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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!!!

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

$$\mu(T_Y^{-1}B)=\mu(B)$$

First,

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

Now,

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

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. :)

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  • $\begingroup$ $\varphi= \varphi_Y ?$ $\endgroup$ – user27456 Jul 14 '12 at 4:05
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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.

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