Multiplication operator on $L^2$ and spectral theorem. Let's consider the multiplication operator by the independent variable in $L^2(\mu)$, where $\mu$ is a borel regular measure on $\mathbb{C}$: $Mf(z)=zf(z)$.
I want to show that if $\phi$ is a borel bounded function, then, with the borel functional calculus for normal operators $\phi(M)=M_\phi=$ the multiplication operator by $\phi(z)$.
Intuitively it's true: the spectral measures of the borel sets in $L^2(\mu)$ are the characteristic functions of the sets. So $\phi(M)f=\int\phi(z) \delta_z dz f= \phi f$. But I don't feel that this approach is very rigorous.
Conway suggests to use the uniqueness of the functional calculus to solve this exercise.
 A: Define $E(S)f = \chi_{S}f$ for any $f \in L^{2}_{\mu}$ and for any Borel subset $S$ of $\mathbb{C}$. For any Borel subsets $S$, $T$,
$$
                     E(S)=E(S)^{2}=E(S)^{\star},\\
                     E(S)E(T)=E(T)E(S)=E(T\cap S),\\
                     E(\emptyset)=0,\;\;\; E(\mathbb{C})=I.
$$
So, $E$ is a spectral measure. And $E(S)M=ME(S)$ so that the ranges of every $E(S)$ are invariant subspaces of $M$. Notice that
$$
               (E(S)f,g) = \int_{S}f\overline{g}\,d\mu,\;\;\; f,g\in L^{2}_{\mu}.
$$
If $f \in \mathcal{D}(M)$ and $g \in L^{2}_{\mu}$, then
$$
                   \int \lambda d(E(\lambda)f,g)=\int_{S}\lambda f\overline{g}\,d\mu =(Mf,g).
$$
So $\int\lambda dE(\lambda)f = Mf$. By uniqueness of the Spectral Theorem, $E$ must be the spectral resolution of the identity for $M$. If $F$ is bounded Borel function, then
$$
       (F(M)f,g) = \int F(\lambda)d(E(\lambda)f,g) = \int Ff\overline{g}\,d\lambda = (M_{F}f,g),\;\;\; f,g\in L^{2}_{\mu}.
$$
Therefore, $F(M)=M_{F}$ is the operator of multiplication by $F$. This is true for every bounded Borel function.
