Itroductio to Ergodic Theory ad its Applicatios to Number Theory Karma Dajai October 8, 204
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Cotets Itroductio ad prelimiaries 5. What is Ergodic Theory?..................... 5.2 Measure Preservig Trasformatios.............. 6.3 Basic Examples.......................... 0.4 Recurrece............................. 6.5 Iduced Trasformatios..................... 7.6 Ergodicity............................. 20.7 Other Characterizatios of Ergodicity.............. 22.8 Examples of Ergodic Trasformatios.............. 24 2 Ergodic Theorems 33 2. The Poitwise Ergodic Theorem ad its cosequeces..... 33 2.2 The Mea Ergodic Theorem................... 46 2.3 Mixig............................... 49 3 Measure Preservig Isomorphisms ad Factor Maps 53 3. Measure Preservig Isomorphisms................ 53 3.2 Factor Maps............................ 58 3.3 Natural Extesios........................ 59 4 Cotiued Fractios 63 4. Itroductio ad Basic Properties................ 63 4.2 Ergodic Properties of Cotiued Fractio Map......... 68 4.3 Natural Extesio ad the Doebli-Lestra Cojecture.... 72 5 Etropy 77 5. Radomess ad Iformatio.................. 77 5.2 Defiitios ad Properties.................... 78 3
4 5.3 Calculatio of Etropy ad Examples.............. 85 5.4 The Shao-McMilla-Breima Theorem........... 89 5.5 Lochs Theorem.......................... 96 6 Ivariat Measures for Cotiuous Trasformatios 0 6. Existece............................. 0 6.2 Uique Ergodicity ad Uiform Distributio.......... 08 Bibliography 3
Chapter Itroductio ad prelimiaries. What is Ergodic Theory? It is ot easy to give a simple defiitio of Ergodic Theory because it uses techiques ad examples from may fields such as probability theory, statistical mechaics, umber theory, vector fields o maifolds, group actios of homogeeous spaces ad may more. The word ergodic is a mixture of two Greek words: ergo (work) ad odos (path). The word was itroduced by Boltzma (i statistical mechaics) regardig his hypothesis: for large systems of iteractig particles i equilibrium, the time average alog a sigle trajectory equals the space average. The hypothesis as it was stated was false, ad the ivestigatio for the coditios uder which these two quatities are equal lead to the birth of ergodic theory as is kow owadays. A moder descriptio of what ergodic theory is would be: it is the study of the asymptotic average behavior of systems evolvig i time. The collectio of all states of the system form a space, ad the evolutio is represeted by either a trasformatio T :, where T x is the state of the system at time t =, whe the system (i.e., at time t = 0) was iitially i state x. (This is aalogous to the setup of discrete time stochastic processes). if the evolutio is cotiuous or has a spacial structure, the we describe the evolutio by lookig at a group of trasformatios G (like Z 2, R, R 2 ) actig o, i.e., every g G is idetified with a trasformatio T g :, ad T gg = T g T g. 5
6 Itroductio ad prelimiaries The space usually has a special structure, ad we wat T to preserve the basic structure o. For example if is a measure space, the T must be measurable. if is a topological space, the T must be cotiuous. if has a differetiable structure, the T is a diffeomorphism. I this course our space is a probability space (, B, µ), ad our time is discrete. So the evolutio is described by a measurable map T :, so that T A B for all A B. For each x, the orbit of x is the sequece x, T x, T 2 x,.... If T is ivertible, the oe speaks of the two sided orbit..., T x, x, T x,.... We wat also that the evolutio is i steady state i.e. statioary. I the laguage of ergodic theory, we wat T to be measure preservig..2 Measure Preservig Trasformatios Defiitio.2. Let (, B, µ) be a probability space, ad T : measurable. The map T is said to be measure preservig with respect to µ if µ(t A) = µ(a) for all A B. This defiitio implies that for ay measurable fuctio f : R, the process f, f T, f T 2,... is statioary. This meas that for all Borel sets B,..., B, ad all itegers r < r 2 <... < r, oe has for ay k, µ ({x : f(t r x) B,... f(t r x) B }) = µ ( {x : f(t r +k x) B,... f(t r+k x) B } ). We say T is ivertible if it is oe-to-oe ad if T is measurable. Note that i the case T is ivertible, the T is measure preservig if ad oly if µ(t A) = µ(a) for all A B. We ca geeralize the defiitio of measure preservig to the followig case. Let T : (, B, µ ) ( 2, B 2, µ 2 ) be measurable, the T is measure preservig if µ (T A) = µ 2 (A) for all A B 2.
Measure Preservig Trasformatios 7 The followig gives a useful tool for verifyig that a trasformatio is measure preservig. For this we eed the otios of algebra ad semi-algebra. Recall that a collectio S of subsets of is said to be a semi-algebra if (i) S, (ii) A B S wheever A, B S, ad (iii) if A S, the \A = i=e i is a disjoit uio of elemets of S. For example if = [0, ), ad S is the collectio of all subitervals, the S is a semi-algebra. Or if = {0, } Z, the the collectio of all cylider sets {x : x i = a i,..., x j = a j } is a semi-algebra. A algebra A is a collectio of subsets of satisfyig:(i) A, (ii) if A, B A, the A B A, ad fially (iii) if A A, the \ A A. Clearly a algebra is a semi-algebra. Furthermore, give a semi-algebra S oe ca form a algebra by takig all fiite disjoit uios of elemets of S. We deote this algebra by A(S), ad we call it the algebra geerated by S. It is i fact the smallest algebra cotaiig S. Likewise, give a semi-algebra S (or a algebra A), the σ-algebra geerated by S (A) is deoted by σ(s) (σ(a)), ad is the smallest σ-algebra cotaiig S (or A). A mootoe class C is a collectio of subsets of with the followig two properties if E E 2... are elemets of C, the i=e i C, if F F 2... are elemets of C, the i=f i C. The mootoe class geerated by a collectio S of subsets of is the smallest mootoe class cotaiig S. Theorem.2. Let A be a algebra of, the the σ-algebra σ(a) geerated by A equals the mootoe class geerated by A. Usig the above theorem, oe ca get a easier criterio for checkig that a trasformatio is measure preservig. Theorem.2.2 Let ( i, B i, µ i ) be probability spaces, i =, 2, ad T : 2 a trasformatio. Suppose S 2 is a geeratig semi-algebra of B 2. The, T is measurable ad measure preservig if ad oly if for each A S 2, we have T A B ad µ (T A) = µ 2 (A). Proof. Let C = {B B 2 : T B B, ad µ (T B) = µ 2 (B)},
8 Itroductio ad prelimiaries the S 2 C B 2, ad hece A(S 2 ) C. We show that C is a mootoe class. Let E E 2... be elemets of C, ad let E = i=e i. The, T E = i=t E i B, ad µ (T E) = µ ( =T E ) = lim µ (T E ) = lim µ 2 (E ) = µ 2 ( =E ) = µ 2 (E). Thus, E C. A similar proof shows that if F F 2... are elemets of C, the i=f i C. Hece, C is a mootoe class cotaiig the algebra A(S 2 ). By the mootoe class theorem, B 2 is the smallest mootoe class cotaiig A(S 2 ), hece B 2 C. This shows that B 2 = C, therefore T is measurable ad measure preservig. For example if = [0, ) with the Borel σ-algebra B, ad µ a probability measure o B. The a trasformatio T : is measurable ad measure preservig if ad oly if T [a, b) B ad µ (T [a, b)) = µ ([a, b)) for ay iterval [a, b). = {0, } N with product σ-algebra ad product measure µ. A trasformatio T : is measurable ad measure preservig if ad oly if T ({x : x 0 = a 0,..., x = a }) B, ad µ ( T {x : x 0 = a 0,..., x = a } ) = µ ({x : x 0 = a 0,..., x = a }) for ay cylider set. Exercise.2. Recall that if A ad B are measurable sets, the A B = (A B) \ (A B) = (A \ B) (B \ A). Show that for ay measurable sets A, B, C oe has µ(a B) µ(a C) + µ(c B). Aother useful lemma is the followig; see also [KT].
Measure Preservig Trasformatios 9 Lemma.2. Let (, B, µ) be a probability space, ad A a algebra geeratig B. The, for ay A B ad ay ɛ > 0, there exists C A such that µ(a C) < ɛ. Proof. First ote that sice A, the D. Now let A D ad ε > 0. There exists C A such that µ(a C) < ε. Sice C c A ad A C = A c C c, we have µ(a c C c ) < ε ad hece A c D. Fially, suppose (A ) D ad ε > 0. For each, there exists C A such that µ(a C ) < ε/2 +. It is easy to check that so that = A C = = ( ) µ A C = (A C ), = µ(a C ) < ε/2. Sice A is closed uder fiite uios we do ot kow at this poit if = C is a elemet of A. To solve this problem, we proceed as follows. First ote that m = Cc = Cc, hece ( ) ( ) µ A C c = lim µ m A C c, m ad therefore, ( µ A = = = = = = C ) = lim m µ = = ( ( A = = = = = ) m C) c ( A c C ). Hece there exists m sufficietly large so that ( ) ( m ) µ ( A C) c ( A c C ) < µ A C + ε/2. = = Sice = Ac m = C = Ac = C, we get ( ) ( m m µ ( A C) c ( A c C ) < µ A = = = = = = = ) C + ε/2. =
0 Itroductio ad prelimiaries Thus, ( ) m µ ( A C ) < ε, = = ad m = C A sice A is closed uder fiite uios. This shows that = A D. Thus, D is a σ-algebra. By defiitio of D we have D B. Sice A D, ad B is the smallest σ-algebra cotaiig A we have B D. Therefore, B = D..3 Basic Examples Example.3. (Traslatios) Let = [0, ) with the Lebesgue σ-algebra B, ad Lebesgue measure λ. Let 0 < θ <, defie T : by T x = x + θ mod = x + θ x + θ. The, by cosiderig itervals it is easy to see that T is measurable ad measure preservig. Example.3.2 (Multiplicatio by 2 modulo ) Let (, B, λ) be as i Example (a), ad let T : be give by { 2x 0 x < /2 T x = 2x mod = 2x /2 x <. For ay iterval [a, b), ad T [a, b) = [ a 2, b ) [ a + 2 2, b + ), 2 λ ( T [a, b) ) = b a = λ ([a, b)). Although this map is very simple, it has i fact may facets. For example, iteratios of this map yield the biary expasio of poits i [0, ) i.e., usig T oe ca associate with each poit i [0, ) a ifiite sequece of 0 s ad s. To do so, we defie the fuctio a by { 0 if 0 x < /2 a (x) = if /2 x <,
Basic Examples the T x = 2x a (x). Now, for set a (x) = a (T x). Fix x, for simplicity, we write a istead of a (x), the T x = 2x a. Rewritig we get x = a 2 + T x. Similarly, T x = a 2 2 2 + T 2 x. Cotiuig i this maer, 2 we see that for each, x = a 2 + a 2 2 2 + + a 2 + T x 2. Sice 0 < T x <, we get a i x 2 = T x 0 as. i 2 i= Thus, x = a i i=. We shall later see that the sequece of digits a 2 i, a 2,... forms a i.i.d. sequece of Beroulli radom variables. Exercise.3. (Baker s Trasformatio) Cosider the probability space [0, ) 2, B B, λ λ), where B B is the product Borel σ-algebra ad λ λ the product Lebesgue measure. Defie T : [0, ) 2 [0, ) 2 by { (2x, y 2 T (x, y) = ) 0 x < /2 (2x, y+ ) /2 x <. 2 Show that T is ivertible, measurable ad measure preservig. Example.3.3 (Lüroth Series) Aother kid of series expasio, itroduced by J. Lüroth [L] i 883, motivates this approach. Several authors have studied the dyamics of such systems. Take as partitio of [0, ) the itervals [, ) where N. Every umber x [0, ) ca be writte as a + fiite or ifiite series, the so-called Lüroth (series) expasio x = a (x) + a (x)(a (x) )a 2 (x) + + a (x)(a (x) ) a (x)(a (x) )a (x) + ; here a k (x) 2 for each k. How is such a series geerated? Let T : [0, ) [0, ) be defied by ( + )x, x [, ), + T x = 0, x = 0. (.)
2 Itroductio ad prelimiaries.... (a = 2) 0 6 5 4 3 Figure.: The Lüroth Series map T. 2 Let x 0, for k ad T k x 0 we defie the digits a = a (x) by a k (x) = a (T k x), where a (x) = if x [, ), 2. Now (.) ca be writte as T x = a (x)(a (x) )x (a (x) ), x 0, 0, x = 0.
Basic Examples 3 Thus, for ay x (0, ) such that T k x 0, we have ( ) x = a + T x = a (a ) a + a (a ) a 2 + T 2 x a 2 (a 2 ) = a + + T 2 x a (a )a 2 a (a )a 2 (a 2 ). = a + + a (a ) a k (a k )a k + T k x a (a ) a k (a k ). Notice that, if T k x = 0 for some k, ad if we assume that k is the smallest positive iteger with this property, the x = a + + I case T k x 0 for all k, oe gets x = a + a (a ) a k (a k )a k. + + +, a (a )a 2 a (a ) a k (a k )a k where a k 2 for each k. The above ifiite series ideed coverges to x. To see this, let S k = S k (x) be the sum of the first k terms of the sum. The x S k = T k x a (a ) a k (a k ) ; sice T k x [0, ) ad a k 2 for all x ad all k, we fid x S k 2 k 0 as k. From the above we also see that if x ad y have the same Lüroth expasio, the, for each k, x y 2 k ad it follows that x equals y. For ease of otatio we drop the argumet x from the fuctios a k (x).
4 Itroductio ad prelimiaries Exercise.3.2 Show that the map T give i Example.3.3 is measure preservig with respect to Lebesgue measure λ. Example.3.4 (β-trasformatio) Let = [0, ) with the Lebesgue σ- algebra B. Let β = + 5, the golde mea. Notice that β 2 = β +. Defie 2 a trasformatio T : by { βx 0 x < /β T x = βx mod = βx /β x <. The, T is ot measure preservig with respect to Lebesgue measure (give a couterexample), but is measure preservig with respect to the measure µ give by µ(b) = g(x) dx, where B 5 + 3 5 g(x) = 0 5 + 5 0 0 x < /β /β x <. Exercise.3.3 Verify that T is measure preservig with respect to µ, ad show that (similar to Example.3.2) iteratios of this map geerate expasios for poits x [0, ) (kow as β-expasios) of the form x = i= b i β i, where b i {0, } ad b i b i+ = 0 for all i. Example.3.5 (Cotiued Fractios) Cosider ([0, ), B), where B is the Lebesgue σ-algebra. Defie a trasformatio T : [0, ) [0, ) by T 0 = 0 ad for x 0 T x = x x. Exercise.3.4 Show that T is ot measure preservig with respect to Lebesgue measure, but is measure preservig with respect to the so called Gauss probability measure µ give by µ(b) = log 2 + x dx. B
Basic Examples 5 A iterestig feature of this map is that its iteratios geerate the cotiued fractio expasio for poits i (0, ). For if we defie { if x ( 2 a = a (x) =, ) if x (, ], 2, + the, T x = a x ad hece x = a + T x. For, let a = a (x) = a (T x). The, after iteratios we see that x = a + T x =... =. a + a 2 +... + a + T x I fact, i Chapter 4 we will show that if p =, the the q a + a 2 +... + a sequece {q } is mootoically icreasig, ad x p < 0 as. q 2 q The last statemet implies that x = a + a 2 + a 3 +... Example.3.6 (Beroulli Shifts) Let = {0,,... k } Z (or = {0,,... k } N ), F the σ-algebra geerated by the cyliders. Let p = (p 0, p,..., p k ) be a positive probability vector, defie a measure µ o F by specifyig it o the cylider sets as follows µ ({x : x = a,..., x = a }) = p a... p a. Let T : be defied by T x = y, where y = x +. The map T, called the left shift, is measurable ad measure preservig, sice T {x : x = a,... x = a } = {x : x + = a,..., x + = a },.
6 Itroductio ad prelimiaries ad µ ({x : x + = a,..., x + = a }) = p a... p a. Notice that i case = {0,,... k } N, the oe should cosider cylider sets of the form {x : x 0 = a 0,... x = a }. I this case T {x : x 0 = a 0,..., x = a } = k j=0 {x : x 0 = j, x = a 0,..., x + = a }, ad it is easy to see that T is measurable ad measure preservig. Example.3.7 (Markov Shifts) Let (, F, T ) be as i example (e). We defie a measure ν o F as follows. Let P = (p ij ) be a stochastic k k matrix, ad q = (q 0, q,..., q k ) a positive probability vector such that qp = q. Defie ν o cyliders by ν ({x : x = a,... x = a }) = q a p a a +... p a a. Just as i example.3.6, oe sees that T is measurable ad measure preservig..4 Recurrece Let T be a measure preservig trasformatio o a probability space (, F, µ), ad let B F. A poit x B is said to be B-recurret if there exists k such that T k x B. Theorem.4. (Poicaré Recurrece Theorem) If µ(b) > 0, the a.e. x B is B-recurret. Proof. Let F be the subset of B cosistig of all elemets that are ot B-recurret. The, F = {x B : T k x / B for all k }. We wat to show that µ(f ) = 0. First otice that F T k F = for all k, hece T l F T m F = for all l m. Thus, the sets F, T F,... are pairwise disjoit, ad µ(t F ) = µ(f ) for all (T is measure preservig). If µ(f ) > 0, the = µ() µ ( k 0 T k F ) = k 0 µ(f ) =,
Iduced Trasformatios 7 a cotradictio. The proof of the above theorem implies that almost every x B returs to B ifiitely ofte. I other words, there exist ifiitely may itegers < 2 <... such that T i x B. To see this, let The, D = {x B : T k x B for fiitely may k }. D = {x B : T k x F for some k 0} k=0t k F. Thus, µ(d) = 0 sice µ(f ) = 0 ad T is measure preservig..5 Iduced Trasformatios Let T be a measure preservig trasformatio o the probability space (, F, µ). Let A with µ(a) > 0. By Poicaré s Recurrece Theorem almost every x A returs to A ifiitely ofte uder the actio of T. For x A, let (x) := if{ : T x A}. We call (x) the first retur time of x to A. Exercise.5. Show that is measurable with respect to the σ-algebra F A o A. By Poicaré Theorem, (x) is fiite a.e. o A. I the sequel we remove from A the set of measure zero o which (x) =, ad we deote the ew set agai by A. Cosider the σ-algebra F A o A, which is the restrictio of F to A. Furthermore, let µ A be the probability measure o A, defied by µ A (B) = µ(b), for B F A, µ(a) so that (A, F A, µ A ) is a probability space. Fially, defie the iduced map T A : A A by T A x = T (x) x, for x A. From the above we see that T A is defied o A. What kid of a trasformatio is T A? Exercise.5.2 Show that T A is measurable with respect to the σ-algebra F A.
8 Itroductio ad prelimiaries Propositio.5. T A is measure preservig with respect to µ A. Proof. For k, let A k = {x A : (x) = k} B k = {x \ A : T x,..., T k x A, T k x A}. Notice that A = k= A k, ad T A = A B ad T B = A + B +. (.2) Let C F A, sice T is measure preservig it follows that µ(c) = µ(t C). B B B 2 B B 2 B 3 B B 2 B 3 B 4...... A A 2 A 3 A 4 A 5............... T 2 A \ A T A (A, 3) T A \ A (A, 2) A (A, ) Figure.2: A tower. To show that µ A (C) = µ A (T A C), we show that µ(t A C) = µ(t C). Now, hece T A (C) = A k T A C = A k T k C, k= k= µ ( T A (C)) = µ ( A k T k C ). k=
Iduced Trasformatios 9 O the other had, usig repeatedly (.2), oe gets for ay, Sice µ ( T (C) ) = µ(a T C) + µ(b T C) it follows that Thus, = µ(a T C) + µ(t (B T C)) = µ(a T C) + µ(a 2 T 2 C) + µ(b 2 T 2 C). = µ(a k T k C) + µ(b T C). k= ( ) µ B T C = = µ(c) = µ(t C) = µ(b T C), = lim µ(b T C) = 0. k= µ ( A k T k C ) = µ(t A C). This shows that µ A (C) = µ A (T A C), which implies that T A is measure preservig with respect to µ A. Exercise.5.3 Assume T is ivertible. Without usig Propositio.5. show that for all C F A, µ A (C) = µ A (T A C). Exercise.5.4 Let G = + 5, so that G 2 = G +. Cosider the set 2 = [0, G ) [0, ) [ G, ) [0, G ), edowed with the product Borel σ-algebra, ad the ormalized Lebesgue measure λ λ. Defie the trasformatio (Gx, y G ), (x, y) [0, ) [0, ] G T (x, y) = (Gx, + y G ), (x, y) [, ) [0, ). G G
20 Itroductio ad prelimiaries (a) Show that T is measure preservig with respect to λ λ. (b) Determie explicitely the iduced trasformatio of T o the set [0, ) [0, G ). Exercise.5.5 Let θ (0, ) be irratioal. Cosider the probability space ([0, ), B, λ), where B is the Borel σ-algebra ad λ is Lebesgue measure restricted to [0, ). Let T : [0, ) [0, ) be traslatio by θ (0, ), i.e. T x = x + θ mod. Determie explicitly the iduced trasformatio T A of T o the iterval A = [0, θ)..6 Ergodicity Defiitio.6. Let T be a measure preservig trasformatio o a probability space (, F, µ). The map T is said to be ergodic if for every measurable set A satisfyig T A = A, we have µ(a) = 0 or. Theorem.6. Let (, F, µ) be a probability space ad T : measure preservig. The followig are equivalet: (i) T is ergodic. (ii) If B F with µ(t B B) = 0, the µ(b) = 0 or. (iii) If A F with µ(a) > 0, the µ ( =T A) =. (iv) If A, B F with µ(a) > 0 ad µ(b) > 0, the there exists > 0 such that µ(t A B) > 0. Remark.6.. I case T is ivertible, the i the above characterizatio oe ca replace T by T. 2. Note that if µ(b T B) = 0, the µ(b \ T B) = µ(t B \ B) = 0. Sice B = ( B \ T B ) ( B T B ), ad T B = ( T B \ B ) ( B T B ),
Ergodicity 2 we see that after removig a set of measure 0 from B ad a set of measure 0 from T B, the remaiig parts are equal. I this case we say that B equals T B modulo sets of measure 0. 3. I words, (iii) says that if A is a set of positive measure, almost every x evetually (i fact ifiitely ofte) will visit A. 4. (iv) says that elemets of B will evetually eter A. Proof of Theorem.6. (i) (ii) Let B F be such that µ(b T B) = 0. We shall defie a measurable set C with C = T C ad µ(c B) = 0. Let C = {x : T x B i.o. } = = k= T k B. The, T C = C, hece by (i) µ(c) = 0 or. Furthermore, ( ) ( ) µ(c B) = µ T k B B c + µ T k B c B = k= = k= ( ) ( ) µ T k B B c + µ T k B c B k= µ ( T k B B ). k= Usig iductio (ad the fact that µ(e F ) µ(e G) +µ(g F )), oe ca show that for each k oe has µ ( T k B B ) = 0. Hece, µ(c B) = 0 which implies that µ(c) = µ(b). Therefore, µ(b) = 0 or. (ii) (iii) Let µ(a) > 0 ad let B = = T A. The T B B. Sice T is measure preservig, the µ(b) > 0 ad k= µ(t B B) = µ(b \ T B) = µ(b) µ(t B) = 0. Thus, by (ii) µ(b) =. (iii) (iv) Suppose µ(a)µ(b) > 0. By (iii) ( ) ( ) µ(b) = µ B T A = µ (B T A) > 0. = =
22 Itroductio ad prelimiaries Hece, there exists k such that µ(b T k A) > 0. (iv) (i) Suppose T A = A with µ(a) > 0. If µ(a c ) > 0, the by (iv) there exists k such that µ(a c T k A) > 0. Sice T k A = A, it follows that µ(a c A) > 0, a cotradictio. Hece, µ(a) = ad T is ergodic..7 Other Characterizatios of Ergodicity We deote by L 0 (, F, µ) the space of all complex valued measurable fuctios o the probability space (, F, µ). Let L p (, F, µ) = {f L 0 (, F, µ) : f p dµ(x) < }. We use the subscript R wheever we are dealig oly with real-valued fuctios. Let ( i, F i, µ i ), i =, 2 be two probability spaces, ad T : 2 a measure preservig trasformatio i.e., µ 2 (A) = µ (T A). Defie the iduced operator U T : L 0 ( 2, F 2, µ 2 ) L 0 (, F, µ ) by U T f = f T. The followig properties of U T are easy to prove. Propositio.7. The operator U T has the followig properties: (i) U T is liear (ii) U T (fg) = U T (f)u T (g) (iii) U T c = c for ay costat c. (iv) U T is a positive liear operator (v) U T B = B T = T B for all B F 2. (vi) U T f dµ = 2 f dµ 2 for all f L 0 ( 2, F 2, µ 2 ), (where if oe side does t exist or is ifiite, the the other side has the same property). (vii) Let p. The, U T L p ( 2, F 2, µ 2 ) L p (, F, µ ), ad U T f p = f p for all f L p ( 2, F 2, µ 2 ).
Other Characterizatios of Ergodicity 23 Exercise.7. Prove Propositio.7.. Exercise.7.2 Let (, F, µ) be a probability space, ad T : a measure preservig trasformatio. Let f L (, F, µ). Show that if f(t x) f(x) µ a.e., the f(x) = f(t x) µ a.e. Exercise.7.3 Let (, F, µ) be a probability space, ad T : a measure preservig ad ergodic trasformatio. Suppose f 0 is a measurable fuctio. Show that if the set A = {x : f(x) > 0} has positive µ measure, the for µ a.e. x, oe has f(t x) =. = I the followig theorem, we give a ew characterizatio of ergodicity Theorem.7. Let (, F, µ) be a probability space, ad T : measure preservig. The followig are equivalet: (i) T is ergodic. (ii) If f L 0 (, F, µ), with f(t x) = f(x) for all x, the f is a costat a.e. (iii) If f L 0 (, F, µ), with f(t x) = f(x) for a.e. x, the f is a costat a.e. (iv) If f L 2 (, F, µ), with f(t x) = f(x) for all x, the f is a costat a.e. (v) If f L 2 (, F, µ), with f(t x) = f(x) for a.e. x, the f is a costat a.e. Proof. The implicatios (iii) (ii), (ii) (iv), (v) (iv), ad (iii) (v) are all clear. It remais to show (i) (iii) ad (iv) (i).
24 Itroductio ad prelimiaries (i) (iii) Suppose f(t x) = f(x) a.e. ad assume without ay loss of geerality that f is real (otherwise we cosider separately the real ad imagiary parts of f). For each ad k Z, let I k, = {x : k 2 f(x) < k + 2 }. The, T I k, I k, {x : f(t x) f(x)} which implies that µ ( T I k, I k, ) ) = 0. By ergodicity of T, µ(i k, ) = 0 or, for each k Z. O the other had, for each, we have = k Z I k, (disjoit uio). Hece, for each, there exists a uique iteger k such that µ (I k,) =. I fact, I k, I k2,2..., ad { k } is a bouded icreasig sequece, hece 2 lim k /2 exists. Let Y = I k,, the µ(y ) =. Now, if x Y, the 0 f(x) k /2 < /2 for all. Hece, f(x) = lim k /2, ad f is a costat o Y. (iv) (i) Suppose T A = A ad µ(a) > 0. We wat to show that µ(a) =. Cosider A, the idicator fuctio of A. We have A L 2 (, F, µ), ad A T = T A = A. Hece, by (iv), A is a costat a.e., hece A = a.e. ad therefore µ(a) =..8 Examples of Ergodic Trasformatios Example.8. (Irratioal Rotatios) Cosider ([0, ), B, λ), where B is the Lebesgue σ-algebra, ad λ Lebesgue measure. For θ (0, ), cosider the trasformatio T θ : [0, ) [0, ) defied by T θ x = x + θ (mod ). We have see i Example.3. that T θ is measure preservig with respect λ. Whe is T θ ergodic? If θ is ratioal, the T θ is ot ergodic. Cosider for example θ = /4, the the set A = [0, /8) [/4, 3/8) [/2, 5/8) [3/4, 7/8) is T θ -ivariat but µ(a) = /2.
Examples of Ergodic Trasformatios 25 Exercise.8. Suppose θ = p/q with gcd(p, q) =. Fid a o-trivial T θ -ivariat set. Coclude that T θ is ot ergodic if θ is a ratioal. Claim. T θ is ergodic if ad oly if θ is irratioal. Proof of Claim. ( ) The cotrapositive statemet is give i Exercise.8. i.e., if θ is ratioal, the T θ is ot ergodic. ( ) Suppose θ is irratioal, ad let f L 2 (, B, λ) be T θ -ivariat. Write f i its Fourier series f(x) = Z a e 2πix. Sice f(t θ x) = f(x), the f(t θ x) = Z a e 2πi(x+θ) = Z = f(x) = Z a e 2πix. a e 2πiθ e 2πix Hece, Z a ( e 2πiθ )e 2πix = 0. By the uiqueess of the Fourier coefficiets, we have a ( e 2πiθ ) = 0, for all Z. If 0, sice θ is irratioal we have e 2πiθ 0. Thus, a = 0 for all 0, ad therefore f(x) = a 0 is a costat. By Theorem.7., T θ is ergodic. Exercise.8.2 Cosider the probability space ([0, ), B B, λ λ), where as above B is the Lebesgue σ-algebra o [0, ), ad λ ormalized Lebesgue measure. Suppose θ (0, ) is irratioal, ad defie T θ T θ : [0, ) [0, ) [0, ) [0, ) by T θ T θ (x, y) = (x + θ mod(), y + θ mod()). Show that T θ T θ is measure preservig, but is ot ergodic. Exercise.8.3 Cosider the probability space ([0, ] [0, ], B B, λ λ), where B B is the two-dimesioal Borel σ-algebra ad λ λ is the twodimesioal Lebesgue measure restricted to [0, ] [0, ]. Prove that the trasformatio S : [0, ] [0, ] [0, ] [0, ] give by S(x, y) = (x + θ mod, x + y mod ) with θ irratioal is measure preservig ad ergodic with respect to λ λ. (Hit: The Fourier series,m c,me 2πi(x+my) of a fuctio f L 2 ([0, ] [0, ], B B, λ λ) satisfies,m c,m 2 <.
26 Itroductio ad prelimiaries Example.8.2 (Oe (or two) sided shift) Let = {0,,... k } N, F the σ-algebra geerated by the cyliders, ad µ the product measure defied o cylider sets by µ ({x : x 0 = a 0,... x = a }) = p a0... p a, where p = (p 0, p,..., p k ) is a positive probability vector. Cosider the left shift T defied o by T x = y, where y = x + (See Example (e) i Subsectio.3). We show that T is ergodic. Let E be a measurable subset of which is T -ivariat i.e., T E = E. For ay ɛ > 0, by Lemma.2. (see subsectio.2), there exists A F which is a fiite disjoit uio of cyliders such that µ(e A) < ɛ. The µ(e) µ(a) = µ(e \ A) µ(a \ E) µ(e \ A) + µ(a \ E) = µ(e A) < ɛ. Sice A depeds o fiitely may coordiates oly, there exists 0 > 0 such that T 0 A depeds o differet coordiates tha A. Sice µ is a product measure, we have Further, µ(a T 0 A) = µ(a)µ(t 0 A) = µ(a) 2. µ(e T 0 A) = µ(t 0 E T 0 A) = µ(e A) < ɛ, ad Hece, µ ( E (A T 0 A) ) µ(e A) + µ(e T 0 A) < 2ɛ. µ(e) µ((a T 0 A)) µ ( E (A T 0 A) ) < 2ɛ. Thus, µ(e) µ(e) 2 µ(e) µ(a) 2 + µ(a) 2 µ(e) 2 = µ(e) µ((a T 0 A)) + (µ(a) + µ(e)) µ(a) µ(e) < 4ɛ. Sice ɛ > 0 is arbitrary, it follows that µ(e) = µ(e) 2, hece µ(e) = 0 or. Therefore, T is ergodic.
Examples of Ergodic Trasformatios 27 Example.8.3 (Iduced trasformatios of ergodic trasformatios) Let T be a ergodic measure preservig trasformatio o the probability space (, F, µ), ad A F with µ(a) > 0. Cosider the iduced trasformatio T A o (A, F A, µ A ) of T (see subsectio.5). Recall that T A x = T (x) x, where (x) := if{ : T x A}. Let (as i the proof of Propositio.5.), A k = {x A : (x) = k} B k = {x \ A : T x,..., T k x A, T k x A}. Propositio.8. If T is ergodic o (, F, µ), the T A is ergodic o (A, F A, µ A ). Proof. Let C F A be such that T A C = C. We wat to show that µ A (C) = 0 or ; equivaletly, µ(c) = 0 or µ(c) = µ(a). Sice A = k A k, we have C = T A C = k A k T k C. Let E = k B k T k C, ad F = E C (disjoit uio). Recall that (see subsectio.5) T A = A B, ad T B k = A k+ B k+. Hece, T F = T E T C = [ (Ak+ B k+ ) T (k+) C ] [ (A B ) T C ] k = (A k T k C) k T k k (B k C) = C E = F. Hece, F is T -ivariat, ad by ergodicity of T we have µ(f ) = 0 or. If µ(f ) = 0, the µ(c) = 0, ad hece µ A (C) = 0. If µ(f ) =, the µ( \ F ) = 0. Sice \ F = (A \ C) (( \ A) \ E) A \ C, it follows that µ(a \ C) µ( \ F ) = 0. Sice µ(a \ C) = µ(a) µ(c), we have µ(a) = µ(c), i.e., µ A (C) =.
28 Itroductio ad prelimiaries Exercise.8.4 Show that if T A is ergodic ad µ ( k T k A ) =, the, T is ergodic. The followig lemma provides, i some cases, a useful tool to verify that a measure preservig trasformatio defied o ([0, ), B, µ) is ergodic, where B is the Lebesgue σ-algebra, ad µ is a probability measure equivalet to Lebesgue measure λ (i.e., µ(a) = 0 if ad oly if λ(a) = 0). Lemma.8. (Kopp s Lemma) If B is a Lebesgue set ad C is a class of subitervals of [0, ), satisfyig (a) every ope subiterval of [0, ) is at most a coutable uio of disjoit elemets from C, (b) A C, λ(a B) γλ(a), where γ > 0 is idepedet of A, the λ(b) =. Proof. The proof is doe by cotradictio. Suppose λ(b c ) > 0. Give ε > 0 there exists by Lemma.2. a set E ε that is a fiite disjoit uio of ope itervals such that λ(b c E ε ) < ε. Now by coditios (a) ad (b) (that is, writig E ε as a coutable uio of disjoit elemets of C) oe gets that λ(b E ε ) γλ(e ε ). Also from our choice of E ε ad the fact that λ(b c E ε ) λ(b E ε ) γλ(e ε ) γλ(b c E ε ) > γ(λ(b c ) ε), we have that γ(λ(b c ) ε) < λ(b c E ε ) < ε, implyig that γλ(b c ) < ε + γε. Sice ε > 0 is arbitrary, we get a cotradictio. Example.8.4 (Multiplicatio by 2 modulo ) Cosider ([0, ), B, λ), ad let T : be give by { 2x 0 x < /2 T x = 2x mod = 2x /2 x <, We have see that T is measure preservig. We will use Lemma.8. to show that T is ergodic. Let C be the collectio of all itervals of the form
Examples of Ergodic Trasformatios 29 [k/2, (k + )/2 ) with ad 0 k 2. Notice that the the set {k/2 :, 0 k < 2 } of dyadic ratioals is dese i [0, ), hece each ope iterval is at most a coutable uio of disjoit elemets of C. Hece, C satisfies the first hypothesis of Kopp s Lemma. Now, T maps each dyadic iterval of the form [k/2, (k + )/2 ) liearly oto [0, ), (we call such a iterval dyadic of order ); i fact, T x = 2 x mod(). Let B B be T -ivariat, ad assume λ(b) > 0. Let A C, ad assume that A is dyadic of order. The, T A = [0, ) ad λ(a B) = λ(a T B) = λ(a) λ(t A B) = λ(b) = λ(a)λ(b). 2 Thus, the secod hypothesis of Kopp s Lemma is satisfied with γ = λ(b) > 0. Hece, λ(b) =. Therefore T is ergodic. Example.8.5 (Lüroth series revisited) Cosider the map T of Example.3.3. I Exercise.3.2 we saw that T is measure preservig with respect to Lebesgue measure λ. We ow show that T is ergodic with respect to Lebesgue measure λ usig Kopp s Lemma. The collectio C cosists i this case of all fudametal itervals of all raks. A fudametal iterval of rak k is a set of the form where (i, i 2,..., i k ) = (i ) T (i 2 ) (i k ) = {x : a (x) = i, a 2 (x) = i 2,..., a k (x) = i k }. Notice that (i, i 2,..., i k ) is a iterval with ed poits P k Q k P k /Q k = i + ad P k + Q k i (i ) i k (i k ), + +. i (i )i 2 i (i ) i k (i k )i k Furthermore, T k ( (i, i 2,..., i k )) = [0, ), ad T k restricted to (i, i 2,..., i k ) has slope i (i ) i k (i k )i k (i k ) = λ( (i, i 2,..., i k )).
30 Itroductio ad prelimiaries Sice lim k diam( (i, i 2,..., i k )) = 0 for ay sequece i, i 2,, the collectio C geerates the Borel σ-algebra. Now let A be a T -ivariat Borel set of positive Lebesgue measure, ad let E be ay fudametal iterval of rak, the λ(a E) = λ(t A E) = λ(e)λ(a). By Kopp s Lemma with γ = λ(a) we get that λ(a) = ; i.e. T is ergodic with respect to λ. Exercise.8.5 Let λ be the ormalized Lebesque measure o ([0, ), B), where B is the Lebesgue σ-algebra. Cosider the trasformatio T : [0, ) [0, ) give by { 3x 0 x < /3 T x = 3 x /3 x <. 2 2 For x [0, ) let ad { 3 0 x < /3 s (x) = 3 /3 x <, 2 { 0 0 x < /3 h (x) = /3 x <, 2 a (x) = { 0 0 x < /3 /3 x <. Let s = s (x) = s (T x), h = h (x) = h (T x) ad a = a (x) = a (T x) for. h k (a) Show that for ay x [0, ) oe has x =. s s 2 s k (b) Show that T is measure presevig ad ergodic with respect to the measure λ. (c) Show that for each ad ay sequece i, i 2,..., i {0, } oe has k= λ ({x [0, ) : a (x) = i, a 2 (x) = i 2,..., a (x) = i }) = 2k 3, where k = #{ j : i j = }.
Examples of Ergodic Trasformatios 3 Exercise.8.6 Let β > be a o-iteger, ad cosider the trasformatio T β : [0, ) [0, ) give by T β x = βx mod() = βx βx. Use Lemma.8. to show that T β is ergodic with respect to Lebesgue measure λ, i.e. if T β A = A, the λ(a) = 0 or.
32 Itroductio ad prelimiaries
Chapter 2 Ergodic Theorems 2. The Poitwise Ergodic Theorem ad its cosequeces The Poitwise Ergodic Theorem is also kow as Birkhoff s Ergodic Theorem or the Idividual Ergodic Theorem (93). This theorem is i fact a geeralizatio of the Strog Law of Large Numbers (SLLN) which states that for a sequece Y, Y 2,... of i.i.d. radom variables o a probability space (, F, µ), with E Y i < ; oe has lim Y i = EY (a.e.). i= For example cosider = {0, } N, F the σ-algebra geerated by the cylider sets, ad µ the uiform product measure, i.e., µ ({x : x = a, x 2 = a 2,..., x = a }) = /2. Suppose oe is iterested i fidig the frequecy of the digit. More precisely, for a.e. x we would like to fid lim #{ i : x i = }. Usig the Strog Law of Large Numbers oe ca aswer this questio easily. Defie {, if x i =, Y i (x) := 0, otherwise. 33
34 Ergodic Theorems Sice µ is product measure, it is easy to see that Y, Y 2,... form a i.i.d. Beroulli process, ad EY i = E Y i = /2. Further, #{ i : x i = } = i= Y i(x). Hece, by SLLN oe has lim #{ i : x i = } = 2. Suppose ow we are iterested i the frequecy of the block 0, i.e., we would like to fid lim #{ i : x i = 0, x i+ =, x i+2 = }. We ca start as above by defiig radom variables {, if x i = 0, x i+ =, x i+2 =, Z i (x) := 0, otherwise. The, #{ i : x i = 0, x i+ =, x i+2 = } = Z i (x). It is ot hard to see that this sequece is statioary but ot idepedet. So oe caot directly apply the strog law of large umbers. Notice that if T is the left shift o, the Y = Y T ad Z = Z T. I geeral, suppose (, F, µ) is a probability space ad T : a measure preservig trasformatio. For f L (, F, µ), we would like to kow uder what coditios does the limit lim f(t i x) exist a.e. If it does exist what is its value? This is aswered by the Poitwise Ergodic Theorem which was origially proved by G.D. Birkhoff i 93. Sice the, several proofs of this importat theorem have bee obtaied; here we preset a recet proof give by T. Kamae ad M.S. Keae i [KK]. Theorem 2.. (The Poitwise Ergodic Theorem) Let (, F, µ) be a probability space ad T : a measure preservig trasformatio. The, for ay f i L (µ), lim f(t i (x)) = f (x) i=
exists a.e., is T -ivariat ad f dµ = f dµ. If moreover T is ergodic, the f is a costat a.e. ad f = f dµ. 35 For the proof of the above theorem, we eed the followig simple lemma. Lemma 2.. Let M > 0 be a iteger, ad suppose {a } 0, {b } 0 are sequeces of o-egative real umbers such that for each = 0,, 2,... there exists a iteger m M with a + + a +m b + + b +m. The, for each positive iteger N > M, oe has a 0 + + a N b 0 + + b N M. Proof of Lemma 2.. Usig the hypothesis we recursively fid itegers m 0 < m < < m k < N with the followig properties The, m 0 M, m i+ m i M for i = 0,..., k, ad N m k < M, a 0 + + a m0 b 0 + + b m0, a m0 + + a m b m0 + + b m,. a mk + + a mk b mk + + b mk. a 0 + + a N a 0 + + a mk b 0 + + b mk b 0 + + b N M. Proof of Theorem 2.. Assume with o loss of geerality that f 0 (otherwise we write f = f + f, ad we cosider each part separately). Let
36 Ergodic Theorems f (x) = f(x) +... + f(t f (x) x), f(x) = lim sup, ad f(x) = f (x) lim if. The f ad f are T -ivariat, sice f(t x) = lim sup = lim sup = lim sup f (T x) [ f+ (x) + + f + (x) + = f(x). f(x) ] Similarly f is T -ivariat. Now, to prove that f exists, is itegrable ad T -ivariat, it is eough to show that f dµ f dµ f dµ. For sice f f 0, this would imply that f = f = f. a.e. We first prove that fdµ f dµ. Fix ay 0 < ɛ <, ad let L > 0 be ay real umber. By defiitio of f, for ay x, there exists a iteger m > 0 such that f m (x) mi(f(x), L)( ɛ). m Now, for ay δ > 0 there exists a iteger M > 0 such that the set 0 = {x : m M with f m (x) m mi(f(x), L)( ɛ)} has measure at least δ. Defie F o by { f(x) x 0 F (x) = L x / 0. Notice that f F (why?). For ay x, let a = a (x) = F (T x), ad b = b (x) = mi(f(x), L)( ɛ) (so b is idepedet of ).We ow show that {a } ad {b } satisfy the hypothesis of Lemma 2.. with M > 0 as above. For ay = 0,, 2,... - if T x 0, the there exists m M such that f m (T x) m mi(f(t x), L)( ɛ) = m mi(f(x), L)( ɛ) = b +... + b +m.
37 Hece, a +... + a +m = F (T x) +... + F (T +m x) f(t x) +... + f(t +m x) = f m (T x) b +... + b +m. If T x / 0, the take m = sice a = F (T x) = L mi(f(x), L)( ɛ) = b. Hece by Lemma 2.. for all itegers N > M oe has F (x) +... + F (T N x) (N M) mi(f(x), L)( ɛ). Itegratig both sides, ad usig the fact that T is measure preservig, oe gets N F (x) dµ(x) (N M) mi(f(x), L)( ɛ) dµ(x). Sice F (x) dµ(x) = f(x) dµ(x) + Lµ( \ 0 ), 0 oe has f(x) dµ(x) = f(x) dµ(x) 0 F (x) dµ(x) Lµ( \ 0 ) (N M) mi(f(x), L)( ɛ) dµ(x) Lδ. N Now lettig first N, the δ 0, the ɛ 0, ad lastly L oe gets together with the mootoe covergece theorem that f is itegrable, ad f(x) dµ(x) f(x) dµ(x). We ow prove that f(x) dµ(x) f(x) dµ(x).
38 Ergodic Theorems Fix ɛ > 0, ad δ 0 > 0. Sice f 0, there exists δ > 0 such that wheever A F with µ(a) < δ, the A fdµ < δ 0. Note that for ay x there exists a iteger m such that f m (x) m Now choose M > 0 such that the set (f(x) + ɛ). Y 0 = {x : m M with f m (x) m (f(x) + ɛ)} has measure at least δ. Defie G o by { f(x) x Y0 G(x) = 0 x / Y 0. Notice that G f. Let b = G(T x), ad a = f(x)+ɛ (so a is idepedet of ). Oe ca easily check that the sequeces {a } ad {b } satisfy the hypothesis of Lemma 2.. with M > 0 as above. Hece for ay M > N, oe has G(x) + + G(T N M x) N(f(x) + ɛ). Itegratig both sides yields (N M) G(x)dµ(x) N( f(x)dµ(x) + ɛ). Sice µ( \ Y 0 ) < δ, the ν( \ Y 0 ) = \Y 0 f(x)dµ(x) < δ 0. Hece, f(x) dµ(x) = G(x) dµ(x) + f(x) dµ(x) \Y 0 N (f(x) + ɛ) dµ(x) + δ 0. N M Now, let first N, the δ 0 (ad hece δ 0 0), ad fially ɛ 0, oe gets f(x) dµ(x) f(x) dµ(x). This shows that f dµ f dµ f dµ,
hece, f = f = f a.e., ad f is T -ivariat. I case T is ergodic, the the T -ivariace of f implies that f is a costat a.e. Therefore, f (x) = f (y)dµ(y) = f(y) dµ(y). Remarks () Let us study further the limit f i the case that T is ot ergodic. Let I be the sub-σ-algebra of F cosistig of all T -ivariat subsets A F. Notice that if f L (µ), the the coditioal expectatio of f give I (deoted by E µ (f I)), is the uique a.e. I-measurable L (µ) fuctio with the property that f(x) dµ(x) = E µ (f I)(x) dµ(x) A A for all A I i.e., T A = A. We claim that f = E µ (f I). Sice the limit fuctio f is T -ivariat, it follows that f is I-measurable. Furthermore, for ay A I, by the ergodic theorem ad the T -ivariace of A, ad lim (f A )(T i x) = A (x) lim f(t i x) = A (x)f (x) a.e. This shows that f = E µ (f I). f A (x) dµ(x) = f A (x) dµ(x). (2) Suppose T is ergodic ad measure preservig with respect to µ, ad let ν be a probability measure which is equivalet to µ (i.e. µ ad ν have the same sets of measure zero so µ(a) = 0 if ad oly if ν(a) = 0), the for every f L (µ) oe has ν a.e. lim f(t i (x)) = f dµ. 39 Exercise 2.. (Kac s Lemma) Let T be a measure preservig ad ergodic trasformatio o a probability space (, F, µ). Let A be a measurable
40 Ergodic Theorems subset of of positive µ measure, ad deote by the first retur time map ad let T A be the iduced trasformatio of T o A (see sectio.5). Prove that (x) dµ =. Coclude that (x) L (A, µ A ), ad that almost everywhere o A. A lim (T i A(x)) = µ(a), Exercise 2..2 Let β = + 5, ad cosider the trasformatio T β : 2 [0, ) [0, ), give by T β x = βx mod() = βx βx. Defie b o [0, ) by { 0 if 0 x < /β b (x) = if /β x <, Fix k 0. Fid the a.e. value (with respect to Lebesgue measure) of the followig limit lim #{ i : b i = 0, b i+ = 0,..., b i+k = 0}. Exercise 2..3 Let (, F, µ) be a probability space ad f L (µ). Suppose {T t : t R} is a family of trasformatios T t : satisfyig (i) T 0 = id ad T t+s = T t T s (ii) T t is measurable, measure preservig ad ergodic w.r.t. µ. (iii) The map G : R give by G(x, t) = f(t t (x)) is measurable, where R is edowed with the product σ algebra F B ad product measure µ λ, with B the Borel σ-algebra o R ad λ is Lebesgue measure.
(a) Show that for all s 0, s 0 f(t t (x)) dµ(x) dλ(t) = s (b) Show that for all s 0, s 0 f(t t(x)) dλ(t) < µ a.e. 0 4 f(t t (x)) dλ(t) dµ(x) = s f(x)dµ(x). (c) Defie F : R by F (x) = f(t 0 t(x)) dλ(t), ad cosider the trasformatio T correspodig to t =. Show that for ay oe has F (T k (x)) = f(t t (x)) dλ(t), k=0 ad F dµ = f dµ. (d) Show that for µ a.e. x oe has lim 0 0 f(t t (x)) dλ(t) = f dµ. Usig the Ergodic Theorem, oe ca give yet aother characterizatio of ergodicity. Corollary 2.. Let (, F, µ) be a probability space, ad T : a measure preservig trasformatio. The, T is ergodic if ad oly if for all A, B F, oe has lim µ(t i A B) = µ(a)µ(b). (2.) Proof. Suppose T is ergodic, ad let A, B F. Sice the idicator fuctio A L (, F, µ), by the ergodic theorem oe has lim A (T i x) = A (x) dµ(x) = µ(a) a.e.
42 Ergodic Theorems The, lim T i A B(x) = lim T i A(x) B (x) = B (x) lim A (T i x) = B (x)µ(a) a.e. Sice for each, the fuctio lim T i A B is domiated by the costat fuctio, it follows by the domiated covergece theorem that lim µ(t i A B) = lim T i A B(x) dµ(x) = B µ(a) dµ(x) = µ(a)µ(b). Coversely, suppose (2.) holds for every A, B F. Let E F be such that T E = E ad µ(e) > 0. By ivariace of E, we have µ(t i E E) = µ(e), hece lim µ(t i E E) = µ(e). O the other had, by (2.) lim µ(t i E E) = µ(e) 2. Hece, µ(e) = µ(e) 2. Sice µ(e) > 0, this implies µ(e) =. Therefore, T is ergodic. To show ergodicity oe eeds to verify equatio (2.) for sets A ad B belogig to a geeratig semi-algebra oly as the ext propositio shows. Propositio 2.. Let (, F, µ) be a probability space, ad S a geeratig semi-algebra of F. Let T : be a measure preservig trasformatio. The, T is ergodic if ad oly if for all A, B S, oe has lim µ(t i A B) = µ(a)µ(b). (2.2)
Proof. We oly eed to show that if (2.2) holds for all A, B S, the it holds for all A, B F. Note that (2.2) holds for elemets i the algebra geerated by S. Let ɛ > 0, ad A, B F. The, by Lemma.2. (i Subsectio.2) there exist sets A 0, B 0 each of which is a fiite disjoit uio of elemets of S such that µ(a A 0 ) < ɛ, ad µ(b B 0 ) < ɛ. Sice, it follows that (T i A B) (T i A 0 B 0 ) (T i A T i A 0 ) (B B 0 ), µ(t i A B) µ(t i A 0 B 0 ) µ [ (T i A B) (T i A 0 B 0 ) ] Further, µ(t i A T i A 0 ) + µ(b B 0 ) < 2ɛ. µ(a)µ(b) µ(a 0 )µ(b 0 ) µ(a) µ(b) µ(b 0 ) + µ(b 0 ) µ(a) µ(a 0 ) µ(b) µ(b 0 ) + µ(a) µ(a 0 ) µ(b B 0 ) + µ(a A 0 ) < 2ɛ. Hece, ( ) µ(t i A B) µ(a)µ(b) ( 43 µ(t i A 0 B 0 ) µ(a 0 )µ(b 0 )) µ(t i A B) + µ(t i A 0 B 0 ) µ(a)µ(b) µ(a0 )µ(b 0 ) < 4ɛ. Therefore, lim [ µ(t i A B) µ(a)µ(b) ] = 0.
44 Ergodic Theorems Theorem 2..2 Suppose µ ad µ 2 are probability measures o (, F), ad T : is measurable ad measure preservig with respect to µ ad µ 2. The, (i) if T is ergodic with respect to µ, ad µ 2 is absolutely cotiuous with respect to µ, the µ = µ 2, (ii) if T is ergodic with respect to µ ad µ 2, the either µ = µ 2 or µ ad µ 2 are sigular with respect to each other. Proof. (i) Suppose T is ergodic with respect to µ ad µ 2 is absolutely cotiuous with respect to µ. For ay A F, by the ergodic theorem for a.e. x oe has lim A (T i x) = µ (A). Let C A = {x : lim A (T i x) = µ (A)}, the µ (C A ) =, ad by absolute cotiuity of µ 2 oe has µ 2 (C A ) =. Sice T is measure preservig with respect to µ 2, for each oe has A (T i x) dµ 2 (x) = µ 2 (A). O the other had, by the domiated covergece theorem oe has lim A (T i x)dµ 2 (x) = µ (A) dµ 2 (x). This implies that µ (A) = µ 2 (A). Sice A F is arbitrary, we have µ = µ 2. (ii) Suppose T is ergodic with respect to µ ad µ 2. Assume that µ µ 2. The, there exists a set A F such that µ (A) µ 2 (A). For i =, 2 let C i = {x : lim A (T j x) = µ i (A)}. j=0 By the ergodic theorem µ i (C i ) = for i =, 2. Sice µ (A) µ 2 (A), the C C 2 =. Thus µ ad µ 2 are supported o disjoit sets, ad hece µ ad µ 2 are mutually sigular.
We ed this subsectio with a short discussio that the assumptio of ergodicity is ot very restrictive. Let T be a trasformatio o the probability space (, F, µ), ad suppose T is measure preservig but ot ecessarily ergodic. We assume that is a complete separable metric space, ad F the correspodig Borel σ-algebra (i order to make sure that the coditioal expectatio is well-defied a.e.). Let I be the sub-σ-algebra of T -ivariat measurable sets. We ca decompose µ ito T -ivariat ergodic compoets i the followig way. For x, defie a measure µ x o F by µ x (A) = E µ ( A I)(x). The, for ay f L (, F, µ), f(y) dµ x (y) = E µ (f I)(x). Note that µ(a) = E µ ( A I)(x) dµ(x) = µ x (A) dµ(x), ad that E µ ( A I)(x) is T -ivariat. We show that µ x is T -ivariat ad ergodic for a.e. x. So let A F, the for a.e. x µ x (T A) = E µ ( A T I)(x) = E µ (I A I)(T x) = E µ (I A I)(x) = µ x (A). Now, let A F be such that T A = A. The, A is T -ivariat, ad hece I-measurable. The, Hece, for a.e. x ad for ay B F, µ x (A) = E µ ( A I)(x) = A (x) a.e. µ x (A B) = E µ ( A B I)(x) = A (x)e µ ( B I)(x) = µ x (A)µ x (B). I particular, if A = B, the the latter equality yields µ x (A) = µ x (A) 2 which implies that for a.e. x, µ x (A) = 0 or. Therefore, µ x is ergodic. (Oe i fact eeds to work a little harder to show that oe ca fid a set N of µ-measure zero, such that for ay x \ N, ad ay T -ivariat set A, oe has µ x (A) = 0 or. I the above aalysis the a.e. set depeded o the choice of A. Hece, the above aalysis is just a rough sketch of the proof of what is called the ergodic decompositio of measure preservig trasformatios.) 45
46 Ergodic Theorems 2.2 The Mea Ergodic Theorem I the previous sectio, we studied the poitwise behaviour of the ergodic averages f(t i x) for f L (µ). I this sectio, we will restrict our attetio to f L 2 (µ), ad study the L 2 (µ) covergece of the ergodic averages. The result that we will be provig is due to vo Neuma (932). Before we start let us recall few facts over the Hilbert space L 2 (µ) ad the operator U T defied i sectio.7. We begi with the space L 2 (µ) = L 2 (, F, µ) which is a Hilbert space equipped with the ier product (f, g) = fg dµ, where g is the complex cojugate of g. The ier product iduces the L 2 (µ) orm defied by f 2 = (f, f) /2. If S is a closed liear subspace of L 2 (µ), the the orthogoal complemet of S is defied by S = {h L 2 (µ) : (h, g) = 0 for all g S}. The followig well kow Theorem states that each elemet f L 2 (µ) ca be uiquely writte as a sum f = g + h, where g S ad h S. We state it without a proof, ad refer the reader to ay stadard book o fuctioal aalysis. Theorem 2.2. Let S be a closed liear subspace of L 2 (µ). The for ay elemet f L 2 (µ), there exists a uique elemet g S satisfyig if{ f f 2 : f S} = f g 2. Furthermore, if P : L 2 (µ) S is defied by P (f) = g, the every elemet f L 2 (µ) ca be writte uiquely as f = P (f) + h, where h S. The trasformatio P is called the orthogoal projectio of L 2 (µ) oto S. Now suppose (, F, µ) is a probability space ad T : is a measure preservig trasformatio. Cosider the operator U T defied i sectio.7, but restricted to L 2 (µ), U T (f) = f T, for f L 2 (µ). Note that U T is a isometry sice (U T (f), U T (f)) = (f, f). Associated with U T, oe defies the adjoit operator U T : L2 (µ) L 2 (µ) satisfyig (U T (f), g) = (f, U T (g)), for all f, g L 2 (µ).
Sice U T is a isometry, oe gets that UT U T = I L 2 (µ), where I L 2 (µ) is the idetity operator. Cosider the set I = {f L 2 (µ) : f = U T (f) = f T }. It is easy to check that I is a closed liear subspace of L 2 (µ). By Theorem 2.2., we ca write each elemet f L 2 (µ) as f = g + h, where g = P f I ad h I. I the followig Lemma, we idetify explicitly the elemets of I. Lemma 2.2. Let B = {U T (g) g : g L 2 (µ)}. The B = I, where B is the closure of B i the L 2 (µ) orm. Proof. Equivaletly, we will show that B = I. Let f I, the U T (f) = f ad for ay (U T (g) g) B, we have (f, U T (g) g) = (U T (f), U T (g)) (f, g) = (f, g) (f, g) = 0. Thus f is orthogoal to every elemet of B. We show this is true for elemets of B as well. Let h B, ad (h i = U T (g i ) g i ) a sequece i B covergig i L 2 (µ) to h. The, (f, h i ) = 0 for all i, ad by Cauchy Schwartz iequality, (f, h) = lim i (f, h i ) = 0. Thus f B. Coversly, suppose f B. For every g L 2 (µ) ((f, U T (g) g) = 0), we have (f, g) = (f, U T (g)) = (U T (f), g). This implies that f = U T (f). We ow show that f = U T (f). To this ed cosider U T (f) f 2 2 = (U T (f) f, U T (f) f) Thus, f I. = U T (f) 2 (U T (f), f) (f, U T (f)) + f 2 = f 2 (f, U T (f)) (U T (f), f) + f 2 = 0 We are ow ready to prove the Mea Ergodic Theorem. Theorem 2.2.2 Let (, F, µ, T ) be a measure preservig system, ad let P T deote the orthogoal projectio oto the closed subspace I. The, for ay f L 2 (µ) the sequece ( f T i ) coverges i L 2 (µ) to P T (f). 47
48 Ergodic Theorems Proof. By Theorem 2.2. ad Lemma 2.2., ay f L 2 (µ) ca be writte as f = P T (f) + h with h B. Now P T (f) I, hece P T (f) T = U T (P T (f)) = P T (f). This implies that P T (f) T i coverges i L 2 (µ) to P T (f). Now let (h j = U T (g j ) g j ) be a sequece i B covergig to h i L 2 (µ). Note that h j T i = (g j T g j ) = (U T (g j ) g j ). Thus, h j T i 2 = UT i (h j ) 2 = U T (g j ) g j 2 2 g j 2. This shows that h j T i coverges i L 2 (µ) to 0 for all j. Fially, for ay j sufficietly large, h T i 2 (h h j ) T i 2 + h j T i 2 = (h h j ) 2 + h j T i 2 = (h h j ) 2 + h j T i 2. Takig the limit as we see that h T i coverges to 0 i L 2 (µ). Sice f T i = P T (f) T i + h T i, we get the required result.
49 2.3 Mixig As a corollary to the Poitwise Ergodic Theorem we foud a ew defiitio of ergodicity; amely, asymptotic average idepedece. Based o the same idea, we ow defie other otios of weak idepedece that are stroger tha ergodicity. Defiitio 2.3. Let (, F, µ) be a probability space, ad T : a measure preservig trasformatio. The, (i) T is weakly mixig if for all A, B F, oe has lim µ(t i A B) µ(a)µ(b) = 0. (2.3) (ii) T is strogly mixig if for all A, B F, oe has lim µ(t i A B) = µ(a)µ(b). (2.4) Notice that strogly mixig implies weakly mixig, ad weakly mixig implies ergodicity. This follows from the simple fact that if {a } is a sequece of real umbers such that lim a = 0, the lim a i = 0, ad hece lim a i = 0. Furthermore, if {a } is a bouded sequece of 0-egative real umbers, the the followig are equivalet (see [W] for the proof): (i) lim a i = 0 (ii) lim a i 2 = 0 (iii) there exists a subset J of the itegers of desity zero, i.e. lim such that lim,/ J a = 0. # ({0,,..., } J) = 0,