Extreme points of the set of positive regular borel measures on a compact Hausdorff space I have some troubles with a specific proof of a (Bochner-type) theorem in Rudin's book "Functional Analysis".
More specifically, let $X$ denote a compact Hausdorff-Space and let $M$ denote the set of positive regular Borel measures $\mu$ on $X$ satisfying $\mu(X)\leq 1$. 
It is stated that the unit masses concentrated at the different points of $X$ are the extreme points of $M$. I can see that these are indeed extreme points - however, I am stuck arguing that these are the only extreme points of $M$. 

If $0<\mu(X)<1$, then the decomposition $\mu=\frac12((1+\epsilon)\mu+(1-\epsilon)\mu)$, with $\epsilon$ sufficiently small, shows that $\mu$ is not extreme.
I'm not sure if you allow $\mu(X)=0$ but if you do, this measure is also an extreme point.
Finally, suppose $\mu(X)=1$. If there is a set $A\subset X$ with $0<\mu(A)<1$, write
$$\mu=\frac12\left\{[(1+\epsilon)\mu_{|A}+(1-\delta)\mu_{|A^c}]+[(1-\epsilon)\mu+(1+\delta)\mu_{|A^c}]\right\} $$
where $\epsilon,\delta$ are small and are chosen so that the total measure in each square bracket is $1$.
It remains to consider the case when  $\mu(A)\in \{0,1\}$ for every measurable set $A$. Let $S$ be the set of points $x$ such that every neighborhood of $x$ has positive measure. By assumption, this measure is $1$. It follows that $S$ cannot have more than one point. It cannot be empty either, for then $\mu$ would be identically zero. Thus, $S$ is a single point $\{x_0\}$. Use regularity of the measure to obtain $\mu(\{x_0\})=1$.
