Question about domains of unbounded operators This is a part of a theorem in Rudin's Functional Analysis, in the chapter on unbounded operators. Let $\mathcal M$ be a $\sigma$-algebra in a set $\Omega$, $H$, a Hilbert space and $E:\mathcal M\rightarrow \mathscr B(H)$, a resolution of identity. Let $f:\Omega\rightarrow \mathbb C$ be measurable and let $\mathscr D_{f}$={$x\in H:\int_D |f|^{2}dE_{x,x}< \infty$}, where $E_{x,y}(\omega)=<E(\omega)x,y>$ for each $\omega \in \mathcal M$.
The question is to show that $\mathscr D_{f\overline{f}}\subseteq \mathscr D_{f}$. 
I would be grateful for hints on this.

You want to show that $\int_\Omega|f|^4\,dE_{x,x}<\infty$ implies $\int_\Omega|f|^2\,dE_{x,x}<\infty$. and
$$
\int_\Omega|f|^2dE_{x,x}=\int_{\Omega\cap\{|f|<1\}}|f|^2dE_{x,x}
+\int_{\Omega\cap\{|f|\geq1\}}|f|^2dE_{x,x}\leq E_{x,x}(\Omega)+\int_{\Omega\cap\{|f|\geq1\}}|f|^4dE_{x,x}\\<\|x\|^2+\int_{\Omega}|f|^2dE_{x,x}<\infty,
$$
since the measure $E_{x,x}$ is finite (because $E(\Omega)=I$).
