How to prove the Picard recurrence theorem studying for the ergodic theory course, I have the following recurrence theorem:

Theorem: Let $(X,\mathcal{F},\mu,T)$ a system that preserves measure and let
$A\in\mathcal{F}$ such that $\mu(A) > 0$. Then exists
$n\in\mathbb{N}\setminus\{0\}$ such that $\mu(A\cap T^{-n}(A)) > 0$.

Proving the theorem and understanding it was not a problem for me,but now I have the following recurrence result which has complicated me

Theorem: Under the same hypotheses of the previous theorem, for
$\mu$-a.e., all $x\in A$, exists $n\in\mathbb{N}\setminus\{0\}$ such that $T^{n}x\in A$.

Hint to proof: The hint is that
\begin{equation*}
        A\cap T^{-k}\left(A\bigcap_{k\in\mathbb{N}\setminus\{0\}} T^{-n}(A^{c})\right)
    \end{equation*}
is the set of points that return in time k, but never return after that.
How can I understand this set and then use? I honestly don't see the meaning that the indication gives.
 A: I don’t quite know how to interpret your hint. Also, I don’t know how to do this if the space is not finite measure. But, I think I can prove this statement in a similar way to the hint if I can assume $X$ is a finite measure space.
(I’ve never studied dynamical systems but this seems false if $X$ is not a finite measure space. If $X = \mathbb{R}$ with lebesgue measure, $A = [0,1]$ and $T(x) = x+1$ then $T$ is measure preserving but none of the points of $A$ return to $A$.)
Now to prove the question!! Note that $$ f^n(x) \in A \iff x \in f^{-n}(A)$$ and so we see that the set of points which never return to $A$ is given by $$ B:= \{ x \in A :\forall n \in \mathbb{N},\;  f^n(x) \notin A \} = A \setminus \bigcup_n f^{-n}(A) = A \cap \bigcap_n f^{-n}(A^c) .$$
$B$ is the points in $A$ which never return to $A$. Suppose $B$ has positive measure. Since $f$ is a measure preserving transformation $$\mu( f_n(B) ) = \mu (B) $$ for all $n$. However, we also see that for all $n$, the sets $f^n(B)$ and $f^m(B)$ are pairwise disjoint for $n \neq m $. This is because \begin{align}
& \;f^{n}(B) \cap f^m(B) &= \emptyset \\
\iff & \; f^{n-m}(B) \cap B& = \emptyset .\end{align}
However, we know that $f^{n-m}(B) \cap B = \emptyset$ since $B \subset A$ and $B$ is the points which never return to $A$. Hence, $f^n(B)$ and $f^m(B)$ are disjoint for all positive $m \neq n$.
Therefore, since $f^n(B) \subset X$ for all $n$ and $\{f^n(B)\} $ are pairwise disjoint, we have from countable additivity of measure and the subset properties of measure that $$  \sum_n \mu(B) = \sum_{n} \mu(f^n(B)) = \mu\big(\bigcup_n f^n(B)\big) \leq \mu(X) .$$ But, the left hand side blows up if $B$ has positive measure contradicting the fact that $X$ has finite measure. Hence, $B$ has zero measure. Or stated equivalently, almost every point of $A$ returns to $A$.
