Defining Rectangle Content on Finite Unions

From Rectangles to a Well-Defined Content

Let \((X,\mathcal{M},\mu)\) and \((Y,\mathcal{N},\nu)\) be measure spaces with finite measures, so that \(\mu(X)<\infty\) and \(\nu(Y)<\infty\). A measurable rectangle is a set \(A\times B\) with \(A\in\mathcal{M}\) and \(B\in\mathcal{N}\). Its proposed content is \(\mu(A)\nu(B)\). The challenge is that a finite union of rectangles can have more than one disjoint rectangle decomposition. Before this formula can define a set function, we must show that all such decompositions give the same value.

Write \(\mathcal{R}\) for the collection of finite unions of measurable rectangles in \(X\times Y\), including the empty union \(\varnothing\). The product sigma-algebra \(\mathcal{M}\otimes\mathcal{N}\) contains every such set, since it contains the measurable rectangles and is closed under finite unions. A ring of sets is a collection closed under finite unions and set differences; the earlier result on the ring generated by measurable rectangles gives that \(\mathcal{R}\) is such a ring. Here we will use finite partitions of the factors to make the decompositions and their comparisons explicit.

Every \(E\in\mathcal{R}\) has such a decomposition. Indeed, take any finite representation \(E=\bigcup_{i=1}^k(A_i\times B_i)\). Form the finite partitions of \(X\) and \(Y\) into the nonempty sets obtained by choosing, for each coordinate set \(A_i\) or \(B_i\), either that set or its complement, and intersecting all the choices. Each resulting set is measurable. The product of one partition set from \(X\) and one from \(Y\) is either wholly contained in \(E\) or disjoint from \(E\): membership in every \(A_i\times B_i\) is constant on that product. The products contained in \(E\) are a finite disjoint rectangle decomposition. If the original union is empty, there are no such products.

Why the Decomposition Does Not Matter

Proof. First consider a single disjoint decomposition \(E=\bigcup_{i=1}^k(A_i\times B_i)\). Form finite measurable partitions \(\{C_r\}_{r=1}^R\) of \(X\) and \(\{D_s\}_{s=1}^S\) of \(Y\) using all the coordinate sets \(A_i\) and \(B_i\) in this decomposition and their complements, omitting empty partition sets. Every \(A_i\) is a union of some \(C_r\), and every \(B_i\) is a union of some \(D_s\). Therefore each rectangle \(A_i\times B_i\) is the union of the cells \(C_r\times D_s\) for which \(C_r\subseteq A_i\) and \(D_s\subseteq B_i\).

For each \(i\), finite additivity of the factor measures and the distributive law for finite sums give

Since the rectangles in the decomposition are pairwise disjoint, each cell \(C_r\times D_s\) contained in \(E\) lies in exactly one of them. In fact, the partitions cover \(X\) and \(Y\), so every point of that cell has the same membership in each rectangle; if the cell belonged to two rectangles, it would be a nonempty subset of their intersection, contradicting disjointness. Summing the displayed identity over \(i\) therefore expresses the decomposition sum as the sum of \(\mu(C_r)\nu(D_s)\) over exactly the cells contained in \(E\).

Now take two disjoint decompositions of the same \(E\). Make new finite partitions using all the coordinate sets from both decompositions and their complements. The preceding argument applies to the first decomposition with these partitions: its sum is the sum over all cells contained in \(E\). It also applies to the second decomposition with these same partitions: its sum is that very same sum over cells. The two decomposition sums are equal. If \(E=\varnothing\), both sums are empty sums and equal \(0\). All terms are finite because \(\mu(X)\) and \(\nu(Y)\) are finite, so \(c(E)\in[0,\infty)\). \(\square\)

The alignment in this argument matters: the partitions used to compare two decompositions must include the coordinate sets from both. A partition formed from only one decomposition need not split the rectangles in the other decomposition into cells.

Finite Additivity, and What Has Not Yet Been Proved

Proof. Choose a disjoint rectangle decomposition of \(E\) and one of \(F\), which exist by the finite-partition construction. Since \(E\cap F=\varnothing\), combining the two decompositions gives a disjoint rectangle decomposition of \(E\cup F\). Its sum is the sum for \(E\) plus the sum for \(F\). By the theorem, these sums equal \(c(E\cup F)\), \(c(E)\), and \(c(F)\), respectively. If either set is empty, its decomposition is empty and contributes \(0\), so the same equality holds. \(\square\)

Finite additivity is a consequence of having a well-defined content, not a substitute for proving that it is well-defined. At this stage no countable additivity is asserted: that requires a separate argument. In particular, the function \(c\) has been defined on the ring \(\mathcal{R}\) of finite unions, not yet on the whole product sigma-algebra \(\mathcal{M}\otimes\mathcal{N}\). The finite factor measures ensure that every rectangle has finite content, but they do not by themselves extend \(c\) to a product measure.

A Common Mistake: Using a Grid That Is Not Common

A tempting but invalid comparison is to form a grid using only one representation and then claim that every grid cell in the union lies in one rectangle of the other representation. For \(E=[0,1]\times[0,1]\), a grid formed from the single-rectangle decomposition has just the cell \(E\). But in the split decomposition from the previous example, this cell crosses the boundary at \(x=1/2\): it is contained in neither \([0,\tfrac12)\times[0,1]\) nor \([\tfrac12,1]\times[0,1]\). So that grid cannot be used to sum the split decomposition rectangle by rectangle.

The correction is to form a common grid using the coordinate sets from both representations. Then each rectangle in either decomposition is a union of grid cells. Disjointness ensures that every cell contained in \(E\) belongs to exactly one rectangle in each decomposition, and finite additivity of \(\mu\) and \(\nu\) makes both sums equal to the same cell sum. This is the precise reason the decomposition-independence proof works.