Caratheodory Construction of Finite Product Measures

From Rectangle Content to a Measure

For finite measures \(\mu\) and \(\nu\), the intended value of a measurable rectangle \(A\times B\) is \(\mu(A)\nu(B)\). The challenge is to turn this rule into a countably additive measure on the entire product sigma-algebra: rectangles alone need not form a sigma-algebra. The construction proceeds in two stages. First, the rectangle values define a premeasure on the ring of finite unions of measurable rectangles. Then Caratheodory’s extension theorem constructs a measure on the sigma-algebra generated by that ring.

Let \((X,\mathcal{M},\mu)\) and \((Y,\mathcal{N},\nu)\) be measure spaces, with \(\mu(X)<\infty\) and \(\nu(Y)<\infty\). No completeness assumption is made on either factor. Write \(\mathcal{M}\otimes\mathcal{N}\) for the sigma-algebra generated by the measurable rectangles \(A\times B\), where \(A\in\mathcal{M}\) and \(B\in\mathcal{N}\). Let \(\mathcal{R}\) be the ring of finite unions of such rectangles. A ring of sets is closed under finite unions and differences; it need not be a sigma-algebra.

The premeasure assertion is the needed input from the earlier result on countable additivity of rectangle content; it is not an assertion that \(\rho\) is already a measure on \(\mathcal{M}\otimes\mathcal{N}\). To extend it, cover arbitrary subsets of \(X\times Y\) by countably many sets from \(\mathcal{R}\), and take the infimum of the total costs.

Because \(X\times Y\in\mathcal{R}\), every subset has at least one such cover. The set function \(\rho^*\) is an outer measure: in particular, it is monotone and countably subadditive. The Caratheodory extension theorem states that the sets satisfying the displayed measurability criterion form a sigma-algebra, and that \(\rho^*\) restricted to this sigma-algebra is a measure. For a premeasure on a ring, the theorem also gives that every set in the ring is Caratheodory measurable and that the restriction to \(\sigma(\mathcal{R})\) extends the premeasure. We use this standard extension theorem here; its general proof belongs to outer-measure theory.

Why Rectangle Sets Are Measurable

The measurability criterion can be checked directly for any \(Q\in\mathcal{R}\). This calculation explains why splitting a cover along a ring set is compatible with the outer measure, and in particular applies to every measurable rectangle.

Proposition. Every \(Q\in\mathcal{R}\) is Caratheodory measurable for \(\rho^*\).

Proof. Fix \(Q\in\mathcal{R}\) and an arbitrary subset \(S\subseteq X\times Y\). Outer-measure subadditivity, applied to \(S=(S\cap Q)\cup(S\setminus Q)\), gives \(\rho^*(S)\le \rho^*(S\cap Q)+\rho^*(S\setminus Q)\). For the reverse inequality, take any countable cover \(S\subseteq\bigcup_{n=1}^{\infty}R_n\) with \(R_n\in\mathcal{R}\). Since \(\mathcal{R}\) is a ring, both \(R_n\cap Q\) and \(R_n\setminus Q\) belong to \(\mathcal{R}\). The sets \(R_n\cap Q\) cover \(S\cap Q\), and the sets \(R_n\setminus Q\) cover \(S\setminus Q\). Therefore \(\rho^*(S\cap Q)+\rho^*(S\setminus Q)\le\sum_{n=1}^{\infty}\bigl(\rho(R_n\cap Q)+\rho(R_n\setminus Q)\bigr)\). The two pieces partition \(R_n\), so finite additivity of the premeasure gives \(\rho(R_n\cap Q)+\rho(R_n\setminus Q)=\rho(R_n)\) for each \(n\). Thus the left side is at most \(\sum_{n=1}^{\infty}\rho(R_n)\) for every ring cover of \(S\). Taking the infimum over those covers gives \(\rho^*(S\cap Q)+\rho^*(S\setminus Q)\le\rho^*(S)\). The two inequalities prove the criterion, including when \(S\) is empty. \(\square\)

Proof. Apply the Caratheodory extension theorem to the premeasure \(\rho\) on \(\mathcal{R}\). It gives a measure extending \(\rho\) on \(\sigma(\mathcal{R})\). Since \(\mathcal{R}\) is generated by measurable rectangles, \(\sigma(\mathcal{R})=\mathcal{M}\otimes\mathcal{N}\). Each rectangle \(A\times B\) lies in \(\mathcal{R}\), so extension of \(\rho\) gives \(\lambda(A\times B)=\rho(A\times B)=\mu(A)\nu(B)\). Taking \(A=X\) and \(B=Y\) gives the stated finite total mass. If either factor measure is zero, these identities still hold: the rectangle values and total mass are zero. \(\square\)

The conclusion concerns the product sigma-algebra \(\mathcal{M}\otimes\mathcal{N}\), not automatically its completion. A set contained in a \(\lambda\)-null product-measurable set need not belong to this sigma-algebra unless additional completeness conditions are imposed or the measure is completed. Finiteness is the hypothesis used here; the construction with sigma-finite factors is a separate case.

Worked Applications of the Construction

A Common Pitfall and the Scope of the Result

A frequent error is to calculate the measure of a non-rectangular set by multiplying the measures of its coordinate projections. For the diagonal in the preceding example, both coordinate projections are \(\mathbb{N}\), so this incorrect rule would give \(\mu(\mathbb{N})\nu(\mathbb{N})=1\cdot1=1\). But \(D\) is not the rectangle \(\mathbb{N}\times\mathbb{N}\): that rectangle contains every pair of natural numbers. The correct calculation uses the disjoint decomposition into singleton rectangles and gives \(\lambda(D)=1/3\). The product formula applies to rectangles; for other sets, use additivity when a suitable disjoint decomposition is available, or the measure constructed on the product sigma-algebra.

The construction matters because it turns prescribed values on rectangles into a genuine measure on a sigma-algebra, where countable unions and complements can be handled. It also fixes the domain of the result: here it is \(\mathcal{M}\otimes\mathcal{N}\). The proof establishes existence and the rectangle values needed for later product-measure arguments; it does not rely on a uniqueness theorem or on integration-by-sections results.