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

I was looking for some results on Infinite Ergodic Theory and I found this proposition. Do you guys know how to prove the last item (iii)?

I managed to prove (i) and (ii) but I can't do (iii).

Let $(X,\Sigma,\mu,T)$ be a $\sigma$-finite space with $T$ presearving the measure $\mu$, $Y\in\Sigma$ sweep-out s.t. $0<\mu(Y)<\infty$. Making $$\varphi(x)= \operatorname{min}\{n\geq0; \ T^n(x)\in Y\}$$ and also $$T_Y(x) = T^{\varphi(x)}(x)$$ if $T$ is conservative then

(i) $\mu|_{Y\cap\Sigma}$ under the action of $T_Y$ on $(Y,Y\cap\Sigma,\mu|_{Y\cap\Sigma})$;

(ii) $T_Y$ is conservative;

(iii) If $T$ is ergodic, then $T_Y$ is ergodic on $(Y,Y\cap\Sigma,\mu|_{Y\cap\Sigma})$.

Any ideas?

Thank you guys in advance!!!

share|cite|improve this question
up vote 2 down vote accepted

Let be $B \subset Y$. To prove the invariance of $\mu_A$ it is sucient to prove that,



$$\mu(T_Y^{-1}B)=\sum_{n=1}^{\infty}\mu(Y\cap\{\varphi_Y =n \}\cap T^{-n}B)$$


$$\{ \varphi_Y\leq n\}\cap T^{-n-1}B=T^{-1}( \{\varphi_Y\leq n-1\}\cap T^{-n}B)\cup T^{-1}(Y \cap\{\varphi_Y =n \}\cap T^{-n}B ) $$

This gives by invariance of the measure

$$\mu(Y\cap \cap\{\varphi_Y =n \}\cap T^{-n}B)=\mu(B_n)-\mu(B_{n-1}) $$

where $B_n=\{ \varphi_Y\leq n\}\cap T^{-n-1}B.$ We have $\mu(B_n)\to\mu(B)$ as $n\to \infty$, thus $$\mu(T_Y^{-1} B)=\lim\mu(B_n)=\mu(B).~~~~ :)$$

Let us assume now the ergodicity of the original system. Let $B\subset Y$ be a measurable $T_Y$-invariant subset. For any $x \in B$ , the first iterate $T^n x~~~(n\geq 1)$ that belongs to $Y$ also belongs to $B$ , which means that $\varphi_B=\varphi_Y$ on $B.$ But if, $\mu(B)\neq 0$ , Kac's lemma gives that

$$\int_{B}\varphi_B d\mu=1=\int_{Y}\varphi_{Y} d\mu$$ which implies that $\mu(B \setminus A) = 0$, proving ergodicity. :)

share|cite|improve this answer
$\varphi= \varphi_Y ?$ – user27456 Jul 14 '12 at 4:05

Kac's lemma can be used if $\mu (X) = 1$, which is not the case here. In fact, $\phi_B$ and $\phi_Y$ need not be integrable.

Instead, to prove ergodicity one may argue by contradiction. Assume that $(Y, T_Y, \mu)$ is not ergodic. Then there is an invariant set $E \subset Y$ such that $0 < \mu (E) < \mu (Y)$. This will then lead to a conclusion that $(X, T, \mu)$ is not ergodic.

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.