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I need some background before asking the question:

Let $\mathcal{P}$ is called a measurable partition if there is a measurable set $M_0\subset M$ with full probability measure such that, restric to $M_0$ $$\mathcal{P}=\bigvee_{n=1}^\infty \mathcal{P}_n$$ for some crescent sequence $\mathcal{P}_1 \prec \dots \prec \mathcal{P}_n \prec \dots$ of enumerable partitions, where $\mathcal{P}_i\prec \mathcal{P}_{i+1}$ means that every element of $\mathcal{P}_{i+1}$ is contained in some element of $\mathcal{P}_i$ (we say that $\mathcal{P}_i$ is less thin than $\mathcal{P}_{n+1}$) and also $\bigvee_{n=1}^\infty \mathcal{P}_n$ is the thinnest such that $$\mathcal{P}_n \prec \bigvee_{n=1}^\infty \mathcal{P}_n, \forall n$$

It's elements are of the form $\cap_{n=1}^\infty P_n$, where $P_n \in \mathcal{P}_n$.

Questions: i) Show that a partition $\mathcal{P}$ is a measurable partition iff there exists measurable subsets $M_0, E_1, E_2,\dots,E_n,\dots$ such that $\mu(M_0)=1$ and, restrict to $M_0$, $$\mathcal{P}= \bigvee_{n=1}^\infty \{E_n, M\setminus E_n\}$$

ii) If $\mu$ is an ergodic probability for a transformation $f$ and $k\geq2$. How is the ergodic decomposition of $f^k$?

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And your thoughts on this are? –  Did Feb 12 '13 at 19:42
    
First, thank you for being interested!! Wel... for i), $(\Leftarrow)$ I have a clue about how it works. For the other implication, I was planning on, for each $E_i$, maybe taking it from the previews $\mathcal{P}_{i-1}$ but i couldn't manage to do it wisely. For ii), I can see that the decomposition of $f^k$ is "bigger" because we every measure that is ergodic for $f$ is ergodic for $f^k$, but I don't have a clue how to control those other measures. –  user35682 Feb 12 '13 at 20:22
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These should be in the question. –  Did Feb 12 '13 at 20:43
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1 Answer

up vote 1 down vote accepted

For i), we can suppose each partition is finite, i.e., $\mathcal P_n=\{P_1,\cdots,P_{k_n}\}$, then define $E_{k_i}=P_{k_i}$, then one gets $\mathcal P_n\prec \bigvee_{i=1}^{k_n}\{E_{k_i},M\backslash E_{k_i}\}$.Thus, taking all the $E_j$ in this form, we have $\bigvee_n \mathcal P_n\prec \bigvee_j \{E_j,M\backslash E_j\}$.

For ii), Consider $k=2$. There are two cases. Case 1. For any $A\subset M$ such that $f^{-2}(A)=A$, $\mu(A)>0$, we also have $f^{-1}(A)=A$, this implies that $\mu(A)=0$ or $1$. In this case, we have $(f^2,\mu)$ is also ergodic.

Case two. there exists some $A$ such that $f^{-2}(A)=A$, but $\mu(A\backslash f^{-1}(A))>0$, in this case,

$f^{-1} (A\backslash f^{-1}(A)\cup f^{-1}(A)\backslash A)=A\backslash f^{-1}(A)\cup f^{-1}(A)\backslash A$ and therefore, $\mu(A\Delta f^{-1}(A))=1$, which implies that $M=A\cup f^{-1}(A)$.

Now define $\nu_1$ such that $\nu_1(E)=2\mu(E\cap A)$ and $\nu_2(E)=2\mu(E\cap A^c)$, then both $\nu_2$ and $\nu_2$ are ergodic and $\mu=\frac{1}{2} (\nu_1+\nu_2)$.

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So questions i) and ii) are not related ? –  Stéphane Laurent Feb 13 '13 at 22:40
    
don't think so. –  Xiaochuan Feb 14 '13 at 12:03
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