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

Then the following are equivalent:
 * $(1):$ $T$ is ergodic
 * $(2):$ For all $\BB$-measurable function $f$:
 * if $\map {f \circ T} x = \map f x$ for all $x \in X$, then $f$ is constant $\mu$-a.e.
 * $(3):$ For all $\BB$-measurable function $f$:
 * if $\map {f \circ T} x = \map f x$ for $\mu$-a.a. $x \in X$, then $f$ is constant $\mu$-a.e.
 * $(4):$ For all $f \in \map {\LL ^2} \mu$:
 * if $\map {f \circ T} x = \map f x$ for all $x \in X$, then $f$ is constant $\mu$-a.e.
 * $(5):$ For all $f \in \map {\LL ^2} \mu$:
 * if $\map {f \circ T} x = \map f x$ for $\mu$-a.a. $x \in X$, then $f$ is constant $\mu$-a.e.

$(1) \implies (3)$
This is clear, since $(3)$ is exactly.

$(3) \implies (2)$
This is trivial in view of.

Indeed, if $\map {f \circ T} x = \map f x$ for all $x \in X$, then:

$(2) \implies (4)$
This is trivial in view of.

Indeed, if $f \in \map {\LL ^2} \mu$, then $f$ is $\BB$-measurable.

$(4) \implies (1)$
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 $x \in X$:

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

The claim follows from this, since by :
 * $A = \set {x \in X : \map {\chi_A} x = 1}$

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

$(3) \implies (5)$
This is trivial in view of.

See the proof of $(3) \implies (2)$.

$(5) \implies (4)$
This is trivial in view of.

See the proof of $(2) \implies (4)$.