Finite disjoint unions of measurable rectangles form an algebra generating the product sigma-algebra

By [L1] and [A1], the intersection of two measurable rectangles is again a measurable rectangle, and the complement of a measurable rectangle is a finite union of measurable rectangles.

Let $\mathcal{R}$ be the family of finite disjoint unions of measurable rectangles. It contains $\mathrm{\varnothing }$ and $X\times Y$. If $E,F\in \mathcal{R}$, then [A2] disjointifies the finite union $E\cup F$ into finitely many pairwise disjoint measurable rectangles, so $E\cup F\in \mathcal{R}$. Likewise $E\setminus F=E\cap F^c$ belongs to $\mathcal{R}$ by step 1.1. Thus $\mathcal{R}$ is an algebra.

Every measurable rectangle belongs to $\mathcal{R}$, so [L2] gives $\mathcal{A}\otimes \mathcal{B}=\sigma (\text{measurable rectangles})\subseteq \sigma (\mathcal{R})\mathrm{.}$ The reverse inclusion holds because every member of $\mathcal{R}$ is a finite union of measurable rectangles and hence lies in $\mathcal{A}\otimes \mathcal{B}$. Therefore $\sigma (\mathcal{R})=\mathcal{A}\otimes \mathcal{B}$, and $\mathcal{R}$ is an algebra generating the product sigma-algebra.