Problem
Given a Hilbert space $\mathcal{H}$.
Consider a spectral measure: $$E:\mathcal{B}(\mathbb{C})\to\mathcal{B}(\mathcal{H})$$
Then the criterion holds: $$\mathcal{D}f(E)\subseteq\mathcal{D}g(E)\iff|g|^2\leq R^2(|f|^2+1)$$
How can I prove this?
Application
See the thread: Dominated Convergence
Criterion
Denote for shorthand: $$S:=f(E)\quad T:=g(E)$$
Consider the operator: $$\Gamma:\mathcal{G}(S)\to\mathcal{G}(T):(x,Sx)\mapsto(x,Tx)$$
It is well defined since: $$\mathcal{D}(S)\subseteq\mathcal{D}(T)$$
And it is closed since: $$\varphi_n\to\varphi,S\varphi_n\to\psi,\,T\varphi_n\to\chi\implies S\varphi=\psi,T\psi=\chi$$
Also its domain is complete: $$\mathcal{D}(\Gamma)=\mathcal{G}(S):\quad\mathcal{D}(\Gamma)=\overline{\mathcal{D}(\Gamma)}$$
By closed graph theorem: $$\varphi\in\mathcal{D}(S):\quad\|T\varphi\|^2\leq\|T\varphi\|^2+\|\varphi\|^2\leq\|\Gamma\|^2\left(\|S\varphi\|^2+\|\varphi\|^2\right)$$
By mean value theorem: $$\int_A|g|^2\mathrm{d}\nu_\varphi\leq\int_A\|\Gamma\|^2\left(|f|^2+1\right)\mathrm{d}\nu_\varphi\implies|g|^2\leq\|\Gamma\|^2\left(|f|^2+1\right)\mod\nu_\varphi$$
So one obtains that: $$|g|^2\leq\|\Gamma\|^2\left(|f|^2+1\right)\mod E$$
Conversely one has: $$\int|g|^2\mathrm{d}\nu_\varphi\leq R^2\int|f|^2\mathrm{d}\nu_\varphi+R^2\|\varphi\|^2$$
Concluding the assertion.
Null Set
Remind that it holds: $$\nu_\varphi(A)=0\iff E(A)\varphi=0$$
Regard the null set:* $$N_0:=\bigcap_{\varphi\in\mathcal{H}}N_\varphi:\quad\nu_\varphi(N_\varphi)=0$$
Indeed one has: $$E(N_0)\varphi\quad(\varphi\in\mathcal{H})\implies E(N_0)=0$$
That was desired!
Separable Spaces
For separable spaces: $$N_0:=\bigcup_{\alpha\in\mathcal{A}}N_\alpha:\quad N_\alpha\in\mathcal{B}(\mathbb{C})\implies N_0\in\mathcal{B}(\mathbb{C})$$
By density one has: $$E(N_0)\alpha=0\quad(\alpha\in\mathcal{A})\implies E(N_0)=0$$
That was desired.
*There's a flaw: It may fail to be Borel!