Poincaré's recurrence theorem

Step 1 (points that never return). Let $A_0=\{x\in A: T^nx\notin A\ \forall n\ge1\}$ be the points of $A$ that never come back to $A$. The sets $A_0, T^{-1}A_0, T^{-2}A_0,\dots$ are pairwise disjoint: if $x\in T^{-i}A_0\cap T^{-j}A_0$ with $i<j$, then $T^ix\in A_0$ but also $T^j x = T^{j-i}(T^ix) \in A$ with $j-i\ge1$, contradicting that $T^ix\in A_0$ never returns to $A$. So $T^{-i}A_0 \cap T^{-j}A_0=\varnothing\ (i\ne j)$.

Step 2 (the never-return set has measure zero). Since $T$ is measure-preserving, $\mu(A_0)=\mu(T^{-i}A_0)$ for every $i$. If $\mu(A_0)>0$, the countably many pairwise disjoint sets $T^{-i}A_0$ ($i=0,1,2,\dots$) would all have this same positive measure, so their union would have infinite total measure — impossible since $\mu(X)<\infty$ (indeed $\mu(X)=1$). Hence $\mu(A_0)=0$.

Step 3 (finitely-often visitors also have measure zero). Let $B=\{x\in A: x\ \text{returns to}\ A\ \text{only finitely often}\}$. Writing $B$ as a countable union over $k$ of (essentially) the never-return set of $T^kA$ under the shifted dynamics, $B=\bigcup_{k\ge0} T^{-k}\{x\in T^kA: x\ \text{never returns to}\ T^kA\}$, each term has measure zero by exactly the Step 1–2 argument applied to $T^kA$ in place of $A$ (using $\mu(T^kA)=\mu(A)$). By countable subadditivity, $\mu(B)=0$.

Step 4 (conclusion). Every $x\in A\setminus B$ (which has full measure in $A$, since $\mu(B)=0$) returns to $A$ infinitely often by definition of $B$. This is exactly the statement of the theorem.