Infinite distributive laws in atomless free sigma-algebra

Question

Let $\frak{A}$ be the free $\sigma$-algebra on $\omega_1$ free $\sigma$-generators. Then $\frak{A}$ is not completely distributive because it is atomless. However, is it $\omega$-distributive in the sense that

$\bigwedge_{m\in\omega}\bigvee_{n\in\omega} A_{m,n}=\bigvee_{\alpha\in\omega^\omega}\bigwedge_{m\in\omega}A_{m,\alpha(m)}$

where $\omega^\omega$ denotes the set of all mappings of $\omega$ into $\omega$?

Of course, since in every Boolean algebra, we always have

$\bigwedge_{m\in\omega}\bigvee_{n\in\omega} A_{m,n}\geq\bigvee_{\alpha\in\omega^\omega}\bigwedge_{m\in\omega}A_{m,\alpha(m)}$

the question reduces to ask whether

$\bigwedge_{m\in\omega}\bigvee_{n\in\omega} A_{m,n}\leq\bigvee_{\alpha\in\omega^\omega}\bigwedge_{m\in\omega}A_{m,\alpha(m)}$

holds in the free $\sigma$-algebra on $\omega_1$ free $\sigma$-generators.

Answer 1

This holds because $\mathfrak{A}$ is a concrete $\sigma$-algebra, being the Baire $\sigma$-algebra of $2^{\omega_1}$. In fact, the cardinality of $\omega_1$ plays no role whatsoever and $\omega_1$ could be replaced by any set.

It is easy to prove using a couple of lemmas.

Lemma 1: Let $\Sigma$ be a concrete $\sigma$-algebra on a set $X$, and $(S_i)_{i \in I}$ a family of subsets (not necessarily countable). If $\bigcup_{i \in I} S_i$ is in $\Sigma$, then $\bigcup_{i \in I}S_i = \bigvee_{i \in I}S_i$ in $\Sigma$.

This is easily proved using the definition of supremum. The second lemma is:

Lemma 2: Let $X$ be a set, $(S_{i,j})_{i \in I,j \in J}$ a family of subsets of $X$. Then $$ \bigcap_{i \in I}\bigcup_{j \in J} a_{i,j} = \bigcup_{f \in J^I} \bigcap_{i \in I} a_{i,f(i)} $$

This is proved by showing that an element of $X$ is in the left hand side iff it is in the right hand side (using the axiom of choice to construct a suitable function at the right moment).

The proof that $\mathfrak{A}$ is $\omega$-distributive then goes like this: $$ \bigwedge_{m \in \omega} \bigvee_{n \in \omega} a_{m,n} = \bigcap_{m \in \omega} \bigcup_{n \in \omega} a_{m,n} = \bigcup_{f \in \omega^\omega}\bigcap_{m \in \omega} a_{m,f(m)} = \bigvee_{f \in \omega^\omega} \bigwedge_{m \in \omega} a_{m,f(m)} $$

One last thing - as $\mathfrak{A}$ is only $\sigma$-complete, the statement of $\omega$-distributivity is actually that if $(a_{m,n})_{m,n \in \omega}$ is a family of elements of $\mathfrak{A}$, then $\bigvee_{f \in \omega^\omega}\bigwedge_{m \in \omega}a_{m,f(m)}$ exists and is equal to $\bigwedge_{m \in \omega}\bigvee_{n \in \omega} a_{m,n}$.