Birkhoff's theorem for an ergodic probability system

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: $\int (\alpha f+\beta g)d\mu =\alpha \int fd\mu +\beta \int gd\mu (\alpha ,\beta \in \mathbb{C},f,g\in L^1(\mu ))\mathrm{.}$

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 convention of rem-zero-times-infinity-convention-for-pointwise-products, the pointwise product $fg$ is measurable.

Sequential suprema, infima, limsup, liminf, and pointwise limits of measurable functions are measurable: Let $(X,\mathcal{A})$ be a measurable space and let $f_n:X\to \bar{\mathbb{R}}$ be measurable for every $n\in \mathbb{N}$. Then the functions

${\sup }_n{f}_n,{\inf }_n{f}_n,{\mathrm{lim\hspace{0.17em}sup}}_n{f}_n,{\mathrm{lim\hspace{0.17em}inf}}_n{f}_n$

are measurable. The set

$\{x:{\lim }_n{f}_n(x)\text{ exists in }\bar{\mathbb{R}}\}$

is measurable. In particular, if $f_n\to f$ pointwise, then $f$ is measurable.

The maximal ergodic inequality on a probability space: Let $T$ preserve a probability measure $P$, and let $f$ be integrable, real-valued and measurable. Put $S_{k}f={\sum }_{j=0}^{k-1}f\circ T^j$, $M_N=\max (0,S_{1}f,\dots ,S_{N}f)$ and $E_N=\{M_N>0\}$ for $N\ge 1$. Then ${\int }_{E_N}fdP\ge 0$, and also ${\int }_{E}fdP\ge 0$ for $E=\{{\sup }_{k\ge 1}{S}_{k}f>0\}$.

Ergodicity relative to an invariant measure: A measure-preserving system is ergodic for $\mu$ if each $E\in \mathcal{I}$ has $\mu (E)=0$ or $\mu (X\setminus E)=0$, with $\mathcal{I}$ as in def-strict-and-mod-null-invariant-$\sigma$-algebras. For a probability system this means $\mu (E)\in \{0,1\}$. The definition is relative to the invariant measure; no probability assumption is implicit in the general null/conull formulation.

Finite and countable subadditivity of measures: Let $\mu$ be a measure and let $(E_k{)}_{k\in \mathbb{N}}$ be measurable. Then

$\mu \left({\bigcup }_{k\in \mathbb{N}}{E}_k\right)\le {\sum }_{k=0}^{\mathrm{\infty }}\mu (E_k)\mathrm{.}$

For every $m\in \mathbb{N}$ one also has

$\mu \left({\bigcup }_{k<m}{E}_k\right)\le {\sum }_{k<m}\mu (E_k),$

including $m=0$, where both sides are $0$.

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 $\int f_{n}d\mu ⟶\int fd\mu \mathrm{.}$

The modulus of an integral is bounded by the integral of the modulus: If $f\in L^1(\mu )$, then $\mid \int fd\mu \mid \le \int \mathrm{\mid }f\mathrm{\mid }d\mu \mathrm{.}$

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.