In proving that the product of two $\sigma-$algebras, say $\mathcal{A}$ and $\mathcal{B}$, is not necessarily a $\sigma-$algebra, i.e. $\mathcal{A}\times \mathcal{B}$ is not a $\sigma-$algebra. And we prove this by getting to a statement of the form:
$[0,1]\times [0,1] \cup [2,3]\times [2,3]\notin \mathcal{A}\times \mathcal{B}$
Why is it so clear that the union of the product sets cannot lie in the product of the $\sigma-$algebras?
Is it because if $[0,1]\times [0,1] \cup [2,3]\times [2,3]\in \mathcal{A}\times \mathcal{B}$, then by definition of the product $(0,2)\in[0,1]\times [0,1] \cup [2,3]\times [2,3]$ which obviously is not true? Or am I using the wrong reason?
Best Answer
Do you agree that if $(x, y) \in A\times B$ and $(z, w) \in A \times B$ we will then have $(x, w)\in A\times B$? Can you think of two points for which this implication doesn't hold for the set in question?