Suppose $T$ is a self-adjoint operator and $E_T$ is the spectral measure of $T$. Let $a_n=E_T[-n, n]$.
We know the fact that $T=\int_{\sigma(T)}\iota dE_T$ and $E_T[-n, n]= \int_{\sigma(T)}\chi_{[-n, n]} dE_T$, where $\iota(z)=z, z\in \sigma(T)$, $\sigma(T)$ is the spectrum of $T$.
Is $TE_T[-n, n]=\int_{-n}^n\iota dE_T$ ? If the equation holds, how to prove it?
Any book that does the Spectral Theorem surely does this too. One can show that the map $$\tag1f\longmapsto \int_{\sigma(T)} f(\lambda)\,dE_T(\lambda)$$ is a $*$-homomorphism from $B(\sigma(T))$ to $W^*(T)$ that takes $p$ to $p(T)$ for any polynomial.
That $\int_{\sigma(T)}p(\lambda)\,dE_T(\lambda)=p(T)$ for any polynomial $T$ can be shown by using the definition of the integral.
The map in $(1)$ has the property that if $\int_{\sigma(T)} f_j\,d\mu\to0$ for all Radon measures on $\sigma(T)$, then $\int_{\sigma(T)}f_j(\lambda)\,dE_T(\lambda)\to0$ wot. As any continuous function is a norm limit of polynomials, one gets that $\int_{\sigma(T)} f(\lambda)\,dE_T(\lambda)=f(T)$ for any continuous function.
Because the map in $(1)$ is a $*$ homomorphism, we have \begin{align} \int_{\sigma(T)\cap[-n,n]} \lambda\,dE_T(\lambda) &=\int_{\sigma(T)} \lambda\,1_{[-n,n]}(\lambda)\,dE_T(\lambda)\\[0.3cm] &=\bigg(\int_{\sigma(T)} \lambda\,dE_T(\lambda)\bigg)\bigg(\int_{\sigma(T)} 1_{[-n,n]}(\lambda)\,dE_T(\lambda)\bigg)\\[0.3cm] &=T\,E_{[-n,n]}. \end{align}