Spectral Measures: Multi Version (III) This question is only Q&A!
Problem
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:
$$E:\mathcal{B}(\mathbb{C})\to\mathcal{B}(\mathcal{H}):\quad N=\int\lambda\mathrm{d}E(\lambda)$$
Denote for shorthand:
$$\mathcal{S}_\varphi:=\overline{\langle\{E(A)\varphi_0:A\in\mathcal{B}(\mathbb{C})\}\rangle}$$
By the previous thread:
$$\mathcal{A}\subseteq\mathcal{H}:\quad\mathcal{H}=\sum_{\alpha\in\mathcal{A}}\mathcal{S}_\alpha=:\sum_\alpha\mathcal{S}_\alpha$$
Denote for shorthand:
$$\nu_\alpha:=\nu_{\alpha}(A)=\|E(A)\alpha\|^2$$
Introduce the measure space:*
$$\Omega:=\coprod_\alpha\mathbb{C}_\alpha\cong\mathbb{C}\times\mathcal{A}:\quad\nu(A):=\sum_\alpha\nu_\alpha(\iota_\alpha^{-1}A)$$

Then one has:
  $$M_\eta:\mathcal{D}(M_\eta)\to\mathcal{L}^2(\nu):\quad N=U^{-1}M_\eta U$$

How can I prove this?
Reference
This is the final thread!
*See the thread: Borel Measures: Coproduct
 A: Embedding
Consider the embeddings:
$$J_\alpha:\mathcal{S}_\alpha\to\mathcal{H}:\quad J_\beta^*J_\alpha=\delta_{\beta\alpha}1_\alpha\quad J_\alpha J_\alpha^*=P_\alpha$$
Construct the unitary map:
$$U\varphi:=(J_\alpha^*\varphi)_\alpha\quad V(\varphi_\alpha)_\alpha:=\sum_\alpha J_\alpha\varphi_\alpha$$
Indeed they are inverses:*
$$VU\varphi=\sum_\alpha J_\alpha J_\alpha^*\varphi=\sum_\alpha P_\alpha\varphi=\varphi$$
$$(UV(\varphi_\alpha)_\alpha)_\beta=(J_\beta^*\sum_\alpha J_\alpha\varphi_\alpha)_\beta=(\varphi_\beta)_\beta$$
(Continuity can been used!)
Reducibility
Denote for readability:
$$N_\alpha:=J_\alpha^*NJ_\alpha\quad E_\alpha(A):=J_\alpha^*E(A)J_\alpha$$
By reducibility:
$$E(A)P_\alpha=P_\alpha E(A)\implies N_\alpha=\int\lambda_\alpha\mathrm{d}E_\alpha(\lambda_\alpha)$$
By the previous thread:
$$M_\mathrm{id}^\alpha:\mathcal{D}(M_\mathrm{id}^\alpha)\to\mathcal{L}^2(\nu_\alpha):\quad N_\alpha=U_\alpha^{-1}M_\mathrm{id}^\alpha U_\alpha$$
By reducibility:
$$E(A)P_\alpha=P_\alpha\implies P_\alpha N\subseteq NP_\alpha$$
That gives also:*
$$\mathcal{D}(N)=\sum_\alpha\mathcal{D}(N)\cap\mathcal{S}_\alpha:\quad N\mathcal{S}_\alpha\subseteq\mathcal{S}_\alpha$$
So one obtains:*
$$(UNV(\varphi_\alpha)_\alpha)_\beta=(J_\beta^*N\sum_\alpha J_\alpha\varphi_\alpha)_\beta=(J_\beta^*NJ_\beta\varphi_\beta)_\beta=(N_\beta\varphi_\beta)_\beta$$
In terms of operators:
$$UNV=\bigoplus_\alpha N_\alpha=\bigoplus_\alpha U_\alpha^{-1}M_\mathrm{id}^\alpha U_\alpha=\left(\bigoplus_\alpha U_\alpha^{-1}\right)\left(\bigoplus_\alpha M_\mathrm{id}^\alpha\right)\left(\bigoplus_\alpha U_\alpha\right)$$
(This is a checkpoint!)
Identification
Consult the unitary map:
$$\Phi:\mathcal{L}^2(\Omega,\nu)\to\bigoplus_\alpha\mathcal{L}^2(\mathbb{C}_\alpha,\nu_\alpha):\quad h\mapsto(h\circ\iota_\alpha)_\alpha$$
Moreover it is algebraic:
$$\Phi(hh')=\Phi(h)\Phi(h')\quad\Phi(\overline{h})=(\overline{\Phi(h)_\alpha})_\alpha$$
Define the function:
$$\eta:=\Phi^{-1}(\mathrm{id_\alpha})_\alpha:\quad M_\eta h=\Phi^{-1}\left(\bigoplus_\alpha M_\mathrm{id}^\alpha\right)\Phi h$$
All together gives:
$$N=V\left(\bigoplus_\alpha U_\alpha^{-1}\right)\Phi M_\eta\Phi^{-1}\left(\bigoplus_\alpha U_\alpha\right)U$$
Concluding the assertion.
*See the thread: Reducing Spaces: Decomposition
