For an invertible measure preserving system, show that $\lim_NA_f^T(N)=\lim_N A_f^{T^{-1}}(N)$.
Here we consider the measure preserving system $(X,\mathcal A,\mu,T)$ where $T$ is invertible and $\mu$ is a probability measure. For any $f\in \mathcal L_\mu^1$ define $A_f^T(N)=\dfrac{1}{N}\sum_{n=0}^{N-1}f\circ T^n$. Then we know that by Birkhoff ergodic theorem and Mean ergodic theorem, $A_f^T(N)\to f'\in\mathcal L_\mu^1$ with $f'$ $T-$invariant i.e. $f'\circ T=f'$ and for any $T-$invariant measurable set $A,$ we have $\int_Af'd\mu=\int_Af d\mu$.
Now I want to show that if in particular, $T$ is invertible, then $\lim_N\dfrac{1}{N}\sum_{n=0}^{N-1}f(T^n(x))=\lim_N \dfrac{1}{N}\sum_{n=0}^{N-1}f(T^{-n}(x))$ a.e.
I observed that the second limit is $T$ invariant and if I call it $g$ then for any $T-$invariant m'ble set $A$, $\int_Agd\mu=\int_Afd\mu=\int_Af'd\mu$.
But nothing more than that. I want to show that $f'=g$ a.e.
Hint: note that two invariant integrable functions are equal almost everywhere if and only if their integrals are equal on every invariant set; why? :)