given dynamic system $(X, \mathcal{B}, F, \mu), \mu \circ F^{-1}=\mu, F $ is mixing, $ A \in \mathcal{B}, s.t. \mu(A) >0 $. 


consider dynamic system $(X\times X, \mathcal{B}\otimes \mathcal{B}, F\times F, \mu \times \mu)$, then $ F \times F $ is mixing too, hence ergodic, $(\mu \times \mu )(A \times A )>0$.


fixed delay time $n\_0 \in \mathbb{N}$, define $R(x)=\min \{n: F^n(x) \in A \} $, and stopping times recursively: for all $ (x, x') \in X \times X$,


$ \tau\_1(x,x')= n\_0 +R(F^{n\_0}x) $


$ \tau\_2(x,x')= \tau\_1+n\_0 +R(F^{n\_0+\tau\_1}x') $


$ \tau\_3(x,x')= \tau\_2+n\_0 +R(F^{n\_0+\tau\_2}x) $


$ \tau\_4(x,x')= \tau\_3+n\_0 +R(F^{n\_0+\tau\_3}x') $ ect.


then we know $ F^{\tau\_1} \in A \times X, F^{\tau\_2} \in X \times A$ ect.


can we prove:


for almost $ (x, x') $ wrt $ \mu\times \mu$, $ \exists \tau\_n=\tau\_n(x,x'), s.t. F^{\tau\_n} (x,x') \in A \times A$? Thanks!



time delay ergodic theorem
Here is a revised answer, having correctly understood the question. 


Let $A\_1=A$ and $A\_0=A^c$. For a sequence $\mathbf w=w\_0,\ldots,w\_{k-1}$, set 
$A\_{\mathbf w}=A\_{w\_0}\cap F^{-1}A\_{w\_1}\cap\ldots\cap F^{-(k-1)}A\_{w\_{k-1}}$. That is, $A\_{\mathbf w}$ is the set of points that is inside or outside of $A$ for $k$ steps according to $\mathbf w$. 


Let $S=\{\mathbf w\colon |\mathbf w|=n\_0+1;\, w\_{n\_0}=1\}$. 
Notice that $\bigcup\_{\mathbf w\in S}A\_{\mathbf w}=F^{-n\_0}A$, so that there exists $\mathbf w\in S$ with $\mu(A\_{\mathbf w})>0$. Now $\mu\times\mu$-almost every pair $(x,x')$ visits $A\_{\mathbf w}\times A\_{\mathbf w}$ infinitely often. On each visit, there is at least one common return to $A$. To see this, if $(F^mx,F^mx')\in A\_{\mathbf w}\times A\_{\mathbf w}$, then let $\tau\_j<m$ be the last term in the sequence of $\tau$'s that is smaller than $m$. Then $\tau\_j+{n\_0}$ is inside the block $A\_{\mathbf w}\times A\_{\mathbf w}$ that starts at $m$, so that the first occurrence of $A$ is common.


