This is not an answer to the question, but this is too long for a comment.
I give what looks like an example where $f$ is not $L^1$ but where $f(T^n\cdot)/n \to 0$ a.e. I will state it in a probabilistic way (sorry about that).
Let $\mu$ be a probability measure on ${1,2,\dots}$. I assume that $m=\int k\mu(dk)$ is finite. I define a new probability measure by $\hat{\mu}(dk)=k/m.\mu(dk)$. Let us consider a Markov chain $((A_n,B_n))$ on $\Omega={(a,b) \Omega=\{(a,b) : a \ge 1, 1 \le b \le aa\} \subset N \times N$ with the following transitions probabilities :
1) From $(a,b)$ with $b>1$ one goes to $(a,b-1)$.
2) From $(a,1)$ one goes to $(c,c)$ where $c$ is chosen according to $\mu$.
Now let $\nu$ be a probability measure on $\Omega$ defined by $\nu({(a,b)})=\mu({a})/m$. \nu(\{(a,b)\})=\mu(\{a\})/m$. The above Markov chain admits $\nu$ as a stationary distribution and I consider it under this distribution. Note that $A_n$ is distributed according to $\hat{\mu}$. Now set $X_n=g(A_n)$ where $g$ is in $L^1(\mu)$ but not in $L^1(\hat{\mu})$. I think that $(X_n)_n$ is a counterexample (consider the Markov Chain seen at the times at wich which it jumps from $(a,\cdot)$ to $(c,c)$, then the first coordinates make a sequence of i.i.d.r.v. distributed according to $\mu$).
Note : where did my { and } escaped ?
