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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!

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2 Answers 2

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A somewhat disappointing answer: assuming that

  1. $\mu\times\nu$ is Radon and outer regular,
  2. you complete the measure space $(X \times Y, \mathcal{B}(X\times Y), \mu\times \nu)$, i.e. add the sets of $\mu\times\nu$-measure zero to the Borel sigma-algebra,

then the product of any locally Borel sets is locally Borel, see proof below. Point 2 seems unavoidable but not very costly.

I am not sure how to prove or disprove point 1 in general. It seems to be true at least when $\mu$ and $\nu$ are $\sigma$-finite, see this thread and the book cited within. Maybe you can use the representation of Radon measures as elements of the dual of the space of continuous functions with compact supports; see point 6.5 of [Godfrey, M. C., & Sion, M. (1969). On product of Radon measures. Canadian Mathematical Bulletin, 12(4), 427-444] (and the reference to Bourbaki therein for completeness) together with the Riesz-Markov-Kakutani representation theorem.

Proof. Let $C \subset X\times Y$ Borel with finite measure. By point 1, for every $\epsilon>0$ there exists $\tilde O \subset O \supset C$ with $O$ open and $\tilde O$ $\sigma$-compact such that $$ \mu\times\nu(\tilde O) = \mu\times\nu(O) \leq \epsilon + \mu\times\nu(C).$$ Using similar arguments as in your answer, $\tilde O \cap (E\times F)$ is Borel, thus (using the completion in point 2) so is $O \cap (E\times F)$, and finally so is $C\cap(E\times F)$. $\square$

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In case that $G:=X=Y$ is a locally compact group and $\lambda:=\mu= \nu$ Haar measure, the answer is positive.

Let $A\subseteq G \times G$ be a Borel set with $(\lambda\times \lambda)(A)<\infty$. Then there exist $\sigma$-compact subsets $K,L\subseteq G$ with $A\subseteq K\times L$. Therefore, $$A\cap (E\times F)= A \cap (K\times L)\cap (E\times F) = A\cap [(K\cap E)\times (L\cap F)]$$ which is Borel since $K\cap E$ is Borel and $L\cap F$ is Borel.

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