Does almost every point in Euclidean space have unique projection on any given set?

Question

I am stuck at one point in my research, where I need to prove something which appears trivial to me, but cannot find a rigorous proof. I describe it below. Whenever I will say projection, I will mean the $L^2$ projection in Euclidean spaces.

We all know that the projection of a point on a closed, convex subset of the Euclidean space is unique. Now, if a set $A \subseteq \mathbb{R}^n$ is closed, we can easily show that the projection of a point $p \in \mathbb{R}^n$ on $A$ exists, but can find counterexamples, such that it is not unique. So, here is my first question (assume henceforth, that $A$ is a closed set):

1. Is it true that the set of all points $p \in \mathbb{R}^n$ that have more than one projection on the set $A$ (let us call that set $\mathcal{P}(A)$), has Lebesgue measure $0$?

It seems too much for question 1 to have an affirmitive answer for any arbitrary closed set $A$ (I do not even know whether the set $\mathcal{P}(A)$ is measurable, but if not, I can work with outer measures). So, here comes a simpler question.

2. If $A$ is the union of two convex sets, is it true that $\mathcal{P}(A)$ has Lebesgue measure $0$?

Even if this seems too much, I would really be happy to have an affirmitive answer (with a proof) to the following even simpler question:

3. If $A$ is the union of two polyhedra (a polyhedron is a finite intersection of half-spaces, and hence, is convex), does $\mathcal{P}(A)$ have Lebesgue measure $0$?

I intuitively feel that 3 must be correct, the reason being as follows. If a point $p$ has two distinct projections in $A$, then these projections must lie on two different faces of the union of the polyhedra, must be projections of $p$ on the respective faces too, and must be equidistant from $p$. The set of such point seems to be a lower dimensional hyperplane, which has Lebesgue measure $0$. But I cannot make this further rigorous. The problem is, I cannot seek help from linear algebra, as these faces are flats, and not even subspaces.

Any help (at least with answering question 3) will be greatly appreciated!

Answer 1

That is true. The set of points with non-unique projection has measure zero. The proof is a beautiful application of the Rademacher theorem. You can find comments and the link to a proof here:

Set of points with a unique closest point in a compact set.

See also: https://mathoverflow.net/a/324877/121665.