Birkhoff average, non-ergodic case 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!
 A: 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.
