Lecture 2 Measures. Measure spaces. µ(a n ), for n N, and pairwise disjoint A 1,..., A n, we say that the. (S, S) is called

Transcription

1 Lecture 2: Measures 1 of 17 Course: Theory of Probability I Term: Fall 2013 Istructor: Gorda Zitkovic Lecture 2 Measures Measure spaces Defiitio 2.1 (Measure). Let (S, S) be a measurable space. A mappig µ : S [0, ] is called a (positive) measure if 1. µ( ) = 0, ad 2. µ( A ) = N µ(a ), for all pairwise disjoit {A } N i S. A triple (S, S, µ) cosistig of a o-empty set, a σ-algebra S o it ad a measure µ o S is called a measure space. Remark A mappig whose domai is some oempty set A of subsets of some set S is sometimes called a set fuctio. 2. If the requiremet 2. i the defiitio of the measure is weakeed so that it is oly required that µ(a 1 A ) = µ(a 1 ) + + µ(a ), for N, ad pairwise disjoit A 1,..., A, we say that the mappig µ is a fiitely-additive measure. If we wat to stress that a mappig µ satisfies the origial requiremet 2. for sequeces of sets, we say that µ is σ-additive (coutably additive). Defiitio 2.3 (Termiology). A measure µ o the measurable space (S, S) is called 1. a probability measure, if µ(s) = 1, 2. a fiite measure, if µ(s) <, 3. a σ-fiite measure, if there exists a sequece {A } N i S such that A = S ad µ(a ) <, 4. diffuse or atom-free, if µ({x}) = 0, wheever x S ad {x} S. A set N S is said to be ull if µ(n) = 0.

2 Lecture 2: Measures 2 of 17 Example 2.4 (Examples of measures). Let S be a o-empty set, ad let S be a σ-algebra o S. 1. Measures o coutable sets. Suppose that S is a fiite or coutable set. The each measure µ o S = 2 S is of the form µ(a) = p(x), x A for some fuctio p : S [0, ] (why?). I particular, for a fiite set S with N elemets, if p(x) = 1/N the µ is a probability measure called the uiform measure 1 o S. 2. Dirac measure. For x S, we defie the set fuctio δ x o S by 1, x A, δ x (A) = 0, x A. 1 I the fiite case, it has the well-kow property that µ(a) = #A #S, where # deotes the cardiality (umber of elemets). It is easy to check that δ x is ideed a measure o S. Alteratively, δ x is called the poit mass at x (or a atom o x, or the Dirac fuctio, eve though it is ot really a fuctio). Moreover, δ x is a probability measure ad, therefore, a fiite ad a σ-fiite measure. It is atom free oly if {x} S. 3. Coutig Measure. Defie a set fuctio µ : S [0, ] by #A, A is fiite, µ(a) =, A is ifiite, where, as above, #A deotes the umber of elemets i the set A. Agai, it is ot hard to check that µ is a measure - it is called the coutig measure. Clearly, µ is a fiite measure if ad oly is S is a fiite set. µ could be σ-fiite, though, eve without S beig fiite. Simply take S = N, S = 2 N. I that case µ(s) =, but for A = {}, N, we have µ(a ) = 1, ad S = A. Fially, µ is ever atom-free ad it is a probability measure oly if #S = 1. Example 2.5 (A fiitely-additive set fuctio which is ot a measure). Let S = N, ad S = 2 S. For A S defie µ(a) = 0 if A is fiite ad µ(a) =, otherwise. For A 1,..., A S, either 1. all A i is fiite, for i = 1,...,. The i=1 A i is also fiite ad so 0 = µ( i=1 A i) = µ(a i ), or i=1 2. at least oe A i is ifiite. The i=1 A i is also ifiite ad so = µ( i=1 A i) = µ(a i ). i=1

3 Lecture 2: Measures 3 of 17 O the other had, take A i = {i}, for i N. The µ(a i ) = 0, for Note: It is possible to costruct very each i N, ad, so, simple-lookig fiite-additive measures i N µ(a i ) = 0, but µ( i A i ) = µ(n) =. which are ot σ-additive. For example, there exist {0, 1}-valued fiitely-additive measures o all subsets of N, which are Propositio 2.6 (First properties of measures). Let (S, S, µ) be a measure space. 1. For A 1,..., A S with A i A j =, for i = j, we have 2. If A, B S, A B, the µ(a i ) = µ( i=1 A i) i=1 (Fiite additivity) ot σ-additive. Such objects are called ultrafilters ad their existece is equivalet to a certai versio of the Axiom of Choice. µ(a) µ(b) (Mootoicity of measures) 3. If {A } N i S is icreasig, the µ( A ) = lim µ(a ) = sup µ(a ). (Cotiuity with respect to icreasig sequeces) 4. If {A } N i S is decreasig ad µ(a 1 ) <, the µ( A ) = lim µ(a ) = if µ(a ). (Cotiuity with respect to decreasig sequeces) 5. For a sequece {A } N i S, we have µ( A ) µ(a ). N (Subadditivity) Proof. 1. Note that the sequece A 1, A 2,..., A,,,... is pairwise disjoit, ad so, by σ-additivity, µ( i=1 A i) = µ( i N A i ) = µ(a i ) = i=1 µ(a i ) + i N i=+1 µ( ) = µ(a i ). i=1 2. Write B as a disjoit uio A (B \ A) of elemets of S. By (1) above, µ(b) = µ(a) + µ(b \ A) µ(a). 3. Defie B 1 = A 1, B = A \ A 1 for > 1. The {B } N is a pairwise disjoit sequece i S with k=1 B k = A for each N (why?). By σ-additivity we have µ( A ) = µ( B ) = µ(b ) = lim µ(b k ) N k=1 = lim µ( k=1 B k) = lim µ(a ).

4 Lecture 2: Measures 4 of Cosider the icreasig sequece {B } N i S give by B = A 1 \ A. By De Morga laws, fiiteess of µ(a 1 ) ad (3) above, we have µ(a 1 ) µ( A ) = µ(a 1 \ ( A )) = µ( B ) = lim µ(b ) = lim µ(a 1 \ A ) = µ(a 1 ) lim µ(a ). Subtractig both sides from µ(a 1 ) < produces the statemet. 5. We start from the observatio that for A 1, A 1 S the set A 1 A 2 ca be writte as a disjoit uio so that A 1 A 2 = (A 1 \ A 2 ) (A 2 \ A 1 ) (A 1 A 2 ), µ(a 1 A 2 ) = µ(a 1 \ A 2 ) + µ(a 2 \ A 1 ) + µ(a 1 A 2 ). O the other had, ad so µ(a 1 ) + µ(a 2 ) = (µ(a 1 \ A 2 ) + µ(a 1 A 2 )) ( ) + µ(a 2 \ A 1 ) + µ(a 1 A 2 ) = µ(a 1 \ A 2 ) + µ(a 2 \ A 1 ) + 2µ(A 1 A 2 ), µ(a 1 ) + µ(a 2 ) µ(a 1 A 2 ) = µ(a 1 A 2 ) 0. Iductio ca be used to show that µ(a 1 A ) µ(a k ). k=1 Sice all µ(a ) are oegative, we ow have µ(a 1 A ) α, for each N, where α = µ(a ). N The sequece {B } N give by B = k=1 A k is icreasig, so the cotiuity of measure with respect to icreasig sequeces implies that µ( A ) = µ( B ) = lim µ(b ) = lim µ(a 1 A ) α. Remark 2.7. The coditio µ(a 1 ) < i the part (4) of Propositio 2.6 caot be sigificatly relaxed. Ideed, let µ be the coutig measure o N, ad let A = {, + 1,... }. The µ(a ) = ad, so lim µ(a ) =. O the other had, A =, so µ( A ) = 0.

5 Lecture 2: Measures 5 of 17 I additio to uios ad itersectios, oe ca produce other importat ew sets from sequeces of old oes. More specifically, let {A } N be a sequece of subsets of S. The subset lim if A of S, defied by lim if A = B, where B = k A k, is called the limit iferior of the sequece A. It is also deoted by lim A or {A, ev.} (ev. stads for evetually 2 ). Similarly, the subset lim sup A of S, defied by lim sup A = B, where B = k A k, is called the limit superior of the sequece A. It is also deoted by lim A or {A, i.o.} (i.o. stads for ifiitely ofte 3 ). Clearly, we have lim if A lim sup A. 2 the reaso for the use of the word evetually is the followig: lim if A is the set of all x S which belog to A for all but fiitely may values of the idex, i.e., from some value of the idex owards. 3 i words, lim sup A is the set of all x S which belog A for ifiitely may values of. Problem 2.1. Let (S, S, µ) be a fiite measure space. Show that Hit: For the secod part, a measure space with fiite (ad small) S will do. µ(lim if A ) lim if µ(a ) lim sup µ(a ) µ(lim sup A ), for ay sequece {A } N i S. Give a example of a (sigle) sequece {A } N for which all iequalities above are strict. Propositio 2.8 (Borel-Catelli Lemma I). Let (S, S, µ) be a measure space, ad let {A } N be a sequece of sets i S with the property that N µ(a ) <. The µ(lim sup A ) = 0. Proof. Set B = k A k, so that {B } N is a decreasig sequece of sets i S with lim sup A = B, ad so µ(lim sup A ) µ(b ), for each N. Usig the subadditivity of measures of Propositio 2.6, part 5., we get µ(b ) µ(a ). (2.1) k= Sice N µ(a ) coverges, the right-had side of (2.1) ca be made arbitrarily small by choosig large eough N. Extesios of measures ad the coi-toss space Example 1.19 has itroduced the measurable space ({ 1, 1} N, S), with S = 2 { 1,1} beig the product σ-algebra o { 1, 1} N. The purpose

6 Lecture 2: Measures 6 of 17 of the preset sectio is to tur ({ 1, 1} N, S) ito a measure space, i.e., to defie a suitable measure o it. It is easy to costruct just ay measure o { 1, 1} N, but the oe we are after is the oe which will justify the ame coi-toss space. The ituitio we have about tossig a fair coi ifiitely may times should help us start with the defiitio of the coi-toss measure - deoted by µ C - o cyliders. Sice the coordiate spaces { 1, 1} are particularly simple, each product cylider is of the form C = { 1, 1} N or C = C 1,..., k ;b 1,...,b k, where } C 1,..., k ;b 1,...,b k = {s = (s 1, s 2,... ) { 1, 1} N : s 1 = b 1,..., s k = b k, for some k N, ad a choice of 1 1 < 2 < < k N of coordiates ad the correspodig values b 1, b 2,..., b k { 1, 1}. I the laguage of elemetary probability, each cylider correspods to the evet whe the outcome of the i -th coi is b i { 1, 1}, for i = 1,...,. The measure (probability) of this evet ca oly be give by µ C (C 1,..., k ;b 1,...,b k ) = = 2 k. }{{} k times (2.2) The hard part is to exted this defiitio to all elemets of S, ad ot oly cyliders. For example, i order to state the law of large umbers later o, we will eed to be able to compute the measure of the set { s { 1, 1} N } 1 : lim s k = 0, k=1 which is clearly ot a cylider. Problem 1.9 states, however, that cyliders form a algebra ad geerate the σ-algebra S. Luckily, this puts us close to the coditios of the followig importat theorem of Caratheodory. Theorem 2.9 (Caratheodory s Extesio Theorem). Let S be a o- Note: I words, a σ-additive measure empty set, let A be a algebra of its subsets ad let µ : A [0, ] be a o a algebra A ca be exteded to a σ-additive measure o the σ-algebra set-fuctio with the followig properties: geerated by A. It is clear that the σ- additivity requiremet of Theorem 2.9 is 1. µ( ) = 0, ad ecessary, but it is quite surprisig that it is actually sufficiet. 2. µ(a) = =1 µ(a ), if {A } N A is a partitio of A. The, there exists a measure µ o σ(a) with the property that µ(a) = µ(a) for A A. Of Theorem 2.9. PART I. We start by defiig a measure-like object, Note: Ituitively, we try all differet called a outer measure, µ : 2 S [0, ] i the followig way: coutable covers of B with elemets of A ad miimize the total µ. { µ (B) = if µ(a ) : B =1 =1 } A, A A for all N.

7 Lecture 2: Measures 7 of 17 Eve though we do t expect the ifimum i the defiitio of µ to be attaied, µ has the followig properties: 1. µ ( ) = 0 (otriviality), 2. for B C, µ (B) µ (C) (mootoicity), ad 3. µ ( k B k ) k=1 µ (B k ) (subadditivity) Parts 1. ad 2. are immediately clear, while, to show 3., we pick ε > 0 ad k N ad fid a coutable cover {A k } N with elemets of A such that µ (B k ) µ(a k ) + ε. 2 k =1 Usig {A k } k N, N as a cadidate cover for k B k, we coclude that µ ( k B k ) k=1 µ (B k ) + ε. This beig true for each ε > 0 implies 3. We remark at this poit that µ ad µ coicide o A. By usig (A,,,... ) as a cadidate coutable cover of A A, we ca coclude that µ (A) µ(a), for all A A. Coversely, suppose that A N A with A A. Give that the elemets of A form a algebra, we ca assume that A A, for all N ad that {A } N are pairwise disjoit, as ay sequece coverig A ca be trasformed ito such a sequece without icreasig µ(a ). The assumed coutable Note: It is, probably, iterestig to ote additivity of µ o A ow comes ito play sice, for partitio {A that this is the oly place i the etire } N proof where the coutable additivity of of A ito elemets i A, we ecessarily have µ(a ) = µ(a), ad, µ o A is used. so µ (A) µ(a). PART II. The set-fuctio µ is, i geeral, ot a measure, but comes with the advatage of beig defied o all subsets of S. The cetral idea of the proof is to recover coutable additivity by restrictig its domai a little. We say that a subset M S is Caratheodory-measurable or µ -measurable if µ (B) = µ (B M) + µ (B M c ) for all B S, (2.3) with the family of all µ -measurable subsets of S deoted by M. We ote that, by subadditivity, the equality sig i the defiitio of the measurability ca be replaced by ; this will be used below. The first thig we eed to establish about M is that it is a algebra ad that µ is a fiitely-additive measure o M. Clearly A ad the complemet axiom follows directly from the symmetry i (2.3). Oly the closure uder fiite uios eeds some discussio, ad, by iductio, we oly eed to cosider two-elemet uios; for that, we pick M, N M, ad itroduce the followig otatio M 00 = M c N c, M 01 = M c N, M 10 = M N c M 11 = M N. (2.4)

8 Lecture 2: Measures 8 of 17 By the measurability of M ad N, for ay B S, we have µ (B) = µ (B N c ) + µ (B N) = µ (M 00 B) + µ (M 10 B) + µ (M 01 B) + µ (M 11 B) O the other had, M 01 M 10 M 11 = M N, so that, by subadditivity ad (2.4), we have µ ( (M N) c B ) + µ ( (M N) B ) = µ (M 00 B) + µ ( ) (M 01 B) (M 10 B) (M 11 B) µ (M 00 B) + µ (M 10 B) + µ (M 01 B) + µ (M 11 B) = µ (B), which implies that that M N M. Whe M N = a applicatio of measurability of N to B = M N yields the fiite additivity of µ o M : µ (M N) = µ ((M N) N) + µ ((M N) N c ) = µ (N) + µ (M). PART III. We ow tur to the closure of M uder coutable uios ad the σ-additive property of µ. Sice M already kow to be a algebra, it will be eough to show that it is closed uder pairwisedisjoit uios, i.e., that M M wheever {M } N are pairwise disjoit elemets i M with M = M. For N, we set L = k=1 M k so that, for B S, we have so that µ (B) = µ (B L ) + µ (B L c ) k=1 µ (B M k ) + µ (B M c ), µ (B) µ (B M c ) + k N µ (B M k ) µ (B M c ) + µ ( k (B M k )) + µ (B M c ) = µ (B M) + µ (B M c ). Sice all the iequalities above eed to be equalities, we immediately coclude that, with B = S, µ (M) = µ (M k ), k i.e., that µ is a coutably-additive measure o M. Sice A M, we have M σ(a) ad µ = µ σ(a) is the required σ-additive extesio of µ. Back to the coi-toss space. I order to apply Theorem 2.9 i our situatio, we eed to check that µ is ideed a coutably-additive measure o the algebra A of all cyliders. The followig problem will help pipoit the hard part of the argumet:

9 Lecture 2: Measures 9 of 17 Problem 2.2. Let A be a algebra o a o-empty set S, ad let µ : A [0, ] be a fiite (µ(s) < ) ad fiitely-additive set fuctio o S with the followig, additioal, property: lim µ(a ) = 0, wheever A. (2.5) The µ satisfies the coditios of Theorem 2.9. The part about fiite additivity is easy (perhaps a bit messy) ad we leave it to the reader: Problem 2.3. Show that the set-fuctio µ C, defied by (2.2) o the product cylliders ad exteded by additivity to the algebra A of cyliders, is fiitely additive. Lemma 2.10 (Coditios of Caratheodory s theorem). The set-fuctio µ C, defied by (2.2), ad exteede by additivity to the the algebra A of cyliders, has the property (2.5). Proof. By Problem 1.10, cyliders are closed sets, ad so {A } N is a sequece of closed sets whose itersectio is empty. The same problem states that { 1, 1} N is compact, so, by the fiite-itersectio property 4, we have A 1... A k =, for some fiite collectio 1,..., k 4 The fiite-itersectio property refers to of idices. Sice {A } N is decreasig, we must have A =, for all k, ad, cosequetly, lim µ(a ) = 0. Propositio 2.11 (Existece of the coi-toss measure). There exists a measure µ C o ({ 1, 1} N, S) with the property that (2.2) holds for all cyliders. Proof. Thaks to Lemma 2.10, Theorem 2.9 ca ow be used. I order to prove uiqueess, we will eed the celebrated π-λ Theorem of Eugee Dyki: the followig fact, familiar from real aalysis: If a family of closed sets of a compact topological space has empty itersectio, the it admits a fiite subfamily with a empty itersectio. Theorem 2.12 (Dyki s π-λ Theorem). Let P be a π-system o a oempty set S, ad let Λ be a λ-system which cotais P. The Λ also cotais the σ-algebra σ(p) geerated by P. Proof. Usig the result of part 4. of Problem 1.1, we oly eed to prove that λ(p) (where λ(p) deotes the λ-system geerated by P) is a π- system. For A S, let G A deote the family of all subsets of S whose itersectios with A are i λ(p): G A = {C S : C A λ(p)}.

10 Lecture 2: Measures 10 of 17 Claim: G A is a λ-system for A λ(p). Sice A λ(p), clearly S G A. For a icreasig family {C } N i G A we have ( C ) A = (C A). Each C A is i Λ, ad the family {C A} N is icreasig, so ( C ) A Λ. Fially, for C 1, C 2 G with C 1 C 2, we have (C 2 \ C 1 ) A = (C 2 A) \ (C 1 A) Λ, because C 1 A C 2 A. P is a π-system, so for ay A P, we have P G A. Therefore, λ(p) G A, because G A is a λ-system. I other words, for A P ad B λ(p), we have A B λ(p). That meas, however, that P G B, for ay B λ(p). Usig the fact that G B is a λ-system we must also have λ(p) G B, for ay B λ(p), i.e., A B λ(p), for all A, B λ(p), which shows that λ(p) is π-system. Propositio 2.13 (Measures which agree o a π-system). Let (S, S) be a measurable space, ad let P be a π-system which geerates S. Suppose that µ 1 ad µ 2 are two measures o S with the property that µ 1 (S) = µ 2 (S) < ad µ 1 (A) = µ 2 (A), for all A P. The µ 1 = µ 2, i.e., µ 1 (A) = µ 2 (A), for all A S. Proof. Let L be the family of all subsets A of S for which µ 1 (A) = µ 2 (A). Clearly P L, but L is, potetially, bigger. I fact, it follows easily from the elemetary properties of measures (see Propositio 2.6) ad the fact that µ 1 (S) = µ 2 (S) < that it ecessarily has the structure of a λ-system 5. By Theorem 2.12 (the π-λ Theorem), L cotais the σ-algebra geerated by P, i.e., S L. O the other had, by defiitio, L S ad so µ 1 = µ 2. Propositio 2.14 (Uiqueess of the coi-toss measure). The measure µ C is the uique measure o ({ 1, 1} N, S) with the property that (2.2) holds for all cyliders. 5 It seems that the structure of a λ-system is defied so that it would exactly describe the structure of the family of all sets o which two measures (with the same total mass) agree. The structure of the π-system correspods to the miimal assumptio that allows Propositio 2.13 to hold. Proof. The existece is the cotet of Propositio To prove uiqueess, it suffices to ote that algebras are π-systems ad use Propositio Problem 2.4. Defie D 1, D 2 { 1, 1} N by

11 Lecture 2: Measures 11 of D 1 = {s { 1, 1} N : lim sup s = 1}, 2. D 2 = {s { 1, 1} N : N N, s N = s N+1 = s N+2 }. Show that D 1, D 2 S ad compute µ(d 1 ), µ(d 2 ). Our ext task is to probe the structure of the σ-algebra S o { 1, 1} N a little bit more ad show that S = 2 { 1,1}N. It is iterestig that such a result (which deals exclusively with the structure of S) requires a use of a measure i its proof. Example 2.15 (A o-measurable subset of { 1, 1} N (*)). Sice σ- algebras are closed uder coutable set operatios, ad sice the product σ-algebra S for the coi-toss space { 1, 1} N is geerated by sets obtaied by restrictig fiite collectios of coordiates, oe is tempted to thik that S cotais all subsets of { 1, 1} N. That is ot the case. We will use the axiom of choice, together with the fact that a measure µ C ca be defied o the whole of { 1, 1} N, to show to costruct a example of a o-measurable set. Let us start by costructig a relatio o { 1, 1} N i the followig 6 way: we set s 1 s 2 if ad oly if there exists N such that s 1 k = s2 k, for k (here, as always, si = (s1 i, si 2,... ), i = 1, 2). It is easy to check that is a equivalece relatio ad that it splits { 1, 1} N ito disjoit equivalece classes. Oe of the may equivalet forms of the axiom of choice states that there exists a subset N of { 1, 1} N which cotais exactly oe elemet from each of the equivalece classes. Let us suppose that N is a elemet i S ad see if we ca reach a cotradictio. For each oempty = { 1,..., k } 2 N f i, where 2N f i deotes the family of all fiite subsets of N, let us defie the mappig T : { 1, 1} N { 1, 1} N i the followig 7 maer: s l, l, T = Id ad (T (s)) l = for N. s l, l, 6 I words, s 1 ad s 2 are related if they oly differ i a fiite umber of coordiates. 7 T flips the sigs of the elemets of its argumet o the positios correspodig to. Sice is fiite, T preserves the -equivalece class of each elemet. Cosequetly (ad usig the fact that N cotais exactly oe elemet from each equivalece class) the sets N ad T (N) = {T (s) : s N} are disjoit. Similarly ad more geerally, the sets T (N) ad T (N) are also disjoit wheever =. O the other had, each s { 1, 1} N is equivalet to some ŝ N, i.e., it ca be obtaied from ŝ by flippig a fiite umber of coordiates. Therefore, the family forms a partitio of { 1, 1} N. N = {T (N) : 2 N f i }

12 Lecture 2: Measures 12 of 17 The mappig T has several other ice properties. First of all, it is immediate that it is ivolutory, i.e., T T = Id. To show that it is (S, S)-measurable, we eed to prove that its compositio with each projectio map π k : S { 1, 1} is measurable. This follows immediately from the fact that for k N (π k T ) 1 C k;1, k, ({1}) = C k; 1, k, where, for b { 1, 1}, we recall that C k;b = {s { 1, 1} N : s k = b} is a product cylider. If we combie the ivolutivity ad measurability of T, we immediately coclude that T (A) S for each A S. I particular, N S. I additio to preservig measurability, the map T also preserves the measure 8 the i µ C, i.e., µ C (T (A)) = µ C (A), for all A S. To prove that, let us pick F ad cosider the set-fuctio µ : S [0, 1] give by µ (A) = µ C (T (A)). It is a simple matter to show that µ is, i fact, a measure o (S, S) with µ (S) = 1. Moreover, thaks to the simple form (2.2) of the actio of the measure µ C o cyliders, it is clear that µ = µ C o the algebra of all cyliders. It suffices to ivoke Propositio 2.13 to coclude that µ = µ C o the etire S, i.e., that T preserves µ C. The above properties of the maps T, F ca imply the followig: N is a partitio of S ito coutably may measurable subsets of equal measure. Such a partitio {N 1, N 2,... } caot exist, however. Ideed if it did, oe of the followig two cases would occur: 1. µ(n 1 ) = 0. I that case µ(s) = µ( k N k ) = µ(n k ) = 0 = 0 = 1 = µ(s). 8 Actually, we say that a map f from a measure space (S, S, µ S ) to the measure space (T, T, µ T ) is measure preservig if it is measurable ad µ S ( f 1 (A)) = µ T (A), for all A T. The ivolutivity of the map T implies that this geeral defiitio agrees with our usage i this example. 2. µ(n 1 ) = α > 0. I that case µ(s) = µ( k N k ) = µ(n k ) = α = = 1 = µ(s). Therefore, the set N caot be measurable 9 i S. 9 Somewhat heavier set-theoretic machiery ca be used to prove that most of the subsets of S are ot i S, i the sese The Lebesgue measure As we shall see, the coi-toss space ca be used as a sort of a uiversal measure space i probability theory. We use it here to costruct the Lebesgue measure o [0, 1]. We start with the otio somewhat dual to the already itroduced otio of the pull-back i Defiitio 1.8. We leave it as a exercise for the reader to show that the set fuctio f µ from Defiitio 2.16 is ideed a measure. that the cardiality of the set S is strictly smaller tha the cardiality of the set 2 S of all subsets of S

13 Lecture 2: Measures 13 of 17 Defiitio 2.16 (Push-forwards). Let (S, S, µ) be a measure space ad let (T, T ) be a measurable space. The measure f µ o (T, T ), defied by f µ(b) = µ( f 1 (B)), for B T, is called the push-forward of the measure µ by f. Let f : { 1, 1} N [0, 1] be the mappig give by f (s) = k=1 ( ) 1+sk 2 2 k, s { 1, 1} N. The idea is to use f to establish a correspodece betwee all real umbers i [0, 1] ad their expasios i the biary system, with the codig 1 0 ad 1 1. It is iterestig to ote that f is ot oe-to-oe 10 as it, for example, maps s 1 = (1, 1, 1,... ) ad s 2 = ( 1, 1, 1,... ) ito the same value - amely 2 1. Let us show, first, that the map f is cotiuous i the metric d defied by part (1.2) of Problem 1.9. Ideed, we pick s 1 ad s 2 i { 1, 1} N ad remember that for d(s 1, s 2 ) 2, the first 1 coordiates of s 1 ad s 2 coicide. Therefore, f (s 1 ) f (s 2 ) 2 k = 2 +1 = 2d(s 1, s 2 ). k= Hece, the map f is Lipschitz ad, therefore, cotiuous. The cotiuity of f (together with the fact that S is the Borel σ- algebra for the topology iduced by the metric d) implies that f : ({ 1, 1} N, S) ([0, 1], B([0, 1])) is a measurable mappig. Therefore, the push-forward λ = f (µ) is well defied o ([0, 1], B([0, 1])), ad we call it the Lebesgue measure o [0, 1]. 10 The reaso for this is, poetically speakig, that [0, 1] is ot the Cator set. Propositio 2.17 (Ituitive properties of the Lebesgue measure). The Lebesgue measure λ o ([0, 1], B([0, 1])) satisfies λ([a, b)) = b a, λ({a}) = 0 for 0 a < b 1. (2.6) Proof. 1. Cosider a, b of the form b = 2 k ad b = k+1 2, for N ad k < 2. For such a, b we have f 1 ([a, b)) = C 1,...,;c1,c 2,...,c, where c 1 c 2... c is the base-2 expasio of k (after the recodig 1 0, 1 1). By the very defiitio of λ ad the form (2.2) of the actio of the coi-toss measure µ C o cyliders, we have ( ) ( λ [a, b) = µ C f 1( [a, b) )) = µ C (C 1,...,;c1,c 2,...,c ) = 2 = k k. Therefore, (2.6) holds for a, b of the form b = 2 k ad b = 2 l, for N, k < 2 ad l = k + 1. Usig (fiite) additivity of λ, we

14 Lecture 2: Measures 14 of 17 immediately coclude that (2.6) holds for all k, l, i.e., that it holds for all dyadic ratioals. A geeral a (0, 1] ca be approximated by a icreasig sequece {q } N of dyadic ratioals from the left, ad the cotiuity of measures with respect to decreasig sequeces implies that ( ) ( ) ( ) λ [a, p) = λ [q, p) = lim λ [q, p) = lim(p q ) = (p a), wheever a (0, 1] ad p is a dyadic ratioal. I order to remove the dyadicity requiremet from the right limit, we approximate it from the left by a sequece {p } N of dyadic ratioals with p > a, ad use the cotiuity with respect to icreasig sequeces to get, for a < b (0, 1), ( ) ( ) ( ) λ [a, b) = λ [a, p ) = lim λ [a, p ) = lim(p a) = (b a). The Lebesgue measure has aother importat property: Problem 2.5. Show that the Lebesgue measure is traslatio ivariat. More precisely, for B B([0, 1]) ad x [0, 1), we have Hit: Use Propositio 2.13 for the secod part. 1. B + 1 x = {b + x (mod 1) : b B} is i B([0, 1]) ad 2. λ(b + 1 x) = λ(b), where, for a [0, 2), we defie a, a 1, a (mod 1) = a 1, a > 1. Geometrically, the set x + 1 B is obtaied from B by traslatig it to the right by x ad the shiftig the part that is stickig out by 1 to the left. Fially, the otio of the Lebesgue measure is just as useful o the etire R, as o its compact subset [0, 1]. For a geeral B B(R), we ca defie the Lebesgue measure of B by measurig its itersectios with all itervals of the form [, + 1), ad addig them together, i.e., λ(b) = ( (B ) ) λ [, + 1). = Note how we are overloadig the otatio ad usig the letter λ for both the Lebesgue measure o [0, 1] ad the Lebesgue measure o R. It is a quite tedious, but does ot require ay ew tools, to show that may of the properties of λ o [0, 1] trasfer to λ o R: Problem 2.6. Let λ be the Lebesgue measure o (R, B(R)). Show that 1. λ([a, b)) = b a, λ({a}) = 0 for a < b,

15 Lecture 2: Measures 15 of λ is σ-fiite but ot fiite, 3. λ(b + x) = λ(b), for all B B(R) ad x R, where B + x = {b + x : b B}. Additioal Problems Problem 2.7 (Local separatio by costats). Let (S, S, µ) be a measure space ad let f, g L 0 (S, S, µ) satisfy µ ( {x S : f (x) < g(x)} ) > 0. Prove or costruct a couterexample for the followig statemet: There exist costats a, b R such that µ ( {x S : f (x) a < b g(x)} ) > 0. Problem 2.8 (A pseudometric o sets). Let (S, S, µ) be a fiite measure space. For A, B S defie d(a, B) = µ(a B), where deotes the symmetric differece: A B = (A \ B) (B \ A). Show that d is a pseudometric o S, ad for A S describe the set of all B S with d(a, B) = 0. Problem 2.9 (Complete measure spaces). A measure space (S, S, µ) is called complete if all subsets of ull sets are themselves i S. For a (possibly icomplete) measure space (S, S, µ) we defie the completio (S, S, µ ) i the followig way: S = {A N : A S ad N N for some N S with µ(n) = 0}. For B S with represetatio B = A N we set µ (B) = µ(a). 1. Show that S is a σ-algebra. Note: Let X be a oempty set. A fuctio d : X X [0, ) is called a pseudometric if 1. d(x, y) + d(y, x) d(x, z), for all x, y, z X, 2. d(x, y) = d(y, x), for all x, y X, ad 3. d(x, x) = 0, for all x X. Note how the oly differece betwee a metric ad a pseudometric is that for a metric d(x, y) = 0 implies x = y, while o such requiremet is imposed o a pseudometric. Note: Ufortuately, the same otatio µ is ofte used for the completio of the measure µ ad the outer measure associated with µ as i the proof of Theorem 2.9. Fortuately, it ca be show that these two object coicide o the domai of the completio. 2. Show that the defiitio µ (B) = µ(a) above does ot deped o the choice of the decompositio B = A N, i.e., that µ(â) = µ(a) if B =  ˆN is aother decompositio of B ito a set  i S ad a subset ˆN of a ull set i S. 3. Show that µ is a measure o (S, S ) ad that (S, S, µ ) is a complete measure space with the property that µ (A) = µ(a), for A S. Problem 2.10 (The Cator set). The Cator set is defied as the collectio of all real umbers x i [0, 1] with the represetatio x = c 3, where c {0, 2}. =1 Show that it is Borel-measurable ad compute its Lebesgue measure.

16 Lecture 2: Measures 16 of 17 Problem 2.11 (The uiform measure o a circle). Let S 1 be the uit circle, ad let f : [0, 1) S 1 be the widig map ( ) f (x) = cos(2πx), si(2πx), x [0, 1). 1. Show that the map f is (B([0, 1)), S 1 )-measurable, where S 1 deotes the Borel σ-algebra o S 1 (with the topology iherited from R 2 ). 2. For α (0, 2π), let R α deote the (couter-clockwise) rotatio of R 2 with ceter (0, 0) ad agle α. Show that R α (A) = {R α (x) : x A} is i S 1 if ad oly if A S Let µ 1 be the push-forward of the Lebesgue measure λ by the map Note: The measure µ 1 is called the uiform measure (or the uiform distribu- α (A) ). f. Show that µ 1 is rotatio-ivariat, i.e., that µ 1 (A) = µ 1( R tio) o S 1. Problem 2.12 (Asymptotic desities). We say that the subset A of N admits asymptotic desity if the limit d(a) = lim #(A {1, 2,..., }), exists (remember that # deotes the umber of elemets of a set). Let D be the collectio of all subsets of N which admit asymptotic desity. 1. Is D a algebra? A σ-algebra? 2. Is the map A d(a) fiitely-additive o D? A measure? Problem 2.13 (A subset of the coi-toss space). A elemet i { 1, 1} N (i.e., a sequece s = (s 1, s 2,... ) where s { 1, 1} for all N) is said to be evetually periodic if there exists N 0, K N such that s = s +K for all N 0. Let P { 1, 1} N be the collectio of all evetually-period sequeces. Show that P is measurable i the product σ-algebra S ad compute µ C (P). Problem 2.14 (Regular measures). The measure space (S, S, µ), where (S, d) is a metric space ad S is a σ-algebra o S which cotais the Borel σ-algebra B(d) o S is called regular if for each A S ad each ε > 0 there exist a closed set C ad a ope set O such that C A O ad µ(o \ C) < ε. 1. Suppose that (S, S, µ) is a regular measure space, ad that the measure space (S, B(d), µ B(d) ) is obtaied from (S, S, µ) by restrictig the measure µ oto the σ-algebra of Borel sets. Show that S B(d), where ( S, B(d), (µ B(d) ) ) is the completio (i the sese of Problem 2.9) of (S, B(d), µ B(d) ) 2. Suppose that (S, d) is a metric space ad that µ is a fiite measure o B(d). Show that (S, B(d), µ) is a regular measure space.

17 Lecture 2: Measures 17 of 17 Hit: Cosider a collectio A of subsets A of S such that for each ε > 0 there exists a closed set C ad a ope set O with C A O ad µ(o \ C) < ε. Argue that A is a σ-algebra. The show that each closed set ca be writte as a itersectio of ope sets; use (but prove, first) the fact that the map x d(x, C) = if{d(x, y) : y C}, is cotiuous o S for ay oempty C S. 3. Show that (S, B(d), µ) is regular if µ is ot ecessarily fiite, but has the property that µ(a) < wheever A B(d) is bouded, i.e., whe sup{d(x, y) : x, y A} <. Hit: Pick a poit x 0 S ad, for N, defie the family {R } N of subsets of S as follows: R 1 = {x S : d(x, x 0 ) < 2}, ad R = {x S : 1 < d(x, x 0 ) < + 1}, for > 1, as well as a sequece {µ } N of set fuctios o B(d), give by µ (A) = µ(a R ), for A B(d). Uder the right circumstaces, eve coutable uios of closed sets are closed. 4. Coclude that the Lebesgue measure o ( R, B(R ) ) is regular.