Spectral families of commuting operators Consider two self-adjoint bounded operators $A$ and $B$ on a separable Hilbert space. According to the spectral theorem we can write
$$
A=\int_{-\infty}^{\infty} x d E^{A}_x, \quad B=\int_{-\infty}^{\infty} y d E^{A}_y
$$
where $E^{A}_x$ and $E^{B}_y$ are the spectral families of projectors of $A$ and $B$ respectively. Is there a simple way to prove that if $[A,B]=AB-BA=0$, then $[E^{A}_x,E^{B}_y]=0$ for all $x,y$?

What you need is a way to construct $E_A$ and $E_B$ directly from $A,B$. This is accomplished through Stone's Formula:
$$
       \frac{1}{2}\left(E(a,b)x+E[a,b]x\right) \\
= \lim_{\epsilon\downarrow 0}\frac{1}{2\pi i} \int_{a}^{b}(A-(r+i\epsilon)I)^{-1}x-(A-(r-i\epsilon)I)^{-1}x dr
$$
This is a contour integral around $[a,b]$ with the vertical pieces missing. Using strong limits you can isolate $E(a,b)x$ and $E[a,b]x$ through limits in $a$, $b$. You can actually do this in a constructive way using the $\tan^{-1}$ function to explicitly integrate and take the limit in the strong topology.
If $AB=BA$, then
$$
         (A-\lambda I)B = B(A-\lambda I) \\
          (A-\lambda I)(B-\mu I)=(B-\mu I)(A-\lambda I) \\
          (B-\mu I)(A-\lambda I)^{-1} = (A-\lambda I)^{-1}(B-\mu I) \\
           (A-\lambda I)^{-1}(B-\mu I)^{-1}=(B-\mu I)^{-1}(A-\lambda I)^{-1}.
$$
From this and Stone's formula, you have a constructive proof that the spectral measures of $A$ and $B$ commute. I think it helps to see how the spectral measure is constructively determined on intervals of $\mathbb{R}$.
