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

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

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

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

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

the question reduces to ask whether

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

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

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.