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Consider $\mathbb R^n$ with the usual topology and the Borel sigma-algebra. Let $A$ be open and $B$ be closed sets, respectively, in $\mathbb R^n$. Let $C$ be a compact set. Is the set $(A \cap B) \oplus C$ always Borel measurable?

Comments: The set $A \oplus C$ is open (and therefore measurable), $B \oplus C$ is closed (and measurable). In general, the Minkowski sum of a Borel measurable set with a compact set need not be Borel measurable.

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  • $\begingroup$ Sorry, $A \oplus B = \{a+b | a \in A, b \in B\}$ -- the Minkowski sum of $A$ and $B$. $\endgroup$
    – VSJ
    Mar 14, 2021 at 16:34

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$(A \cap B) \oplus C$ must be $F_\sigma$ (and therefore Borel measurable).

Indeed, since $A$ is open it is $F_\sigma$, and say $A=\cup_{n<\omega} F_n$ where each $F_n$ is closed.

Then $F_n\cap B$ is closed and hence $(F_n\cap B) \oplus C$ must be closed (where we use that $C$ is compact). Clearly $(A \cap B) \oplus C=\cup_{n<\omega}\big((F_n\cap B) \oplus C\big)$, showing that $(A \cap B) \oplus C$ is $F_\sigma.$

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