# Characterization of Ergodicity in terms of Koopman Operator

## Theorem

Let $\struct {X, \BB, \mu}$ be a probability space.

Let $T: X \to X$ be a measure-preserving transformation.

Let $\map \MM {X, \R}$ be the set of $\BB$-measurable functions.

Let $\map {\LL^2} \mu$ denote the Lebesgue $2$-space.

Let $U_T : \map \MM {X, \R} \to \map \MM {X, \R}$ be the Koopman operator:

$U_T : f \mapsto f \circ T$

Then the following are equivalent:

$(1):$ $T$ is ergodic
$(2):$ For all $f \in \map \MM {X, \R}$:
if $\map {U_T} f = f$, then $f$ is constant $\mu$-a.e.
$(3):$ For all $f \in \map \MM {X, \R}$:
if $\map {U_T} f = f$ for $\mu$-a.e., then $f$ is constant $\mu$-a.e.
$(4):$ For all $f \in \map {\LL ^2} \mu$:
if $\map {U_T} f = f$, then $f$ is constant $\mu$-a.e.
$(5):$ For all $f \in \map {\LL ^2} \mu$:
if $\map {U_T} f = f$ for $\mu$-a.e., then $f$ is constant $\mu$-a.e.

## Proof

### $(1) \implies (3)$

This is clear, since $(3)$ is exactly the definition of ergodicity.

$\Box$

### $(3) \implies (2)$ and $(5) \implies (4)$

These are a direct consequence of Definition of Almost Everywhere.

If $\map {U_T} f = f$, the same holds especially $\mu$-almost everywhere, since:

 $\ds \map \mu {\set { x \in X : \map {\map {U_T} f} x \ne \map f x} }$ $=$ $\ds \map \mu \empty$ $\ds$ $=$ $\ds 0$ Measure of Empty Set is Zero

$\Box$

### $(2) \implies (4)$ and $(3) \implies (5)$

In view of Definition of $\map {\LL ^2} \mu$:

$\map {\LL ^2} \mu \subseteq \map \MM {X, \R}$

$\Box$

### $(4) \implies (1)$

Let $A \in \BB$ be such that $T^{-1} \sqbrk A = A$.

Let $\chi_A : X \to \set {0, 1}$ be the characteristic function of $A$.

Note that $\chi_A^2 = \chi_A$, as $0^2=0$ and $1^2=1$.

In particular, $\chi_A \in \map {\LL^2} \mu$, since:

$\ds \int \chi_A^2 \rd \mu = \map \mu A < + \infty$

Moreover:

$\chi_A \circ T = \chi_A$

since for all $x \in X$:

 $\ds \map {\chi_A \circ T} x$ $=$ $\ds 1$ $\ds \leadstoandfrom \ \$ $\ds \map T x$ $\in$ $\ds A$ Definition of $\chi_A$ $\ds \leadstoandfrom \ \$ $\ds x$ $\in$ $\ds T^{-1} \sqbrk A$ Definition of Preimage of Mapping $\ds \leadstoandfrom \ \$ $\ds x$ $\in$ $\ds A$ as $T^{-1} \sqbrk A = A$ $\ds \leadstoandfrom \ \$ $\ds \map {\chi_A} x$ $=$ $\ds 1$ Definition of $\chi_A$

That is, $\map {U_T} f = f$ by Definition of Koopman Operator.

Therefore by hypothesis, $\chi_A$ is constant $\mu$-almost everywhere.

The claim follows from this, since by Definition of $\chi_A$:

$A = \set {x \in X : \map {\chi_A} x = 1}$

and:

$X \setminus A = \set {x \in X : \map {\chi_A} x = 0}$

$\blacksquare$