I am studying von Neumann algebras. In the wiki article [abelian von Neumann algebras](https://en.wikipedia.org/wiki/Abelian_von_Neumann_algebra), 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?



Questions about Maharam's classification theorem
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](https://mathoverflow.net/questions/23408/reference-for-the-gelfand-duality-theorem-for-commutative-von-neumann-algebras/360088#360088).


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.


