Let $f: X \rightarrow X$ be measurable (but not necessarily ergodic), $\mu$ an invariant probability measure, and $A \subseteq X$ be a set of positive measure. Prove that $$\limsup_n \frac{\vert \{ j \leq n \mid f^j(x) \in A\}\vert}{n} > 0$$ for almost every $x \in A$.
I am struggling to find a simple proof of this. That is, a proof which doesn't also show that the limit inferior is also positive. I've tried using Kac's return time lemma, but to no avail. I appreciate it!
Let $S_n^A=\sum_{j=1}^n {\bf 1}_A\circ f^j$ so that $S_n^A(x)= \vert \{ j \in [1,n] \,: \, f^j(x) \in A\}\vert$, and define $$B:=\Bigl\{x \in A: \limsup_{n \to \infty} \frac{S_n^A}{n} =0 \Bigr\} \,. $$ Our goal is to show that $\mu(B)=0$, so suppose that $\mu(B)>0$.
Since $\mu$ is an invariant measure, $$ \int_X S_n^B \, d\mu= \sum_{j=1}^n \int_X {\bf 1}_B\circ f^j \, d\mu =n \mu(B) \tag{*} \,.$$ Let $\psi=\limsup_{n \to \infty} S_n^B/n \,.$ Observe that $${\bf 1}_B+(S_n^B\circ f) =S_{n+1}^B \,. $$ Dividing by $n$ and taking limsup yields $\psi\circ f =\psi$, so $\psi\circ f^k=\psi$ for all $k \ge 1$.
By Fatou's lemma applied to the non-negative functions $1-S_n^B/n$, we have $$\int_X \psi \, d\mu \ge \limsup_{n \to \infty} \int_X S_n^B/n \, d\mu=\mu(B)\,.$$ Since $\psi$ vanishes outside $\cup_{k \ge 1} f^{-k}(B)$, there must exist some $k$ such that $$0<\int_X \psi \cdot {\bf 1}_{f^{-k}(B)} \, d\mu= \int_X (\psi \circ f^k) \cdot ({\bf 1}_{B}\circ f^k) \, d\mu= \int_X \psi \cdot {\bf 1}_{B} \, d\mu \,.$$ But $B \subset A$, so for all $x \in B$, $$0 \le \psi(x) \le \limsup_{n \to \infty} \frac{S_n^A(x)}{n}=0 \,,$$ using the definition of $B$ in the last step. This yields the desired contradiction.