Birkhoff's pointwise ergodic theorem

Step 1 (invariant limsup and liminf). Define $f^*(x)=\limsup_{n\to\infty}\frac{S_nf(x)}{n}$ and $f_*(x)=\liminf_{n\to\infty}\frac{S_nf(x)}{n}$. Since $S_{n+1}f(x) = f(x) + S_nf(Tx)$, dividing by $n+1$ and letting $n\to\infty$ shows $f^*(Tx)=f^*(x)$ and $f_*(Tx)=f_*(x)$: both are $T$-invariant functions.

Step 2 (the maximal ergodic theorem). For $\lambda\in\mathbb R$, let $A_\lambda=\{x: \sup_n S_nf(x)/n>\lambda\}$ be the set where the running time average ever exceeds $\lambda$. The key technical lemma (proved by considering $g=f-\lambda$ and the maximum of the partial sums $M_n(x)=\max(0,S_1g(x),\dots,S_ng(x))$, then using $M_n(Tx)\ge S_kg(Tx)$ termwise and integrating over the set where $M_n>0$) gives $\int_{A_\lambda} f\,d\mu \ge \lambda\,\mu(A_\lambda)$.

Step 3 (squeezing f and f_ together). Suppose for contradiction that $f^*>f_*$ on a set of positive measure; then there exist rationals $\alpha<\beta$ with $E=\{x: f_*(x)<\alpha<\beta<f^*(x)\}$ of positive measure. $E$ is $T$-invariant (since $f^*,f_*$ are), so we may restrict attention to $E$. Applying the maximal inequality to $f-\beta$ on $E$ forces $\int_E f\,d\mu\ge\beta\mu(E)$, and applying it to $\alpha-f$ similarly forces $\int_E f\,d\mu\le\alpha\mu(E)$; since $\alpha<\beta$ and $\mu(E)>0$ these contradict each other. Hence $f^*=f_*$ almost everywhere, so the limit $\bar f(x)=\lim_n S_nf(x)/n$ exists a.e. and is $T$-invariant.

Step 4 (matching the integrals, and the ergodic case). A dominated-convergence argument (truncating $f$ and controlling the tails using the maximal inequality again) shows $\int_X \bar f\,d\mu = \int_X f\,d\mu$. Finally, if $T$ is ergodic, the $T$-invariant function $\bar f$ must be constant almost everywhere (by the very definition of ergodicity applied to its level sets $\{\bar f\le c\}$, each of which is $T$-invariant and hence has measure $0$ or $1$); combined with the equality of integrals, that constant must be $\int_X f\,d\mu$, giving $\bar f(x)=\int_X f\,d\mu\ \text{a.e.}$.