We know that a abelian von Neumann algebras $\mathcal{A}$ is isomorphic to $\mathscr{L}^{\infty}(X)$ for some special measure space $(X,\Sigma,\mu)$. For that reasone one can see an arbitrary von Neumann algebra as a noncommutative measure space. My question is the following: is there a possibility to compare a measure space with a von Neumann algebra? In particular, what is an "integral" on such $\mathcal{A}$? Is it just a positive linear functional? What is the equivalent to the $\sigma$-algebra or the measure? Thank you very much for your inspiration.
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5$\begingroup$ There is no canonical equivalent of "the" $\sigma$-algebra. Indeed, one cannot recover the underlying measure space from $\mathcal{L}^\infty(X)$, only the measure algebra, which is what one gets by taking quotients by the ideal of sets of measure zero. There are two natural ways to obtain representing measure spaces. One uses the Stone-space of the measure algebra, the other one employs the Maharam representation theorem of measure algebras. $\endgroup$– Michael GreineckerCommented Apr 13, 2017 at 12:38
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Given a localizable measure space $(X,\mu)$, the space $\mathcal{A} = L^\infty(X,\mu)$ is a von Neumann algebra. The integral on $L^\infty(X,\mu)$ is abstractly understood as a weight on $\mathcal{A}$ and I suppose the measure corresponds to the restriction of this weight to the projections in $\mathcal{A}$.
answered Apr 13, 2017 at 14:11