Let $M:=\displaystyle\int_0^{+\infty} \lambda \,\mathrm{d}E_\lambda$ be the spectral decomposition of a positive operator $M$ on a complex Hilbert space $F$.
For $x\in F$, we set
$$x_n=\displaystyle\int_{\{\lambda>1/n\}}\lambda^{-1/2}E_{\lambda}(x)\,\mathrm{d}E_\lambda.$$
Why
$$Mx_n=\int_{\{\lambda> 1/n\}}\lambda\lambda^{-1/2}E_\lambda(x) \,\mathrm{d}E_\lambda?$$
Best Answer
The Borel functional calculus $f\mapsto \int_{\sigma(M)}f(\lambda) \,d E_\lambda$ is a $*$-homomorphism from the set of bounded Borel measurable functions to the set of bounded operators on a Hilbert space. So, you have: $$\int_{\sigma(M)}f(\lambda) \,d E_\lambda\int_{\sigma(M)}g(\lambda) \,d E_\lambda=\int_{\sigma(M)}f(\lambda)g(\lambda) \,d E_\lambda$$ for any bounded Borel measurable functions on $\sigma(M)$.