Constructing Product Measures for Finite Spaces

Why Begin with Finite Measures?

A product measure should assign to a measurable rectangle the product of the measures of its sides. The challenge is to turn that prescription into a measure on the entire product σ-algebra. When both factor measures are finite, one can first define a set function on finite unions of rectangles, prove it is a premeasure, and then use the measure extension theorem.

Here, “finite” refers to the total measure, not necessarily to the number of points in the underlying space. The spaces may be infinite, as long as their whole-space measures are finite. This distinction matters: the construction below works for general measurable spaces with finite measures, not just for finite sets.

Rectangles and Their Algebra

A measurable rectangle is a set \(A\times B\), where \(A\in\mathcal A\) and \(B\in\mathcal B\). The product σ-algebra \(\mathcal A\otimes\mathcal B\) is the σ-algebra on \(X\times Y\) generated by all such rectangles. Before constructing a measure on this σ-algebra, consider the smaller collection \(\mathcal R\) of finite unions of measurable rectangles, including the empty union \(\varnothing\).

The collection \(\mathcal R\) is an algebra of sets: it contains \(X\times Y\), and it is closed under finite unions and complements relative to \(X\times Y\). To see why, finite intersections of rectangles are rectangles, since \((A_1\times B_1)\cap(A_2\times B_2)=(A_1\cap A_2)\times(B_1\cap B_2)\). Complements and finite combinations can then be expressed as finite unions of rectangles by partitioning each coordinate space according to the finitely many sets involved.

Any finite collection of rectangles can be refined into a finite disjoint collection of rectangles. In each coordinate, partition the space into the atoms formed by the finitely many measurable sets and their complements. The products of the resulting coordinate pieces form a disjoint grid. This finite refinement lets us assign a weight to a finite union without depending on how it was originally written.

In particular, \(\lambda(A\times B)=\mu(A)\nu(B)\). The grid refinement also shows finite additivity: if finitely many members of \(\mathcal R\) are disjoint, the weight of their union is the sum of their weights. Finite additivity alone is not enough for measure extension, however. We need countable additivity whenever a countable disjoint union belongs to \(\mathcal R\).

The Key Countable-Additivity Step

The following rectangle-partition argument is the point where the finiteness of the factor measures makes a direct construction possible. We use the standard monotone convergence theorem for nonnegative integrals.

Proof. Each rectangle in the partition is contained in \(A\times B\), so \(A_n\subseteq A\) and \(B_n\subseteq B\). Fix \(x\in A\). The sets \(B_n\) for which \(x\in A_n\) are pairwise disjoint and have union \(B\): this follows because the rectangles partition \(A\times B\). Countable additivity of \(\nu\) therefore gives \[ \nu(B)=\sum_{n=1}^{\infty}\mathbf 1_{A_n}(x)\nu(B_n). \] Integrate this identity over \(x\in A\) with respect to \(\mu\). The summands are nonnegative, so the monotone convergence theorem permits integration of the series term by term. Since \(A_n\subseteq A\), the result is \[ \mu(A)\nu(B) =\sum_{n=1}^{\infty}\nu(B_n)\int_A\mathbf 1_{A_n}(x)\,d\mu(x) =\sum_{n=1}^{\infty}\mu(A_n)\nu(B_n). \] This proves the lemma. \(\square\)

We can now verify the premeasure property on the whole algebra. Suppose \(E,E_1,E_2,\ldots\in\mathcal R\), the sets \(E_n\) are pairwise disjoint, and \(E=\bigcup_{n=1}^{\infty}E_n\). Write \(E\) as a finite disjoint union of rectangles \(R_1,\ldots,R_m\). For each \(n\) and \(j\), the set \(E_n\cap R_j\) belongs to \(\mathcal R\), and can be written as a finite disjoint union of rectangles. For fixed \(j\), all these pieces, over every \(n\), form a countable disjoint rectangle partition of \(R_j\). Applying the lemma to \(R_j\) gives \[ \lambda(R_j)=\sum_{n=1}^{\infty}\lambda(E_n\cap R_j). \] Finite additivity within each \(E_n\), followed by summation over the finitely many \(R_j\), yields \[ \lambda(E)=\sum_{j=1}^m\lambda(R_j) =\sum_{n=1}^{\infty}\sum_{j=1}^m\lambda(E_n\cap R_j) =\sum_{n=1}^{\infty}\lambda(E_n). \] Thus \(\lambda\) is countably additive whenever a disjoint union and its terms belong to \(\mathcal R\). That is precisely the premeasure property on an algebra.

Extending the Premeasure

We use the standard measure extension theorem in the following form: a premeasure on an algebra extends to a measure on the σ-algebra generated by that algebra if it is σ-finite. Here, σ-finiteness means that the underlying space is a countable union of algebra sets with finite premeasure.

The algebra \(\mathcal R\) generates \(\mathcal A\otimes\mathcal B\), because it contains every measurable rectangle, and every member of \(\mathcal R\) is a finite union of such rectangles. Moreover, the whole product is itself in \(\mathcal R\), with \(\lambda(X\times Y)=\mu(X)\nu(Y)<\infty\). So the premeasure is finite on a single set covering the product and is, in particular, σ-finite. The extension theorem applies.

Proof. Define \(\lambda\) on \(\mathcal R\) by disjoint rectangle decompositions. Finite grid refinement makes the definition well-defined and finitely additive. The rectangle-partition lemma and its application to finite disjoint rectangle decompositions show that \(\lambda\) is a premeasure on \(\mathcal R\). It is finite on \(X\times Y\), so the measure extension theorem gives a measure \(\rho\) on \(\sigma(\mathcal R)=\mathcal A\otimes\mathcal B\) extending \(\lambda\). Since each rectangle \(A\times B\) belongs to \(\mathcal R\), the extension preserves its premeasure value: \[ \rho(A\times B)=\lambda(A\times B)=\mu(A)\nu(B). \] This proves existence and the required rectangle values. \(\square\)

Worked Examples

What the Construction Does—and Does Not—Say

The construction establishes existence on the product σ-algebra and verifies the prescribed values on measurable rectangles. It does not calculate the measure of every product-measurable set by a new formula; such sets can be more complicated than finite unions of rectangles. The extension theorem supplies a measure on the generated σ-algebra while preserving the premeasure where it was already defined.

A common mistake is to assign \(\mu(A)\nu(B)\) to rectangles and assume that this prescription automatically defines a measure on all sets. A measure must be defined on a σ-algebra and must satisfy countable additivity. The premeasure argument is the necessary bridge: it verifies that the rectangle prescription is consistent with countable disjoint decompositions inside the algebra before the extension theorem is invoked.

The finiteness assumption is used directly to ensure that the premeasure is finite on the entire product, which gives σ-finiteness for the extension step. The proof of the rectangle-partition lemma uses countable additivity in one factor and monotone convergence for a nonnegative series. More general product constructions require additional care; the finite case provides a clean setting in which the extension mechanism is visible.