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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' 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.

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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.

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  • $\begingroup$ Is this true when $M$ is finite and $\tau$ is a tracial state? $\endgroup$
    – John
    Commented Feb 16, 2023 at 15:58
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    $\begingroup$ Yes, then the property holds. Defining the spectral projections $e_n = e_{(1/n,\infty)}(m)$, you get that $e_n \to e_{(0,\infty)}(m)$ and $m^{-1/2} e_n$ is well defined and bounded for all $n$. $\endgroup$
    – Stefaan Vaes
    Commented Feb 16, 2023 at 16:06
  • $\begingroup$ Sorry, I didn't get. Could you please elaborate how does that imply the $\tau$-measurability of $m^{-1/2}e_{(0,\infty)}(m)$? $\endgroup$
    – John
    Commented Feb 16, 2023 at 16:15
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    $\begingroup$ Write $e = e_{(0,\infty)}(m)$, $e_n = e_{(1/n,\infty)}(m)$ and $p_n = (1-e) + e_n$. Since $e_n \to e$ strongly, we get that $\tau(1-p_n) \to 0$. Also, $m^{-1/2} e p_n = m^{-1/2} e_n$ has operator norm at most $\sqrt{n}$. $\endgroup$
    – Stefaan Vaes
    Commented Feb 16, 2023 at 16:20

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