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 ===
=== Definition 2 ===
=== Definition 3 ===
=== Definition 4 ===
=== Definition 5 ===
== Proof ==
=== 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:
:$\ds \map \mu {\set {\map \Re f \circ T \ne \map \Re f} } = 0$
and:
:$\ds \map \mu {\set {\map \Im f \circ T \ne \map \Im f} } = 0$
, we may therefore assume $f : X \to \R$.
Consider the increasing function $ F : \R \to \R$ defined by:
:$\map F c := \map \mu { \set {f \le c} }$
Observe that:
:$\ds \lim _{x \mathop \to -\infty} \map F x = \map \mu {\set {f = -\infty} } = 0$
and:
:$\ds \lim _{x \mathop \to +\infty} \map F x = \map \mu {\set {f < +\infty} } = 1$
Moreover, for each $c \in \R$, since:
we have:
:$\map \mu {\set { f \circ T \le c } \symdif \set {f \le c} } = 0$
so that:
:$\map F c \in \set {0,1}$.
Therefore 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:
:$\map \mu {\set {f = c} } = 1$
Since:
:$\map \mu {\set {f = 0} } + \map \mu {\set {f = 1} } = 1$
it follows that:
:$\map \mu A = \map \mu {\set {f = 1} } = \begin{cases} 0 &: c=0 \\ 1 &: c=1 \end{cases}$
Category:Ergodic Theory
Category:Ergodic Measure-Preserving Transformations