Any example of measurable spaces where the measurable rectangles form an algebra on the product space?

Question

As far as I know, for any $(\Omega_{i},\sigma_{i})$ i=1,2, $\textit{A}=\{A_{1}\times A_{2}: A_{1}\in\sigma_{1}, A_{2}\in\sigma_{2}\} $ is not an algebra in general on $\Omega_{1}\times\Omega_{2}$ in general because it is not closed for finite unions nor complementaries. I'm trying to think about an example of particular spaces where this is true, but the only one that appears to work is choosing the trivial $\sigma$-algebra $\sigma_{i}=\{\emptyset,\Omega_{i}\}$ for one of the spaces. Any other idea on how to proof A is an algebra in other spaces, because when I try with $\sigma$-álgebra with more than one set different from the empty one, I don't find a way to write the union of measurable rectangles as a measurable rectangle. Thanks.

Answer 1

There is no example where there exist sets $A$ and $B$ in $\Omega_1$ and $\Omega_2$ respectively which are non-empty and also not equal to the whole space. Suppose $(A \times B)^{c} =C\times D$ for some $C$ and $D$. Then $C=\Omega_1$ and $D=\Omega_2$ : for any $x \in \Omega_1$ we can pick $y \notin B$ to see that $(x,y) \in (A \times B)^{c} =C\times D$ and this implies $x \in C$. Similarly $D=\Omega_2$. But now if we pick $x \in A$ and $y \in B$ we get $(x,y) \in C \times D$ but $(x,y) \notin (A \times B)^{c}$ a contradiction.