# An introduction to ergodic theory - Walters P.

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(3) There arc examples of weak-mixing T which are not strong-mixing. Kakutani [1] ha.*> an example constructed by combinatorial methods and Maruyama [1] constructed an example using Gaussian processes. Chacon and Katok and Stepin ([1] p. 94) also have examples. If is a probability spacc let t(A') denote the collection of all invertible measure-preserving transformations of (A", ;iti, m). If we topologize r(X) with the “weak” topology (see Halmos [1]), the class of weak-mixing transformations is of second category while the class of strong-mixing transformations is of first category. So from the point of view of this topology most transformations arc weak-mixing blit not strong-mixing.

(4) Intuitive descriptions of ergodicity and strong-mixing can be given as follows. To say T is strong-mixing means that for any set A the sequence of .sets T~"A becomes, asymptotically, independent of any other set B. Ergodicity means T~ "A becomes independent of В on the average, for each pair of sets A, BeJ). We shall give a similar description of weak-mixing after Theorem 1.22.

$1.7 Mixing

41

The following theorem gives a way of checking the mixing properties lor examples by reducing the computations to a class of sets we can manipulate with. For example, it implies we need only consider measurable rectangles wnen dealing with the mixing properties of shifts.

Theorem 1.17. Lei (Abe a measure space and let ff be a semi-algebra that generates .16. Let T: X -» A' be a measure-preserving transformation. Then

(i) T is ergodic iff V/l, В e .cf

1

lim - ^ m(T ‘A n B) = m[A)m(B),

и-» (X- ^ i — 0

(ii) T is шик-mixing iJJ V/l, Be.'/’

j и - i

lim - £ \m(T ‘A n В) — т(А)т(В)\ = 0, and

II -* «< H 1 = 0

(iii) T is strong-mixing iff V/l, В e Sf

lim m(T~"A n B) = m(A)m\B),

Proof. Since each member of the algebra, generated by ff can be

written as a finite disjoint union of members of У it follows that if any of the three convergence properties hold for all members of if then they hold for all members of

Let £ > 0 be given and let A, В e J?. Choose A0, B0e with

m(A Д A0) < e and ni(B Д B0) < e. For any i > 0, (T~'A n В) Д (T~lA0 n B0) с (Т~‘А Д Т~‘А0) и(ВД B0), so we have m({T~"A пВ)Д (T~‘A0 n B0)) < 2l, and therefore \m(T~‘A n B) — m(T~‘A0 n B0)| < 2e. Therefore

\т(Т~‘А n B) - m(A)m(B)\ <, \m(T~‘A n В) — т(Т~‘А0 n B0)|

+ jт(Т~1А0 n B0) - m(/l0)m(B0)|

+ \m(A0)m{B0) - m(A)m(B0)\

+ \m(A)m{B0) — m(A)m(B)\

< 4e + \m(T~iA0 n B0) - m(/l0>?z(B0)|.

This inequality together with the known behaviour of the righthand term proves (ii) and (iii). To prove (i) one can easily obtain

1 n_1

- У m(T ‘A n B) — m(A)m(B) n i = о

< 4c

+

l";1

- £ m(T ‘^0 n Bo) - m(A0)m(B0}

and then use the known behav юиг of the right-hand side.

□

.42

I Measure Preserving Truncformaiions

As an application of this result we shall prove the result about ergodicity of Markov shifts mentioned in §1.5. To do this we shall use the following

Lemma 1.18. Let P be a stochastic matrix, hewing a strictly positive probability lector p with, pP = p. Then Q = limlV_., (1/&’)]Г» = о P" exists. The matrix Q js also stochastic and QP = PQ = Q. Any eigenvector of P for the eigenvalue

1 is also an eigenvector of Q Also Q2 = Q.

Pkooi . Let in denote the (p, P) Markov measure and T be two-sided (p. P) Markov shift. Let /, denote the characteristic function of the cylinder0[/]0 = {(*„)- t.|*o = '}■ ВУ BirkhofHs ergodic theorem (1 /N) Y.»=o 'Aj(Tnx) -> y*(x) a.e., and by multiplying by хДх) and using the dominated convergence theorem we have (1 /Л') X« = o P-tif dm(x). So Q = (q:j) is given

by 4,j = (l/Pi)\y*(x)Zj(x)dni(K) The other properties arc clear. □

Theorem 1.19. Let T denote the (p, P) Markov shift (either one-sided or two-sided). We can assume pt > 0 for each i where p = (p0,..., pk_,). Let Q be the matrix obtained in Lemma 1.18. The following are equivalent:

(i) T is ergodic.

(ii) All rows of the matrix Q are identical.

(iii) Every entry in Q is strictly positive.

(iv) P is irreducible.

(v) 1 is a simple eigenvalue of P.

Pkooi-. Let m denote the (p, P) Markov measure.

(i) (ii). A.-, in the proof of Lemma 1.1S liirkholT’s ergodic theorem gives 0/N)£?:,j M„[<]0 о „[/]„) = Pi4ij. Since T is crgodic the limit is w(0[i]o)-w,(n[y’]n) = PiPj- Therefore q4 = pt and so the rows of Q are identical.

(ii) => (iii). If the rows of Q are identical then pQ = p implies qi} = p} and so q4 > 0.

(iii) => (iv). Fix i,j. Since (1 /N) Y£=o P^f Qij > 0. then p-"' > 0 for some

n.

(iv) => (iii). Fix i and let St denote the collection of all states j with qi} > 0. Since Q = QP we have > quptj for each I. Therefore if I e 5, and p,j > 0 then tjij > 0 so j e S,. This implies that if I e S, then ZjeS, ptj = 1. Since P is irreducible we must have that S, is the whole state space and so qtj > 0 for

all j.

(iii) => (ii). Fix j and put qj = max,- qi}. We know Q2 = Q. If qtj < qj for some i then

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