Let $(\Omega,\mathcal{A},\mu,T)$ be an ergodic dynamical system and $f\in L_{\mu}^1$. Then it is a.s. $$ \lim_{n\to\infty}\frac{1}{n}\sum_{k=0}^{n-1}f\circ T^k=\int f\, d\mu. $$
Now I am trying to understand the proof of this.
Here is one point I do not understand.
For $\varepsilon > 0$ define $$ E:=\left\{\limsup_{n\to\infty}\frac{1}{n}\sum_{k=0}^{n-1}f\circ T^k>\varepsilon\right\}. $$ Now the proof says: It's obvious that $E$ is $T$-invariant.
For me it's not obvious. Can you explain me why $E$ is $T$-invariant, i.e. why $$ T^{-1}(E)=E\text{ a.s.}? $$
First assume that $x\in E$, and we will prove that $T(x)\in E$ as well, so that $x\in T^{-1}(E)$. By definition, $$\varepsilon<\limsup_{n\to \infty}\frac{1}{n}\sum_{k=0}^{n-1}f\circ T^k(x) = \limsup_{n\to\infty}\frac{1}{n}f(x) + \frac{1}{n}\sum_{k=0}^{n-2}f\circ T^{k}(T(x)).$$ Of course $\frac{1}{n}f(x)\to 0$ for almost every $x$, so the right hand side of this expression is just$$\limsup_{n\to \infty}\frac{1}{n}\sum_{k=0}^{n-2}f\circ T^k(T(x)) = \limsup_{n\to \infty}\frac{n-1}{n}\frac{1}{n-1}\sum_{k=0}^{n-2}f\circ T^k(T(x)).$$ In the limit $(n-1)/n\to 1$, leaving us with $$ = \limsup_{m\to \infty}\frac{1}{m}\sum_{k=0}^{m-1} f\circ T^k(T(x)).$$ This proves that $T(x)\in E$.
Next assume that $x\in T^{-1}(E)$, and we will prove that $x\in E$. By definition $T(x)\in E$, so $$\varepsilon<\limsup_{n\to \infty}\frac{1}{n}\sum_{k=0}^{n-1}f\circ T^k(T(x)) = \limsup_{n\to \infty}\frac{1}{n}\sum_{k=0}^{n-1}f\circ T^{k+1}(x).$$ Similar to before, we rewrite the sum on the right hand side as $$\frac{1}{n}\sum_{k=0}^{n-1}f\circ T^{k+1}(x) = -\frac{f(x)}{n} + \frac{n+1}{n}\frac{1}{n+1}\sum_{k=0}^{n}f\circ T^k(x).$$ Since $f(x)/n\to 0$ for almost every $x$ and $(n+1)/n\to 1$, we see that $$\varepsilon<\limsup_{n\to \infty}\frac{1}{n}\sum_{k=0}^{n-1}f\circ T^{k+1}(x) = \limsup_{n\to \infty}\frac{1}{n+1}\sum_{k=0}^{n}f\circ T^k(x) = \limsup_{m\to \infty}\sum_{k=0}^{m-1}f\circ T^m(x).$$ Therefore, $x\in E$.