Pure states on $C(X)$ are exactly evaluations

Question

Let $X$ be a compact Hausdorff space. I want to show that pure states are of the form $ \phi (f) =f(x)$.
By Reisz Represenation Theorem states on $C(X)$ are of the form $\phi (f)= \int fd\mu$ where $\mu$ is a probability regular borel measure. I'm also familiar with the equivalent defenition: $\phi$ is pure iff for every positive functional $\psi$ satisfying $\psi \le \phi$ there exists $t \in [0,1]$ s.t. $\psi = t \phi$.
$\Rightarrow$Suppose $\phi$ is an evaluation and show it's pure: write $\phi = t \psi_1 + (1-t) \psi_2$ for $t \in [0,1]$, and $\psi_1, \psi_2 \in S(A)$ so $\forall f\in C(X)$ we have $f(x)=\phi(f) = t \psi_1(f) + (1-t) \psi_2(f)=t\int fd\mu_1+ (1-t)\int fd\mu_2$ for some $\mu_1, \mu_2$ borel, regular probability measures.
If I could take $f=\chi_{\{x\}}$ I think I can finish the argument. Maybe I should approximate this indicator by continuous functions in $L^1$, but I'm not sure it's the right direction.
For the second direction I have nothing to present...
I would like to get some hints.
Thank you.

Answer 1

In your attempt to prove that $\phi$ is pure you are mixing the definition you gave ($\psi\leq\phi$ implies $\psi=t\phi$) with being an extreme point (the two things are equivalent, but I think you need to make up your mind).

For the converse, you need to show that any measure that has support with at least two points does not satisfy the definition you gave (or is not an extreme point, depending which definition you choose). So you need to use that the support of $\phi$ has more than one point to construct $\psi\leq\phi$ that is not a multiple of $\phi$.

No approximations are needed (and there are not that many approximation results in an arbitrary compact Hausdorff $X$).