There is an exercise about invertibility of spectral integral.
$H$ is a Hilbert space and $(X,\mathscr{M},E)$ is a spectral measure. $f$ is a bounded measurable function on $X$. Show that $\int f \, \mathrm{d} E$ is invertible if and only if there is an $S \in \mathscr{M}$ and $\delta>0$ such that $E(S)=0$ and $\lvert f \rvert \ge \delta$ on $S^c$.
I can prove the "if" part. For the "only if" part, we only need to prove that there is a $\delta>0$ such that $E(\lvert f \rvert \le \delta)=0$. What I know is that $E(\lvert f \rvert \le 1/n) \overset{s}{\to} E(f=0)$ and $E(f=0)=0$ following from the fact that \begin{equation*} E(f=0)\int f \, \mathrm{d} E=0. \end{equation*} But then I don't know how to continue.
Thanks for your help!
Suppose $\int f dE$ is invertible. Then there exists $\epsilon > 0$ such that $$ \epsilon^2\|x\|^2 \le \int |f|^2 d\|Ex\|^2, \;\;\; x \in H. $$ Let $x=E(|f| \le \delta)y$ to obtain $$ \epsilon^2 \|E(|f|\le \delta)y\|^2 \le \int_{|f|\le \delta} |f|^2d\|Ex\|^2 \le \delta^2\|E(|f| \le \delta)y\|^2 $$ Therefore, if $\delta < \epsilon$, it follows that $E(|f|\le \delta)y=0$ for all $y$ and, hence $E(|f|\le \delta)=0$ for all $\delta < \epsilon$.