Let $X$ be a locally compact Hausdorff space with a fixed Radon measure (= Borel measure that is finite on compact subsets, inner regular on open subsets and outer regular on Borel sets) $\mu$ . 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 $(\mu\times \nu)(A) < \infty$, why is $A \cap (E\times F)$ a Borel subset? Or is this not true at all? I am mainly interested in the case where $\mu = \nu$ is Haar measure on a locally compact Hausdorff group.

Note that if $K\subseteq X \times Y$ is compact, then $K\subseteq K_1\times K_2$ where $K_1, K_2$ are the projections of $K$ onto its components. Hence, $$K \cap (E\times F)= K\cap ((K_1\cap E)\times (K_2\cap F))$$ and $K_1\cap E, K_2\cap F$ are Borel. This does not depend on the Radon measure, so will not give a general proof.

Thanks in advance for your help and/or valuable suggestions!