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

Question

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.)

Answer 1

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.