Birkhoff pointwise ergodic theorem

Suppose first that $f$ is real. Write $u={\mathrm{lim\hspace{0.17em}sup}}_n{A}_{n}f$ and $\mathrm{\ell }={\mathrm{lim\hspace{0.17em}inf}}_n{A}_{n}f$. The exact identity $A_{n}f(Tx)=\frac{n+1}{n}{A}_{n+1}f(x)-\frac{1}{n}f(x)$ shows that $u\circ T=u$ and $\mathrm{\ell }\circ T=\mathrm{\ell }$, with extended values allowed. The divergence set $\{\mathrm{\ell }<u\}$ is the union of the sets $E_{\alpha ,\beta }$ over rational $\beta <\alpha$.

If $f=g$ a.e., let $N=\{f\mathrm{\ne }g\}$. Outside ${\bigcup }_{k\ge 0}{T}^{-k}N$, a null set by preservation and [F4], every summand in $S_{n}f$ equals the corresponding summand in $S_{n}g$. Thus the limits agree a.e.; the construction descends to the $L^1$ class.

Fix such $\beta <\alpha$ and put $E=E_{\alpha ,\beta }$. By [F1], $E$ is strictly invariant and $\mu (E)<\mathrm{\infty }$. For $g=(f-\alpha ){\mathbf{1}}_E$, strict invariance gives $S_{n}g={\mathbf{1}}_E(S_{n}f-n\alpha )\mathrm{.}$ Every $x\in E$ has $S_{n}f(x)>n\alpha$ for some $n$, so $E=\{{\sup }_n{S}_{n}g>0\}$. Applying [F2] gives ${\int }_{E}fd\mu \ge \alpha \mu (E)\mathrm{.}$

Apply the same argument to $-f$ and the thresholds $-\beta >-\alpha$. Its oscillation set is again $E$, so ${\int }_{E}fd\mu \le \beta \mu (E)\mathrm{.}$ Because $f$ is integrable and $E$ has finite measure, both inequalities are finite. Thus $(\alpha -\beta )\mu (E)\le 0$, and $\mu (E)=0$.

There are only countably many rational pairs. Hence [F4] and steps 1.1–3.1 show that $A_{n}f$ has an extended-real limit off a null set. On that set, $\mathrm{\mid }{f}^{\ast }\mathrm{\mid }={\mathrm{lim\hspace{0.17em}inf}}_n\mathrm{\mid }{A}_{n}f\mathrm{\mid },$ and contractivity gives $\int \mathrm{\mid }{A}_{n}f\mathrm{\mid }\le \int \mathrm{\mid }f\mathrm{\mid }$. Fatou therefore yields $\int \mathrm{\mid }{f}^{\ast }\mathrm{\mid }\le \int \mathrm{\mid }f\mathrm{\mid }<\mathrm{\infty }$, which both makes $f^{\ast }$ finite a.e. and proves integrability.

Define $f^{\ast }=0$ on the exceptional null set. The identity in step 1.1 shows that the convergence set and the limit are invariant wherever the averages converge; after adding the null exceptional orbit set if necessary, $f^{\ast }\circ T=f^{\ast }$ a.e.

For complex $f$, apply the real result to $\mathrm{Re}f$ and $\mathrm{Im}f$ and combine their two conull convergence sets. Their limits give the complex limit, its invariance, and its representative independence. Fatou applied directly to $\mathrm{\mid }{A}_{n}f\mathrm{\mid }$ gives the stated complex $L^1$ bound. Only these two determined components are used, so no choice principle enters.