I am studying von Neumann algebras. In the wiki article abelian von Neumann algebras, it mentions that every abelian von Neumann algebras acting on a separable Hilbert space is *-isomorphic to $L^{\infty}[0,1]$, $l^{\infty}(\mathbb{N})$ or their direct sum. I am wondering why $[0,1]$ is special in this case, why not other measure space such as $S^{1}$ or $[0,1]^{2}$? Also, if we consider $L^{\infty}([0,1]^{2})$ acting on $L^{2}([0,1]^{2})$, isn't that a von Neumann algebra acting on a separable Hilbert space? Does that mean that $L^{\infty}[0,1]$ and $L^{\infty}([0,1]^{2})$ are *-isomorphic?
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3$\begingroup$ Yes, the probability spaces $[0,1]$ and $[0,1]^2$ (with Lebesque measure) are isomorphic and hence, so are the $L^\infty$-spaces over them. You might wish to search for the notion "standard probability space". $\endgroup$– Jochen GlueckCommented Oct 10, 2021 at 10:40
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2$\begingroup$ @JochenGlueck, another for the Lebesque column. 😁 $\endgroup$– LSpiceCommented Oct 10, 2021 at 15:42
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2$\begingroup$ @LSpice: Arrgh, and I was so proud that I finally got the 's' at the right position... (I had often misspelled the name as "Lesbegue" - or "Lesbeque"? - in the past.) $\endgroup$– Jochen GlueckCommented Oct 10, 2021 at 16:03
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The spaces $[0,1]$, $[0,1]^2$, and $S^1$ are all isomorphic as measurable spaces, including their sets of measure 0, as required by the Gelfand-type duality for measurable spaces.
For instance, the isomorphism $[0,1]→S^1$ is given by identifying $S^1=[0,1]/(0∼1)$. Since the points $0$ and $1$ have measure 0, and maps that differ on a set of measure 0 are identified, the above map is indeed an isomorphism.
Likewise, the isomorphism $[0,1]^2→[0,1]$ is given by identifying $[0,1]$ with $\{0,1\}^N$ (e.g., using binary expansions), where $N$ is a countable set, and then taking the isomorphism $[0,1]^2≅\{0,1\}^{N⊔N}≅\{0,1\}^N≅[0,1]$, where $N⊔N$ is isomorphic to $N$ since both are countable sets.
Indeed, the very point of Maharam's theorem is that there are very few measurable spaces up to an isomorphism: any measurable space is isomorphic to the disjoint union of measurable spaces of the form $\{0,1\}^S$, where $S$ is a set of arbitrary cardinality, and this gives a complete classification of measurable spaces up to isomorphism: simply count the (infinite) number of summands for each possible cardinality of $S$, and also count the number of isolated points.
answered Oct 10, 2021 at 15:25
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$\begingroup$ "spaces [...] are all isomorphic as measurable spaces, including their sets of measure $0$, [...] isomorphism $[0,1]^2\to[0,1]$ is given by [...]" Do you have a simple proof of this, or is this just a consequence of something given in you article? For example, the map $[0,1]\to[0,1]$ given by $t\mapsto\frac12(ct+t)$ where $c$ is a "Cantor staircase" is even a homeomorphism but it does not map null sets to null sets. $\endgroup$– TaQCommented Oct 12, 2021 at 15:46
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$\begingroup$ @TaQ: An arbitrary continuous map does not preserve or reflect sets of measure 0. Rather, the specific map that is constructed in the answer does preserve and reflect sets of measure 0. This follows from the description of sets of measure 0 in [0,1] and in $\{0,1\}^N$, for example. $\endgroup$ Commented Oct 12, 2021 at 17:05
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$\begingroup$ "the description of sets of measure $0$" — that is? Can you be more specific? $\endgroup$– TaQCommented Oct 12, 2021 at 18:25
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$\begingroup$ @TaQ: Once you remove 1 from [0,1] and sequences $P$ ending in infinitely many 1's from $\{0,1\}^N$, sending a real number r∈[0,1] to its binary expansion establishes a measure-preserving bijection from [0,1) to $\{0,1\}^N∖P$, so in particular, set of measure 0 are the same. To see that the bijection is measure preserving, it suffices to observe that the Borel algebra is generated by sets of sequences with a fixed binary digit in some position $k$. In [0,1], this set corresponds to the union of $2^{k-1}$ half-open intervals of length $2^{-k}$, so the total length is $1/2$ in both cases. $\endgroup$ Commented Oct 13, 2021 at 0:09
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$\begingroup$ Thank you for the clarification. So the idea is that we have between some subsets of full measure a bijection that preserves the measures of sets in some sets generating the $\sigma$-algebras whence it follows that the function preserves measures of all measurable subsets. $\endgroup$– TaQCommented Oct 13, 2021 at 21:32