Proof. Let $\lambda \in \sigma_{A}(x)$. By Proposition 33.8.2, there exists $\phi \in \Omega(A[x])$ such that $\dpn{x, \phi}{A[x]}= \lambda$. By Proposition 34.10.5, $\phi \in P(A[x])$. The pure state extension theorem implies that there exists $\Phi \in P(A)$ such that $\Phi|_{A[x]}= \phi$. Thus $\Phi$ is a pure state with $\dpn{x, \Phi}{A}= \lambda$, and $\sigma_{A}(x) \subset \bracs{\dpn{x, \Phi}{A}|\Phi \in P(A)}$.

Let $\Phi \in S(A)$ and $\phi = \Phi|_{A[x]}$, then $\phi \in S(A[x])$ as well. By the Gelfand-Naimark Theorem, the Spectral Theorem, and the Riesz Representation Theorem, $\phi$ takes the form of a Radon probability measure $\mu$ on $\sigma_{A}(x)$. In which case,

Finally, since $S(A)$ is compact and convex by Proposition 34.10.4,

by Proposition 5.16.3 and Proposition 5.5.3.$\square$