# Minkowski sum of the intersection of a closed and an open set with a compact set

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.

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

$$(A \cap B) \oplus C$$ 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 $$F_\sigma.$$