A Proof of Birkhoff s Ergodic Theorem

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1 A Proof of Birkhoff s Ergodic Theorem Joseph Hora September 2, 205 Itroductio I Fall 203, I was learig the basics of ergodic theory, ad I came across this theorem. Oe of my supervisors, Athoy Quas, showed me this proof, as commuicated to him by a colleague, Mate Wierdl. It s hard to fid it i prit, so here it is. 2 The Theorem ad the Proof Theorem 2. (Birkhoff). Let (, B, µ, T ) be a measure-preservig system o a σ-fiite measure space, ad let f be a itegrable fuctio. The lim f(t k (x)) k=0 coverges for almost every x, ad the resultig fuctio f is i L, with f T = f almost everywhere ad f f. If µ() <, the f dµ = f dµ. Corollary 2.2. If (, B, µ, T ) is ergodic, the f as obtaied above is almost everywhere costat, ad if µ() <, the f = f dµ. µ() Proof of Birkhoff s Ergodic Theorem. We split the proof ito two parts: first, assumig the almost everywhere existece of the limit of the ergodic averages, we prove that it has the requisite properties. Secod, we prove that the limit exists for all L fuctios. So for ow, let f L (µ) ad assume that the limit f(x) = lim f(t i (x)) exists for almost every x. As a limit of a sum of measurable fuctios, f is measurable, ad by Fatou s lemma, we have: f dµ = lim f(t i (x) dµ(x) (Fatou s Lemma) lim if f(t i (x) dµ(x) ivariace = lim if = lim if f(x) dµ(x) f = f.

2 Hece f is itegrable, with f f. Precomposig by T gives us: f(t (x)) = lim = lim = f(x). f(x) 0 f(t i+ (x)) = lim f(t i (x)) i= f(t i (x)) So we have that f is T -ivariat. Lastly for this part, suppose that µ() <. We will show that for ay bouded f L (µ), we have f dµ = f dµ, ad bootstrap to show it for ubouded f. So let f L (µ) be bouded, with f C <. The we have f(t i (x)) C = C, ad C dµ = C dµ = Cµ() <, so g C is a itegrable majorat for this sequece, which allows us to apply the Lebesgue Domiated Covergece Theorem: f dµ = = lim lim f(t i (x)) dµ f(t i (x)) dµ = lim f dµ = f dµ, usig T -ivariace i the last steps. Now let f be ubouded, ad let ɛ > 0; we may fid f B L (µ) which is bouded, with f f B < ɛ 2. By the above argumet, f f B f f B < ɛ 2. We the approximate the differece of the itegrals: f dµ f dµ f dµ f B dµ + f B dµ = f f B dµ + f B f dµ f f B dµ + f B f dµ f f B + f f B < ɛ 2 + ɛ 2 = ɛ. f B dµ + f B dµ f dµ 2

3 Sice ɛ was arbitrary, we have f dµ = as desired. We ow oly have to prove that for every f L (µ), the poitwise limit of the ergodic averages coverges almost everywhere. There are four mai steps of the proof, together with some mior argumets.. Prove a maximal ergodic lemma for l (Z). f dµ, 2. Use this lemma to prove a maximal ergodic lemma for L (, µ). 3. Show that the subset of L fuctios satisfyig the BET is closed i L. 4. Fid a dese subset of L for which the BET holds. Everythig will be doe for real-valued fuctios; oce it is prove, the result holds for a complex-valued fuctio by writig the the fuctio as a sum of a real fuctio ad aother real fuctio multiplied by the imagiary uit. As well, we will do the mai proof i the case of a probability space, ad the show how to exted the result to a σ-fiite space. Or at least, we would, if I kew how to do it. See Halmos s Lectures o Ergodic Theory or similar for a proof that works i the σ-fiite case. Sorry. Step : This lemma is of a more combiatorial ature. Lemma 2.3 (Maximal Ergodic Lemma for l (Z)). Let a l (Z), ad deote The we have, for ay positive real umber λ, j M a () = sup a( + i). j j : M a () λ a λ. Proof. First, assume that the statemet holds whe λ =. Now, let λ > 0 be fixed, ad let ã = λa. The we have: : M a () λ = : Mã() ã = λ a. Hece we eed oly prove the claim for λ =. Note that because a l (Z), a a(i) <, so M a () is actually a maximum, because the averagig factor j forces the terms towards 0. Let 0 < ɛ <. Agai because a l (Z), we fid M R ad idices k k 2 Z such that a > M > a ɛ, ad M = k 2 i=k a(i). This says that almost all of the mass is cotaied betwee two idices. We observe that for < k (M +), there is a boud o the value of M a (). For j = 0,..., k, we have j a( + i) < j j ɛ <, ad for j > k, we have that j > k > M +, by defiitio of, so that j a( + i) < M + (ɛ + M) < j M + M + =. Hece M a () < for < k. For > k 2, the we clearly have M a () <, as less tha ɛ of the weight is after k 2. Therefore : M a () is fiite. Deote A = : M a (). 3

4 Let 0 = mi A, ad defie j 0 = max j : j j m=0 a( 0 + m) which we ca do because M a ( 0 ) is a hoest maximum. The iductively defie i = mi A, > i + j i, ad j j i = max j : a( i + m), j for i =... k, sice A is fiite. The sets i,..., i + j i are disjoit, which meas à = m=0 k i,..., i + j i is a disjoit uio, ad we see that A Ã. This implies: ad so we are doe. a k j i m=0 a( i + m) j i k k j i = i,..., i + j i = à A = : M a (), Step 2: This is a so-called trasferece priciple, where we obtai the result i a more complicated space by bootstrappig from the l (Z) case. Lemma 2.4 (Maximal Ergodic Lemma for L (, µ)). Let (, B, µ, T ) be a measure-preservig system, ad let f be a itegrable fuctio. Defie, for each x, The for ay positive real umber λ, we have Mf(x) = sup j j j f(t i (x)). µ(x : Mf(x) λ) f λ. Proof. The same proof as i the l (Z) case applies to show that it suffices to prove the lemma for λ =. So assume λ =. We shall perform two trucatios, to allow us to deal with fiite summatios. Let K N +, ad defie f x,k : Z R by f(t (x)) 0 K, f x,k () = 0 < 0, > K. The for µ-almost every x, f x,k l (Z) as it is a fiite sum (it is t summable everywhere because f could be ifiite o a set B of measure zero), ad for a fixed, f,k () = f(t ( )) is both measurable ad itegrable (the latter by T -ivariace of µ). Because Z is coutable, we see that f,k ( ) is a measurable fuctio o Z. For ay fuctio g L µ, ad N N +, we defie M N g : R, by M N g(x) = max j N j 4 j f(t i (x)).,

5 Note that as a maximum of a sum of measurable fuctios, M N f is measurable. We also have that Mf(x) = lim N M Nf(x), by defiitio, so that Mf is measurable. Now, observe that for N < N 2, we have M N f(x) M N2 f(x), so that we have a icreasig chai of subsets of, give by x : M N f(x). If we were able to prove that µ(x : M N f(x) ) f for all N, the we would be able to show the result for Mf, by cotiuity of the measure µ alog chais: µ(x : Mf(x) ) = µ( x : lim M N f(x) ) N = lim N µ(x : M Nf(x) ) lim N f = f. So it is ow our goal to show that the result holds for ay N. If we let N >> K, the we see that because a large umber of the terms i the sum are zero, the followig idetity is true: M N f x,k () = max j N j j f(t ((T i (x))) = max j N j j f(t +i (x)) = M fx,k (), where we use the otatio from the previous lemma for summable fuctios. This allows us to apply Lemma 2.3, to obtai (writig c for the orm o l (Z)): We may ow prove the lemma: f x,k c : M fx,k () = : MN f x,k (). K f = = (by Fubii) = (by Fubii) = (T ivariace of µ) = K =0 K = =0 K =0 f(x) dµ(x) = K =0 f x,k () dµ(x) = f x,k () dµ(x) = f(t (x)) dµ(x) (by T ivariace) Z f x,k c : M N f x,k () dµ(x) K =0 K =0 K =0 : MN f x,k () () dµ(x) x : MN (f T )(x) (x) dµ(x) x : MN f(x) (x) dµ(x) = Kµ(x : M N f(x) ). f x,k () dµ c(x, ) dµ(x) Hece we obtai µ(x : M N f(x) ) f for all N >> K, ad thus we get by the earlier remark. µ(x : Mf(x) ) f 5

6 Step 3: We ow show that the set of L fuctios satisfyig the BET is closed. Suppose that f k k= L (µ) with each f k satisfyig the BET, with the correspodig T -ivariat fuctio deoted f k, ad f k f i L. Deote Ā f (x) = lim sup f(t i (x)), A f (x) = lim if f(t i (x)). k Lemma 2.5. We have µ( x : Ā f (x) A f (x) > 0 ) = 0, ad hece the limit fuctio f exists almost everywhere o. Proof. We may boud the differece of Ā f (x) ad A f (x). ivolvig limit supremums ad ifimums: For ay k, we have, by applyig iequalities Ā f (x) A f (x) = lim sup = lim sup lim if f(t i (x)) lim if f(t i (x)) ( ) f k (T i (x)) + (f f k )(T i (x)) ( ) f k (T i (x)) + (f f k )(T i (x)) = f k (x) + lim sup lim sup sup 2(f fk )(T i (x)) 2(f fk )(T i (x)) = M(2(f f k ))(x). (f f k )(T i (x)) f k (x) lim if (f f k )(T i (x)) From here, fix ɛ > 0, ad pick δ > 0 such that for all sufficietly large k, f k f < ɛδ 2. We may apply Lemma 2.4 i the case of 2(f f k ), where k is large eough to obtai: µ( x : Ā f (x) A f (x) > ɛ ) µ(x : M(2(f f k ))(x) > ɛ) 2(f f k) ɛ We picked δ arbitrarily, so that ad ɛ was also arbitrary, so that µ( x : Ā f (x) A f (x) > ɛ ) = 0, < 2ɛδ 2ɛ = δ. Hece Āf = A f µ-almost everywhere, so that exists µ-almost everywhere. µ( x : Ā f (x) A f (x) > 0 ) = 0. f(x) = lim f(t i (x)) Hece the set of L fuctios which satisfy the BET is closed i the L sese. Step 4: Fially, we show that there is a L dese set of fuctios which satisfy the BET. We shall do this i two smaller steps: first we show that the square-itegrable fuctios are dese i L (µ), the we explicitly fid our desired set. 6

7 Lemma 2.6. Supposig that µ() <, we have L 2 (µ) L (µ), ad moreover L 2 (µ) is dese i the L sese i L (µ). Proof. We are still assumig that µ() <. We first follow the proof of a result I should ve remembered from my itro measure theory class from udergrad, but i the special case of L 2 ad L. Let f L 2 (µ); the f L 2 (µ) also. We apply Cauchy-Schwarz (a special case of Hölder s Iequality): f, = f dµ f 2 2 = f 2 µ() 2 <. Thus f L (µ), so L 2 (µ) L (µ). I additio, the orm iequality meas covergece i the L 2 sese implies covergece i the L sese. Furthermore, if f L (µ), the for each N defie f : R, by f(x), f (x) = f(x) < f(x) <, f(x). The f (x) f(x), so f L (µ), ad moreover f is bouded. This meas f 2 is bouded, ad hece itegrable o, so f L 2 (µ). Fially, f f almost everywhere, so f (x) f(x) 0 for almost every x. We also have, by ispectio of the defiitio of f, that f(x) f(x) f (x) f(x) = 0 f(x) < f(x), so that f is a itegrable majorat for f f, ad so we obtai: lim f f = lim f f dµ LDCT = lim f f dµ = Hece f coverges to f i the L sese, so we are doe. 0 dµ = 0. The previous lemma serves a very useful purpose: if a sequece of L 2 fuctios f k k= coverges i the L 2 sese to a L 2 fuctio f, ad BET holds for each f k, the we see that this covergece happes i the L sese, ad so by two lemmas ago, BET holds for f also. The fial step, the, is to fid a subset of L o which BET holds, ad the L -closure of which cotais L 2. Lemma 2.7. BET holds for the subspace of L 2 coboudaries, C = f f T L 2 -closure of C, C L2. Proof. For a coboudary f f T, we have: (f f T )(T i (x)) = (f(x) f(t (x))). By Markov s Iequality, we obtai: µ x : f(t (x)) Hece for almost every x, we have 2 3 f T ( ) = f lim (f f T )(T i (x)) = 0, 0. ad so BET holds for C. By the above covergece result, BET also holds for C L2. : f L 2, ad thus for the 7

8 Lemma 2.8. The orthogoal complemet of C is C = g L 2 (µ) : hg dµ = 0, h C = g L 2 (µ) : g = g T, the set of T -ivariat fuctios i L 2, ad BET holds for this subspace. Proof. Let g C. The g g T C, by defiitio, ad we have: Addig, we obtai: g, g g T = g, g g, g T = g T, g T g, g T = g T g, g T = g T, g g T = g T, g g T = 0. 0 = = g, g g T + g T, g g T = g g T, g g T = g g T 2 2, which implies that g = g T almost everywhere. Coversely, if g is (almost) T -ivariat, ad f L 2, the we have: g, f f T = g, f g, f T = g T, f T g T, f T = 0. Hece g C, ad we get that C = g L 2 (µ) : g = g T. To coclude, observe that for g C, lim so BET holds for elemets of C. g(t i (x)) = lim g(x) g(x) = lim = g(x), We ow observe that L 2 (µ) = C L2 2 C, so that every elemet of L2(µ) may be writte as a sum of a limit of a coboudary ad a T -ivariat fuctio. If BET holds for two fuctios, it holds for their sum (by limit laws), so that BET ow holds for all of L 2 (µ). but we already kow that the L -closure of L 2 (µ) is all of L (µ), so we see that BET holds for all fuctios i L, thus completig the proof i the fiite measure case. We give the proof of the corollary of Birkhoff s theorem, Corollary 2.2. Proof. Let (µ, T ) be ergodic, let f L (µ), ad let f be the limit of the ergodic averages. We kow that f is T -ivariat, so f is almost everywhere costat. If µ() =, the sice f is itegrable, f 0 almost everywhere. If µ() <, the we have: f dµ = fµ() = f dµ, so that as desired. f(x) = f dµ, µ() Fially, we show that f is the coditioal expectatio of f with respect to the σ-algebra of T -ivariat measurable sets, T. Proof. Let A T ad let f L (µ). The we have A = T (A) = A T, ad A f L (µ), so we have: A f(x) = lim A (T i (x))f(t i (x)) = lim f(t i (x)) A (x) = f(x) A (x). 8

9 Hece, we obtai: A f dµ = f A dµ = A f dµ = f A dµ = ad so by a theorem about coditioal expectatios, f is the coditioal expectatio of f with respect to T (equality of itegrals over ay set i the σ-algebra is sufficiet ad ecessary). A f dµ, 9