Birkhoff's ergodic theorem in infinite measure case

Question

I've been reading ergodic theory from Krengel's book, "Ergodic theorems" and am in the section where the proof of Birkhoff's pointwise ergodic theorem is given. Although, he states the theorem for measure preserving systems $(\Omega,\mathscr{A},\mu,\tau)$ without any restriction of finiteness or $\sigma-$finiteness of the measure. I've read the whole proof and i came up to the conclusion that i only need to prove the following maximal inequality for arbitrary measures:

Let $(\Omega,\mathscr{A},\mu,\tau)$ a measure preserving system $(\mu(\tau^{-1}(A))=\mu(A)$ for every $A\in \mathscr{A}$). Denote $A_k(f)=k^{-1}\sum_{i=0}^{k-1}f\circ \tau^i,\ M_n(f)=\max\{ A_1(f),...,A_n(f)\}$. Then, $$\int_{\Omega}|f|d\mu \geq \alpha \mu(M_n(f)>\alpha)$$ for every real valued $f\in L_1(\mu)$.

I can prove the above inequality in the $\sigma$-finite case. In the general case my idea was to construct a proper subsystem $(B,\mathscr{A}',\mu',\tau')$ where $\mu'$ is $\sigma-$finite on $(B,\mathscr{A}')$. I defined $A_k=\{|f|\geq \frac{1}{k}\},\ A=\bigcup_{k=1}^{\infty}A_k$, since every $f\circ \tau^n$ is integrable we have $\mu(\tau^{-n}A_k)<\infty$ for every $k,n$. I set $B=\bigcup_{n=0}^{\infty}\tau^{-n}A,\ \mathscr{A}'=\{A\subseteq B:A\in \mathscr{A}\}$ and $\mu'(A)=\mu(A)$ for every $A\in \mathscr{A}'$, but my problem was $\tau$, i couldnt restrict $\tau$ to $B$ since i dont know if $\tau(B)\subseteq B$.

Do you have any ideas how to treat this case? Thanks in advance!

Answer 1

I proved it, am posting the answer just for completion.

I can prove it directly without proving it first for finite,$\sigma$-finite and then for general measure spaces. Am using the maximal ergodic theorem: $$\int\limits_{\{M_n(f)\geq 0\}}fd\mu\geq 0$$ for every real valued integrable $f$. Let $A_k,A,B$ the sets i defined above. Then $$B=\bigcup_{n=0}^{\infty}\bigcup_{k=1}^{\infty}\tau^{-n}(A_k)$$ let $B_N=\bigcup_{n=0}^{N}\bigcup_{k=1}^{N}\tau^{-n}(A_k)$, then $B_{N}\subseteq B_{N+1},\mu(B_N)<\infty$ and $\bigcup_{N=1}^{\infty}B_N=B$. Also, for $\omega \notin B$ we have that $f\circ \tau^{n}(\omega)=0$ for every $n=0,1,...\ .$ Hence, $\{M_n(f)\geq \alpha\}\subseteq B$. So, in order to prove $$\int\limits_{\Omega}|f|d\mu\geq \alpha \mu(M_n(f)\geq \alpha)$$ we only need to prove that $$\int\limits_{\Omega}|f|d\mu\geq \alpha \mu(\{M_n(f)\geq \alpha\}\cap B_N)$$ For $E_N=\{M_n(f)\geq \alpha \}\cap B_N$ we have $\mu(E_N)<\infty$, hence the function $g=f-\alpha \boldsymbol{1}_{E_N}$ is integrable. Now, for $\omega \in \{M_n(f)\geq \alpha\}$ there exists $1\leq k\leq n$ such that $A_k(f)(\omega)\geq \alpha$. So, $$\sum_{i=0}^{k-1}f\circ \tau^{i}(\omega)\geq k\alpha$$ and because $k\alpha \geq \sum_{i=0}^{k-1}\alpha\boldsymbol{1}_{\tau^{-i}(E_N)}$ we get $$\sum_{i=0}^{k-1}f\circ \tau^{i}(\omega)\geq \sum_{i=0}^{k-1}\alpha \boldsymbol{1}_{\tau^{-i}(E_N)}\implies A_k(g)(\omega)\geq 0$$ and this shows that $\{M_n(f)\geq \alpha\}\subseteq \{M_n(g)\geq 0\}$. But from the maximal ergodic theorem $$\int\limits_{\{M_n(g)\geq 0\}}fd\mu \geq \alpha \mu(\{M_n(g)\geq 0\}\cap E_N)$$ and now using $\{M_n(g)\geq 0\}\cap E_N=E_N=\{M_n(f)\geq \alpha\}\cap B_N$ we get $$\int\limits_{\Omega}|f|d\mu \geq \int\limits_{\{M_n(g)\geq 0\}}fd\mu\geq \alpha \mu(\{M_n(f)\geq \alpha\}\cap B_N)$$ as desired.