TITLE: Proving Stone's Formula for Constructively obtaining the Spectral Measure for $A=A^\star$
QUESTION [5 upvotes]: Let $A$ be a bounded or unbounded selfadjoint linear operator on a complex Hilbert space $H$ with spectral representation $A=\int_{\sigma}\lambda \, dE(\lambda)$ given by the Spectral Theorem for Selfadjoint Operators. Stone's Formula gives a way to constructively obtain the spectral measure $E[a,b]$ of a finite interval as a strong (vector) limit:
$$
\frac{1}{2}\{E[a,b]+E(a,b)\}x=\lim_{\varepsilon\downarrow 0}\frac{1}{2\pi i}\int_a^b \{R(u+i\varepsilon)-R(u-i\varepsilon)\}x\,du,\;\;\; x \in H,
$$
where $R$ is the resolvent operator $R(\lambda)=(A-\lambda I)^{-1}$.
Question: What is the easiest way to prove that Stone's Formula is valid?
Application: This may be viewed as a generalized residue of the resolvent function. If $A$ has isolated spectrum, then such an integral may be evaluated using residues of the resolvent.

REPLY [3 votes]: I think that an easy method is delivered by functional calculus. 
Consider the right-hand side of your equation as
$$ \lim_{ \varepsilon \downarrow 0} \frac{1}{2\pi i}f_{\varepsilon, a, b}(A)x, $$
where $$f_{\varepsilon, a, b}(t) = \int_{a}^b \left(\frac{1}{t-(u+i\varepsilon)}  - \frac{1}{t-(u-i\varepsilon)}\right) \mbox{d}u.$$
We can show that (by doing calculations)
$$\lim_{ \varepsilon \downarrow 0}  \frac{1}{2\pi i}\int_{a}^b \left(\frac{1}{t-(u+i\varepsilon)}  - \frac{1}{t-(u-i\varepsilon)}\right) \mbox{d}u = \begin{cases} 1 & \mbox{if } t \in (a,b) \\ \frac{1}{2} &\mbox{if } t = a \\
\frac{1}{2} &\mbox{if } t = b \\
0 &\mbox{if }t\notin [a, b]
  \end{cases}$$
If now you apply functional calculus to the right-hand side of above experssion you will obtain excatly
$$ \frac{1}{2}\left(E[a,b] + E(a,b)\right)x.$$