Suppose $\mathrm{E}$ is a spectral measure for some normal operator $N$ and $A\in\mathcal{B}(\sigma(N))$. I am unable to find an expression of the spectral measure of $A^c:=\sigma(N)\setminus A$ in terms of $\mathrm{E}(A)$.
My conjecture is that the correct expression would be $$\mathrm{E}(A^c) ~=~ I-\mathrm{E}(A),$$ since this would imply
- $\mathrm{E}(\sigma(N))=\mathrm{E}(A\sqcup A^c)=\mathrm{E}(A)+\mathrm{E}(A^c)=\mathrm{E}(A)+I-\mathrm{E}(A)=I$
as well as
- $\mathrm{E}(\emptyset)=\mathrm{E}(A\cap A^c)=\mathrm{E}(A)\mathrm{E}(A^c)=\mathrm{E}(A)(I-\mathrm{E}(A))=\mathrm{E}(A)-\mathrm{E}(A)^2=\mathsf{0}$, since $\mathrm{E}(A)$ is a projection,
both of which we know to hold for $\mathrm{E}$.
Is my thinking correct?