Resolution of identity of self adjoint belongs to von Neumann Algebra

Question

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..

Answer 1

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).