In his proof of the ergodic theorem Smorodinsky (https://link.springer.com/chapter/10.1007/BFb0066089), assuming $(\Omega,\mathcal{F},\mu)$ a probability space and $f\in L^1(\Omega,\mathcal{F},\mu)$ argues as follows: Let $a<b$ a pair of rational numbers. Put: $$ A(a,b)=\{\omega\in\Omega:\lim\inf_{n\rightarrow\infty}\frac{1}{n}S_n(\omega)<a<b<\lim\sup_{n\rightarrow\infty}\frac{1}{n}S_n(\omega)\}. $$ There are a countable number of sets $A(a,b)$ and $\lim_{n\rightarrow\infty}\frac{1}{n}S_n(\omega)$ exists a.e. if all have measure zero. Observe that the set $A(a,b)$ is invariant under $T$ i.e. if $\omega\in A(a,b)$ then $T\omega\in A(a,b)$. Thus we can restrict ourselves to $A(a,b)$ and consider the function $g(\omega)=f(\omega)-b$ on this space. Applying the maximal ergodic theorem to this situation gives $$ \int_N g(\omega)d\mu=\int_N f(\omega)-b d\mu\geq 0 $$ where $$ N=\{\omega:\sup_{n\in\mathbb{N}}:\frac{1}{n}\sum_{i=0}^{n-1}f(T^i\omega)-b>0\}=\{\omega:\sup_{n\in\mathbb{N}}:\frac{1}{n}S_n(\omega)>b\}. $$ Obviously every point in $A(a,b)$ is in $N$ and since $A(a,b)$ is the whole space we get $$ \int_{A(a,b)}f(\omega)d\mu\geq b\mu(A(a,b)). $$ By a similar argument we get $$ \int_{A(a,b)}f(\omega)d\mu\leq a\mu(A(a,b)). $$ Therefore $\mu(A(a,b))=0$. Now denote by $f^*$ the limit function, since $$ |\int_{\Omega}\frac{1}{n}S_n(\omega)d\mu|\leq \frac{1}{n}\sum_{i=0}^{n-1}\int_{\Omega}|f(T^i\omega)|d\mu=\int_{\Omega}|f(\omega)|d\mu $$ by Fatou s lemma the limit function is in $L^1(\Omega,\mathcal{F},\mu)$. Let us prove the identity $\int_{\Omega}f(\omega)d\mu=\int_{\Omega}f^*(\omega)d\mu$. Consider the set $B(a,b)=\{\omega\in\Omega:a<f^*(\omega)<b\}$ which is $T$-invariant. By using the maximal ergodic theorem once more we can establish $$ a\mu(B(a,b))\leq\int_{B(a,b)}f^*(\omega)\leq b\mu(B(a,b)) $$ and an easy continuity argument will establish that $$ a\mu(B(a,b))\leq\int_{B(a,b)}f(\omega)\leq b\mu(B(a,b)). $$ Partition the space $\Omega$ by the sets $B(\frac{k}{2^n},\frac{k}{2^{n+1}})$ where $n$ is fixed and $k=0,\pm 1,\pm 2,...$. Then $$ \int_{\Omega}f(\omega)d\mu=\sum_{k\in\mathbb{Z}}\int_{B(\frac{k}{2^n},\frac{k}{2^{n+1}})} f(\omega)d\mu, \int_{\Omega}f^*(\omega)d\mu=\sum_{k\in\mathbb{Z}}\int_{B(\frac{k}{2^n},\frac{k}{2^{n+1}})} f^*(\omega)d\mu. $$ Therefore $$ |\int_{\Omega}f(\omega)d\mu-\int_{\Omega}f^*(\omega)d\mu|\leq\frac{1}{2^n}\mu(\Omega). $$ Now taking the limit $n\rightarrow\infty$ completes the proof.
$\bf Question:$ I have the impression that the $T$-invariance of the sets $A(a,b)$ and $B(a,b)$ are consequences of the $T$-invariance of the limit function $f^*$ and in the following Corollary, where Smorodinsky assumes $T$ to be ergodic the proof starts:"Since the limit function $f^*$ is invariant...". However this is not shown though stated in the Theorem. Of course $$f^*(T\omega)-f^*(\omega)=\lim_{n\rightarrow\infty}\frac{1}{n}(f(T^n\omega)-f(\omega)),$$ but how does this imply the $T$- invariance of $f^*$? Any help will be appreciated! Many thanks in advance!
For all $\omega$, $f\left(\omega\right)/n$ goes to zero so it suffices to prove that for almost every $\omega\in\Omega$, $f\left(T^n\omega\right) /n\to 0$. Define for a positive $\varepsilon$ the event $$ A_n:=\left\{\omega\in\Omega\mid \left\lvert f\left(T^n\omega\right)\right\rvert \gt n\varepsilon \right\}. $$ Since $T$ preserves $\mu$, the measure of $A_n$ equals that of $B_n$, where $$ B_n:=\left\{\omega\in\Omega\mid \left\lvert f\left(\omega\right)\right\rvert \gt n\varepsilon \right\}. $$ Since $f$ is integrable, the series $\sum_{n\geqslant 1}\mu\left(B_n\right)$ converges hence so does $\sum_{n\geqslant 1}\mu\left(A_n\right)$. By the Borel-Cantelli lemma, it follows that for almost every $\omega\in\Omega$, there exists $N\left(\omega\right)$ such that for all $n\geqslant N\left(\omega\right) $, $\omega\notin A_n$, that is, $\left\lvert f\left(T^n\omega\right)\right\rvert \leqslant n\varepsilon$.