Proposition 23.7.4.label Let $X$ be an LCH space and $\cm \subset \overline{B_{M_R(X; \complex)}(0, 1)}$ be a compact convex set such that:
- (a)
For each $\mu \in \cm$ and $A, B \in \cb_{X}$, let $\mu_{A}(B) = \mu(A \cap B)$, then $\mu_{A} \in \cm$.
- (b)
For each $\mu \in \cm \setminus \bracs{0}$ and $t \in [0, 1/\norm{\mu}_{\text{var}}]$, $t\mu \in \cm$.
then for any $\mu \in \cm \setminus \bracs{0}$, the following are equivalent:
- (1)
$\norm{\mu}_{\text{var}}= 1$ and $\mu$ takes on exactly two distinct values.
- (2)
There exists $x \in X$ and $\lambda \in \partial B_{\complex}(0, 1)$ such that $\mu = \lambda \delta_{x}$.
- (3)
$\mu$ is an extreme point of $\cm$.
Moreover, $\cm$ is the closed convex hull of $(\bracsn{\lambda \delta_x|x \in X, \lambda \in \partial B_\complex(0, 1)}\cup\bracs{0}) \cap \cm$.
Proof. (1) $\Rightarrow$ (2): Assume without loss of generality that $\mu$ is positive and $\mu(\cb_{X}) = \bracs{0, 1}$. By inner regularity on open sets, there exists at least one compact set $K \subset X$ such that $\mu(K) = 1$.
Let $\mathcal{F}= \bracs{K \subset X|K \text{ compact}, \mu(K) = 1}$, then $\mathcal{F}$ is a $\pi$-system that does not contain $\emptyset$, and as such satisfies the finite intersection property. Thus $A = \bigcap_{K \in \mathcal{F}}K \ne \emptyset$.
Let $U \in \cn_{X}(A)$ and $K \in \cf$, then $K \setminus U$ is compact. Since $K \setminus U \cap A = \emptyset$, $K \setminus U \not\in \cf$, and $\mu(K \setminus U) = 0$. Thus $\mu(U) = \mu(K \cap U) = 1$. As this holds for all $U \in \cn_{X}(A)$, $\mu(A) = 1$ by outer regularity.
Finally, let $x \in A$ and $U \in \cn_{X}(x)$, then $A \setminus U \subsetneq A$, so $A \setminus U \not\in \cf$. As such, $A \subset A \cap \ol U$ for all $U \in \cn_{X}(x)$. Since $\bigcap_{U \in \cn_X(x)}\ol{U}= \bracs{x}$, $A = \bracs{x}$, and $\mu = \delta_{x}$.
(2) $\Rightarrow$ (3): Assume without loss of generality that $\mu = \delta_{x}$.
Let $\nu, \rho \in \cm$ and $t \in (0, 1)$ such that $\mu = (1 - t)\nu + t\rho$, then $1 = \mu(\bracs{x}) = (1 - t)\nu(\bracs{x}) + t\rho(\bracs{x})$. Since $\mu(\bracs{x}) = 1$ and $|\nu(\bracs{x})|, |\rho(\bracs{x})| \le 1$, $\nu(\bracs{x}) = \rho(\bracs{x}) = 1$. As $\norm{\nu}_{\text{var}}, \norm{\rho}_{\text{var}}\le 1$, $\nu = \rho = \delta_{x} = \mu$. Therefore $\mu$ is an extreme point of $\cm$.
(3) $\Rightarrow$ (1): If $\norm{\mu}_{\text{var}}\in (0, 1)$, then $\mu$ is a convex combination of $0$ and $\mu/\norm{\mu}_{\text{var}}$, so $\norm{\mu}_{\text{var}}$ must be $1$.
Suppose that $\mu$ takes on at least three distinct values, then there exists $A \in \cb_{X}$ such that $|\mu|(A), |\mu|(X \setminus A) > 0$. For each $B \in \cb_{X}$, let $\nu(B) = \mu(B \cap A)$ and $\rho(B) = \mu(B \setminus A)$, then $\mu = \nu + \rho$, $\nu, \rho \ne 0$, $\nu \perp \rho$, and $\norm{\nu}_{\text{var}}+ \norm{\rho}_{\text{var}}= \norm{\mu}_{\text{var}}$. In which case,
is a convex combination of $\mu$ in terms of two other elements of $\cm$.
Finally, by the Krein-Milman Theorem, $\cm$ is the closed convex hull of $(\bracsn{\lambda \delta_x|x \in X, \lambda \in \partial B_\complex(0, 1)}\cup\bracs{0}) \cap \cm$.$\square$
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