A Short Introduction to Ergodic Theory of Numbers. Karma Dajani

Transcription

1 A Short Introduction to Ergodic Theory of Numbers Karma Dajani June 3, 203

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3 Contents Motivation and Examples 5 What is Ergodic Theory? 5 2 Number Theoretic Examples 6 2 Measure Preserving, Ergodicity and the Ergodic Theorem 5 2 Measure Preserving Transformations 5 22 Ergodicity 7 23 The Ergodic Theorem 9 24 Mixing 28 3 Examples Revisited 3 4 Natural Extensions 4 4 Natural Extensions of m-adic and β-expansions 4 42 Natural Extension of Continued Fractions The Doeblin-Lenstra Conjecture Some Diophantine spinoff A proof of the Lenstra-Doeblin Conjecture 50 5 Random β-expansions 5 5 Random β-expansions 5 52 Basic Properties of Random β-transformations An ergodic measure for random β-expansions 56 3

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5 Chapter Motivation and Examples The aim of these short lecture notes is to show how one can use basic ideas in ergodic theory in order to understand the global behaviour of a family of series expansions of numbers in a given interval This is done by showing that the expansions under study can be generated by iterations of an appropriate map which will be shown to be measure preserving and ergodic What is Ergodic Theory? It is not easy to give a simple definition of Ergodic Theory because it uses techniques and examples from many fields such as probability theory, statistical mechanics, number theory, vector fields on manifolds, group actions of homogeneous spaces and many more The word ergodic is a mixture of two Greek words: ergon (work) and odos (path) The word was introduced by Boltzmann (in statistical mechanics) regarding his hypothesis: for large systems of interacting particles in equilibrium, the time average along a single trajectory equals the space average The hypothesis as it was stated was false, and the investigation for the conditions under which these two quantities are equal lead to the birth of ergodic theory as is known nowadays A modern description of what ergodic theory is would be: it is the study of the long term average behavior of systems evolving in time The collection of all states of the system form a space, and the evolution is represented by either a transformation T :, where T x is the state of the system at time t =, when the system (ie, at time t = 0) was initially in state x (This is analogous to the setup of discrete time stochastic processes) if the evolution is continuous or if the configurations have spacial structure, then we describe the evolution by looking at a group of transformations 5

6 6 Motivation and Examples G (like Z 2, R, R 2 ) acting on, ie, every g G is identified with a transformation T g :, and T gg = T g T g The space usually has a special structure, and we want T to preserve the basic structure on For example if is a measure space, then T must be measurable if is a topological space, then T must be continuous if has a differentiable structure, then T is a diffeomorphism In these lectures our space is a probability space (, B, µ), and our time is discrete So the evolution 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 sequence x, T x, T 2 x, If T is invertible, then one speaks of the two sided orbit, T x, x, T x, Before we go any further with ergodic theory, let us see the connection of the above setup with a certain collection of number theoretic expansions of points in the unit interval 2 Number Theoretic Examples Example 2 (Binary Expansion) Let = [0, ) with the Lebesgue σ-algebra B, and Lebesgue measure λ Define T : be given by { 2x 0 x < /2 T x = 2x mod = 2x /2 x < Using T one can associate with each point in [0, ) an infinite sequence of 0 s and s To do so, we define the function a by { 0 if 0 x < /2 a (x) = if /2 x <, then T x = 2x a (x) Now, for n set a n (x) = a (T n x) Fix x, for simplicity, we write a n instead of a n (x), then T x = 2x a Rewriting we get x = a 2 + T x a2 2 Similarly, T x = 2 + T 2 x 2 Continuing in this manner, we see that for each n, x = a 2 + a a n 2 n + T n x 2 n Since 0 < T n x <, we get x n a i 2 i = T n x 0 as n 2n i=

7 Number Theoretic Examples 7 Thus, x = i= We shall later see that the sequence of digits a, a 2, forms an iid sequence of Bernoulli random variables Example 22 (m-ary Expansion) If we replace in the above example the transformation by T x = mx mod, and a (x) = k if x [ k m, k + m ), k = 0,,, m, then for each n one has a i 2 i x = a m + a 2 m a n m n + T n x 2 n, and since 0 < T n x <, taking limits one gets the m-array expansion of x = a i m i with digits a i {0,,, m } i= In the above two examples, we looked at series expansions in integer bases In these cases, the expansion obtained by the above maps is essentially unique The exceptional set consists of all points of the form k which have exactly two mn expansions, one ending in zeros and the other ending in the digit m In case the base is not an integer, then the situation is completely reversed, typically there are uncountably many algorithms generating expansions in noninteger base We mention here the two extreme cases Example 23 (Greedy Expansions) Let β > be a noninteger, define T β : [0, β /(β )) : [0, β /(β )) by βx (mod ), 0 x <, T β (x) = βx β, x < β /(β ), see also Figure Similar to the above examples, we define the digits of x as { i if i β a (x) = a = x < i+ β, i = 0,, β β if β β x < β β, and a n (x) = a n = a (T n β ), m 2 One easily sees that for any n, x = a β + a 2 β a n β n + T n β x β n a n Taking limits, lead to the greedy expansion of x = If for some n one βn n= has Tβ n x = 0, then x has a finite expansion, and we do not need to take limits It is not hard to see that for each n, the digits a n is the largest element in {0,,, β } satisfying n a i i= β x i

8 8 Motivation and Examples β β 0 β 2 β β β Figure : The greedy map T β (here β = 2 + ) Example 24 (Lazy Expansions) Consider the map S β : (0, β /(β )] (0, β /(β )] by where and Since (d) = S β (x) = βx d, (0) = ( 0, for x (d), ] β β(β ) ( β β(β ) + d β, β β(β ) + d ], d {, 2,, β } β β β(β ) = β β β β, () one has that ( β (d) = β β d + β, β β β d ], d {, 2,, β } β (2) Hence, to get the defining partition one starts from β /(β ) by taking β intervals of length /β from right to left The last interval with endpoints 0 and ( β + β)/β(β ), corresponding to the lazy digit 0, is longer than the rest; see also Figure 2 One can easily see that the interval A β = (( β + β)/(β ), β /(β )] is an attractor, in the sense that for any x there exists n 0 such that Sβ mx A β for all m n Defining now the digits of x by a (x) = a = d if x (d), and a n (x) = a n = a (S n x) for n 2 It is easily seen that x = n= a n, where the summation can be finite βn β

9 Number Theoretic Examples 9 β β β β - 0 β 2 β +β β β β(β ) β(β ) β β Figure 2: The lazy map S β Example 25 (Lüroth Series) Another kind of series expansion, introduced by J Lüroth [L] in 883, motivates this approach Several authors have studied the dynamics of such systems Take as partition of [0, ) the intervals [ n+, n ) where n N Every number x [0, ) can be written as a finite or infinite series, the so-called Lüroth (series) expansion x = a (x) + a (x)(a (x) )a 2 (x) + + a (x)(a (x) ) a n (x)(a n (x) )a n (x) + ; here a k (x) 2 for each k How is such a series generated? Let T : [0, ) [0, ) be defined by n(n + )x n, x [ n+, n ), T x = 0, x = 0 (3) Let x 0, for k and T k x 0 we define the digits a n = a n (x) by a k (x) = a (T k x), where a (x) = n if x [ n, n ), n 2 Now (3) can be written as a (x)(a (x) )x (a (x) ), x 0, T x = 0, x = 0

10 0 Motivation and Examples (a = 2) 0 Figure 3: The Lüroth Series map T Thus, for any 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, and if we assume that k is the smallest positive integer with this property, then x = a + + In case T k x 0 for all k, one gets x = a + a (a ) a k (a k )a k + + +, a (a )a 2 a (a ) a k (a k )a k For ease of notation we drop the argument x from the functions a k (x)

11 Number Theoretic Examples where a k 2 for each k Let us convince ourselves that this last infinite series indeed converges to x Let S k = S k (x) be the sum of the first k terms of the sum Then x S k = T k x a (a ) a k (a k ) ; since T k x [0, ) and a k 2 for all x and all k, we find x S k 2 k 0 as k From the above we also see that if x and y have the same Lüroth expansion, then, for each k, x y 2 k and it follows that x equals y Example 26 (Generalized Lüroth Series) Consider any partition I = {[l n, r n ) : n D} of [0, ) where D Z + is finite or countable and n D (r n l n ) = We write L n = r n l n and I n = [l n, r n ) for n D Moreover, we assume that i, j D with i > j satisfy 0 < L i L j < D is called the digit set; see also Figure 4 [ ) [ ) [ ) 0 l 3 r 3 l r l 2 r 2 Figure 4: The partition I We will consider the following transformation T on [0, ): x r n l l n, x I n r n l n, n D, n T x = 0, x I = [0, ) \ n D I n ; (4) see also Figure 5 We want to iterate T in order to generate a series expansion of points x in [0, ), in fact of points x whose T -orbit never hits I We will show that the set of such points has measure We first need some notation For x [l n, r n ), n D, we write s(x) = so that T x = xs(x) h(x) Now let r n l n and h(x) = l n r n l n, s k (x) = { s(t k x), if T k x n D I n,, otherwise,

12 2 Motivation and Examples 0 r 3 l r l 2 r 2 l 3 Figure 5: The GLS-map T h k (x) = { h(t k x), if T k x n D I n,, otherwise (thus s(x) = s (x), h(x) = h (x)) From these definitions we see that for x n D I n (0, ) such that T k x n D I n (0, ) for all k, one has x = h (x) s (x) + s T x (x) = h s + T s x = h ( ) s + s h2 s 2 + T 2 x s 2 = h s + h 2 s s 2 + T 2 x s s 2 = h s + h 2 s s h k s s 2 s k + T k x s s 2 s k = h s + h 2 s s h k s s 2 s k + We refer to the above expansion as the GLS(I) expansion of x with a specified digit set D Such an expansion converges to x Moreover, it is unique To prove the first statement we define the nth GLS-convergent P k /Q k of x by then P k Q k = h s + h 2 s s x P k = x P k = Q k Q k h k s s 2 s k ; T k x s s 2 s k (5)

13 Number Theoretic Examples 3 Notice that s k = length of the interval that T k x belongs to Let L := max n D L n Then, x P k Lk 0 as k Q k For the proof of the second statement, use (5) and the triangle inequality Example 27 (Continued Fraction) Define a transformation T : [0, ) [0, ) by T 0 = 0 and for x 0 T x = x x ; see Figure Figure 6: The continued fraction map T An interesting feature of this map is that its iterations generate the continued fraction expansion for points in (0, ) For if we define a = a (x) = { if x ( 2, ) n if x ( n+, n ], n 2,

14 4 Motivation and Examples then, T x = x a and hence x = a + T x For n, let a n = a n (x) = a (T n x) Then, after n iterations we see that x = a + T x = = a + a a n + T n x In fact, if p n, q n Z, with gcd(p n, q n ) =, and q n > 0 are such, that p n q n = a + a a n then we will show (in Section 422) that the q n are monotonically increasing, and x p n < 0 as n (6) q n The last statement implies that q 2 n x = a + a 2 + a 3 + In view of (6) we define for every real number x and every n 0 the so-called approximation coefficients Θ n (x) by Θ n (x) = qn 2 x p n (7) It immediately follows from (6) that Θ n (x) < for all irrational x and all n 0 We will return to these approximation coefficients in Chapter 4 q n,

15 Chapter 2 Measure Preserving, Ergodicity and the Ergodic Theorem The basic setup of all the examples in the previous chapter consists of a probability space (, B, µ), where is a set consisting of all possible outcomes, B is a σ-algebra, and µ is a probability measure on B The evolution is given by a transformation T : which is measurable, ie T A = {x : T x A} B for any A B We want also that the evolution is in steady state ie stationary In the language of ergodic theory, we want T to be measure preserving 2 Measure Preserving Transformations Definition 2 Let (, B, µ) be a probability space, and T : measurable The map T is said to be measure preserving with respect to µ if µ(t A) = µ(a) for all A B In case T is invertible, then T is measure preserving if and only if µ(t A) = µ(a) for all A B We can generalize the definition of measure preserving to the following case Let T : (, B, µ ) ( 2, B 2, µ 2 ) be measurable, then T is measure preserving if µ (T A) = µ 2 (A) for all A B 2 Recall that a collection S of subsets of is said to be a semi-algebra if (i) S, (ii) A B S whenever A, B S, and (iii) if A S, then \A = n i= E i is a disjoint union of elements of S For example if = [0, ), and S is the collection of all subintervals, then S is a semi-algebra Or if = {0, } Z, then the collection of all cylinder sets {x : x i = a i,, x j = a j } is a semi-algebra An algebra A is a collection of subsets of satisfying:(i) A, (ii) if A, B A, then A B A, and finally (iii) if A A, then \ A A Clearly an algebra is a semi-algebra Furthermore, given a semi-algebra S one can form an algebra by taking all finite disjoint unions of elements of S We denote this algebra by 5

16 6 Measure Preserving, Ergodicity and the Ergodic Theorem A(S), and we call it the algebra generated by S It is in fact the smallest algebra containing S Likewise, given a semi-algebra S (or an algebra A), the σ-algebra generated by S (A) is denoted by σ(s) (σ(a)), and is the smallest σ-algebra containing S (or A) A monotone class C is a collection of subsets of with the following two properties if E E 2 are elements of C, then i= E i C, if F F 2 are elements of C, then i= F i C The monotone class generated by a collection S of subsets of is the smallest monotone class containing S; for a proof see [H] Theorem 2 Let A be an algebra of, then the σ-algebra σ(a) generated by A equals the monotone class generated by A Using the above theorem, one can get an easier criterion for checking that a transformation is measure preserving Theorem 22 Let ( i, B i, µ i ) be probability spaces, i =, 2, and T : 2 a transformation Suppose S 2 is a generating semi-algebra of B 2 Then, T is measurable and measure preserving if and only if for each A S 2, we have T A B and µ (T A) = µ 2 (A) Proof Let C = {B B 2 : T B B, and µ (T B) = µ 2 (B)}, then S 2 C B 2, and hence A(S 2 ) C We show that C is a monotone class Let E E 2 be elements of C, and let E = i= E i Then, T E = i= T E i B, and µ (T E) = µ ( n=t E n ) = lim n µ (T E n ) = lim n µ 2(E n ) = µ 2 ( n=e n ) = µ 2 (E) Thus, E C A similar proof shows that if F F 2 are elements of C, then i= F i C Hence, C is a monotone class containing the algebra A(S 2 ) By the monotone class theorem, B 2 is the smallest monotone class containing A(S 2 ), hence B 2 C This shows that B 2 = C, therefore T is measurable and measure preserving For example if

17 Ergodicity 7 = [0, ) with the Borel σ-algebra B, and µ a probability measure on B Then a transformation T : is measurable and measure preserving if and only if T [a, b) B and µ ( T [a, b) ) = µ ([a, b)) for any interval [a, b) = {0, } N with product σ-algebra and product measure µ A transformation T : is measurable and measure preserving if and only if and T ({x : x 0 = a 0,, x n = a n }) B, µ ( T {x : x 0 = a 0,, x n = a n } ) = µ ({x : x 0 = a 0,, x n = a n }) for any cylinder set Another useful lemma is the following Lemma 2 Let (, B, µ) be a probability space, and A an algebra generating B Then, for any A B and any ɛ > 0, there exists C A such that µ(a C) < ɛ Proof Let D = {A B : for any ɛ > 0, there exists C A such that µ(a C) < ɛ} Clearly, A D B By the Monotone Class Theorem (Theorem 2), we need to show that D is a monotone class To this end, let A A 2 be a sequence in D, and let A = n= A n, notice that µ(a) = lim µ(a n) Let n ɛ > 0, there exists an N such that µ(a A N ) = µ(a) µ(a N ) < ɛ/2 Since A N D, then there exists C A such that µ(a N C) < ɛ/2 Then, µ(a C) µ(a A N ) + µ(a N C) < ɛ Hence, A D Similarly, one can show that D is closed under decreasing intersections so that D is a monotone class containg A, hence by the Monotone Class Theorem B D Therefore, B = D, and the theorem is proved 22 Ergodicity Definition 22 Let T be a measure preserving transformation on a probability space (, F, µ) The map T is said to be ergodic if for every measurable set A satisfying T A = A, we have µ(a) = 0 or Theorem 22 Let (, F, µ) be a probability space and T : measure preserving The following are equivalent: (i) T is ergodic (ii) If B F with µ(t B B) = 0, then µ(b) = 0 or (iii) If A F with µ(a) > 0, then µ ( n=t n A) =

18 8 Measure Preserving, Ergodicity and the Ergodic Theorem (iv) If A, B F with µ(a) > 0 and µ(b) > 0, then there exists n > 0 such that µ(t n A B) > 0 Remark 22 In case T is invertible, then in the above characterization one can replace T n by T n 2 Note that if µ(b T B) = 0, then µ(b \ T B) = µ(t B \ B) = 0 Since and B = ( B \ T B ) ( B T B ), T B = ( T B \ B ) ( B T B ), we see that after removing a set of measure 0 from B and a set of measure 0 from T B, the remaining parts are equal In this case we say that B equals T B modulo sets of measure 0 3 In words, (iii) says that if A is a set of positive measure, almost every x eventually (in fact infinitely often) will visit A 4 (iv) says that elements of B will eventually enter A Proof of Theorem 22 (i) (ii) Let B F be such that µ(b T B) = 0 We shall define a measurable set C with C = T C and µ(c B) = 0 Let C = {x : T n x B io } = n= k=n T k B Then, T C = C, hence by (i) µ(c) = 0 or Furthermore, ( ) ( ) µ(c B) = µ T k B B c + µ T k B c B n= k=n n= k=n ( ) ( ) µ T k B B c + µ T k B c B k= µ ( T k B B ) k= Using induction (and the fact that µ(e F ) µ(e G) + µ(g F )), one can show that for each k one has µ ( T k B B ) = 0 Hence, µ(c B) = 0 which implies that µ(c) = µ(b) Therefore, µ(b) = 0 or (ii) (iii) Let µ(a) > 0 and let B = n= T n A Then T B B Since T is measure preserving, then µ(b) > 0 and k= µ(t B B) = µ(b \ T B) = µ(b) µ(t B) = 0

19 The Ergodic Theorem 9 Thus, by (ii) µ(b) = (iii) (iv) Suppose µ(a)µ(b) > 0 By (iii) ( ) ( ) µ(b) = µ B T n A = µ (B T n A) > 0 n= n= Hence, 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, then by (iv) there exists k such that µ(a c T k A) > 0 Since T k A = A, it follows that µ(a c A) > 0, a contradiction Hence, µ(a) = and T is ergodic The following lemma provides, in some cases, a useful tool to verify that a measure preserving transformation defined on ([0, ), B, µ) is ergodic, where B is the Lebesgue σ-algebra, and µ is a probability measure equivalent to Lebesgue measure λ (ie, µ(a) = 0 if and only if λ(a) = 0) Lemma 22 (Knopp s Lemma) If B is a Lebesgue set and C is a class of subintervals of [0, ), satisfying (a) every open subinterval of [0, ) is at most a countable union of disjoint elements from C, (b) A C, λ(a B) γλ(a), where γ > 0 is independent of A, then λ(b) = Proof The proof is done by contradiction Suppose λ(b c ) > 0 Given ε > 0 there exists by Lemma 2 a set E ε that is a finite disjoint union of open intervals such that λ(b c E ε ) < ε Now by conditions (a) and (b) (that is, writing E ε as a countable union of disjoint elements of C) one gets that λ(b E ε ) γλ(e ε ) Also from our choice of E ε and the fact that we have that λ(b c E ε ) λ(b E ε ) γλ(e ε ) γλ(b c E ε ) > γ(λ(b c ) ε), γ(λ(b c ) ε) < λ(b c E ε ) < ε, implying that γλ(b c ) < ε + γε Since ε > 0 is arbitrary, we get a contradiction 23 The Ergodic Theorem The Ergodic Theorem is also known as Birkhoff s Ergodic Theorem or the Individual Ergodic Theorem (93) This theorem is in fact a generalization of the Strong Law of Large Numbers (SLLN) which states that for a sequence

20 20 Measure Preserving, Ergodicity and the Ergodic Theorem Y, Y 2, of iid random variables on a probability space (, F, µ), with E Y i < ; one has n lim Y i = EY (ae) n n i= For example consider = {0, } N, F the σ-algebra generated by the cylinder sets, and µ the uniform product measure, ie, µ ({x : x = a, x 2 = a 2,, x n = a n }) = /2 n Suppose one is interested in finding the frequency of the digit More precisely, for ae x we would like to find lim n n #{ i n : x i = } Using the Strong Law of Large Numbers one can answer this question easily Define {, if x i =, Y i (x) := 0, otherwise Since µ is product measure, it is easy to see that Y, Y 2, form an iid Bernoulli process, and EY i = E Y i = /2 Further, #{ i n : x i = } = n i= Y i(x) Hence, by SLLN one has lim n n #{ i n : x i = } = 2 Suppose now we are interested in the frequency of the block 0, ie, we would like to find lim n n #{ i n : x i = 0, x i+ =, x i+2 = } We can start as above by defining random variables {, if x i = 0, x i+ =, x i+2 =, Z i (x) := 0, otherwise Then, n #{ i n : x i = 0, x i+ =, x i+2 = } = n n Z i (x) It is not hard to see that this sequence is stationary but not independent So one cannot directly apply the strong law of large numbers Notice that if T is the left shift on, then Y n = Y T n and Z n = Z T n In general, suppose (, F, µ) is a probability space and T : a measure preserving transformation For f L (, F, µ), we would like to know under what conditions does the limit lim n n i= n f(t i x) exist ae If it does exist i=0

21 The Ergodic Theorem 2 what is its value? This is answered by the Ergodic Theorem which was originally proved by GD Birkhoff in 93 Since then, several proofs of this important theorem have been obtained; here we present a recent proof given by T Kamae and MS Keane in [KK] Theorem 23 (The Ergodic Theorem) Let (, F, µ) be a probability space and T : a measure preserving transformation Then, for any f in L (µ), n lim f(t i (x)) = f (x) n n i=0 exists ae, is T -invariant and f dµ = f dµ If moreover T is ergodic, then f is a constant ae and f = f dµ For the proof of the above theorem, we need the following simple lemma Lemma 23 Let M > 0 be an integer, and suppose {a n } n 0, {b n } n 0 are sequences of non-negative real numbers such that for each n = 0,, 2, there exists an integer m M with a n + + a n+m b n + + b n+m Then, for each positive integer N > M, one has a a N b b N M Proof of Lemma 23 Using the hypothesis we recursively find integers m 0 < m < < m k < N with the following properties Then, m 0 M, m i+ m i M for i = 0,, k, and N m k < M, a a m0 b b m0, a m0 + + a m b m0 + + b m, a mk + + a mk b mk + cldots + b mk a a N a a mk b b mk b b N M Proof of Theorem 23 Assume with no loss of generality that f 0 (otherwise we write f = f + f, and we consider each part separately) Let f n (x) =

22 22 Measure Preserving, Ergodicity and the Ergodic Theorem f(x)+ +f(t n f n (x) x), f(x) = lim sup n n, and f(x) = lim inf n Then f and f are T -invariant, since f(t x) = lim sup n = lim sup n = lim sup n f n (T x) n [ fn+ (x) n + n + f(x) n n f n+ (x) n + = f(x) ] f n (x) n Similarly f is T -invariant Now, to prove that f exists, is integrable and T - invariant, it is enough to show that f dµ f dµ f dµ For since f f 0, this would imply that f = f = f ae We first prove that fdµ f dµ Fix any 0 < ɛ <, and let L > 0 be any real number By definition of f, for any x, there exists an integer m > 0 such that f m (x) min(f(x), L)( ɛ) m Now, for any δ > 0 there exists an integer M > 0 such that the set 0 = {x : m M with f m (x) m min(f(x), L)( ɛ)} has measure at least δ Define F on by { f(x) x 0 F (x) = L x / 0 Notice that f F (why?) For any x, let a n = a n (x) = F (T n x), and b n = b n (x) = min(f(x), L)( ɛ) (so b n is independent of n)we now show that {a n } and {b n } satisfy the hypothesis of Lemma 23 with M > 0 as above For any n = 0,, 2, Hence, - if T n x 0, then there exists m M such that f m (T n x) m min(f(t n x), L)( ɛ) = m min(f(x), L)( ɛ) = b n + + b n+m a n + + a n+m = F (T n x) + + F (T n+m x) f(t n x) + + f(t n+m x) = f m (T n x) b n + + b n+m

23 The Ergodic Theorem 23 If T n x / 0, then take m = since a n = F (T n x) = L min(f(x), L)( ɛ) = b n Hence by Lemma 23 for all integers N > M one has F (x) + + F (T N x) (N M) min(f(x), L)( ɛ) Integrating both sides, and using the fact that T is measure preserving, one gets N F (x) dµ(x) (N M) min(f(x), L)( ɛ) dµ(x) Since one has f(x) dµ(x) F (x) dµ(x) = f(x) dµ(x) + Lµ( \ 0 ), 0 = f(x) dµ(x) 0 F (x) dµ(x) Lµ( \ 0 ) (N M) min(f(x), L)( ɛ) dµ(x) Lδ N Now letting first N, then δ 0, then ɛ 0, and lastly L one gets together with the monotone convergence theorem that f is integrable, and f(x) dµ(x) f(x) dµ(x) We now prove that f(x) dµ(x) f(x) dµ(x) Fix ɛ > 0, and δ 0 > 0 Since f 0, there exists δ > 0 such that whenever A F with µ(a) < δ, then A fdµ < δ 0 Note that for any x there exists an integer 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 Define G on by { f(x) x Y0 G(x) = 0 x / Y 0

24 24 Measure Preserving, Ergodicity and the Ergodic Theorem Notice that G f Let b n = G(T n x), and a n = f(x)+ɛ (so a n is independent of n) One can easily check that the sequences {a n } and {b n } satisfy the hypothesis of Lemma 23 with M > 0 as above Hence for any M > N, one has G(x) + + G(T N M x) N(f(x) + ɛ) Integrating both sides yields (N M) G(x)dµ(x) N( f(x)dµ(x) + ɛ) Since µ( \ Y 0 ) < δ, then ν( \ Y 0 ) = \Y 0 f(x)dµ(x) < δ 0 Hence, 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, then δ 0 (and hence δ 0 0), and finally ɛ 0, one gets f(x) dµ(x) f(x) dµ(x) This shows that f dµ f dµ f dµ, hence, f = f = f ae, and f is T -invariant In case T is ergodic, then the T -invariance of f implies that f is a constant ae Therefore, f (x) = f (y)dµ(y) = f(y) dµ(y) Remark 23 () Let us study further the limit f in the case that T is not ergodic Let I be the sub-σ-algebra of F consisting of all T -invariant subsets A F Notice that if f L (µ), then the conditional expectation of f given I (denoted by E µ (f I)), is the unique ae I-measurable L (µ) function with the property that f(x) dµ(x) = E µ (f I)(x) dµ(x) A A for all A I ie, T A = A We claim that f = E µ (f I) Since the limit function f is T -invariant, it follows that f is I-measurable Furthermore, for any A I, by the ergodic theorem and the T -invariance of A, n lim n n i=0 (f A )(T i x) = A (x) lim n n n i=0 f(t i x) = A (x)f (x) ae

25 The Ergodic Theorem 25 and This shows that f = E µ (f I) f A (x) dµ(x) = f A (x) dµ(x) (2) Suppose that T is ergodic and measure preserving with respect to µ, and let ν be a probability measure which is equivalent to µ (ie µ and ν have the same sets of measure zero so µ(a) = 0 if and only if ν(a) = 0), then for every f L (µ) one has ν ae n lim f(t i (x)) = n n i=0 f dµ Corollary 23 Let (, F, µ) be a probability space, and T : a measure preserving transformation Then, T is ergodic if and only if for all A, B F, one has n lim µ(t i A B) = µ(a)µ(b) (2) n n i=0 Proof Suppose T is ergodic, and let A, B F Since the indicator function A L (, F, µ), by the ergodic theorem one has Then, n lim A (T i x) = n n i=0 A (x) dµ(x) = µ(a) ae n n lim n n T i A B(x) = lim n n T i A(x) B (x) i=0 i=0 n = B (x) lim A (T i x) n n i=0 = B (x)µ(a) ae n Since for each n, the function lim n n i=0 T i A B is dominated by the constant function, it follows by the dominated convergence theorem that lim n n n µ(t i A B) = i=0 = n lim n n T i A B(x) dµ(x) i=0 B µ(a) dµ(x) = µ(a)µ(b)

26 26 Measure Preserving, Ergodicity and the Ergodic Theorem Conversely, suppose (2) holds for every A, B F Let E F be such that T E = E and µ(e) > 0 By invariance of E, we have µ(t i E E) = µ(e), hence n lim µ(t i E E) = µ(e) n n On the other hand, by (2) i=0 n lim µ(t i E E) = µ(e) 2 n n i=0 Hence, µ(e) = µ(e) 2 Since µ(e) > 0, this implies µ(e) = Therefore, T is ergodic To show ergodicity one needs to verify equation (2) for sets A and B belonging to a generating semi-algebra only as the next proposition shows Proposition 23 Let (, F, µ) be a probability space, and S a generating semi-algebra of F Let T : be a measure preserving transformation Then, T is ergodic if and only if for all A, B S, one has n lim µ(t i A B) = µ(a)µ(b) (22) n n i=0 Proof We only need to show that if (22) holds for all A, B S, then it holds for all A, B F Let ɛ > 0, and A, B F Then, by Lemma 2 (in Subsection 2) there exist sets A 0, B 0 each of which is a finite disjoint union of elements of S such that Since, it follows that Further, µ(a A 0 ) < ɛ, and µ(b B 0 ) < ɛ (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 ) ] 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

27 The Ergodic Theorem 27 Hence, ( ) n µ(t i A B) µ(a)µ(b) n i=0 ( n n i=0 µ(t i A 0 B 0 ) µ(a 0 )µ(b 0 )) n µ(t i A B) + µ(t i A 0 B 0 ) µ(a)µ(b) µ(a0 )µ(b 0 ) n < 4ɛ i=0 Therefore, lim n [ n n i=0 µ(t i A B) µ(a)µ(b) ] = 0 Theorem 232 Suppose µ and µ 2 are probability measures on (, F), and T : is measurable and measure preserving with respect to µ and µ 2 Then, (i) if T is ergodic with respect to µ, and µ 2 is absolutely continuous with respect to µ, then µ = µ 2, (ii) if T is ergodic with respect to µ and µ 2, then either µ = µ 2 or µ and µ 2 are singular with respect to each other Proof (i) Suppose T is ergodic with respect to µ and µ 2 is absolutely continuous with respect to µ For any A F, by the ergodic theorem for ae x one has Let n lim A (T i x) = µ (A) n n i=0 n C A = {x : lim A (T i x) = µ (A)}, n n then µ (C A ) =, and by absolute continuity of µ 2 one has µ 2 (C A ) = Since T is measure preserving with respect to µ 2, for each n one has n n i=0 i=0 A (T i x) dµ 2 (x) = µ 2 (A) On the other hand, by the dominated convergence theorem one has n lim A (T i x)dµ 2 (x) = n n i=0 µ (A) dµ 2 (x)

28 28 Measure Preserving, Ergodicity and the Ergodic Theorem This implies that µ (A) = µ 2 (A) Since A F is arbitrary, we have µ = µ 2 (ii) Suppose T is ergodic with respect to µ and µ 2 Assume that µ µ 2 Then, there exists a set A F such that µ (A) µ 2 (A) For i =, 2 let n C i = {x : lim A (T j x) = µ i (A)} n n j=0 By the ergodic theorem µ i (C i ) = for i =, 2 Since µ (A) µ 2 (A), then C C 2 = Thus µ and µ 2 are supported on disjoint sets, and hence µ and µ 2 are mutually singular 24 Mixing As a corollary to the ergodic theorem we found a new definition of ergodicity; namely, asymptotic average independence Based on the same idea, we now define other notions of weak independence that are stronger than ergodicity Definition 24 Let (, F, µ) be a probability space, and T : a measure preserving transformation Then, (i) T is weakly mixing if for all A, B F, one has lim n n n µ(t i A B) µ(a)µ(b) = 0 (23) i=0 (ii) T is strongly mixing if for all A, B F, one has lim µ(t i A B) = µ(a)µ(b) (24) n Notice that strongly mixing implies weakly mixing, and weakly mixing implies ergodicity This follows from the simple fact that if {a n } is a sequence of real n numbers such that lim n a n = 0, then lim n a i = 0, and hence n n lim n a i = 0 Furthermore, if {a n } is a bounded sequence, then the n i=0 following are equivalent (see [W] for the proof): n (i) lim n a i = 0 n i=0 n (ii) lim n a i 2 = 0 n i=0 i=0

29 The Ergodic Theorem 29 (iii) there exists a subset J of the integers of density zero, ie lim n such that lim n,n/ J a n = 0 # ({0,,, n } J) = 0, n Using this one can give three equivalent characterizations of weakly mixing transformations Proposition 24 Let (, F, µ) be a probability space, and T : a measure preserving transformation Let S be a generating semi-algebra of F (a) If Equation (23) holds for all A, B S, then T is weakly mixing (b) If Equation (24) holds for all A, B S, then T is strongly mixing

30 30 Measure Preserving, Ergodicity and the Ergodic Theorem

31 Chapter 3 Examples Revisited In this chapter, we will study the ergodic behavior of the examples given in Chapter For each map we will give an invariant ergodic measure absolutely continuous with respect to Lebesgue measure For all the examples, the invariance of the measure is verified on intervals (see Theorem 22), and the ergodicity is shown using Knopp s Lemma (Lemma 22) Example 30 (Binary expansion revisited) Consider the map of example 2 We will show that Lebesgue measure λ is T -invariant (or that T is measure preserving with respect to λ) For any interval [a, b), [ a T [a, b) = 2, b ) [ a + 2 2, b + ), 2 and λ ( T [a, b) ) = b a = λ ([a, b)) hence, by Theorem 22 we see that λ is T -invariant For ergodicity we use Knopp s Lemma To this end, let C be the collection of all intervals of the form [k/2 n, (k + )/2 n ) with n and 0 k 2 n Notice that the the set {k/2 n : n, 0 k < 2 n } of dyadic rationals is dense in [0, ), hence each open interval is at most a countable union of disjoint elements of C Hence, C satisfies the first hypothesis of Knopp s Lemma Now, T n maps each dyadic interval of the form [k/2 n, (k + )/2 n ) linearly onto [0, ), (we call such an interval dyadic of order n); in fact, T n x = 2 n x mod() Let B B be T -invariant, and assume λ(b) > 0 Let A C, and assume that A is dyadic of order n Then, T n A = [0, ) and λ(a) = 2 n Furthermore, for any Lebesgue measurable set C, λ(t n C A) = 2 n λ(c) Hence, by invariance of of the set B we have λ(b A) = λ(t n B A) = 2 n λ(b) = λ(b)λ(a), 3

32 32 Examples Revisited Thus, the second hypothesis of Knopp s Lemma is satisfied with γ = λ(b) > 0 Hence, λ(b) = Therefore T is ergodic Example 302 (m-ary expansion revisited) Consider the map T of example 22 A slight modification of the arguments used in the above example show that T is measure preserving and ergodic with respect to Lebesgue measure λ Example 303 (Greedy expansion revisited) Consider the transformation of example 23 It is easy to see that Lebesgue measure is not invariant We are seeking a T β -invariant measure µ of the form µ β (A) = A h β(x)dλ(x) for any Borel set A It is not hard to see that the interval [0, ) is an attractor in the sense that for any x [0, β β ), there exists n 0 such that T β m x [0, ) for all m n Independently, AO Gel fond (in 959) [G] and W Parry [P] (in 960) showed that h β (x) = { F (β) n=0 β n [0,T n β ())(x) x [0, ) 0 x [, [0, β β ), where F (β) = 0 ( x<tβ n() β )dx is a normalizing constant The T n β -invariance of µ β follows from the equality (proven by Parry) βh β (x) = h β (y) y:t β y=x To prove ergodicity, we need few definitions first From now on, we will consider T β as a map on [0, ] We define fundamental intervals (of rank n) in the usual way: the intervals of rank are (i) = {x : a (x) = i} = I i, for i {0, }, and the intervals of rank n, for n 2 are (i,, i n ) = (i ) T β (i 2) T (n ) β (i n ) = {x : a (x) = i,, a n (x) = i n } = x : x = n i j j= β j + T β nx β n, A fundamental interval (i,, i n ) is full if T n (i,, i n ) = [0, ), ie λ (T n ( (i,, i n ))) =, here λ denotes Lebesgue measure on [0, ] From the above we see that if (i,, i n ) is full, it is equal to the interval [ n j= i n j β j, i j β j + β n ) j= Let C be the collection of all full fundamental intervals of all ranks We show that C generates the Borel σ-algebra To this end, let B n be the collection of non-full intervals of rank n that are not subsets of full intervals of lower rank Note that ( β ) is the only member of B Suppose that (i,, i n ) is an element of B n, then (i,, i j ) B j for j n We claim that (i,, i n ) contains at most one element of B n+ There are two cases:

33 Examples Revisited 33 If Tβ n = k β for some k =,, β, then all (n + ) order fundamental intervals are full, and B n+ is empty If Tβ n = (k)o (interior) for some k = 0,,, β, then (i,, i n, j), j = 0,, k are full, and (i,, i n, k) is non-full, and hence in B n+ Since B =, it thus follows by induction from the above that B n for all n Denote by n the unique element of B n (note that n could be empty) We are now ready to show that the collection C of full intervals generates the Borel σ-algebra on [0, ] Let F n be the collection of all full intervals of rank n, and let D n be the collection of full intervals of rank n that are not subsets of full intervals of lower rank, ie, D n = { (i,, i n ) F n : (i,, i j ) F j for any j n } the union of all full intervals that are not subsets of full intervals of lower rank has full Lebesgue measure, ie, for any N, ( ) N λ [0, ) \ (j,, j n ) = λ ( n)) < β N n= D n Taking the limit as N tends to infinity, we get ( ) λ [0, ) \ (j,, j n ) = 0 n= D n So applying a similar procedure to any interval, we find that any interval can be covered by a countable disjoint union of full intervals, so C generates, and C satisfies condition (i) of Knopp s Lemma Now let B be a T β -invariant set, and let E be a full interval of rank n Note that λ(e) = β n, and for any Lebesgue measurable set C, λ(t n β C E) = β n λ(c) Hence, by invariance of of the set B we have λ(b E) = λ(t n β B E) = β n λ(b) = λ(b)λ(e), By Knopp s Lemma with γ = λ(b) we get that λ(b) =, and hence µ β (B) = (since λ and µ β are equivalent on [0, ]) Therefore, T β is ergodic with respect to µ β Example 304 (Lazy expansions revisited) Consider the map S β of Example 24 The dynamical behaviour of this map is essentially the same as that of T β in the previous example In the language of ergodic theory these two maps are isomorphic To be more precise, consider the map ψ : [0, β /(β )) (0, β /(β )] defined by ψ(x) = β β x,

34 34 Examples Revisited then ψ is continuous, hence Borel measurable, and ψt β = S β ψ The latter equality implies that the absolutely continuous measure ρ β defined by ρ β (A) = µ β (ψ (A)), (A a Borel set) is S β - invariant The ergodicity of S β with respect to the measure ρ β follows again from the commuting relation ψt β = S β ψ For if A is an S β - invariant Borel set, then ψ A is a T β invariant set By ergodicity of T β we have µ β (ψ (A)) equals 0 or Since ρ β (A) = µ β (ψ (A)), ergodicity of S β follows Example 305 (Lüroth series revisited) Consider the map T of Example 25 We show that T is measure preserving and ergodic with respect to Lebesgue measure λ Using the definition of T, for any interval [a, b) of [0, ) one has T [a, b) = k=2 [ k + a k(k ), k + ) b, k(k ) Hence λ(t [a, b)) = λ([a, b)), and T is measure preserving with respect to λ Ergodicity follows again from Knopp s Lemma The collection C consists in this case of all fundamental intervals of all ranks A fundamental interval of rank n is a set of the form (i, i 2,, i k ) = (i ) T (i 2 ) (i n ) = {x : a (x) = i, a 2 (x) = i 2,, a k (x) = i k } Notice that (i, i 2,, i k ) is an interval with end points P k Q k and P k + Q k i (i ) i k (i k ), where P k /Q k = i i (i )i 2 i (i ) i k (i k )i k Furthermore, T n ( (i, i 2,, i k )) = [0, ), and T n restricted to (i, i 2,, i k ) has slope i (i ) i k (i k )i k = λ( (i, i 2,, i k )) Since lim k diam( (i, i 2,, i k )) = 0 for any sequence i, i 2,, the collection C generates the Borel σ-algebra Now let A be a T -invariant Borel set of positive Lebesgue measure, and let E be any fundamental interval of rank n, then λ(a E) = λ(t n A E) = λ(e)λ(a) By Knopp s Lemma with γ = λ(a) we get that λ(a) = ; ie T is ergodic with respect to λ

35 Examples Revisited 35 Example 306 (Generalized Lüroth series revisited) Our transformation is as given in Example 26 We will show again that T is measure preserving and ergodic with respect to Lebesgue measure λ For any interval [a, b) of [0, ), ( T [a, b) = T [a, b) ) I n ( T ) [a, b) I n = n ((r n l n )a + l n, (r n l n )b + l n ) ( T (a, b) I ) Since λ(i ) = 0, it follows that λ ( T [a, b) ) = n (r n l n )(b a) = b a = λ([a, b)) So λ is T -invariant Before we prove ergodicity, we introduce few notations similar to those in the above examples A GLS expansion is identified with the partition I and the index (or digit) set D Let x have an infinite GLS(I) expansion, given by x = h + h 2 h k s s s 2 s s 2 s k Now h k and s k are identified once we know in which partition element T k x lies (h k and s k are constants determined by partition elements) Therefore, to determine the GLS-expansion of x (for a given I and D) we only need to keep track of which partition elements the orbit of x visits For x [0, ) we define the sequence of digits a n = a n (x), n, as follows a n = k T n x I k, k D { } Thus the values of the digits of points x [0, ) are elements of D; this is why D was called the digit set Notice that every GLS expansion determines a unique sequence of digits, and conversely So x = k= h k s s 2 s k =: [ a, a 2, ] We can now define fundamental intervals (or cylinder sets) in the usual way Setting (i) = {x : a (x) = i} if i D { }, then For i, i 2,, i n D { }, define (i) = [l i, r i ) if i D, and ( ) = I (i, i 2,, i n ) = {x : a (x) = i, a 2 (x) = i 2,, a n (x) = i n }

36 36 Examples Revisited Notice that, if i j = for some j n, then (i, i 2,, i n ) is a subset of a set of measure zero, namely the set consisting of all points in (0, ) whose orbit hits I Let us determine the cylinder sets (i,, i k ), for i, i 2,, i n D All points x with the same first k digits have the same first k terms in their GLS expansion Let us call the sum of the first k terms p k /q k ; then x = p k + T k x, q k s s k where s j = /L ij and T k x can vary freely in [0, ) This implies that from which we clearly have (i,, i k ) = [ p k, p k ) +, q k q k s s k λ( (i,, i k )) = s s k, where s s k is the slope of the restriction of T k to the fundamental interval (i,, i k ) Since L ij = /s j for each j, we find that λ( (i,, i k )) = L i L i2 L ik = λ( (i ))λ( (i 2 )) λ( (i k )) Hence the digits are independent If we let C be the collection of all fundamental intervals of all rank, by a similar reasoning as in the above examples, the collection C generates the Borel σ-algebra let A be a T -invariant Borel set of positive Lebesgue measure, and let E be any fundamental interval of rank n, then λ(a E) = λ(t n A E) = λ(e)λ(a) By Knopp s Lemma with γ = λ(a) we get that λ(a) = ; ie, T is ergodic with respect to λ Example 307 (Continued fractions revisited) Consider the map T of Example 27 One can easily see that Lebesgue measure λ is not T -invariant However, there exists a T -invariant measure µ which is equivalent to Lebesgue measure on the interval [0, ) This invariant measure was found by Gauss in 800, and is known nowadays as the Gauss measure which is given by µ(a) = log 2 A + x dx for all Borel sets A [0, ), where log refers to the natural logarithm; see Figure 33 Nobody knows how Gauss found µ, and his achievement is even more remarkable if we realize that modern probability theory and ergodic theory started almost a century later! In general, finding the invariant measure is a difficult

37 Examples Revisited 37 log 2 2 log 2 0 Figure 3: The densities of Lebesgue measure λ and Gauss measure µ task The T invariance of µ can be verified on intervals of the form [a, b) Easy calculations show that ( ] T [a, b) = n + b,, n + a n= and µ([a, b)) = µ(t [a, b)) Ergodicity is again proved by Knopp s Lemma We first define the notion of fundamental intervals similar to the above examples A fundamental interval of order n is a set of the form (a,, a n ) := {x [0, ) : a (x) = a,, a n (x) = a n }, where a j N for each j n When a,, a n are fixed, we sometimes write n instead of (a, a 2,, a n ) We list few properties of these sets without proofs, and we refer to [DK] for more details (i) (a, a 2,, a k ) is an interval in [0, ) with endpoints p k q k and p k + p k q k + q k, where p n q n = a + a a n (ii) The sequences (p n ) n and (q n ) n satisfy the following recurrence relations p := ; p 0 := a 0 ; p n = a n p n + p n 2, n, q := 0; q 0 := ; q n = a n q n + q n 2, n (3) Furthermore, p n (x) = q n (T x) for all n 0, and x (0, ) A proof of these recurrence formulas will be given in Section 422

38 38 Examples Revisited (iii) and λ ( (a, a 2,, a k )) = q k (q k + q k ), µ( (a k, a k,, a )) = µ( (a, a 2,, a k )) (iv) If 0 a < b, that {x : a T n x < b} n equals when n is even, and equals [ pn a + p n, p n b + p ) n q n a + q n q n b + q n ( pn b + p n, p n a + p ] n q n b + q n q n a + q n for n odd Here n = n (a,, a n ) is a fundamental interval of rank n This leads to Since λ(t n q n (q n + q n ) [a, b) n ) = λ([a, b))λ( n ) (q n b + q n )(q n a + q n ) 2 < q n q n + q n < q n (q n + q n ) (q n b + q n )(q n a + q n ) < q n(q n + q n ) qn 2 Therefore we find for every interval I, that 2 λ(i)λ( n) < λ(t n I n ) < 2λ(I)λ( n ) < 2 Let A be a finite disjoint union of such intervals I Since Lebesgue measure is additive one has 2 λ(a)λ( n) λ(t n A n ) 2λ(A)λ( n ) (32) The collection of finite disjoint unions of such intervals generates the Borel σ-algebra It follows that (32) holds for any Borel set A (v) For any Borel set A one has hence by (32) and (33) one has λ(a) µ(a) λ(a), (33) 2 log 2 log 2 µ(t n A n ) log 2 4 µ(a)µ( n) (34)

39 Examples Revisited 39 Now let C be the collection of all fundamental intervals n Since the set of all endpoints of these fundamental intervals is the set of all rationals in [0, ), it follows that condition (a) of Knopp s Lemma is satisfied Now suppose that B is invariant with respect to T and µ(b) > 0 Then it follows from (34) that for every fundamental interval n µ(b n ) log 2 4 µ(b)µ( n) So condition (b) from Knopp s Lemma is satisfied with γ = log 2 4 µ(b); thus µ(b) = ; ie T is ergodic We now use the ergodic Theorem to give simple proofs of old and famous results of Paul Lévy; see [Le] Proposition 302 (Paul Lévy, 929) For almost all x [0, ) one has lim n n log q n = π 2 2 log 2, (35) lim n n log(λ( n)) = π2, 6 log 2 and (36) lim n n log x p n = π2 q n 6 log 2 (37) Proof By property (ii) above, for any irrational x [0, ) one has q n (x) = Taking logarithms yields p n (x) p n (T x) q n (x) q n (T x) q n 2 (T 2 x) p2(t n 2 x) q (T n x) = p n(x) p n (T x) q n (x) q n (T x) p(t n x) q (T n x) log q n (x) = log p n(x) q n (x) + log p n (T x) q n (T x) + + log p (T n x) q (T n x) (38) For any k N, and any irrational x [0, ), p k(x) q k (x) is a rational number close to x Therefore we compare the right-hand side of (38) with We have log x + log T x + log T 2 x + + log(t n x) log q n (x) = log x + log T x + log T 2 x + + log(t n x) + R(n, x) In order to estimate the error term R(n, x), we recall from property (i) that x lies in the interval n, which has endpoints p n /q n and (p n +p n )/(q n +q n ) Therefore, in case n is even, one has 0 < log x log p n q n = (x p n ) q n ξ q n (q n + q n ) <, p n /q n q n

40 40 Examples Revisited where ξ (p n /q n, x) is given by the mean value theorem Let F, F 2, be the sequence of Fibonacci,, 2, 3, 5, (these are the q i s of the small golden ratio g = /G) It follows from the recurrence relation for the q i s (property (ii)) that q n (x) F n A similar argument shows that in case n is odd Thus and since we have q n < log x log p n q n, R(n, x) F n + F n + + F, F n = Gn + ( ) n+ g n 5 it follows that F n 5 G n, n Thus F n + F n + + F is the nth partial sum of a convergent series, and therefore Hence for each x for which R(n, x) F n + + F n= F n := C lim n n (log x + log T x + log T 2 x + + log(t n x)) exists, lim n n log q n(x) exists too, and these limits are equal Now lim n n (log x + log T x + log T 2 x + + log(t n x)) is ideally suited for the Ergodic Theorem; we only need to check that the conditions of the Ergodic Theorem are satisfied and to calculate the integral This is left as an exercise for the reader This proves (35) thus It follows from Property (iii) above that λ( n (a,, a n )) = q n (q n + q n ) ; log 2 2 log q n < log λ( n ) < 2 log q n Now apply (35) to obtain (36) Finally (37) follows from (35) and < 2q n q n+ x p n <, n q n q n+ In Section 422 we give a proof of this last statement q n

41 Chapter 4 Natural Extensions In this Chapter we will show how one constructs an invertible system associated with a given non-invertible system in such a way that all the dynamical properties of the original system are preserved To this end, suppose (Y, G, ν, S) is a non-invertible measure-preserving dynamical system An invertible measure-preserving dynamical system (, F, µ, T ) is called a natural extension of (Y, G, ν, S) if there exists a measurable surjective (ae) map ψ : Y such that (i) ψ T = S ψ, (ii) ν = µ ψ, and (iii) m=0t m ψ G = F, where k=0 T k ψ G is the smallest σ-algebra containing the σ-algebras T k ψ G for all k 0 Natural extensions were first introduced by Rohlin in the early 60 s (see [Ro]) He gave a canonical way of constructing a natural extension, and he showed that his construction is unique up to isomorphism In many examples the canonical construction may not be the easiest version to work with, especially if one is seeking an invariant measure of the original system that is absolutely continuous with respect to Lebesgue measure In this chapter, we will construct natural extensions that are planar, easy to work with and to deduce properties of the original system 4 Natural Extensions of m-adic and β-expansions Example 4 (m-adic) For simplicity, we consider the binary map as given in Example 2, T : [0, ) [0, ) given by { 2x 0 x < /2 T x = 2x mod = 2x /2 x < A natural extension of T is the well know Baker s transformation T : [0, ) 2 [0, ) 2 by { (2x, y T (x, y) = 2 ) 0 x < /2 (2x, y+ 2 ) /2 x < 4

42 42 Natural Extensions It is easy to see that T is measurable with respect to product Lebesgue σ- algebra B B, and is measure preserving with respect to λ λ Furthermore, it is straightforward to see that the map ψ : [0, ) 2 [0, ) given by ψ(x, y) = x satisfies conditions (i) and (ii) in the definition of the natural extension It remains to verify that T m (B [0, )) = B B m 0 T m π B = m 0 For this it suffices to show that m 0 T m (B [0, )) contains all the twodimensional cylinders (k,, k n ) (l,, l m ), where (k,, k n ) = {x : a (x) = k,, a n (x)}, with a n (x) the n th binary digit of x, and k i {0, } A closer look at the action of T shows that (k,, k n ) (l,, l m ) = T m ( (l m,, l, k,, k n ) [0, )) T m (B [0, )) Example 42 (Greedy β-expansions) Consider the transformation of example 23 T β x = βx (mod ) Note that here we restrict the domain to the interval [0, ) which as we saw in Example 303 is an attractor We also saw that T β is invariant with respect to the measure µ β with density h β To build a convenient natural extension of T β, we first look at the case β is a pseudo-golden mean (or what it commonly know as an mbonacci number, ie β is the positive root of the polynomial x m x m x Then, = β + β β m, so that has a finite β-expansion Note that in the β-expansion of any x [0, ), one can have at most m consecutive digits equal to The underlying space of the natural extension is the set = m k=0 [ T m k β, T m k β ) [ 0, Tβ k ) equipped with the Lebesgue σ-algebra L restricted to, and the two dimensional Lebesgue measure λ restricted to On we consider the transformation T given by T β (x, y) := (T β x, β ) ( βx + y) It is easy to see that the map T is measure preserving with respect to λ If one considers the map ψ : [0, ) given by ψ(x, y) = x, then a proof similar to that used in the previous example shows that ψ satisfies conditions (i), (ii),