**Problem:** Let $M$ be a semifinite von Neumann algebra with a faithful semifinite normal trace $\tau$. Let $m$ be a positive element in $M$ and let $e\_{(0,\infty)}(m)$ be the spectral projection of $m$ on $(0,\infty)$. How can we show that $e\_{(0,\infty)}(m)m^{-1/2}$ is $\tau$-measurable?


I got stuck with this problem while reading $\tau$-measurable operators from the book '[Lectures on Selected Topics in von Neumann Algebras](https://drive.google.com/file/d/1H6ki8gjUPY5dsWE-CrGn9WFNhMEfEkIv/view?usp=sharing)' by Hiwi. Here I recall the definition of $\tau$-measurable operator.


**Definition 1:** For each $\epsilon,\delta>0$, define $$\mathscr{O}(\epsilon,\delta)=\{m\text{ affiliated to } M:eH\subseteq \mathcal{D}(m),\,\|me\|\leq \epsilon \text{ and }\tau(1-e)\leq\delta \text{ for some } e\in Proj(M)\}.$$
Let $m$ be a densely defined closed operator such that $m$ is affiliated to $M$. We say that $m$ is $\tau$-measurable if for any $\delta >0$, there exists an $\epsilon >0$ such that $m\in\mathscr{O}(\epsilon,\delta)$. We denote by $\widetilde{M}$ the set of such $\tau$-measurable operators.


Thanks in advance for any help or suggestion.



$\tau$-measurable operator
I think that this is simply not true. Take $M = \ell^\infty(\mathbb{N})$ with the semifinite trace $\tau(F) = \sum\_n F(n)$. When $p \in M$ is a nonzero projection, we have $\tau(p) \geq 1$. So the only $\tau$-measurable operators are the elements of $M$ itself. Taking $m \in M$ given by $m(k) = 1/k$, the element $m^{-1/2}$ is not $\tau$-measurable.


