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Let $(X,A,\nu)$ be a probability space and $T\colon X\to X$ a measure preserving transformation $\nu$. Take a measurable partition $P=\{P_0,\dots,P_{k-1}\}$. Let$I$ be a set of all possible itineraries, that is, $I=\{(i_1,\dots,i_n,\dots)\in k^N;$ there is a $x\in X$, such that $T^n(x)\in P_{i_n}$ for all $n\in\Bbb N$.

Suppose that $I$ is countably infinite.

Is true that the entropy of $T$ with respect to $P$ is $0$ ($h(T,P)=0$)?

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Do you mean "denumerable", i.e. countably infinite? – Robert Israel Dec 28 '12 at 20:32
If $\nu$ is ergodic, I know that is true. – user52188 Dec 28 '12 at 20:33
yes, countably infinite – user52188 Dec 28 '12 at 20:36
this question has been solved here: – Stéphane Laurent Jan 25 '13 at 22:52

Let ${(Z_n)}_{n \geq 0}$ be the stationary (shift-invariant) process defined by $Z_n(x)=P(T^n(x))$ where $P(x)$ denotes the element of the partition $P$ to which $x$ belongs. The assumption of the question is equivalent to say that the $\sigma$-field ${\cal F}_0:=\sigma(Z_n; n \geq 0)$ is (purely) atomic (modulo $\nu$). The entropy of $T$ with respect to $P$ is $h(T,P)=\lim \frac{H(Z_1,\ldots,Z_n)}{n}$ and the question is to prove that $h(T,P)=0$.

  • As mentioned by Davide, a first proof has been given at MathOverflow. The answer says that the law of the process ${(Z_n)}_{n \geq 0}$ is a certain convex combination of shift-invariant periodic measures and this implies $h(T,P)=0$.

  • A second proof has been given as an answer to this question which is the more general question obtained by dropping the stationarity assumption.

  • Now let me mention a third proof in the stationary case. Denoting by ${\cal F}_1=\sigma(Z_n; n \geq 1)$, the equality $h(T,P)=H({\cal F}_0 \mid {\cal F}_1)$ is a well-known result in ergodic theory. But I claim that ${\cal F}_1={\cal F_0}$ under the assumption of the question, therefore $H({\cal F}_0 \mid {\cal F}_1)=0$. My claim is a consequence of the following lemma whose proof is left to the reader:

Lemma. Let ${\cal F_0}$ and ${\cal F_1}$ be two $\sigma$-fields such that

  • ${\cal F_0} \supset {\cal F_1}$;
  • ${\cal F_0}$ is isomorphic to ${\cal F_1}$;
  • ${\cal F_0}$ and ${\cal F_1}$ are purely atomic.

Then ${\cal F_0} = {\cal F_1}$.

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