The maximal ergodic inequality on a probability space

Measure-preserving transformations and systems: Let $(X,\mathcal{A},\mu )$ be a measure space. A measurable self-map $T:X\to X$ is measure preserving if $\mu (T^{-1}E)=\mu (E)$ for every $E\in \mathcal{A}$. The quadruple $(X,\mathcal{A},\mu ,T)$ is a measure-preserving system; it is a probability system if $\mu (X)=1$. Here $T^{-1}E=\{x:T(x)\in E\}$ denotes an inverse image, whether or not $T$ is invertible. Neither completeness nor finiteness is implicit. The measure-space and measurable-map conventions are def-measure-space and def-measurable-function-between-measurable-spaces.

Arithmetic and lattice operations preserve measurability whenever they are defined: Let $(X,\mathcal{A})$ be a measurable space and let $f,g:X\to \bar{\mathbb{R}}$ be measurable. Then:

1. $cf$ is measurable for every real scalar $c$;
2. $\max (f,g)$, $\min (f,g)$, $\mathrm{\mid }f\mathrm{\mid }$, $f^{+}$, and $f^{-}$ are measurable;
3. if $f+g$ is pointwise defined, then $f+g$ is measurable;
4. with the total function-product convention declared in that theorem, the pointwise product $fg$ is measurable.

Integral invariance under measure-preserving maps: If $T$ preserves $\mu$ and $f:X\to [0,\mathrm{\infty }]$ is measurable, then $\int f\circ Td\mu =\int fd\mu$, allowing infinity. If $f$ is integrable real or complex valued, $f\circ T$ is integrable and the same equality holds. Conversely, for a measurable self-map, equality for every measurable indicator implies measure preservation.

The Lebesgue integral is linear on $L^1(\mu )$: The class $L^1(\mu )$ is a complex vector space, and the Lebesgue integral is complex-linear on it: $\int (\alpha f+\beta g)d\mu =\alpha \int fd\mu +\beta \int gd\mu (\alpha ,\beta \in \mathbb{C},f,g\in L^1(\mu ))\mathrm{.}$

Dominated convergence: Let $f$ and $(f_n)$ be measurable complex-valued functions such that $f_n\to f$ almost everywhere and $\mathrm{\mid }{f}_n\mathrm{\mid }\le g$ almost everywhere for a single nonnegative measurable function $g$ with $\int gd\mu <+\mathrm{\infty }$. Then $f\in L^1(\mu )$, $\int \mathrm{\mid }{f}_n-f\mathrm{\mid }d\mu ⟶0,$ and hence $\int f_{n}d\mu ⟶\int fd\mu \mathrm{.}$