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Consider the following theorem from Takesaki's first volume "Theory of operator algebras":

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In $(i)$, it is claimed that $L^\infty(\Gamma,\mu)$ is an abelian von Neumann algebra. How does Takesaki define "Radon" measure for this to be true?

For example, it is well-known that semi-finiteness of a (Radon) measure is equivalent with the canonical map $L^\infty(\Gamma, \mu)\to B(L^2(X,\mu))$ being isometric. So if $\mu$ is not semi-finite, then $L^\infty(\Gamma, \mu)$ is no von Neumann algebra!

Could anyone give some insight in the matter? Maybe Takesaki only works with finite measures?

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  • $\begingroup$ It is probably best to provide a precise reference for the “well-known” claim in the second-to-the-last paragraph, including the definition of L^∞ and L^2 for nonsemifinite measures, since the claim appears to be false at least for some definitions. (As a trivial counterexample, take Γ={*} with μ({*})=∞. With certain definitions of L^∞ and L^2 for nonsemifinite measures, we can have L^∞(Γ,μ)=L^2(Γ,μ)=0.) $\endgroup$
    – Dmitri Pavlov
    Commented Oct 3, 2022 at 23:19
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    $\begingroup$ Also, why is it that $L^\infty(X,\mu)$ is a von Neumann algebra on $L^2(X)$ if $\mu$ is Radon on the locally compact space $X$? Maybe you would be willing to write an answer to clarify the technicalities? $\endgroup$
    – Andromeda
    Commented Oct 4, 2022 at 7:04
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    $\begingroup$ Let me just make one comment: certainly Takesaki does not just work with finite measures. Your other queries are interesting though, IMHO. $\endgroup$
    – Matthew Daws
    Commented Oct 4, 2022 at 8:55
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    $\begingroup$ A Radon measure is required to be locally finite, and therefore finite on every compact subset. Inner regularity with respect to compact sets then implies semifiniteness. But in fact a Radon measure on a locally compact space is not just semifinite, but strictly localizable. One important point - the measure defined on the spectrum of a commutative W$^*$-algebra $A$ by a faithful normal weight $\phi$ on $A$ won't be locally finite unless it's finite. Although it's always inner regular, it's not a Radon measure except in the aforementioned case. $\endgroup$
    – Robert Furber
    Commented Oct 4, 2022 at 16:10
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    $\begingroup$ This is why Takesaki picks a partition of the spectrum of a commutative W$^*$-algebra into clopen sets of finite measure - their union is then a locally compact space on which this measure is locally finite. $\endgroup$
    – Robert Furber
    Commented Oct 4, 2022 at 16:20

1 Answer 1

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Making no claims of originality, one possible proof can be obtained by combining Example 4.60, Lemma 5.11, and Lemma 3.14 in arXiv:2005.05284.

This shows that for any Radon measure its algebra of equivalences classes of bounded complex-valued measurable functions modulo equality almost everywhere is a von Neumann algebra.

A Radon measure is a measure on a Hausdorff topological space that is locally finite and inner regular with respect to compact subsets and all open sets are measurable.

As a side remark, a reasonable property of $\def\L{{\rm L}}\L^p$-spaces that is worth preserving is that $\L^p(X,μ)$ only depends (up to an isomorphism of topological vector spaces) on the σ-ideal of μ-negiligible sets and the semifinite support of μ. Thus, the definition of $\L^∞(X,μ)$ for a nonsemifinite measure μ should take equivalence classes of bounded complex-valued functions on the semifinite support of μ, whereas for $\L^{1/p}(X,μ)$ with $\Re p≠0$ this is automatic.

With this definition, the spaces $\L^{1/p}(X,μ)$ form a $\def\C{{\bf C}}\C$-graded (by $p∈\C$) complex *-algebra, and for all $p∈\C_{\Re≥0}$, the space $\L^{1/p}(X,μ)$ is a faithful module over the von Neumann algebra $\L^∞(X,μ)$ and is generated as a topological module by a single element. See the answer to Is there an introduction to probability theory from a structuralist/categorical perspective? for more details.

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