Given a Hilbert space $\mathcal{H}$.
Consider a normal operator: $$N:\mathcal{D}(N)\to\mathcal{H}:\quad N^*N=NN^*$$
Construct the operator: $$Q:=(1+N^*N)^{-1}:\quad Z:=N\sqrt{Q}$$
By the previous proof: $$Z\in\mathcal{B}(\mathcal{H}):\quad Z^*Z=ZZ^*$$
Construct the operator: $$R:=1-Z^*Z:\quad W:=Z\sqrt{R}^{-1}$$
Then one obtains: $$N=W=Z\sqrt{1-Z^*Z}^{-1}$$
Especially one has: $$Z=\int\lambda\mathrm{d}F(\lambda):\quad F(\mathbb{D})=1\quad F(\mathbb{S})=0$$
How can I prove this from scratch?
Identity
Regard dense elements: $$\overline{\mathcal{D}N^*}=\mathcal{H}:\quad\varphi\in\mathcal{D}N^*$$
A calculation gives: $$ZZ^*\varphi=NQN^*\varphi=NN^*Q\varphi$$
The operator is closed: $$Q\in\mathcal{B}(\mathcal{H}):\quad NN^*=\overline{NN^*}\implies NN^*Q=\overline{NN^*Q}$$
By closed graph theorem: $$\mathcal{D}(NN^*Q)=\mathcal{H}\implies NN^*Q\in\mathcal{B}(\mathcal{H})$$
By uniform extension: $$ZZ^*\restriction_{\mathcal{D}N^*}=NN^*Q\restriction_{\mathcal{D}N^*}\implies ZZ^*=NN^*Q$$
So on operator level: $$R=1-NN^*Q=1-(1+NN^*)Q+Q=Q$$
And on operator level: $$Z\sqrt{R}^{-1}=N\sqrt{Q}\sqrt{Q}^{-1}=N$$
Concluding identity.
Support
For the spectrum: $$\|Z\|\leq1\implies r(Z)\leq1\implies\sigma(Z)\subseteq\overline{\mathbb{D}}\implies F(\overline{\mathbb{D}})=1$$
Define the function: $$\eta\in\mathcal{B}(\overline{\mathbb{D}}):\quad\eta(\lambda):=1-|\lambda|^2$$
As it is bounded: $$Z\in\mathcal{B}(\mathcal{H}):\quad\eta(Z)=1-Z^*Z=Q$$
So for the boundary: $$\mathcal{N}\eta(Z)=\mathcal{N}Q=(0)\implies F(\mathbb{S})=F\{\eta=0\}=(0)$$
Concluding support.