Maximal ergodic theorem

Remarks

Proof idea, after Garsia: let $F_N = \max(0, S_1 f, \dots, S_N f)$. For $1 \le n \le N$ one has $S_n f \le f + F_N \circ T$, so $F_N \le f + F_N \circ T$ on the set $E_N = \set{F_N > 0}$. Integrating over $E_N$ and using $F_N \ge 0$ and the invariance of $\mu$ gives $\int_{E_N} f \, d\mu \ge \int_X F_N \, d\mu - \int_X F_N \circ T \, d\mu = 0$. The sets $E_N$ increase to $E$, and dominated convergence finishes the proof.

Applied to $f - \lambda$ for a real number $\lambda > 0$, it gives the maximal inequality $$\lambda \, \mu\left(\set{x : \sup_{n \ge 1} \tfrac{1}{n} S_n f(x) > \lambda}\right) \le \int_X \abs{f} \, d\mu.$$ The inequality $\int_E f \, d\mu \ge 0$ holds, with the same proof, for a measure-preserving map of any measure space and any integrable $f$.

Example: if $f \ge 0$ is not almost everywhere $0$, then $E = \set{x : f(T^k(x)) > 0 \text{ for some } k \ge 0}$, and the conclusion $\int_E f \, d\mu \ge 0$ is clear; the content of the theorem lies in functions of both signs.

Going deeper

For a student

This is the key lemma in the standard proof of Birkhoff's pointwise ergodic theorem. Through the maximal inequality, it controls the supremum of all Birkhoff averages at once. This control is what is needed to pass from a dense class of functions to all integrable functions. It plays the same role as the Hardy–Littlewood maximal inequality in the proof of the Lebesgue differentiation theorem.