What will be the spectral measure corresponding to $M_{\varphi}$ if $\varphi$ is real-valued $\mu$ a.e.? Let $(X,\mathcal A, \mu)$ be a measure space. Let $\varphi \in L^{\infty} (\mu)$ be such that $\varphi$ is real-valued $\mu$ a.e. and $\mathcal H = L^2 (\mu).$ Then I know that the multiplication operator $M_{\varphi}$ on $\mathcal H$ is self-adjoint. Now by Spectral Theorem for Self Adjoint Operators  we know that given a self-adjoint operator $A$ we can get hold of a unique spectral measure $E_A$ defined on the Borel-$\sigma$-algebra of subsets of the spectrum $\sigma (A)$ of $A$ such that $$A = \int_{\sigma (A)} t\ dE_A.$$
Now my question is the following $:$

What will be the unique spectral measure associated to $M_{\varphi}$ if $\varphi \in L^{\infty} (\mu)$ be such that $\varphi$ is real-valued $\mu$ a.e.?

Could anybody give me some idea or suggestion in this regard? Thanks!
 A: I write $T = M_{\varphi}$ for simplicity.
For $z \in \rho(T)$ (the resolvent set of $T$), I write $R(z,T) = (zI - T)^{-1}$ the resolvent of $T$ at point $z$. For a self-adjoint operator on a complex Hilbert space, it is known that $\mathbb C \setminus \mathbb R \subset \rho(T)$.
Moreover, in the complex Hilbert space case, the spectral theorem gives the following inversion formula :
$$\langle y, (E(b)-E(a))x \rangle = \lim \limits_{\delta \to 0^+} \lim \limits_{\epsilon \to 0^+}  \dfrac{1}{2\pi i} \int_{a+\delta}^{b+\delta} \langle y, (R(t + i \epsilon,T) - R(t - i \epsilon,T))x \rangle dt \tag{1}$$
Since $\varphi$ is real-valued, solving $$(zI - T)f = (z - \varphi)f = g$$ yields $R(z,T)g =(z-\varphi)^{-1}g$ for $z \in \mathbb C \setminus \mathbb R$, i.e. $R(z,T)$ is the multiplication operator by $(z-\varphi)^{-1}$.
For your multiplication operator $T$, you can compute $(1)$ explicitely. In your case, it will give
\begin{align*}\lim \limits_{\delta \to 0^+} \lim \limits_{\epsilon \to 0^+} \pi^{-1} \int_{X} y(z) \overline{x(z)} \left( \arctan\left( \dfrac{b+\delta-\varphi(z)}{\epsilon} \right) - \arctan\left( \dfrac{a+\delta-\varphi(z)}{\epsilon} \right) \right) d\mu(z)
\end{align*}
where we used Fubini-Tonelli to interchange the order of integration. Here, I assumed that the measure space was sigma-finite.
By dominated convergence, you can take the limits inside the integral and you should get
$$ \langle y, (E(b)-E(a))x \rangle = \int_{X} y(z) \overline{x(z)} \chi_{(a,b]}(\varphi(z)) d\mu(z)$$
Taking the limit $a \to -\infty$, we deduce that $E(t)$ is the multiplication operator by $$\chi_{(-\infty,t]}(\varphi(z))$$
Here, $\chi_{A}$ denotes the characteristic function of the set $A$.
