Choquet theory

Remarks

Every point of $K$ is represented by the Dirac measure at that point; the theory asks for measures on the extreme points. Choquet's theorem: if $K$ is metrizable, then $\operatorname{ext} K$ is a $G_\delta$ set, and every $x \in K$ is represented by a probability measure $\mu$ with $\mu(\operatorname{ext} K) = 1$. The Bishop–de Leeuw theorem extends this to non-metrizable $K$ in a weaker form. For metrizable $K$, the measure on $\operatorname{ext} K$ is unique for every $x$ exactly when $K$ is a Choquet simplex.

Example: let $K = [0, 1] \subseteq \R$, with $\operatorname{ext} K = \set{0, 1}$. The measure $\mu = (1 - x) \delta_0 + x \delta_1$ represents $x$. The check: for $\ell(t) = c t$, one gets $\int \ell \, d\mu = c x = \ell(x)$.

Non-example: for $x = 1/2$, the measure $\delta_0$ is concentrated on $\operatorname{ext} K$ but does not represent $x$, since its barycenter is $0$.

For the square $K = [0, 1]^2$, the centre is represented by $\frac12(\delta_{(0,0)} + \delta_{(1,1)})$ and by $\frac12(\delta_{(1,0)} + \delta_{(0,1)})$, so uniqueness fails.