Proposition 3

— Let K be a compact convex subset of a Hausdorff locally convex space E, x a point of K. Every measure $\mu$ on K, positive, of total mass 1, and admitting x as barycenter, is in the vague closure of the set of discrete positive measures of total mass 1 that admit x as barycenter.

Let U be a neighborhood of $\mu$ for the vague topology; we can suppose that U consists of the measures $\nu$ on K such that

for a finite number of functions $f_i \in \mathscr{C}(K; \mathbf{C})$ ($1 \leq i \leq p$) and a number $\delta > 0$. For every point $a \in K$, there exists a closed convex neighborhood $V_a$ of 0 in E such that

for $1 \leq i \leq p$ and for every $y \in W_a = K \cap (a + V_a)$. Since $K$ is compact, there exists a finite number of points $a_j$ ($1 \leq j \leq r$) of $K$ such that the $W_{a_j}$ form a covering of $K$ ($1 \leq j \leq r$). Consider a continuous partition of unity $(g_j)_{1 \leq j \leq r}$ on $K$, subordinate to the covering $(W_{a_j})$, and set $\alpha_j = \mu(g_j)$ for all $j$; if $\alpha_j \neq 0$ set $\mu_j = \alpha_j^{-1} g_j \cdot \mu$, and if $\alpha_j = 0$ set $\mu_j = \varepsilon_{a_j}$. Each of the measures $\mu_j$ is positive, of total mass 1, and its support is contained in the compact convex set $W_{a_j}$; moreover, by definition,

since $g_j \cdot \mu = 0$ if $\mu(g_j) = 0$; the $\alpha_j$ are $\geq 0$ and

Let $x_j$ be the barycenter of $\mu_j$, which belongs to $W_{a_j}$ (No. 1, Prop. 1), and consider the discrete measure $\nu = \sum_{j=1}^r \alpha_j \varepsilon_{x_j}$; it is positive and of total mass 1, and its barycenter is $\sum_{j=1}^r \alpha_j x_j$, which is also the barycenter of $\mu$ by virtue of (3), thus is equal to $x$. Moreover, by (2), $|f_i(y) - f_i(a_j)| \leq \delta/2$ for all $y \in W_{a_j}$ and for all $i$, whence, since $\operatorname{Supp}(\mu_j) \subset W_{a_j}$, $|\mu_j(f_i) - f_i(a_j)| \leq \delta/2$ for $1 \leq i \leq p$. On the other hand, since $x_j \in W_{a_j}$, one also has

for $1 \leq i \leq p$, whence $|\mu_j(f_i) - \varepsilon_{x_j}(f_i)| \leq \delta$ for all $i$ and $j$. Since the $\alpha_j$ are $\geq 0$ and have sum 1, it follows from (3) and the definition of $\nu$ that $\nu$ satisfies the inequality (1).

Q.E.D.