Let $A:\mathscr{D}(A)\subset\mathscr{H}\to\mathscr{H}$ be an unbounded self-adjoint operator and $(E_{\lambda})_{\lambda\in\mathbb{R}}$ its spectral family and $E_{\lambda}=E((-\infty,\lambda])$, with $E$ the resolution of the identity on $\mathbb{R}$.
I'm trying to show that one can write $E_{\lambda}=I-E^{+}_{\lambda}$, whereby $E^{+}_{\lambda}$ is the orthogonal projection on $\mathscr{N}((A-\lambda)-|A-\lambda|)$.
Thus far I have followed a hint which suggests to express $(A-\lambda)-|A-\lambda|$ as an integral using the spectral theorem. Namely for fixed $\eta\in\mathbb{R}$ and we can define the function $$f(\eta)=(\eta-\lambda)-|\eta-\lambda|,$$ so that by the spectral theorem: $$ \begin{aligned}\langle f(A)x,x\rangle&=\langle ((A-\lambda)-|A-\lambda|)x,x\rangle\\&=\int_{-\infty}^{\infty} f(\eta)\,d\langle E_{\eta}x,x\rangle\\&=\int_{-\infty}^{\infty}f(\eta)\,dF_{x}(\eta)=\langle\Psi(f)x,x\rangle,\end{aligned} $$ where $F_{x}(\eta)$ is the generalised function defined by $$ F_{x}(\eta):=\int_{-\infty}^{\eta}\,d\langle E_{\eta}x,x\rangle. $$
However, I really don't know where to go from this hint.
The function $f_{\lambda}(\eta)=(\eta-\lambda)-|\eta-\lambda|$ is $0$ for $\eta \le \lambda$ and is $2(\eta-\lambda)$ for $\eta < \lambda$. Therefore, $$ (A-\lambda I)x-|A-\lambda I|x=\int_{-\infty}^{\lambda}2(\eta-\lambda)dE(\eta)x, \\ \|(A-\lambda I)x-|A-\lambda I|x\|^2=4\int_{-\infty}^{\lambda}|\eta-\lambda|^2d\|E(\eta)x\|^2 $$ It follows that $x\in\mathcal{N}((A-\lambda I)-|A-\lambda I|)$ iff $$ \|E(-\infty,\lambda)x\|=0. $$ (The open interval on the right is important.) And $E(-\infty,\lambda)=I-E[\lambda,\infty)$.