Spectral Measures: Reducibility Given a Hilbert space $\mathcal{H}$.
Consider a normal operator:
$$N:\mathcal{D}(N)\to\mathcal{H}:\quad N^*N=NN^*$$
And its spectral measure:
$$E:\mathcal{B}(\mathbb{C})\to\mathcal{B}(\mathcal{H}):\quad N=\int\lambda\mathrm{d}E(\lambda)$$
Regard a projection:
$$P\in\mathcal{B}(\mathcal{H}):\quad P^2=P=P^*$$

Then for adjoint:
  $$PN\subseteq PN\iff PN^*\subseteq N^*P$$
And one has also:
  $$PN\subseteq NP\implies PE(A)=E(A)P\implies P\eta(N)\subseteq\eta(N)P$$

How can I prove this?
 A: Meanwhile I got it...
Adjoint
Denote for readability:
$$\mathcal{D}:=\mathcal{D}(N)=\mathcal{D}(N^*)=:\mathcal{D}^*$$
By reducibility one has:*
$$N(\mathcal{S}\cap\mathcal{D})\subseteq\mathcal{S}\quad N(\mathcal{S}^\perp\cap\mathcal{D})\subseteq\mathcal{S}^\perp\quad(P\mathcal{D}\subseteq\mathcal{D})$$
So for the domain:
$$P\mathcal{D}^*=P\mathcal{D}\subseteq\mathcal{D}=\mathcal{D}^*$$
And for invariance:
$$\varphi\in\mathcal{S}\cap\mathcal{D}^*:\quad\langle N^*\varphi,\chi\rangle=\langle\varphi,N\chi\rangle=0\quad(\chi\in\mathcal{S}^\perp\cap\mathcal{D})$$
$$\psi\in\mathcal{S}^\perp\cap\mathcal{D}^*:\quad\langle N^*\psi,\chi\rangle=\langle\psi,N\chi\rangle=0\quad(\chi\in\mathcal{S}\cap\mathcal{D})$$
But they were dense:**
$$\overline{\mathcal{S}\cap\mathcal{D}}=\mathcal{S}\quad\overline{\mathcal{S}^\perp\cap\mathcal{D}}=\mathcal{S}^\perp$$
Concluding adjoint.*
Equality
Consider the projections:
$$N_\Re:=\frac{1}{2}\{N+N^*\}\quad N_\Im:=\frac{1}{2i}\{N-N^*\}$$
Their resolvents reduce:
$$R_\alpha(z)P=R_\alpha(z)P(z-N_\alpha)R_\alpha(z)
\subseteq R_\alpha(z)(z-N_\alpha)PR_\alpha(z)=PR_\alpha(z)$$
By Stone's formula:
$$PE_\alpha(-\infty,\lambda_\alpha]\varphi=P\lim_{\delta\to0^+}\lim_{\varepsilon\to0^+}\frac{1}{2\pi i}\int_{-\infty}^{\lambda_\alpha+\delta}\Delta R_\alpha(s\pm i\varepsilon)\varphi\mathrm{d}s\\
=\lim_{\delta\to0^+}\lim_{\varepsilon\to0^+}\frac{1}{2\pi i}\int_{-\infty}^{\lambda_\alpha+\delta}\Delta R_\alpha(s\pm i\varepsilon)P\varphi\mathrm{d}s=E_\alpha(-\infty,\lambda_\alpha]P\varphi$$
By construction one has:
$$E_\Re(A)=E(A\times\mathbb{R})\quad E_\Im(B)=E(\mathbb{R}\times B)$$
Therefore one obtains:
$$PE(-\infty,\lambda]=PE\bigg((-\infty,\lambda_\Re]\times\mathbb{R}\bigg)E\bigg(\mathbb{R}\times(-\infty,\lambda_\Im]\bigg)\\
=E\bigg((-\infty,\lambda_\Re]\times\mathbb{R}\bigg)E\bigg(\mathbb{R}\times(-\infty,\lambda_\Im]\bigg)P=E(-\infty,\lambda]P$$
By Dynkin one derives:
$$S\in\mathcal{B}(\mathbb{C}):\quad PE(S)=E(S)P$$
Concluding equality.
Inclusion
For the domain issue:
$$\int_\mathbb{R}|\eta(\lambda)|^2\mathrm{d}\|E(\lambda)P\varphi\|^2=\int_\mathbb{R}|\eta(\lambda)|^2\mathrm{d}\|PE(\lambda)\varphi\|^2
\leq\int_\mathbb{R}|\eta(\lambda)|^2\mathrm{d}\|E(\lambda)\varphi\|^2<\infty$$
And they act the same:
$$\langle P\eta(H)\varphi,\chi\rangle=\langle\eta(H)\varphi,P\chi\rangle=\int_\mathbb{R}\eta(\lambda)\mathrm{d}\langle E(\lambda)\varphi,P\chi\rangle\\
=\int_\mathbb{R}\eta(\lambda)\mathrm{d}\langle PE(\lambda)\varphi,\chi\rangle=\int_\mathbb{R}\eta(\lambda)\mathrm{d}\langle E(\lambda)P\varphi,\chi\rangle=\langle\eta(H)P\varphi,\chi\rangle$$
Concluding inclusion.
Strictness
Consider the case:
$$N:\mathcal{D}(N)\to\mathcal{H}:\quad\mathcal{D}(N)\subsetneq\mathcal{H}$$
Then one obtains:
$$P=0:\quad\mathcal{D}(PN)=\mathcal{D}(N)\subsetneq\mathcal{H}=\mathcal{D}(0)=\mathcal{D}(NP)$$
Concluding strictness.
*See the thread: Characterization
**See the thread: Denseness
