In the lecture notes of Peterson https://math.vanderbilt.edu/peters10/teaching/spring2013/vonNeumannAlgebras.pdf he proves the following theorem THEOREM 2.7.5: Let $\mathcal H$ be a Hilbert space and $A\subseteq B(\mathcal H)$ is an abelian unital $C^*$-algebra. Then there is a unique spectral measure $E$ on $\sigma(A)$ (Gelfand spectrum) such that $x=\int\Gamma(x) dE.$ $\Gamma$ is the Gelfand transform.
Proof:Using the linear map $f\mapsto \langle \Gamma^{-1}(f)\zeta,\eta\rangle ,$ $\zeta,\eta\in\mathcal H$ we get a Radon measure $E_{\zeta,\eta}$ corresponding to the linear functional. Using properties of $\Gamma$ it is easy to see that $fdE_{\zeta,\eta}=dE_{\Gamma^{-1}(f)\zeta,\eta}=dE_{\zeta,\Gamma(\bar{f})\eta}.$ therefore, for each Borel se $B\subseteq \sigma(A)$ we can define a bounded sesquilinear form $(\zeta,\eta)\mapsto \int 1_{B}dE_{\zeta,\eta}$ ($1_B$ is characteristic function).This we get a linear map $E(B)$ such that $\langle E(B)\zeta,\eta\rangle=\int 1_BdE_{\zeta,\eta}.$ One can see that for all $f\in C(\sigma(A))$ we have $\langle \Gamma^{-1}(f)E(B)\zeta,\eta\rangle=\int 1_BdE_{\zeta,\Gamma(\bar{f})\eta}=\int f1_BdE_{\zeta,\eta}$ Now Peterson argues that he last identity implies $B\mapsto E(B)$ is a spectra measure. Can anyone tell mte how to prove this?
