Equivalence of Defintions of Ergodic Measure-Preserving Transformation

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

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

Definition 1 implies Definition 2
First, we claim that for each $j \in \N$:
 * $\map \mu {T^{-j} \sqbrk A \symdif A} = 0$.

Indeed, applying Symmetric Difference is Subset of Union of Symmetric Differences, inductively:

so that:

Next, let:
 * $\ds A_\infty := \bigcap _{N \mathop \ge 1} \bigcup _{j \mathop \ge N} T^{-j} \sqbrk A$

Then $A_\infty \in \set {0, 1}$, since:
 * $T^{-1} \sqbrk {A_\infty} = A_\infty$

Therefore, we shall show $\map \mu A = \map \mu {A_\infty}$.

Observe:

Therefore:

Definition 2 implies Definition 3
Let:
 * $\ds B := \bigcup_{n \mathop = 1}^\infty T^{-n} \sqbrk A$

Then:
 * $T^{-1} \sqbrk B = \bigcup_{n \mathop = 2}^\infty T^{-n} \sqbrk A \subseteq B$

so that:
 * $\map \mu {T^{-1} \sqbrk B \symdif B} = \map \mu {T^{-1} \sqbrk B} - \map \mu B = 0$

Thus $\map \mu B \in \set {0,1}$.

As $\map \mu B > 0$, we have $\map \mu B = 1$.

Definition 3 implies Definition 4
Let $\map \mu A \map \mu B > 0$.

As $\map \mu A > 0$, we have:
 * $\ds \map \mu {\bigcup_{n \mathop = 1}^\infty T^{-n} \sqbrk A} = 1$

Thus:

Therefore there must exist an $n \ge 1$ such that:
 * $\map \mu { T^{-n} \sqbrk A \cap B } > 0$.

Definition 4 implies Definition 1
Let $A \in \BB$ be such that $T^{-1} \sqbrk A = A$.

Let $B := X \setminus A$ so that:
 * $\map \mu B = 1 - \map \mu A$

Since:
 * $\forall n \ge 1 : T^{-n} \sqbrk A \cap B = A \cap B = \O$

we have :
 * $\map \mu A \map \mu B = 0$

That is:
 * $\map \mu A \paren {1 - \map \mu A} = 0$

Therefore:
 * $\map \mu A \in \set {0,1}$

Definition 2 implies Definition 5
Since:
 * $\ds \set {\map \Re f \circ T \ne \map \Re f} \cup \set {\map \Im f \circ T \ne \map \Im f} \subseteq \set {f \circ T \ne f}$

the assumption: $\ds \map \mu {\set {f \circ T \ne f} } = 0$ implies:
 * $\ds \map \mu {\set {\map \Re f \circ T \ne \map \Re f} } = 0$
 * $\ds \map \mu {\set {\map \Im f \circ T \ne \map \Im f} } = 0$

, we may therefore assume $f : X \to \R$.

Let $ F : \R \to \R$ be defined by:
 * $\map F c := \map \mu { \set {f \le c} }$

For each $c \in \R$, since:

we have:
 * $\map \mu {\set { f \circ T \le c } \symdif \set {f \le c} } = 0$

so that:
 * $\map F c \in \set {0,1}$.

As:
 * $F$ is monotone
 * $\lim _{x \to -\infty} \map F x = \map \mu {\set {f = -\infty} } = 0$
 * $\lim _{x \to +\infty} \map F x = \map \mu {\set {f < +\infty} } = 1$

there is a $c \in \R$ such that:
 * $\ds c = \sup \set {x \in \R : \map F x = 0} = \inf \set {x \in \R : \map F x = 1}$

For each $n \in \N_{>0}$:
 * $\map \mu {\set { c - \dfrac 1 n < f \le c + \dfrac 1 n } } = \map F {c + \dfrac 1 n} - \map F {c - \dfrac 1 n} = 1$

By $n \to \infty$:
 * $\map \mu {\set { f = c } } = 1$

Definition 5 implies Definition 1
Let $T^{-1} \sqbrk A = A$.

Let $f := \chi _A$ be the characteristic function of $A$ so that:
 * $f \circ T = f$.

Then there is a $c \in \set {0,1}$ such that:
 * $\map \mu {\set {f = c} } = 1$

As:
 * $\map \mu {\set {f = 0} } = \map \mu {X \setminus A} = 1 - \map \mu A$
 * $\map \mu {\set {f = 1} } = \map \mu A$

the claim follows.