Perhaps this is a simple fact that I'm missing. In a paper I'm reading ("Nuclear C*-Algebras and the Approximation Property" by M. D. Choi and E. G. Effros) they claim that if $A$ is a C*-algebra (treated as its image under its own universal representation, so that $A'' = A^{**}$ is its enveloping von Neumann algebra) then every normal state on $A''$ is a vector state. They claim this without proof, but I haven't been able to come up with an argument for it, nor have I seen it stated anywhere.
Certainly every normal state is the infinite linear combination of vector states (in any von Neumann algebra). I also see that every state on $A$ is a vector state (by virtue of treating $A$ as its image under its universal representation), and every state on $A$ extends to a normal vector state on $A^{**}$. I'm not sure why every normal state on $A^{**}$ is the unique extension of a state on $A$ - is it perhaps a $\sigma$-weak density argument? Come to think of it this might be the case, but I still thought it was worth posting this question.
Is there a simple reason why this should be true? Have I stumbled across the right answer there in the second paragraph?
A state on $A^{**}$ induces a state on $A$, hence a representation of $A$ and thus a vector state on $A$ in the universal representation.
The normal states are the $\sigma$-weak continuous states and $A$ is $\sigma$-weak-dense in $A^{**}$, so if the state was normal then it is determined by its action on $A$. But we have just seen that on $A$ it agrees with a vector state — hence the normal state agrees with a vector state on all of $A^{**}$.