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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.

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No, let $\tau$ be integration against counting measure on $l^\infty$ and let $x_n$ be the characteristic function of $\{i: i \geq n\}$.

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