Let $X$ be a locally compact Hausdorff space with a fixed Radon measure $\mu$. A subset $E\subseteq X$ is called locally Borel iff $E \cap A$ is a Borel subset of $X$ for every Borel subset $A$ of finite $\mu$-measure.
Suppose now that also $Y$ is a locally compact Hausdorff space and $F\subseteq Y$ is locally Borel with respect to a fixed Radon measure $\nu$ on $Y$.
Is it true that $E\times F$ is locally Borel in the product space $X\times Y$ with respect to the product Radon measure $\mu\times \nu$? I can prove this in the $\sigma$-compact or second countable case, but the general case is still unclear to me.
In other words, given $A\subseteq X \times Y$ with $(\mu\times \nu)(E) < \infty$, why is $A \cap (E\times F)$ a Borel subset? Or is this not true at all?
Thanks in advance for your help/valuable suggestions!