The topological duals of spaces of finite measures

Question

In volume 1 of "Linear Operators", Dunford and Schwartz say that (footnote F1, page 374)

"No completely satisfactory representation for the conjugate space of $ba(S, \Sigma)$, $ca(S, \Sigma)$ or $rca(S, \Sigma)$ [...] seems to be known.

In the above, $ba(S, \Sigma)$ is the space of finite, finitely additive measures on an algebra $\Sigma$, $ca(S, \Sigma)$ is the space of finite, countably additive measures on a $\sigma$-algebra $\Sigma$, and $rca(S, \Sigma)$ is the space of finite, regular, countably additive measures on a $\sigma$-algebra $\Sigma$ (all of them understood as complex measures). One may also consider the space $rba(S, \Sigma)$ with the obvious meaning.

This was in 1957. Today, 60 years later, has our understanding improved?

(For my work, I am searching for a space $X$ with a known dual $X^*$, such that $X^*$ should contain $C(S)$, the space of bounded continuous functions on $S$. For the moment, I am working with $X = L^1 (S)$, but the elements of $X^* = L^\infty (S)$ are not regular enough for my needs. I was hoping that the dual of a space of measures might be closer to $C(S)$.)

Answer 1

The bidual of $C(S)$ can be described as the direct product of von Neumann algebras generated by the cyclic representations associated with positive elements of $C(S)^*$. For a positive finite measure $\mu$ the von Neumann algebra generated by the representation associated with $\mu$ is just $L^\infty(S, \mu)$, so if the elements of that space aren't regular enough for you, then an indescribable direct product of them probably won't be any better.

Because $C(S)^{**}$ is a commutative C$^*$-algebra, it is $*$-isomorphic to $C(K)$ for some compact $K$, but as far as I know, no description of the space $K$ has ever occurred. There is a new book coming out on the subject of spaces of continuous functions as dual spaces, which will probably be a lot easier to read than the other literature on the subject.