Let be $(X,\mathcal{A},\mu)$ a probability space, $T:X\to X$ a measurable tranformation preserving $\mu$ and $f: X \rightarrow \mathbb C$ a measurable function. Show that for almost every $x \in X$ either:
\begin{equation} \lim_{N \rightarrow \infty} \frac{1}{N} \sum_{n=0}^{N-1} |f(T^n(x))| = \infty \end{equation}
or
\begin{equation} \lim_{N \rightarrow \infty} \frac{1}{N} \sum_{n=0}^{N-1} f(T^n(x)) \quad\text{exists and it's finite.} \end{equation}
My attempt: Since $f$ isn't necessarily integrable, i tried to applicate Birkhoff ergodic theorem to $f_m = f 1_{|f| \leq m} + m 1_{|f| > m}$ which is an integrable function. The problem that i have is that i can't seem to get an upper bound for $\int_E f $, where $E = \{ x | \sup_N S_N^{|f|}(x) < + \infty \}$ where $S_N^f$ are the Birkhoff averages of $f$. I have shown that $E$ an invariant set.
Does $S_N^f(x)$ converge to $\int_E f d\mu$ or i am on a bad track?
Suppose we have the result when $f$ is real. We deduce the result for $f$ complex. For a.e. $x$, either (1) $\lim_N \frac{1}{N}\sum_{n \le N} Re[f](T^nx)$ exists and is finite or (2) $\lim_N \frac{1}{N}\sum_{n \le N} |Re[f](T^nx)| = +\infty$. Same with $Im[f]$. For a given $x$, if (1) occurs for $Re[f]$ and $Im[f]$ at $x$, then $\lim_N \frac{1}{N}\sum_{n \le N} f(T^nx)$ exists and is finite. And if (2) occurs for either $Re[f]$ or $Im[f]$ at $x$, then $\lim_N \frac{1}{N}\sum_{n \le N} |f(T^nx)| = +\infty$. Since a.e. $x$ has (1) or (2) occurring for each $Re[f]$ and $Im[f]$ at $x$, we're done.
The proof above allows us to deduce the result for $f$ real if we have the result for $f$ non-negative.
So suppose $f$ is non-negative. For $M \ge 1$, let $E_M := \{x \in X : \liminf_N S_N^f(x) \le M\}$. Since $E_M$ is obviously $T$-invariant, we may consider the probability space $(E_M,\mathcal{A}\cap E_M, \mu|_{E_M})$ and apply Birkhoff to $f_k := f1_{f \le k} \in L^1(E_M)$ for each $k \ge 1$ to get that for each $k \ge 1$, $S_N^{f_k}(x) \to f_k^*(x)$ for a.e. $x \in E_M$, for some $f_k^* \in L^1(E_M)$ with $\frac{1}{\mu(E_M)}\int_{E_M} f_k^*(x)d\mu(x) = \frac{1}{\mu(E_M)}\int_{E_M} f_k(x)d\mu(x)$ (this is a version of Birkhoff applying to measure preserving systems, not necessarily ergodic ones). The point is that, by the definition of $E_M$, $\frac{1}{\mu(E_M)}\int_{E_M} f_k(x)d\mu(x) \le M$ for each $k$, so by MCT ($f_k$'s increase to $f$), $\frac{1}{\mu(E_M)}\int_{E_M} f(x)d\mu(x) \le M$, i.e., $f$ is integrable on $E_M$! Therefore, we can use Birkhoff on $f$ to get that for a.e. $x \in E_M$, $\lim_N S_N^f(x)$ exists, and is finite since it is at most $M$. Therefore, for a.e. $x$ for which $\liminf_N S_N^f(x) < \infty$, the limit $\lim_N S_N^f(x)$ exists and is finite. And of course for every $x$ for which $\liminf_N S_N^f(x) = +\infty$ we have $\lim_N S_N^{|f|}(x) = +\infty$ (recall $f$ is non-negative).