This statement in probability seems definitely true, but I cannot work out how to prove it.
Let $X, Y$ be measure spaces, $U \subset X$, $V \subset Y$ open sets, and $f: X \times Y \to X$ a measurable function. We then define two sequences of random variables in $X$ and $Y$ accordingly: Let $(y^k)_{k\in\mathbb{N}}$ be a sequence of i.i.d. random variables in $Y$, such that $\mathbb{P}(y^k \in V) = p > 0$, and define $(x^k)_{k\in\mathbb{N}} \subset X$ via $$ x^{k+1} := f(x^k, y^k), \qquad \mbox{where } k \in \mathbb{N}, \mbox{ and } x^0 \in X \mbox{ is given}. $$ The statement I want to prove is then: $$ \mathbb{P}\left( (x^k \in U) \cap (y^k \in V) \mbox{ infinitely often } \; | \; x^k \in U \mbox{ infinitely often} \right) = 1. $$
Any suggestions on how to prove this would be appreciated. I have tried to consider Borel-Cantelli type results, but without pairwise independence of $(x^k)_{k\in\mathbb{N}}$, I wonder if that is a dead-end.
At first I also thought that this must be true, but it isn’t, actually. Consider $f$ such that $f(x,y)=a\not\in U$ for $(x\in U)\cap(y\in V)$ and $f(a,y)=a$ for all $y\in Y$. Then $(x^k\in U)\cap(y^k\in V)$ at most once, and yet $x^k\in U$ can occur infinitely often (albeit with probability $0$, as it requires $y^k\not\in V$ whenever $x^k\in U$).