, 2] is neither νf -positive (νf (]1, 2]) = −1 < 0) nor νf -negative (νf ([0, 1]) = 1 > 0). The set B ∈ A is said to be ν-null if it is both ν-positive and ν-negative. B is ν-null if and only if ν(C) = 0, for every C ∈ A(B). The empty set ∅ is ν-positive and ν-negative, hence ν-null, for every signed measure ν. If ν is a positive measure on A, then every set of A is ν-positive; since the positive measures are monotonic, a set B ∈ A is ν-null if and only if ν(B) = 0.

160

Chapter 6 • Signed Measures

Example 6.1.5 Let f ∈ M(X) with f − ∈ L1+ (X) and let νf be the signed measure generated by f . The set X+ = (f ≥ 0) is νf -positive, X− = (f ≤ 0) is νf -negative, and (f = 0) is νf -null. Let us remember that (f ≥ 0) = {x ∈ X : f (x) ≥ 0}; (f ≤ 0) and (f = 0) are similarly defined. Moreover, X+ is a maximal set with the property mentioned in the sense that for any other νf -positive set B ∈ A, νf (B \ X+ ) = 0. Indeed, because B \ X+ ∈ A, B \ X+ ⊆ B, νf (B \ X+ ) ≥ 0; if we suppose that νf (B \ X+ ) > 0, then 0 < νf (B \ X+ ) =

6

%

% B\X+

f dγ =

B\X+

(−f − )dγ ≤ 0,

which is absurd. We can even notice that B \ X+ is νf -null set. A similar property of maximality can be formulated for the set X− : for every νf -negative set B ∈ A, νf (B \ X− ) = 0. Proposition 6.1.6 Let ν : A → (−∞, +∞] be a signed measure and let E ∈ A with ν(E) < 0. There exists a ν-negative set F ∈ A, F ⊆ E such that ν(F ) < 0. Proof If E is ν-negative, then the set F = E verifies the conclusion of the proposition. Let us suppose that E is not ν-negative; then E has strictly positive measure subsets. Let n1 be the smallest natural number for which there exists E1 ∈ A(E) with ν(E1 ) ≥ n11 . If F = E \ E1 is ν-negative, then, since ν(E) = ν(E1 ) + ν(F ) < 0, it follows that ν(F ) < 0, and therefore F verifies the conclusion. If E \ E1 is not ν-negative, then it has strictly positive measure subsets. Let n2 be the smallest natural number for which there exists E2 ∈ A(E \ E1 ) with ν(E2 ) ≥ n12 and so on. If Fk = E \ (E1 ∪ · · · ∪ Ek−1 ) is ν-negative, then F = Fk verifies the conclusion. If Fk is not ν-negative, then it has strictly positive measure subsets, and we consider nk the smallest natural number for which there exists Ek ∈ A(Fk ) with ν(Ek ) ≥ n1k and so on. If the above construction does not produce a solution ∞F of the problem after a finite  ∞ number of steps, then we consider F = n=1 Fk = E \ Ek . Since 0 > ν(E) > −∞, n=1 ∞  ν(Ek ) ≥ 0, and ν(E) = ν(F ) + ∞ n=1 ν(Ek ), it follows that ν(F ) ∈] − ∞, 0[. Then n=1  ∞ 1 ∞ k=1 nk ≤ k=1 ν(Ek ) < +∞, from where nk → +∞. Let us show that F verifies the conclusion of proposition. If we suppose that F is not ν-negative, then there exists G ∈ A(F ) such that ν(G) > 0. Let N ∈ N such that ν(G) > N1 and k ∈ N with N < nk . The fact that G ⊆ Fk = E \ (E1 ∪ · · · ∪ Ek−1 ) and ν(G) > N1 contradicts the property that nk is the smallest natural number for which there exists Ek ⊆ Fk with ν(Ek ) ≥ n1k . Therefore, F is ν-negative and, since ν(F ) < 0, F verifies the conclusion of proposition. 

The previous proposition proves its usefulness in the demonstration of the Hahn decomposition theorem.

161 6.1 · Decomposition Theorems

Theorem 6.1.7 (Hahn Decomposition Theorem) For every signed measure ν : A → (−∞, +∞], there exists two sets E, F ∈ A, E ν-negative, F ν-positive such that X = E ∪ F and E ∩ F = ∅. What would be another pair of sets E , F with the same properties, the set EE = FF is ν-null.

Proof Let a = inf{ν(B) : B ∈ A, B ν-negative} and let (Bn )n≥1 ⊆ A be a sequence of ν-negative  sets such that ν(Bn ) ↓ a. Let us denote E = ∞ n=1 Bn , and let us consider (Cn )n≥1 the disjoint sequence associated with (Bn )n≥1 : C1 = B1 , Cn = Bn \ (B1 ∪ · · · ∪ Bn−1 ), for any n ≥ 2. The sequence (Cn )n≥1 is composed of pairwise disjoint ν-negative sets and  E= ∞ n=1 Cn .  ∞ For every C ∈ A(E), ν(C) = ν(C ∩ ∞ n=1 Cn ) = n=1 ν(C ∩ Cn ) ≤ 0. Therefore, E is ν-negative. Then, for any n ∈ N∗ , ν(E \Bn ) ≤ 0 and so ν(E) = ν(Bn )+ν(E \Bn ) ≤ ν(Bn ). Passing to the limit for n → +∞, we obtain ν(E) ≤ limn ν(Bn ) = a. According to the definition of a, it follows that ν(E) = a. Let F = A \ E; if we suppose that F is not ν-positive, then there exists B ∈ A(F ) such that ν(B) < 0. Now we use Proposition 6.1.6; there exists C ∈ A(B) ν-negative such that ν(C) < 0. Then, E ∪ C is ν-negative and ν(E ∪ C) = ν(E) + ν(C) = a + ν(C) < a, which contradicts the definition of a. Therefore, F is ν-positive and then the pair of sets E, F verifies the conclusion of the theorem. Let now E , F another pair of sets, E ν-negative, F ν-positive, X = E ∪ F , and

E ∩ F = ∅. It is clear that E \ E = F c \ (F )c = F c ∩ F = (F ) \ F (F c = X \ F is the complement of F ). For every B ∈ A, B ⊆ E \ E = (F ) \ F, ν(B) ≤ 0 (E is ν-negative) and ν(B) ≥ 0 (F is ν-positive). It follows that ν(B) = 0. Similarly, F \ F = (E ) \ E and, for every B ∈ A, B ⊆ F \ F = (E ) \ E, ν(B) = 0. Hence, EE = FF is a ν-null set. 

Definition 6.1.8 Let ν : A →] − ∞, +∞] be a signed measure; a pair of sets (E, F ) ∈ A × A is said to be a Hahn decomposition of X with respect to ν if E is ν-negative, F is ν-positive, X = E ∪ F , and E ∩ F = ∅. The previous theorem showed us that the set X admits a Hahn decomposition with respect to any signed measure ν; two decompositions coincide up to a ν-null set. Let f ∈ M(X) with f − ∈ L1+ (X); so as we observed it in Example 6.1.5, a Hahn decomposition of X with respect to the generated measure νf is (X− , X+ ), where X− = (f < 0) = {x ∈ X : f (x) < 0} and X+ = (f ≥ 0) = {x ∈ X : f (x) ≥ 0}.

6

162

Chapter 6 • Signed Measures

Definition 6.1.9 Two signed measures ν1 , ν2 : A →] − ∞, +∞] on a measurable space (X, A) are said to be mutually singular (or simple singular) if there exists a set E ∈ A, such that E is ν1 -null and E c = X \ E is ν2 -null. We denote this by ν1 ⊥ ν2 . If ν1 and ν2 are two positive measures, then ν1 ⊥ ν2 if and only if there exists E ∈ A such that ν1 (E) = 0 = ν2 (X \ E).

6

Let γ be a positive, complete, and σ -finite measure on X, and let f, g ∈ M+ (X) such that f = 0 γ -a.e. on E ∈ A and g = 0 γ -a.e. on X \ E; then νf ⊥ νg , where νf and νg are the signed measures generated by f , respectively by g. The following theorem allows us to decompose any real measure into a difference of positive measures. Theorem 6.1.10 (Jordan Decomposition Theorem) For every signed measure ν : A →] − ∞, +∞], there exist two positive measures, ¯ + , such that ν = ν + − ν − , ν + ⊥ ν − , and ν − is finite. ν+, ν− : A → R This decomposition of ν as the difference of two positive orthogonal measures is unique.

Proof Let (E, F ) be a Hahn decomposition of the set X with respect to ν; therefore E is ν-negative, ¯ + be defined by F is ν-positive, X = E ∪ F , and E ∩ F = ∅. Let ν + , ν − : A → R ν + (B) = ν(F ∩ B), ν − (B) = −ν(E ∩ B), for every B ∈ A. It is easy to see that ν + , ν − are two positive measures on X, that ν − is finite, and that, for every B ∈ A, ν + (B) − ν − (B) = ν(F ∩ B) + ν(E ∩ B) = ν((E ∪ F ) ∩ B) = ν(B). Moreover, ν + (E) = ν(F ∩ E) = 0 = −ν(E ∩ F ) = ν − (F ) = ν − (X \ E); hence + ν ⊥ ν− . Let now ν = ν1 − ν2 be another decomposition of ν as difference of positive singular measures. Since ν1 ⊥ ν2 , there exists E1 ∈ A such that ν1 (E1 ) = ν2 (X \ E1 ) = 0. Then (E1 , X \ E1 ) is another Hahn decomposition of X with respect to ν. Indeed, for every B ∈ A(E1 ), ν(B) = ν1 (B) − ν2 (B) = 0 − ν2 (B) ≤ 0 (ν1 is a positive measure; hence, it is monotonic: 0 ≤ ν1 (B) ≤ ν1 (E1 ) = 0). Then, E1 is ν-negative. Similarly, it is shown that X\E1 is ν-positive. According to Theorem 6.1.7, EE1 = FF1 is a ν-null set (we denoted F1 = X \ E1 ). Then, for every B ∈ A, ν((F \ F1 ) ∩ B) = 0 = ν((F1 \ F ) ∩ B) from where, ν + (B) = ν(F ∩ B) = ν((F \ F1 ) ∩ B) + ν((F ∩ F1 ) ∩ B) = = ν((F1 ∩ F ) ∩ B) = ν((F1 ∩ F ) ∩ B) + ν((F1 \ F ) ∩ B) = = ν(F1 ∩ B) = ν1 (F1 ∩ B) − ν2 (F1 ∩ B) = ν1 (F1 ∩ B) = ν1 (B). It is also shown that ν − (B) = ν2 (B).



6

163 6.1 · Decomposition Theorems

Definition 6.1.11 The decomposition ν = ν + − ν − with ν + and ν − positive, singular measures and ν − finite is said to be the Jordan decomposition of the measure ν; the mapping ¯ + , defined by |ν| = ν + + ν − , is a positive measure, and it is called the |ν| : A → R total variation of ν.

Remark 6.1.12 Note that the measure ν is finite if and only if ν + is finite or, equivalently, if and only if |ν| is finite.

Let f ∈ M(X) with f − ∈ L1+ (X); a Hahn decomposition of X with respect to the measure generated by f , νf , is (X− , X+ ), where X− = (f < 0) = {x ∈ X : f (x) < 0} and X+ = (f ≥ 0) = {x ∈ X : f (x) ≥ 0} (see Definition 6.1.8). Note that x ∈ X+ + + if and only if f (x) = sup{f (x), 0} = f + (x); hence & X = {x ∈ X & : f+(x) = f (x)}. + + Then, for every B ∈ A, νf (B) = νf (X ∩ B) = X+ ∩B f dγ = B f dγ . Similarly, & νf− (B) = B f − dγ . It follows that νf+ is the measure generated by f + and νf− is the measure generated by f − . The total variation of νf is νf+ + νf− —the positive measure generated by f + + f − = |f |. The explanation of the name of total variation can be found in the following theorem.

Theorem 6.1.13 Let ν : A →] − ∞, +∞] be a signed measure; the total variation of ν, |ν|, has the following properties: 'n ( ∗ (1) |ν|(B) = sup k=1 |ν(Bk )| : n ∈ N , {B1 , . . . , Bn } ∈ B , where B is the family of all finite partitions of B with sets of A. (2) sup{|ν(C)| : C ∈ A(B)} ≤ |ν|(B) ≤ 2 · sup{|ν(C)| : C ∈ A(B)}, for every B ∈ A. (3) |ν| is the smallest of the positive measures σ for which |ν(B)| ≤ σ (B), for every B ∈ A.

Proof (1) Let {B1 , . . . , Bn } ∈ B be arbitrary; then |ν|(B) =

n  k=1

≥

n  k=1

|ν|(Bk ) =

n  [ν + (Bk ) + ν − (Bk )] ≥ k=1

|ν + (Bk ) − ν − (Bk )| =

n  k=1

|ν(Bk )|.

(∗)

164

Chapter 6 • Signed Measures

On the other hand, if (E, F ) is a Hahn decomposition of X with respect to ν, then {E ∩ B, F ∩ B} ∈ B and |ν(E ∩ B)| + |ν(F ∩ B)| = −ν(E ∩ B) + ν(F ∩ B) = ν − (B) + ν + (B) = |ν|(B). It follows that  |ν|(B) ≤ sup

n 

 ∗

|ν(Bk )| : n ∈ N , {B1 , . . . , Bn } ∈ B ,

k=1

6

which, with the inequality (∗), demonstrates the equality of point (1). (2) For every B ∈ A and every C ∈ A(B), {C, B \ C} ∈ B and then |ν|(B) ≥ |ν(C)| + |ν(B \ C)| ≥ |ν(C)|. Therefore, sup{|ν(C)| : C ∈ A(B)} ≤ |ν|(B). If (E, F ) is a Hahn decomposition of X, then |ν|(B) = ν + (B) + ν − (B) = ν(B ∩ F ) − ν(B ∩ E) = = |ν(B ∩ E)| + |ν(B ∩ F )| ≤ 2 · sup{|ν(C)| : C ∈ A(B)}. (3) Using the first inequality from the previous point, it turns out that |ν| is a positive measure such that |ν(B)| ≤ |ν|(B), for every B ∈ A. ¯ + be an arbitrary positive measure with the property |ν(B)| ≤ σ (B), Let σ : A → R for every B ∈ A, and let (E, F ) be a Hahn decomposition of X with respect to ν. Then, for every B ∈ A, |ν|(B) = ν(B ∩ F ) − ν(B ∩ E) = |ν(B ∩ E)| + |ν(B ∩ F )| ≤ ≤ σ (B ∩ E) + σ (B ∩ F ) = σ (B). Therefore, |ν| ≤ σ .  Remark 6.1.14 Let ν : A →] − ∞, +∞] be a signed measure and let |ν| be the total variation of ν. Then, a set B ∈ A is ν-null if and only if |ν|(B) = 0. Indeed, according to Definition 6.1.4, B is ν-null if and only if ν(C) = 0, for every C ∈ A(B), which is equivalent to sup{|ν(C)| : C ∈ A(B)} = 0. Point (2) of the previous theorem shows us that it comes down to |ν|(B) = 0.

6

165 6.2 · Radon-Nikodym Theorem

6.2

Radon-Nikodym Theorem

¯ + be a positive, σ -finite measure Let A be a σ -algebra on X and let γ : A → R − 1 on X. For every f ∈ M(X) with f ∈ L+ (X), let& νf : A →] − ∞, +∞] be the measure & generated by f ; we defined νf (B) = B f dγ and we observed that |νf |(B) = B |f |dγ (see Definition 6.1.11). Point (2) of Theorem 3.6.8 assures us that if γ (B) = 0, then |νf |(B) = 0; according to Remark 6.1.14, B is νf -null set. We will show in this paragraph that this property characterizes the measures generated by the measurable functions: any signed measure whose total variation is zero on the γ -null sets is generated by a measurable function. Definition 6.2.1 A signed measure ν : A →] − ∞, +∞] is said to be absolutely continuous with respect to a positive measure γ if, for every B ∈ A with γ (B) = 0, ν(B) = 0. We denote this by ν  γ .

Remark 6.2.2 ν  γ if and only if |ν|(B) = 0, for every B ∈ A with γ (B) = 0. The sufficiency of condition results from point (2) of Theorem 6.1.13. To show that this condition is necessary, we observe that because γ is a positive measure and γ (B) = 0, then B is ν-null; Remark 6.1.14 assures us that |ν|(B) = 0. Proposition 6.2.3 Let ν be a finite signed measure; ν  γ if and only if, for every ε > 0, there exists δ > 0 such that, for every B ∈ A with γ (B) < δ, |ν|(B) < ε. Proof The sufficiency: let us suppose that, for every ε > 0, there exists δ > 0 such that, for every C ∈ A with γ (C) < δ, |ν|(C) < ε and let B ∈ A with γ (B) = 0. Then, γ (B) < δ and so |ν|(B) < ε. Since ε is arbitrary, it follows that |ν|(B) = 0 and therefore ν(B) = 0. The necessity: We reason by reduction to the absurd; we suppose that ν  γ and that there is ε0 > 0 so that, for any k ∈ N, there exists Bk ∈ A with γ (Bk ) < 21k = δ and  ∞ |ν|(Bk ) ≥ ε0 . Let B = lim supn Bn = ∞ n=1 k=n Bk ∈ A. For any n ∈ N, γ (B) ≤ γ

∞ 

Bk

k=n

≤

∞ 

γ (Bk )
k) ∈ A. If we suppose that γ (A) = a > 0, then, for any k ∈ N∗ , there is nk ∈ N∗ such that γ (gnk > k) ≥ a and then % k · a ≤ k · γ ((gnk > k)) ≤

%

(gnk >k)

gnk dγ ≤

X

gnk dγ ≤ L, for any k ∈ N∗ ,

which is absurd. Therefore, γ (A) = 0 and, since  ν  γ , ν(A) = 0. limn gn (x) , x ∈ X \ A, Let then f : X → R+ , defined by f (x) = 0 , x ∈ A. The function f is measurable (see Exercise (4) of 2.5) and positive; on the set X \ A, gn ↑ f . Using the theorem of monotone convergence on X \ A, we obtain & & gn dγ ↑ X\A f dγ ≤ L. On the other hand, according to Theorem 3.6.8, &X\A & & & 1 A f dγ =& 0 = A gn dγ . It follows that f ∈ L+ (X) and X gn dγ ↑ X f dγ . Therefore, X f dγ = L. Using again the theorem of monotone convergence on B \ A, we obtain %

% f dγ = B

% f dγ = lim n

B\A

gn dγ ≤ ν(B \ A) = ν(B), for every B ∈ A. B\A

Therefore, f ∈ F . We prove that f is the desired function. & Let us suppose that there exists B0 ∈ A such that B0 f dγ < ν(B0 ). Note that γ (B0 ) > 0 (if we suppose that γ (B0 ) = 0, then ν(B0 ) = 0 and the above strict inequality could not happen). Then, there is ε > 0 such that % η(B0 ) = ν(B0 ) −

f dγ − ε · γ (B0 ) > 0. B0

The mapping η : A → R, defined by % η(B) = ν(B) −

f dγ − ε · γ (B), for every B ∈ A, B

is a signed measure and η(B0 ) > 0. The Hahn decomposition theorem assures us that there exist two sets E, F ∈ A, such that E ∩ F = ∅ and X = E ∪ F and E is η-negative and F is η-positive. Since 0 < η(B0 ) = η(B0 ∩ E) + η(B0 ∩ F ), it follows that η(B0 ∩ F ) > 0 and then η(F ) > 0. For every B ∈ A, η(B ∩ F ) ≥ 0, hence % f dγ + ε · γ (B ∩ F ) ≤ ν(B ∩ F ). B∩F

6

168

Chapter 6 • Signed Measures

Then %

% B

(f + ε · χF )dγ = %

f dγ + ε · γ (B ∩ F ) = B

%

=

f dγ + B\F

f dγ + ε · γ (B ∩ F ) ≤ ν(B \ F ) + ν(B ∩ F ) = ν(B). B∩F

It follows that f + ε · χF ∈ F and then

6

%

%

L≥ X

(f + ε · χF )dγ =

f dγ + ε · γ (F ) = L + ε · γ (F ). X

From the above relation, γ (F ) = 0 and, since ν  γ , ν(F ) = 0. It follows that η(F ) = 0, which contradicts η(F ) > 0. The contradiction we have reached shows that the hypothesis that there is B0 ∈ A & such that B0 f dγ < ν(B0 ) is false. & Therefore, for every B ∈ A, ν(B) = B f dγ which shows that the function f verifies the conclusion of the theorem: ν = νf . (2) Now consider the case where ν is a positive finite measure and γ is a positive σ -finite  measure. Let (Xn )n≥1 ⊆ A be an increasing sequence of sets such that ∞ n=1 Xn = X and γ (Xn ) < +∞, for any n ≥ 1. For any n ≥ 1, we define γn , νn : A → R+ by γn (B) = γ (B ∩ Xn ) and νn (B) = ν(B ∩ Xn ), for every B ∈ A. Since νn  γn , for any n ∈ N∗ , we can apply the previous point of the demonstration to find the positive and γn -integrable functions fn on X such that %

fn dγn , for any n ∈ N∗ and for every B ∈ A.

νn (B) = B

We remark that γn (B) = γ (B), νn (B) = ν(B), for every B ∈ A, B ⊆ Xn and νn (B) = γn (B) = 0, for every B ⊆ X \ Xn ; it follows that %

%

%

(fn+1 − fn )dγ = B

fn+1 dγn+1 − B

fn dγn = B

= νn+1 (B) − νn (B) = ν(B) − ν(B) = 0, for every B ∈ A, B ⊆ Xn ⊆ Xn+1 . Using point (c) of Theorem 3.6.9, fn = fn+1 γ -a.e. on the set Xn . Every γn -integrable function is γ -integrable (see Exercise (21) of 3.7). It follows that (fn )n≥1 ⊆ L1+ (X). We then define consistently f : X → R by f (x) = fn (x), if x ∈ Xn . For any  a ∈ R, f −1 (] − ∞, a[) = n≥1 fn−1 (] − ∞, a[) ∈ A. It follows that the function thus defined is measurable and positive. Using the property of continuity from below of

6

169 6.2 · Radon-Nikodym Theorem

measures on increasing sequences (see (6) of Theorem 1.4.7), we obtain %

%

%

f dγ = lim n

B

f dγ = lim

fn dγn = lim νn (B ∩ Xn ) =

n

B∩Xn

B∩Xn

n

= lim ν(B ∩ Xn ) = ν(B), n

& for every B ∈ A. Particularly, X f dγ = ν(X) < +∞ and so f ∈ L1+ (X). (3) Let ν be a finite signed measure absolutely continuous with respect to a positive and σ -finite measure, γ . Let ν = ν + − ν − be the Jordan decomposition of ν. According to (ii) of Remark 6.2.4, it follows that ν +  γ and ν −  γ . We can now apply the previous case of the demonstration: there exist g, h ∈ L1+ (X) such that ν + (B) =

%

gdγ and ν − (B) =

B

% hdγ , for every B ∈ A. B

The function f = g − h is integrable with respect to γ and verifies the conclusion of the theorem. Let us now deal with the uniqueness. Let f1 , f2 ∈ L1 (X, A, γ ) be two functions such that, for every B ∈ A, %

%

ν(B) =

f1 dγ = B

f2 dγ . B

& Then, B (f1 − f2 )dγ = 0, for every B ∈ A. Theorem 3.6.9 assures us that f1 = f2 , γ -almost everywhere.  Remarks 6.2.6 (i) The theorem can be easily extended to the case where ν is a σ -finite signed measure absolutely continuous with respect to the positive σ -finite measure γ ; in this case, however, the function f will have the integral, without necessarily being γ -integrable. (ii) If we take into account point (iii) of Remark 6.2.4 and the Radon-Nikodym theorem, we can say that if ν is a finite signed measure, then ν  γ if and only if there exists f ∈ L1 (X, A, γ ) tel que ν = f · γ .

Definition 6.2.7 The function f ∈ L1 (X) whose existence has been proven in the Radon-Nikodym theorem is called the Radon-Nikodym derivative of ν with respect to γ , and it is dν denoted by f = . dγ

170

Chapter 6 • Signed Measures

A demonstration similar to that of the Radon-Nikodym theorem can be made for the following Lebesgue decomposition theorem.

Theorem 6.2.8 (Lebesgue Decomposition Theorem) For every finite signed measure, ν : A → R and every σ -finite positive measure γ there exist two signed measures η, ρ : A → R such that η  γ , ρ ⊥ γ , and ν = η + ρ.

6

We conclude this paragraph with a result which expresses the integral with respect to ν  γ using the integral with respect to γ . Theorem 6.2.9 Let ν : A → R+ be a finite positive measure, absolutely continuous with respect to a dν positive σ -finite measure γ , and let f = ∈ L1+ (X, A, γ ) be its Radon-Nikodym dγ derivative. Then, %

%



%

gdν =

g · f dγ =

X

X

g· X

dν dγ

 dγ , for every function g ∈ L1 (X, A, ν).

Proof First consider g = χB , where B ∈ A; then %

% gdν = ν(B) = X

% f dγ =

B

X

% χB · f dγ =

g · f dγ . X

We can immediately extend this result to any positive simple function g ∈ E+ (X). Let now g ∈ M+ (X); then there exists a sequence (gn )n ⊆ E+ (X) such that gn ↑ g. The monotone convergence theorem allows us to write %

%

%

gdν = lim X

n

%

gn dν = lim X

n

gn · f dγ = X

g · f dγ . X

& & & Let g be a ν-integrable function; then X g + dν < +∞, X g − dν < +∞, X g + dν = & + & − & − X g · f dγ , and X g dν = X g · f dγ . By subtracting the last two relationships, we get the desired result. 

6.3

The Integral and the Derivative

In this paragraph, we return to integration on the set of real numbers. The Riemann integral gives a partial answer to the following two questions: (1) Is the integral differentiable with respect to its upper limit?

6

171 6.3 · The Integral and the Derivative

If the function f : [a, b] → R is continuous on [a, b], then %



x

f (t)dt

= f (x), for every x ∈ [a, b].

a

But what if f is only integrable? We will show (see Theorem 6.3.16) that if the function f is Lebesgue integrable and if we replace the Riemann integral by the Lebesgue integral, then the above formula is satisfied for almost all points in the interval [a, b]. (2) If f : [a, b] → R is differentiable at every point of [a, b], then is its derivative f integrable? If we assume& that the derivative f is integrable, then f can be found x by the formula: f (x) = a f (x)dx + k? We recall that if the derivative of f , f , is Riemann integrable on [a, b], then the Leibniz-Newton formula works: % x f (x) = f (x)dx + f (a), for every x ∈ [a, b]. (L − N) a

The formula remains true and in the more general hypothesis that f is Lebesgue integrable on [a, b] (see Corollary 6.3.23). We mention that there are differentiable functions whose derivative, although bounded, is not Riemann integrable but is Lebesgue integrable (see the example of Pompeiu in  Sect. 7.2). But if the function f is differentiable almost everywhere on [a, b], then the formula (L − N) is no longer true (see Cantor staircase 6.3.17). We will see (Theorem 6.3.22) that the Leibniz-Newton formula (L−N) remains true in the case where f is absolutely continuous and the Riemann integral will be replaced by the Lebesgue integral. We denoted by L the σ -algebra of Lebesgue measurable subsets of R and by λ : L → R¯ + the Lebesgue measure on R. A set A ⊆ R is a null set if, for every ε > 0,  there exists a sequence of open intervals (In )n∈N ⊆ I such that A ⊆ ∞ n=0 In and ∞ |I | < ε. We recall that any null set is Lebesgue measurable. A property P (x) n=0 n is satisfied almost for all x ∈ A ⊆ R if the set of points x ∈ A for which P (x) is not satisfied is a null set. We will present below two important notions: that of function with bounded variation and that of absolutely continuous function. Definition 6.3.1 (1) Let f : [a, b] → R and let  = {x0 , x1 , · · · , xn } be a partition of the interval [a, b] (a = x0 < x1 < · · · < xn = b). The variation of f with respect to  is (Continued )

172

Chapter 6 • Signed Measures

Definition 6.3.1 (continued) V (f ) =

n 

|f (xk ) − f (xk−1 )|

k=1

and the total variation of f on [a, b] is Vab (f ) = sup{V (f ) :  ∈ D([a, b])} ∈ [0, +∞],

6

where D([a, b]) is the set of all partitions of [a, b]. We say that f has bounded variation on [a, b] if Vab (f ) < +∞. Let BV[a,b] be the set of all bounded variation functions on [a, b]. (2) The function f is absolutely continuous on [a, b] if, for every ε > 0, there exists δ > 0 such that, for every finite family of nonoverlapping subintervals  of [a, b], {[xk , yk ] : k = 1, · · · , n}, with nk=1 (yk − xk ) < δ we have n k=1 |f (yk ) − f (xk )| < ε. Let AC[a,b] be the set of all absolutely continuous functions on [a, b]. It is obvious that any absolutely continuous function is uniformly continuous; the reverse is not true (see Remark 6.3.21).

Proposition 6.3.2 (1) Every absolutely continuous function on [a, b] has bounded variation on [a, b]. (2) Every monotonic function f : [a, b] → R has bounded variation on [a, b] and Vab (f ) = |f (b) − f (a)|. (3) Every Lipschitz function on [a, b] is absolutely continuous on [a, b]. Proof (1) The demonstration is the same as that of Proposition 5.3.4 of [1]. Let f : [a, b] → R be an absolutely continuous function. We take δ > 0 corresponding to ε = 1 in the definition of absolutely continuous functions. Let 0 = {a0 , a1 , · · · , ap } ∈ D([a, b]) be a fixed partition of mesh 0  < δ. For every partition  = {x0 , x1 , · · · , xn } ∈ D([a, b]), let 1 = 0 ∪  = {y0 , y1 , · · · , ym } ∈ D([a, b]);  ⊆ 1 and then V (f ) ≤ V1 (f ). But V1 (f ) =

p 



|f (yk+1 ) − f (yk )|.

j =1 yk+1 ∈]aj −1 ,aj ]

 Since, for any j = 1, · · · , p, yk+1 ∈]aj −1 ,aj ] (yk+1 − yk ) ≤ aj − aj −1 ≤ 0  < δ,  yk+1 ∈]aj −1 ,aj ] |f (yk+1 − f (yk )| < 1 and then V (f ) < p, for every partition  ∈

D([a, b]). Therefore, Vab (f ) ≤ p. (2) Let f : [a, b] → R be an increasing function. For every partition  =   {x0 , x1 , · · · , xn } ∈ D([a, b]), V (f ) = nk=1 |f (xk ) − f (xk−1 )| = nk=1 (f (xk ) − f (xk−1 )) = f (b) − f (a) = |f (b) − f (a)|.

6

173 6.3 · The Integral and the Derivative

(3) If f : [a, b] → R is a L-Lipschitz function, then |f (x) − f (y)| ≤ L · |x − y|, for every x, y ∈ [a, b]. For every ε > 0, let δ = Lε > 0 and let {[xk , yk ] : k = 1, · · · , n} be a  finite family of nonoverlapping subintervals of [a, b] with nk=1 (yk − xk ) < δ. Then, n n k=1 |f (yk ) − f (xk )| ≤ L · k=1 (yk − xk ) < ε. 

An important class of absolutely continuous functions on the interval [a, b] is the set of differentiable functions at every point of the interval and whose derivative is Lebesgue integrable on [a, b]. Firstly, we need to prove two lemmas. Lemma 6.3.3 Let A ⊆ [a, b] be an arbitrary set, let f : [a, b] → R be a function differentiable at every point of A, and let M = supx∈A |f (x)|. Then, λ∗ (f (A)) ≤ M · λ∗ (A). Proof Recall that λ∗ is the Lebesgue outer measure. If M = +∞, then the inequality is obvious. Suppose now that M < +∞ and let an arbitrary ε > 0. For every n ∈ N∗ , we denote   1 . An = x ∈ A : |f (x) − f (y)| ≤ (M + ε) · |x − y|, ∀y ∈ [a, b], |x − y| < n Obviously, An ⊆ An+1 ⊆ A and then f (An ) ⊆ f (An+1 ) ⊆ f (A), for any n ∈ N∗ . On the (x) other hand, for every x ∈ A, there exists limy→x f (y)−f = f (x); hence there is n ∈ N∗ $ y−x $ $ f (y) − f (x) $ such that, for every y ∈ [a, b] with |y − x| < n1 , $$ − f (x)$$ < ε. From here it y−x follows that $ $ $ f (y) − f (x) $ $ < ε + |f (x)| ≤ ε + M or |f (y) − f (x)| ≤ (M + ε) · |y − x| $ $ $ y −x  ∞ and therefore x ∈ An . Then, A = ∞ n=1 An and then f (A) = n=1 f (An ). Since the outer measure is continuous from below (see Proposition 1.3.13): 

λ∗ (A) = limn λ∗ (An ) and λ∗ (f (A)) = limn λ∗ (f (An )).

(6.1)

According to Proposition 1.2.8, for any n ∈ N∗ , there exist the closed intervals (Ikn )k such that An ⊆

∞  k=1

Ikn and

∞  k=1

|Ikn | < λ∗ (An ) + ε.

(6.2)

174

Chapter 6 • Signed Measures

Eventually replacing Ikn with Ikn ∩ [a, b], we can assume that Ikn ⊆ [a, b], for all k and n. Possibly dividing the intervals Ikn , we can yet assume that |Ikn | < n1 , for all k, n ∈ N∗ . For all k, n ∈ N∗ and for every x, y ∈ An ∩ Ikn , |x − y| < n1 and then |f (x) − f (y)| ≤ (M + ε) · |x − y| ≤ (M + ε) · |Ikn |. From the last inequalities, it follows that diam(f (An ∩ Ikn )) ≡ sup{|f (x) − f (y)| : x, y ∈ An ∩ Ikn } ≤ (M + ε) · |Ikn |,

6

and then λ∗ (f (An ∩ Ikn )) ≤ (M + ε) · |Ikn |, for all k, n ∈ N∗ .

(6.3)

Using relations (6.3) and (6.2), we obtain λ∗ (f (An )) = λ∗

∞ 

f (An ∩ Ikn ) ≤

k=1

≤ (M + ε) ·

∞ 

∞ 

λ∗ (f (An ∩ Ikn )) ≤

k=1

|Ikn | ≤ (M + ε) · (λ∗ (An ) + ε).

k=1

From the last inequality and from (6.1), we obtain λ∗ (f (A)) ≤ (M + ε) · (lim λ∗ (An ) + ε) = (M + ε) · (λ∗ (A) + ε). n

Because ε is arbitrarily positive, λ∗ (f (A)) ≤ M · λ∗ (A).



Remarks 6.3.4 (i) If A is a null set, then f (A) is also a null set. Indeed, for every n ∈ N, let An = {x ∈ A : |f (x)| ≤ n}. Then λ∗ (f (An )) ≤ n · λ∗ (An ) = 0, for any n ∈ N, and so  ∞ ∗ λ∗ (f (A)) = λ∗ ( ∞ n=0 f (An )) ≤ n=0 λ (f (An )) = 0. A function which maps null sets in null sets has the Lusin property. Therefore, any differentiable function has the Lusin property. (ii) If we replace the differentiability with the differentiability almost everywhere, the Lusin property is no longer preserved. We will present below Cantor’s function (see Cantor Staircase 6.3.17). This function has the derivative 0 almost everywhere on [0, 1] but does not have the Lusin property. Lemma 6.3.5 Let A ∈ L([a, b]) be a measurable set, and let f ∈ L([a, b]) be a function differentiable at every point of A. Then, f (A) ∈ L and % λ(f (A)) ≤ A

|f |dλ.

6

175 6.3 · The Integral and the Derivative

Proof & According to Exercise (6) of 2.5, f ∈ L([a, b]) and so there exists the integral A |f |dλ. From the proof of Theorem 1.3.20, it follows that any measurable set A is of the form  A = B ∪N = ( ∞ n=1 Fn ) ∪ N, where Fn ⊆ A are closed sets and N is a null set. In our case, A is bounded; therefore, Fn are compact sets. f being continuous on A (it is differentiable), f (Fn ) is compact, so that f (Fn ) ∈ L, for any n ∈ N. Moreover, from the (i) of the previous  remark, f (N) is negligible, so that it is measurable. So f (A) = ∞ n=1 f (Fn ) ∪ f (N) ∈ L. For every ε > 0 and any n ∈ N∗ , let An = {x ∈ A : (n − 1)ε ≤ |f (x)| < nε}. Then,  An ∈ L, An ∩ Ak = ∅, for all n, k ∈ N∗ , n = k, and A = ∞ n=1 An . It follows that λ(A) =

∞ 

λ(An ).

n=1

Taking into account the previous lemma, ∗

∗

λ (f (A)) = λ

∞ 

f (An ) ≤

n=1

=

∞ 

λ∗ (f (An )) ≤

n=1

%

n=1

nε · λ(An ) =

n=1

∞ ∞ ∞ %    (n − 1)ε · λ(An ) + ε · λ(An ) ≤ n=1

∞ 

|f |dλ + ε · λ(A) =

n=1 An

|f |dλ + ε · λ(A).

= A

Because λ(A) < +∞ and ε is arbitrarily positive, we obtain the announced result.



Theorem 6.3.6 Let f : [a, b] → R be a differentiable function at every point of [a, b]. If f ∈ L1 ([a, b]), then f ∈ AC[a,b] .

Proof According to the property of absolute continuity of integral (see (2) of Theorem 3.3.1), for every ε > 0, there exists δ > 0 such that % |f |dλ < ε, for every B ∈ L([a, b]) with λ(B) < δ. (∗) B

Let {[xk , yk ] : k = 1, · · · , n} be a family of nonoverlapping subintervals of [a, b] with n n k=1 (yk − xk ) < δ, and let B = k=1 [xk , yk ]. Then, λ(B) < δ and therefore from (∗), n %  k=1 [xk ,yk ]

|f |dλ =

%

|f |dλ < ε. B

(∗∗)

176

Chapter 6 • Signed Measures

Since f is continuous on [a, b], for any k for which f (xk ) ≤ f (yk ), we have [f (xk ), f (yk )] ⊆ f ([xk , yk ]); if f (xk ) ≥ f (yk ), then [f (yk ), f (xk )] ⊆ f ([xk , yk ]). Therefore, from the previous lemma and from (∗∗), n 

|f (yk ) − f (xk )| ≤

k=1

n 

λ∗ (f ([xk , yk ])) ≤

k=1

n %  k=1 [xk ,yk ]

|f |dλ < ε. 

6

Remark 6.3.7 Any differentiable function with bounded derivative on an interval [a, b] is a Lipschitz function; according to (3) of Proposition 6.3.2, it will be absolutely continuous. We note that the result is also a consequence of the previous theorem, because, in this case, the derivative is Lebesgue integrable on [a, b].

The following theorem shows that any functions with bounded variation (and therefore also any absolutely continuous functions) is a difference of two monotonic functions.

Theorem 6.3.8 (Jordan) Every bounded variation function on [a, b] is the difference of two increasing functions on [a, b].

Proof We suppose that f : [a, b] → R has bounded variation and let g : [a, b] → R defined by g(x) = Vax (f ), for every x ∈ [a, b] (if f ∈ BV[a,b] , then, for any x ∈ [a, b], f ∈ BV[a,x] ). The function g is monotonically increasing on [a, b]. Indeed, for every x, y ∈ [a, b] with x < y and for every partition  = {a = y0 , y1 , · · · , yn = x} ∈ D([a, x]), 1 = {a = y0 , y1 , · · · , yn = x, yn+1 = y} is a partition of [a, y]. Then V (f ) ≤

n 

y

|f (yk ) − f (yk−1 )| + |f (y) − f (x)| = V1 (f ) ≤ Va (f ).

k=1 y

Therefore Vax (f ) = sup∈D ([a,x]) V (f ) ≤ Va (f ) which means that g is monotonically increasing. Let h : [a, b] → R, defined by h(x) = f (x) + Vax (f ); then h is increasing also. Indeed, for every x, y ∈ [a, b] with x < y we have y

y

h(x) − h(y) = f (x) − f (y) + Vax (f ) − Va (f ) = f (x) − f (y) − Vx (f ) ≤ 0. In the above relations we took into account the additivity property of the total variation: y y y Va (f ) = Vax (f ) + Vx (f ), and the fact that Vx (f ) ≥ |f (y) − f (x)|. It is obvious that f = h − g. 

6

177 6.3 · The Integral and the Derivative

Then, we give some consequences of the Lebesgue derivation theorem (see Theorem 7.3.5), of Proposition 6.3.2 and of the Jordan theorem (6.3.8). Corollary 6.3.9 (1) Every monotonic function on an interval I is differentiable almost everywhere on I . (2) Every bounded variation function on an interval [a, b] is differentiable almost everywhere on [a, b]. (3) Every absolutely continuous function on [a, b] is differentiable almost everywhere on [a, b].

A first attempt to obtain the Leibniz-Newton formula is presented in the following proposition: Proposition 6.3.10 (1) For every monotonically increasing function f : [a, b] → R, f ∈ L1+ ([a, b]) and &

[a,b] f dλ ≤ f (b) − f (a). (2) For every bounded variation function (so for any absolutely continuous one) f : [a, b] → R, f ∈ L1 ([a, b]). Proof (1) We extend f to ]b, b + 1] by putting f (x) = f (b), for every x ∈]b, b + 1]; the function f remains monotonically increasing on [a, b + 1]. According to (1) of the previous corollary, f is differentiable almost everywhere on [a, b + 1]. Then, f is continuous almost everywhere on [a, b+1] and, according to (2) of Theorem 2.1.13, f ∈ L([a, b+ 1]).     1 Let fn : [a, b] → R+ , defined by fn (x) = n · f x + − f (x) , for any n n ∈ N∗ . The sequence (fn )n ⊆ L+ ([a, b]) converges a.e. on [a, b] to f . It follows that f ∈ L+ ([a, b]). We apply to (fn )n Fatou’s lemma (see Corollary 3.1.12), and, since lim infn fn = f a.e. on [a, b], we obtain % [a,b]

f dλ ≤ lim inf n

% [a,b]

(6.1)

fn dλ.

We calculate the integral of fn by applying the change variable formula (Theorem 3.5.6) with g(x) = x + n1 (we note that, f being increasing, fn is the Borel function— see Proposition 2.1.5). %

% [a,b]

fn dλ = n ·

[a,b]

f

   % 1 dλ(x) − x+ f (x)dλ(x) = n [a,b]

,% =n·

-

% [a+ n1 ,b+ n1 ]

f (x)dλ(x) −

[a,b]

f (x)dλ(x) =

(6.2)

178

Chapter 6 • Signed Measures

,% =n·

-

% [b,b+ n1 ]

f (x)dλ(x) −

[a,a+ n1 ]

f (x)dλ(x) ≤

    1 1 1 · − n · f (a) · = f (b) − f (a). ≤ n·f b+ n n n

6

& From (6.1) and (6.2), we obtain [a,b] f dλ ≤ f (b) − f (a). According to the previous inequality, it follows that f ∈ L1 ([a, b]). (2) Let f ∈ BV[a,b] ; according to the Jordan theorem (6.3.8), f = f1 − f2 , where f1 , f2 are monotonically increasing functions on [a, b]. The previous point assures us that f1 , f2 ∈ L1+ ([a, b]) and so f ∈ L1 ([a, b]).  Remarks 6.3.11 (i) The inequality of the previous proposition may be strict. Indeed, we notice that we can modify the function f by giving it in b any value greater than f (b); this change leaves the first member of the inequality unchanged, while the right member can become as large as possible. (ii) The derivative of a function over an interval is a measurable function over this interval (see Exercise (6) of 2.5) but is not always integrable. An example of a differentiable function for which the derivative, although bounded, is not Riemann integrable is given in  Sect. 7.2 (Pompeiu’s function). Here is a simple example of a differentiable function with an unbounded and non-integrable Lebesgue derivative.  0, x=0 Let be the function f : [0, 1] → R, f (x) = , f (x) = x 2 cos x12 , x ∈]0, 1]  0, x=0 Then, f (x) = u(x) + v(x), where u(x) = 2x cos x12 + x2 sin x12 , x ∈]0, 1].   0, x=0 0, x=0 and v(x) = 2 The function u is continuous 1 2x cos x12 , x ∈]0, 1] sin , x ∈]0, 1]. x x2 on [0, 1]; then it is Riemann integrable and so it is Lebesgue integrable on [0, 1]. By 1 changing the variable x = y − 2 , the generalized Riemann integral of the function & +∞ sin y v on ]0, 1] becomes 2 1 y dy. This integral is simply convergent (see (v) of Remark 3.4.3), and then, according to Theorem 3.4.2, v ∈ / L1 ([0, 1]). Therefore, f ∈ / L1 ([0, 1]).

Definition 6.3.12 Let f ∈ L1 ([a, b]); for every x ∈ [a, b], f ∈ L1 ([a, x]), and so we can define the & function F : [a, b] → R by F (x) = [a,x] f dλ, for every x ∈ [a, b]. The function F is called an indefinite integral of f or the integral depending on the upper limit.

6

179 6.3 · The Integral and the Derivative

Proposition 6.3.13 Let f ∈ L1 ([a, b]) and let F : [a, b] → R, F (x) = indefinite integral. Then, F is absolutely continuous on [a, b].

& [a,x]

f dλ be its

Proof For the function f being integrable on [a, b], we can apply point (2) of Theorem 3.3.1—the absolute continuity property of the integral (not by coincidence so-called!). Then, for every & ε > 0, there exists δ > 0 such that, for every A ∈ L([a, b]) with λ(A) < δ, A |f |dλ < ε. Let now {[xk , yk ] : k = 1, . . . , n} be a finite family of closed nonoverlapping subintervals  of [a, b] with nk=1 (yk −xk ) < δ. Then, A = ∪nk=1 [xk , yk ] ∈ L([a, b]) and λ(A) < δ. Using property (4) of Theorem 3.3.1, it follows that n 

n $%  $ $ |F (yk ) − F (xk )| = $

k=1

k=1

]xk ,yk ]

$  n % $ f dλ$$ ≤

%

k=1 ]xk ,yk [

|f |dλ =

|f |dλ < ε. A

 Remark 6.3.14 According to the previous proposition and to Corollary 6.3.9, it follows that F —the indefinite integral of f —is differentiable a.e. on [a, b]. If the function f is positive, then F is monotonically increasing, and so F ≥ 0 a.e.

We will show that the derivative of F is equal to f almost everywhere on [a, b]. First we need a lemma. Lemma 6.3.15 Let f ∈ L1 ([a, b]) and let F be its indefinite integral; if F (x) = 0, for every x ∈ [a, b], then f = 0 a.e. on [a, b]. Proof & For every subinterval I = |c, d| ⊆ [a, b], I f dλ = F (d) − F (c) = 0. Let D ∈ τu be an open set. According to Theorem 1.1.3, D admits a unique representation as a union of a countable family of pairwise disjoint open intervals {In : n ∈ N∗ }. Then, % f dλ = D∩[a,b]

∞ %  n=1 In ∩[a,b]

f dλ = 0.

Let A ∈ L([a, b]); according to (2) of Theorem 3.3.1), for every ε > 0, there exists & δ > 0 such that, for every B ∈ L([a, b]) with λ(B) < δ, B |f |dλ < ε. Using the Lebesgue measurability definition of the subsets of R, there exists D ∈ τu such that A ⊆ D and & λ(D \ A) < δ; then λ(D ∩ [a, b] \ A) < δ and so D∩[a,b]\A |f |dλ < ε. It follows that $% $ $% $ $ $ $ f dλ$ = $ $ $ $ A

% f dλ − D∩[a,b]

D∩[a,b]\A

$ % $ f dλ$$ ≤

|f |dλ < ε.

D∩[a,b]\A

& ε > 0 being arbitrary, it follows that A f dλ = 0, for every A ∈ L([a, b]). Point (c) of Theorem 3.2.5 assures us that f = 0 a.e. on [a, b]. 

180

Chapter 6 • Signed Measures

The following theorem answers to the first question at the beginning of this paragraph: if f is Lebesgue integrable over an interval, then its indefinite integral has a derivative equal to f almost everywhere.

Theorem 6.3.16 Let f ∈ L1 ([a, b]) and let F be its indefinite integral; then F = f a.e. on [a, b] or % [a,x]

6

 f dλ = f (x), for almost every x ∈ [a, b].

Proof We noted above that F exists almost everywhere on [a, b]. (1) First, we suppose that f is a bounded function; let k > 0 such that |f (x)| ≤ k, for every x ∈ [a, b].     1 Let us define the sequence (Fn )n , letting Fn (x) = n · F x + − F (x) = n % n·

]x,x+ n1 ]

f dλ, for any n ∈ N∗ and for every x ∈ [a, b]. The function F is continuous

on [a, b]; therefore, Fn are continuous functions and then they are measurable on [a, b]. The sequence (Fn )n converges almost everywhere on [a, b] to F . Moreover, for any n ∈ N∗ and every x ∈ [a, b], |Fn (x)| ≤ n · k · n1 = k. The bounded convergence theorem can be applied (see Corollary 3.3.10), and, repeating the calculation performed in the demonstration of Proposition 6.3.10, it turns out that for all c ∈ [a, b], %

    1 F dλ = lim Fn dλ = lim n · F x+ − F (x) dλ = n n n [a,c] [a,c] [a,c] %

,

%

c+ n1

= lim n · n

%

%

a+ n1

F (x)dx − n ·

c

F (x)dx =

a

% = lim[F (cn ) − F (an )] = F (c) − F (a) = n

[a,c]

f dλ

(since F is continuous, we have applied a mean-value theorem for the Riemann integral; & here cn ∈]c, c + n1 [ and an ∈]a, a + n1 [). It follows that [a,c] (F − f )dλ = 0, for every c ∈ [a, b]. The above lemma assures us that F = f , a.e. on [a, b]. (2) Let us now suppose that f ∈ L1+ ([a, b]); then its indefinite integral F is an increasing function on [a, b]. According to Proposition 6.3.10, F ∈ L1+ ([a, b]) and % [a,b]

F dλ ≤ F (b) − F (a).

(6.1)

6

181 6.3 · The Integral and the Derivative

For any n ∈ N, fn = min{f, n} : [a, b] → [0, n], defined by fn (x) =  f (x), if f (x) ≤ n, Then, fn ∈ L1+ ([a, b]) and fn is bounded. Let Fn be the n, if f (x) > n. indefinite integral of fn ; according to the first step of the demonstration, Fn = fn a.e. on [a, b], for any n ∈ N. We remark that F − Fn is the indefinite integral of positive function f − fn ((F − & Fn )(x) = [a,x] (f − fn )dλ). It follows that F − Fn is an increasing function and then (F − Fn ) ≥ 0 a.e. But (F − Fn ) = F − Fn = F − fn ≥ 0 a.e. Then, for any n ∈ N, F ≥ fn a.e. Since fn ↑ f , it follows that F ≥ f a.e. on [a, b]. F and f are integrable on [a, b]; it follows that % [a,b]

F dλ ≥

% [a,b]

f dλ = F (b) − F (a).

(6.2)

From (6.1) and (6.2), we deduce that % [a,b]

F dλ = F (b) − F (a) =

% [a,b]

f dλ.

& Therefore, [a,b] (F − f )dλ = 0, and, since F − f ≥ 0 a.e., it follows that F = f a.e. (see Proposition 3.1.13). (3) Let f ∈ L1 ([a, b]) be an arbitrary function and let f + , f − ∈ L1+ ([a, b]) be the positive, respectively the negative part of f (see Definition 2.1.21). We denote by F+ and F− the indefinite integrals of f + and respectively of f − . From the second step of the demonstration, F+ = f + and F− = f − , a.e. on [a, b]. Then, the derivative of F = F+ − F− is F = F+ − F− = f + − f − = f a.e. on [a, b]. 

We will now try to answer the second question posed at the beginning of this section. It is possible that a function can be differentiable almost everywhere over an interval, and its derivative can be Lebesgue integrable, yet the formula (L − N) is not verified. One such example is offered by the Cantor staircase. The Cantor staircase 6.3.17 (“Devil’s Staircase”) We recall some of the notations of   Cantor 1 2 , , Ternary Set 1.3.16, where we built the triadic set of Cantor. So I = [0, 1], J1 =            3 3 1 2 7 8 1 2 7 8 19 20 25 26 J2 = 2 , 2 ∪ 2 , 2 , J3 = 3 , 3 ∪ 3 , 3 ∪ 3 , 3 ∪ 3 , 3 , and so 3 3 3 3 3 3 3 3 3 3 3 3 on. Jn will be a union of 2n−1 pairwise disjoint intervals of length 3ln each. We have denote  J = ∞ n=1 Jn and D = I \ J and we have shown that λ(J ) = 1. Using Exercise (16) of 1.5, it follows that J is dense in I (for every x, y ∈ I with x < y, there exists z ∈ J such that x < z < y).

182

6

Chapter 6 • Signed Measures

We will build a function F : [0, 1] → [0, 1] as follows. First we will give the values of F on the set J :   1 2 1 , , ▬ F (x) = , if x ∈ 2 3 3 ⎧  1 2 1 ⎪ ⎪ ⎨ 2 , if x ∈ 2 , 2 , 3 3  ▬ F (x) = 23 7 8 ⎪ ⎪ ⎩ 2 , if x ∈ 2 , 2 , 2 3 3  ⎧  1 2 1 ⎪ ⎪ , if x ∈ 3 , 3 , ⎪ ⎪ ⎪ 23 3 3 ⎪   ⎪ ⎪ 7 8 3 ⎪ ⎪ ⎨ 3 , if x ∈ 3 , 3 , 3 3  ▬ F (x) = 25 19 20 ⎪ ⎪ , 3 , ⎪ 3 , if x ∈ ⎪ 3 ⎪ 2 ⎪ 3 3  ⎪ ⎪ 7 25 26 ⎪ ⎪ , , ⎩ 3 , if x ∈ 2 33 33 and so on. 2k − 1 Generally, F (x) = , if x ∈ Jk,n where, for any n ∈ N∗ , Jk,n is the k-th 2n interval between the 2n−1 intervals composing Jn . The graph of the function F on the set J roughly resembles the following ⊡ Fig. 6.1:  Note that F is increasing on J = ∞ n=1 Jn . Since J is a dense subset of I , we will define F on D = I \ J by: F (0) = 0 and F (x) = sup{F (t) : t ∈ J, t < x}, for every x ∈]0, 1] \ J. ⊡ Fig. 6.1 Graph of Cantor staircase

6 1 7 23 3 22 5 23

1 2 3 23 1 22 1 23

1 32

2 32

1 3

2 3

7 32

8 32

6

0

1

183 6.3 · The Integral and the Derivative

Let us show that F is an increasing function on I . We suppose by absurd that there are two points x, y ∈ I with x < y and F (x) > F (y). As we mentioned, F is monotonically increasing on J ; so there cannot be x, y ∈ J . It remains to study the following cases: (1) x, y ∈]0, 1] \ J , (2) x ∈ J, y ∈]0, 1] \ J , and (3) x ∈]0, 1] \ J and y ∈ J. (1) Since x ∈]0, 1] \ J and F (y) < F (x), there is t1 ∈ J such that t1 < x and F (y) < F (t1 ); let t2 ∈]x, y[∩J . Then, F (t2 ) ≤ F (y) < F (t1 ) which is absurd (F is increasing on J ). (2) Since x ∈ J and x < y, F (x) ≤ F (y), which is absurd. (3) Since x ∈]0, 1] \ J and F (y) < F (x), there is t ∈ J, t < x < y such that F (y) < F (t) which contradicts y ∈ J . The hypothesis adopted leads to contradictions in all cases; it turns out that F is increasing over I . : n ∈ N∗ , k = 1, 2, · · · , 2n−1 } is dense in I , Since the set {F (t) : t ∈ J } = { 2k−1 2n the increasing it is continuous on I .   function F can’t have jumps; therefore 2 2 8 2 8 20 26 , , , , , , , · · · , then the Cantor set is C = D ∪ E. If E = 3 32 32 33 33 33 33 We can immediately notice that the set E is formed by the right ends of the intervals Jk,n ; then, since F is constant over each open interval Jk,n \ E, it follows that F (x) = 0  2n−1 on Jk,n \ E. The open set J \ E = ∞ n=1 k=1 Jk,n \ E has the Lebesgue measure equal to 1 (λ(E) = 0) and F (x) = 0, for every x ∈ J \ E. Therefore, F = 0, almost everywhere and then F ∈ L1 ([0, 1]). However, F does not satisfy the formula (L − N)! Moreover, we can see that this function does not have the Lusin property (see (i) of Remark 6.3.4). Indeed, the Lebesgue measure of the Cantor set (see 1.3.16) is zero, but its image by F has the measure of 1. From the above, we can see that the Cantor function has a derivative equal to 0 almost everywhere on [0, 1], but it is not constant on [0, 1]. We will prove that the absolutely continuous functions whose derivative is 0 almost everywhere are constant. To achieve this result, we need the concept of Vitali cover. Definition 6.3.18 Let E ⊆ R and let V be a family of nondegenerate closed intervals (which are not reduced to a point). V is said to be a Vitali cover for E if, for every ε > 0 and every x ∈ E, there exists I ∈ V such that x ∈ I and |I | < ε.

Lemma 6.3.19 (Vitali) Let E ⊆ [a, b] be a bounded measurable set and let V be a Vitali cover for E. Then, for every ε > 0, there exist I1 , · · · , In ∈ V pairwise disjoint such that  λ(E \ nk=1 Ik ) < ε.

6

184

6

Chapter 6 • Signed Measures

Proof From the definition, for every x ∈ E, there exists I ∈ V such that x ∈ I and |I | < 1; then I ⊆ [a − 1, b + 1]. So we can assume, possibly abandoning some of the sets of V , that the  set {I : I ∈ V } is bounded. Let I0 ∈ V ; if E ⊆ I0 , then the demonstration is over. Let us suppose that there is x ∈ E \ I0 ; then there exists ε0 > 0 such that ]x − ε0 , x + ε0 [∩I0 = ∅, and so there is I ∈ V such that x ∈ I and |I | < ε0 . It follows that I ⊆ ]x − ε0 , x + ε0 [ and then I ∩ I0 = ∅.  Let k1 = sup{|I | : I ∈ V , I ∩ I0 = ∅}; since {I : I ∈ V } is bounded, k1 < +∞. Let then I1 ∈ V , I0 ∩ I1 = ∅ and |I1 | >

k1 . 2

(1)

If E ⊆ I0 ∪ I1 , then the demonstration is over. Let us suppose that there is x ∈ E \ (I0 ∪ I1 ); then there exists ε1 > 0 such that ]x − ε1 , x + ε1 [∩(I0 ∪ I1 ) = ∅ and so there is I ∈ V such that x ∈ I and |I | < ε1 . It follows that I ⊆]x − ε1 , x + ε1 [ and then I ∩ (I0 ∪ I1 ) = ∅. Let k2 = sup{|I | : I ∈ V , I ∩ (I0 ∪ I1 ) = ∅} < +∞. Let then I2 ∈ V , I2 ∩ (I0 ∪ I2 ) = ∅ and |I2 | >

k2 . 2

(2)

Suppose that we built by induction I0 , I1 , · · · , In , · · · ∈ V pairwise disjoint such that, for any n ∈ N∗ , ⎛ In ∩ ⎝

n−1 

⎞ Ij ⎠ = ∅ and |In | >

j =0

kn , 2

(n)

where kn = sup{|I | : I ∈ V , I ∩ (I0 ∪ · · · ∪ In−1 ) = ∅}.  If there is n ∈ N∗ such that E ⊆ nj=0 Ij , then the demonstration is over.  Otherwise, for any n ∈ N, there exists x ∈ E \ nj=0 Ij . ∞   Since n=1 In ⊆ {I : I ∈ V } is bounded, it follows that ∞ n=1 |In | < +∞. Let ε > 0 be arbitrary and let N ∈ N such that ∞  n=N +1

|In |
0, there exists δ > 0 such that, for every finite family of closed nonoverlapping intervals {[ak , bk ] ⊆ [a, b] : k = 1, · · · , n} such that n k=1 (bk − ak ) < δ, n 

|F (bk ) − F (ak )|
0 such that ε |F (x + h) − F (x)| < ≡ α, for any h ∈]0, hx [. h 2(c − a)

(6.3)

Then V = {[x, x + h] : x ∈ E, h ∈]0, hx [} is a Vitali cover for E. Since E is bounded, according to Vitali lemma, there exists a finite family of pairwise disjoint intervals {I1 , · · · , In } ⊆ V such that λ E\

n 

Ik

(6.4)

< δ.

k=1

Suppose that the intervals {Ik = [xk , xk + hk ] : k = 1, · · · , n} are numbered so that a < x1 < x1 + h1 < x2 < x2 + h2 < · · · < xn < xn + hn < c. According to (6.1), (6.4), n  λ ]a, c[\ Ik = λ(]a, x1 [∪]x1 + h1 , x2 [∪ · · · ∪]xn + hn , c[) < δ. k=1

From (6.2) it follows that |F (x1 ) − F (a)| + |F (x2 ) − F (x1 + h1 )| + · · · + |F (c) − F (xn + hn )|
0 such that

6

% |g|dλ < ε, for every B ∈ L([a, b]) with λ(B) < δ.

(∗)

B

Let {[xk , yk ] : k = 1, · · · , n} be a finite family of nonoverlapping closed subintervals of   [a, b] with nk=1 (yk − xk ) < δ, and let B = nk=1 [xk , yk ] ∈ L([a, b]). Then, λ(B) < δ and so, from (∗), n 

|f (yk ) − f (xk )| =

k=1

n $%  $ $ $ k=1

]xk ,yk ]

$ n % $  gdλ$$ ≤

%

k=1 [xk ,yk ]

|g|dλ =

|g|dλ < ε. B



As an immediate consequence of Theorems 6.3.6 and 6.3.22, we have Corollary 6.3.23 Let f : [a, b] → R be a function differentiable at every point of [a, b]. If f ∈ L1 ([a, b]), then, for every x ∈ [a, b], % f (x) = f (a) +

[a,x]

f dλ.

Another consequence of Theorem 6.3.22 is the formula for integration by parts.

Theorem 6.3.24 Let f, g ∈ AC[a,b] ; then fg , f g ∈ L1 ([a, b]) and % [a,b]

(f · g )dλ = f (b) · g(b) − f (a) · g(a) −

% [a,b]

(f · g)dλ.

Proof According to Theorem 6.3.22, f and g are differentiable almost everywhere on [a, b] and f , g ∈ L1 ([a, b]). Since f, g are continuous on [a, b], they are bounded; let M > 0 such that, for every x ∈ [a, b], |f (x)| ≤ M and |g(x)| ≤ M. Then, |fg | ≤ M|g | and |f g| ≤ M|f |. By Theorem 3.2.6, fg , f g ∈ L1 ([a, b]).

6

189 6.4 · Exercises

Now f · g ∈ AC[a,b] ; indeed, since f, g ∈ AC[a,b] , for every ε > 0, there exists δ > 0 such that for every family of closed nonoverlapping subintervals of [a, b], {[xk , yk ] : k =  1, · · · , n} with nk=1 (yk − xk ) < δ, n ε |f (yk ) − f (xk )| < 2M , k=1 n ε |g(y ) − g(x )| < . k k k=1 2M Then n 

n 

|f (yk )g(yk ) − f (xk )g(xk )| ≤

k=1

+

|f (yk ) − f (xk )| · |g(yk )|+

k=1

n 

|f (xk )| · |g(yk ) − g(xk )| ≤ M ·

k=1

ε ε +M · = ε. 2M 2M

According to (2) of Theorem 6.3.22, for every x ∈ [a, b], % f (x) · g(x) = f (a) · g(a) +

[a,x]

(f · g) dλ.

Then % f (b) · g(b) − f (a) · g(a) =

[a,b]

(f · g)dλ +

% [a,b]

(f · g )dλ. 

6.4

Exercises

& 2 (1) Let ν : L(R) → R be defined by ν(B) = B x · e−x dλ(x), for every B ∈ L(R). Determine the ν-positive and ν-negative sets. Find a Hahn decomposition of the set R with respect to ν.  (2) Let (an )n ⊆ R be an absolute summable sequence ( ∞ n=0 |an | < +∞), and let  a , for every A ⊆ N, A = ∅ and ν : P (N) → R, defined by ν(A) = n n∈A ν(∅) = 0. Show that ν is well defined and that it is a finite signed measure on N (the measure generated by the sequence (an )n ). Determine a Hahn decomposition of N with respect to ν. Find Jordan’s decomposition of ν. What is the total variation |ν| of ν? &

Indication: Let γ be the counting measure on N (see (ii) of Example 1.4.6). It will be noted that ν(A) = − + A f dγ where f : N → R, f (n) = an (see also Remark 4.5.5). Then N = (f < 0), N = (f ≥ 0) (see

Definition 6.1.8). For the Jordan decomposition, see Remark 6.1.12.

(3) Let γ be the counting measure on N (see (ii) of Example 1.4.6). Prove that any finite signed measure ν : P (N) → R is absolutely continuous with respect to γ . dν ? Which is the Radon-Nikodym derivative f = dγ

190

Chapter 6 • Signed Measures

Indication: It will be shown that f (n) = ν({n}), for every n ∈ N.

6

(4) Let γ : A → R be a positive σ -finite measure on X, and let ν1 , ν2 : A → R be the two finite signed measures absolutely continuous with respect to γ ; show that ν1 + ν2  γ and |ν1 |  γ . $ $ $ 1$ 1| 1 +ν2 ) 1 2 = dν + dν and d|ν = $ dν . Show that d(νdγ dγ dγ dγ dγ $ (5) Let γ , ν : A → R+ be the two finite positive measures on X such that ν  γ and dν · dγ = 1 a.e. on X. γ  ν. Show that dγ dν (6) Let f : [a, b] → R be a differentiable function with the derivative bounded on [a, b]. & (a) Show that limε→0 ε1 [c,c+ε] f dλ = f (c), for every c ∈ [a, b]. & (b) Show that f ∈ L1 ([a, b]) and [a,b] f dλ = f (b) − f (a) (see Exercise (6) of 2.5).

191

Appendices

7.1

Riemann Integral

The concepts and the results concerning the Riemann integral necessary to read this book are presented in this appendix. For some demonstrations we used section 8.6 of [9]. A finite set  = {x0 , x1 , · · · , xn } ⊆ [a, b] is a partition of the interval [a, b] if a = x0 < x1 < · · · < xn = b;  = max{xk − xk−1 : k = 1, · · · , n} is the mesh of partition . We denote with D([a, b]) the set of all partitions of [a, b]. Another partition  of [a, b] is said to be a refinement of  if  ⊆  . Let f : [a, b] → R be a bounded function and let  = {x0 , x1 , · · · , xn } be a partition of [a, b]; for any k = 1, · · · , n, let Mk = sup{f (x) : x ∈ [xk−1 , xk ]} and mk = inf{f (x) : x ∈ [xk−1 , xk ]}.   The sums S = nk=1 Mk (xk − xk−1 ) and s = nk=1 mk (xk − xk−1 ) are called the upper, respectively lower, Darboux sums of f corresponding to the partition . If  is a refinement of , alors s ≤ s ≤ S ≤ S . The upper Darboux integral and , respectively, the lower Darboux integral of f are I¯f = inf{S :  ∈ D([a, b])}, I f = sup{s :  ∈ D([a, b])}.

Definition 7.1.1 The bounded function f : [a, b] → R is Riemann integrable on [a, b] if, for every ε > 0, there exists  ∈ D([a, b]) such that S − s < ε. We denote with R[a,b] the set of all Riemann integrable functions on [a, b].

The following theorem presents, without proof, several conditions equivalent to Riemann’s integrability. © The Author(s), under exclusive license to Springer Nature Switzerland AG 2021 L. C. Florescu, Lebesgue Integral, Compact Textbooks in Mathematics, https://doi.org/10.1007/978-3-030-60163-8_7

7

192

7

Chapter 7 • Appendices

Theorem 7.1.2 Let f : [a, b] → R be a bounded function; the following conditions are equivalent: 1. f is Riemann integrable on [a, b]. &b 2. I¯f = I f (the common value I = a f (x)dx is the Riemann integral of f on [a, b]). 3. For every ε > 0, there exists ε ∈ D([a, b]) such that, for every  ⊇ ε , S − s < ε. 4. For every ε > 0, there exists δ > 0 such that, for every  ∈ D([a, b]) with  < δ, S − s < ε. 5. There is I ∈ R such that, for every ε > 0, there exists ε ∈ D([a, b]) such that, for every  = {x0 , · · · , xn } ⊇ ε and any {c1 , · · · , cn } with x0 ≤ c1 ≤ x1 ≤ c2 ≤ x2 ≤ · · · ≤ xn−1 ≤ cn ≤ xn , $ $ n $ $  $ $ f (ck )(xk − xk−1 )$ < ε $I − $ $ k=1

&b (the real number I is unique and I = a f (x)dx). 6. There is I ∈ R such that, for every ε > 0, there exists δ > 0 such that, for every  = {x0 , · · · , xn } ∈ D([a, b]) with  < δ and any {c1 , · · · , cn } with x0 ≤ c1 ≤ x1 ≤ c2 ≤ x2 ≤ · · · ≤ xn−1 ≤ cn ≤ xn , $ $ n $ $  $ $ f (ck )(xk − xk−1 )$ < ε. $I − $ $ k=1

In his famous doctoral dissertation (“Intégrale, longueur, aire”, 1902),. H. Lebesgue presents a Riemann integrability criterion which shows to what point the integrable Riemann functions can be discontinuous. To demonstrate this criterion, we must introduce the notion of oscillation of a function and present some preliminary results. Definition 7.1.3 Let I be an interval, J ⊆ I be a subinterval, and f : I → R be a bounded function. The oscillation of f on J is ωf (J ) = sup {|f (x) − f (y)| : x, y ∈ J } = sup f (x) − inf f (y). y∈J

x∈J

Let x ∈ I ; the oscillation of f at x is -

, ωf (x) = inf ωf (]x − δ, x + δ[∩I ) = inf δ>0

δ>0

sup u,v∈]x−δ,x+δ[∩I

|f (u) − f (v)| .

7

193 7.1 · Riemann Integral

The demonstration of the following proposition is an elementary exercise. Proposition 7.1.4 The function f is continuous at x ∈ I if and only if ωf (x) = 0. Lemma 7.1.5 Let I be a closed interval and let f : I → R be a bounded function. For every α ≥ 0, Dα = {x ∈ I : ωf (x) ≥ α} is a closed set. Proof Let us suppose that there is a sequence (xn )n ⊆ Dα , xn → x such that x ∈ / Dα . Since I is a closed interval, it is a closed set (1.1.7) and then x ∈ I . Therefore ωf (x) < α, and then there is β and δ > 0 such that sup{|f (u) − f (v)| : u, v ∈]x − δ, x + δ[∩I } < β < α. Let p ∈ N such that |xp − x| < 2δ . Then, for every u, v ∈]xp − 2δ , xp + 2δ [∩I, |u − x| < δ, and |v − x| < δ and so |f (u) − f (v)| < β from where we deduce that ωf (xp ) = infδ>0 supu,v∈]xp − δ ,xp + δ [∩I |f (u) − f (v)| ≤ β < α which contradicts xp ∈ Dα . 2

2



The following topological lemma is formulated for the compacts of R; however, the demonstration can be reproduced with small modifications for the subsets of R2 , R3 , etc. Lemma 7.1.6 A set K ⊆ R is compact if and only if for every open cover of K, {Dγ : γ ∈   } ⊆ τu , (K ⊆ γ ∈ Dγ ), there is γ1 , · · · , γn ∈  such that K ⊆ ni=1 Dγi (any open cover of K, has a finite subcover ). Proof Let’s remember that a set K ⊆ R is compact if it is closed and bounded or, equivalently, if every sequence of K has a subsequence convergent to a point of K. Let K be a compact set and let A = {Dγ : γ ∈ } be an open cover of K. The set of rational numbers, Q, is countable, and then in the family D = {]q, r[: q, r ∈ Q, there is   γ ∈  such that ]q, r[⊆ Dγ } is also countable. It is easy to see that γ ∈ Dγ = D∈D D. ∞ Then D = {D1 , · · · , Dn , · · · } is a countable open cover of K: K ⊆ n=1 Dn . Let’s show that D has a finite subcover. If not, for any n ∈ N∗ and every {D1 , · · · , Dn } ⊆  D, there exists xn ∈ K \ ni=1 Di . The sequence (xn )n ⊆ K has a subsequence (xkn )n which converges to x ∈ K. Let p ∈ N∗ such that x ∈ Dp . Since Dp is open, there is n0 ∈ N∗ such n that xkn ∈ Dp , for any n ≥ n0 . Let n ≥ n0 such that n > p; then xkn ∈ K \ ki=1 Di , and, since p < n < kn , it follows that xkn ∈ / Dp , which is a contradiction.  Therefore there is n ∈ N∗ such that K ⊆ ni=1 Di . The way we built the family D, for any i = 1, · · · n, there is γi ∈  such that Di ⊆ Dγi . Then {Dγ1 , · · · , Dγn } is a finite subcover of A. Conversely, suppose that any open cover of K has a finite subcover. Then, since K ⊆  p ∗ R= ∞ n=1 ] − n, n[, there is p ∈ N such that K ⊆ n=1 ] − n, n[=] − p, p[, and so K is bounded.

194

Chapter 7 • Appendices

Let us suppose that K is not closed; and then there exists a sequence (xn )n ⊆ K, xn → x, #  r r " and x ∈ / K. For every y ∈ K, ry = |x − y| > 0 and K ⊆ y∈K y − 3y , y + 3y . This cover

 ry ry has a finite subcover; let y1 , · · · yn ∈ K such that K ⊆ ni=1 yi − 3i , yi + 3i . There is

ry ry i ∈ {1, · · · , n} and kn ↑ +∞ such that the sequence (xkn )n ⊆ yi − 3i , yi + 3i . Then

r r r x ∈ yi − 3yi , yi + 3yi and so ryi = |x − yi | < 2 · 3yi , which is absurd. Therefore K is bounded and closed, hence a compact set. 

7

Theorem 7.1.7 (Lebesgue criterion for Riemann integrability) The bounded function f : [a, b] → R is Riemann integrable if and only if f is continuous almost everywhere on [a, b].

Proof Let D be the set of discontinuities of f on [a, b]; according to Proposition 7.1.4, D = {x ∈  1 [a, b] : ωf (x) > 0}. We remark that D = ∞ p=1 Dp where Dp = {x ∈ [a, b] : ωf (x) ≥ p }, ∗ ∗ for any p ∈ N . The theorem says that f ∈ R[a,b] if and only if λ (D) = 0. Necessity. Let f ∈ R[a,b] . We will prove that, for any p ∈ N∗ , λ∗ (Dp ) = 0 and then it will result that λ∗ (D) = 0. Let so p ∈ N∗ ; for every ε > 0, there exists  = {x0 , · · · , xn } ∈ D([a, b]) such that S − s