The Role of a Finite Exhaustion
The uniqueness argument for product measures has two distinct stages. First, rectangle agreement gives equality on a family of sets that generates the product sigma-algebra. The finite-measure uniqueness theorem can then extend equality from that family to the whole sigma-algebra—but only on a region where both measures are finite. Second, an exhaustion by such regions allows equality to pass to the entire product space. Sigma-finiteness of the factors supplies natural candidate regions; it does not, by itself, guarantee that two arbitrary measures agreeing on rectangles are finite there.
Let \((X,\mathcal{M},\mu)\) and \((Y,\mathcal{N},\nu)\) be measure spaces. The product sigma-algebra \(\mathcal{M}\otimes\mathcal{N}\) is the sigma-algebra on \(X\times Y\) generated by measurable rectangles \(A\times B\), where \(A\in\mathcal{M}\) and \(B\in\mathcal{N}\). In the standard product-measure setting, \(\mu\) and \(\nu\) are sigma-finite: each space has an increasing sequence of measurable sets of finite measure whose union is the whole space. We will track what that assumption contributes to uniqueness.
The finiteness assumption on the blocks is automatic in the usual construction if each measure is known to give the prescribed rectangle value \(\mu(A)\nu(B)\): in particular, \(\Lambda_i(X_n\times Y_n)=\mu(X_n)\nu(Y_n)<\infty\). But it does not follow merely from the facts that the factors are sigma-finite and that \(\Lambda_1,\Lambda_2\) agree with each other on rectangles. That distinction is essential.
Equality on One Finite Block
Fix \(n\) and put \(C_n=X_n\times Y_n\). Consider the trace sigma-algebra on this block, consisting of the sets \(E\cap C_n\) for \(E\in\mathcal{M}\otimes\mathcal{N}\). The restrictions \(E\mapsto\Lambda_i(E\cap C_n)\) are finite measures on this trace sigma-algebra, because their total masses are \(\Lambda_i(C_n)<\infty\).
Use the family \(\mathcal{P}_n=\{(A\cap X_n)\times(B\cap Y_n):A\in\mathcal{M}, B\in\mathcal{N}\}\). It is a pi-system: it contains \(C_n\), and the intersection of two of its members is again in the family. It generates the trace sigma-algebra on \(C_n\), because measurable rectangles generate \(\mathcal{M}\otimes\mathcal{N}\) and intersection with \(C_n\) commutes with taking complements relative to \(C_n\) and countable unions. For any member of this pi-system, rectangle agreement gives \(\Lambda_1((A\cap X_n)\times(B\cap Y_n))=\Lambda_2((A\cap X_n)\times(B\cap Y_n))\).
The finite-measure pi-lambda uniqueness theorem says that two finite measures on a sigma-algebra which agree on a generating pi-system containing the whole space agree everywhere. This is the finite uniqueness result used in the earlier material on product measures. Its finiteness condition matters: it is what permits the lambda-system argument to compare complements by subtracting from the finite total mass. Applying it to the two restricted measures yields equality on every measurable subset of \(C_n\).
Proof of the theorem. For each \(n\), the preceding argument gives \(\Lambda_1(E\cap C_n)=\Lambda_2(E\cap C_n)\) for every \(E\in\mathcal{M}\otimes\mathcal{N}\). Since both exhaustions are increasing, the blocks \(C_n=X_n\times Y_n\) are increasing and \(\bigcup_n C_n=X\times Y\). Thus \(E\cap C_n\) increases to \(E\). By continuity from below for measures, \(\Lambda_i(E)=\lim_{n\to\infty}\Lambda_i(E\cap C_n)\), with infinity allowed. The terms in the two limits agree for every \(n\), so \(\Lambda_1(E)=\Lambda_2(E)\). This holds for every measurable \(E\), proving equality of the measures. \(\square\)
Worked Example: Finite Blocks for Lebesgue Factors
Take \(X=Y=\mathbb{R}\), \(\mathcal{M}=\mathcal{N}=\mathcal{B}_{\mathbb{R}}\), and let \(\mu=\nu=m\), Lebesgue measure restricted to the Borel sets. Use \(X_n=Y_n=[-n,n]\) for \(n\ge1\). These sets increase to \(\mathbb{R}\), and \(m([-n,n])=2n<\infty\). If two measures on \(\mathcal{B}_{\mathbb{R}}\otimes\mathcal{B}_{\mathbb{R}}\) have the prescribed rectangle values \(m(A)m(B)\) whenever both factor measures are finite, their common value on each exhaustion block is \(m([-n,n])m([-n,n])=(2n)(2n)=4n^2\), which is finite. The finite-block uniqueness argument therefore applies on every \([-n,n]^2\), and continuity from below gives equality on the full product sigma-algebra.
Check: at \(n=1\), each interval has measure \(2\), so the block value is \(2\cdot2=4\). At \(n=3\), each interval has measure \(6\), so the block value is \(6\cdot6=36\). In each case the block has finite measure, as required; the fact that these values grow with \(n\) does not affect finiteness of any individual block.
Worked Example: Comparing Restrictions on a Finite Block
Let \(X=Y=\{0,1\}\), with each power set as its sigma-algebra, and let \(\mu,\nu\) be counting measure. Take \(X_1=Y_1=\{0,1\}\). Suppose \(\Lambda_1,\Lambda_2\) are finite measures on the product and agree on every rectangle. The rectangle \(X_1\times Y_1\) is the whole space, so the common total mass is finite. For any \(A,B\subseteq\{0,1\}\), the rectangle \(A\times B\) belongs to the generating pi-system; agreement there and finite-measure uniqueness give equality on every subset of \(X\times Y\).
For an explicit check, take the rectangle \(A=\{0\}\), \(B=\{0,1\}\). Its points are \((0,0)\) and \((0,1)\). Agreement says both measures assign the same mass to this two-point set. The same reasoning applies to each rectangle, including the empty rectangle, whose measure is zero. Because every subset of this finite product is a finite union of singleton rectangles, these rectangle values determine all subset values. This example illustrates the finite step directly; it does not require a limiting argument because one block already covers the whole space.
What Fails Without the Required Finiteness
If there are no sigma-finite exhaustions of the factors, the standard block construction may not cover the product. For example, let \(X=Y\) be an uncountable set with the power-set sigma-algebra and counting measure. A set has finite counting measure exactly when it is finite. Any countable union of finite subsets is countable, so no sequence of finite-measure sets can exhaust this uncountable space. The same holds for \(Y\). Thus there is no sequence of finite rectangles \(X_n\times Y_n\) covering \(X\times Y\) obtained from factor exhaustions.
This identifies the exact gap: the proof can compare restrictions only on regions where both restricted measures are finite. Without a countable covering by such regions, the final continuity-from-below step cannot be invoked to recover equality on an arbitrary measurable set of the full product. This observation is a limitation of the argument, not a claim that uniqueness must fail for every particular pair of measures on non-sigma-finite spaces. A different hypothesis or argument could still establish uniqueness in a special case.
Worked Example: A Concrete Failure of the Rectangle-Agreement Shortcut
Here is why sigma-finiteness of the factors alone does not repair a theorem whose only assumption on the product measures is that they agree with each other on rectangles. Let \(X=Y=\mathbb{R}\) with Borel sigma-algebras and Lebesgue measure \(m\) on each factor. The factors are sigma-finite. Let \(D=\{(x,x):x\in\mathbb{R}\}\), a Borel subset of \(\mathbb{R}^2\). Define measures on \(\mathcal{B}_{\mathbb{R}^2}\) by \(\Lambda_1(S)=0\) if \(S\) is countable, and \(\Lambda_1(S)=\infty\) otherwise; define \(\Lambda_2(S)=0\) if \(S\setminus D\) is countable, and \(\Lambda_2(S)=\infty\) otherwise.
Both set functions are countably additive. For \(\Lambda_1\), a countable union of countable sets is countable, so if a disjoint union is uncountable, at least one member is uncountable and the sum of the measures is infinity. If the union is countable, all its members are countable and both the union and the sum have measure zero. For \(\Lambda_2\), apply the same reasoning to the parts outside \(D\): if the union has an uncountable part outside \(D\), at least one member does; if its part outside \(D\) is countable, each disjoint member has countable part outside \(D\). Thus \(\Lambda_2\) is also countably additive.
For a Borel rectangle \(A\times B\), if it is countable, its part outside \(D\) is countable, so both measures give zero. If it is uncountable, its part outside \(D\) must be uncountable. Indeed, if that part were countable, the diagonal part would imply that \(A\cap B\) is uncountable; then \((A\cap B)\times(A\cap B)\) would contain uncountably many off-diagonal points, a contradiction. Therefore both measures give infinity to an uncountable rectangle. They agree on every measurable rectangle.
Nevertheless, \(D\) is uncountable, so \(\Lambda_1(D)=\infty\), while \(D\setminus D=\varnothing\) is countable, so \(\Lambda_2(D)=0\). This does not contradict the theorem: these measures do not have finite mass on the exhaustion blocks. For example, \([-1,1]\times[-1,1]\) is uncountable and has uncountable part off the diagonal, so both measures assign it infinity. Also, their shared rectangle values are not the prescribed product values: on this block the Lebesgue product value would be \(2\cdot2=4\), not infinity.
Correction to the tempting shortcut: from rectangle agreement one may infer \(\Lambda_1(X_n\times Y_n)=\Lambda_2(X_n\times Y_n)\). One may not infer that this common value is \(\mu(X_n)\nu(Y_n)\) unless that product-value hypothesis is separately known. In this example the two measures agree on the block, but its common value is infinity, so the finite-measure uniqueness theorem cannot be applied there.
What the Argument Does—and Does Not—Say
The useful logical chain is precise:
- Factor sigma-finiteness supplies increasing finite-measure exhaustions \(X_n\) and \(Y_n\).
- A separate product-measure condition, or a direct assumption, ensures that both measures are finite on the corresponding blocks \(X_n\times Y_n\).
- Rectangle agreement then gives agreement on the generating pi-system inside each block; finite-measure uniqueness gives agreement on the whole block.
- The blocks increase to \(X\times Y\), so continuity from below extends equality to every measurable subset of the product.
Each link has a distinct role. Sigma-finiteness of \(\mu\) and \(\nu\) does not state anything by itself about the values of arbitrary \(\Lambda_i\). Rectangle agreement between \(\Lambda_1\) and \(\Lambda_2\) does not identify their shared values with \(\mu(A)\nu(B)\). And without finite restrictions, the finite uniqueness theorem cannot be applied. When the factors lack sigma-finite exhaustions, the familiar proof has no countable family of finite blocks to carry out the final step. These are the exact points at which the argument depends on its hypotheses.
Check Your Understanding
For each question, separate the role of rectangle agreement from the role of finiteness.
- On a fixed block \(C_n=X_n\times Y_n\), which two conditions allow finite-measure uniqueness to be applied to the restricted measures?
- Why does \(\Lambda_1(A\times B)=\Lambda_2(A\times B)\) not imply \(\Lambda_i(X_n\times Y_n)=\mu(X_n)\nu(Y_n)\)? State the additional type of hypothesis that would justify that value.
- Explain why increasing finite-measure exhaustions of both factors give increasing blocks whose union is \(X\times Y\).
- For counting measure on an uncountable set with the power-set sigma-algebra, why can no countable sequence of finite-measure sets exhaust the space?
- If the standard finite-block proof cannot be completed, what conclusion is justified: that uniqueness fails, or only that this proof does not establish uniqueness?