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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 $A_{f}$?

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The intersection of uncountably many sets of probability 1 can be the empty set. – azarel Jul 17 '12 at 15:01
up vote 5 down vote accepted

Given any $\omega \in \Omega$, define a function $f_\omega$ by $f_\omega(T^k(\omega))=1$, $f_\omega(\alpha)=0$ for all other $\alpha$. Then $$\int f_\omega = \sum_{k=0}^\infty P(T^k(\omega))=\sum_{k=0}^\infty P(\omega)$$ by translation-invariance. Since $\Omega$ is a probability space, this last sum cannot diverge and thus must vanish.

But by construction $\frac{1}{N}\sum_k T^k(\omega)=1$ for all $N$. So $\omega \not\in A_{f_{\omega}}$. Since $\omega$ was arbitrary, $\cap A_f$ is empty.

In other words, you can't remove the $f$-dependence from $A_f$; given any point, you can always concoct some function that's badly behaved (ergodically speaking) at that point.

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Thank you for your help. I wish to ask one more question: What if we restrict $f$ to be continuous and integrable? – Yamamoto Jul 17 '12 at 15:36
@DavideGiraudo: Thanks, now I see what you were getting at! I've updated that bit of the proof. – Micah Jul 17 '12 at 16:01
@Haruo: We're in an arbitrary probability space, which need not even have a specified topology. It wouldn't surprise me if there's some result like what you're looking for in sufficiently nice situations, but I'm far enough out of my field that I couldn't tell you what the right version of "sufficiently nice" might be. – Micah Jul 17 '12 at 16:24

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