# Is the algebra of universally integrable functions a von Neumann algebra?

I would like to continue this discussion.

Let $X$ be a compact space. Let us call a function $f:X\to {\mathbb C}$ universally integrable if it is integrable with respect to each regular Borel measure $\mu$ on $X$ (one can imagine $\mu$ as an arbitrary positive continuous functional on ${\mathcal C}(X)$). We denote by ${\mathcal U}(X)$ the space of all universally integrable functions on $X$.

Nate Eldredge noticed here, that ${\mathcal U}(X)$ is a $C^*$-algebra with respect to the sup-norm: $$||f||=\sup_{x\in X}|f(x)|.$$ Question:

Is ${\mathcal U}(X)$ a von Neumann algebra with respect to this norm?

• Why not say "bounded universally measurable" instead of "universially integrable"? Mar 23, 2016 at 18:33
• Yes, this is the same. Do you think it will be better for understanding this notion? Mar 23, 2016 at 18:36
• @JonasMeyer what are the rules, who is supposed to accept the answer? It's you or me who have to push the button? Mar 28, 2017 at 4:57
• @SergeiAkbarov: You can accept in the normal way. If you're asking about the bounty, that is awarded by whoever starts it, so me in this case. Accept and bounty are independent. Mar 28, 2017 at 6:34
• Thank you, Jonas, for drawing people's attention to this question! Mar 28, 2017 at 7:39

The following is a slight elaboration on Martin Argerami's not-quite-complete (now deleted) answer. Let $E\subset X$ be any set which is not universally measurable (i.e., its characteristic function is not universally integrable). For each finite subset $F\subset E$, note that the characteristic function $1_F$ is in $\mathcal{U}(X)$. These characteristic functions form a bounded increasing net $(1_F)$ in $\mathcal{U}(X)$. If $\mathcal{U}(X)$ were a von Neumann algebra, it would be monotone-complete, and so $(1_F)$ would have a supremum $f\in\mathcal{U}(X)$. Note that for each $x\in X\setminus E$, $1_{X\setminus\{x\}}\in\mathcal{U}(X)$ is an upper bound for each $1_F$, so we have $f\leq 1_{X\setminus\{x\}}$ for all such $x$. The only function $f$ on $X$ which satisfies $1_F\leq f$ whenever $F\subset E$ is finite and $f\leq 1_{X\setminus\{x\}}$ wheneve $x\in X\setminus E$ is $f=1_E$. But $1_E\not\in\mathcal{U}(X)$, so no such supremum $f\in\mathcal{U}(X)$ can exist. Thus $\mathcal{U}(X)$ is not a von Neumann algebra, at least whenever $X$ is nontrivial enough that such a set $E$ exists.
More generally, a similar argument shows that if $A$ is a *-subalgebra of the algebra of all bounded functions on a set $X$ which contains all characteristic function of singletons, then if $A$ is a von Neumann algebra it must actually be the entire algebra of bounded functions.