Counterexample on product of measurable spaces Let $(X_\alpha, A_\alpha)_{\alpha \in \mathcal{A}}$ be a family of measurable spaces, where $\mathcal{A}$ is an index set, and $D_\alpha \in A_\alpha, \forall \alpha \in \mathcal{A}$. It's possible to show that if $\mathcal{A}$ is countable, then $\prod_{\alpha \in \mathcal{A}} D_\alpha \in \prod_{\alpha \in \mathcal{A}} A_\alpha$, as we can write $$\prod_{\alpha \in \mathcal{A}} D_\alpha = \bigcap_{\alpha \in \mathcal{A}} \pi_\alpha^{-1}(D_\alpha)$$ where $\pi_\alpha$ are projections, and, as the right-hand side is a countable intersection of measurable sets, we have the result.
Is it the case if $\mathcal{A}$ is uncountable? If not, would there be a simple counterexample?
 A: This is not true if $\mathcal{A}$ is uncountable; almost any example you might try to write down is a counterexample.  If $\mathcal{B}\subseteq\mathcal{A}$, let $\pi_\mathcal{B}$ be the projection map $\prod_{\alpha\in\mathcal{A}} A_\alpha\to \prod_{\alpha\in\mathcal{B}}A_\alpha$.  Now let $C$ denote the set of all subsets of $\prod_{\alpha\in A} A_\alpha$ which are of the form $\pi_\mathcal{B}^{-1}(X)$ for some countable subset $\mathcal{B}\subset\mathcal{A}$ and some subset $X\subseteq\prod_{\alpha\in\mathcal{B}}A_\alpha$.  Then it is straightforward to verify that $C$ is a $\sigma$-algebra.  Furthermore, the definition of the measurable subsets of $\prod_{\alpha\in\mathcal{A}}A_\alpha$ is as the $\sigma$-algebra generated by certain sets, and all of these sets are in $C$.  Thus every measurable set is in $C$.
On the other hand, suppose $\mathcal{A}$ is uncountable and for every $\alpha$, $D_\alpha$ is not empty and is not all of $X_\alpha$.  Then clearly $\prod_{\alpha\in\mathcal{A}} D_\alpha$ is not in $C$.  Thus $\prod_{\alpha\in\mathcal{A}} D_\alpha$ is not measurable.
A: For each $\alpha$ let $X_\alpha=\{0,1\}$ and $A_\alpha=\wp(X_\alpha)$. Let $X=\prod_\alpha X_\alpha$, and let $A$ be the product $\sigma$-algebra. If $|\mathscr{A}|=2^\omega$, an easy transfinite induction shows that $|A|=\omega_1\cdot 2^\omega=2^\omega$. However, there are $2^{2^\omega}>2^\omega$ sets of the form $\prod_\alpha D_\alpha$, so they cannot all be in $A$.
In fact it's clear that either all singletons or none are in $A$, and since there are $2^{2^\omega}$ singletons, no singleton is in $A$.
