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Problem

Given a Hilbert space $\mathcal{H}$.

Consider a normal operator: $$N:\mathcal{D}(N)\to\mathcal{H}:\quad N^*N=NN^*$$

By the previous threads: $$Z=N\sqrt{(1+N^*N)^{-1}}\quad N=Z\left(\sqrt{1-Z^*Z}\right)^{-1}$$

Especially one had: $$Z=\int\lambda\mathrm{d}F:\quad F(\overline{\mathbb{D}})=1\quad F(\mathbb{S})=0$$

Define the function: $$\eta\in\mathcal{B}(\mathbb{D}):\quad\eta(\lambda):=\frac{\lambda}{\sqrt{1-|\lambda|^2}}$$

Construct as measure: $$E(A):=F_\eta(A):=F(\eta^{-1}A)$$

Then one obtains: $$N=\int\lambda\mathrm{d}F_\eta(\lambda)=:\int\lambda\mathrm{d}E(\lambda)$$

How can I prove this?

Reference

This builds up on: Tranform, Retransform

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Meanwhile I got it...

Define the functions: $$f(\lambda):=\lambda\quad \iota(\lambda):=\frac{1}{\lambda}\quad\chi(\lambda):=\sqrt{\lambda}\quad g(\lambda):=1-|\lambda|^2$$

As it was bounded: $$Z\in\mathcal{B}(\mathcal{H}):\quad 1-Z^*Z=g(Z)$$

For compositions:* $$\iota(\chi(g(Z))=(\iota\circ\chi\circ g)(Z)=:g'(Z)$$

By measurable calculus: $$N=Z\sqrt{1-Z^*Z}^{-1}=f(Z)g'(Z)\subseteq(fg')(Z)=\eta(Z)$$

For normal extensions: $$N\subseteq\eta(Z)\implies N=\eta(Z)$$

So one arrives at:* $$N=\eta(Z)=\int\eta(\lambda)\mathrm{d}F(\lambda)=\int\lambda\mathrm{d}F_\eta(\lambda)=\int\lambda\mathrm{d}E(\lambda)$$

Concluding existence.

*See the thread: Pushforward

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