Extreme points of a subset of dual space of continuous function $K$ is compact Hausdorff, and $C(K)$ denotes the space of continuous functions on $K$. Let $\mathbb{1}\in C(K)$ denotes the constant function taking value 1, and let $S$ be the subset of $C(K)^*$ consisting of positive linear bounded functionals on $C(K)$ that map $\mathbb{1}$ to $1$. Show that the extreme points of $S$ are the point evaluation maps, $f\to f(x)$.
I know what $C(K)^*$ is like by Riesz Representation Theorem, and for any $T\in C(K)^*$, $T(f)=\int_K fd\mu$ for some $\mu$. It is also easy to check that $S$ is convex.
Can anyone give a hint for this? Thanks.
 A: Edit (deleted original content as it was a bad answer, as pointed out by the asker)
Hint:


*

*Following what the asker pointed out in the comments to my former answer. Let $x\in K$, let $T_x$ be the evaluation functional $x\mapsto T_x(f)=f(x)$. Assume there is $\lambda \in (0,1)$ and $T_1\neq T_2\in S$ s.t. 
$$
f(x)=T_x(f)=\lambda T_1+(1-\lambda)T_2=\lambda\int_K f d\mu_1+(1-\lambda)\int_K f d\mu_2 
$$
Since $T_i\neq T$ for $i=1,2$, there is $y_i\neq x$ in the support of $\mu_i$. Can you use Urysohn's lemma here in a clever way to derive a contradiction?

*Let $T\in S$. Then there is some $\mu$ s.t. 
$$
T(f)=\int_K fd\mu
$$
Assume $T$ is not an evaluation functional, which implies there are $x\neq y$ in the support of $\mu$. Since $T\in S$, $\mu(K)=1$. Then, there are $K_x\subset K$ and $K_y=K\setminus K_x \subset K$, with $x\in K_x$, $y\in K_y$ measurable with $\mu(K_x),\mu(K_y)>0$. Then
$$
T(f)=T_{K_x}(f)+T_{K_y}(f)=\int_K fd\mu_1+\int_K fd\mu_2
$$
with $\mu_1=\mu|_{K_x}$, $\mu_2=\mu|_{K_y}$. How to conclude $T$ is not an extreme point of $S$ using this?
