Added Krein-Milman for measures.

src/fa/order/positive.tex

@@ -28,7 +28,7 @@
 \begin{proof}[Proof, {{\cite[V.5.4]{SchaeferWolff}}}. ]
     (1) $\Rightarrow$ (2): Let $U = \bracs{\text{Re}(\Phi) < 1}$, then $U \in \cn_\topo(0)$ is convex. Since $\Phi$ is positive, for any $x \in U$ and $y \in C$, $\text{Re}\dpn{x - y, \Phi}{E} \le \text{Re}\dpn{x, \Phi}{E} < 1$. 
 
-    (2) $\Rightarrow$ (1): Assume without loss of generality that $K = \real$. Let $\alpha > 0$ such that $F \cap (U - C) \subset \bracs{\phi < \alpha}$. Since $U$ is convex and open, and $C$ is convex, $U - C \subset E$ is an open convex set, disjoint from the convex set $\bracs{\phi = \alpha}$. By the \autoref[Hahn-Banach Theorem]{theorem:hahn-banach-geometric-1}, there exists $\Phi \in E^*$ such that $U - C \subset \bracs{\Phi < \alpha}$, and $\bracs{\phi = \alpha} \subset \bracs{\Phi \ge \alpha}$. 
+    (2) $\Rightarrow$ (1): Assume without loss of generality that $K = \real$. Let $\alpha > 0$ such that $F \cap (U - C) \subset \bracs{\phi < \alpha}$. Since $U$ is convex and open, and $C$ is convex, $U - C \subset E$ is an open convex set, disjoint from the convex set $\bracs{\phi = \alpha}$. By the \hyperref[Hahn-Banach Theorem]{theorem:hahn-banach-geometric-1}, there exists $\Phi \in E^*$ such that $U - C \subset \bracs{\Phi < \alpha}$, and $\bracs{\phi = \alpha} \subset \bracs{\Phi \ge \alpha}$. 
 
     After rescaling, assume without loss of generality that $\bracs{\phi = \alpha} \subset \bracs{\Phi = \alpha}$, then $\Phi \in E^*$ is an extension of $\phi$. For each $x \in C$ and $\lambda > 0$, $-\lambda x \in U - C$, and $-\lambda\dpn{x, \Phi}{E} < \alpha$. As this holds for all $\lambda > 0$, $\dpn{x, \Phi}{E} \ge 0$, so $\Phi$ is the desired extension.
 \end{proof}

src/measure/vector/fin.tex

@@ -117,3 +117,46 @@ Despite not covering the full dual space, the bounded Borel functions still form
     (2) $\Rightarrow$ (1): By the \hyperref[Dominated Convergence Theorem]{theorem:dct-bochner-vector}. 
 \end{proof}
 
+\begin{proposition}
+\label{proposition:space-of-measures-extreme-points}
+    Let $X$ be a LCH space and $\cm \subset \overline{B_{M_R(X; \complex)}(0, 1)}$ be a compact convex set such that:
+    \begin{enumerate}[label=(\alph*)]
+        \item For each $\mu \in \cm$ and $A, B \in \cb_X$, let $\mu_A(B) = \mu(A \cap B)$, then $\mu_A \in \cm$.
+        \item For each $\mu \in \cm \setminus \bracs{0}$ and $t \in [0, 1/\norm{\mu}_{\text{var}}]$, $t\mu \in \cm$. 
+    \end{enumerate}
+
+    then for any $\mu \in \cm \setminus \bracs{0}$, the following are equivalent: 
+    \begin{enumerate}
+        \item $\norm{\mu}_{\text{var}} = 1$ and $\mu$ takes on exactly two distinct values. 
+        \item There exists $x \in X$ and $\lambda \in \partial B_\complex(0, 1)$ such that $\mu = \lambda \delta_x$. 
+        \item $\mu$ is an extreme point of $\cm$. 
+    \end{enumerate}
+
+    Moreover, $\cm$ is the closed convex hull of $(\bracsn{\lambda \delta_x|x \in X, \lambda \in \partial B_\complex(0, 1)} \cup\bracs{0}) \cap \cm$. 
+\end{proposition}
+\begin{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, 
+    \[
+    \mu = \frac{\norm{\nu}_{\text{var}}}{\norm{\mu}_{\text{var}}} \cdot \underbrace{\frac{\norm{\mu}_{\text{var}} \cdot \nu}{\norm{\nu}_{\text{var}}}}_{\in \cm} + \frac{\norm{\rho}_{\text{var}}}{\norm{\mu}_{\text{var}}} \cdot \underbrace{\frac{\norm{\mu}_{\text{var}} \cdot \rho}{\norm{\rho}_{\text{var}}}}_{ \in \cm}
+    \]
+
+    is a convex combination of $\mu$ in terms of two other elements of $\cm$. 
+
+    Finally, by the \hyperref[Krein-Milman Theorem]{theorem:krein-milman}, $\cm$ is the closed convex hull of $(\bracsn{\lambda \delta_x|x \in X, \lambda \in \partial B_\complex(0, 1)} \cup\bracs{0}) \cap \cm$. 
+\end{proof}
+
+