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Originally posted on Maths StackExchange, but repositing here because of getting no answer there. Not a research question really - I'm just confused by implications between various ergodic theorems. So I'll happily close the question if deemed inappropriate.

Let $G$ be a group and let $F_i$ be a sequence of finite subsets of $G$. Suppose $G$ acts on a probability measure space $(X,\mu)$ in a measure preserving way, and suppose that this action is ergodic.

Let us say that $F_i$ satisfies pointwise ergodic theorem iff for almost all $x\in X$, and all $f\in L^1(X)$ we have that the limit of $$ \frac{1}{|F_i|} \sum_{g\in F_i} f(g.x) $$ exists and is equal to $\int_X f\, d\mu$.

Let us say that $F_i$ satisfies mean sojourn time theorem iff for every measurable $U\subset X$ and almost every $x\in X$ we have that the limit of $$ \frac{1}{|F_i|} |\lbrace g\in F_i\colon g.x \in U\rbrace | $$ exists and is equal to $\mu(U)$.

Question 1: It is easy to see that if $F_i$ satisfies pointwise ergodic theorem then it also satisfies mean sojourn time theorem. Is it also the other way around?

A reference would be most appreciated (I imagine that the answer in the general case is the same as in the case when $G$ is the infinite cyclic group, so a reference for the latter case would also be fine.)

A related question:

Question 2: Is there a proof of the mean sojourn theorem for say $\mathbb Z$ which doesn't use the pointwise ergodic theorem?

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Link to the question on math.SE: math.stackexchange.com/q/297233 –  Martin Feb 10 '13 at 0:18

2 Answers 2

I can show that $$\int_X f(x) \ d \mu= \liminf_{k \to \infty} \frac{1}{|F_k|} \sum_{g \in F_k} f(g.x)$$ for $f \in L^1(X)$ with $f \geq 0$ and almost all $x \in X$.

The conclusion is obvious if $f \in L^{\infty}(X)$, since any such $f$ is a uniform limit of step-functions. Now, set $$f_n(x) = \min \lbrace n,f(x)\rbrace.$$

The set of points, where point-wise convergence holds for $f_n$ has measure $1$. Hence, taking the countable intersection of these sets, there is a set of measure $1$, where it holds for all $f_n$ at the same time. I denote the mean of $f$ over $F$ at $x$ by $m(F,x,f)$. Now, for all those $x \in X$, we get:

$$g(x):=\liminf_{k \to \infty} m(F_k,x,f) \geq \sup_{n} \lim_{k \to \infty} m(F_k,x,f_n) = \sup_{n} \int_X f_n(x) \ d \mu = \int_X f(x)\ d\mu$$

However, $\int_X m(F_k,x,f) d\mu = \int_X f(x) d\mu$ for all $k$ and hence $$\int g(x) \ d\mu = \int_X \liminf_{k \to \infty} m(F_k,x,f) \ d\mu \leq \liminf_{k \to \infty} \int_X m(F_k,x,f) \ d\mu = \int_{X} f(x) \ d\mu$$ by Fatou's lemma. It follows that $g(x)=\int_X f(x)\ d\mu$ almost everywhere.

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Why mean sojourn time theorem is not the pointwise ergodic theorem with $f=\chi_U?$ And mean sojourn time theorem iff ergodicity. Or in other words, we cannot prove the pointwise ergodic theorem from the sojourn time theorem, but the converse is obvious. And of course there are many proves of mean sojourn time theorem that do not use pointwise ergodic theorem.

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