From jacob-Protter: Ergodic Strong Law of Large Numbers
Let $\tau$ be a one-to-one measure preserving transformation of $\Omega$ onto itself. Assume the only $\tau$-invariant sets are sets of probability $0$ or $1$. If $X \in L^1$ then
\begin{equation*}
\lim_{n \to \infty} \frac{1}{n} \sum_{j=1}^n X(\tau^j(\omega)) = E\{X\}
\end{equation*}
a.s. and in $L^1$.
It is said there that it is a consequence of the Birkhoff ergodic theorem.
Birkhoff Ergodic Theorem from 'Probability Theory A Comprehensive Course (Universitext) (Klenke, Achim)' :
Let $f = X_0 \in L^1(P)$. Then
$$\frac{1}{n} \sum_{k=0}^{n-1} X_k = \frac{1}{n} \sum_{k=0}^{n-1} f \circ \tau^k \to E\{X_0|\mathcal{I}\} ~~ P-a.s. $$
In showing the ergodic strong law of large numbers, we only have to show that $E\{X_0|\mathcal{I}\} = E\{X_0\}$. Can you please show how prove this.
Best Answer
Denote by $Y$ the random variable $E\{X_0|\mathcal{I}\}$. We know that $Y$ is $\mathcal I$-measurable, hence for each $t$, the set $\{Y\leqslant t\}$ has probability zero or one. Therefore, we only need the following:
To see this, let $t_0:=\inf\left\{t\in\mathbb R, \mathbb P\{Y\leqslant t\}=1\right\}$. Note that $t_0$ is finite because $\mathbb P\{Y\leqslant t\}=1$ for $t$ large enough. Also, $\mathbb P\{Y\leqslant t_0+1/n\}=1$ by definition of supremum hence $\mathbb P\{Y\leqslant t_0\}=1$. Moreover, $\mathbb P\{Y\leqslant t_0-1/n\}=0$ for each $n$ hence $\mathbb P\{Y\lt t_0\}=0$/