Borel functional calculus on reduced von Neumann algebra Let $M$ be an von Neumann algebra, $x\in M_+$ is a positive element, $e\in M$ is a projection ($e^2=e,e^*=e$). $f$ is a bounded Borel function on spectral set $\sigma(x)$ and $f(0)=0$. If $exe = x$, prove that $f(x)\in eMe$.
My attempt: I know that $f(x) \in M$ by Borel functional calculus, but on the reduced von Neumann algebra $eMe$, we can't say that $f(x)\in eMe$, because the spectral set $\sigma(x)$ in $M$ is different from spectral set in $eMe$ even if $exe=x$. I think that $0\notin \sigma(exe)$ but may be belongs to $\sigma(x)$. so this is why we need that $f(0)=0$, but I don't know how to use this condition.
Thanks in advance!
 A: You don't have to re-do the Borel functional calculus on $eMe$. All you need is the assertion that $f(x)\in eMe$.
Since $exe=x$, for any $n\in\mathbb N$
$$
x^n=x\,x^{n-1}=exe\,x^{n-1}=e\,exe\,x^{n-1}=ex^n. 
$$
Do the same on the right, or take adjoints, and $x^n=ex^ne$. Thus $p(x)=ep(x)e$ for any polynomial $p$ with $p(0)=0$. The continuous functional calculus then gives you
$$\tag1
f(x)=ef(x)e,\qquad f\in C(\sigma(x))\ \text{ such that $f(0)=0$}. 
$$A typical way to construct the spectral measure for $x$ is to define, for each $\xi\in H$,
$$\tag2
f\longmapsto \langle f(x)\xi,\xi\rangle. 
$$
This is a bounded positive linear functional on $C(\sigma(x))$, and so by Riesz-Markov there exists a regular Borel measure $\mu_\xi$ on $\sigma(x)$ such that
$$\tag3
\langle f(x)\xi,\xi\rangle=\int_{\sigma(x)}f\,d\mu_\xi,\qquad f\in C(\sigma(x)).
$$
For $f$ bounded Borel, one defines an operator $f(x)\in M$ by
$$
\langle f(x)\xi,\xi\rangle=\int_{\sigma(x)}f\,d\mu_\xi. 
$$
Combining $(1)$ and $(3)$ one gets that  $\mu_{e\xi}(\Delta)=\mu_\xi(\Delta)$ for all $\xi\in H$, as long as $0\not\in\Delta$ (since they agree on all continuous $f$ with $f(0)=0$). Then, as long as $f$ is bounded Borel with $f(0)=0$,
$$
\langle ef(x)e\xi,\xi\rangle=\langle f(x)e\xi,e\xi\rangle=\int_{\sigma(x)\setminus\{0\}}f\,d\mu_{e\xi}=\int_{\sigma(x)\setminus\{0\}}f\,d\mu_{\xi}=\langle f(x)\xi,\xi\rangle, 
$$
which shows that $f(x)=ef(x)e\in eMe$.
