Definition talk:Product Sigma-Algebra/Binary Case

I claim that $\sigma \left( \Sigma_{1} \times \Sigma_{2} \right)$ is the incorrect definition of $\Sigma_{1} \otimes \Sigma_{2}$. My complaint is that we are trying to define $\Sigma_{1} \otimes \Sigma_{2}$ to be a $\sigma$-algebra on the set $X_{1} \times X_{2}$, so our generating sets should be subsets of $X_{1} \times X_{2}$. However, $\Sigma_{1} \times \Sigma_{2}$ is not a set of subsets of $X_{1} \times X_{2}$. Rather, it is a set containing ordered pairs $\left( S_{1}, S_{2} \right)$ where $S_{1} \in \Sigma_{1}$ and $S_{2} \in \Sigma_{2}$. The ordered pair $\left( S_{1}, S_{2} \right)$ is not the same as the Cartesian product $S_{1} \times S_{2} \subseteq X_{1} \times X_{2}$.

It is worth mentioning that the given source by Schilling defines $\Sigma_{1} \otimes \Sigma_{2}$ by $\sigma \left( \Sigma_{1} \times \Sigma_{2} \right)$, but only after a paragraph where he clarifies $\Sigma_{1} \times \Sigma_{2}$ to denote not the literal Cartesian product, but rather the collection of Cartesian products $\set{S_{1} \times S_{2} : S_{1} \in \Sigma_{1} \wedge S_{2} \in \Sigma_{2}}$. I'm OK with this abuse of notation, but only if it is explicitly stated. The book referenced in this article mentions it, this article itself (and other ProofWiki articles) does not.

In case it was not clear from the above, I claim the correct definition of $\Sigma_{1} \otimes \Sigma_{2}$ should be $\sigma \left( \set{S_{1} \times S_{2} : S_{1} \in \Sigma_{1} \wedge S_{2} \in \Sigma_{2}} \right)$.

I made this change, but it was reverted by User:Lord Farin. I do not believe my original edit comment made my complaint fully understood; hopefully this longer post is able to explain the issue I have with this subtlety. --Jumpythehat (talk) 01:26, 20 November 2019 (EST)


 * I did revert because I misinterpreted your comment to mean that $\Sigma_1\times\Sigma_2$ does not make sense; obviously I was mistaken, and your longer explanation makes clear that the regular "set-wise" abuse of notation does not make sense in the same way that $f[X]$ normally generalises $f(x)$. (Hence the $\otimes$ instead of the $\times$, I now realise.)
 * Thanks for taking the time to notice and for the patience to explain it! &mdash; Lord_Farin (talk) 15:13, 20 November 2019 (EST)