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

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?
 A: 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.
