# Sigma-distributivity of the algebra of Baire sets modulo meagre sets

Let $$A$$ be the free $$\sigma$$-algebra with $$\omega_1$$ free $$\sigma$$-generators, $$X$$ its Stone space, and $$Ba(X)/M$$ the algebra of Baire subsets of $$X$$ modulo meagre sets. Then $$A$$ is $$\sigma$$-isomorphic to $$Ba(X)/M$$ by the Loomis-Sikorski theorem.

Let $$2^{\omega_1}$$ be the Cantor cube of weight $$\omega_1$$ (i.e. the Stone space of the free Boolean algebra with $$\omega_1$$ free generators) and $$Ba(2^{\omega_1})$$ the $$\sigma$$-field of Baire subsets of $$2^{\omega_1}$$. Then it can be shown that $$A$$ is also $$\sigma$$-isomorphic to $$Ba(2^{\omega_1})$$.

Now, since $$Ba(2^{\omega_1})$$ is a $$\sigma$$-field of sets, it is $$\sigma$$-distributive. Hence, since $$A$$ is $$\sigma$$-isomorphic to $$Ba(2^{\omega_1})$$, $$A$$ is an example of an atomless $$\sigma$$-distributive $$\sigma$$-algebra.

My question: Since $$A$$ is also $$\sigma$$-isomorphic to $$Ba(X)/M$$, is $$Ba(X)/M$$ also $$\sigma$$-distributive? (I have a doubt.)

• Baire subsets are the $\sigma$-algebra generated by the clopen subsets, correct? The first $B(X)$ should also be $\text{Ba}(X)$ ? Commented Aug 3, 2020 at 17:42
• @HennoBrandsma Yes, Baire sets are elements of the sigma-algebra generated by the clopen sets. And also, yes, $B(X)$ should be read $Ba(X)$. I corrected. Commented Aug 3, 2020 at 17:52

Yes, it should be clear that two if two Boolean Algebras are isomorphic by an isomorphism that preserves all countable sups (i.e. $$\sigma$$-isomorphic), and one is $$\sigma$$-distributive, then the other one is too.