Let $\Gamma$ be differentiable Jordan curve in $\rho(A)$. Show that $$\chi_\Omega(A)=\int_\Gamma R_A(z)dz$$
Where $\Omega$ is the intersection of the interior of $\Gamma$ with $\mathbb{R}$.
My attempt: My idea was to emulate the proof of Stone's formula. I considered the function $r_x(z)=\frac{1}{x-z}$. Then i write
$$ \int_\Gamma \frac{1}{x-z}dz=\begin{cases} 2\pi i \quad &\text{if} \quad x\in\text{region covered by }\Gamma \\ 0 &\text{if} \quad x\notin\text{region covered by }\Gamma
\end{cases}$$
Now if i empoze that $x\in\mathbb{R}$ then i would get my answer $2\pi i \chi_\Omega(x)$. By assuming $A$ is self-adjoint (Which justifies my choice that $x\in\mathbb{R}$(??)). I could use spectral theorem to get $2\pi i \chi_\Omega(A)$.
Note: This is from "Teschl G. – Mathematical methods in quantum mechanics" Problem 4.1. By writing (??) what i mean is i'm not sure.
Best Answer
If $A$ is self-adjoint with spectral measure $E$ we have by Fubini's theorem $$ \int_{\Gamma} (zI-A)^{-1} dz = \int_{\Gamma} \int_{\mathbb{R}} (z-\lambda)^{-1} dE(\lambda) dz = \int_{\mathbb{R}} \int_{\Gamma} (z-\lambda)^{-1} dz dE(\lambda). $$ Applying what you already wrote $$ \int_{\mathbb{R}} \int_{\Gamma} (z-\lambda)^{-1} dz dE(\lambda) = 2\pi i \int_{\mathbb{R}}\chi_{\text{int}({\Gamma})}(\lambda) dE(\lambda) = 2\pi i \chi_{\Omega}. $$