Events in a measure preserving system satisfying $\mu (A \cap T^{-n}B) = \mu(A)\mu(B)$ for all $n$ large imply $\mu(A) \in \{0,1\}$ Let $(X, \mathscr B_X, \mu, T)$ be a measure-preserving system such that the condition in the title is satisfied for all $n$ larger than $N \in \Bbb N$. How can we conclude that $A$ is trivial? Any help\hints are appreciated.
 A: Here is a more elementary proof.
Take any $B$ with $0 < \mu(B) < 1$. Let $N_1$ be such that $\mu(B \cap T^{-N_1} B) = \mu(B)^2$. With $N_1,\dots,N_k$ chosen, choose $N_{k+1} > N_k$ such that $$\mu\left(\bigcap_{j \in \Sigma_k} T^{-N_j}B \cap T^{-N_{k+1}} B\right) = \mu(B)^{1+|\Sigma_k|}$$ for each $\Sigma_k \subseteq \{0,N_1,\dots,N_k\}$. Let $$A = \bigcup_{k \ge 1} [B \cap T^{-N_1} B \cap \dots \cap T^{-N_{k-1}}B \cap T^{-N_k}B^c \cap T^{-N_{k+1}}B].$$ Then, since the union is a disjoint one, $$\mu(A) = \sum_{k \ge 1} \mu(B)^{k+1}(1-\mu(B)) = \mu(B)^2,$$ so $$\mu(A)\mu(B) = \mu(B)^3,$$ while for $j \ge 3$, $$\mu(A \cap T^{-N_j} B) = \sum_{k=1}^{j-2} \mu(B)^{k+2}(1-\mu(B))+\mu(B)^j(1-\mu(B))+\sum_{k=j+1}^\infty \mu(B)^{k+1}(1-\mu(B))$$ $= \mu(B)^3+(1-\mu(B))^2\mu(B)^j.$
A: Actually, there is no non-trivial dynamical system $(X,\mathcal B,\mu)$ for which $\mu(A\cap T^{-n}B)=\mu(A)\mu(B)$ holds eventually for any $A$ and $B$. This is a consequence of Baire theorem. Indeed, fix a measurable set $B$ and consider the complete pseudo-metric space $(\mathcal B,\rho)$ where $\rho(A,B)=\mu(A\Delta B)$. The map $F_n\colon A\mapsto \mu(A\cap T^{-n}B)-\mu(A)\mu(B)$ is continuous. Defining $F_N:=\bigcap_{n\geqslant N}F_n^{-1}\{0\}$, we obtain the existence of $N_0$, a positive $\delta$ and a measurable set $A_0$ such that if $\rho(A,A_0)\lt \delta$ then for each $n\geqslant N_0$, $F_n(A)=0$. 
Using successively this property with $A\cup A_0$ and $A_0\setminus A$ instead of $A$, we obtain that if $\mu(A)\lt \delta$, then for each $n\geqslant N_0$, 
$F_n(A)=0$. 
We now use a proposition in Bogachev's book: if $\varepsilon$ is a positive number and $(X,\mathbb B,\mu)$ is a finite measure space, then there exists a finite partition $(B_i)_{1\leqslant i\leqslant N}$ such that either $\mu(B_i)\lt\varepsilon$ or $B_i$ is a $\mu$-atom of measure greater than $\varepsilon$. Consequently, we obtain that $\mu(T^{-n}B\cap T^{-n}B)=\mu(B)^2$, hence $\mu(B)$ is $0$ or $1$. 
