Spectral Measures: Multi Version (I) This question is only Q&A!
Problem
Given a Hilbert space $\mathcal{H}$.
Consider a spectral measure:
$$E:\mathcal{B}(\mathbb{C})\to\mathcal{B}(\mathcal{H}):\quad N=\int\lambda\mathrm{d}E(\lambda)$$
Regard the spaces:
$$\mathcal{S}_\varphi:=\overline{\langle\{E(A)\varphi:A\in\mathcal{B}(\mathbb{C})\}\rangle}$$
All of them reduce:
$$\varphi\in\mathcal{H}:\quad E(A)\mathcal{S}_\varphi\subseteq\mathcal{S}_\varphi$$

Then they decompose:
  $$\mathcal{A}_\infty\subseteq\mathcal{H}:\quad\mathcal{H}=\sum_{\alpha\in\mathcal{A}_\infty}\mathcal{S}_\alpha$$

How can I prove this?
Reference
This is a start-up for: Multi Version (II)
 A: Consider the POSET:
$$\lambda\in\Lambda:\quad\mathcal{S}\in\lambda\implies\mathcal{S}=\mathcal{S}_\varphi$$
$$\lambda\in\Lambda:\quad\mathcal{S},\mathcal{S}'\in\lambda\implies\mathcal{S}\perp\mathcal{S}'$$
It is nonempty:
$$\mathcal{S}_0=(0):\quad\{\mathcal{S}_0\}\in\Lambda$$
And admits upper bounds:
$$\Delta\subseteq\Lambda:\quad\lambda_\Delta:=\bigcup_{\delta\in\Delta}\delta\in\Lambda$$
Indeed for chains:
$$\mathcal{S}\in\lambda_\Delta\implies\mathcal{S}\in\delta\implies\mathcal{S}=\mathcal{S}_\varphi$$
$$\mathcal{S},\mathcal{S}'\in\lambda_\Delta\implies\mathcal{S},\mathcal{S}'\in\delta\wedge\delta'\implies\mathcal{S}\perp\mathcal{S}'$$
By Zorn's lemma:
$$\lambda_\infty\in\Lambda:\quad\lambda\geq\lambda_\infty\implies\lambda=\lambda_\infty$$
Consider the closed space:
$$\mathcal{S}_\infty:=\overline{\langle\bigcup_{\mathcal{S}\in\lambda_\infty}\mathcal{S}\rangle}=\sum_{\alpha\in\mathcal{A}_\infty}\mathcal{S}_\alpha$$
Regard the complement:
$$\varphi\in\mathcal{S}_\infty^\perp\implies\varphi\in\mathcal{S}_\alpha^\perp\quad(\alpha\in\mathcal{A}_\infty)$$
On dense elements:
$$\langle E(A)\varphi,E(A')\alpha\rangle=\langle\varphi,E(A\cap A')\alpha\rangle=0$$
So one obtains:
$$\mathcal{S}_\varphi\perp\mathcal{S}_\alpha\quad(\alpha\in\mathcal{A}_\infty)\implies\lambda_\infty\cup\{\mathcal{S}_\varphi\}\in\Lambda$$
But maximality implies:
$$\lambda\cup\{\mathcal{S}_\varphi\}\geq\lambda_\infty\implies\mathcal{S}_\varphi\in\lambda_\infty$$
So one derives at:
$$\varphi\in\mathcal{S}_\varphi\subseteq\mathcal{S}_\infty\implies\varphi=0$$
Thus as desired:
$$\mathcal{S}_\infty^\perp=(0)\implies\mathcal{S}_\infty=\mathcal{H}$$
Concluding the assertion.
