In Ergodic Theory with a view towards Number Theory book, of Einsiedler and Ward, there is a proof of Poincaré Recurrence.
Theorem: Let $T:X\to X$ be a measure-preserving transformation on a probability space $(X, \Sigma, \mu)$, and let $E\subseteq X$ be a measurable set. Then there is a measurable set $F\subseteq E$ such that $\mu(F)=\mu(E)$ and that for every $x\in F$ there exist integers $0<n_1<n_2<\dots$ with $T^{n_i}x\in E$ for all $i\geq 1$.
Proof: We consider a set
$$B=E\cap T^{-1}(X\setminus E)\cap T^{-2}(X\setminus E)\cap \dots .$$ So, for any $n\geq 1$, we have
$$T^{-n}B=T^{-n}E\cap T^{-n-1}(X\setminus E)\cap\dots .$$
All these $T^{-n}B$ have the same measure $\mu(B)$, since $T$ is $\mu$-preserving. Thus $\mu(B)=0$ (and I guess this is because $\mu(X)=1<\infty$).
My question now is: what is the subset $F$ of $E$ that will satisfy the Theorem? Or what are the sets $F_n$ that I should consider so that their countable intersection is a valid candidate?
Best Answer
The set $$F=E \:\backslash \:\Biggl(\:\bigcup_{n=0}^{\infty}T^{-n}B\Biggr)$$ satisfies the conditions given.
B is the set of elements of $E$ that immediately leave $E$ and never return. Because of this, we must have $$T^{-i}B\cap T^{-j}B=\emptyset \quad\quad\quad \forall \:i,j\geq 0,\quad i\neq j $$ If this failed for some $i < j$, we would have for some $x$, $$T^i x\in B$$$$T^j x=T^{j-i}(T^i x)\in B$$ and so the set $T^i x$ would be elements of $B$ that return to $B$ in $j-i$ steps, which contradicts the definition of $B$.
We then have by measure preservation of $T$, $$\mu \Biggl(\bigcup_{n=0}^{\infty}T^{-n}B\Biggr)=\sum_{n=0}^{\infty}\mu(B)\leq 1 \implies \mu(B)=\mu \Biggl(\bigcup_{n=0}^{\infty}T^{-n}B\Biggr)=0$$ So, $\mu(F)=\mu(E)$.
F satisfies the second condition given in the theorem, because the condition is equivalent to $T^n x\in E$ for infinitely many $n$. If for an $x\in F$, we had $T^n x\in E$ for finitely many $n$, we would have $T^m x\in B$ for a finite $m$, and so, $$x\in \bigcup_{n=0}^{\infty}T^{-n}B\implies x \notin F$$ a contradiction.