Corollary

— Let $K'$ be a compact subset of $K$ such that $K$ is the closed convex envelope of $K'$. In order that $x \in K'$ be an extremal point of $K$, it is necessary and sufficient that $\varepsilon_x$ be the only positive measure on $K'$, of total mass 1, having $x$ as barycenter.

Suppose $x$ is an extremal point of $K$; to prove that $\varepsilon_x$ is the only positive measure on $K'$, of total mass 1, having $x$ as barycenter, it suffices, by Prop. 3, to see that the set of discrete measures $\nu$ on $K'$ that are positive, of total mass 1, and have $x$ as barycenter, reduces to $\varepsilon_x$. But such a measure $\nu$ is of the form $\sum_{i=1}^r \lambda_i \varepsilon_{x_i}$ with $\lambda_i > 0$ for $1 \leq i \leq r$ and $\sum_{i=1}^{r} \lambda_i = 1$, and the hypothesis that $x$ is the barycenter of $\nu$ may be written $x = \sum_{i=1}^{r} \lambda_i x_i$. Since $x$ is extremal, this implies that $x_i = x$ for all $i$, whence $\nu = \varepsilon_x$.

Conversely, let us assume that $\varepsilon_x$ is the only positive measure on $K'$, of total mass 1, having $x$ as barycenter, and let us show that $x$ is extremal. In the contrary case, there would exist two distinct points $x', x''$ of $K$ and a real number $\lambda$ such that $0 < \lambda < 1$ and $x = \lambda x' + (1 - \lambda) x''$. By Prop. 1, $x'$ (resp. $x''$) is the barycenter of a positive measure $\mu'$ (resp. $\mu''$) on $K'$ of total mass 1. Then $x$ is the barycenter of $\lambda \mu' + (1 - \lambda) \mu''$. Therefore $\lambda \mu' + (1 - \lambda) \mu'' = \varepsilon_x$. Therefore $\mu'$ and $\mu''$ are proportional to $\varepsilon_x$, whence $x' = x'' = x$, which is absurd.