Expectation over an expected value in ergodic process From ergodic theorem we can see that:
$\frac{1}{N} \sum_{i}f(x_{i}) \to {E}[f(x)]$   as $n \to \infty$
Does this imply that in ergodic process we can similarly write: 
$\frac{1}{N} \sum_{i}E[f(x_{i})] \to \lim_{n \to \infty}{E}[f(x_{n})]$
If this statement is right, then I am a little bit confused about how can we write this? 
 A: Fist of all, I think you should correct your writing. At first I was a little confused, because I thought by $x_n$ you mean random variables, but now I think I got what you mean. Please, for the next time write every infromaiton you have, this makes it much easier for others to answer.

We assume $f\in\mathcal{L}^1(P)$, $x\in X$ ($(X,\sigma,P)$ is your probability space) and $T$ is a measure-preserving transformation. We write $x_i$ for $T^ix$. Then from ergodic theorem follows $$\frac{1}{N} \sum_{i}f(x_{i}) \to {E}[f]\qquad (N\to\infty)\qquad a.s.$$
Define $f^n(y):=f(T^n y)$. Since $T$ is a measure-preserving transformation we have $$E[f^n]=E[f].$$ I think this is what you mean by the rhs of your second equation. The expression $E[f(x_n)]$ doesn't make any sense to me, since we are not talking about random variables. So what you want to show is $$\frac{1}{N} \sum_{i}E[f^i] \to \lim_{n\to\infty}E[f^n]\qquad(\ N\to\infty).$$ 
But on the lhs you have $\frac{1}{N} \sum_{i}E[f^i]=\frac{1}{N} \sum_{i}E[f]=E[f]$ and on the rhs $E[f^n]=E[f]$. So there is nowhere the need of the limit.
