I asked the question on MSE.
The answer I got, however, seems disputed. I just thought that someone here could answer the question for sure. Many thanks.
I asked the question on MSE.
The answer I got, however, seems disputed. I just thought that someone here could answer the question for sure. Many thanks.
You got a wrong answer on Math Stackexchange from Daron. The free Boolean algebra on countably many generators is the Boolean algebra of clopens of $2^\omega$ (topologized with the product topology), and the free $\sigma$-algebra on countably many generators is the Baire $\sigma$-algebra of $2^\omega$, which, as $2^\omega$ is metrizable, is the same as the Borel $\sigma$-algebra. A good source for this is Halmos's book Lectures on Boolean Algebras.
To see that Daron's claim is wrong, observe that $\mathcal{P}(\mathbb{N})$ is a complete Boolean algebra, but the Borel $\sigma$-algebra of $2^\omega$ is not, because the least upper bound of the family of all singletons contained in a non-Borel subset of $2^\omega$ does not exist. The algebras $\mathcal{P}(\mathbb{N})$ and $\mathrm{Borel}(2^\omega)$ cannot be isomorphic because any Boolean algebra isomorphic to a complete Boolean algebra is complete.
The "$\sigma$-Stone space" of $\mathrm{Borel}(2^\omega)$, i.e. the set of ultrafilters closed under countable intersection, or equivalently the $\{0,1\}$-valued countably additive measures, does have a nice description, as it is isomorphic to $2^\omega$ itself.
Unfortunately, the ordinary Stone space of $\mathrm{Borel}(2^\omega)$ does not have a nice description, except tautologous rephrasings of the usual one, such as saying it is the space of $\{0,1\}$-valued finitely-additive Borel measures on $2^\omega$. I think this follows from the fact that there exist models of ZF where the axiom of choice fails and there are no nonprincipal ultrafilters on $\omega$. Any ultrafilter on $\mathrm{Borel}(2^\omega)$ that is not closed under countable intersections can be used to define a non-principal ultrafilter on $\omega$.