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



> 
> $\bigwedge\_{m\in\omega}\bigvee\_{n\in\omega} A\_{m,n}=\bigvee\_{\alpha\in\omega^\omega}\bigwedge\_{m\in\omega}A\_{m,\alpha(m)}$
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> 
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where $\omega^\omega$ denotes the set of all mappings of $\omega$ into $\omega$?


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



> 
> $\bigwedge\_{m\in\omega}\bigvee\_{n\in\omega} A\_{m,n}\geq\bigvee\_{\alpha\in\omega^\omega}\bigwedge\_{m\in\omega}A\_{m,\alpha(m)}$
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> 
> 


the question reduces to ask whether



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> $\bigwedge\_{m\in\omega}\bigvee\_{n\in\omega} A\_{m,n}\leq\bigvee\_{\alpha\in\omega^\omega}\bigwedge\_{m\in\omega}A\_{m,\alpha(m)}$
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holds in the free $\sigma$-algebra on $\omega\_1$ free $\sigma$-generators.



Infinite distributive laws in atomless free sigma-algebra
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 $\Sigma$.
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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 $$ \bigcap\_{i \in I}\bigcup\_{j \in J} a\_{i,j} = \bigcup\_{f \in J^I} \bigcap\_{i \in I} a\_{i,f(i)}
> $$
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> 


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 $\bigwedge\_{m \in \omega}\bigvee\_{n \in \omega} a\_{m,n}$. 


