We want to prove the strong law of large numbers with Birkhoff's ergodic theorem.
Let $X_k$ be an i.i.d. sequence of $\mathcal{L}^1$ random variables. This is a stochastic process with measure-preserving operation $\theta$ (the shift operator). From Birkhoff's ergodic theorem, we obtain $\frac{X_0 + \dotsb + X_{n-1}}{n} \to Y$ a.s., with $Y=\mathbb{E}[X_1 \mid \mathcal{J}_{\theta}]$ a.s.
Now, if $Y$ constant a.s., $Y= \mathbb{E}[X_1]$ a.s., and we would have the desired result. But why is $Y$ constant a.s.?
Best Answer
The transformation $\theta$ on $\Omega^{\Bbb N}$ is ergodic. Indeed, it's enough to show that for each cylinder $A$ and $B$, we have $$\frac 1n\sum_{k=0}^{n-1}\mu(\theta^{-k}A\cap B)\to \mu(A)\mu(B),$$ where $\mu$ is the measure on the product $\sigma$-algebra. If $A=\prod_{j=0}^NA_j\times \Omega\times\dots$ and $B=\prod_{j=0}^NB_j\times \Omega\times\dots$, we have for $k>N$ \begin{align} \theta^{-k}A\cap B&=\{(x_j)_{j\geq 0}, (x_{j+k})_{j\geq 0}\in A, (x_j)_{j\geq 0}\in B\}\\ &=\{(x_j)_{j\geq 0},x_{j+k}\in A_j, 0\leq j\leq N, x_j\in B_j,0\leq j\leq N\}\\ &=B_0\times \dots\times B_N\times \Omega\times\dots\times \Omega\times A_0\times\dots\times A_n\times \Omega\times\dots, \end{align} and we use the definition of product measure $\mu$ on cylinders (the $N$ first terms doesn't matter).
Since $\theta$ is ergodic, $\mathcal J_{\theta}$ consists only of events of measure $0$ or $1$. The conditional expectation with respect such a $\sigma$-algebra is necessarily constant.