# PART TWO. Measure, Integration, and Differentiation

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1 PART TWO Measure, Itegratio, ad Differetiatio

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 σ-algebra, it follows that A is a σ-algebra. 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 σ-algebra. 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.

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 σ-algebra 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 σ-algebra 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 σ-algebra 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 σ-algebra 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 σ-algebra 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 σ-algebra; 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 σ-algebra 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 Heie-Borel 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.

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.

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 ]

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 σ-algebra 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 σ-algebra. 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 σ-algebra. Cosequetly, sice B is the smallest σ-algebra 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.

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.

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