Hi all,

I've just come up with something that's relevant to the question I asked here. I'm bumping the thread partly in case anyone else cares, and partly in case (as is more likely) I've made an error and someone can point it out. Anyway: I believe I can prove that the $\sigma$-algebra generated by a Boolean algebra $A$ (in the sense of Todd's answer, i.e. the image of a left adjoint to the forgetful functor from $\sigma$-algebras to Boolean algebras) has a rather natural representation, namely as the $\sigma$-field generated by the double dual of $A$, i.e. the smallest $\sigma$-field containing all the clopen subsets of the dual space of $A$. Here is the proof:

Let $A$ be a Boolean algebra, let $A^\star$ be its dual Boolean space and $A^{\star \star}$ the dual algebra of its dual space, i.e. the set of clopen subsets of $A^\star$. Let $\bar{A}$ be the $\sigma$-algebra of Baire sets in $A^\star$, i.e. the $\sigma$-field of subsets of $A^\star$ generated by $A^{\star \star}$. Let $\alpha: A \cong A^{\star \star}$ be the canonical isomorphism, and let $\eta: A \to \bar{A}$ be the composition of $\alpha$ with the inclusion.

Suppose given a $\sigma$-algebra $B$ and a homomorphism (of Boolean algebras) $h: A \to B$. Define $B^\star$, $B^{\star \star}$, $\bar{B}$ and $\beta: B \cong B^{\star \star}$ as before. By Theorem 41, p. 376 of [1], $B^\star$ is a $\sigma$-space, i.e. the closure of every open Baire set is open. By Theorem 42, p. 381, there is a $\sigma$-homomorphism $\phi: \bar{B} \to B^{\star \star}$ such that $\phi$ maps ever clopen set to itself.

$\beta h \alpha^{-1}$ is a homomorphism $A^{\star \star} \to B^{\star \star}$, so by duality there is a unique continuous function $f: B^\star \to A^\star$ such that $f^{-1} P = \beta h \alpha^{-1} (P)$ for every $P \in A^{\star \star}$. It is easy to see that $f^{-1} S$ is a Baire set whenever $S$ is, so define

$f^\star : \bar{A} \to \bar{B}$;
$S \mapsto f^{-1} S$.

$f^\star$ is clearly a $\sigma$-homomorphism. Let

$\bar{h} \equiv \beta^{-1} \phi f^\star: \bar{A} \to B$.

Then one may check, using the defining property of $f$ and the fact that $\phi$ maps clopen sets to themselves, that $\bar{h} \eta = h$. The uniqueness of $\bar{h}$ with this property follows from the fact that the range of $\eta$ generates $\bar{A}$.

So there's the alleged proof; I can't see anything wrong with it but the result strikes me as being "too good to be true", and if it is true then I'm surprised I didn't see any reference to it online before I started this thread. So I'll be grateful if anyone can spot a mistake.

[1] Steven Givant and Paul Halmos, *Introduction to Boolean Algebras*, Springer 2009