I need a some help with understanding the last step of the proof of the mean ergodic theorem in Walter's An introduction to ergodic theory (Corollary 1.14.1). The statement of the theorem is as follows:
Let $1\leq p<\infty$ and let $T$ be a measure-preserving transformation of the probablility space $(X,\mathscr{B},m)$. If $f\in L^p(m)$ there exists $f^*\in L^p(m)$ with $f^*\circ T=f^*$ a.e. and $\lVert(1/n)\sum_{i=0}^{n-1}f(T^ix)-f^*\rVert_p\to0$.
In the very last step of the proof, he writes the following:
We have $f^*=f^*\circ T$ a.e. because $$\left(\frac{n+1}{n}\right)(S_{n+1}f)(x)-(S_nf)(x)(Tx)=\frac{f(x)}{n}.$$
Here, he has defined $$(S_nf)(x)=\frac{1}{n}\sum_{k=0}^{n-1}f(T^kx),$$ and he has shown that $(S_nf)_{n\in\mathbb{N}}$ is a Cauchy sequence in $L^p(m)$ and defined $f^*\in L^p(m)$ to be the function that satisfies $$\lim_{n\to\infty}\lVert S_nf-f^*\rVert_p=0.$$ I assume that we want something along these lines: \begin{align*} f^*(x)-f^*(Tx)&=\lim_{n\to\infty}((S_nf)(x)-(S_nf)(Tx))\\ &=\lim_{n\to\infty}\left(\frac{n+1}{n}(S_{n+1}f)(x)-(S_nf)(Tx)\right)\\ &=\lim_{n\to\infty}\frac{f(x)}{n}=0 \end{align*} for almost all $x\in X$. However, I don't see why the first equality in this computation should hold. I know that $S_nf\to f^*$ in the $p$th mean implies that there exists a sequence $(n_k)_{k\in\mathbb{N}}$ so that $S_{n_k}f\to f^*$ a.e. but if we replace $S_n$ by $S_{n_k}$, then I don't see why the second equality holds (since we could have that $n_{k+1}/n_k\not\to1$). Any help is appreciated!
You are right, but it suffices to say the following.
The simultaneous convergence of $S_{n_k}(x)$ and $S_{n_k}(Tx)$ holds for $x$ and $Tx$ in some full measure set, say $E$. That is, both converge provided that $x\in E$, and $Tx\in E\Leftrightarrow x\in T^{-1}E$.
Now note that the set $E\cap T^{-1}E$ has full measure (since the measure is invariant) and so $S_{n_k}(x)$ and $S_{n_k}(Tx)$ converge simultaneous for $x$ in $F$.