# An introduction to ergodic theory - Walters P.

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Theorem 4.30. Every Bernoulli automorphism is a Kolmogorov automorphism.

Prooi Let the stale spacc for T be (Y, &, /.<)■ If Fe.f, let F = {e A':.v0 e F} e .!S. Let 'S = {F:F e .■?}> which is called the time-0 a-algebra. Let Ж = \J°= _ ^ Tw/J. We now verify that Ж satisfies the conditions for a Kolmogorov automorphism.

о 1

(i) Ж = V T</i c V Т‘У = тж-

i — — oo i = — оо

x эо n ac

(ii) V Jn^r = V V Г‘У = V T'^ = by definition of S.

*i — 0 n = 0 t = — t - t

(iii) We have to show Т~аЖ = . ^ = [Х,ф}. Fix A e Pjq Т~"Ж = Гin= о V-" Let В e Vfci’3 F^'/J, for some fixed j e Z. Since A e V*<j Т1У, A and В are independent, and therefore m(A n В) = /w(y4)/n(B). The collection of all sets В for which m(A n B) = m(A)m(B) is a monotone class, and, by the above, contains (JyL \/Г= j 7*#. Therefore VB e т(Л n B) = ш(/4)ш(В). Put В = A, then m(A) = m(A)2 which implies m(A) = 0 or 1. Hence

JO

П = ж.

n — 0

□

4 Lr.lropy

li was an open problem from 1958 to 1969 :ь to whether the converse of Theorem 4.30 was true, i.e., whether a Kolmogorov automorphism acting on a Lebesgue space is a Bernoulli automorphism. This was shown to be false by Ornstein.

Theorem 4.31 (Ornstein). There is an example of a Kolmogorov automorphism T which is not a Bernoulli automorphism.

Corollary 4.31.1. Entropy is not a complete invariant for the class of Kolmogorov automorphisms.

Proof. Let T be the example of Ornstein. By Corollary 4.14.4 li(T) > 0. Choose ft Bernoulli automorphism 5 with h(S) = h[T). S and T are not isomorphic. □

The following results show that the class of Kolmogorov automorphisms does not share all the properties the class of Bernoulli automorphisms enjoys. The proofs are given in the references cited.

Theorem 4.32

(i) There are uncountably many non-conjugate Kolmogorov automorphisms with the same entropy (Ornstein and Shields [1]).

(ii) There is a Kolmogorov automorphism T not conjugate to its inverse T~1 (Ornstein and Shields [l]).

(iii) There is a Kolmogorov automorphism which has no n-th roots for any n > 2 (Clark [1]).

(iv) There are non-conjugate Kolmogorov automorphisms T, S with T2 = S2 (Rudolf [1]).

(v) There are two non-conjugate Kolmogorov automorphisms each of which i s a jailor of the oilier (Polit [1] and Rudolf [2]).

Remarks

(1) Statement (ii) of Theorem 4.32 contrasts with the behaviour of ergodic transformations with pure point spectrum (see Corollary 3.4.1).

(2) Ornstein’s example for Theorem 4.31 is defined by induction and so ь fairly complicated to describe. It is therefore important to check whether the more "natural” examples of Kolmogorov automorphisms are Bernoulli automorphisms or not. We consider some of these at the end of this section and give an (easy to describe) example of a Kolmogorov automorphism that was recently shown not to be a Bernoulli automorphism.

(3) Sinai has proved that if T is an ergodic invertible measure-preserving transformation of a Lebesgue space (Aг.,эв,т) with h(T) > 0 and if .S’ is « Bernoulli automorphism with h(S) < h{T) then there exis's a measure-preserving transformation ф such that ф'Г = Бф, i.e., S is a factor of T isee Rohlin [3], p. 45).

jj-J.'J Bernoulli Automorphisms and Kolmogorov Auiomorph'sms

109

The next theorem shows that all Kolmogorov automorphisms are spectrally the same

Theorem 4.33 (Rohlin). If (X,Jd,m) is a probability space wiili a countable basis then any Kolmogorov automorphism T: X -* X has countable Lebesgue spectrum.

Proof. Recall that we arc assuming J? Ф {X, ф) — Ж. We have (i) Ж с ТЖ,

(ii) \]Т"Ж = 0, (iii) (~)Т~"Ж = . Г. Wc split the proof into three parts:

(a) We first show that Ж has no atoms, i.e., if С e Ж and m(C) > 0 then 3D e X with D cz С and m{D) < m(C).

Suppose С is an atom of Ж with m(C) > 0. Then TC is an atom of ТЖ and since Ж cz ТЖ either TC с С or m(C гл TC) = 0. If ТС с С then 'ГС = С since both sets have the same measure so that С 6 P),%0 Т~"Ж and therefore m(C) = 1. Hence Ж = Ж so = Ж, a contradiction. On the other hand, suppose m(TC n C) = 0. Then either for some к > 0 TkC <t С (and we use the above proof to get a contradiction) or m(TkC о С) = 0 V/c > 0 and then С u TC u T2C u ■ has infinite measure, a contradiction.

(b) Let Ж = {/6 L2(m)\f is .>r-mebsurable}. Then 11ТЖ' с: Ж. Let .// = V® итЖ. From U-J-"Ж = U'TV Q uyxX\n,m > 0) it follows that L2(m) = ©! , U'\ V © С where С is the subspace of constants. It suffices to show V is infinite-dimensional since if {/b/2,/3,. ..} is a basis for V, then {/0 s 1, Unr fj'.n e Z,j > 0} is a basis for L2(/h)-

(c) We now show V is infinite-dimensional Since ТЖ ^ Ж (we are assuming Л Ф i I wc know V Ф '0}. Lei g e V, g ф 0 and then G = [x:g{а) Ф 0} satisfies m(G) > 0. Since g is .^-measurable we have G e Ж and using (a) we know /СЖ = is infinite-dimensional. Also /СЖ = V' © /с, Ur.4C‘ where V с V so either V' is infinite-dimensional (and hence V is) or ylt UTJF is infinite-dimensional. In this second case there is a linearly independent sequence of functions {/GUrf„} where the f, are bounded functions in Ж. Then {gUTf,} are linearly independent in Ж. It suffices to show these functions are in V. But if/e Ж then

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