Here is the result I'm covering:
Let $X$ be a compact Hausdorff space and $H$ a Hilbert space, and suppose that $\varphi: C(X) \longrightarrow B(H)$ is a unital $*$ homomorphism. Then there exists a unique spectral measure $E$ with respect to the pair $(X,H)$ such that: $$\varphi(f) = \int_{X} f \: dE \: \: \forall \: f \in C(X).$$
Proof: Let $x,y \in H$. Then the map $\tau_{x,y}: C(X) \longrightarrow \mathbb{C}$ by $\tau_{x,y}(f) = \langle \varphi(f)(x),y\rangle $ is linear and $||\tau_{x,y}|| \leq ||x|| \: ||y||$. Hence By Reisz Kakutani, there exists a unique measure $\mu_{x,y} \in M(X) = \{\mu: \Sigma(X) \longrightarrow \mathbb{C} \cup \{\infty\} \: | \: \text{$\mu$ is a complex regular Borel measure with $|\mu|(X) = ||\mu|| < \infty$}\}$ such that: $$\tau_{x,y}(f) = \int_{X} f \: d\mu_{x,y} \: \: \forall f \in C(X).$$ Also we have that $||\tau_{x,y}|| = ||\mu_{x,y}||$...
My question is, why do we have that $||\tau_{x,y}|| = ||\mu_{x,y}||$? I know from above that $$|\tau_{x,y}(1)| = |\mu_{x,y}(X)| \leq |\mu_{x,y}|(X) = ||\mu_{x,y}||$$ but I dont see how the equality is achieved unless $|\mu_{x,y}(X)| = |\mu_{x,y}|(X)$, which only would be true if $\mu_{x,y}$ was nonnegative, which we don't know.
Does $\mu_{x,y}$ being inner regular imply $|\mu_{x,y}(X)| = |\mu_{x,y}|(X)$ or something?