From Rectangles to a Ring
Measurable rectangles are the basic sets used to build product sigma-algebras. They are not, in general, closed under unions or differences: combining rectangles can produce a set that is no longer a single rectangle. The natural finite-stage collection is a ring, whose members are finite unions of rectangles. A finite partition of each coordinate space provides both a proof of this description and a practical way to break any such union into disjoint rectangular pieces.
Let \((X,\mathcal{M})\) and \((Y,\mathcal{N})\) be measurable spaces, with no measure assumed on either space. The product sigma-algebra \(\mathcal{M}\otimes\mathcal{N}\) is the sigma-algebra on \(X\times Y\) generated by the measurable rectangles. A measurable rectangle is a set \(A\times B\), where \(A\in\mathcal{M}\) and \(B\in\mathcal{N}\). We allow \(A\) or \(B\) to be empty, so the empty set is a rectangle.
A ring need not contain \(X\times Y\), and it need not be closed under countable unions. We will show that the ring generated by all measurable rectangles has a particularly direct description: its members are precisely the finite unions of measurable rectangles. The key step is to partition the coordinate spaces according to membership in the finitely many coordinate sets involved.
Finite Coordinate Partitions
Given finitely many measurable sets \(A_1,\ldots,A_m\in\mathcal{M}\), form, for each subset \(I\subseteq\{1,\ldots,m\}\), the set
An intersection over no indices means the whole space \(X\). Each \(P_I\) is measurable, the sets \(P_I\) are pairwise disjoint, and their union is \(X\). Empty atoms are permitted and can be discarded. A point \(x\in P_I\) belongs to exactly those \(A_i\) whose indices lie in \(I\). Constructing the analogous atoms \(Q_J\) from finitely many sets in \(\mathcal{N}\) partitions \(Y\). The product cells \(P_I\times Q_J\) are then pairwise disjoint measurable rectangles partitioning \(X\times Y\).
Proof. Form the atoms \(P_I\) from \(A_1,\ldots,A_m\) and \(Q_J\) from \(B_1,\ldots,B_m\), using the same index set \(\{1,\ldots,m\}\) for both coordinate partitions. On a product cell \(P_I\times Q_J\), membership in each rectangle \(A_i\times B_i\) is constant: every point of the cell belongs to \(A_i\times B_i\) exactly when \(i\in I\) and \(i\in J\). Consequently, the whole cell is contained in \(E\) exactly when there is at least one index \(i\) belonging to both \(I\) and \(J\). Retain precisely those cells for which \(I\cap J\ne\varnothing\), omitting empty cells. The retained cells are measurable rectangles, are pairwise disjoint because the atoms in each coordinate are pairwise disjoint, and their union is \(E\) by the membership criterion. There are at most \(2^m2^m\) cells, so the union is finite. If \(m=0\), then \(E=\varnothing\), which is the empty disjoint union. \(\square\)
Characterizing the Generated Ring
Proof. Let \(\mathcal{R}\) be the collection of finite unions of measurable rectangles, including the empty union \(\varnothing\). We first show that \(\mathcal{R}\) is a ring. It contains \(\varnothing\) and is closed under finite unions by concatenating the finite lists of rectangles. To prove closure under differences, take \(E=\bigcup_{i=1}^m(A_i\times B_i)\) and \(F=\bigcup_{j=1}^n(C_j\times D_j)\). Use the finite coordinate partitions formed from all the \(A_i,C_j\) in \(X\) and all the \(B_i,D_j\) in \(Y\). On each product cell, membership in \(E\) and in \(F\) is constant. Retain a cell exactly when it is contained in \(E\) and not contained in \(F\). The retained cells are finitely many measurable rectangles, are pairwise disjoint, and their union is \(E\setminus F\). Thus \(E\setminus F\in\mathcal{R}\), including when either set is empty. Hence \(\mathcal{R}\) is a ring containing every measurable rectangle.
Conversely, any ring containing every measurable rectangle must contain every finite union of rectangles, by closure under finite unions and the inclusion of the empty set. Therefore the smallest such ring is exactly \(\mathcal{R}\). The proposition gives the finite disjoint rectangular decomposition for every member of \(\mathcal{R}\). \(\square\)
Worked Example: Disjoint Decomposition of an Overlapping Union
In \((\mathbb{R},\mathcal{B}_{\mathbb{R}})\times(\mathbb{R},\mathcal{B}_{\mathbb{R}})\), consider \(E=([0,2)\times[0,2))\cup([1,3)\times[1,3))\). Here all intervals are Borel sets. We will write \(E\) as a disjoint union.
Keep the first rectangle \(R_1=[0,2)\times[0,2)\). The part of the second rectangle \(R_2=[1,3)\times[1,3)\) not already in \(R_1\) consists of two pieces: \([2,3)\times[1,3)\), where the first coordinate lies outside \([0,2)\), and \([1,2)\times[2,3)\), where the second coordinate lies outside \([0,2)\). Thus
The first and second pieces are disjoint because \([0,2)\cap[2,3)=\varnothing\) in the first coordinate. The first and third are disjoint because \([0,2)\cap[2,3)=\varnothing\) in the second coordinate. The second and third are disjoint because \([2,3)\cap[1,2)=\varnothing\) in the first coordinate. The two added pieces cover \(R_2\setminus R_1\): its first-coordinate part \([2,3)\) gives the second piece, and its remaining first-coordinate part \([1,2)\) requires second coordinate \([2,3)\), giving the third. Together with \(R_1\), they therefore cover exactly \(E\).
Check: Each listed rectangle is contained in \(E\), and every point of either original rectangle is either in \(R_1\) or in one of the two added pieces. No boundary point is lost: the half-open intervals place \(2\) in \([2,3)\), not in \([0,2)\).
Worked Example: Subtracting an Inner Rectangle
In the same Borel product space, let \(E=[0,3)\times[0,3)\) and \(F=[1,2)\times[1,2)\). The set difference removes the inner rectangle. Partition each coordinate interval at \(1\) and \(2\). The four remaining rectangles are
To verify the equality, a point of \(E\) has first coordinate in exactly one of \([0,1)\), \([1,2)\), or \([2,3)\). If it is in \([0,1)\) or \([2,3)\), it is outside \(F\) regardless of its second coordinate, giving the first or second rectangle. If its first coordinate is in \([1,2)\), it is outside \(F\) exactly when its second coordinate is in \([0,1)\) or \([2,3)\), giving the third or fourth rectangle. These cases are exhaustive and mutually exclusive, so the decomposition is disjoint.
Check: Every listed rectangle lies in \(E\). None meets \(F\), since each has at least one coordinate in an interval disjoint from \([1,2)\). The omitted cell \([1,2)\times[1,2)\) is exactly \(F\), so the listed cells leave no other part of \(E\) uncovered.
Worked Example: A Ring Member That Is Not a Rectangle
Let \(X=\{a,b\}\) and \(Y=\{u,v\}\), each equipped with its power-set sigma-algebra. Consider \(G=\{(a,u),(a,v),(b,u)\}\). It is a finite union of disjoint measurable rectangles:
The pieces are disjoint because \(\{a\}\cap\{b\}=\varnothing\). The first contains exactly \((a,u)\) and \((a,v)\); the second contains exactly \((b,u)\), so their union is \(G\). Yet \(G\) is not itself a rectangle. Its coordinate projections are both full sets, \(\{a,b\}\) and \(\{u,v\}\); any rectangle containing all three listed pairs and formed from those projections also contains \((b,v)\), which is not in \(G\). More directly, a rectangle containing \((a,u),(a,v),(b,u)\) must have first factor containing \(a,b\) and second factor containing \(u,v\), and so must contain \((b,v)\).
Check: The missing pair \((b,v)\) confirms that \(G\ne X\times Y\), while the displayed finite union proves \(G\) belongs to the ring generated by rectangles.
The Ring Is Not Usually the Product Sigma-Algebra
Every measurable rectangle belongs to \(\mathcal{M}\otimes\mathcal{N}\), so every finite union of such rectangles belongs to that product sigma-algebra. This gives an inclusion of the generated ring in the product sigma-algebra, not generally equality: a sigma-algebra also allows countable unions.
For a concrete distinction, let \(X=Y=\mathbb{N}\), both with the power-set sigma-algebra. The diagonal \(D=\{(n,n):n\in\mathbb{N}\}\) is a countable union of singleton rectangles, so it belongs to the product sigma-algebra. It cannot be a finite union of rectangles. Indeed, any nonempty rectangle \(A\times B\) contained in \(D\) must be a singleton: if \(a\in A\) and \(b\in B\), then \(a=b\); fixing \(b\) forces every element of \(A\) to equal \(b\), and fixing \(a\) forces every element of \(B\) to equal \(a\). A finite union of such rectangles is finite, whereas \(D\) is infinite. Thus \(D\) lies in the product sigma-algebra but not in the ring.
A common mistake is to assume that a union of rectangles is itself a rectangle. The finite example \(G\) above disproves this: its coordinate projections would force the extra pair \((b,v)\). The correct finite-stage conclusion is that a union belongs to the ring and can be decomposed into disjoint rectangles. This distinction matters when constructing set functions on finite unions: first one must know what the ring contains and how its members can be partitioned. No measure or rectangle content has been assigned here; those require additional arguments.
Check Your Understanding
Use finite coordinate partitions to justify each answer, paying attention to whether the rectangles in a decomposition are disjoint.
- For measurable \(A,C\in\mathcal{M}\) and \(B,D\in\mathcal{N}\), describe a finite rectangular decomposition of \((A\times B)\setminus(C\times D)\).
- Explain why finite unions of measurable rectangles form a ring, but the proof of closure under differences needs more than simply taking differences of the coordinate sets separately.
- In the finite example \(G=\{(a,u),(a,v),(b,u)\}\), why does any rectangle containing all three points also contain \((b,v)\)?
- For \(X=Y=\mathbb{N}\) with power-set sigma-algebras, prove that the diagonal is in the product sigma-algebra but not in the ring of finite unions of rectangles.
- Does the ring generated by measurable rectangles have to contain \(X\times Y\)? Give a condition under which it does.