Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 ($\mathbb{P} \circ T = \mathbb{P}$) ?

share|cite|improve this question
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.

share|cite|improve this answer

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.