PART TWO. Measure, Integration, and Differentiation


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1 PART TWO Measure, Itegratio, ad Differetiatio
2 Émile FélixÉdouardJusti Borel ( Émile Borel was bor at SaitAffrique, Frace, o Jauary 7, 1871, the third child of Hooré Borel, a Protestat miister, ad Emilie TeissiSolier. Émile had two older sisters who were 14 ad 16 years old whe he was bor. From a early age, Borel exhibited a strog proclivity for mathematics. At the age of 11, he wet to live with his eldest sister, Madame Lebeau. This move allowed Borel to atted the Lycée at Motaube, where he showed extraordiary talets. Several years later, Borel wet to Paris, where he took courses at the Lycée LouisleGrad i preparatio for takig the examiatios to eter the École Polytechique ad the École Normale. He raked first i both of these examiatios ad could choose either of the two uiversities. I 1890, Borel etered the École Polytechique i Paris, where he graduated i He received his doctorate from the École Normale Supérieure i Borel s most importat research was doe i the 1890s whe he worked o probability, ifiitesimal calculus, diverget series, ad measure theory. I 1896, he provided the proof of Picard s theorem, a proof that mathematicias had bee seekig for early 20 years. Although Joh vo Neuma is credited as the fouder of game theory, Borel completed a series of papers o the subject betwee 1921 ad 1927, thus beig the first to defie games of strategy. After WW I, Borel developed a iterest i politics, servig as Miister of the Navy from He was arrested ad briefly imprisoed by the Vichy regime i 1940, after which he worked i the Résistace. His hoors icluded the Résistace medal i 1945, the Croix de Guerre i 1918, the Grad Cross of the Légio d Hoeur i 1950, ad the first gold medal of the Cetre Natioal de la Recherche Scietifique i Borel was appoited to the faculty of the École Normale Supérieure i 1896, held the Chair i Fuctio Theory at the Sorboe from 1909 util 1940, ad was director ad foudig member of the Heri Poicaré Istitute from 1928 util his death o February 3, 1956, i Paris. 80
3 3 Lebesgue Measure o the Real Lie I Chapter 2, we discussed ope sets, cotiuous fuctios, ad the Riema itegral. Those classical cocepts have served mathematics ad its applicatios well. However, for the purposes of moder mathematics, a more geeral ad sophisticated framework is required. I this chapter ad the ext, we will take the first steps toward obtaiig that framework. Here we will first expad the collectio of cotiuous fuctios to the collectio of Borel measurable fuctios, the smallest algebra of fuctios that cotais the cotiuous fuctios ad is closed uder poitwise limits. I doig so, we will be led to cosider the collectio of Borel sets, the smallest σalgebra of subsets of R that cotais the ope sets. Next we will geeralize the cocept of legth so that it applies to all Borel sets. I fact, that geeralizatio will be to a eve larger collectio of sets tha the Borel sets, amely, the Lebesgue measurable sets. 3.1 BOREL MEASURABLE FUNCTIONS AND BOREL SETS A Course i Real Aalysis. Secod editio. Copyright 1999, 2013, Elsevier Ic. All rights reserved. I Chapter 2, we showed that the collectio of cotiuous, realvalued fuctios forms a algebra but is ot closed uder poitwise limits. As this latter property is essetial i moder mathematical aalysis, we will elarge the collectio of cotiuous fuctios to a collectio of fuctios that is closed uder poitwise limits. Specifically, we will cosider the smallest algebra of (realvalued fuctios that cotais the cotiuous fuctios ad is closed uder poitwise limits. Such a algebra of fuctios exists it is the itersectio of all algebras of fuctios that cotai the cotiuous fuctios ad are closed uder poitwise limits. 81
4 82 Chapter 3 Lebesgue Measure o the Real Lie Note: The aforemetioed itersectio is ot vacuous because the collectio of all realvalued fuctios is a algebra that cotais the cotiuous fuctios ad is closed uder poitwise limits. As we will see presetly, the coditio of beig a algebra is superfluous. That is, the smallest collectio of fuctios that cotais the cotiuous fuctios ad is closed uder poitwise limits is ecessarily a algebra of fuctios. Thus, we make the followig defiitio: DEFINITION 3.1 Borel Measurable Fuctios We deote by Ĉ the smallest collectio of realvalued fuctios o R that cotais the collectio of cotiuous fuctios ad is closed uder poitwise limits. The members of Ĉ are called Borel measurable fuctios. THEOREM 3.1 The collectio Ĉ of Borel measurable fuctios forms a algebra. That is, if f ad g are Borel measurable ad α R, the a f + g is Borel measurable. b αf is Borel measurable. c f g is Borel measurable. We prove oly part (a; parts (b ad (c are left as exercises. First of all, let g C (the collectio of cotiuous fuctios o R ad set D = { f Ĉ : f + g Ĉ }. If f C, the f Ĉ ad f + g C Ĉ. Thus, D C. Now suppose that {f } D ad that f f poitwise. The f Ĉ ad f + g Ĉ for all N ad f + g f + g poitwise. Sice Ĉ is closed uder poitwise limits, we coclude that f Ĉ ad f + g Ĉ. Hece, f D. Therefore, we see that D is closed uder poitwise limits. The previous paragraph shows that D cotais the cotiuous fuctios ad is closed uder poitwise limits. Sice, by defiitio, Ĉ is the smallest such collectio of fuctios, it follows that D Ĉ. But, by the defiitio of D, D Ĉ. Thus, D = Ĉ; i other words, f + g Ĉ wheever f Ĉ ad g C. Next, let f Ĉ ad set E = { g Ĉ : f + g Ĉ }. It follows from what we just proved that E cotais the cotiuous fuctios. Moreover, the same argumet that we used to show that D is closed uder poitwise limits shows that E is closed uder poitwise limits. Hece, E = Ĉ; that is, f + g Ĉ wheever f Ĉ ad g Ĉ. Borel Sets I Chapter 2, we discovered that there is a atural associatio betwee the cotiuous fuctios ad the collectio of ope sets: A fuctio is cotiuous
5 3.1 Borel Measurable Fuctios ad Borel Sets 83 if ad oly if the iverse image of each ope set is ope. Now we ask which collectio of sets correspods aturally to the Borel measurable fuctios. As we will see, it is the collectio of sets whose characteristic fuctios are Borel measurable fuctios. DEFINITION 3.2 Borel Sets A set B Ris called a Borel set if its characteristic fuctio is Borel measurable. The collectio of all Borel sets is deoted B. So, B = { B R: χ B Ĉ }. To begi, we will prove that the Borel sets form a σalgebra of subsets of R. I order to accomplish this, we will eed several lemmas. The proof of the first lemma is left as a exercise for the reader (see Exercise 3.2. LEMMA 3.1 Let h deote the absolute value fuctio, that is h(x = x, x R. The there is a sequece {p } of polyomials such that p h poitwise. Next we itroduce the otatio used for the maximum ad miimum of two fuctios ad prove that the Borel measurable fuctios are closed uder those two operatios. DEFINITION 3.3 Maximum ad Miimum of Two Fuctios Let f ad g be realvalued fuctios o R. The we defie f g = max{f,g} ad f g = mi{f,g}. That is, ad (f g(x = max{f(x,g(x} (f g(x = mi{f(x,g(x}. LEMMA 3.2 If f ad g are Borel measurable fuctios, the so are f g ad f g. More geerally, if f 1, f 2,..., f are Borel measurable fuctios, the so are f 1 f ad f 1 f. We first ote that the followig idetities hold: f g = 1 ( f + g + f g 2 (3.1 ad f g = 1 2( f + g f g. (3.2
6 84 Chapter 3 Lebesgue Measure o the Real Lie Next we show that F Ĉ wheever F Ĉ. From Lemma 3.1, we ca choose a sequece of polyomials {p } such that p (x x for all x R. Because Ĉ is a algebra of fuctios (Theorem 3.1, it follows that p F Ĉ for all N. However, p F F poitwise ad, cosequetly, as Ĉ is closed uder poitwise limits, F Ĉ. Now suppose that f, g Ĉ. The f g Ĉ (why?. Usig (3.1 ad (3.2 ad the fact that Ĉ is a algebra, we deduce that f g Ĉ ad f g Ĉ. The remaiig coclusios of the lemma follow by mathematical iductio. LEMMA 3.3 If {f } is a sequece of Borel measurable fuctios with {f (x} bouded for each x R, the sup f ad if f are Borel measurable. By Lemma 3.2, if f 1, f 2,..., f are Borel measurable, the so are f 1 f ad f 1 f. But, ad sup f = lim f 1 f if f = lim f 1 f. The lemma ow follows because Ĉ is closed uder poitwise limits. Now that we have established Lemmas , we ca prove that the collectio B of Borel sets is a σalgebra of subsets of R. THEOREM 3.2 The collectio of Borel sets B = { B R: χ B Ĉ } is a σalgebra of subsets of R. We first show that the collectio of Borel sets is closed uder complemetatio. Assume B B. The, by defiitio, χ B Ĉ. Sice 1 Ĉ ad Ĉ is a algebra, we coclude that 1 χ B Ĉ. But 1 χ B = χ B c ad, cosequetly, B c B. Next we show that the collectio of Borel sets is closed uder coutable uios. Suppose that B B, for N. The χ B Ĉ for N ad, therefore, by Lemma 3.3, sup χ B Ĉ. But sup χ B = χ B. Hece, B B. Further Properties of Borel Sets ad Borel Measurable Fuctios It is left as a exercise for the reader to show that if O is a ope set, the χ O is a Borel measurable fuctio. I other words, every ope set is a Borel set. We will prove shortly that, i fact, B is the smallest σalgebra that cotais all the ope sets. But first, we will justify our cotetio that it is atural to associate the Borel sets with the Borel measurable fuctios, as we do the ope sets with the cotiuous fuctios. Specifically, we will show that a fuctio is Borel
7 3.1 Borel Measurable Fuctios ad Borel Sets 85 measurable if ad oly if the iverse image of each ope set is a Borel set. I order to accomplish this, we will itroduce aother collectio of fuctios which, as we will see, turs out to be idetical to the collectio of Borel measurable fuctios. LEMMA 3.4 Let F = { f : f 1 (O Bfor all ope sets O }. The F cotais the cotiuous fuctios. Suppose that f is a cotiuous fuctio o R. The, by Theorem 2.5 o page 55, f 1 (O is ope wheever O is ope. But every ope set is a Borel set (Exercise 3.3. Therefore, f 1 (O B wheever O is ope. This shows that F cotais the cotiuous fuctios. LEMMA 3.5 f F if either of the followig coditios hold: a For each a R, f 1( (,a B. b For each a R, f 1( (a, B. We will prove part (a. The proof of part (b is similar ad is left as a exercise. So, suppose that f satisfies the coditio i part (a. We claim that f F; that is, f 1 (O Bfor all ope sets O. Set A = { A R: f 1 (A B}. Because f 1 (A c =[f 1 (A] c, f 1 ( A = f 1 (A, ad B is a σalgebra, it follows that A is a σalgebra. Now, by assumptio, A cotais all sets of the form (,α, where α R.If a R, the we ca write (,a]= (,a+1/; therefore, (,a] A because A is a σalgebra. This i tur implies that (a, b Afor each a, b R, sice (a, b =(,b (,a] c. It ow follows easily that A cotais all ope itervals. But, by Propositio 2.13 o page 48, every ope set is a coutable uio of ope itervals. Cosequetly, A cotais all ope sets. This meas that f 1 (O Bfor all ope sets O; that is, f F. LEMMA 3.6 F is closed uder poitwise limits. Suppose that {g } F ad let g = sup g. The, for each a R, we have g 1( (a, = ( g 1 (a, B. Therefore, by the precedig lemma, sup g F. Similarly, if g F. Now suppose that {f } F ad that f f poitwise. The, for each x R, lim f (x =f(x ad, so, lim sup f (x =f(x. Cosequetly, f = if {sup k f k }. Let g = sup k f k. The the previous paragraph shows i tur that g F for all N ad if g F. Hece, f F. COROLLARY 3.1 F cotais the Borel measurable fuctios.
8 86 Chapter 3 Lebesgue Measure o the Real Lie By Lemmas 3.4 ad 3.6, F cotais the cotiuous fuctios ad is closed uder poitwise limits. Sice the collectio of Borel measurable fuctios Ĉ is the smallest collectio of fuctios that cotais the cotiuous fuctios ad is closed uder poitwise limits, it must be that F Ĉ. LEMMA 3.7 Let f F. The there is a sequece {f } of Borel measurable fuctios such that f f poitwise. First of all, ote that if a, b R, the f 1 ([a, b B(why?. For N, let { E k = x : k } ([ k +1 k f(x < = f 1, k +1 for k =0,±1, ±2,... The E k Bad so χ Ek Ĉ. Sice Ĉ is a algebra of fuctios ad is closed uder poitwise limits, the fuctio f = k= k χ E k = lim k k j= k j χ E j is i Ĉ. It is easy to see that f(x f (x < 1/ for all x R. Hece f f poitwise (i fact, the covergece is uiform. Usig the precedig results, it is ow evidet that F ad the collectio of Borel measurable fuctios are idetical. Specifically, we have the followig theorem. THEOREM 3.3 A fuctio f is Borel measurable if ad oly if the iverse image of each ope set uder f is a Borel set; that is, if ad oly if f 1 (O Bfor all ope sets O. By Corollary 3.1, F Ĉ. Coversely, suppose that f F. The, by Lemma 3.7, f is the poitwise limit of fuctios i Ĉ. Sice Ĉ is closed uder poitwise limits, this implies f Ĉ. Thus, Ĉ F. We metioed previously that the collectio of Borel sets B is the smallest σalgebra of sets that cotais the ope sets. Here is a proof of that result. THEOREM 3.4 The collectio of Borel sets B is the smallest σalgebra of subsets of R that cotais all the ope sets. We already kow that B is a σalgebra that cotais all the ope sets. Let A be ay σalgebra that cotais all the ope sets. We claim that B A. Let G = { f : f 1 (O Afor all ope sets O }. The argumets used for Lemmas 3.4 ad 3.6 deped oly o the fact that B is a σalgebra cotaiig the ope sets. It follows that G cotais the cotiuous fuctios ad is closed uder poitwise limits; thus, G Ĉ. This last fact implies that χ B Gfor all B B. But the, for each B B, B = χ 1 ( B (1/2, 3/2 A.ThusB A.
9 3.1 Borel Measurable Fuctios ad Borel Sets 87 Remarks: I may texts, the collectio of Borel sets B is defied to be the smallest σalgebra of sets that cotais the ope sets; ad a fuctio is defied to be Borel measurable if the iverse image of each ope set is a Borel set. As we see from Theorems 3.3 ad 3.4, the defiitios preseted here (Defiitios 3.1 ad 3.2 are equivalet to those. It seems more atural, though, to itroduce the Borel measurable fuctios i a way that is motivated by a defect i the collectio of cotiuous fuctios; amely, the defect of ot beig closed uder poitwise limits. Oce this itroductio is accomplished, however, it may ideed be easier to thik of Borel measurable fuctios via Theorem 3.3 ad Borel sets via Theorem 3.4. Moreover, those characterizatios will be used as a meas to defie Borel sets ad Borel measurable fuctios i more geeral cotexts. Here ow are several examples that illustrate Borel measurable fuctios ad Borel sets. We have left some of the justificatios as exercises for the reader. EXAMPLE 3.1 Illustrates Borel Measurable Fuctios ad Borel Sets a Because B is a σalgebra cotaiig the ope sets, it follows that all ope sets, closed sets, ad itervals are Borel sets. b If B is a coutable set, the B B; i particular, Q B. From this, it also follows that the set of irratioal umbers R\Q is i B. c By defiitio, ay cotiuous fuctio is Borel measurable. d χ Q is Borel measurable because Q B. Ideed, if B B, the χ B is Borel measurable by the defiitio of B. e If B 1,..., B Bad α 1,..., α R, the f = α kχ Bk is Borel measurable. This fact follows immediately from part (d ad the fact that Ĉ is a algebra of fuctios. I particular, all step fuctios are Borel measurable. f Every mootoe fuctio is Borel measurable, as the reader ca easily verify by applyig Lemma 3.5. g A realvalued fuctio f o R that is 0 except o a coutable set is Borel measurable. Ideed, suppose K is coutable ad f(x =0forx/ K. Let {x } be a eumeratio of K. The f = f(x χ {x}. IfK is fiite, the f is Borel measurable by parts (a ad (e. If K is ifiite, the f is the poitwise limit of the Borel measurable fuctios f(x kχ {xk }, N, ad, hece, is itself Borel measurable. Borel Measurable Fuctios ad Borel Sets o Subsets of R We coclude this sectio with a brief discussio of Borel measurable fuctios ad Borel sets whe the uderlyig space is some subset D R. The pertiet defiitios ad theorems are obvious modificatios of those discussed earlier. DEFINITION 3.4 Borel Measurable Fuctios o D We deote by Ĉ(D the smallest collectio of realvalued fuctios o D that cotais the cotiuous fuctios o D ad is closed uder poitwise limits. The members of Ĉ(D are called Borel measurable fuctios o D.
10 88 Chapter 3 Lebesgue Measure o the Real Lie DEFINITION 3.5 Borel Sets of D A set B D is called a Borel set of D if its characteristic fuctio is a Borel measurable fuctio o D. The collectio of all Borel sets of D is deoted B(D. Thus, B(D ={ B D : χ B Ĉ(D }. Usig argumets similar to those used earlier, we ca obtai the followig theorems: THEOREM 3.5 A fuctio f is Borel measurable o D if ad oly if the iverse image of each ope set uder f is a Borel set i D, that is, f 1 (O B(D for all ope sets O. THEOREM 3.6 The collectio of Borel sets of D, B(D, is the smallest σalgebra of subsets of D that cotais all ope sets i D. A iterestig ad useful characterizatio of B(D is give by the followig theorem. Note the aalogy with ope sets i D (see Theorem 2.3 o page 51. THEOREM 3.7 B B(D if ad oly if there is a A B such that B = D A. That is, the Borel sets of D are precisely the Borel sets (of R itersected with D. Let A = { D A : A B}. We claim that A = B(D. It is easy to see that A is a σalgebra of subsets of D ad, sice B cotais all ope sets of R, A cotais all ope sets of D. Thus, by Theorem 3.6, A B(D. Now, let C be ay σalgebra of subsets of D that cotais the ope sets of D ad set D = { A B: D A C}. The D cotais the ope sets of R because C cotais the ope sets of D. Also, D is a σalgebra because B ad C are. Cosequetly, D B. But, by defiitio, D B. Hece, D = B. It ow follows that A Cad, sice C was a arbitrary σalgebra of subsets of D that cotais the ope sets, we coclude that A B(D. This last result ad the previous paragraph show that A = B(D. Exercises for Sectio 3.1 Note: A deotes a exercise that will be subsequetly refereced. 3.1 Prove parts (b ad (c of Theorem 3.1 o page Let h(x = x. Prove Lemma 3.1 o page 83 by proceedig as follows: a Show that there exists a sequece of polyomials that coverges uiformly to h o [ 1, 1]. Hit: Cosider the Taylor series expasio for (1 t 1/2 o [0, 1]. b Use part (a to coclude that for each compact subset K of R, there exists a sequece of polyomials that coverges uiformly to h o K. Hit: If b>0, we ca write x = b x/b.
11 3.2 Lebesgue Outer Measure 89 c Use part (b to coclude that there exists a sequece of polyomials that coverges to h uiformly o each compact subset of R. d Deduce Lemma 3.1 from part (c. 3.3 Prove that every ope set is a Borel set by showig that for each ope set O, χ O is a Borel measurable fuctio. Hit: Begi by showig that χ I is Borel measurable for each ope iterval I. 3.4 Verify part (b of Lemma 3.5 o page Show that f is Borel measurable if ad oly if f 1 (B Bfor all Borel sets B. 3.6 Let D be a dese subset of R. Show that f is Borel measurable if either of the followig coditios holds: a For each d D, f 1( (,d B. b For each d D, f 1( (d, B. 3.7 Show that all closed sets ad all itervals are Borel sets. 3.8 Prove that every mootoe fuctio is Borel measurable. 3.9 Prove Theorems Verify (3.1 ad (3.2 o page Show that ay coutable subset of R is a Borel set For subsets A ad B of R, defie A + B = { a + b : a A ad b B }. Suppose that B is a Borel set. Prove that A + B is a Borel set if A is a coutable. b ope Most fuctios ecoutered i a calculus course ca be obtaied from the idetity fuctio i(x = x by usig the stadard operatios of algebra (sums, products, quotiets, ad the extractio of roots together with the operatio of passig to the limit i a sequece of fuctios. For example, e x1/2 = lim k=0 (x 1/2 k. k! Explai why ay fuctio obtaied usig the foremetioed operatios is a Borel measurable fuctio. 3.2 LEBESGUE OUTER MEASURE I the previous sectio, we elarged our basic collectio of fuctios from the cotiuous fuctios to the Borel measurable fuctios. Although both of those collectios of fuctios are algebras, the latter collectio has the advatage of beig closed uder poitwise limits. Our ext goal is to exted the Riema itegral to a itegral that applies to all Borel measurable fuctios. The extesio is ot trivial sice there are Borel measurable fuctios that are ot Riema itegrable. Ideed, as we leared i Theorem 2.7 o page 72, a bouded fuctio is Riema itegrable if ad oly if it is cotiuous except o a set of measure zero. There are certaily Borel measurable fuctios that do ot satisfy this last coditio (e.g., χ Q.
12 90 Chapter 3 Lebesgue Measure o the Real Lie Referrig to Sectio 2.6, begiig o page 67, we see that the Riema itegral is developed by first defiig the itegral of a step fuctio h = a kχ Ik o [a, b] tobe b h(x dx = a k l(i k, a where l(i deotes the legth of a iterval I. Therefore, the defiitio of the Riema itegral ultimately depeds o the cocept of legth, which applies oly to itervals of real umbers. To obtai a itegral that applies to all Borel measurable fuctios, we proceed by aalogy with the developmet of the Riema itegral. Specifically, we must first defie the itegral of a Borel measurable fuctio of the form s = a k χ Bk, where the B k s are Borel sets. If the B k s are itervals, the s is a step fuctio ad we simply defie the itegral to equal the Riema itegral a k l(b k. If the B k s are ot itervals, the what? It seems that we eed to geeralize the cocept of legth so that it applies to arbitrary Borel sets. The Defiitio of Lebesgue Outer Measure The cocept of legth will be exteded ad replaced by that of measure. As we will see, this is by o meas a simple procedure. Let us deote the required measure by the Greek letter μ, ad the collectio of subsets of R to which it applies by the letter A. Subsets of R that belog to A are called measurable sets. We will ow list some properties that μ ad A should satisfy. Sice measure is to be a geeralizatio of legth, we require that the measure of a iterval be its legth; that is, μ(i = l(i for all itervals I. Also, for purely mathematical reasos, we require that A be a σalgebra; ad as we wat all Borel sets to be measurable, we require that A B. Now, clearly, the measure of the uio of two disjoit itervals should be the sum of their legths (measures. More geerally, the, we require that the measure of the uio of two disjoit measurable sets be the sum of their measures. That is, if A, B Aad A B =, the μ(a B =μ(a+μ(b. (3.3 Usig mathematical iductio, we ca show that the previous coditio implies that if A 1, A 2,..., A Aad A i A j = for i j, the ( μ A k = μ(a k. (3.4 This coditio o μ is called fiite additivity. For purposes of moder mathematical aalysis, we eed to impose a somewhat stroger coditio o our measure tha fiite additivity; amely, that (3.4 hold ot oly for fiite collectios of pairwise disjoit measurable sets but also for coutably ifiite collectios of pairwise disjoit measurable sets. This coditio is called coutable additivity.
13 3.2 Lebesgue Outer Measure 91 I summary, if μ is the required geeralizatio of legth ad A is the collectio of subsets of R that have a legth i this exteded sese, the the followig coditios should be satisfied: (M1 A is a σalgebra ad A B. (M2 μ(i = l(i, for all itervals I. (M3 If A 1, A 2,... are i A, with A i A j = for i j, the ( μ A = μ(a. Coditios (M1 (M3 provide us with the meas for extedig the otio of legth to all ope sets. First, sice every ope set is a Borel set, Coditio (M1 implies that every ope set should be measurable. Now, let O be a ope set. The O is a coutable uio of disjoit ope itervals, say O = I. Now applyig, i tur, Coditios (M3 ad (M2, we ifer that ( μ(o =μ I = μ(i = l(i. So, we ow see how to exted the otio of legth to all ope sets. For sets that are more complicated tha ope sets, however, it is ot at all obvious what to do. I fact, defiig a suitable measure for subsets of R costituted a major problem for mathematicias util the begiig of the twetieth cetury, whe Heri Lebesgue foud the key. His idea was as follows: For a subset A R, cosider all ope sets that cotai A as a subset. The defie the measure of A to be the greatest lower boud of the measures of all those ope sets: if{ μ(o :O ope, O A }. (3.5 With this defiitio, we close dow o A or come at A from the outside, so we call this measure of A its outer measure. Outer measure is defied for all subsets of R. But, as we will see, it is coutably additive (i.e., satisfies Coditio (M3 oly whe restricted to a proper subcollectio of subsets of R. Cosequetly, we will deote outer measure ot by μ, but istead by λ. Below we give a formal defiitio of outer measure. The defiitio that we preset does ot use (3.5 but is equivalet to it. DEFINITION 3.6 Lebesgue Outer Measure For each subset A R, the Lebesgue outer measure of A, deoted by λ (A, is defied by { λ (A = if l(i :{I } ope itervals, } I A. Note: A sequece of ope itervals {I } appearig i Defiitio 3.6 ca be either a fiite or ifiite sequece.
14 92 Chapter 3 Lebesgue Measure o the Real Lie Basic Properties of Lebesgue Outer Measure Some basic properties of Lebesgue outer measure are proved i the ext two propositios. PROPOSITION 3.1 Lebesgue outer measure λ has the followig properties: a λ (A 0, for all A R. (oegativity b λ ( =0. c A B λ (A λ (B. (mootoicity d λ (x + A =λ (A for x R, A R, where x + A = { x + y : y A }. (traslatio ivariace e If {A } is a sequece of subsets of R, the ( λ A λ (A. (3.6 I particular if A, B R, the λ (A B λ (A+λ (B. The relatio i (3.6 is called coutable subadditivity. For each E R, let { S E = l(i :{I } ope itervals, } I E. The, by defiitio, λ (E = if{ x : x S E }. a If A R, the S A [0, ] so that λ (A = if{ x : x S A } 0. b For ɛ>0, the iterval I ɛ =( ɛ/2,ɛ/2 cotais ; so, ɛ = l(i ɛ S. Hece, λ ( = if{ x : x S } ɛ for all ɛ>0. This implies that λ ( =0. c Let u S B. The there is a sequece {I } of ope itervals such that B I ad u = l(i. But B I A I u S A. Therefore, S B S A ad, cosequetly, λ (A = if{ x : x S A } if{ x : x S B } = λ (B. d The proof of this part is left to the reader as a exercise. e If λ (A = for some, the, by part (c, λ ( A = ; cosequetly, (3.6 holds. So, assume that λ (A < for all. Let ɛ>0 be give. For each, choose a sequece {I k } k of ope itervals such that k I k A ad k l(i k <λ (A +ɛ/2. The collectio of itervals {I k },k is coutable (because N Nis coutable ad,k I k = ( k I k A. Hece, ( λ A l(i k = l(i k,k k ( λ (A + ɛ 2 λ (A +ɛ. As ɛ>0 was arbitrary, this proves that λ ( A λ (A.
15 3.2 Lebesgue Outer Measure 93 As we have oted, the domai of λ is P(R; that is, every subset of R has a outer measure. Our questio ow is whether λ is the desired extesio of legth. That is, do Coditios (M1 (M3 hold with μ = λ ad A = P(R? Certaily, Coditio (M1 holds; ad the ext propositio shows that Coditio (M2 holds also. PROPOSITION 3.2 The outer measure of a iterval is its legth. That is, λ (I =l(i for every iterval I. First assume I =[a, b], that is, that I is a bouded ad closed iterval. If ɛ>0, the (a ɛ/2,b+ ɛ/2 [a, b] ad so ( ( λ ([a, b] l a ɛ 2,b+ ɛ = b a + ɛ. 2 Thus, for ay ɛ>0, λ ([a, b] b a + ɛ ad, hece, λ ([a, b] b a. Cosequetly, it remais to establish that λ ([a, b] b a. Let {I } be a sequece of ope itervals such that I [a, b]. We claim that l(i >b a. Sice {I } is a ope cover for [a, b], the HeieBorel theorem implies that there is a fiite subcover, say {I k } N. Now, clearly, N l(i k l(i. So, we eed oly show that N l(i k >b a. Because a [a, b], there must be a iterval, say J 1 =(a 1,b 1, i the collectio {I k } N with a 1 <a<b 1.Ifb<b 1, the N l(i k l(j 1 =b 1 a 1 >b a. Otherwise, b 1 [a, b], so there must be a iterval, say J 2 =(a 2,b 2, i the collectio {I k } N with a 2 <b 1 <b 2. Note that, ecessarily, J 2 J 1.Ifb<b 2, the N l(i k l(j 1 +l(j 2 =(b 1 a 1 +(b 2 a 2 =(b 2 a 1 +(b 1 a 2 >b 2 a 1 >b a. Otherwise, b 2 [a, b], so there must be a iterval, say J 3 =(a 3,b 3, i the collectio {I k } N such that a 3 <b 2 <b 3 ad, ecessarily, J 3 J 2 ad J 3 J 1. This process ca cotiue at most N times. Cosequetly, there is a m N with m N such that J i =(a i,b i {I k } N for 1 i m ad Therefore, a 1 <a, a 2 <b 1 <b 2,..., a m <b m 1 <b m, b<b m. N m l(i k l(j i =(b 1 a 1 +(b 2 a 2 + +(b m a m i=1 =(b m a 1 +(b 1 a 2 +(b 2 a 3 + +(b m 1 a m >b m a 1 >b a.
16 94 Chapter 3 Lebesgue Measure o the Real Lie Exercises for Sectio 3.2 So, if {I } is a sequece of ope itervals with I [a, b], the l(i >b a. But the, by defiitio, λ ([a, b] b a. This last fact ad the previously established reverse iequality show λ ([a, b] = b a. Now, let I be ay fiite (i.e., bouded iterval. The for each ɛ>0, there is a closed iterval J with J I ad l(i <l(j+ɛ (why?. Thus, l(i ɛ<l(j =λ (J λ (I. Sice ɛ>0 was arbitrary, it follows that λ (I l(i. But, o the other had, λ (I λ ( I = l ( I = l(i, so that λ (I l(i. Fially, if I is a ifiite (i.e., ubouded iterval, the for each real umber M, there is a closed iterval J of legth M with J I. It follows that λ (I λ (J =l(j =M. Hece, λ (I =. We have observed that Coditios (M1 ad (M2 are satisfied with μ = λ ad A = P(R. Therefore, our questio ow is: Does Coditio (M3 hold with μ = λ ad A = P(R? If the aswer to this questio were yes, the λ would be the desired extesio of legth ad every subset of R would be measurable. The aswer, however, is o! I fact, as we will discover i the ext sectio, λ is ot eve fiitely additive Let I be ay fiite iterval. Show that for each ɛ>0, there is a closed iterval J with J I ad l(i <l(j+ɛ Prove part (d of Propositio 3.1. That is, show that λ (x + A =λ (A for x R, A R, where x + A = { x + y : y A } Suppose that A is a set with λ (A <. Show that the fuctio g defied by g(x =λ ( A (,x] is uiformly cotiuous o R Show that the Cator set P has Lebesgue outer measure zero Show that, for each E R, there is a sequece of ope sets {O } such that O 1 O 2 E ad λ (E =λ ( O = lim λ (O For A Rad b R, defie ba = { ba : a A }. Show that λ (ba = b λ (A Suppose that f: R Ris differetiable at each poit of R. a If f (x 1 for each x R, prove that for each A R, λ ( f(a λ (A. Hit: Use the meavalue theorem. b Provide a example to show that the precedig iequality may fail to hold if f (x > 1 for some x R. 3.3 FURTHER PROPERTIES OF LEBESGUE OUTER MEASURE Recall that we are tryig to exted the otio of legth so that it applies to all Borel sets. Specifically, we are searchig for a fuctio μ defied o some
17 3.3 Further Properties of Lebesgue Outer Measure 95 collectio A of subsets of R such that (M1 A is a σalgebra ad A B. (M2 μ(i = l(i, for all itervals I. (M3 If A 1, A 2,... are i A, with A i A j = for i j, the ( μ A = μ(a. I Sectio 3.2, we discovered that Coditios (M1 ad (M2 are satisfied with μ = λ (Lebesgue outer measure ad A = P(R. We will prove i this sectio that Coditio (M3 does ot hold with μ = λ ad A = P(R. I fact, we will show that eve fiite additivity does ot obtai. That is, it is possible to fid disjoit subsets A ad B of R such that the equatio λ (A B =λ (A+λ (B (3.7 fails to hold. The idea is to choose A ad B so that they are disjoit but sufficietly itermigled. Fiite Additivity Properties of λ It is best to begi by determiig coditios o disjoit sets A ad B that imply that (3.7 holds. Our first result is that if A ad B are ot oly disjoit but are a positive distace apart, the (3.7 is true. Before provig that fact, we eed some prelimiary defiitios ad lemmas. DEFINITION 3.7 Distace Betwee a Poit ad a Set or Two Sets If x Rad E R, the the distace from x to E, deoted by d(x, E, is defied to be d(x, E = if{ y x : y E }. If E ad F are subsets of R, the the distace from E to F, deoted by d(e, F, is defied to be d(e,f = if{ y x : y E,x F }. It is left as a exercise for the reader to show that (1 for fixed E R, the fuctio d(x, E is cotiuous, (2 d(e,f = if{ d(y, F :y E }, ad (3 if A E ad B F, the d(e,f d(a, B. The proof of the followig lemma is also left to the reader as a exercise. LEMMA 3.8 Suppose that I is a fiite ope iterval ad let ɛ, δ > 0 be give. The there are a fiite umber of ope itervals, say J 1,..., J, such that l(j k <δ, 1 k, I J k, ad l(j k <l(i+ɛ.
18 96 Chapter 3 Lebesgue Measure o the Real Lie LEMMA 3.9 Suppose that A is a subset of R with λ (A <. The for each ɛ, δ>0, there is a sequece {I } of ope itervals such that l(i <δ for all, I A, ad l(i <λ (A+ɛ. Give ɛ>0, there is a sequece {J } of ope itervals such that J A ad l(j <λ (A+ɛ/2. By Lemma 3.8, for each J, there are a fiite umber of ope itervals, say J 1, J 2,..., J k, with l(j j <δ,1 j k, k j=1 J j J, ad k j=1 l(j j <l(j +ɛ/2 +1. Now, the collectio {J j } k j=1 = {J 11, J 12,..., J 1k1, J 21, J 22,..., J 2k2,...} is coutable, beig a coutable uio of fiite collectios. We have l(j j <δ, for each ad j, ad l(j j =,j k l(j j ( l(j + ɛ 2 +1 j=1 ɛ 2 +1 λ (A+ɛ. <λ (A+ ɛ 2 + Also,,j J j = ( k j=1 J j J A. If we ow reidex the collectio {J 11, J 12,..., J 1k1, J 21, J 22,..., J 2k2,...} by usig a sigle subscript ad obtai {I }, the this sequece satisfies the coclusios of the lemma. THEOREM 3.8 Suppose that A ad B are subsets of R that are a positive distace apart; that is, d(a, B > 0. The λ (A B =λ (A+λ (B. Let δ = d(a, B. If λ (A B =, the it follows from Propositio 3.1(e that the coclusio of the theorem holds. So, assume that λ (A B <. Let ɛ>0 be give. By Lemma 3.9, there is a sequece {I } of ope itervals such that l(i <δfor all, I A B, ad l(i <λ (A B+ɛ. Now, let {J } deote the members of {I } that cotai a poit of A ad let {K } deote the oes that do ot cotai a poit of A. AsA A B I, it follows that A J. Also, because d(a, B =δ ad l(i <δfor all, there ca be o poits of B i ay J. Therefore, because B A B I, it must be that B K. Usig the defiitio of outer measure, we coclude that λ (A+λ (B l(j + l(k = l(i <λ (A B+ɛ. Because ɛ>0 was arbitrary, λ (A+λ (B λ (A B. The reverse iequality is true by Propositio 3.1(e.
19 3.3 Further Properties of Lebesgue Outer Measure 97 We ca, i fact, improve o Theorem 3.8. Ideed, suppose that A ad B are two subsets of R with the property that there is a ope set O with A O ad B O c. The the coclusio of Theorem 3.8 obtais. Roughly speakig, the reaso is as follows. Sice O is ope, it ca be writte as a coutable uio of disjoit ope itervals. Because the poits of A must lie withi these ope itervals ad the poits of B must lie outside of them, the sets A ad B caot be too itermigled. Before we ca provide a rigorous proof of the improvemet of Theorem 3.8, we eed two more lemmas. LEMMA 3.10 Let O be a proper ope subset of R (i.e., O is ope, oempty, ad ot equal to R. For each N, let { O = x : d(x, O c > 1 }. The, a O is ope ad O O for all N. b O 1 O 2 ad O = O. c If O, the d(o,o c =1/. d If O, the d(o,o c +1 =1/( +1. The proofs of parts (a ad (b are left to the reader. c Sice d(o,o c = if{ d(x, O c :x O }, we see that d(o,o c 1/. To prove the reverse iequality, we first ote that because O is ope, it ca be expressed as a coutable uio of disjoit ope itervals, say the itervals {I j } j. Now, by assumptio, O. This meas that there is a y O such that d(y, O c > 1/. Sice y O, there is a k such that y I k. Clearly, the distace from y to O c is the same as the distace from y to the earest edpoit of I k. Therefore, if we write I k =(a k,b k, the y (a k +1/, b k 1/. It follows that (a k +1/, b k 1/ O. Note that at least oe of the two umbers a k ad b k must be fiite. We will assume that a k is fiite. (If a k is ifiite, a similar argumet holds. Sice (a k +1/, b k 1/ O ad a k O c, we coclude by applyig Exercise 3.21(c that ( ( d(o,o c d a k + 1,b k 1, {a k } = 1. This completes the proof of part (c. d We first show that d(o,o c +1 1/( + 1. Suppose y O ad z O c +1. By defiitio, d(y, O c > 1/ ad d(z,o c 1/( + 1. Let ɛ>0 be give. The there is a w O c with w z < 1/( +1+ɛ. Also, w O c implies that w y > 1/. Thus, z y w y w z > ɛ = 1 ( +1 ɛ. As z ad y do ot deped o ɛ, we coclude that z y 1/( + 1. Cosequetly, because y O, z O c +1 were arbitrary, d(o,o c +1 1/( +1.
20 98 Chapter 3 Lebesgue Measure o the Real Lie To prove the reverse iequality, let I k =(a k,b k be as i the proof of part (c, ad assume as before that a k is fiite. Because a k O c, we have a k +1/( +1 O c +1. Therefore, by Exercise 3.21(c, ( ( d(o,o+1 c d a k + 1,b k 1 {, a k + 1 } 1 = +1 ( +1. This completes the proof of part (d. LEMMA 3.11 Suppose that A Rad λ (A <. Assume there is a proper ope subset O of R with A O. Let O = { x : d(x, O c > 1/ }. The λ (A = lim λ (A O. Let A = A O. The, by Lemma 3.10(b, A 1 A 2 ad, cosequetly, λ (A 1 λ (A 2. Also, sice A A for all, λ (A λ (A for all. By assumptio, λ (A <. Thus, {λ (A } is a mootoe odecreasig, bouded sequece of real umbers; ad, hece, coverges to a real umber, say α. Clearly, α λ (A. Now, let B = A \ A ad C = A +1 \ A. The we have A = A B ad B = C C +1. Thus, λ (A λ (A +λ (B (3.8 ad λ (B λ (C +λ (C (3.9 Now, for 2, A +1 = A C A 1 C, so that λ (A 1 C λ (A +1. (3.10 Also, A 1 O 1 ad C O. c So, by Lemma 3.10(d, d(a 1,C d(o 1,O=1/( c 1 > 0. Therefore, Theorem 3.8 implies that λ (A 1 C =λ (A 1 +λ (C. Usig (3.10 ad this last equatio, we coclude that, for 2, λ (C λ (A +1 λ (A 1. The (3.9 implies λ (B [λ (A +k λ (A +k 2 ]
21 3.3 Further Properties of Lebesgue Outer Measure 99 so that by (3.8 But, λ (A λ (A +λ (B λ (A + λ (A + [λ (A +k λ (A +k 2 ]. [λ (A +k λ (A +k 2 ] { = lim λ (A + m m } [λ (A +k λ (A +k 2 ] = lim m { λ (A 1 +λ (A +m 1 +λ (A +m } = λ (A 1 +2α. Cosequetly, we have show that λ (A λ (A 1 +2α for all. Lettig, reveals that λ (A α. As we kow, however, α λ (A. Thus, we have λ (A =α = lim λ (A, as required. THEOREM 3.9 Suppose that A ad B are subsets of R with the property that there is a ope set O with A O ad B O c. The λ (A B =λ (A+λ (B. If either λ (A orλ (B is ifiite, the the result is trivial. So, assume that both are fiite. If O =, the A = ; ad if O = R, the B =. I either of those cases, the result is also trivial. Cosequetly, we ca assume that O is a proper ope subset of R. As before, let O = { x : d(x, O c > 1/ } ad A = A O. Because A O ad B O c, Lemma 3.10(c implies that d(a,b d(o,o c =1/ ad, thus, by Theorem 3.8, λ (A B =λ (A +λ (B. Sice A B A B, we have λ (A B λ (A B. Therefore, for each N, λ (A B λ (A B =λ (A +λ (B. Lettig ad applyig Lemma 3.11, we get that λ (A B λ (A+λ (B. Propositio 3.1(e shows that the reverse iequality holds. Lebesgue Outer Measure Is Not Fiitely Additive We have ow see that, uder certai coditios, λ (A B =λ (A+λ (B (3.11 for disjoit subsets A ad B of R. Our ext theorem, which we will state ad prove shortly, shows that (3.11 does ot hold for every pair of disjoit subsets A ad B of R, that is, that λ is ot fiitely additive. I view of Theorem 3.9, it is clear that if (3.11 fails to hold for disjoit subsets A ad B, the those sets must be cosiderably itermigled. To obtai this itermiglig, we proceed as follows.
22 100 Chapter 3 Lebesgue Measure o the Real Lie LEMMA 3.12 For x, y R, defie x y if ad oly if y x Q. The is a equivalece relatio ad, hece, partitios R ito disjoit equivalece classes. Moreover, there is a set S [0, 1 cotaiig exactly oe elemet from each equivalece class. That is a equivalece relatio is left as a exercise for the reader. By the axiom of choice (see page 14, we ca select a set T R that cotais exactly oe elemet from each equivalece class. Let us set S = { x [x] :x T } where [x] deotes the greatest iteger i x. Because for each x, x [x] [0, 1 ad x [x] x, the proof is complete. LEMMA 3.13 Let S be the set defied i Lemma 3.12 ad W =( 1, 1 Q. The a {S + r} r Q forms a collectio of pairwise disjoit sets. b (0, 1 r W (S + r ( 1, 2. a Suppose q, r Q ad (S + q (S + r. Let y (S + q (S + r. The there exist u, v S such that y = u + q = v + r. Hece, u v. Sice S cotais oly oe elemet from each equivalece class, we must have u = v, which, i tur, implies q = r. b Let x (0, 1. The there is a u S such that x u. Put r = x u. The r Q ad x S + r. Moreover, sice x (0, 1 ad S [0, 1, 1 <r<1. Thus, (0, 1 r W (S + r. That r W (S + r ( 1, 2 follows immediately from the fact that S [0, 1. Note: Lemma 3.13(a shows that the sets {S + r} r Q are pairwise disjoit. They are also cosiderably itermigled as is show i Exercise THEOREM 3.10 Lebesgue outer measure λ is ot fiitely additive. Suppose to the cotrary that λ is fiitely additive. Let {q } be a eumeratio of the ratioals i ( 1, 1 ad set E = S + q. Usig the assumed fiite additivity of λ, Propositio 3.2, Lemma 3.13, ad Propositio 3.1(c ad (e, we coclude that 1=λ ( (0, 1 ( λ E λ (E = lim λ (E k = lim λ ( E k λ ( ( 1, 2 =3. This shows 1 lim λ (E k 3. But, by Propositio 3.1(d, λ (E k = λ (S + q k = λ (S =λ (S. (3.12 Cosequetly, 1 lim λ (S 3, which is impossible (why?. Hece, λ is ot fiitely additive.
23 3.4 Lebesgue Measure 101 Exercises for Sectio 3.3 COROLLARY 3.2 Lebesgue outer measure λ is ot coutably additive. That is, Coditio (M3 does ot hold with μ = λ ad A = P(R Prove the followig facts: a For fixed E R, the fuctio d(x, E is cotiuous. b If E ad F are subsets of R, the d(e,f = if{ d(y, F :y E }. c If A E ad B F, the d(e,f d(a, B. d d(e,f =d(e,f Prove the followig facts: a Suppose that F is a closed subset of R, K is a compact subset of R, ad F K =. The d(f, K > 0. b Show that part (a is ot true if it is assumed oly that K is closed Prove Lemma 3.8 o page Verify parts (a ad (b of Lemma 3.10 o page Suppose that O is ope. Prove that λ (W =λ (W O+λ (W O c for all subsets W of R Defie x y if ad oly if x y Q. Show that is a equivalece relatio Let N be a positive iteger, {r } a eumeratio of Q, ad S as i Lemma For each N, defie S = S + r. Prove that there is o ope set O with the property that N S O ad =N+1 S Oc Suppose that 0 a<b 1. Prove that it is possible to select the elemets of the set S i Lemma 3.12 so that S (a, b Provide a detailed justificatio for each step i ( LEBESGUE MEASURE For ease i referece, we repeat oce more that we are searchig for a fuctio μ defied o some collectio A of subsets of R such that (M1 A is a σalgebra ad A B. (M2 μ(i = l(i, for all itervals I. (M3 If A 1, A 2,... are i A, with A i A j = for i j, the ( μ A = μ(a. As we have see, Coditios (M1 ad (M2 hold with μ = λ ad A = P(R, but Coditio (M3 does ot (Corollary 3.2. Note, however, that we do ot eed to have our measure μ defied for all subsets of R; Coditio (M1 requires oly that it be defied o a σalgebra A of subsets of R that cotais the Borel sets.
24 102 Chapter 3 Lebesgue Measure o the Real Lie Thus, oe way to get Coditio (M3 to hold might be to restrict λ to some proper subcollectio of subsets of R; that is, select A to be a proper subset of P(R. Ad, to do that, we eed to idetify a criterio for decidig whether a subset of R is measurable, that is, is a member of A. By Coditio (M1, we must have B A; so, i particular, A must cotai all ope sets. Hece, the criterio we select must be satisfied by all ope sets. The Carathéodory Criterio Theorem 3.9 states that if A ad B are subsets of R with the property that there is a ope set O with A O ad B O c, the λ (A B =λ (A+λ (B. As a cosequece of Theorem 3.9, we obtai the followig propositio. PROPOSITION 3.3 Let O be a ope set. The λ (W =λ (W O+λ (W O c (3.13 for every subset W of R. For every subset W of R, we have W =(W O (W O c. Sice W O O ad W O c O c, we see that (3.13 is a simple cosequece of Theorem 3.9. Equatio (3.13 provides a additivity relatio for Lebesgue outer measure that is satisfied by all ope sets. That relatio shows the way to the required criterio for decidig whether a subset of R is measurable. DEFINITION 3.8 Carathéodory Criterio A set E Ris said to satisfy the Carathéodory criterio if λ (W =λ (W E+λ (W E c (3.14 for all subsets W of R. We deote by M the collectio of all subsets of R that satisfy the Carathéodory criterio. Note: By Propositio 3.1(e, the iequality λ (W λ (W E+λ (W E c always holds. Cosequetly, to prove that a subset E of R is a member of M, it suffices to establish the iequality λ (W λ (W E+λ (W E c (3.15 for all subsets W of R. The ext theorem demostrates that Coditio (M1 holds for the collectio M of subsets of R that satisfy the Carathéodory criterio.
25 3.4 Lebesgue Measure 103 THEOREM 3.11 M is a σalgebra ad M B. That M is closed uder complemetatio is clear. First we prove that M is closed uder fiite uios. So, assume A, B M. We claim that A B M. Let W R. The, we must show that λ (W λ ( W (A B + λ ( W (A B c. (3.16 (See the ote followig Defiitio 3.8. Now, we ca write W (A B =(W A (W A c B ad, hece, by the subadditivity of λ, Cosequetly, λ ( W (A B λ (W A+λ (W A c B. λ ( W (A B + λ ( W (A B c λ (W A+λ (W A c B+λ ( W (A B c = λ (W A+ [ λ ( (W A c B + λ ( (W A c B c]. Because B M, the quatity betwee the square brackets i the previous expressio equals λ (W A c. Thus, λ ( W (A B + λ ( W (A B c λ (W A+λ (W A c. This last sum equals λ (W because A M. Hece, (3.16 holds. We have ow established that M is a algebra of sets. Next, we show that M is closed uder coutable uios. To that ed, let {E } M. We must prove that E M. To begi, we disjoitize the sets E, =1, 2,... Let A 1 = E 1, A 2 = E 2 \ E 1, A 3 = E 3 \ (E 1 E 2, ad, i geeral, A = E \ ( 1 E k. The, see Exercise 3.30, Ai A j =, for i j, ad A = E. Moreover, because M is a algebra of sets ad E Mfor N, it follows that A Mfor N. Now, let W be ay subset of R ad set E = E = A. We must show that λ (W λ (W E+λ (W E c. By the subadditivity of λ, ( λ (W E =λ (W A ( (3.17 = λ (W A λ (W A. For each N, set B = A k. The, because M is a algebra, B M for all N. Cosequetly, for all, λ (W =λ (W B +λ (W B c. (3.18
26 104 Chapter 3 Lebesgue Measure o the Real Lie Because B m=1 A m = E, it follows that E c B c. This last fact ad (3.18 imply that λ (W λ (W B +λ (W E c. (3.19 We will ow prove by iductio that for all N, λ (W B = λ (W A k. (3.20 The equatio holds trivially whe = 1. So, assume that it holds for. Sice A +1 M,wehave λ (W B +1 =λ ( (W B +1 A +1 + λ ( (W B +1 A c ( Because the A k s are pairwise disjoit, we have W B +1 A +1 = W A +1 ad W B +1 A c +1 = W B. Thus, by (3.21 ad the iductio hypothesis, λ (W B +1 =λ (W A +1 +λ (W B +1 = λ (W A +1 + λ (W A k = λ (W A k, as required. Employig (3.19 ad (3.20, we coclude that λ (W λ (W A k +λ (W E c for all N ad, cosequetly, λ (W λ (W A +λ (W E c. Applyig (3.17 to the previous iequality, we deduce that λ (W λ (W E+λ (W E c. This shows E M. We have ow established that M is a σalgebra. It remais to prove that M B. By Propositio 3.3, M cotais all ope sets ad, as we have just see, M is a σalgebra. Cosequetly, sice B is the smallest σalgebra that cotais all ope sets, it must be that M B. Our ext theorem shows that Coditio (M3 is satisfied whe Lebesgue outer measure λ is restricted to M. We deote by λ the restrictio of Lebesgue outer measure to M; that is, λ: M Ris defied by λ(e =λ (E.
27 3.4 Lebesgue Measure 105 THEOREM 3.12 If A 1, A 2,... are i M, with A i A j = for i j, the ( λ A = λ(a. We first prove that λ is fiitely additive o M. So, let A, B Mwith A B =. Set W = A B. The W A = A ad W A c = B. Cosequetly, as A M, we have by (3.14 that λ(a B =λ(w =λ (W =λ (W A+λ (W A c = λ (A+λ (B =λ(a+λ(b. This shows that λ is fiitely additive. Suppose ow that {A } Mwith A i A j = for i j. Usig the fact that λ is fiitely additive o M ad the mootoicity of Lebesgue outer measure, we coclude that m ( m λ(a k =λ ( A k λ A for all m N. Lettig m gives λ(a λ ( A. The reverse iequality obtais because of the coutable subadditivity of Lebesgue outer measure. Lebesgue Measurable Sets ad Lebesgue Measure From Propositio 3.2 o page 93 ad Theorems 3.11 ad 3.12, we see that Coditios (M1 (M3 are satisfied with μ = λ ad A = M; that is, (L1 M is a σalgebra ad M B. (L2 λ(i = l(i, for all itervals I. (L3 If A 1, A 2,... are i M, with A i A j = for i j, the ( λ A = λ(a. Cosequetly, the set fuctio λ: M Ris the required extesio of legth. We will employ the followig termiology: DEFINITION 3.9 Lebesgue Measurable Sets ad Lebesgue Measure The members of M are called Lebesgue measurable sets. That is, E is a Lebesgue measurable set if ad oly if for every subset W of R, λ (W =λ (W E+λ (W E c. The restrictio of Lebesgue outer measure to M is deoted by λ ad is called Lebesgue measure. I the ext three propositios, we establish some additioal properties of Lebesgue measure ad Lebesgue measurable sets.
28 106 Chapter 3 Lebesgue Measure o the Real Lie PROPOSITION 3.4 A subset of R with Lebesgue outer measure zero is a Lebesgue measurable set; that is, λ (E =0 E M. Suppose that λ (E =0. LetW be a arbitrary subset of R. AsW E E, the mootoicity of Lebesgue outer measure implies that λ (W E λ (E =0. Usig the fact that W E c W, we ow coclude that λ (W λ (W E c =λ (W E+λ (W E c. This last iequality shows that E M. PROPOSITION 3.5 Every coutable subset of R has Lebesgue measure zero. Let E Rbe coutable, say E = {x }. The we ca write E = {x }. Note that if a R, the, by (L2, λ({a} =λ([a, a] = a a = 0. Therefore, applyig (L3, we coclude that ( λ(e =λ {x } = λ({x }=0, as required. The ext propositio shows that the coverse of Propositio 3.5 does ot hold. PROPOSITION 3.6 The Cator set P has Lebesgue measure zero. Let G =[0, 1] \ P. From Chapter 2 (page 62, we kow that G ca be writte as a coutable uio of disjoit ope itervals {I } with the property that l(i = 1. Hece, by (L3 ad (L2, λ(g = λ(i = l(i =1. Clearly, P ad G are disjoit ad P G =[0, 1]. Therefore, which shows that λ(p =0. 1=λ(P +λ(g =λ(p +1, Aother useful result is the followig. THEOREM 3.13 If {E } is a sequece of Lebesgue measurable sets with E 1 E 2, the ( λ E = lim λ(e.
29 3.4 Lebesgue Measure 107 If λ(e = for some, the both sides of the previous equatio equal. So, assume λ(e < for all. We begi by disjoitizig the E s as follows. Let A 1 = E 1 ad, for 2, let A = E \ E 1. The it is easy to see that {A } M, A i A j = for i j, ad A = E. Therefore, by coutable additivity, ( λ ( E = λ A = λ(a. As E 1 E, we have λ(a =λ(e \ E 1 =λ(e λ(e 1 for 2. Cosequetly, ( λ as required. E = λ(a = lim = lim ( λ(e 1 + λ(a k [λ(e k λ(e k 1 ] k=2 = lim λ(e, The Relatio Betwee B ad M We close this sectio by discussig the relatioship betwee the collectio of Borel sets B ad the collectio of Lebesgue measurable sets M. By Theorem 3.11, B M. The questio ow is: Does B = M? I other words, is every Lebesgue measurable set a Borel set or are there Lebesgue measurable sets that are ot Borel sets? It is ot easy to aswer that questio. I fact, Lebesgue ad Borel argued the questio without fidig the aswer. It turs out that the aswer to the questio is o there are Lebesgue measurable sets that are ot Borel sets. I other words, we have the followig theorem: THEOREM 3.14 The σalgebra of Borel sets B is a proper subcollectio of the σalgebra of Lebesgue measurable sets M. See Exercise Exercises for Sectio Let {E } be a sequece of subsets of R. Defie A1 = E1 ad A = E \ ( 1 E k for 2. Prove that A i A j =, for i j, ad that A = E I Chapter 2, we itroduced the cocept of measure zero. Prove that this cocept is equivalet to that of Lebesgue measure zero. I other words, show that a subset E R has measure zero i the sese of Defiitio 2.19 o page 70 if ad oly if λ(e = Verify that if A M, λ(a =0,adB A, the B Mad λ(b = Let A, B Mwith A B ad λ(a <. Show that λ(b \ A =λ(b λ(a.
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