I am reading J. Weidmann's book, Spectral Theory of Ordinary Differential Operators, particularly on deriving the spectral matrix $\rho$ for the spectral representation of a self-adjoint Sturm-Liouville operator. Basically, to derive $\rho$, we use the Weyl-Titchmarsh-Kodaira formula (Theorem 9.4)
$$\rho(\lambda) = \frac{1}{2\pi i}\lim_{\delta \to 0^+} \lim_{\epsilon \to 0^+} \int_\delta^{\lambda + \delta} (m_{jk}(t+i\epsilon) - m_{jk}(t-i\epsilon))\,dt$$ where $m_{jk}$ is obtained by computing the resolvent kernel via a fundamental system.
How does one show that $\rho$ has purely absolutely continuous spectrum? I am sure we can start by showing that the self-adjoint operator has no eigenvalues, but how about the singular continuous part? Does the above form of $\rho$ automatically implies absolutely continuous with respect to the Lebesgue measure?
What you're essentially doing is reconstructing a positive measure $\rho$ on $\mathbb{R}$ from the following holomorphic function: $$ m(\lambda)=\int_{-\infty}^{\infty}\frac{1}{t-\lambda}d\rho(t),\;\;\;\lambda\in\mathbb{C}\setminus\mathbb{R}. $$ This can be done by integrating around the singularities of $m$ on an interval in a positively-oriented direction: $$ \frac{1}{2}(\rho(a,b)+\rho[a,b]) = \lim_{\epsilon\downarrow 0}\frac{1}{2\pi i}\int_{a}^{b}m(u-i\epsilon)-m(u+i\epsilon)du. $$