Suppose $\mathbb{B}$ is the boolean algebra formed by the borel subsets of $\omega^{\omega}$ modulo the meager sets (also called the category boolean algebra), that is, if $B, C$ are borel and $B \Delta C$ is meager, then $B \sim C$. A very minor step towards the construction of the solovay model is to show that $\mathbb{B}$ is complete for when we force over it. But I'm unable to prove this apparently simple step.
Since $\{[N_x]: x \in \omega^\omega\} \subseteq \mathbb{B}$ is countable + dense, we can conlude that $\mathbb{B}$ has the ccc. I think we're supposed to leverage this property to show that $\mathbb{B}$ is complete.
Suppose $\kappa$ a cardinal, and $[A_{\alpha}] \in \mathbb{B}$ for all $\alpha \in \kappa$, I would like to show that $\sum [A_{\alpha}]$ exists.
First if $\kappa = \omega_1$ this is easy: Let $B_\alpha = A_\alpha \setminus \cup_{\beta < \alpha}A_\beta$ be the disjointifications. So since $\alpha < \omega_1$ these $B_\alpha$ are still Borel (but this may not hold if $\alpha > \omega_1$, and this step is where I struggle if $\kappa > \omega_1$). Then by ccc there should be some $\alpha_0$ beyond which for all $\alpha \geq \alpha_0$, $[B_\alpha] = [\emptyset]$, and then $\sum [A_{\alpha}] = \sum [B_\alpha] = \sum_{\alpha < \alpha_0} [B_\alpha] \in \mathbb{B}$.
So now what if $\kappa > \omega_1$?
I have some ideas for regular or cardinals with uncountable cofinality, but I'm not sure about when $cf(\kappa) = \omega$.
The obvious idea is to use that for every borel set $B$ there is an open set $U$ such that $[B] = [U]$ and use the fact that open sets are closed under arbitrary unions, but I'm not sure that's enough.