Let $H$ be a $\mathbb R$-Hilbert space and $(\mathcal D(A),A)$ be a self-adjoint linear operator on $H$. We know that there is an unique spectral family$^1$ on $H$ with $$\left(\mathcal D(\text{id}_\pi),\text{id}_\pi\right)=(\mathcal D(A),A).$$ For each $x\in H$, let $\mu_x$ denote the Lebesgue-Stieltjes measure associated with $(1)$ (see below).
How can we show that the resolvent set $\rho(A)$ of $A$ is a $\mu_x$-null set for all $x\in H$?
$^1$ $(\pi_\lambda)_{\lambda\in\mathbb R}$ is called spectral family on $H$ if
- $\pi_\lambda$ is an orthogonal projection on $H$ for all $\lambda\in\mathbb R$;
- $(\pi_\lambda)_{\lambda\in\mathbb R}$ is nondecreasing, i.e. $$\langle\pi_\lambda x,x\rangle_H\le\langle\pi_\mu x,x\rangle_H\;\;\;\text{for all }x\in H\text{ and }\lambda\le\mu$$ and $$\mathbb R\ni\lambda\mapsto\pi_\lambda\tag2$$ is right-continuous with respect to the strong operator topology;
- $\pi_\lambda\xrightarrow{\lambda\to-\infty}0$ and $\pi_\lambda\xrightarrow{\lambda\to\infty}\operatorname{id}_H$ with respect to the strong operator topology.
If $u:\mathbb R\to\mathbb R$ is Borel measurable, $$u_\pi x:=\int u(\lambda)\:{\rm d}\pi_\lambda x\;\;\;\text{for }x\in\mathcal D(u_\pi):=\left\{x\in H:u\in L^2(\pi_\lambda x)\right\}$$ is self-adjoint (note that for each $x\in H$ $$\mathbb R\ni\lambda\mapsto\pi_\lambda x\tag1$$ is right-continuous and of bounded variation and hence the Lebesgue-Stieltjes measure associated with $(1)$ is well-defined).