Theorem 3 (Choquet)

Let $E$ be a Hausdorff weak locally convex space, $C$ a weakly complete proper convex cone with vertex 0 in $E$, $G$ the union of the extremal generators of $C$, $K$ a compact convex subset of $C$, $\lambda$ and $\lambda'$ positive measures of mass 1 on $K$, admitting the same barycenter, such that $\lambda^*(K - (K \cap G)) = {\lambda'}^*(K - (K \cap G)) = 0$. Assume that $C$ is lattice-ordered. Then, for every lower semi-continuous, positively homogeneous convex function $f \geq 0$ on $C$, $\lambda^*(f|K) = {\lambda'}^*(f|K)$.

Let $\mathscr{A}$ (resp. $\mathscr{A}'$) be the set of restrictions to $C$ of the continuous linear forms (resp. affine functions) on $E$. We know (TVS, II, §5, No. 4, Remark 2) that $f$ is the upper envelope of the set of elements of $\mathscr{A}$ that are $\leq f$. The set of functions of the form $\sup(f_1, \ldots, f_p)$, where $f_1, \ldots, f_p$ belong to $\mathscr{A}$, $f_1 \geq 0, \ldots, f_p \geq 0$, is an increasing directed set and has $f$ as its upper envelope. Taking into account §1, No. 1, Th. 1, it suffices to verify the equality $\lambda(f|K) = \lambda'(f|K)$ when $f$ is of the preceding form.

Define $\overline{f}$ as in Lemma 5. It is clear that $\overline{f}(y) = f(y)$ if $y \in G$. Since $\lambda^*(K - (K \cap G)) = 0$, we have $\lambda(f|K) = \lambda(\overline{f}|K)$. By Lemma 6, $\overline{f}$ is affine and upper semi-continuous. Therefore $\overline{f}|K$ is the lower envelope of a decreasing directed set of restrictions of elements of $\mathscr{A}'$ to K (TVS, II, §5, No. 4, Prop. 6). Let $x \in K$ be the barycenter of $\lambda$. If $g \in \mathscr{A}$ then $\lambda(g|K) = g(x)$. Therefore $\lambda(\overline{f}|K) = \overline{f}(x)$ (§ 4, No. 4, Cor. 2 of Prop. 5). Thus $\lambda(f|K) = \overline{f}(x)$, and one sees similarly that $\lambda'(f|K) = \overline{f}(x)$.