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