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Let $(\Omega,\cal F,\mathbb P)$ a $\sigma $-finite measure space, $T:\Omega\to\Omega$ measure preserving and $A\in\cal F$ with $\mathbb P(A \triangle T^{-1}(A))=0$. Let $B:=\{x\in\Omega\mid x\in T^{-n}(A)\, \text{ infinitly often}\}$.

Show B is $T$-invariant.

$T:\Omega\to\Omega$ is a measure preserving transformation iff $\mathbb P(C)=\mathbb P(T^{-1}(C))$ for all $C$ from a $\cap$-stable generator from $\cal F$.

Therefore it will suffice to show $B$ is from a $\cap$-stable generator from $\cal F$, right? But how does one show that?

Help is highly appreciated.

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  • $\begingroup$ $A$ need not be $T$-invariant, as a non-empty set can have probability zero. $\endgroup$ Dec 8, 2017 at 22:45

1 Answer 1

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To show that $B$ is $T$-invariant, you must show that $T^{-1}(B)=B$. This essentially follows from the definition of $B$:

If $x\in B$ then there are infinitely many $n$ such that $T^nx\in A$. But if $n\geq 2$ and $T^nx\in A$ then $T^{n-1}(Tx)\in A$. Thus $x\in B$ implies $Tx\in B$, or $x\in T^{-1}(B)$.

On the other hand, if $x\in T^{-1}(B)$ then $Tx\in B$, so there are infinitely many $n$ such that $T^n(Tx)\in A$, hence infinitely many $m$ such that $T^mx\in A$.

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