TITLE: Which complete Boolean algebras arise as the algebras of projections of commutative von Neumann algebras?
QUESTION [21 upvotes]: Projections in an arbitrary commutative von Neumann algebra form a complete Boolean algebra.
Moreover, a morphism of commutative von Neumann algebras induces
a continuous morphism of the corresponding complete Boolean algebras.
Thus we have a fully faithful functor F from the category of commutative von Neumann algebras
to the category of complete Boolean algebras and their continuous morphisms.
The category of complete Boolean algebras and their continuous morphisms is a full subcategory
of the opposite category of the category of locales.
Thus the functor F can be seen as implementing the Gelfand-Neumark duality
for commutative von Neumann algebras.
However, to obtain a satisfactory statement of the duality we still need to characterize
in topological terms objects in the essential image of F,
which we call measurable spaces (or locales, think of a localic version of point-set measurable spaces).
What additional topological conditions do we need to impose on a complete Boolean algebra
to ensure that it is the algebra of projections of some von Neumann algebra,
i.e., a measurable space?
It is relatively easy to pin down non-topological conditions.
For example, a complete Boolean algebra comes from a von Neumann algebra
if and only if it admits sufficiently many normal positive measures.
The reason for requiring additional conditions to be topological
is that the resulting definition of a measurable space should be easy
to relate to other parts of general topology.
For example, consider the forgetful functor that sends
a commutative von Neumann algebra to its underlying C*-algebra.
Applying the Gelfand-Neumark duality to both sides we obtain
the forgetful functor from the category of measurable spaces
to the category of compact regular locales
(or compact Hausdorff spaces, if we have the axiom of choice).
A topological definition of a measurable space should allow
for an explicit description of this forgetful functor in terms of open sets.
Other potential applications include functors that send a locale (or a topological space)
to its underlying measurable space, or a smooth manifold to its underlying measurable space.
More speculatively, one could use this definition to replace
ad hoc techniques of classical point-set measure theory with standard tools of general topology.

REPLY [3 votes]: A pair $(\mathcal{A}, \mu)$ of a $\sigma$-complete boolean algebra $\mathcal{A}$ and a functional $\mu : \mathcal{A} \to [0, \infty]$ is called a measure algebra if $\mu$ is strictly positive and countably additive on disjoint sequences. A measure algebra is semifinite if whenever $\mu(a) = \infty$ there exists a $b < a$ such that $0 < \mu(b) < \infty$. A measure algebra is localizable if it is complete and semifinite.
A measure algebra can be constructed from a measure space by taking the boolean algebra of equivalence classes of measurable sets modulo the null sets. In the reverse direction, by the Loomis-Sikorski Theorem, every $\sigma$-complete boolean algebra is isomorphic to the quotient of the $\sigma$-algebra $\{ A \bigtriangleup B : A \text{ clopen }, B \text{ meager } \}$ by the $\sigma$-ideal of meager sets. An ordinary measure can be defined on the quotient in the natural way, and the concrete measure algebra of the resulting measure space is isomorphic to the original measure algebra.
The preceding construction can also be used to show that the localizable measure algebras are precisely the Boolean algebras of projections of commutative von Neumann algebras. Note the similarities between the definitions of a localizable measure algebra and a normal semifinite weight on a von Neumann algebra.
Due to the semifiniteness of the measure, the problem of characterizing the Boolean algebras that are measure algebras can be reduced to the case of finite measure. Call a $\sigma$-complete boolean algebra $\mathcal{A}$ finitely measurable when there exists a functional $\mu : \mathcal{A} \to [0, \infty)$ making it a measure algebra. Then a complete boolean algebra $\mathcal{A}$ has a functional $\mu : \mathcal{A} \to [0, \infty]$ making it a localizable measure algebra precisely when the set $\{ a \in \mathcal{A} : \mathcal{A}_a \text{ is finitely measurable } \}$ is order-dense in $\mathcal{A}$, where $\mathcal{A}_a$ is the principal ideal generated by $a$.
Unfortunately, even in the finite measure case there is no great solution to this problem. There’s a characterization by Kelley that reduces it to the existence of a strictly positive finitely additive measure and a combinatorial condition (what he calls weakly countably distributive). He also characterizes the existence of a finitely additive measure in terms of intersection/covering numbers. Gaifman wrote a survey paper on this problem, and Jech proved a game-theoretic characterization.
A good reference for most of the facts mentioned about measure algebras is volume 3 of Fremlin’s tome Measure Theory, particularly chapters 32 and 39.