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Birkhoff's ergodic theorem goes as follows: Let $T$ be a measure-preserving transformation on $(\Omega,\mathcal{F}, \mathbb{P})$ and $X \in \mathcal{L}^1$. Then $S_n := \frac{1}{n} \sum_{i=0}^{n-1} X \circ T^{i}$ converges almost surely towards some $\mathcal{J}_T$-measurable random variable $Y$ satisfying $\mathbb{E}[X] = \mathbb{E}[Y]$.

One way to prove this is by showing:

$$\int \bar{X} \mathbb{P} \le \int X d\mathbb{P} \le \int \underline{X} d\mathbb{P}$$

where $\bar{X} := \lim \sup S_n$ and $\underline{X} := \lim \inf S_n$.

You can show that $\bar{X} \in \mathcal{J}_T$ and $\underline{X} \in \mathcal{J}_T$, where $\mathcal{J}_T$ is the set of $T$-invariant measurable sets in $\mathcal{F}$. This follows because if we set $\frac{n+1}{n}S_{n+1}(\omega) = S_n(T\omega) + \frac{1}{n}S_n(\omega)$ we can take the $\lim \sup$ and $\lim \inf$ respectively and derive the desired property.

My question is why can't I do this just with $\lim$ as well? Is it because we don't know a priori if this limit exists or have I overlooked something else?

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    $\begingroup$ My answer was delete so I repost as a comment. Yes you are right it is just because we do not know if the limit exists. $\endgroup$
    – EtienneBfx
    Apr 20 at 7:53

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