I have a trouble with understanding a proof about one-dimensional Schrodinger operator.
$\mathfrak{h}_{0}$ is the Hilbert space of $\mathbb{C}^2$-valued functions $\Phi(\lambda) = (\varphi_{1}(\lambda), \varphi_{2} (\lambda))^{t}$ on $[0, \infty) \subseteq \mathbb{R}$ with the norm
$$ \| \Phi \|_{0}^2 = \int_{0}^{\infty} (|\varphi_{1}(\lambda)|^2 + | \varphi_{2}(\lambda)|^2 ) d \sigma (\lambda)$$ where $d \sigma (\lambda) = d\lambda / (2 \sqrt{\lambda})$ and $\sqrt{\lambda} \ge 0$ for $\lambda \ge 0$.
For $\mu \in \mathbb{C} \setminus [0, \infty)$, $R_{\mu}^{0}$ is the resolvent of the multiplication by $\lambda$ operator in $\mathfrak{h}_{0}$. (If we denote the multiplication by $\lambda$ operator by $Q$, then $Q$ is defined by $Q \psi (\lambda) = \lambda \psi (\lambda)$, and $R_{\mu}^{0} = (Q - \mu I)^{-1}$. Note that $Q$ is self-adjoint.)
According to Lemma 2.2 of this book, $Q$ has an absolutely continuous spectrum $\mathbb{R}$.
What I'm not following is:
If $A \subseteq \mathfrak{h}_{0}$ is a closed subspace and an invariant subspace for $R_{\mu}^{0}$ for any $\mu \in \mathbb{C} \setminus [0, \infty)$, then there exists Borel subsets $E_{1}, E_{2} \subseteq [0, \infty)$ such that $$ A = \{ \tilde{\Phi} = (\chi_{E_{1}} \varphi_{1}, \chi_{E_{2}} \varphi_{2})^{t} : \Phi = (\varphi_{1}, \varphi_{2})^{t} \in \mathfrak{h}_{0} \}$$ where $\chi_{E_{i}}$ are characteristic functions of $E_{i}$, $i=1,2$.
It seems to be related with the projection-valued operator, since for any Borel subset $E \subseteq [0, \infty)$, we have $(P(E) \varphi)(q) = \chi_{E}(q) \varphi(q)$ for the operator $Q$.
However, I couldn't find any results that imply such claim.
Any help will be appreciated. Thank you.