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I see this question on the math stack exchange. I found it interesting and still there is no solution there

Let $(X,A,\nu)$ be a probability space and $T:X\to X$ a measure preserving transformation $\nu$. Take a measurable partiton $P=\{P_0,...,P_{k-1}\}$. Let I be a set of all possible itineraries, that is, I=$\{(i_1,...,i_n,...)\in k^{N}$; there is a $x\in X$, such that $T^n(x)\in P_{i_n}$ for all n$\in$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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Can you say something about where the question comes from? It sounds a bit like a homework assignment... –  Anthony Quas Jan 13 '13 at 4:26
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2 Answers

This condition is much more restrictive than just having zero entropy. In fact you are talking about a purely atomic invariant measure of the shift on $A^{\mathbb Z_+}$, where $A$ is a finite alphabet. Any such measure is a (possibly infinite) convex combination of measures equidistributed on orbits of periodic points.

EDIT As the exposition of Daniel contains a number of mistakes, let me expand my answer.

  1. The partition $P$ gives rise to a symbolic coding of the original transformation by assigning to any point $x\in X$ the sequence of $P$-names of $(T^n x)_{n\ge 0}$, i.e., $$ x \mapsto (i_0,i_1,\dots) \;, $$ where $i_n$ is determined by the condition $T^n x \in P_{i_n}$. The image of the original measure on $X$ under this transformation is a shift invariant measure.

  2. Now one can forget about the original measure and deal just with the quotient shift invariant measure on $A^{\mathbb Z_+}$ (where $A$ is the set of elements of the partition $P$). By the assumption it is purely atomic. I claim that any purely atomic probability measure $m$ invariant under a certain transformation $T$ is a convex combination of uniform measures on periodic orbits. Indeed, let $x$ be a maximal weight atom of $m$. Then $m(Tx)\le m(x)$ because the weight of $x$ is maximal; on the other hand, $m(Tx)\ge m(x)$ because $x\in T^{-1}(Tx)$ and $T$ is measure preserving. Therefore $m(Tx)=m(x)>0$, so that $x$ must be a periodic point. Now one can remove the orbit of $x$ and apply this argument again, etc.

  3. The entropy of a periodic measure preserving transformation is zero.

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You can drill a little more your answer? –  Juan Valdez Jan 12 '13 at 21:50
What concretely? –  R W Jan 12 '13 at 22:20
I like this answer because it answers what the question should have been. –  Daniel Mansfield Jan 13 '13 at 8:39
I have asked the similar question without the stationarity assumption here: math.stackexchange.com/questions/285709/… –  Stéphane Laurent Jan 25 '13 at 22:51
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I think that I can expand on RW's answer

If $I$ is countably infinite, then we can cover $X$ with the countable partition $$ X = \vee_{n=0}^\infty T^{-n}P $$ Take any element of this partition: $$Q = P_{n_0} \vee T^{-1}P_{n_1} \vee \cdots \vee T^{-i}P_{n_i} \vee \cdots $$ where $n_i \in [1,k]$. Then $$ TQ = TP_{n_0} \vee P_{n_1} \vee T^{-1}P_{n_2} \vee \cdots \vee T^{-i+1}P_{n_i} \vee \cdots $$ is contained within partition element $$ P_{n_1} \vee T^{-1}P_{n_2} \vee \cdots \vee T^{-i+1}P_{n_i} \vee \cdots $$ so $TQ$ (or more generally $T^iQ$) is either disjoint from $Q$ (when it is contained within another partition), or equal to $Q$

There are two possibilities for the measure of $Q$:

(1) the measure of $Q$ is zero, or

(2) the measure of $Q$ is positive, and $T^nQ=T^{n+k}Q$ for some $n,k < \infty$. Otherwise $\{T^iQ\}$ is an infinite sequence of disjoint sets, each of constant measure $\mu(Q)$, and

$$ \infty = \sum_{i=1}^\infty \mu(Q) = \sum_{i=1}^\infty \mu(T^iQ) \leq \mu(X) = 1$$

Hence there are (possibly infinitely many) $Q$'s of finite measure, each with finite orbit.

Each $Q$ represents the set of points that are indistinguishable in your topology, so we can say that the space you are talking about is really just a (countable) union of finite orbits with an atomic probability measure.

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It's not clear a priori why $TQ$ should be another element of the partition... –  Anthony Quas Jan 13 '13 at 4:24
Thank you Anthony, $TQ$ is not necessarily an element of the partition. I have modified the answer in response to your comment. –  Daniel Mansfield Jan 13 '13 at 5:48
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