I'm using Royden & Fitzpatrick's Real Analysis, and I came across something that, for the life of me, I cannot make sense of. Background: Start by letting $(X,\mathcal{A},\mu)$ and $(Y,\mathcal{B},\nu)$ be two reference measure spaces. A measurable rectangle is a subset $A\times B \subseteq X\times Y$ where $A\in \mathcal{A}$, $B\in\mathcal{B}$, and $\mu(A)<\infty$, $\nu(B)<\infty$.
Now for the issue:
Lemma 1 $\quad$Let $\{A_{k}\times B_{k}\}_{k=1}^{\infty}$ be a countable disjoint collection of measurable rectangles whose union also is a measurable rectangle $A\times B$. Then $$\mu(A)\times \nu(B) = \sum_{k=1}^{\infty}\mu(A_{k})\times\nu(B_{k}).$$
Okay fine, the lemma seems reasonable in and of itself. As does most of the proof, except for this step in the beginning:
Proof$\quad$ Fix a point $x\in A$. For each $y\in B$, the point $(x,y)$ belongs to exactly one $A_{k}\times B_{k}$. Therefore we have the following disjoint union: $$B=\bigcup_{\{k\,:\,x\in A_{k}\}}B_{k}\qquad\qquad(1)$$ ...
This doesn't quite make sense to me. Take for example the disjoint rectangles with $m$ the Lebesgue measure on $\mathbb{R}$: $A_{1}\times B_{1} = [1,3]\times [1,3]$ and $A_{2}\times B_{2} = [4,5]\times [1,4]$ (see picture -- Note that I've done the natural thing and $A_{1},A_{2}$ are on the $x$-axis and $B_{1},B_{2}$ are on the $y$-axis.)
Now say I start by fixing a point, say $2\in A$. Interpreting the union (1) as written, the only set involved in the union is $B_{1}$, since $2$ is only in $A_{1}$. Interpreting the union this way, I don't get all of $B$, as clearly $B$ contains the point $y=4$ for instance, and $4\notin B_{1}$.
On the other hand, if I interpret the $x$ as not really "fixed" as they say, but rather as moving along with $k$, then the union is not disjoint as claimed, because then $B=B_{1}\cup B_{2}$ yet $B_{1}\cap B_{2}\neq\emptyset$.
Now, I am aware that $(A_{1}\times B_{1})\cup (A_{2}\times B_{2}) \subseteq (A_{1}\cup A_{2})\times (B_{1}\cup B_{2})$. But basically my questions are:
(1) How should I interpret this union? Am I interpreting this wrong or is there some issue with the notation? (2) Is there a better way to denote this that gets the point across, i.e., the expression of $B$ as a countable disjoint union? If so, what is it, and how can I transform that into what is given here?
I suspect I am missing something fundamental here, but I am not sure. Any commentary or intuition is welcome, and thank you in advance.
The result does not apply to your example, since your set $A_1\times B_1\cup A_2\times B_2$ is not a measurable rectangle. If it were a measurable rectangle, that would mean there are sets $A$ and $B$ such that $$A_1\times B_1\cup A_2\times B_2=A\times B.$$ No such sets $A$ and $B$ exist, essentially by your argument: we would have to have $2\in A$ and $4\in B$, but $(2,4)\not\in A_1\times B_1\cup A_2\times B_2$.
As for why $$B=\bigcup_{\{k\,:\,x\in A_{k}\}}B_{k}\qquad\qquad(1)$$ is true, it is for exactly the reason that is stated: for each $y\in B$, the point $(x,y)$ is in exactly one $A_k\times B_k$. Note that since by assumption the union is equal to $A\times B$ (this is what fails in your example), for each $y\in B$ there must exist some $k$ such that $(x,y)\in A_k\times B_k$, so that $y\in B_k$ for that value of $k$ (and $x\in A_k$ as well). The union is disjoint because there is only one value of $k$ such that $x\in A_k$ and $y\in B_k$, since the sets $A_k\times B_k$ are disjoint (only one of them can contain $(x,y)$.