Let $(\Omega, \mathcal{F})$ be a measurable space. Let $E: \mathcal{F}\to B(H)$ be a regular resolution of the identity on the Hilbert space $H$, see e.g. Rudin's functional analysis book.


Suppose that $g: \Omega \to \mathbb{C}$ is Borel-measurable and that $\xi \in \mathscr{D}\left(\int\_\Omega g dE\right)=: \mathscr{D}\_g$. We can then define
$$\eta:= \left(\int\_\Omega g dE\right)\xi\in H.$$
Is it true that
$$dE\_{\eta, \eta}= |g|^2 dE\_{\xi, \xi}?$$
I.e., if $A\in \mathcal{F}$, do we have
$$E\_{\eta, \eta}(A)= \int\_{A}|g|^2 dE\_{\xi, \xi}.$$


I believe I can show this if $g$ is bounded. But I am mostly interested in the case where $g$ is unbounded.


Thanks in advance for your help!



Identity for spectral resolution: $dE_{\xi, \xi}= |g|^2 dE_{\eta, \eta}$
If $g=\sum\_k\beta\_k\,1\_{X\_k}$ is simple,
$$
\Big\langle\Big(\int\_\Omega g\,dE\Big)\xi,\xi\Big\rangle=\int\_\Omega g\,dE\_{\xi,\xi}=\sum\_k\beta\_k\,E\_{\xi,\xi}(X\_k)=\Big\langle\sum\_k\beta\_k\,E(X\_k)\xi,\xi\Big\rangle. 
$$


Since $g$ is measurable, then there exist simple functions $g\_n$ such that $g\_n\to g$ and $|g\_n|^2\nearrow |g|^2$ (this is easily achievable by finding first simple functions that increase to $(\operatorname{Re}g)^+$, $(\operatorname{Re}g)^-$, $(\operatorname{Im}g)^+$, $(\operatorname{Im}g)^-$, respectively).


For any $\nu$ we have, using monotone convergence in each of the sixteen integrals coming from decomposing $g$ and $dE\_{\xi,\nu}$ as a linear combinations of positive functions/measures,


$$
\langle \eta,\nu\rangle=\int\_\Omega g\,dE\_{\xi,\nu}=\lim\_n\int\_\Omega g\_n\,dE\_{\xi,\nu}=\lim\_n\Big\langle\Big(\int\_\Omega g\_n\,dE\Big)\xi,\nu\Big\rangle
$$


So, writing $g\_n=\sum\_k\beta\_{n,k}\,1\_{X\_{n,k}}$,
\begin{align}
E\_{\eta,\eta}(A)
&=\langle E(A)\eta,\eta\rangle
=\Big\langle E(A)\Big(\int\_\Omega g\,dE\Big)\xi,\eta\rangle\\[0.3cm]
&=\lim\_n\Big\langle E(A)\Big(\int\_\Omega g\,dE\Big)\xi,\Big(\int\_\Omega g\_n\,dE\Big)\xi\Big\rangle\\[0.3cm]
&=\lim\_n\sum\_k\overline{\beta\_{n,k}}\Big\langle E(A)\Big(\int\_\Omega g\,dE\Big)\xi,E(X\_{n,k})\xi\Big\rangle\\[0.3cm]
&=\lim\_n\sum\_k\overline{\beta\_{n,k}}\Big\langle \Big(\int\_\Omega g\,dE\Big)\xi,E(A)E(X\_{n,k})\xi\Big\rangle\\[0.3cm]
&=\lim\_n\sum\_k\overline{\beta\_{n,k}}\Big\langle \Big(\int\_\Omega g\,dE\Big)\xi,E(A\cap X\_{n,k})\xi\Big\rangle\\[0.3cm]
&=\lim\_n\lim\_m\sum\_k\sum\_j\overline{\beta\_{n,k}}\beta\_{m,j}\Big\langle E(X\_{m,j})\xi,E(A\cap X\_{n,k})\xi\Big\rangle\\[0.3cm]
&=\lim\_n\lim\_m\sum\_k\sum\_j\overline{\beta\_{n,k}}\beta\_{m,j}\Big\langle E(A\cap X\_{n,k})E(X\_{m,j})\xi,\xi\Big\rangle\\[0.3cm]
&=\lim\_n\lim\_m\Big\langle \Big(\int\_A \overline{g\_n}g\_m\,dE\Big)\xi,\xi\Big\rangle\\[0.3cm]
&=\lim\_n\Big\langle \Big(\int\_A \overline{g\_n}g\,dE\Big)\xi,\xi\Big\rangle\\[0.3cm]
&=\Big\langle \Big(\int\_A \overline{g}g\,dE\Big)\xi,\xi\Big\rangle\\[0.3cm]
&=\int\_A|g|^2\,dE\_{\xi,\xi}.
\end{align}


