I have a question to an exercise which was already posted (but I'm not allowed to comment it).
"Let $X$ be a compact metric space with metric $d$, endow $X$ with the Borel $\sigma$-algebra and a probability measure $\mu$. Let $T\colon X\to X$ be a continuous map which is $\mu$-preserving. Then for $\mu$-almost every $x\in X$ there is a sequence $n_k\to\infty$ in $\mathbb{N}$ such that $T^{n_k}x\to x$ as $k\to\infty$."
I don't understand this proof given in this thread:
"Let $\mathcal P_1,\mathcal P_2,\ldots$ be a sequence of refining partitions of decreasing diameter (converging to 0) of $X$. Let $\delta_n$ be the diameter of $\mathcal P_n$. Let $\mathcal P_n(x)$ be the element of $\mathcal P_n$ containing $x$. For $\mu$-a.e. $x$, $\mu(\mathcal P_n(x))\ne 0$."
Define $A:=\{x:\mu(P_n(x))=0\}$ => $B\subset\cup_{x \in A}(P_n(x))$. Now I can just deduce $\mu(A)=0$ if A is countable.
"Now the Poincaré recurrence theorem applies to each element (with positive measure) of $\mathcal P_n$, saying that for almost every $z\in B\in\mathcal P_n$, $z$ returns to $B$ infinitely often. Hence there exist infinitely many $k^{(n)}_i$ such that $d(T^{k^{(n)}_i}z,z)\le \delta_n$. Let the union of the exceptional sets for each $B$ in $\mathcal P_n$ be $E_n$. Let $E$ be the union of the $E_n$. Then $E$ has measure 0."
I don't see why $\mu(E_n)=0$ … it's clear that the measure of the exceptional set for every B is 0, but why should the measure of the union be 0? If $P_n$ is uncountable I can't use $\sigma$-additivity (same problem as before).
"Now just put it all together. Let $E'$ be the set of $x'$ such that $\mu(\mathcal P_n(x))=0$ for some $n$. This is another set of measure 0."
Again what happens if $E'$ is uncountable??
"Now if $x\in X\setminus(E\cup E')$, we're in business: $\mu(\mathcal P_n(x))>0$ for all $n$ and $x$ returns to $\mathcal P_n(x)$ infinitely often. A diagonal argument finishes things off."
Okay $\mu(X\setminus(E\cup E'))=\mu(X)$ (for what I need $\mu(\mathcal P_n(x))>0$)?) We have sequences $(n_{m_k})$ such that $T^{m_k}(x) \in P_m(x)$. Since $diam(P_n)\to 0$ it follows that $T^{n_{k_k}}(x)\to x$ for $k\to\infty$ (do I need compactness here?)