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In ergodic theory, why does the defintion of an ergodic transformation $T$, why do I have to claim that it is measure-preserving? E.g.

$T$ is ergodic if $\mathbb{P}(A) \in \{0,1\}$ for all $A$ with $T^{-1} (A) = A$

Couldn't I also have this definition without $T$ being measure-preserving ($\mathbb{P} \circ T = \mathbb{P}$) ?

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2 Answers 2

up vote 2 down vote accepted

You could certainly have your definition without measure preservation, but we generally don't. One reason is that we don't need it: the Krylov-Bogolyubov theorem gives a construction of an invariant (Borel, probability) measure for a continuous function from any reasonable topological space into itself. So we can rewrite all kinds of dissipative systems and the like as measure-invariant maps or flows. Then we can apply great stuff like Birkhoff's theorem.

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Consider $\Omega:=\{a,b\}$ with $a\neq b$, and $P(\{a\})=1/3$, $P(\{b\})=2/3$, $T(a)=b$, $T(b)=a$. Then $P(T^{-1}(\{b\}))=P(\{a\})=1/3\neq P(\{b\})$ hence $T$ doesn't preserve measure. But the only invariant sets are $\emptyset$ and $\Omega$, and their measure is respectively $0$ and $1$.

So in the definition of ergodicity, we have to require $P$ to be measure-preserving, in order to have for example ergodic theorems.

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