Finite union of measurable rectangles can be written as union of pairwise disjoint measurable rectangles? Suppose we have two measure spaces $(X, \Sigma, \mu)$ and $(Y, \tau, \nu)$.  $R \subseteq X \times Y$ is called a measurable rectangle if $R = A \times B$ with $A \in \Sigma$ and $B \in \tau$.
I have the following idea proved and at my disposal:  If $R_{1}$ and $R_{2}$ are measurable rectangles, then $R_{1} \cup R_{2}$ and $R_{1} \setminus R_{2}$ can both be written as finite unions of pairwise disjoint measurable rectangles.
The proof that $R_{1} \cup R_{2}$ can be written as a finite union of pairwise disjoint rectangles is as follows:  $R_{1} \cup R_{2} = (R_{1} \setminus R_{2}) \cup (R_{1} \cap R_{2}) \cup (R_{2} \setminus R_{1})$.
Can someone help me prove that $R_{1} \cup R_{2} \cup R_{3}$ can be written as a finite union of pairwise disjoint measurable rectangles?  I tried and failed to do this.
 A: Let $\cal R$ be the collection of sets that are expressible as the finite union of disjoint rectangles, and $\bar{\cal R}$ be the collection of sets that are expressible as the finite or countable disjoint union of rectangles. Obviously ${\cal R} \subset \bar{\cal R}$.
Suppose $R\in {\cal R}$ and $D$ is a rectangle. Then $R =\cup_{k=1}^n D_k$, where $D_k$ are pairwise disjoint rectangles. Then $D \cup D_k = 
D \cup (D_k \setminus D) $, and so $R = D \cup \left( \cup_{k=1}^n D_k \setminus D \right)$. You know that each $D_k \setminus D$ can be written as the union of pairwise disjoint rectangles, hence so can $R \cup D$.
It follows by induction that if $R_k \in {\cal R}$ then $\cup_{k=1}^n R_k \in {\cal R}$.
If $D_1, D_2$ are rectangles, then you know that $D_1 \setminus D_2 \in { \cal R}$. Suppose $R\in {\cal R}$ and $D$ is a rectangle, then $R =\cup_{k=1}^n D_k$, where $D_k$ are pairwise disjoint rectangles. Then
$R \setminus D = \cup_{k=1}^n \left(  D_k \setminus D \right)$, and so $R \setminus D \in {\cal R}$.
It follows by induction that if $R_1, R_2 \in {\cal R}$, then $R_1 \setminus R_2 \in {\cal R}$.
Also, if $R_k \in {\cal R}$, then since $\cup_{k=1}^n R_k \in {\cal R}$ for all $n$, we have 
$\cup_{k=1}^\infty R_k \in \bar{\cal R}$.
A: I think you can use an indirect approach. If D is the algebra of finite disjoint unions and F is the algebra of finite unions, then D is contained in F, both contains rectangles and F is minimal with this property, so D=F.
Now the problem is to show that F and D specially are algebras. Insted of proving that D is close under finite unions, try yo prove that it is close under finite intersections.
This is problem 10.D un Bartle's book Measure Theory.
