Part II Probability and Measure

Transcription

1 Part II Probability ad Measure Based o lectures by J. Miller Notes take by Dexter Chua Michaelmas 2016 These otes are ot edorsed by the lecturers, ad I have modified them (ofte sigificatly) after lectures. They are owhere ear accurate represetatios of what was actually lectured, ad i particular, all errors are almost surely mie. Aalysis II is essetial Measure spaces, σ-algebras, π-systems ad uiqueess of extesio, statemet *ad proof* of Carathéodory s extesio theorem. Costructio of Lebesgue measure o R. The Borel σ-algebra of R. Existece of o-measurable subsets of R. Lebesgue-Stieltjes measures ad probability distributio fuctios. Idepedece of evets, idepedece of σ-algebras. The Borel Catelli lemmas. Kolmogorov s zero-oe law. [6] Measurable fuctios, radom variables, idepedece of radom variables. Costructio of the itegral, expectatio. Covergece i measure ad covergece almost everywhere. Fatou s lemma, mootoe ad domiated covergece, differetiatio uder the itegral sig. Discussio of product measure ad statemet of Fubii s theorem. [6] Chebyshev s iequality, tail estimates. Jese s iequality. Completeess of L p for 1 p. The Hölder ad Mikowski iequalities, uiform itegrability. [4] L 2 as a Hilbert space. Orthogoal projectio, relatio with elemetary coditioal probability. Variace ad covariace. Gaussia radom variables, the multivariate ormal distributio. [2] The strog law of large umbers, proof for idepedet radom variables with bouded fourth momets. Measure preservig trasformatios, Beroulli shifts. Statemets *ad proofs* of maximal ergodic theorem ad Birkhoff s almost everywhere ergodic theorem, proof of the strog law. [4] The Fourier trasform of a fiite measure, characteristic fuctios, uiqueess ad iversio. Weak covergece, statemet of Lévy s covergece theorem for characteristic fuctios. The cetral limit theorem. [2] 1

2 Cotets II Probability ad Measure Cotets 0 Itroductio 3 1 Measures Measures Probability measures Measurable fuctios ad radom variables Measurable fuctios Costructig ew measures Radom variables Covergece of measurable fuctios Tail evets Itegratio Defiitio ad basic properties Itegrals ad limits New measures from old Itegratio ad differetiatio Product measures ad Fubii s theorem Iequalities ad L p spaces Four iequalities L p spaces Orthogoal projectio i L Covergece i L 1 (P) ad uiform itegrability Fourier trasform The Fourier trasform Covolutios Fourier iversio formula Fourier trasform i L Properties of characteristic fuctios Gaussia radom variables Ergodic theory Ergodic theorems Big theorems The strog law of large umbers Cetral limit theorem Idex 94 2

3 0 Itroductio II Probability ad Measure 0 Itroductio I measure theory, the mai idea is that we wat to assig sizes to differet sets. For example, we might thik [0, 2] R has size 2, while perhaps Q R has size 0. This is kow as a measure. Oe of the mai applicatios of a measure is that we ca use it to come up with a ew defiitio of a itegral. The idea is very simple, but it is goig to be very powerful mathematically. Recall that if f : [0, 1] R is cotiuous, the the Riema itegral of f is defied as follows: (i) Take a partitio 0 = t 0 < t 1 < < t = 1 of [0, 1]. (ii) Cosider the Riema sum f(t j )(t j t j 1 ) j=1 (iii) The Riema itegral is f = Limit of Riema sums as the mesh size of the partitio 0. y 0 t 1 t 2 t 3 t k t k+1 1 x The idea of measure theory is to use a differet approximatio scheme. Istead of partitioig the domai, we partitio the rage of the fuctio. We fix some umbers r 0 < r 1 < r 2 < < r. We the approximate the itegral of f by r j ( size of f 1 ([r j 1, r j ]) ). j=1 We the defie the itegral as the limit of approximatios of this type as the mesh size of the partitio 0. 3

4 0 Itroductio II Probability ad Measure y x We ca make a aalogy with bakers If a Riema baker is give a stack of moey, they would just add the values of the moey i order. A measuretheoretic baker will sort the bak otes accordig to the type, ad the fid the total value by multiplyig the umber of each type by the value, ad addig up. Why would we wat to do so? It turs out this leads to a much more geeral theory of itegratio o much more geeral spaces. Istead of itegratig fuctios [a, b] R oly, we ca replace the domai with ay measure space. Eve i the cotext of R, this theory of itegratio is much much more powerful tha the Riema sum, ad ca itegrate a much wider class of fuctios. While you probably do t care about those pathological fuctios ayway, beig able to itegrate more thigs meas that we ca state more geeral theorems about itegratio without havig to put i fuy coditios. That was all about measures. What about probability? It turs out the cocepts we develop for measures correspod exactly to may familiar otios from probability if we restrict it to the particular case where the total measure of the space is 1. Thus, whe studyig measure theory, we are also secretly studyig probability! 4

5 1 Measures II Probability ad Measure 1 Measures I the course, we will write f f for f coverges to f mootoically icreasigly, ad f f similarly. Uless otherwise specified, covergece is take to be poitwise. 1.1 Measures The startig poit of all these is to come up with a fuctio that determies the size of a give set, kow as a measure. It turs out we caot sesibly defie a size for all subsets of [0, 1]. Thus, we eed to restrict our attetio to a collectio of ice subsets. Specifyig which subsets are ice would ivolve specifyig a σ-algebra. This sectio is mostly techical. Defiitio (σ-algebra). Let E be a set. A σ-algebra E o E is a collectio of subsets of E such that (i) E. (ii) A E implies that A C = X \ A E. (iii) For ay sequece (A ) i E, we have that A E. The pair (E, E) is called a measurable space. Note that the axioms imply that σ-algebras are also closed uder coutable itersectios, as we have A B = (A C B C ) C. Defiitio (Measure). A measure o a measurable space (E, E) is a fuctio µ : E [0, ] such that (i) µ( ) = 0 (ii) Coutable additivity: For ay disjoit sequece (A ) i E, the ( ) µ A = µ(a ). Example. Let E be ay coutable set, ad E = P (E) be the set of all subsets of E. A mass fuctio is ay fuctio m : E [0, ]. We ca the defie a measure by settig µ(a) = x A m(x). =1 I particular, if we put m(x) = 1 for all x E, the we obtai the coutig measure. 5

6 1 Measures II Probability ad Measure Coutable spaces are ice, because we ca always take E = P (E), ad the measure ca be defied o all possible subsets. However, for bigger spaces, we have to be more careful. The set of all subsets is ofte too large. We will see a cocrete ad also importat example of this later. I geeral, σ-algebras are ofte described o large spaces i terms of a smaller set, kow as the geeratig sets. Defiitio (Geerator of σ-algebra). Let E be a set, ad that A P (E) be a collectio of subsets of E. We defie σ(a) = {A E : A E for all σ-algebras E that cotai A}. I other words σ(a) is the smallest sigma algebra that cotais A. This is kow as the sigma algebra geerated by A. Example. Take E = Z, ad A = {{x} : x Z}. The σ(a) is just P (E), sice every subset of E ca be writte as a coutable uio of sigletos. Example. Take E = Z, ad let A = {{x, x + 1, x + 2, x + 3, } : x E}. The agai σ(e) is the set of all subsets of E. The followig is the most importat σ-algebra i the course: Defiitio (Borel σ-algebra). Let E = R, ad A = {U R : U is ope}. The σ(a) is kow as the Borel σ-algebra, which is ot the set of all subsets of R. We ca equivaletly defie this by à = {(a, b) : a < b, a, b Q}. The σ(ã) is also the Borel σ-algebra. Ofte, we would like to prove results that allow us to deduce properties about the σ-algebra just by checkig it o a geeratig set. However, usually, we caot just check it o a arbitrary geeratig set. Istead, the geeratig set has to satisfy some ice closure properties. We are ow goig to itroduce a buch of may differet defiitios that you eed ot aim to remember (except whe exams are ear). Defiitio (π-system). Let A be a collectio of subsets of E. The A is called a π-system if (i) A (ii) If A, B A, the A B A. Defiitio (d-system). Let A be a collectio of subsets of E. The A is called a d-system if (i) E A (ii) If A, B A ad A B, the B \ A A (iii) For all icreasig sequeces (A ) i A, we have that A A. The poit of d-systems ad π-systems is that they separate the axioms of a σ-algebra ito two parts. More precisely, we have Propositio. A collectio A is a σ-algebra if ad oly if it is both a π-system ad a d-system. 6

7 1 Measures II Probability ad Measure This follows rather straightforwardly from the defiitios. The followig defiitios are also useful: Defiitio (Rig). A collectio of subsets A is a rig o E if A ad for all A, B A, we have B \ A A ad A B A. Defiitio (Algebra). A collectio of subsets A is a algebra o E if A, ad for all A, B A, we have A C A ad A B A. So a algebra is like a σ-algebra, but it is just closed uder fiite uios oly, rather tha coutable uios. While the ames π-system ad d-system are rather arbitrary, we ca make some sese of the ames rig ad algebra. Ideed, a rig forms a rig (without uity) i the algebraic sese with symmetric differece as additio ad itersectio as multiplicatio. The the empty set acts as the additive idetity, ad E, if preset, acts as the multiplicative idetity. Similarly, a algebra is a boolea subalgebra uder the boolea algebra P (E). A very importat lemma about these thigs is Dyki s lemma: Lemma (Dyki s π-system lemma). Let A be a π-system. The ay d-system which cotais A cotais σ(a). This will be very useful i the future. If we wat to show that all elemets of σ(a) satisfy a particular property for some geeratig π-system A, we just have to show that the elemets of A satisfy that property, ad that the collectio of thigs that satisfy the property form a d-system. While this use case might seem rather cotrived, it is surprisigly commo whe we have to prove thigs. Proof. Let D be the itersectio of all d-systems cotaiig A, i.e. the smallest d-system cotaiig A. We show that D cotais σ(a). To do so, we will show that D is a π-system, hece a σ-algebra. There are two steps to the proof, both of which are straightforward verificatios: (i) We first show that if B D ad A A, the B A D. (ii) We the show that if A, B D, the A B D. The the result immediately follows from the secod part. We let D = {B D : B A D for all A A}. We ote that D A because A is a π-system, ad is hece closed uder itersectios. We check that D is a d-system. It is clear that E D. If we have B 1, B 2 D, where B 1 B 2, the for ay A A, we have (B 2 \ B 1 ) A = (B 2 A) \ (B 1 A). By defiitio of D, we kow B 2 A ad B 1 A are elemets of D. Sice D is a d-system, we kow this itersectio is i D. So B 2 \ B 1 D. Fially, suppose that (B ) is a icreasig sequece i D, with B = B. The for every A A, we have that ( B ) A = (B A) = B A D. 7

8 1 Measures II Probability ad Measure Therefore B D. Therefore D is a d-system cotaied i D, which also cotais A. By our choice of D, we kow D = D. We ow let D = {B D : B A D for all A D}. Sice D = D, we agai have A D, ad the same argumet as above implies that D is a d-system which is betwee A ad D. But the oly way that ca happe is if D = D, ad this implies that D is a π-system. After defiig all sorts of thigs that are weaker versios of σ-algebras, we ow defied a buch of measure-like objects that satisfy fewer properties. Agai, o oe really remembers these defiitios: Defiitio (Set fuctio). Let A be a collectio of subsets of E with A. A set fuctio fuctio µ : A [0, ] such that µ( ) = 0. Defiitio (Icreasig set fuctio). A set fuctio is icreasig if it has the property that for all A, B A with A B, we have µ(a) µ(b). Defiitio (Additive set fuctio). A set fuctio is additive if wheever A, B A ad A B A, A B =, the µ(a B) = µ(a) + µ(b). Defiitio (Coutably additive set fuctio). A set fuctio is coutably additive if wheever A is a sequece of disjoit sets i A with A A, the ( ) µ A = µ(a ). Uder these defiitios, a measure is just a coutable additive set fuctio defied o a σ-algebra. Defiitio (Coutably subadditive set fuctio). A set fuctio is coutably subadditive if wheever (A ) is a sequece of sets i A with A A, the ( ) µ A µ(a ). The big theorem that allows us to costruct measures is the Caratheodory extesio theorem. I particular, this will help us costruct the Lebesgue measure o R. Theorem (Caratheodory extesio theorem). Let A be a rig o E, ad µ a coutably additive set fuctio o A. The µ exteds to a measure o the σ-algebra geerated by A. Proof. (o-examiable) We start by defiig what we wat our measure to be. For B E, we set { µ (B) = if µ(a ) : (A ) A ad B } A. 8

9 1 Measures II Probability ad Measure If it happes that there is o such sequece, we set this to be. This measure is kow as the outer measure. It is clear that µ (φ) = 0, ad that µ is icreasig. We say a set A E is µ -measurable if for all B E. We let We will show the followig: (i) M is a σ-algebra cotaiig A. µ (B) = µ (B A) + µ (B A C ) M = {µ -measurable sets}. (ii) µ is a measure o M with µ A = µ. Note that it is ot true i geeral that M = σ(a). However, we will always have M σ(a). We are goig to break this up ito five ice bite-size chuks. Claim. µ is coutably subadditive. Suppose B B. We eed to show that µ (B) µ (B ). We ca wlog assume that µ (B ) is fiite for all, or else the iequality is trivial. Let ε > 0. The by defiitio of the outer measure, for each, we ca fid a sequece (B,m ) m=1 i A with the property that B m B,m ad µ (B ) + ε 2 m µ(b,m ). The we have B B,m B,m. Thus, by defiitio, we have µ (B) µ (B,m ),m ( µ (B ) + ε 2 ) = ε + µ (B ). Sice ε was arbitrary, we are doe. Claim. µ agrees with µ o A. I the first example sheet, we will show that if A is a rig ad µ is a coutably additive set fuctio o µ, the µ is i fact coutably subadditive ad icreasig. Assumig this, suppose that A, (A ) are i A ad A A. The by subadditivity, we have µ(a) µ(a A ) µ(a ), 9

10 1 Measures II Probability ad Measure usig that µ is coutably subadditivity ad icreasig. Note that we have to do this i two steps, rather tha just applyig coutable subadditivity, sice we did ot assume that A A. Takig the ifimum over all sequeces, we have µ(a) µ (A). Also, we see by defiitio that µ(a) µ (A), sice A covers A. So we get that µ(a) = µ (A) for all A A. Claim. M cotais A. Suppose that A A ad B E. We eed to show that µ (B) = µ (B A) + µ (B A C ). Sice µ is coutably subadditive, we immediately have µ (B) µ (B A) + µ (B A C ). For the other iequality, we first observe that it is trivial if µ (B) is ifiite. If it is fiite, the by defiitio, give ε > 0, we ca fid some (B ) i A such that B B ad µ (B) + ε µ(b ). The we have B A (B A) B A C (B A C ) We otice that B A C = B \ A A. Thus, by defiitio of µ, we have µ (B A) + µ (B A c ) µ(b A) + µ(b A C ) = = (µ(b A) + µ(b A C )) µ(b ) µ (B ) + ε. Sice ε was arbitrary, the result follows. Claim. We show that M is a algebra. We first show that E M. This is true sice we obviously have µ (B) = µ (B E) + µ (B E C ) for all B E. Next, ote that if A M, the by defiitio we have, for all B, µ (B) = µ (B A) + µ (B A C ). Now ote that this defiitio is symmetric i A ad A C. A C M. So we also have 10

11 1 Measures II Probability ad Measure Fially, we have to show that M is closed uder itersectio (which is equivalet to beig closed uder uio whe we have complemets). Suppose A 1, A 2 M ad B E. The we have µ (B) = µ (B A 1 ) + µ (B A C 1 ) = µ (B A 1 A 2 ) + µ (B A 1 A C 2 ) + µ (B A C 1 ) = µ (B (A 1 A 2 )) + µ (B (A 1 A 2 ) C A 1 ) + µ (B (A 1 A 2 ) C A C 1 ) = µ (B (A 1 A 2 )) + µ (B (A 1 A 2 ) C ). So we have A 1 A 2 M. So M is a algebra. Claim. M is a σ-algebra, ad µ is a measure o M. To show that M is a σ-algebra, we eed to show that it is closed uder coutable uios. We let (A ) be a disjoit collectio of sets i M, the we wat to show that A = A M ad µ (A) = µ (A ). Suppose that B E. The we have µ (B) = µ (B A 1 ) + µ (B A C 1 ) Usig the fact that A 2 M ad A 1 A 2 =, we have = µ (B A 1 ) + µ (B A 2 ) + µ (B A C 1 A C 2 ) = = µ (B A i ) + µ (B A C 1 A C ) i=1 µ (B A i ) + µ (B A C ). i=1 Takig the limit as, we have µ (B) µ (B A i ) + µ (B A C ). i=1 By the coutable-subadditivity of µ, we have Thus we obtai µ (B A) µ (B A i ). i=1 µ (B) µ (B A) + µ (B A C ). By coutable subadditivity, we also have iequality i the other directio. So equality holds. So A M. So M is a σ-algebra. To see that µ is a measure o M, ote that the above implies that µ (B) = (B A i ) + µ (B A C ). i=1 11

12 1 Measures II Probability ad Measure Takig B = A, this gives µ (A) = (A A i ) + µ (A A C ) = µ (A i ). i=1 Note that whe A itself is actually a σ-algebra, the outer measure ca be simply writte as i=1 µ (B) = if{µ(a) : A A, B A}. Caratheodory gives us the existece of some measure extedig the set fuctio o A. Could there be may? I geeral, there could. However, i the special case where the measure is fiite, we do get uiqueess. Theorem. Suppose that µ 1, µ 2 are measures o (E, E) with µ 1 (E) = µ 2 (E) <. If A is a π-system with σ(a) = E, ad µ 1 agrees with µ 2 o A, the µ 1 = µ 2. Proof. Let D = {A E : µ 1 (A) = µ 2 (A)} We kow that D A. By Dyki s lemma, it suffices to show that D is a d-system. The thigs to check are: (i) E D this follows by assumptio. (ii) If A, B D with A B, the B \ A D. Ideed, we have the equatios µ 1 (B) = µ 1 (A) + µ 1 (B \ A) < µ 2 (B) = µ 2 (A) + µ 2 (B \ A) <. Sice µ 1 (B) = µ 2 (B) ad µ 1 (A) = µ 2 (A), we must have µ 1 (B \ A) = µ 2 (B \ A). (iii) Let (A ) D be a icreasig sequece with A = A. The So A D. µ 1 (A) = lim µ 1(A ) = lim µ 2(A ) = µ 2 (A). The assumptio that µ 1 (E) = µ 2 (E) < is ecessary. The theorem does ot ecessarily hold without it. We ca see this from a simple couterexample: Example. Let E = Z, ad let E = P (E). We let A = {{x, x + 1, x + 2, } : x E} { }. This is a π-system with σ(a) = E. We let µ 1 (A) be the umber of elemets i A, ad µ 2 = 2µ 1 (A). The obviously µ 1 µ 2, but µ 1 (A) = = µ 2 (A) for A A. Defiitio (Borel σ-algebra). Let E be a topological space. We defie the Borel σ-algebra as B(E) = σ({u E : U is ope}). We write B for B(R). 12

13 1 Measures II Probability ad Measure Defiitio (Borel measure ad Rado measure). A measure µ o (E, B(E)) is called a Borel measure. If µ(k) < for all K E compact, the µ is a Rado measure. The most importat example of a Borel measure we will cosider is the Lebesgue measure. Theorem. There exists a uique Borel measure µ o R with µ([a, b]) = b a. Proof. We first show uiqueess. Suppose µ is aother measure o B satisfyig the above property. We wat to apply the previous uiqueess theorem, but our measure is ot fiite. So we eed to carefully get aroud that problem. For each Z, we set µ (A) = µ(a (, + 1])) µ (A) = µ(a (, + 1])) The µ ad µ are fiite measures o B which agree o the π-system of itervals of the form (a, b] with a, b R, a < b. Therefore we have µ = µ for all Z. Now we have µ(a) = Z µ(a (, + 1]) = Z µ (A) = Z µ (A) = µ(a) for all Borel sets A. To show existece, we wat to use the Caratheodory extesio theorem. We let A be the collectio of fiite, disjoit uios of the form A = (a 1, b 1 ] (a 2, b 2 ] (a, b ]. The A is a rig of subsets of R, ad σ(a) = B (details are to be checked o the first example sheet). We set µ(a) = (b i a i ). We ote that µ is well-defied, sice if the i=1 A = (a 1, b 1 ] (a, b ] = (ã 1, b 1 ] (ã, b ], (b i a i ) = i=1 ( b i ã i ). Also, if µ is additive, A, B A, A B = ad A B A, we obviously have µ(a B) = µ(a) + µ(b). So µ is additive. Fially, we have to show that µ is i fact coutably additive. Let (A ) be a disjoit sequece i A, ad let A = i=1 A A. The we eed to show that µ(a) = =1 µ(a ). Sice µ is additive, we have i=1 µ(a) = µ(a 1 ) + µ(a \ A 1 ) = µ(a 1 ) + µ(a 2 ) + µ(a \ A 1 A 2 ) ( ) = µ(a i ) + µ A \ A i i=1 i=1 13

14 1 Measures II Probability ad Measure To fiish the proof, we show that ( ) µ A \ A i 0 as. i=1 We are goig to reduce this to the fiite itersectio property of compact sets i R: if (K ) is a sequece of compact sets i R with the property that m=1 K m for all, the m=1 K m. We first itroduce some ew otatio. We let B = A \ m=1 A m. We ow suppose, for cotradictio, that µ(b ) 0 as. Sice the B s are decreasig, there must exist ε > 0 such that µ(b ) 2ε for every. For each, we take C A with the property that C B ad µ(b \C ) ε 2. This is possible sice each B is just a fiite uio of itervals. Thus we have ( ) ( ) µ(b ) µ C m = µ B \ C m m=1 m=1 ( ) µ (B m \ C m ) ε. m=1 µ(b m \ C m ) m=1 m=1 ε 2 m O the other had, we also kow that µ(b ) 2ε. ( ) µ C m ε m=1 for all. We ow let that K = m=1 C m. The µ(k ) ε, ad i particular K for all. Thus, the fiite itersectio property says K B =. =1 =1 This is a cotradictio. So we have µ(b ) 0 as. So doe. Defiitio (Lebesgue measure). The Lebesgue measure is the uique Borel measure µ o R with µ([a, b]) = b a. Note that the Lebesgue measure is ot a fiite measure, sice µ(r) =. However, it is a σ-fiite measure. 14

15 1 Measures II Probability ad Measure Defiitio (σ-fiite measure). Let (E, E) be a measurable space, ad µ a measure. We say µ is σ-fiite if there exists a sequece (E ) i E such that E = E ad µ(e ) < for all. This is the ext best thig we ca hope after fiiteess, ad ofte proofs that ivolve fiiteess carry over to σ-fiite measures. Propositio. The Lebesgue measure is traslatio ivariat, i.e. for all A B ad x R, where µ(a + x) = µ(a) A + x = {y + x, y A}. Proof. We use the uiqueess of the Lebesgue measure. We let µ x (A) = µ(a + x) for A B. The this is a measure o B satisfyig µ x ([a, b]) = b a. So the uiqueess of the Lebesgue measure shows that µ x = µ. It turs out that traslatio ivariace actually characterizes the Lebesgue measure. Propositio. Let µ be a Borel measure o R that is traslatio ivariat ad µ([0, 1]) = 1. The µ is the Lebesgue measure. Proof. We show that ay such measure must satisfy µ([a, b]) = b a. By additivity ad traslatio ivariace, we ca show that µ([p, q]) = q p for all ratioal p < q. By cosiderig µ([p, p + 1/]) for all ad usig the icreasig property, we kow µ({p}) = 0. So µ(([p, q)) = µ((p, q]) = µ((p, q)) = q p for all ratioal p, q. Fially, by coutable additivity, we ca exted this to all real itervals. The the result follows from the uiqueess of the Lebesgue measure. I the proof of the Caratheodory extesio theorem, we costructed a measure µ o the σ-algebra M of µ -measurable sets which cotais A. This cotais B = σ(a), but could i fact be bigger tha it. We call M the Lebesgue σ-algebra. Ideed, it ca be give by M = {A N : A B, N B B with µ(b) = 0}. If A N M, the µ(a N) = µ(a). The proof is left for the example sheet. It is also true that M is strictly larger tha B, so there exists A M with A B. Costructio of such a set was o last year s exam (2016). O the other had, it is also true that ot all sets are Lebesgue measurable. This is a rather fuy costructio. 15

16 1 Measures II Probability ad Measure Example. For x, y [0, 1), we say x y if x y is ratioal. This defies a equivalece relatio o [0, 1). By the axiom of choice, we pick a represetative of each equivalece class, ad put them ito a set S [0, 1). We will show that S is ot Lebesgue measurable. Suppose that S were Lebesgue measurable. We are goig to get a cotradictio to the coutable additivity of the Lebesgue measure. For each ratioal r [0, 1) Q, we defie S r = {s + r mod 1 : s S}. By traslatio ivariace, we kow S r is also Lebesgue measurable, ad µ(s r ) = µ(s). Also, by costructio of S, we kow (S r ) r Q is disjoit, ad r Q S r = [0, 1). Now by coutable additivity, we have 1 = µ([0, 1)) = µ S r = µ(s r ) = µ(s), r Q r Q r Q which is clearly ot possible. Ideed, if µ(s) = 0, the this says 1 = 0; If µ(s) > 0, the this says 1 =. Both are absurd. 1.2 Probability measures Sice the course is called probability ad measure, we d better start talkig about probability! It turs out the otios we care about i probability theory are very aturally just special cases of the cocepts we have previously cosidered. Defiitio (Probability measure ad probability space). Let (E, E) be a measure space with the property that µ(e) = 1. The we ofte call µ a probability measure, ad (E, E, µ) a probability space. Probability spaces are usually writte as (Ω, F, P) istead. Defiitio (Sample space). I a probability space (Ω, F, P), we ofte call Ω the sample space. Defiitio (Evets). I a probability space (Ω, F, P), we ofte call the elemets of F the evets. Defiitio (Probaiblity). I a probability space (Ω, F, P), if A F, we ofte call P[A] the probability of the evet A. These are exactly the same thigs as measures, but with differet ames! However, thikig of them as probabilities could make us ask differet questios about these measure spaces. For example, i probability, oe is ofte iterested i idepedece. Defiitio (Idepedece of evets). A sequece of evets (A ) is said to be idepedet if [ ] P A = P[A ] J for all fiite subsets J N. J 16

17 1 Measures II Probability ad Measure However, it turs out that talkig about idepedece of evets is usually too restrictive. Istead, we wat to talk about the idepedece of σ-algebras: Defiitio (Idepedece of σ-algebras). A sequece of σ-algebras (A ) with A F for all is said to be idepedet if the followig is true: If (A ) is a sequece where A A for all, them (A ) is idepedet. Propositio. Evets (A ) are idepedet iff the σ-algebras σ(a ) are idepedet. While provig this directly would be rather tedious (but ot too hard), it is a immediate cosequece of the followig theorem: Theorem. Suppose A 1 ad A 2 are π-systems i F. If P[A 1 A 2 ] = P[A 1 ]P[A 2 ] for all A 1 A 1 ad A 2 A 2, the σ(a 1 ) ad σ(a 2 ) are idepedet. Proof. This will follow from two applicatios of the fact that a fiite measure is determied by its values o a π-system which geerates the etire σ-algebra. We first fix A 1 A 1. We defie the measures ad µ(a) = P[A A 1 ] ν(a) = P[A]P[A 1 ] for all A F. By assumptio, we kow µ ad ν agree o A 2, ad we have that µ(ω) = P[A 1 ] = ν(ω) 1 <. So µ ad ν agree o σ(a 2 ). So we have P[A 1 A 2 ] = µ(a 2 ) = ν(a 2 ) = P[A 1 ]P[A 2 ] for all A 2 σ(a 2 ). So we have ow show that if A 1 ad A 2 are idepedet, the A 1 ad σ(a 2 ) are idepedet. By symmetry, the same argumet shows that σ(a 1 ) ad σ(a 2 ) are idepedet. Say we are rollig a dice. Istead of askig what the probability of gettig a 6, we might be iterested istead i the probability of gettig a 6 ifiitely ofte. Ituitively, the aswer is it happes with probability 1, because i each dice roll, we have a probability of 1 6 of gettig a 6, ad they are all idepedet. We would like to make this precise ad actually prove it. It turs out that the otios of occurs ifiitely ofte ad also occurs evetually correspod to more aalytic otios of lim sup ad lim if. Defiitio (limsup ad limif). Let (A ) be a sequece of evets. We defie lim sup A = A m m lim if A = A m. m 17

18 1 Measures II Probability ad Measure To parse these defiitios more easily, we ca read as for all, ad as there exits. For example, we ca write Similarly, we have lim sup A =, m such that A m occurs = {x :, m, x A m } = {A m occurs ifiitely ofte} = {A m i.o.} lim if A =, m such that A m occurs = {x :, m, x A m } = {A m occurs evetually} = {A m e.v.} We are ow goig to prove two obvious results, kow as the Borel Catelli lemmas. These give us ecessary coditios for a evet to happe ifiitely ofte, ad i the case where the evets are idepedet, the coditio is also sufficiet. Lemma (Borel Catelli lemma). If P[A ] <, the Proof. For each k, we have P[A i.o.] = 0. P[A i.o] = P m A m P A m m k 0 as k. So we have P[A i.o.] = 0. P[A m ] m=k Note that we did ot eed to use the fact that we are workig with a probability measure. So i fact this holds for ay measure space. Lemma (Borel Catelli lemma II). Let (A ) be idepedet evets. If P[A ] =, the P[A i.o.] = 1. 18

19 1 Measures II Probability ad Measure Note that idepedece is crucial. If we flip a fair coi, ad we set all the A to be equal to gettig a heads, the P[A ] = 1 2 =, but we certaily do ot have P[A i.o.] = 1. Istead it is just 1 2. Proof. By example sheet, if (A ) is idepedet, the so is (A C ). The we have [ N P m= A C m ] N = P[A C m] m= N = (1 P[A m ]) m= N exp( P[A m ]) m= ( = exp ) N P[A m ] m= 0 as N, as we assumed that P[A ] =. So we have [ ] P = 0. m= By coutable subadditivity, we have [ P m= A C m A C m ] = 0. This i tur implies that [ ] [ P A m = 1 P So we are doe. m= m= A C m ] = 1. 19

20 2 Measurable fuctios ad radom variables II Probability ad Measure 2 Measurable fuctios ad radom variables We ve had eough of measurable sets. As i most of mathematics, ot oly should we talk about objects, but also maps betwee objects. Here we wat to talk about maps betwee measure spaces, kow as measurable fuctios. I the case of a probability space, a measurable fuctio is a radom variable! I this chapter, we are goig to start by defiig a measurable fuctio ad ivestigate some of its basic properties. I particular, we are goig to prove the mootoe class theorem, which is the aalogue of Dyki s lemma for measurable fuctios. Afterwards, we tur to the probabilistic aspects, ad see how we ca make sese of the idepedece of radom variables. Fially, we are goig to cosider differet otios of covergece of fuctios. 2.1 Measurable fuctios The defiitio of a measurable fuctio is somewhat like the defiitio of a cotiuous fuctio, except that we replace ope with i the σ-algebra. Defiitio (Measurable fuctios). Let (E, E) ad (G, G) be measure spaces. A map f : E G is measurable if for every A G, we have f 1 (A) = {x E : f(x) E} E. If (G, G) = (R, B), the we will just say that f is measurable o E. If (G, G) = ([0, ], B), the we will just say that f is o-egative measurable. If E is a topological space ad E = B(E), the we call f a Borel fuctio. How do we actually check i practice that a fuctio is measurable? It turs out we are lucky. We ca simply check that f 1 (A) E for A i ay geeratig set Q of G. Lemma. Let (E, E) ad (G, G) be measurable spaces, ad G = σ(q) for some Q. If f 1 (A) E for all A Q, the f is measurable. Proof. We claim that {A G : f 1 (A) E} is a σ-algebra o G. The the result follows immediately by defiitio of σ(q). Ideed, this follows from the fact that f 1 preserves everythig. More precisely, we have ( ) f 1 A = f 1 (A ), f 1 (A C ) = (f 1 (A)) C, f 1 ( ) =. So if, say, all A A, the so is A. Example. I the particular case where we have a fuctio f : E R, we kow that B = B(R) is geerated by (, y] for y R. So we just have to check that {x E : f(x) y} = f 1 ((, y])) E. 20

21 2 Measurable fuctios ad radom variables II Probability ad Measure Example. Let E, F be topological spaces, ad f : E F be cotiuous. We will see that f is a measurable fuctio (uder the Borel σ-algebras). Ideed, by defiitio, wheever U F is ope, we have f 1 (U) ope as well. So f 1 (U) B(E) for all U F ope. But sice B(F ) is the σ-algebra geerated by the ope sets, this implies that f is measurable. This is oe very importat example. We ca do aother very importat example. Example. Suppose that A E. The idicator fuctio of A is 1 A (x) : E {0, 1} give by { 1 x A 1 A (x) = 0 x A. Suppose we give {0, 1} the o-trivial measure. The 1 A is a measurable fuctio iff A E. Example. The idetity fuctio is always measurable. Example. Compositio of measurable fuctios are measurable. More precisely, if (E, E), (F, F) ad (G, G) are measurable spaces, ad the fuctios f : E F ad g : F G are measurable, the the compositio g f : E G is measurable. Ideed, if A G, the g 1 (A) F, so f 1 (g 1 (A)) E. But f 1 (g 1 (A)) = (g f) 1 (A). So doe. Defiitio (σ-algebra geerated by fuctios). Now suppose we have a set E, ad a family of real-valued fuctios {f i : i I} o E. We the defie σ(f i : i I) = σ(f 1 i (A) : A B, i I). This is the smallest σ-algebra o E which makes all the f i s measurable. This is aalogous to the otio of iitial topologies for topological spaces. If we wat to costruct more measurable fuctios, the followig defiitio will be rather useful: Defiitio (Product measurable space). Let (E, E) ad (G, G) be measure spaces. We defie the product measure space as E G whose σ-algebra is geerated by the projectios E G π 1 π 2. E G More explicitly, the σ-algebra is give by E G = σ({a B : A E, B G}). More geerally, if (E i, E i ) is a collectio of measure spaces, the product measure space has uderlyig set i E i, ad the σ-algebra geerated by the projectio maps π i : j E j E i. This satisfies the followig property: 21

22 2 Measurable fuctios ad radom variables II Probability ad Measure Propositio. Let f i : E F i be fuctios. The {f i } are all measurable iff (f i ) : E F i is measurable, where the fuctio (f i ) is defied by settig the ith compoet of (f i )(x) to be f i (x). Proof. If the map (f i ) is measurable, the by compositio with the projectios π i, we kow that each f i is measurable. Coversely, if all f i are measurable, the sice the σ-algebra of F i is geerated by sets of the form π 1 j alog (f i ) is exactly f 1 j (A) : A F j, ad the pullback of such sets (A), we kow the fuctio (f i ) is measurable. Usig this, we ca prove that a whole lot more fuctios are measurable. Propositio. Let (E, E) be a measurable space. Let (f : N) be a sequece of o-egative measurable fuctios o E. The the followig are measurable: f 1 + f 2, f 1 f 2, max{f 1, f 2 }, mi{f 1, f 2 }, if f, sup f, lim if f, lim sup f. The same is true with real replaced with o-egative, provided the ew fuctios are real (i.e. ot ifiity). Proof. This is a (easy) exercise o the example sheet. For example, the sum f 1 + f 2 ca be writte as the followig compositio. (f 1,f 2) E [0, ] 2 + [0, ]. We kow the secod map is cotiuous, hece measurable. The first fuctio is also measurable sice the f i are. So the compositio is also measurable. The product follows similarly, but for the ifimum ad supremum, we eed to check explicitly that the correspodig maps [0, ] N [0, ] is measurable. Notatio. We will write f g = mi{f, g}, f g = max{f, g}. We are ow goig to prove the mootoe class theorem, which is a Dyki s lemma for measurable fuctios. As i the case of Dyki s lemma, it will soud rather awkward but will prove itself to be very useful. Theorem (Mootoe class theorem). Let (E, E) be a measurable space, ad A E be a π-system with σ(a) = E. Let V be a vector space of fuctios such that (i) The costat fuctio 1 = 1 E is i V. (ii) The idicator fuctios 1 A V for all A A (iii) V is closed uder bouded, mootoe limits. More explicitly, if (f ) is a bouded o-egative sequece i V, f f (poitwise) ad f is also bouded, the f V. The V cotais all bouded measurable fuctios. 22

23 2 Measurable fuctios ad radom variables II Probability ad Measure Note that the coditios for V is pretty like the coditios for a d-system, where takig a bouded, mootoe limit is somethig like takig icreasig uios. Proof. We first deduce that 1 A V for all A E. D = {A E : 1 A V}. We wat to show that D = E. To do this, we have to show that D is a d-system. (i) Sice 1 E V, we kow E D. (ii) If 1 A V, the 1 1 A = 1 E\A V. So E \ A D. (iii) If (A ) is a icreasig sequece i D, the 1 A 1 A mootoically icreasigly. So 1 A is i D. So, by Dyki s lemma, we kow D = E. So V cotais idicators of all measurable sets. We will ow try to obtai ay measurable fuctio by approximatig. Suppose that f is bouded ad o-egative measurable. We wat to show that f V. To do this, we approximate it by lettig f = 2 2 f = k2 1 {k2 f<(k+1)2 }. k=0 Note that sice f is bouded, this is a fiite sum. So it is a fiite liear combiatio of idicators of elemets i E. So f V, ad 0 f f mootoically. So f V. More geerally, if f is bouded ad measurable, the we ca write f = (f 0) + (f 0) f + f. The f + ad f are bouded ad o-egative measurable. So f V. Ufortuately, we will ot have a chace to use this result util the ext chapter where we discuss itegratio. There we will use this a lot. 2.2 Costructig ew measures We are goig to look at two ways to costruct ew measures o spaces based o some measurable fuctio we have. Defiitio (Image measure). Let (E, E) ad (G, G) be measure spaces. Suppose µ is a measure o E ad f : E G is a measurable fuctio. We defie the image measure ν = µ f 1 o G by ν(a) = µ(f 1 (A)). It is a routie check that this is ideed a measure. If we have a strictly icreasig cotiuous fuctio, the we kow it is ivertible (if we restrict the codomai appropriately), ad the iverse is also strictly icreasig. It is also clear that these coditios are ecessary for a iverse to exist. However, if we relax the coditios a bit, we ca get some sort of pseudoiverse (some categorists may call them left adjoits (ad will tell you that it is a trivial cosequece of the adjoit fuctor theorem)). Recall that a fuctio g is right cotiuous if x x implies g(x ) g(x), ad similarly f is left cotiuous if x x implies f(x ) f(x). 23

24 2 Measurable fuctios ad radom variables II Probability ad Measure Lemma. Let g : R R be o-costat, o-decreasig ad right cotiuous. We set g(± ) = lim g(x). x ± We set I = (g( ), g( )). Sice g is o-costat, this is o-empty. The there is a o-decreasig, left cotiuous fuctio f : I R such that for all x I ad y R, we have x g(y) f(x) y. Thus, takig the egatio of this, we have Explicitly, for x I, we defie x > g(y) f(x) > y. f(x) = if{y R : x g(y)}. Proof. We just have to verify that it works. For x I, cosider J x = {y R : x g(y)}. Sice g is o-decreasig, if y J x ad y y, the y J x. Sice g is right-cotiuous, if y J x is such that y y, the y J x. So we have Thus, for f R, we have J x = [f(x), ). x g(y) f(x) y. So we just have to prove the remaiig properties of f. Now for x x, we have J x J x. So f(x) f(x ). So f is o-decreasig. Similarly, if x x, the we have J x = J x. So f(x ) f(x). So this is left cotiuous. Example. If g is give by the fuctio the f is give by 24

25 2 Measurable fuctios ad radom variables II Probability ad Measure This allows us to costruct ew measures o R with ease. Theorem. Let g : R R be o-costat, o-decreasig ad right cotiuous. The there exists a uique Rado measure dg o B such that dg((a, b]) = g(b) g(a). Moreover, we obtai all o-zero Rado measures o R i this way. We have already see a istace of this whe we g was the idetity fuctio. Give the lemma, this is very easy. Proof. Take I ad f as i the previous lemma, ad let µ be the restrictio of the Lebesgue measure to Borel subsets of I. Now f is measurable sice it is left cotiuous. We defie dg = µ f 1. The we have dg((a, b]) = µ({x I : a < f(x) b}) = µ({x I : g(a) < x g(b)}) = µ((g(a), g(b)]) = g(b) g(a). So dg is a Rado measure with the required property. There are o other such measures by the argumet used for uiqueess of the Lebesgue measure. To show we get all o-zero Rado measures this way, suppose we have a Rado measure ν o R, we wat to produce a g such that ν = dg. We set { ν((y, 0]) y 0 g(y) = ν((0, y]) y > 0. The ν((a, b]) = g(b) g(a). We see that ν is o-zero, so g is o-costat. It is also easy to see it is o-decreasig ad right cotiuous. So ν = dg by cotiuity. 2.3 Radom variables We are ow goig to look at these ideas i the cotext of probability. It turs out they are cocepts we already kow ad love! Defiitio (Radom variable). Let (Ω, F, P) be a probability space, ad (E, E) a measurable space. The a E-valued radom variable is a measurable fuctio X : Ω E. By default, we will assume the radom variables are real. Usually, whe we have a radom variable X, we might ask questios like what is the probability that X A?. I other words, we are askig for the size of the set of thigs that get set to A. This is just the image measure! Defiitio (Distributio/law). Give a radom variable X : Ω E, the distributio or law of X is the image measure µ x : P X 1. We usually write P(X A) = µ x (A) = P(X 1 (A)). 25

26 2 Measurable fuctios ad radom variables II Probability ad Measure If E = R, the µ x is determied by its values o the π-system of itervals (, y]. We set F X (x) = µ X ((, x]) = P(X x) This is kow as the distributio fuctio of X. Propositio. We have F X (x) { 0 x 1 x +. Also, F X (x) is o-decreasig ad right-cotiuous. We call ay fuctio F with these properties a distributio fuctio. Defiitio (Distributio fuctio). A distributio fuctio is a o-decreasig, right cotiuous fuctio f : R [0, 1] satisfyig { 0 x F X (x) 1 x +. We ow wat to show that every distributio fuctio is ideed a distributio. Propositio. Let F be ay distributio fuctio. The there exists a probability space (Ω, F, P) ad a radom variable X such that F X = F. Proof. Take (Ω, F, P) = ((0, 1), B(0, 1), Lebesgue). We take X : Ω R to be The we have So we have Therefore F X = F. X(ω) = if{x : ω f(x)}. X(ω) x w F (x). F X (x) = P[X x] = P[(0, F (x)]] = F (x). This costructio is actually very useful i practice. If we are writig a computer program ad wat to sample a radom variable, we will use this procedure. The computer usually comes with a uiform (pseudo)-radom umber geerator. The usig this procedure allows us to produce radom variables of ay distributio from a uiform sample. The ext thig we wat to cosider is the otio of idepedece of radom variables. Recall that for radom variables X, Y, we used to say that they are idepedet if for ay A, B, we have P[X A, Y B] = P[X A]P[Y B]. But this is exactly the statemet that the σ-algebras geerated by X ad Y are idepedet! Defiitio (Idepedece of radom variables). A family (X ) of radom variables is said to be idepedet if the family of σ-algebras (σ(x )) is idepedet. 26

27 2 Measurable fuctios ad radom variables II Probability ad Measure Propositio. Two real-valued radom variables X, Y are idepedet iff P[X x, Y y] = P[X x]p[y y]. More geerally, if (X ) is a sequece of real-valued radom variables, the they are idepedet iff for all ad x j. P[x 1 x 1,, x x ] = P[X j x j ] Proof. The directio is obvious. For the other directio, we simply ote that {(, x] : x R} is a geeratig π-system for the Borel σ-algebra of R. I probability, we ofte say thigs like let X 1, X 2, be iid radom variables. However, how ca we guaratee that iid radom variables do ideed exist? We start with the less ambitious goal of fidig iid Beroulli(1/2) radom variables: Propositio. Let j=1 (Ω, F, P) = ((0, 1), B(0, 1), Lebesgue). be our probability space. The there exists as sequece R of idepedet Beroulli(1/2) radom variables. Proof. Suppose we have ω Ω = (0, 1). The we write ω as a biary expasio ω = ω 2, =1 where ω {0, 1}. We make the biary expasio uique by disallowig ifiite sequeces of zeroes. We defie R (ω) = ω. We will show that R is measurable. Ideed, we ca write R 1 (ω) = ω 1 = 1 (1/2,1] (ω), where 1 (1/2,1] is the idicator fuctio. Sice idicator fuctios of measurable sets are measurable, we kow R 1 is measurable. Similarly, we have R 2 (ω) = 1 (1/4,1/2] (ω) + 1 (3/4,1] (ω). So this is also a measurable fuctio. More geerally, we ca do this for ay R (ω): we have R (ω) = 2 1 j=1 1 (2 (2j 1),2 (2j)](ω). So each R is a radom variable, as each ca be expressed as a sum of idicators of measurable sets. Now let s calculate P[R = 1] = 2 1 j=1 2 ((2j) (2j 1)) = 2 1 j=1 2 =

28 2 Measurable fuctios ad radom variables II Probability ad Measure The we have P[R = 0] = 1 P[R = 1] = 1 2 as well. So R Beroulli(1/2). We ca straightforwardly check that (R ) is a idepedet sequece, sice for m, we have P[R = 0 ad R m = 0] = 1 4 = P[R = 0]P[R m = 0]. We will ow use the (R ) to costruct ay idepedet sequece for ay distributio. Propositio. Let (Ω, F, P) = ((0, 1), B(0, 1), Lebesgue). Give ay sequece (F ) of distributio fuctios, there is a sequece (X ) of idepedet radom variables with F X = F for all. Proof. Let m : N 2 N be ay bijectio, ad relabel Y k, = R m(k,), where the R j are as i the previous radom variable. We let Y = 2 k Y k,. k=1 The we kow that (Y ) is a idepedet sequece of radom variables, ad each is uiform o (0, 1). As before, we defie G (y) = if{x : y F (x)}. We set X = G (Y ). The (X ) is a sequece of radom variables with F X = F. We ed the sectio with a radom fact: let (Ω, F, P) ad R j be as above. The 1 j=1 R j is the average of idepedet of Beroulli(1/2) radom variables. The weak law of large umbers says for ay ε > 0, we have P 1 R j 1 2 ε 0 as. j=1 The strog law of large umbers, which we will prove later, says that P ω : 1 R j 1 = 1. 2 j=1 So almost every umber i (0, 1) has a equal proportio of 0 s ad 1 s i its biary expasio. This is kow as the ormal umber theorem. 28

29 2 Measurable fuctios ad radom variables II Probability ad Measure 2.4 Covergece of measurable fuctios The ext thig to look at is the covergece of measurable fuctios. I measure theory, woderful thigs happe whe we talk about covergece. I aalysis, most of the time we had to require uiform covergece, or eve stroger otios, if we wat limits to behave well. However, i measure theory, the kids of covergece we talk about are somewhat poitwise i ature. I fact, it will be weaker tha poitwise covergece. Yet, we are still goig to get good properties out of them. Defiitio (Covergece almost everywhere). Suppose that (E, E, µ) is a measure space. Suppose that (f ), f are measurable fuctios. We say f f almost everywhere (a.e.) if µ({x E : f (x) f(x)}) = 0. If (E, E, µ) is a probability space, this is called almost sure covergece. To see this makes sese, i.e. the set i there is actually measurable, ote that {x E : f (x) f(x)} = {x E : lim sup f (x) f(x) > 0}. We have previously see that lim sup f f is o-egative measurable. So the set {x E : lim sup f (x) f(x) > 0} is measurable. Aother useful otio of covergece is covergece i measure. Defiitio (Covergece i measure). Suppose that (E, E, µ) is a measure space. Suppose that (f ), f are measurable fuctios. We say f f i measure if for each ε > 0, we have µ({x E : f (x) f(x) ε}) 0 as, the we say that f f i measure. If (E, E, µ) is a probability space, the this is called covergece i probability. I the case of a probability space, this says P( X X ε) 0 as for all ε, which is how we state the weak law of large umbers i the past. After we defie itegratio, we ca cosider the orms of a fuctio f by ( 1/p f p = f(x) dx) p. The i particular, if f f p 0, the f f i measure, ad this provides a easy way to see that fuctios coverge i measure. I geeral, either of these otios imply each other. However, the followig theorem provides us with a coveiet dictioary to traslate betwee the two otios. Theorem. (i) If µ(e) <, the f f a.e. implies f f i measure. 29

30 2 Measurable fuctios ad radom variables II Probability ad Measure (ii) For ay E, if f f i measure, the there exists a subsequece (f k ) such that f k f a.e. Proof. (i) First suppose µ(e) <, ad fix ε > 0. Cosider µ({x E : f (x) f(x) ε}). We use the result from the first example sheet that for ay sequece of evets (A ), we have lim if µ(a ) µ(lim if A ). Applyig to the above sequece says lim if µ({x : f (x) f(x) ε}) µ({x : f m (x) f(x) ε evetually}) µ({x E : f m (x) f(x) 0}) = µ(e). As µ(e) <, we have µ({x E : f (x) f(x) > ε}) 0 as. (ii) Suppose that f f i measure. We pick a subsequece ( k ) such that ({ µ x E : f k (x) f(x) > 1 }) 2 k. k The we have ({ µ x E : f k (x) f(x) > 1 }) 2 k = 1 <. k k=1 k=1 By the first Borel Catelli lemma, we kow µ ({x E : f k (x) f(x) > 1k }) i.o. = 0. So f k f a.e. It is importat that we assume that µ(e) < for the first part. Example. Cosider (E, E, µ) = (R, B, Lebesgue). Take f (x) = 1 [, ) (x). The f (x) 0 for all x, ad i particular almost everywhere. However, we have ({ µ x R : f (x) > 1 }) = µ([, )) = 2 for all. There is oe last type of covergece we are iterested i. We will oly first formulate it i the probability settig, but there is a aalogous otio i measure theory kow as weak covergece, which we will discuss much later o i the course. 30

31 2 Measurable fuctios ad radom variables II Probability ad Measure Defiitio (Covergece i distributio). Let (X ), X be radom variables with distributio fuctios F X ad F X, the we say X X i distributio if F X (x) F X (x) for all x R at which F X is cotiuous. Note that here we do ot eed that (X ) ad X live o the same probability space, sice we oly talk about the distributio fuctios. But why do we have the coditio with cotiuity poits? The idea is that if the resultig distributio has a jump at x, it does t matter which side of the jump F X (x) is at. Here is a simple example that tells us why this is very importat: Example. Let X to be uiform o [0, 1/]. Ituitively, this should coverge to the radom variable that is always zero. We ca compute 0 x 0 F X (x) = x 0 < x < 1/. 1 x 1/ We ca also compute the distributio of the zero radom variable as { 0 x < 0 F 0 = 1 x 0. But F X (0) = 0 for all, while F X (0) = 1. Oe might ow thik of cheatig by cookig up some radom variable such that F is discotiuous at so may poits that radom, urelated thigs coverge to F. However, this caot be doe, because F is a o-decreasig fuctio, ad thus ca oly have coutably may poits of discotiuities. The big theorem we are goig to prove about covergece i distributio is that actually it is very borig ad does t give us aythig ew. Theorem (Skorokhod represetatio theorem of weak covergece). (i) If (X ), X are defied o the same probability space, ad X X i probability. The X X i distributio. (ii) If X X i distributio, the there exists radom variables ( X ) ad X defied o a commo probability space with F X = F X ad F X = F X such that X X a.s. Proof. Let S = {x R : F X is cotiuous}. (i) Assume that X X i probability. Fix x S. We eed to show that F X (x) F X (x) as. We fix ε > 0. Sice x S, this implies that there is some δ > 0 such that F X (x δ) F X (x) ε 2 F X (x + δ) F X (x) + ε 2. 31

32 2 Measurable fuctios ad radom variables II Probability ad Measure We fix N large such that N implies P[ X X δ] ε 2. The F X (x) = P[X x] = P[(X X) + X x] We ow otice that {(X X) + X x} {X x + δ} { X X > δ}. So we have P[X x + δ] + P[ X X > δ] We similarly have F X (x + δ) + ε 2 F X (x) + ε. F X (x) = P[X x] P[X x δ] P[ X X > δ] F X (x δ) ε 2 F X (x) ε. Combiig, we have that N implyig F x (x) F X (x) ε. Sice ε was arbitrary, we are doe. (ii) Suppose X X i distributio. We agai let We let (Ω, F, B) = ((0, 1), B((0, 1)), Lebesgue). X (ω) = if{x : ω F X (x)}, X(ω) = if{x : ω F X (x)}. Recall from before that X has the same distributio fuctio as X for all, ad X has the same distributio as X. Moreover, we have X (ω) x ω F X (x) x < X (ω) F X (x) < ω, ad similarly if we replace X with X. We are ow goig to show that with this particular choice, we have X X a.s. Note that X is a o-decreasig fuctio (0, 1) R. The by geeral aalysis, X has at most coutably may discotiuities. We write Ω 0 = {ω (0, 1) : X is cotiuous at ω0 }. The (0, 1) \ Ω 0 is coutable, ad hece has Lebesgue measure 0. So P[Ω 0 ] = 1. 32