Theorem 1 (Choquet)

— Let E be a Hausdorff locally convex space over $\mathbf{R}$, K a metrizable compact convex subset of E, and M the set of extremal points of K. The set M is the intersection of a countable family of open sets in K, and every point of K is the barycenter of a measure $\mu$ on K such that $\mu(K - M) = 0$.

To prove the first assertion, denote by I the interval $[0, 1]$ of $\mathbf{R}$; since K is compact and metrizable, so is $K \times K \times I$; the subset U of $K \times K \times I$ formed by the triples $(x, y, \lambda)$ such that $x \neq y$ and $0 < \lambda < 1$ is open in $K \times K \times I$, therefore there exists a sequence $(F_n)$ of closed sets in $K \times K \times I$ whose union is U (GT, IX, §2, No. 5, Prop. 7). The mapping $q : K \times K \times I \to K$ defined by $q(x, y, \lambda) = \lambda x + (1 - \lambda) y$ is continuous and, by the definition of extremal point, $K - M = q(U) = \bigcup_n q(F_n)$; but $F_n$ is compact since it is closed in $K \times K \times I$, therefore $q(F_n)$ is compact and therefore closed in K; the set $U_n = K - q(F_n)$ is therefore open in K, and we have $M = \bigcap_n U_n$.

In the continuation of the proof, we shall denote by $u$ a continuous and convex numerical function on K, by $G \subset E \times \mathbf{R}$ the graph of $u$, and by S the closed convex envelope of G in $E \times \mathbf{R}$.