Uniqueness for Finite Product Measures

Why Rectangle Values Determine a Finite Measure

In a product space, measurable rectangles are usually much easier to evaluate than arbitrary measurable sets. The uniqueness question is whether that simpler information determines the measure everywhere on the product sigma-algebra. For finite measures, it does: rectangles form a class closed under intersections, and the pi-lambda theorem propagates equality from that class to the sigma-algebra it generates. Finiteness matters in the proof because it makes subtraction from the total mass legitimate.

Let \((X,\mathcal{M})\) and \((Y,\mathcal{N})\) be measurable spaces. Their product sigma-algebra \(\mathcal{M}\otimes\mathcal{N}\) is generated by the measurable rectangles \(A\times B\), where \(A\in\mathcal{M}\) and \(B\in\mathcal{N}\). In particular, the rectangles form a generating class, but they are not in general a sigma-algebra. The argument below uses their closure under finite intersections, not closure under arbitrary countable unions.

For the rectangle class \(\mathcal{P}=\{A\times B:A\in\mathcal{M},\,B\in\mathcal{N}\}\), intersections satisfy \((A_1\times B_1)\cap(A_2\times B_2)=(A_1\cap A_2)\times(B_1\cap B_2)\), so \(\mathcal{P}\) is a pi-system. It contains \(X\times Y\), and it generates \(\mathcal{M}\otimes\mathcal{N}\). The next result explains why these facts suffice.

The Pi-Lambda Theorem

Proof. Let \(\mathcal{L}\) be the intersection of all lambda-systems on \(S\) that contain \(\mathcal{P}\). Such systems exist because the power set of \(S\) is one, and an intersection of lambda-systems is a lambda-system. Thus \(\mathcal{L}\) is the smallest lambda-system containing \(\mathcal{P}\). Since \(\mathcal{D}\) is one such system, \(\mathcal{L}\subseteq\mathcal{D}\). It is enough to prove that \(\mathcal{L}\) is a sigma-algebra.

First fix \(A\in\mathcal{P}\), and define \(\mathcal{L}_A=\{B\in\mathcal{L}:A\cap B\in\mathcal{L}\}\). This is a lambda-system: \(S\in\mathcal{L}_A\) because \(A\in\mathcal{L}\); if \(B\in\mathcal{L}_A\), then \(A\cap B\subseteq A\), and \(A\cap(S\setminus B)=A\setminus(A\cap B)\in\mathcal{L}\), by closure of a lambda-system under differences of nested members; and intersections with \(A\) preserve disjointness, so \(\mathcal{L}_A\) is closed under countable disjoint unions. The nested-difference property follows from the definition: if \(F\subseteq E\) and \(E,F\in\mathcal{L}\), then \(E\setminus F=S\setminus(F\cup(S\setminus E))\in\mathcal{L}\), since the union inside the complement is disjoint.

For each \(B\in\mathcal{P}\), \(A\cap B\in\mathcal{P}\subseteq\mathcal{L}\), so \(\mathcal{P}\subseteq\mathcal{L}_A\). Minimality of \(\mathcal{L}\) gives \(\mathcal{L}\subseteq\mathcal{L}_A\). Consequently, \(A\cap B\in\mathcal{L}\) for every \(A\in\mathcal{P}\) and \(B\in\mathcal{L}\).

Now fix \(B\in\mathcal{L}\) and define \(\mathcal{K}_B=\{A\in\mathcal{L}:A\cap B\in\mathcal{L}\}\). The same verification shows that \(\mathcal{K}_B\) is a lambda-system: it contains \(S\) because \(S\cap B=B\in\mathcal{L}\); complements use \(B\setminus(A\cap B)\); and disjoint unions remain disjoint after intersection with \(B\). The preceding paragraph shows \(\mathcal{P}\subseteq\mathcal{K}_B\). Minimality therefore gives \(\mathcal{L}\subseteq\mathcal{K}_B\). We have proved that \(\mathcal{L}\) is closed under finite intersections.

Because \(\mathcal{L}\) contains \(S\) and is closed under complements, it contains finite unions by De Morgan’s laws. For any sequence \(E_1,E_2,\ldots\) in \(\mathcal{L}\), define \(F_1=E_1\) and \(F_n=E_n\setminus\bigcup_{j=1}^{n-1}E_j\) for \(n\ge2\). Finite unions and nested differences show \(F_n\in\mathcal{L}\); the \(F_n\) are disjoint and \(\bigcup_n F_n=\bigcup_n E_n\). Closure under countable disjoint unions now gives \(\bigcup_n E_n\in\mathcal{L}\). Thus \(\mathcal{L}\) is a sigma-algebra, so \(\sigma(\mathcal{P})\subseteq\mathcal{L}\subseteq\mathcal{D}\). \(\square\)

Uniqueness for Finite Product Measures

Proof. Define \(\mathcal{D}=\{E\in\mathcal{M}\otimes\mathcal{N}:\lambda_1(E)=\lambda_2(E)\}\). The whole space \(S=X\times Y\) is a rectangle, so \(\lambda_1(S)=\mu(X)\nu(Y)=\lambda_2(S)\). Hence \(S\in\mathcal{D}\). If \(E\in\mathcal{D}\), finiteness gives \(\lambda_i(S\setminus E)=\lambda_i(S)-\lambda_i(E)\) for \(i=1,2\). Equality of the total masses and of the masses of \(E\) implies equality on \(S\setminus E\), so \(\mathcal{D}\) is closed under complements. If \(E_1,E_2,\ldots\) are disjoint members of \(\mathcal{D}\), countable additivity gives \(\lambda_i(\bigcup_n E_n)=\sum_n\lambda_i(E_n)\) for each \(i\); the two sums agree term by term. Thus \(\mathcal{D}\) is a lambda-system. It contains every measurable rectangle by the assumed agreement, and the rectangles generate \(\mathcal{M}\otimes\mathcal{N}\). The pi-lambda theorem implies \(\mathcal{M}\otimes\mathcal{N}\subseteq\mathcal{D}\). Therefore the measures agree on every measurable set. \(\square\)

A Common Pitfall: Subtracting Infinite Total Masses

The complement step in the uniqueness proof uses \(\lambda_i(S\setminus E)=\lambda_i(S)-\lambda_i(E)\). This identity is valid when \(\lambda_i(S)\) is finite. It does not provide a usable subtraction when both quantities are infinite. For a concrete illustration, counting measure on \(\mathbb{N}\) assigns mass \(\infty\) to \(S=\mathbb{N}\) and also mass \(\infty\) to the even numbers \(E=\{2,4,6,\ldots\}\). The complement, the odd numbers, has mass \(\infty\), not a value obtained by subtracting \(\infty\) from \(\infty\). Thus one cannot use the finite-measure complement argument for arbitrary infinite measures.

A related incorrect argument is to say that agreement on a generating class automatically implies agreement everywhere. The pi-lambda theorem requires a lambda-system containing a pi-system; for equality sets, proving the lambda-system property is an essential step. Here it follows from finite total mass and countable additivity. The subsequent uniqueness result for sigma-finite product measures uses additional structure to address the infinite case.