Spectral Measures: Spectral Spaces (I) Problem
Given a Hilbert space $\mathcal{H}$.
Let the Lebesgue measure be $\lambda$. 
Consider a Borel spectral measure:
$$E:\mathcal{B}(\mathbb{C})\to\mathcal{B}(\mathcal{H})$$
Denote its probability measures by:
$$\nu_\varphi(A):=\|E(A)\varphi\|^2$$
Introduce the spectral space:
$$\mathcal{H}_\parallel:=\{\varphi:\nu_\varphi\ll\lambda\}$$
$$\mathcal{H}_\perp:=\{\varphi:\nu_\varphi\perp\lambda\}$$
Then they decompose:
$$\mathcal{H}=\mathcal{H}_\parallel\oplus\mathcal{H}_\perp$$
How to prove this?
Attention
This thread has been split: Spectral Spaces (II)
 A: If you start with a unit vector $v \in \mathcal{H}$, then you can define a subspace $\mathcal{H}_{v}$ as the closure in $\mathcal{H}$ of all polynomials in $N$, $N^{\star}$ acting on $v$. $\mathcal{H}_{v}$ is invariant under $N$, $N^{\star}$. This defines a unitary map
$$
          \mathcal{F}_{v} : \mathcal{H}_{v}\rightarrow L^{2}_{\mu_{v}}
$$
where $\mu_{v}(S)=\|E(S)v\|^{2}$. And $\mathcal{F}_{v}N\mathcal{F}_{v}^{-1}=M_{\lambda}$ is a multiplication operator. Now decompose the measure $\mu_{v}$ into an atomic part, a part which is absoutely continuous with respect to Lebesgue measure and another which is concentrated on a Lebesgue measure $0$ set. This decomposition is achieved with multiplication operators on $L^{2}_{\mu_{v}}$ by characteristic functions, i.e., projections on $L^{2}_{\mu_{v}}$. The corresponding subspaces are invariant under $N$, $N^{\star}$ and, therefore, commute with the original spectral measure.
Next, find another unit vector $w$ in the orthogonal complement of $\mathcal{H}_{v}$. The resulting space $\mathcal{H}_{w}$ is orthogonal to $\mathcal{H}_{v}$, and you can do the same thing there. If $\mathcal{H}$ is separable, then there are--at nost--countably many of such spaces needed to span the original space, and you find projection operators onto the desired spaces, which you can sum to obtain the final projections required to decompose the space.
If the space is not separable, then there are complications in summing the projections, but nets probably work. However, if you want to decompose the spectral measure directly, and you still want to end up with measurable sets along which to decompose the spectral measure, then you may need to stick with separable spaces.
A: Preparation
For finite measures:
$$\nu_\varphi(A^\complement)=0\iff\nu_\varphi(A)=\nu_\varphi(\Omega)$$
Check equivalences:
$$\nu_\varphi(A)=0\iff E(A)\varphi=0$$
$$\nu_\varphi(A)=\nu_\varphi(\Omega)\iff E(A)\varphi=\varphi$$
Concluding preparation.
Decomposition
By preparation one has:
$$\mathcal{H}_\parallel=\bigcap_{\lambda(N)=0}\mathcal{N}E(N)\quad\mathcal{H}_\perp=\bigcup_{\lambda(N)=0}\mathcal{R}E(N)$$
Remind projections:
$$\mathcal{R}E(A)^\perp=\mathcal{N}E(A)\quad\mathcal{R}E(A)=\mathcal{N}E(A)^\perp$$
Orthogonal complement:*
$$(\mathcal{H}_\perp)^\perp=\left(\bigcup_{\lambda(N)=0}\mathcal{R}E(N)\right)^\perp=\bigcap_{\lambda(N)=0}\mathcal{R}E(N)^\perp=\bigcap_{\lambda(N)=0}\mathcal{N}E(N)=\mathcal{H}_\parallel$$
It is closed subspace:
$$\varphi^{(\prime)}\in\mathcal{H}_\perp\implies\varphi^{(\prime)}\in\mathcal{R}E(N^{(\prime)})\implies\varphi+\varphi'\in\mathcal{R}E(N\cup N')\implies\varphi+\varphi'\in\mathcal{H}_\perp$$
$$\varphi_n\in\mathcal{H}_\perp\implies\varphi_n\in\mathcal{R}E(N_n)\implies\lim_n\varphi_n\in\mathcal{R}E(\cup_n N_n)\implies\lim_n\varphi_n\in\mathcal{H}_\perp$$
Concluding decomposition.
*Reference: Hilbert vs. De Morgan
