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?

  • $\begingroup$ Why not say "bounded universally measurable" instead of "universially integrable"? $\endgroup$
    – GEdgar
    Mar 23, 2016 at 18:33
  • $\begingroup$ Yes, this is the same. Do you think it will be better for understanding this notion? $\endgroup$ Mar 23, 2016 at 18:36
  • $\begingroup$ @JonasMeyer what are the rules, who is supposed to accept the answer? It's you or me who have to push the button? $\endgroup$ Mar 28, 2017 at 4:57
  • $\begingroup$ @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. $\endgroup$ Mar 28, 2017 at 6:34
  • $\begingroup$ Thank you, Jonas, for drawing people's attention to this question! $\endgroup$ Mar 28, 2017 at 7:39

1 Answer 1


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.


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