Let $(\Omega, {\cal F},P)$ be a complete probability space and $T$ a mesure-preserving transformation on $\Omega$ that is ergodic. The point-wise ergodic theorem states that for any $f\in L^1(P)$, $\frac{1}{N}\sum_{j=0}^{N-1}f(T^j \omega) \to \int_{\Omega}f(\omega)P(d\omega) \quad P\text{-a.e.}$ This is equivalent to saying that the set $A_{f}:=\left\{\omega\in\Omega; \frac{1}{N}\sum_{j=0}^{N-1}f(T^j \omega) \to \int_{\Omega}f(\omega)P(d\omega) \;\; \text{holds}\right\}$ has probability $1$. Notice here that the set of probability $1$ may be different for a different $f\in L^1(P)$.
My desired goal is to make a set with probability $1$ for which the above convergence of the Cesaro means holds regardless of the choice of $f\in L^1(P)$.
I think that one way to take for this objective is to take an intersection of $A_{f}$ over $f\in L^1(P)$. Then the dependence on $f$ disappears, i.e., for $\omega\in \cap A_{f}$ we have the above convergence for every $f\in L^1(P)$.
My question is; the intersection $\cap A_{f}$ is m'ble and has probability 1? (I think that if the intersection is taken over a countable index set, then the question is trivial and the answer is affirmative.) Is there any need for appropriate additional assumptions in order to ensure that the set $\cap A_{f}$ is measurable and has probability one?
Or are there other ways to take than taking the intersection of $A_{f}$?