For $\mathcal{A}$ and $\mathcal{B}$ sets of sets, consider $$A\boxtimes B:= \left\{A\times B\colon (A,B)\in \mathcal{A}\times\mathcal{B}\right\}.$$
(Note that this is not the cartesian product of $\mathcal{A}$ and $\mathcal{B}$. Is there a name for this product of sets?)
Now, for $\sigma$-algebras $\mathcal{A}$ and $\mathcal{B}$, $\mathcal{A}\boxtimes\mathcal{B}$ is, in general, no $\sigma$-algebra. My book states this, using as an example for some $A\subseteq X$ the $\sigma$-algebra $\mathcal{A}:=\{\emptyset, A, A^c, X\}$, but I don't get it. Why (and when) would $$\mathcal{A}\boxtimes\mathcal{A}=\{\emptyset, A\times A, A\times A^c, A\times X, A^c\times A, A^c\times A^c, A^c\times X, X\times A, X\times A^c, X\times X\}$$ be no $\sigma$-Algebra? (I did read this, that is I get the statement in general)
If $A$ is neither $\emptyset$ nor $X$, then $$A \times A^c \cup A^c \times A$$ is not an element of $\mathcal{A} \boxtimes \mathcal{A}$.