Let $X$ be a Polish space and $(Y, \mathcal{S})$ a measurable space. Let $F: Y \rightrightarrows X$ be a set-valued map with nonempty closed values (i.e., $F(y) \subset X$ is nonempty and closed for each $y \in Y$) such that for every [open set](/page/Open%20Set) $U \subset X$, the set $\{y \in Y : F(y) \cap U \neq \varnothing\}$ belongs to $\mathcal{S}$. Then $F$ admits a **measurable selection**: there exists a measurable function $f: Y \to X$ such that $f(y) \in F(y)$ for every $y \in Y$.