Hello,


I have a question about trace measurable operators and I think it's not a hard one. However, I'm quite confused because I cannot prove it.


Let $\mathcal{M}$ be a semi-finite von Neumann algebra with a faithful normal semi-finite trace $\tau$. Let $T$ be a $\tau$- measurable operator (densely defined closed (possibly unbounded) operator affiliated with $\mathcal{M}$ such that
$$ \forall\_{\varepsilon >0} \ \exists\_{E - \text{a projection in} \text{M}} \ \mbox{Range}(E) \subset D(T) \ \& \ \tau(1-E) \leq \varepsilon.)$$


Let $E\_{(s,\infty)}(|T|)$ be a spectral projection of $|T|$ corresponding to the interval $(s, \infty)$, $s \geq 0$.


How do we know that $\| |T|E\_{(s,\infty)}(|T|) \| > s$ or $\| |T|E\_{[0,s]}(|T|) \| \leq s$.


Thank you in advance for any help.



trace measurable operators
Like Matthew said, this doesn't look like it has to do with measurability at all. Those two inequalities follow from the operator inequalities 
`\[
|T|\,E_{(s,\infty)}(|T|) \geq s\,E_{(s,\infty)}(|T|), \ \ \ 
|T|\,E_{[0,s]}(|T|) \leq s\,E_{[0,s]}(|T|).
\]`
In turn, these inequalities follow from the corresponding inequalities for real functions, 
`\[
t\,1_{(s,\infty)}(t) \geq s\,1_{(s,\infty)}(t), \ \ \ 
t\,1_{[0,s]}(t)\leq s\,1_{[0,s]}(t),
\]`
and the fact that functional calculus is positive. 


