If $A$ is self-adjoint operator defined in Hilbert space $\mathscr{H}$ and its resolvent family is $(\mathbb{C}, \mathscr{B}, E)$. How to get $$\ker (\lambda_{0}I-A)=E(D)\mathscr{H}, \forall \lambda_{0}\in \sigma_{d}(A), $$ where $\sigma_{d}(A)$ is the discrete spectral of $A$ and $\exists r>$ $[\lambda_{0}-r,\lambda_{0}+r] \cap \sigma(A)=\{\lambda_{0}\}$, define $D=\{ z\in \mathbb{C}: \vert z-\lambda_{0} \vert <r/2\}$.
We can define $$ E(D) =\int_{\sigma(A)} \chi_{D}(z)E(dz)$$
You can show $E(\{\lambda_0\})H=\mathcal{N}(A-\lambda_0I)$. Start with \begin{align} &\|(A-\lambda_0I)E(|\lambda-\lambda_0|<\epsilon)x\|^2 \\ &=\int_{|\lambda-\lambda_0|<\epsilon}|\lambda-\lambda_0|^2d\|E(\lambda)x\|^2 \\ &\le \epsilon^2\|E(|\lambda-\lambda_0|<\epsilon)x\|^2. \end{align} By regularity of the measure, $\|(A-\lambda_0I)E(\{\lambda_0\})x\|=0$ holds for all $x$. (Let $\epsilon\downarrow 0$ in the above.) So $(A-\lambda_0I)E(\{\lambda_0\})=0$. On the other hand, if $(A-\lambda_0I)x=0$ for some $x$, then \begin{align} 0&= \|E(|\lambda-\lambda_0| > \epsilon)(A-\lambda_0I)x\|^2 \\ &=\|(A-\lambda_0I)E(|\lambda-\lambda_0| > \epsilon)x\|^2 \\ &= \int_{|\lambda-\lambda_0| > \epsilon}|\lambda-\lambda_0|^2d\|E(\lambda)x\|^2 \\ & \ge \epsilon^2\|E(|\lambda-\lambda_0|>\epsilon)x\|^2 \ge 0, \end{align} which implies that $E(|\lambda-\lambda_0|>\epsilon)x=0$ for all $\epsilon > 0$. Hence, $E(|\lambda-\lambda_0| > 0)x=0$, which leaves $E(\{\lambda_0\})x=x$. So, $$ E(\{\lambda_0\})H=\mathcal{N}(A-\lambda_0I). $$