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.
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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 $\omega$.
answered Sep 11, 2018 at 0:48
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$\begingroup$ Thank you Robert for your reply. May I ask you what you mean exactly by the "$\sigma$-Stone space" of the free sigma-algebra on $\omega$ generators; I am not familiar with this concept. Is it what Halmos calls a $\sigma$-space (i.e a space in which the closure of open Borel sets is open)? $\endgroup$– BeginnerCommented Sep 12, 2018 at 21:05
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$\begingroup$ Halmos defines a $\sigma$-space to be a Stone space (i.e. a compact Hausdorff zero-dimensional space) in which the closure of every open Baire set is open. A Boolean algebra is $\sigma$-complete iff its Stone space is a $\sigma$-space. Another name for this property is "basically disconnected". $\endgroup$ Commented Sep 13, 2018 at 1:07
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$\begingroup$ @puzzled By the "$\sigma$-Stone space" of a $\sigma$-complete Boolean algebra $A$, I mean the set of $\sigma$-ultrafilters $X$ equipped with the natural $\sigma$-algebra defined on them by $A$ (as a measurable space without any topology). This is something that can be found in Sikorski's Boolean Algebras, section 24 (by specializing the cardinal $\mathfrak{m}$ to $\aleph_0$). $\endgroup$ Commented Sep 13, 2018 at 1:31
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$\begingroup$ @puzzled If you don't mind self-promotion, I describe the construction of this space in Section III of: people.cs.aau.dk/~furber/papers/unrestrictedstone.pdf using the notation $\mathcal{U}^\sigma(A)$ where $A$ is a $\sigma$-complete Boolean algebra (proofs are in the appendix). $\endgroup$ Commented Sep 13, 2018 at 1:33
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$\begingroup$ So, if I understand you correctly, @Robert, $\sigma$-Stone spaces are not $\sigma$-spaces (they are not extremally disconnected). is there any chance I can force it? $\endgroup$– BeginnerCommented Sep 14, 2018 at 16:18