Extreme points of the probability measures are Dirac masses

If $K=\mathrm{\varnothing }$, then $P(K)=\mathrm{\varnothing }$ and there are no Dirac measures, so the equality is empty on both sides. Hence assume $K\mathrm{\ne }\mathrm{\varnothing }$. By [F7], compactness makes $K$ locally compact, and it is Hausdorff by hypothesis, so the LCH regularity convention [F4] applies.

Let $\mu \in P(K)$ and let $A$ be Borel. The restriction ${\mu }_A(E)=\mu (E\cap A)$ is a measure by [F3]. It is regular: for Borel $E$, inner regularity of $\mu$ on $E\cap A$ gives ${\mu }_A(E)=\sup \{\mu (D):D\subseteq E\cap A,D\text{ compact}\}$; every such $D$ is also a compact subset of $E$ with ${\mu }_A(D)=\mu (D)$, while every compact $C\subseteq E$ satisfies ${\mu }_A(C)\le {\mu }_A(E)$. Thus the required supremum over compact $C\subseteq E$ equals ${\mu }_A(E)$.

Suppose that $\mu (E)\in \{0,1\}$ for every Borel $E$. Let $\mathcal{C}$ be all closed $C\subseteq K$ with $\mu (C)=1$. It contains $K$. A finite intersection of its members has measure one because the complement is a finite union of null sets, so $\mathcal{C}$ has the finite-intersection property. By [F5], choose $x\in \bigcap \mathcal{C}$.

Conversely, fix any $x\in K$. By [F2], ${\delta }_x$ is a probability measure. The Hausdorff hypothesis makes the compact singleton $\{x\}$ closed and hence Borel. Thus ${\delta }_x$ is regular: for a Borel set containing $x$, that singleton realizes mass one, while a set not containing $x$ has mass zero. If ${\delta }_x=t\mu +(1-t)\nu$ with $\mu ,\nu \in P(K)$ and $0<t<1$, evaluation on $K\setminus \{x\}$ gives $0=t\mu (K\setminus \{x\})+(1-t)\nu (K\setminus \{x\})$, so nonnegativity gives both masses zero. Thus $\mu (\{x\})=\nu (\{x\})=1$ and $\mu =\nu ={\delta }_x$; [F1] makes ${\delta }_x$ extreme.

If $0<t:=\mu (A)<1$, define ${\mu }_1=t^{-1}{\mu }_A$ and ${\mu }_2=(1-t{)}^{-1}{\mu }_{K\setminus A}$. Step 1.2 makes both regular probabilities, and $\mu =t{\mu }_1+(1-t){\mu }_2$. They are distinct because ${\mu }_1(A)=1$ and ${\mu }_2(A)=0$, so [F1] shows that $\mu$ is not extreme.

Every open neighborhood $U$ of the point $x$ from step 1.3 has measure one. Otherwise the zero-one hypothesis gives $\mu (U)=0$, so the closed complement $K\setminus U$ has measure one and belongs to $\mathcal{C}$, contradicting $x\in U\cap \bigcap \mathcal{C}$.

If $D\subseteq K\setminus \{x\}$ is compact, [F6] gives disjoint open sets $U,V$ with $x\in U$ and $D\subseteq V$. Step 2.2 gives $\mu (U)=1$, hence $\mu (V)=0$ and $\mu (D)=0$. Inner regularity [F4] on the Borel set $K\setminus \{x\}$ now gives $\mu (K\setminus \{x\})=0$, so $\mu ={\delta }_x$.

Let $\mu$ be extreme. Step 2.1 rules out every Borel set of intermediate mass, so $\mu$ is zero-one valued; steps 1.3, 2.2, and 3.1 then give $\mu ={\delta }_x$ for some $x\in K$. Step 1.4 proves the reverse implication, and step 1.1 covers the empty space.