Birkhoff Ergodic Theorem or Counterexample The Birkhoff Ergodic Theorem states:
Let $(X,\mathcal{B},m)$ be a finite or sigma finite measure space.  Suppose  $T:(X,\mathcal{B},m)\to (X,\mathcal{B},m)$ is measure-preserving and $f\in L^1(m)$. Then 
$$\lim_{n\to \infty} \frac{1}{n} \sum_{i=0}^{n-1} f(T^i(x))$$
converges a.e. to a  function $g\in L^1(m)$. 
Can we say anything if $(X,\mathcal{B},m)$ is larger than sigma finite?  I know in applications these sort of spaces are just "too large", but I can't come up with a proof or find a counter-example (i.e. $f$ and $T$ satisfying the hypothesis such that the yielded $g \not \in L^1(m)$.
I've also posted this here: 
https://math.stackexchange.com/questions/785707/birkhoff-ergodic-theorem-counterexample
 A: The support $S:=\{f\neq0\}$ of an integrable function $f$ is in any case $\sigma$-finite, and so is the $T$-invariant set $$X':=\limsup_{j\to\infty} T^{-j}(S):=\bigcap_{k\ge0}\bigcup_{j\ge k}T^{-j}(S). $$  Clearly the time average is zero in the complement of $X'$ (indeed, for these the points $f(T^j(x))$ eventually vanishes), and we are reduced  to the $\sigma$-finite case  of the  measure-preserving transformation $ T_{|X'}:X'\to X'$.
Rmk: Note that both $T(X')\subset X'$ and $T^{-1}(X')\subset X'$ hold true, and are needed, in order that $ T_{|X'}$ be a true measure-preserving transformation on $ X'$, so that the reduction argument  holds plainly.
A: One can wlog reduce to the $\sigma$-finite case by splitting into the part where $f(T^i(x))=0$ for all $i\in\mathbb{N}$ where nothing happens and the $\sigma$-finite subset $\bigcup_{i} \{|f|\circ T^i\neq 0\}$ where there is really something to prove. Note that $\{|h|\neq 0\}$ is $\sigma$-finite for every integrable $h$ (because $\mu\{|h|\geq\epsilon\}\leq \frac{1}{\epsilon}\|h\|_{L^1}$) and all $f\circ T^i$ are integrale.
