Resolution of identity of self adjoint belongs to von Neumann Algebra If $x \in M \subset B(H)$ where $M$ is a Von Neumann Algebra and
\begin{equation}
 x = \int \lambda \, dE(\lambda)
\end{equation}
Is selfadjoint, how can I show that the resolution of Identity
\begin{equation}
\{E(\lambda) | \lambda \in \mathbb{R}\}
\end{equation}
Belongs to $M$?
I know that $E(\lambda)$ are basically characteristic functions (projectors) but I guess the result is easy to deduce..
 A: Let's tackle this from the very basics.


*

*If $p$ is a polynomial, then $p(x)$ is in the algebra generated by $x$ by definition.

*If $f$ is a continuous function on the spectrum of $x$, then it can be approximated by polynomials $p_n$, and by continuous functional calculus, $p_n(x)\to f(x)$ in operator norm. Thus $f(x)$ is in the $C^\ast$-algebra generated by $x$.

*If $I$ is an open subset of the spectrum of $x$, then its characteristic function $1_I$ is the increasing limit of the continuous functions $f_n=\max\{1,nd(\,\cdot \,,I)\}$. Borel functional calculus tells us that $f_n(x)\to 1_I(x)$ weakly (or strongly or $\sigma$-weakly, whatever you like). Thus $1_I(x)$ is in the von Neumann algebra generated by $x$.

*Let $\mathcal{F}=\{E\in \mathcal{B}(\sigma(x))\mid 1_E(x)\in W^\ast(x)\}$. It follows easily from the properties of the Borel functional calculus that $\mathcal{F}$ is a $\sigma$-algebra and by the previous bullet point, $\mathcal{F}$ contains all open sets. Thus $\mathcal{F}=\mathcal{B}(\sigma(x))$, that is, $1_E(x)$ is in the von Neumann algebra generated by $x$ for every Borel subset $E$ of $\sigma(x)$.

*If $f$ is a bounded Borel function on $\sigma(x)$, then it is the uniform limit of $f_n=\sum_{j=-n}^{n}j\frac{\|f\|_\infty}{n}1_{f^{-1}([j\frac{\|f\|_\infty}{n},(j+1)\frac{\|f\|_\infty}{n}))}$. It follows from Borel functional calculus that $f_n(x)\to f(x)$ in operator norm. Thus $f(x)$ is in the von Neumann algebra generated by $x$ (of course youu don't need this part for the spectral measure).

