When does a $W^*$-algebra have a standard Borel spectrum?

Question

EDIT: André Henriques has commented below that the correct separability condition is not weak-* separability as I have written below, but separability of the predual.

This post came out a bit long, but if you're familiar with the topic, you can probably just skim through most of it - I've put the questions in bold. Some terminology:

1. A (commutative, as it shall always be in this post) $W^* $-algebra is a $C^* $-algebra which has a pre-dual as a Banach space. It is well known (see for example Takesaki's Theory of Operator Algebras I) that such algebras can be equivalently characterized as Von-Neumann subalgebras of the algebra of operators on a Hilbert space, or as spaces of the form $L^\infty \left (X, \mu \right) $ where $X$ is a locally compact space and $\mu$ a Radon measure.

2. By a standard Borel space I mean a measure (or sometimes just measurable) space which is Borel isomorphic to a complete separable metric space.

I am studying group actions on $W^* $-algebras, and I am interested in particular in the question:

When can $ \left (X, \mu \right) $ above be chosen to be a standard Borel space?

The question is important to me because if the answer is what I expect it to be, I have a very simple way of constructing concrete actions on compact measure spaces from given actions on the associated $W^* $-algebra. (I can say more about the motivation, but I don't want to burden this post with too many details.)

My hypothesis is that the answer to my question is: Exactly when $A$ is separable in the weak-* topology . Certainly the $L^\infty $ of a standard Borel space is weak-* separable: the Borel $\sigma$-algebra on a standard Borel space is countably generated, so rational linear combinations of the associated indicator functions are dense in $L^1$, making it norm-separable and hence its dual weak-* separable.

Here is how I tried to prove this: assume that $A$ is a weak-* separable $W^* $-algebra, and let $B$ be a norm-separable, norm-closed, weak-* dense sub-$C^* $-algebra (just take the norm-closed subspace generated by some weak-* dense countable subset). I denote $A$ and $B$'s Gelfand spectra (i.e., the spaces of multiplicative linear functionals on these algebras) by $X_A$ and $X_B$. Since $A \simeq C(X_A)$ and $B \simeq C(X_B)$, I know that $X_B$ is a complete separable metric space. $X$ above can be constructed as an open dense subset of $X_A$ on which $\mu$ is supported. I want to prove that there is a null set $X_A ^0$ such that $X_A-X_A ^0 $ is Borel isomorphic to $X_B$. If I prove this, I win, because removing a null subset gives an isomorphic $L^\infty$.

So basically, what I want to prove is:

Given a $W^* $-algebra $A$ and a weak-* dense subalgebra $B$, there exists a null subset $X_A ^0 $ such that $X_A-X_A ^0$ is Borel isomorphic to $X_B$ .

It may be that I also have to remove a null subset of $X_B$ - that's just as good for me, although I think it can be avoided.

To prove the last statement, I tried going through the following: I have a natural map from $X_A$ to $X_B$ given by restricting a multiplicative functional to B. This map is definitely continuous (the topology on the Gelfand spectrum is given by pointwise convergence) and, while not a completely trivial fact, it is well known that it is onto (a multiplicative linear functional on a $C^* $-subalgebra can always be extended to the entire algebra; see for example Kaniuth's A Course in Commutative Banach Algebras , theorem 4.2.17). There's no reason, of course, to think that the restriction is one to one: extensions of functionals are in general highly non-unique. However, I believe that by using weak-* density and taking away a null subset of $X_A$, it can be made onto. That would already make the map a Borel isomorphism. The following observation may or may not be helpful: asking if there is a conull subset of $X_A$ on which the restriction is one to one is the same as asking if there is a conull subset on which elements of $B$, viewed as functions, separate points.

Well, I could say more, but this post is already exceedingly long. I have been thinking and looking for information on these questions for quite some time now, so I would warmly welcome any suggestions or comments.

To recap, my questions are:

1. When can $(X, \mu)$ be chosen to be a standard Borel space?
2. Is it true that 1 is equivalent to $A$ being weak-* separable?
3. Is it true that the spectrum of a $W^* $-algebra is always Borel isomorphic to the spectrum of a weak-* dense subalgebra?

Answer 1

The category of commutative von Neumann algebras is contravariantly equivalent to the category of measurable spaces. Assuming the axiom of choice, isomorphism classes of objects in the above two categories that satisfy a certain countability property are classified by pairs (m,n) of cardinal numbers, with the additional restriction that either n=0 or n is infinite. (Spaces that do not satisfy the countability property cannot give separable von Neumann algebras.)

To a pair (m,n) the above correspondence assigns the measurable space given by the coproduct of m isolated points (atoms) and n copies of R. Equivalently, one can take the product of m copies of the algebra C of complex numbers and n copies of the von Neumann algebra of bounded measurable functions on the measurable space R.

Separability of the predual is equivalent to m and n being at most countable, which also describes standard Borel spaces.

References for the classification of isomorphism classes of measurable spaces:

1) Theorem 3.4 in Irving Segal. Equivalences of measure spaces. American Journal of Mathematics 73:2 (1951), 275–313. http://dx.doi.org/10.2307/2372178.

2) Theorem 1 in Dorothy Maharam. On homogeneous measure algebras. Proceedings of the National Academy of Sciences 28:3 (1942), 108–111. http://dx.doi.org/10.1073/pnas.28.3.108