A semifinite trace $\tau$ on $M\_{+}$ (for a von Neumann algebra $M$) is said to be normal if $\tau(\sup x\_i ) = \sup \tau(x\_i)$ for an bounded increasing net of positive operators $(x\_i)\_{i \in I}$.


Is it true that $\tau(\inf x\_i) = \inf \tau(x\_i)$ for a bounded decreasing net of positive operators $(x\_i)\_{i \in I}$?


If the trace were finite, it is easy enough to see that the above holds using the increasing net $(x\_0 - x\_i)\_{i \in I}$ and the fact that $\tau(x\_0)$ is finite. In the semifinite case, in order for the same strategy to work we must have the following result: if $\tau(\inf x\_i) < \infty$, then there is an index $j \in I$ such that for all $i \ge j$, we have $\tau(x\_i) < \infty$. This seems like something that should be true. But I am unable to come up with a quick proof.


Thank you.



infimum and supremum for normal semifinite trace
No, let $\tau$ be integration against counting measure on $l^\infty$ and let $x\_n$ be the characteristic function of $\{i: i \geq n\}$.


