If $(X,d)$ is a metric space, is $X^{m+1}\to\{1,...,m\}, (x,x_1,...,x_m)\mapsto\min(\operatorname{argmin}_{k\in\{1,...,m\}}d(x,x_k))$ measurable? Given a a metric space $(X,d)$ and $m\in\mathbb{N}$, equip $X\times X^m$ with the product Borel $\sigma$-algebra and $\{1,...,m\}$ with the $\sigma$-algebra $2^{\{1,...,m\}}$.
Is the map $$X\times X^m\to\{1,...,m\}, (x,x_1,...,x_m)\mapsto\min\left(\operatorname{argmin}_{k\in\{1,...,m\}}d(x,x_k)\right)$$ measurable?

Yes. Denoting the map by $f$, we have
\begin{align*}
\{f(x,x_1,\cdots,x_m) = k\}
&= \left( \bigcap_{i < k} \{ d(x,x_i) > d(x, x_k) \} \right) \cap \left( \bigcap_{i > k} \{ d(x,x_i) \geq d(x, x_k) \} \right).
\end{align*}
Since each function $(x,x_1,\cdots,x_m) \mapsto d(x, x_i) - d(x, x_k)$ is continuous, $\{ d(x,x_i) > d(x, x_k) \}$ is open and $\{ d(x,x_i) \geq d(x, x_k) \}$ is closed. So the right-hand side is not only Borel-measurable, it is in fact $F_{\sigma}$-set.
