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:
- $U_T : f \mapsto f \circ T$
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Then the following are equivalent:
- $(1):$ $T$ is ergodic
- $(2):$ For all $f \in \map \MM {X, \R}$:
- $(3):$ For all $f \in \map \MM {X, \R}$:
- $(4):$ For all $f \in \map {\LL ^2} \mu$:
- $(5):$ For all $f \in \map {\LL ^2} \mu$:
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 $\map {\LL ^2} \mu$:
- $\map {\LL ^2} \mu \subseteq \map \MM {X, \R}$
The claims follow from this.
$\Box$
$(4) \implies (1)$
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:
- $X \setminus A = \set {x \in X : \map {\chi_A} x = 0}$
$\blacksquare$
Sources
- 2013: Peter Walters: An Introduction to Ergodic Theory (4th ed.) $1.5$: Ergodicity