BIRKHOFF ERGODIC THEOREM Abstract. We will give a proof of the poitwise ergodic theorem, which was first proved by Birkhoff. May improvemets have bee made sice Birkhoff s orgial proof. The versio we give here is due to Keae ad Peterse, which builds o the Kamae s o-stadard aalysis proof. 1. Itroductio Theorem 1 (Birkhoff). Let (Ω, F, µ) be a proability space, ad let T : Ω Ω be a measure-preservig map. Let I be the sigma-field of ivariat evets; that is, all sets A F for which µ(a T 1 A) = 0. For all f L 1, we have that 1 f T i E(f I), where the covergece is i L 1 ad almost surely. Exercise 2. Show that Theorem 1 implies the strog law of large umbers. Exercise 3. Let X be a irreducible ad aperiodic Markov chai o a fiite state space S, that is started at the statioary distributio. Let s, t S. What is the limit of 1 1[X k = s, X k+1 = t]? k=0 Exercise 4. Cosider the case Ω = {0, 1} Z ad T (ω) i = ω i+1. Let F be the usual product sigma-algebra. Let C 0 = {ω : Ω : ω 0 = 1}. Let µ ad ν be product measures o Ω for which µ(c 0 ) = 2/3 ad ν(c 0 ) = 1/3. Cosider ρ := 3/4µ + 1/4ν, ad the probability space (Ω, F, ρ), with the map T. (a) Show that T preserves the measure ρ. (b) Let I be a set of all evets A I are such that ρ(a T 1 A) = 0. Show that every evet i I has measure i {0, 1, 3/4, 1/4}. (c) Fid a set of four evets A = {A 1, A 2, A 3, A 4 } so that for all B I there exists A i A such that ρ(a i B) = 0.
2. Prelimiaries Let (Ω, F, µ) be a proability space, ad let T : Ω Ω be a measurepreservig map. Let f L 1. Set A f = 1 1 f T k, f N = k=0 sup 1 N A f, f = sup fn, N A(f) = A = lim sup A f ad A = lim if A f. Most of the work is cotaied i the followig theorem, which we will prove i the ext sectio. Theorem 5 (Maximal ergodic theorem). Let g : Ω R be a ivariat fuctio ad g + L 1. Set B = {f > g}, the E(f g)1 B 0. Corollary 6. A f coverges almost surely. We will also prove Corollary 6 i the ext sectio. Let F := lim A f. Exercise 7. Use Fatou s lemma to obtai that F L 1. Exercise 8 (Weak L 1 iequality). Let α > 0. Show that αµ(f > α) f L 1. The previous iequality is elemetary, the later is ot. Show that αµ(f > α) f L 1. I fact Exercise 8 is basically eough to prove Theorem 1. I what follows, we will prove L 1 covergece, assumig Corollary 6. Exercise 9. I the case that f L 2, show that F = E(f I). Exercise 10. Show that i the case that h 0, ad bouded, we have A k h E(h I) i L 1. Lemma 11. A f E(f I) i L 1. Proof. By takig positive ad egative parts, it suffices to cosider the case f 0. Let 0 h be bouded. The triagle iequality gives A f E(f I) 1 = A f A h + E(h I) E(f I) + A h E(h I) 1 A f A h 1 + A h E(h I) 1 + E(h I) E(f I) 1
For the first term, ote that A f A h 1 = A (f h) 1 f h 1. For the third term, recall that coditioal expecatio is a cotractio, so that E(h I) E(f I) 1 f h 1. Recall that there exists a sequece of oegative bouded fuctios with h m f almost surely. Just use g m = mi(m, f). The domiated covergece theorem, also implies that h m f i L 1. Hece result follows from Exercise 10. 3. Proof of the Maximal iequality Proof of Theorem 5. We will cosider the case f C, so that f is bouded. Sice g + L 1, we have that if g1 B L 1, the E(f g)1 B =, ad we are doe. So we may assume that g1 B L 1 ; this implies that g L 1 ; do Exercise 12. Fix a iteger N > 0, ad let B N = {fn > g}. The orbit of ω Ω is the sequece of poits ω, T ω, T 2 ω,.... We wat to partitio orbit of a poit ω Ω with respect to the set B N. Fix ω Ω. Let b 1 0, be the first time for which T b 1 ω B N. Call the iterval [0, b 1 ) the first red stretch; it may be empty. We kow that for some 1 r 2 N that f N(T b 1 ω) = A r2 f(t b 1 (ω)) g(t b 1 (ω)) = g(ω); take the largest such r 2. Thus r 2 1 (f g) T b1+i (ω) 0. Call the iterval [b 1, r 2 ) the first blue stretch. Similarly, defie the secod red stretch, to be [r 2, b 2 ), where b 2 is the first time b r 2 for which T b ω B N, which may be empty. Similarly, defie the secod blue stretch. Cotiue doig this alog the orbit of ω. Notice that you ca have two cosective blue stretches, but we wat to cosider them separate; i geeral we have coloured the itegers red ad blue as follows: [r 1 = 0, b 1 ) [b 1, r 2 ) [r 2, b 2 ) [b 2, r 3 )... ; each iterval of the form [r i, b i ) is red (ad may be empty), ad each iterval of the form [b i, r i+1 ) is blue ad cotais at least oe iteger ad at most N itegers. Note that (f g)1 BN f g,
sice if fn g, the f g. Let m be much larger tha N. Let h = (f g)1 BN. We have by our colourig ad the ivariace of g that m h(t i ω) N(C + g + (ω)); do Exercise 13. Take expecatios o both sides to obtai: me(f g)1 BN N(C + Eg + ). Hece dividig by m ad takig a limit, we have that from which it follows that E(f g)1 BN 0, E(f g)1 B 0; do Exercise 14. We ca exted to all f L 1 by takig limits: do Exercise 15. The followig exercises refer to details i the proof of Theorem 5. Exercise 12. Show that if g1 B L 1, the g L 1. Exercise 13. Show that if f C, the m h(t i ω) N(C + g). Exercise 14. Show that E(f g)1 BN 0 for all N > 0 implies that E(f g)1 B 0 Exercise 15. Usig that you kow about the bouded case, fiish the proof the case of geeral f L 1. Proof of Corollary 6. It suffices to show that EA Ef, sice by cosiderig f, this implies that Ef EA. By defiitio, A A, so we have that EA = EA, from which we deduce that A = A as required. We eed to first argue that A L 1. Cosider f 0. Check that g = mi { A(f), } 1/ is ivariat. Also clearly, g L 1. So {f > g } = B = Ω. Applyig Theorem 5, we obtai that Ef Eg EA. For geeral f, we have that (Af) + A(f + ), so that A L 1. Now that we kow that A L 1, we ca apply Theorem 5, with g = A 1/, to obtai Ef Eg EA.
Refereces [1] M. Keae ad K. Peterse. Easy ad early simultaeous proofs of the ergodic theorem ad maximal ergodic theorem. I Dyamics & stochastics, volume 48 of IMS Lecture Notes Moogr. Ser., pages 248 251. Ist. Math. Statist., Beachwood, OH, 2006. [2] K. Peterse. Ergodic theory, volume 2 of Cambridge Studies i Advaced Mathematics. Cambridge Uiversity Press, Cambridge, 1989. Corrected reprit of the 1983 origial.