Equivalence of some characterization of pure states I'm looking for a reference or a proof for these well-known facts in $C^*$-algebras theory for which, however, I havent found any clearly written proof of the same type of ones I will sketch.

Lemma. Let $\mathscr A$ be a commutative $C^*$-algebra. A state $\omega$ is pure iff $\mbox{supp}\,\mu_\omega= \{x\}$.

In my definition, a state is pure if it is not a convex combination of two different states. For the $(\implies)$ side, the idea would be suppose, on the contrary, that the support of $\mu_\omega$ contains at least two points, $\varphi_1$ and $\varphi_2$. Since the space of states is a weak*-compact Hausdorff topological space, Urysohn Lemma applies, so one can find a continuous $f : [0,1] \to \mathscr S(\mathscr A)$ s.t. $f(0) = \varphi_1$ and $f(1) = \varphi_2$. Hence it is possible to write $\mbox{d}\mu_\omega = f \mbox{d}\mu_\omega + (1-f)\mbox{d}\mu_\omega$ and, by means of Riesz-Markov theorem,
$$\omega(A) = \int \hat A(\varphi) \mbox{d}\mu_\omega(\varphi) = \int \hat A f \mbox{d}\mu_\omega + \int \hat A (1-f)\mbox{d}\mu_\omega = f_1(A) + f_2(A).$$
Here $f_1, f_2$ are positive functional normalizable so that
$$\omega(A) = \left(\int (1-f) \mbox{d}\mu_\omega \right) \omega_1(A) + \left(\int f \mbox{d}\mu_\omega\right) \omega_2(A).$$
Hence $\omega$ decomposes into a proper convex combination of two states, hence it is not pure. Here, I have less or more caught the principle but passages are somewhat unclear.
For the other side, the idea is that $\mu_\omega$ is a Dirac delta (its mass must be 1, since states are normalized) so one can write something like this in spirit (I don't think this is completely well written)
$$\int \hat A(\varphi) \mbox{d}\mu_\omega(\varphi) = \int \hat A(\varphi) \delta_\omega (\varphi) \mbox{d}s = \hat A(\omega) = \omega(A),$$
i.e., $\omega$ is pure since one can't decompose the domain of integration into two parts both of positive measure. 
From this follows easily (in the commutative case) the direct side of

Claim. $\omega$ is pure iff the only state dominated by $\omega$ is $\omega$ itself.

The other side is more mysterious. This claim is part of the proof I know for the 

Segal irreducibility theorem. A state is pure iff the associated GNS representation is irreducible. 

In few words, one sets up a bijection between the set of states dominated by $\omega$ and the set of operators $T$ in the commutant of the representation s.t. $(\xi_\omega, T \xi_\omega) = 1$, where $\xi_\omega$ is the ciclyc vector. This turn out to be an isomorphism of convex sets, hence the conclusion follows. 
If you can recognize these proofs, may you post references or fill the details for the Lemma and/or the claim? (Segal's theorem is clear, once the claim has been established, so don't mind about it. I have included it for help understanding the role of the other results.)
 A: Some notation is missing, like for example you need to write your C$^*$-algebra as $C(X)$ for a locally compact Hausdorff space.
For your converse, if $\phi_\omega$ is the state given by evaluation at $\omega$ and $\phi_\omega=\alpha\varphi+(1-\alpha)\psi$, by Riesz-Markov you get an equality of measures 
$$\mu_\omega=\alpha\mu_\varphi+(1-\alpha)\mu_\psi.$$ Now let $Y\subset X$ be any Borel subset with $\omega\not\in Y$. In that case, 
$$
0=\mu_\omega(Y)=\alpha\mu_\varphi(Y)+(1-\alpha)\mu_\psi(Y).
$$
It follows that $\mu_\varphi(Y)=\mu_\psi(Y)$. As we can do this for any $Y$ with $\omega\not\in Y$, it follows that both measures are concentrated at $\omega$. As they are probability measures (because they come from states), we get $\mu_\varphi=\mu_\psi=\mu_\omega$.

Regarding the claim, the statement is slightly incorrect: a state can never dominate another state: if $\varphi,\psi$ are states and $\varphi\leq\psi$, then $\psi-\varphi$ is a positive functional that is zero at the identity, so it is zero, and then $\varphi=\psi$. The key is to remove the word "state". 
The argument below proves the claim in general. It requires the algebra to be unital, but states in a C$^*$-algebra extend uniquely to the unitization, so nothing is lost by appealing to it. 
If $\omega$ is not pure, then $$\omega=\alpha\varphi+(1-\alpha)\psi$$ with $\alpha\in(0,1)$ and $\varphi\ne\omega$. Then $\alpha\varphi$ is a positive functional that is dominated by $\omega$ and it is not a scalar multiple of it. 
Conversely, if $\omega$ is pure and $\varphi\leq\omega$. Then
$$
\omega=\varphi(1)\,\frac{\varphi}{\varphi(1)}+(\omega(1)-\varphi(1))\,\frac{\omega-\varphi}{\omega(1)-\varphi(1)}
$$
is a convex combination of states. As $\omega$ is pure, we get $\frac{\varphi}{\varphi(1)}=\omega$, i.e.
$$
{\varphi}={\varphi(1)}\,\omega.
$$
