Let $X$ be a second countable Hausdorff topological space, let $T \colon X \to X$ be a Borel-measurable map, define the $\sigma$-algebra $\mathcal{I}=\{A \in \mathcal{B}(X) : T^{-1}(A)=A\}$, and for each non-empty set $S \subset X$ let $\mathcal{I}[S]$ be the $\sigma$-algebra on $S$ induced from $\mathcal{I}$.
Given a $T$-invariant Borel probability measure $\mu$, are the following two statements equivalent?
- There is a $\mu$-full measure set $\tilde{X} \subset X$ such that $\mathcal{I}[\tilde{X}]$ is countably generated.
- $\mu$-almost every $x \in X$ is a fixed or periodic point of $T$.
The direction 2$\Rightarrow$1 should be pretty much immediate: if $\mathcal{U}$ is a countable generator of $\mathcal{B}(X)$ then I think $\mathcal{I}[\mathrm{Per}(T)]$ is generated by the countable collection $$ \left\{ \mathrm{Per}_n(T) \cap \bigcup_{i=0}^{n-1} T^{-i}(U) : U \in \mathcal{U}, \, n \geq 1 \right\}\text{.} $$
I can prove 1$\Rightarrow$2 when $(X,\mathcal{B}(X))$ is a standard Borel space: Since $(X,\mathcal{B}(X))$ is a standard Borel space, we can find a family $(\mu_x)_{x \in X}$ of Borel probability measures $\mu_x$ on $X$ such that for each $A \in \mathcal{B}(X)$, $x \mapsto \mu_x(A)$ is an $\mathcal{I}$-measurable version of $\mathbb{E}_\mu[\mathbf{1}_A|\mathcal{I}]$. Hence in particular, if $A \in \mathcal{I}$ then $\mu_x(A)=\delta_x(A)$ for $\mu$-a.a. $x$. Using the countable generation of $\mathcal{B}(X)$, it is not hard to show that $\mu_x$ is $T$-invariant for $\mu$-a.a. $x$. Let $\tilde{\mathcal{C}} \subset \mathcal{I}$ be a countable collection such that $\{E \cap \tilde{X} : E \in \tilde{\mathcal{C}}\}$ is a generator of $\mathcal{I}[\tilde{X}]$, and let $\mathcal{C}$ be the set of finite intersections of members of $\tilde{\mathcal{C}}$. Let $\tilde{\tilde{X}} \subset \tilde{X}$ be a $\mu$-full measure set such that for all $x \in \tilde{\tilde{X}}$,
- $\mu_x$ is $T$-invariant;
- $\tilde{X}$ is a $\mu_x$-full measure set;
- $\mu_x$ agrees with $\delta_x$ on $\mathcal{C}$.
Using Dynkin's $\pi$-$\lambda$ theorem, it is not hard to show that for all $x \in \tilde{\tilde{X}}$, $\mu_x$ agrees with $\delta_x$ on $\mathcal{I}$, and hence in particular, $\mu_x$ is $T$-ergodic. So, since $$ x \in \bigcup_{m=0}^\infty T^{-m}(\{T^n(x)\}_{n \geq 0}) \in \mathcal{I}\text{,} $$ it follows that $\mu_x(\{T^n(x)\}_{n \geq 0})>0$; but any ergodic measure assigning positive measure to a countable set must be supported on a periodic orbit or fixed point, and so $\tilde{\tilde{X}} \subset \bigcup_{m=0}^\infty T^{-m}(\mathrm{Per}(T))$. But the sequence $T^{-m}(\mathrm{Per}(T))$ is an increasing sequence of sets all of the same measure, and so the fact that $\bigcup_{m=0}^\infty T^{-m}(\mathrm{Per}(T))$ is a $\mu$-full measure set implies that $\mu(\mathrm{Per}(T))=1$.
