I want to prove that if $A\in B(H)$ and $N\in B(H)$ is a normal operator, and $AE(\Delta)=E(\Delta)A$, where $E$ is the spectral measure given by $N$ and $\Delta$ is a Borel subset of $\sigma(N)$, then $AN=NA$.
I tried these first steps:
$\langle ANg,h \rangle =\langle (\int z dE(z))g,A^\ast h \rangle$... but I cannot go on. Maybe writing down the definition of the integral respect to the spectral measure as the limit of the sum of $z*E(\Delta_i)$ taken on a partition $\{\Delta_1...\Delta_n\}$ of the spectrum with every piece of the partition quite small, is possible to achieve the result. Can you suggest me a more straightforward way? Thank you!
The Spectral Theorem tells you that for every $\varepsilon>0$ there exists a partition $\{\Delta_1,\ldots,\Delta_n\}$ of $\sigma(N)$ and complex numbers $\lambda_1,\ldots,\lambda_n$ such that $$ \left\|N-\sum_{j=1}^n\lambda_j\,E(\Delta_j)\right\|<\varepsilon. $$ As $A$ commutes with $\sum_j\lambda_j\,E(\Delta_j)$, you get that $\|AN-NA\|<2\varepsilon$; as $\varepsilon$ is arbitrary, $AN-NA=0$.