[proofplan] We first prove that the set of points in $A$ which never return to $A$ has measure zero. The key idea is that all backward iterates of this non-returning set are pairwise disjoint and have the same measure, which is impossible inside a finite-measure space unless that measure is zero. We then apply the same one-return argument to every iterate system $(X,\mathcal{B},\mu,T^N)$, $N \geq 1$, to force returns after arbitrarily large times. A countable intersection of full-measure subsets of $A$ then gives infinitely many returns for $\mu$-almost every point of $A$. [/proofplan] [step:Show that almost every point of $A$ returns to $A$ at least once] Define the non-returning set $B \in \mathcal{B}$ by \begin{align*} B := A \cap \bigcap_{n=1}^{\infty} T^{-n}(X \setminus A). \end{align*} Equivalently, $x \in B$ if and only if $x \in A$ and $T^n(x) \notin A$ for every integer $n \geq 1$. For each integer $m \geq 0$, define \begin{align*} E_m := T^{-m}(B) \in \mathcal{B}. \end{align*} We claim that the sets $(E_m)_{m \geq 0}$ are pairwise disjoint. Indeed, suppose that $0 \leq j < k$ and that $x \in E_j \cap E_k$. Then $T^j(x) \in B \subseteq A$ and $T^k(x) \in B \subseteq A$. Since $k-j \geq 1$, \begin{align*} T^{k-j}(T^j(x)) = T^k(x) \in A, \end{align*} which contradicts the defining property of $T^j(x) \in B$, namely that no positive iterate of $T^j(x)$ lies in $A$. Because $T$ preserves $\mu$, the equality \begin{align*} \mu(T^{-1}(C)) = \mu(C) \end{align*} holds for every $C \in \mathcal{B}$. Iterating this identity gives \begin{align*} \mu(E_m) = \mu(T^{-m}(B)) = \mu(B) \end{align*} for every integer $m \geq 0$. Since the sets $(E_m)_{m \geq 0}$ are pairwise disjoint subsets of $X$, countable additivity and monotonicity give, for every integer $M \geq 0$, \begin{align*} (M+1)\mu(B) = \sum_{m=0}^{M} \mu(E_m) = \mu\left(\bigcup_{m=0}^{M} E_m\right) \leq \mu(X). \end{align*} Since $\mu(X) < \infty$, this inequality for all $M$ forces $\mu(B)=0$. Therefore $\mu$-almost every point of $A$ returns to $A$ at least once. [/step] [step:Apply the one-return result to every iterate system] For each integer $N \geq 1$, define the map \begin{align*} S_N: X &\to X \\ x &\mapsto T^N(x). \end{align*} Since $T$ is measurable and measure-preserving, $S_N=T^N$ is measurable and measure-preserving: for every $C \in \mathcal{B}$, \begin{align*} \mu(S_N^{-1}(C)) = \mu(T^{-N}(C)) = \mu(C). \end{align*} Thus $(X,\mathcal{B},\mu,S_N)$ is again a measure-preserving system with $\mu(X)<\infty$. Applying the first step to the system $(X,\mathcal{B},\mu,S_N)$ and the same set $A$, define \begin{align*} B_N := A \cap \bigcap_{m=1}^{\infty} S_N^{-m}(X \setminus A). \end{align*} Then $\mu(B_N)=0$. Equivalently, for every $x \in A \setminus B_N$, there exists an integer $m \geq 1$ such that \begin{align*} S_N^m(x)=T^{Nm}(x) \in A. \end{align*} In particular, for every $x \in A \setminus B_N$, there is an integer $n \geq N$ such that $T^n(x) \in A$. [/step] [step:Intersect the full-measure return sets to obtain infinitely many returns] Define the exceptional set \begin{align*} B_{\infty} := \bigcup_{N=1}^{\infty} B_N. \end{align*} Since each $B_N \in \mathcal{B}$ and $\mu(B_N)=0$, countable subadditivity gives \begin{align*} \mu(B_{\infty}) \leq \sum_{N=1}^{\infty} \mu(B_N) = 0. \end{align*} Hence $\mu(B_{\infty})=0$. Let $x \in A \setminus B_{\infty}$. Then for every integer $N \geq 1$, the point $x$ belongs to $A \setminus B_N$, so there exists an integer $n_N \geq N$ such that \begin{align*} T^{n_N}(x) \in A. \end{align*} Therefore the set \begin{align*} R(x) := \{n \in \mathbb{N} : n \geq 1 \text{ and } T^n(x) \in A\} \end{align*} is unbounded in $\mathbb{N}$. An unbounded subset of $\mathbb{N}$ is infinite, so $T^n(x) \in A$ for infinitely many integers $n \geq 1$. Thus, outside the null set $B_{\infty} \subseteq A$, every point of $A$ returns to $A$ infinitely often. This proves that for $\mu$-almost every $x \in A$, the orbit $(T^n x)_{n \geq 1}$ returns to $A$ infinitely many times. [/step]