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Let $(X,\mathcal B,\mu)$ be a atomless probability measure space and $T:X\to X$ be a non-singular transformation such that $\mu\left(\{x\in X: T^n(x)=x\}\right)=0$ for every $n\ge 1.$ Let $A\in \mathcal B$ such that $\mu(A)>0$ and $k\ge 1$ a positive integer. I want to show that there exists a set $E\in \mathcal B$ such that $\mu(E)>0$ with $E\subseteq A$ and $T^{-k}(E)\cap E=\emptyset.$

Let $E=A\setminus T^{-k}(A)$ and $y\in T^{-k}(E)\cap E$. Then $y\in E\implies y\notin T^{-k}(A)$. Also $y\in T^{-k}E\implies T^k(y)\in E\implies T^ky\in A\implies y\in T^{-k}A.$ So, $T^{-k}(E)\cap E =\emptyset$. But I am unable to construct the set $E$ out of the aforesaid hypothesis such that $\mu(E)>0$. Please help me to solve this. Thank you for your time and help.

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  • $\begingroup$ If your set E, which we can call E_1, has zero measure, then $\mu(A)=\mu(A \cap T^{-k}A)$. So, you can define $E_2=\left( A \cap T^{-k}A \right) \setminus T^{-k} \left( A\cap T^{-k}A \right)$. Repeating the same argument, if $\mu(E_2)=0$, then you have that $\mu(A)=\mu(A\cap T^{-k}A \cap T^{-2k}A)$. Either this procedure stops with some $E_n$ of positive measure or you have that $\mu(A)=\mu( \bigcap_{n=0}^{\infty} T^{-nk}A)$, which in particular means that $\bigcap_{n=0}^{\infty} T^{-nk}A$ would have positive measure. $\endgroup$
    – User
    Commented Apr 15 at 14:48
  • $\begingroup$ @User suppose after continuing the steps we didn't get any $E_n$ with positive measure, but we have $\mu\left(\bigcap_{n=0}^\infty T^{-nk}A\right)>0,$ then what is the set $E$? $\endgroup$
    – abcdmath
    Commented Apr 17 at 4:55

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