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Let us call a finite union $P$ of $n$-dimensional compact convex polytopes in $\mathbb{R}^n$ a non-convex polytope. Recall that a dissection of $P$ is a finite collection $T$ of $n$-dimensional simplices $T_i$ such that the interiors of $T_i$ and $T_j$, $i\neq j$, do not intersect, and that $P=\cup_i T_i$ (note that triangulations are particular types of dissections). Each $T$ comes with the set $V(T)=\cup_i V(T_i)$ of vertices, where we denoted by $V(T_i)$ the set of vertices of $T_i$.
It then appears natural to define $V(P)$, the set of vertices of $P$, as $$V(P):=\bigcap_{T\ \text{is a dissection of $P$}} V(T)$$ That is, $V(P)$ consists of points that appear as vertices of simplices in every dissection (note that if we replace dissections with triangulations in this definition, we can end up with more points, see an example of such a non-convex polytope here). We have the following

Conjecture. Let $\mu(P)$ be a uniform measure supported on $P$. Then $\mu(P)$ is an $\mathbb{R}$-linear combination of uniform measures $\mu(S_j)$ supported on $S_j$, where $S_j$ are $n$-simplices with vertices $V(S_j)\subseteq V(P)$. (The $S_j$ involved in the decomposition certainly need not have disjoint interiors.)

While obvious for convex polytopes, it becomes nontrivial for non-convex polytopes that cannot be triangulated without adding extra vertices, e.g. Schonhardt non-convex polytope. (For the latter it is still easy, as Schonhardt can be obtained by removing non-intersecting simplices from its convex hull, but it's not hard to build trickier examples).

We have a proof of this conjecture for the case of $V(P)$ being sufficiently generic -- namely, any $n+2$ points in $V(P)$ span $\mathbb{R}^n$, which uses quite technical analytic and algebraic tools. One obvious application to this might be computing integrals over $P$.
We wonder whether such questions arose before, whether our definition of vertices can be found in the literature, and whether there are any geometric/combinatorial approaches to this, especially to the full generality case of the conjecture. And any other comments are certainly most welcome, too.

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Is $V(P)$ in general uncountably infinite? – Joseph O'Rourke Jul 27 '12 at 11:46
no, $V(P)$ always finite. This is more or less straightforward. Indeed, there always exists a finite triangulation of $P$, right? Thus $V(P)$ will be contained in the (finite) set of vertices of simplices in this triangulation. – Dima Pasechnik Jul 27 '12 at 12:35
Sorry for not understanding, but even a planar rectangle has an infinite number of dissections into triangles. – Joseph O'Rourke Jul 27 '12 at 13:29
sure, but only the vertices of the triangle itself will appear in all dissections. As I take the intersection over (yes, uncountably many) dissections, I only get these 3. (I wrote "any dissection" in the question, perhaps I should rather use "every" ?) – Dima Pasechnik Jul 27 '12 at 14:41
Ah, yes, "any" misled me; "every" is correct. Apologies for misinterpreting. – Joseph O'Rourke Jul 27 '12 at 14:47

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