Is this subset of a closed set Borel measurable?

Question

Let $V \subset [0,\infty)^n$ be closed. I've been studying the following subset of $V$ for research, which has useful properties, but cannot prove or disprove it is Borel measurable.

Introduce the partial ordering $\leq'$ on $[0,\infty)^n$ such that $(x_1,...,x_n) \leq' (y_1,...y_n) \iff x_k \leq y_k, \forall k$. Consider the set $W$ of minimal elements of $V$ under $\leq'$. That is, $$ W = \big\{ x \in V \ \big| \ \nexists y\in V, y<'x\big\}. $$ One can easily show that $W \subset \partial V$ and if $V \neq \emptyset$ then $W \neq \emptyset$ (with Zorn's lemma).

I have been wondering how to prove/disprove $W$ is Borel measurable for months, gave up, and settled with taking its closure only to make it Borel measurable (if the set were measurable, everything would work out nicely).

I believe this set is Borel measurable and that $\overline{W}\backslash W$ is countable (it does not seem easy to construct a counterexample to either of these, because $W$ seems to always have a nice monotone shape). However, a proof of either of these has continuously evaded me.

Answer 1

The set you are asking for is easily $G_\delta$, and thus Borel, if I'm not missing something obvious.

The set $C = \{ (x,y)\in V^2 : x <' y \}$ is $\sigma$-compact, since $$C = \leq' \cap \bigcup_{i \leq n} \{ (x,y) \in V^2 : x_i < y_i \}.$$ The relation $\leq'$ is closed and the set on the right is open (in $V^2$) and thus $\sigma$-compact (in a Polish space open sets are $F_\sigma$, and if the space is $\sigma$-compact, then $F_\sigma$ sets are obviously also $\sigma$-compact). Thus its projection to the second coordinate $W' = \{ y \in V : \exists x \in V ( x < y) \}$ is also $\sigma$-compact (remember that the projection of a compact set is compact). And the complement $W = V \setminus W'$ is $G_\delta$.

By the way, the question you are asking falls exactly into the field known as "Descriptive Set Theory".

Answer 2

This is too long for a comment, I don't have time to finish this right now but it may be useful.

Let $P$ be the power set of $\{ 1,\dots, n \}$, for any $A\in P$, let $C_A= \{ y\in \mathbb R^n: \forall i\in A, y_i>0, \forall i\notin A,y_i=0 \}$, then $\{ C_A \}_{A\in P}$ is a partition of $[0,\infty[^n$ into finitely many measurable cones, let $V_A=V\cap C_A$ and let $W_A=W\cap C_A$. It is clear that $W_A$ is the set of minimas of $V_A$. This means that if we prove that $W_A$ is measurable, then we are done.

Let $V_A'$ be the measurable convex set defined as \begin{align*} V_A'=\bigcap_{x\in C_A} \{ y\in C_A : \inf_{z\in V_A} \langle x,z\rangle\leq \langle x,y\rangle \} \end{align*}

This is some sort of closed convex hull of $V_A$ from below with respect to the order $\leq$ intersected with $C_A$. I think we can prove that $W_A=V_A\cap \partial V_A'$ which makes it measurable.