Let $X = X_1 \times X_2$ and suppose that $(X_1, \mathcal M_1, \mu_1)$ and $(X_2, \mathcal M_2, \mu_2)$ are two measure spaces. Consider the set of all rectangles, i.e., sets of the form $A \times B$, where $A \in \mathcal M_1$ and $B \in \mathcal M_2$. I read that the union of two rectangles is the disjoint union of $6$ rectangles. It is clearly true for $X = \mathbb R \times \mathbb R$. But how to see this general claim is true, please?
Update: The answer provided below seems to suggest that one needs at most $3$ disjoint rectangles. My question is when do we need more than $3$, please? Thank you!
The quick solution would be nine rectangles: multiply any of $A_1\setminus A_2$, $A_2\setminus A_1$, $A_1\cap A_2$ with any of $B_1\setminus B_2$, $B_2\setminus B_1$, $B_1\cap B_2$. But this can certainly be improved to three rectangles: $$(A_1\cap A_2)\times (B_1\cup B_2)\qquad(A_1\setminus A_2)\times B_1\qquad(A_2\setminus A_1)\times B_2.$$ To arrive at six rectangles, just split these up somehow.