Continuity of the extension of a tracial state with respect to the strong operator topology

Question

Problem: Let $M\subseteq B(H)$ be a finite von Neumann algebra with a faithful tracial state $\tau$. Let $\widetilde{M}$ be the $\tau$-measurable operators on $M$ (recalled below). Extend the trace $\tau$ on $\widetilde{M}_+$ by $\tau(a):=\int_0^\infty\lambda\tau(e_\lambda)$ where $a=\int_0^\infty\lambda\,de_\lambda$ is the spectral decomposition (See Equation 4.6 of Hiwi). If $\{x_n\}_{n\in\mathbb{N}}$ be a sequence of positive elements from $M$, which converges strongly to an element $x$ of $\widetilde{M}_+$, then can we say that $\{\tau(x_n)\}_{n\in\mathbb{N}}$ converges to $\tau (x)$?

Or, can we at least say that if $\{x_n\}_{n\in\mathbb{N}}$ be a sequence of positive elements from $M$, which converges strongly to an element $x$ of $\widetilde{M}_+$, then $\{x_n\}_{n\in\mathbb{N}}$ converges to $x$ in the measure topology?

(A positive answer of any one of the above two questions would be sufficient for me.)

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.

Theorem 2: (Theorem 4.12 of Hiwi) The $\widetilde{M}$ is a complete metrizable Hausdorff topological *-algebra with $\{\widetilde{M}\cap\mathscr{O}(\epsilon,\delta):\epsilon,\delta >0\}$ as a neighborhood basis of $0$. Moreover, $M$ is dense in $\widetilde{M}$.

Thanks in advance for any help or suggestion.

Answer 1

With the specific definition of strong convergence in the comment (namely, a sequence $x_n \in M_+$ is said to converge strongly to $x \in \widetilde{M}_+$ if and only if $x_n \xi \to x \xi$ for all $\xi \in D(x)$), both properties indeed hold.

Take such a sequence $x_n \in M_+$ converging strongly to $x \in \widetilde{M}_+$. We prove that $\tau(x_n) \to \tau(x)$ and $x_n - x \to 0$ in measure.

First assume that $\tau(x)=+\infty$. Choose $\kappa > 0$. Since $\tau(x) = +\infty$, we can choose a spectral projection $p = e_{[0,\lambda]}(x)$ such that $\tau(xp) > \kappa + 1$. By assumption, $x_n p \to x p$ strongly in the usual sense. By the uniform boundedness principle, the sequence $x_n p$ is bounded in operator norm and $\tau(x_n p) \to \tau(x p)$. We can thus take $n_0$ such that $|\tau(x_n p) - \tau(xp)| < 1$ for all $n \geq n_0$. Thus, $\tau(x_n p) > \kappa$ for all $n \geq n_0$. Since $$\tau(x_n) \geq \tau(x_n^{1/2} p x_n^{1/2}) = \tau(x_n p) > \kappa \; ,$$ we conclude that $\tau(x_n) > \kappa$ for all $n \geq n_0$. Thus, $\tau(x_n) \to + \infty$.

Next assume that $\tau(x) < +\infty$. We then have a well-defined normal positive functional $\omega$ on $M$ satisfying $$\omega(a) = \tau((1+x)^{1/2} a (1+x)^{1/2}) \quad\text{for all $a \in M$.}$$ By assumption $x_n (1+x)^{-1} \to x (1+x)^{-1}$ strongly in the usual sense. Again by uniform boundedness, $$\omega(x_n (1+x)^{-1}) \to \omega(x (1+x)^{-1}) \; .$$ This precisely says that $\tau(x_n) \to \tau(x)$.

To prove that $x_n - x \to 0$ in measure, first note the following standard result: if $y_n$ is a sequence in $M$ such that $y_n \to 0$ strongly, then $y_n \to 0$ in measure. Indeed, given $y \in M$ and $\delta > 0$, consider the spectral projection $e = e_{[0,\delta]}(y^* y)$. Since $y^* y \geq \delta (1-e)$, we find that $$\delta \, \tau(1-e) \leq \tau(y^* y) \quad\text{and}\quad \|y e \| \leq \delta \; .$$ Writing $\|y\|_2 = \sqrt{\tau(y^* y)}$, taking $\delta = \|y\|_2$ and using the notation $\mathcal{O}(\cdot,\cdot)$ for the basic neighborhoods of $0$ in the measure topology as in the question, it follows that $$y \in \mathcal{O}(\|y\|_2,\|y\|_2) \quad\text{for all $y \in M$.}$$ When $y_n \to 0$ strongly, also $\|y_n\|_2 \to 0$ and thus $y_n \to 0$ in measure.

We now return to the sequence $x_n - x$. Fix $\varepsilon > 0$. Take a spectral projection $p$ of $x$ such that $xp$ is bounded and $\tau(1-p) < \varepsilon/2$. Since $(x_n - x) p \to 0$ strongly in the usual sense, also $(x_n - x)p \to 0$ in measure. So, for all $n$ large enough, we have that $(x_n - x)p \in \mathcal{O}(\varepsilon,\varepsilon/2)$. It follows that $x_n - x \in \mathcal{O}(\varepsilon,\varepsilon)$ for all $n$ large enough. Thus, $x_n \to x$ in measure.