Preface
Dominated convergence: $$f_n(\omega)\to f(\omega)\quad(\omega\in\Omega)\implies f_n(E)\to f(E)$$ (This gives a tool for analysis of operators.)
Problem
Given a Borel space $\Omega$ and a Hilbert space $\mathcal{H}$.
Consider a spectral measure: $$E:\mathcal{B}(\Omega)\to\mathcal{B}(\mathcal{H})$$
And its Borel measures: $$\nu_\varphi(A):=\|E(A)\varphi\|^2$$ $$\mu_{\varphi\psi}(A):=\langle E(A)\varphi,\psi\rangle$$
Regard its measurable calculus: $$\mathcal{D}f(E):=\left\{\varphi\in\mathcal{H}:\int_\Omega|f|^2\mathrm{d}\nu_\varphi\right\}:\quad\langle f(E)\varphi,\psi\rangle:=\int_\Omega f\mathrm{d}\mu_{\varphi\psi}$$
Suppose pointwise convervence: $$f_n(\omega)\to f(\omega)$$
Denote their common domain by: $$\mathcal{D}:=\bigcap_{n\in\mathbb{N}}\mathcal{D}f_n(E)$$
Define their limiting operator by: $$\mathcal{D}T:=\{\varphi\in\mathcal{D}:f_n(E)\varphi\to\psi\}:\quad T\varphi:=\psi$$ (That is the one taken in Stone's theorem.)
Operator Domain
Can it happen that this domain shrinks?
Domain Criterion
Does a common domain suffice: $$\mathcal{D}f_n(E)\supseteq\mathcal{D}f(E):\quad T\supseteq f(E)$$ In that case one obtains: $$\mathcal{D}f_n(E)\supseteq\mathcal{D}f(E)\iff |f_n|^2\leq C_n(|f|^2+1)$$ So one may miss a dominant for: $$\|f(E)\varphi-f_n(E)\varphi\|^2=\|(f-f_n)(E)\varphi\|^2=\int|f-f_n|^2\mathrm{d}\nu_\varphi$$
But what is an explicit nonexample?
Dominant Criterion
Assuming a dominant one gets: $$|f_n|^2\leq C(|f|^2+1):\quad T=f(E)$$
Clearly one has: $$\mathcal{D}f(E)\subseteq\mathcal{D}f_n(E)$$
So the above shows: $$\|f(E)\varphi-f_n(E)\varphi\|^2=\ldots=\int_\Omega|f-f_n|^2\mathrm{d}\nu_\varphi\to0$$
How to prove the converse inclusion?
Operator Core
Does one have as core: $$\mathcal{D}f_n(E)\supseteq\mathcal{D}f(E):\quad\overline{T}=f(E)$$ (This solves the failure above!)
Ok, I think I finally got it...
(However, if anybody finds bugs, typos, loopholes etc. then please let me know. Thanks!)
Framework
Given the natural numbers $\Omega:=\mathbb{N}$ and the Hilbert space $\mathcal{H}:=\ell^2(\mathbb{N})$.
Choose the canonical basis by: $e_n:=\chi_n$
Consider the spectral measure $E(\{n\}):=P_n$.
Operator Domain
Regard the series of measurable functions $f_n:=\sum_{k=1}^nk\chi_{\{k\}}$.
Denote their sum by: $f:=\sum_{k=1}^\infty k\chi_{\{k\}}$
It approaches pointwise its sum: $f_n\to f$
It has a dominant: $|f_n|^2\leq1(|f|^2+1)$
However, not every element converges: $$\varphi:=\sum_{k=1}^\infty\frac{1}{k}e_k:\quad f_n(E)\varphi=\sum_{k=1}^n1e_k\nrightarrow\psi$$
So the domain may shrink: $\mathcal{D}T\subsetneq\mathcal{D}$
Domain-Criterion
Regard the sequence of measurable functions $f_n:=n\chi_{\{n\}}$.
It vanishes pointwise: $f_n\to0$
Now, it fulfils the domain criterion: $$\mathcal{D}f_n(E)=\mathcal{H}\supseteq\mathcal{H}=\mathcal{D}f(E)$$ However, its measurable calculus does not converge: $$\varphi:=\sum_{k=1}^\infty\frac{1}{k}e_k:\quad f_n(E)\varphi=e_n\nrightarrow0=f(E)\varphi$$
So a common domain is not sufficient: $T\subsetneq f(E)$
Dominant-Criterion
Suppose $\varphi\in\mathcal{D}T$ with $T\varphi=\psi$.
By Fatou applied to the measurable calculus one has: $$0\leq\int|f-f_n|^2\mathrm{d}\nu_\varphi\leq\liminf_m\int|f_m-f_n|^2\mathrm{d}\nu_\varphi=\liminf_m\|f_m(E)\varphi-f_n(E)\varphi\|^2\to0$$ But that implies especially: $\varphi\in\mathcal{D}f(E)$
Moreover, one has then: $f(E)\varphi=\lim_nf_n(E)\varphi=\psi=T\varphi$
Concluding that the operators agree: $f(E)=T$
Operator Core
Suppose $\varphi\in\mathcal{D}f(E)$ with $f(E)\varphi=\psi$.
By Egorov's theorem one finds: $$\nu_\varphi(\Omega_N)\to\nu_\varphi(\Omega):\quad\|f-f_n\|_{\Omega_N}\stackrel{n\to\infty}{\to}0$$
Set the approximation to be: $$\varphi_N:=1_N(E)\varphi:=E(\Omega_N)\varphi$$
So on the hand one has: $\varphi_N\to\varphi$
Also note that: $f_n(E)1_N(E)=(f_n1_N)(E)$
Now by uniform convergence: $$|f_n1_N|\leq R_N(|f1_N|+1)\quad|f1_N|\leq R_N(|f_n1_N|+1)$$ So one has a common domain: $\mathcal{D}(f1_N)(E)=\mathcal{D}(f_n1_N)(E)$
But also one has trivially: $$|f1_N|\leq1(|f|+1)$$ Thus they all include: $\mathcal{D}f(E)\subseteq\mathcal{D}(f1_N)(E)$
Hence one can compute: $$\left\|(f1_N)(E)\varphi-(f_n1_N)(E)\varphi\right\|^2=\int_{\Omega_N}|f-f_n|^2\mathrm{d}\nu_\varphi\leq\|f-f_n\|_{\Omega_N}\|\varphi\|^2\stackrel{n\to\infty}{\to}0$$ That is $\varphi_N\in\mathcal{D}T$ with $T\varphi_N=(f1_N)(E)\varphi$.
But by the preceding dominant-criterion: $$|f1_N|\leq1(|f|+1):\quad(f1_N)(E)\varphi\to f(E)\varphi$$ So as was desired: $T\varphi_N\to f(E)\varphi$
Concluding that it is a core: $\overline{T}=f(E)$