Extreme points of subset of set of probability measures

Question

Let $\mathcal{P}$ be a set of Borel probability measures on $[0, 1]$ equipped with the weak$^{*}$-topology as described here: What is the weak*-topology on a set of probability measures?. Consider for $\rho\in [0, 1]$, $$\mathcal{P}(\rho)=\left\{\nu\in \mathcal{P}:\int_{0}^{1}{td\nu(t)}=\rho\right\}.$$ I can show that this set is convex and compact for the weak$^{*}$-topology. I was wondering what the extreme points of $\mathcal{P}(m)$ are and found that the extreme points are of the form $$\nu=p\delta_{x}+q\delta_{y}$$ with $p, q\geq 0$, $x, y\in [0, 1], p+q=1, px+qy=\rho$ where $\delta_{x}$ denotes the Dirac measure at $x$. Can someone explain to me why they are of this form?

Answer 1

This might not satisfy you, but this is covered by a result of Dubins, cited in my answer to a previous MSE question. You have a compact convex set $\mathcal P$ of probability measures, its set of extreme points is the set of point masses. Your $\mathcal P(\rho)$ is the intersection of $\mathcal P$ and a codimension-1 affine space, namely the measures $\nu$ cut out by the single equation $\int_0^1 td\nu(t)=\rho$. (This equation is affine in $\nu$.) The Dubins result then tells us that the extreme points of $\mathcal P(\rho)$ are linear combinations of $1+1=2$ extreme points of $\mathcal P$, that is, of pairs of point masses.

I wrote "might not satisfy you" because this post only tells you that the fact you ask about has been noticed before but does not explain why the result is true. Maybe you can examine your proof and see how it squares with Dubins's?