# $\lim \frac{1}{n} \sum_{i=0}^{n-1} X \circ T^{i}$ T-invariant?

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?

• 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. Apr 20 at 7:53