# Existence of a postive measurable set such that $T^{-k}(E)\cap E=\emptyset$ for a particular $k\ge 1.$

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.

• 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.
– User
Commented Apr 15 at 14:48
• @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 $E$? Commented Apr 17 at 4:55