Let $A= C^*(1,a)$ where $a \geq 0$, that is $A$ is a unital $C^*$-algebra generated by a positive element $a$ and the identity. Let $f: A \to \mathbb{C}$ be a state on $A$. We have a canonical identification $X: A \cong C(\sigma(a))$ where $\sigma(a) \subseteq \Bbb{R}$ is the spectrum of $a$. By Riesz' theorem, there is a Radon Borel measure $\mu$ on $\sigma(a)$ such that $$f(a) = \int_{\sigma(a)} X(a) d \mu.$$ Is it true that $$\|a \| = \sup (\operatorname{supp} \mu)$$?
Attempt:
Clearly, $\operatorname{supp}\mu \subseteq \sigma(a)$ so $\sup \operatorname{supp} \mu \leq \sup \sigma(a) = r(a)= \|a\|$. However, I can't justify why the other inequality is true.
This is not true. Suppose $x\in\sigma(a)$, with $x\neq\sup\sigma(a)$, and let $f$ be the state such that $f\circ X^{-1}\in C(\sigma(a))^*$ is evaluation at $x$, that is, $f(b)=[X(b)](x)$ for $b\in A$. The corresponding measure is $\delta_x$, the Dirac mass at $x$, and $\operatorname{supp}\delta_x=\{x\}$, so $$\sup(\operatorname{supp}\delta_x)=x<\sup\sigma(a)=\|a\|.$$