Why Rectangles Do Not Form a Sigma-Algebra

Rectangles Are Building Blocks, Not Usually a Sigma-Algebra

In a product space, a measurable rectangle has the form \(A\times B\): membership in it is decided separately by the first and second coordinates. Rectangles are convenient elementary sets, but a sigma-algebra must also contain complements and countable unions. Those operations can produce sets whose membership cannot be described by one pair of coordinate conditions. The distinction matters when constructing a product sigma-algebra: the rectangles are its starting family, not usually the entire sigma-algebra.

Let \((X,\mathcal{M})\) and \((Y,\mathcal{N})\) be measurable spaces, with no measures assumed. Write \(\mathcal{R}=\{A\times B:A\in\mathcal{M},\,B\in\mathcal{N}\}\) for the collection of measurable rectangles in \(X\times Y\). A sigma-algebra on \(X\times Y\) is a collection containing \(\varnothing\), closed under complements relative to \(X\times Y\), and closed under countable unions. The product sigma-algebra \(\mathcal{M}\otimes\mathcal{N}\) is, by definition, the smallest sigma-algebra containing \(\mathcal{R}\), equivalently \(\sigma(\mathcal{R})\). Thus \(\mathcal{R}\subseteq\mathcal{M}\otimes\mathcal{N}\), but this inclusion need not be equality.

Rectangles do satisfy some closure properties. Both \(\varnothing\) and \(X\times Y\) are rectangles, since \(\varnothing=\varnothing\times Y\) and \(X\times Y\) is a rectangle. Also, for \(A,C\in\mathcal{M}\) and \(B,D\in\mathcal{N}\),

The right side is again a measurable rectangle. Complement and union behave differently. In fact, the complement of a rectangle has a useful two-piece expression, but that expression need not itself be one rectangle:

This identity follows directly from membership: \((x,y)\notin A\times B\) means \(x\notin A\) or \(y\notin B\). It gives a union of rectangles, not generally a single rectangle.

A Test for Recognizing Nonrectangles

Every rectangle has a simple structural property. If it contains two opposite corners of a pair of coordinate choices, it must contain the other two corners as well. This is a useful way to disprove that a set is a rectangle without having to guess all possible choices of \(A\) and \(B\).

Proof. Since \((x_1,y_1)\in A\times B\), the definition of Cartesian product gives \(x_1\in A\) and \(y_1\in B\). Since \((x_2,y_2)\in A\times B\), it gives \(x_2\in A\) and \(y_2\in B\). Combining these memberships yields \((x_1,y_2)\in A\times B\) and \((x_2,y_1)\in A\times B\). \(\square\)

The argument also gives a general sufficient condition for failure. If \(X\) and \(Y\) each contain at least two distinct points and every singleton in each space is measurable, then \(\mathcal{R}\) is not a sigma-algebra. The proof is to take a singleton rectangle and apply the cross-corner test to its complement.

Proof. The set \(\{(x_0,y_0)\}=\{x_0\}\times\{y_0\}\) is a measurable rectangle. Its complement contains \((x_0,y_1)\) and \((x_1,y_0)\), because \(y_1\ne y_0\) and \(x_1\ne x_0\). Its complement does not contain \((x_0,y_0)\). If that complement were a rectangle, the cross-corner proposition applied to the two points it contains would require it to contain \((x_0,y_0)\), a contradiction. Thus the rectangle family is not closed under complements, so it is not a sigma-algebra. \(\square\)

Countable Unions Can Also Leave the Rectangle Family

Even when every set in a union is a rectangle, the union need not be one rectangle. Countable unions are required in a sigma-algebra, so this failure is another way to see the difference between \(\mathcal{R}\) and \(\sigma(\mathcal{R})\).

What the Generated Sigma-Algebra Adds

The notation \(\sigma(\mathcal{R})\) is not another name for the collection of rectangles. It records all sets that must be included when one closes the rectangle family under the sigma-algebra operations. In particular, it contains complements of rectangles and countable unions of rectangles, even in examples where those sets fail the cross-corner test and therefore are not rectangles. The diagonal example makes the distinction exact: \(D\in\sigma(\mathcal{R})\) but \(D\notin\mathcal{R}\).

There are also degenerate cases where the rectangle family does form a sigma-algebra, so the right conclusion is “generally not,” rather than “never.” If \(X\) is a singleton \(\{x_0\}\) with its power set sigma-algebra, then every rectangle has the form \(\varnothing\) or \(\{x_0\}\times B\) for \(B\in\mathcal{N}\). The map \(B\mapsto\{x_0\}\times B\) preserves complements relative to the product and countable unions, so these rectangles form a sigma-algebra. If either factor is empty, the product has only the empty set and the same conclusion holds. The sufficient condition in the theorem explains why the usual nontrivial spaces with measurable singletons behave differently.

A Common Mistake

A tempting but incorrect calculation is \((A\times B)^c=A^c\times B^c\). It confuses “not in both coordinate sets” with “in neither coordinate set.” For example, with \(A=B=\{0\}\) in \(\{0,1\}\), the incorrect expression gives only \(\{(1,1)\}\). But \((1,0)\) is outside \(A\times B\), since its first coordinate is not in \(A\), and \((0,1)\) is outside \(A\times B\), since its second coordinate is not in \(B\). Both must lie in the complement. The correct condition is “\(x\notin A\) or \(y\notin B\),” which produces the union \((A^c\times Y)\cup(X\times B^c)\).

The practical test is simple: when asked whether a set is a rectangle, test the cross-corner property. When asked whether a collection is a sigma-algebra, test closure under complements and countable unions rather than relying on closure under intersections alone. Rectangles are well-behaved under intersections, but that fact does not supply the missing sigma-algebra closures.