What Fails Without Sigma-Finiteness

Why Sigma-Finiteness Matters for Uniqueness

For finite measures, agreement on measurable rectangles is a powerful uniqueness principle: rectangles generate the product sigma-algebra, so agreement there determines the measure. A standard uniqueness theorem extends this conclusion to sigma-finite measures. Without sigma-finiteness, however, the conclusion can fail.

The failure is not merely a technical gap in a proof. There can be two genuinely different measures that give the same value to every measurable rectangle. We will see this using counting measure on an uncountable space and Lebesgue measure on a second copy of the interval. The diagonal will distinguish the two measures.

Key point: Agreement on rectangles need not imply agreement on the product sigma-algebra when the measures involved are not sigma-finite.

The Spaces and Their Rectangle Values

Let \(X=Y=[0,1]\), each equipped with its Borel sigma-algebra. On \(X\), use counting measure \(\mu\): for a Borel set \(A\), \(\mu(A)\) is its number of points if it is finite, and is infinity if it is infinite. On \(Y\), use Lebesgue measure \(\lambda\), restricted to Borel sets. Let \(\mathcal{P}\) denote the product sigma-algebra \(\mathcal{B}(X)\otimes\mathcal{B}(Y)\).

Definition: A measure is sigma-finite if its space is a countable union of measurable sets, each having finite measure. The counting measure \(\mu\) on \(X\) is not sigma-finite: every set of finite counting measure is finite, and a countable union of finite sets is countable, so it cannot cover \([0,1]\).

For a set \(E\in\mathcal{P}\), write \(E_x=\{y\in Y:(x,y)\in E\}\) for its section at \(x\), and \(E^y=\{x\in X:(x,y)\in E\}\) for its section at \(y\). Sections of product-measurable sets are Borel, so the measures of these sections are defined.

On a rectangle \(A\times B\), with \(A\) Borel in \(X\) and \(B\) Borel in \(Y\), the natural common rectangle value in this example is \(\mu(A)\lambda(B)\) when this product is defined, with value zero when \(\mu(A)=\infty\) and \(\lambda(B)=0\). This last convention matters: the expression \(0\cdot\infty\) is not defined as an extended-real product. Both measures constructed below give this same value on every rectangle.

Two Measures on the Product

The first measure adds the Lebesgue measures of the horizontal sections. The sum over possibly uncountably many \(x\) is defined as the supremum of all finite subsums. The second integrates the counting measure of the vertical sections. For Borel subsets of \(X\times Y\), the function \(y\mapsto\mu(E^y)\) is Lebesgue measurable: for each positive integer \(n\), the set of \(y\) for which \(E^y\) contains at least \(n\) distinct points is a projection of a Borel set, hence is analytic and Lebesgue measurable. Analytic subsets of the interval are Lebesgue measurable.

\nu_1(E)=\sum_{x\in X}\lambda(E_x), \qquad \nu_2(E)=\int_Y \mu(E^y)\,\lambda(dy), \qquad E\in\mathcal{P}.

Here the sum in \(\nu_1\) has its usual nonnegative meaning: \(\sum_{x\in X}a_x=\sup\{\sum_{x\in F}a_x:F\subseteq X\text{ is finite}\}\). The integral defining \(\nu_2\) uses the completed Lebesgue measure; its value on Borel sets is therefore well-defined.

Worked Example: A Finite Rectangle

Let \(A=\{x_1,x_2\}\) and let \(B\) be a Borel set with \(\lambda(B)=1/3\). For \(\nu_1\), only the two sections indexed by \(x_1,x_2\) have positive measure, and both are \(B\). Thus \(\nu_1(A\times B)=1/3+1/3=2/3\). For \(\nu_2\), the vertical section is \(A\) for \(y\in B\), with counting measure \(2\), and is empty otherwise. Hence \(\nu_2(A\times B)=\int 2\mathbf{1}_B(y)\,\lambda(dy)=2/3\).

Worked Example: An Infinite Rectangle with a Null Side

Let \(A\) be an infinite Borel subset of \(X\), and let \(B=\{y_0\}\). Since \(\lambda(B)=0\), every nonempty horizontal section of \(A\times B\) has measure zero, so \(\nu_1(A\times B)=0\). For \(\nu_2\), the section has counting measure \(\infty\) at \(y_0\) and zero elsewhere. A function that is infinite at a single Lebesgue-null point and zero elsewhere has integral zero. Therefore \(\nu_2(A\times B)=0\) as well. This is the rectangle case in which the informal product \(\infty\cdot 0\) must not be used as a numerical calculation.

Theorem: The set functions \(\nu_1\) and \(\nu_2\) are measures on \(\mathcal{P}\). They agree on every measurable rectangle, but they do not agree on every set in \(\mathcal{P}\).

Proof: First consider \(\nu_1\). It is nonnegative and \(\nu_1(\varnothing)=0\). If \(E_1,E_2,\ldots\) are pairwise disjoint product-measurable sets, their sections at each \(x\) are pairwise disjoint. Countable additivity of \(\lambda\) therefore gives \(\lambda((\bigcup_n E_n)_x)=\sum_n\lambda((E_n)_x)\). Summing over \(x\) gives \(\nu_1(\bigcup_n E_n)=\sum_n\nu_1(E_n)\). The interchange of the two nonnegative sums is valid because each is the supremum of its finite subsums. Thus \(\nu_1\) is a measure.

For \(\nu_2\), each function \(y\mapsto\mu(E^y)\) is measurable by the section-measurability fact stated above, and is nonnegative. If the \(E_n\) are pairwise disjoint, their sections at each \(y\) are pairwise disjoint. Countable additivity of \(\mu\) gives \(\mu((\bigcup_n E_n)^y)=\sum_n\mu(E_n^y)\). Integrating this identity and applying the monotone convergence theorem to the increasing partial sums yields \(\nu_2(\bigcup_n E_n)=\sum_n\nu_2(E_n)\). Also \(\nu_2(\varnothing)=0\), so \(\nu_2\) is a measure.

Now take any Borel rectangle \(A\times B\). For \(\nu_1\), the horizontal section is \(B\) for \(x\in A\) and empty otherwise. Thus its value is zero if \(\lambda(B)=0\), is \(n\lambda(B)\) if \(A\) has \(n\) points, and is infinity if \(A\) is infinite and \(\lambda(B)>0\). For \(\nu_2\), the vertical section is \(A\) for \(y\in B\) and empty otherwise, giving exactly the same three cases by integration. Hence the measures agree on every rectangle.

Finally, let \(D=\{(x,y)\in[0,1]^2:x=y\}\), which is Borel. Each horizontal section \(D_x=\{x\}\) has Lebesgue measure zero, so \(\nu_1(D)=0\). Each vertical section \(D^y=\{y\}\) has counting measure one, so \(\nu_2(D)=\int_{[0,1]}1\,\lambda(dy)=1\). The two measures differ on \(D\), completing the proof.

\(\square\)

Worked Example: The Diagonal Separates the Measures

The diagonal is a useful test because its sections behave differently under the two factor measures. Horizontally, there is one point per section, but each such point has Lebesgue measure zero; summing these section measures still gives \(\nu_1(D)=0\). Vertically, each section has one point, whose counting measure is one; integrating that constant value over an interval of length one gives \(\nu_2(D)=1\). The calculation shows explicitly why checking rectangles alone misses a measurable set that separates the measures.

What the Example Does—and Does Not—Say

This construction shows that rectangle values can fail to determine a measure in the absence of sigma-finiteness. In particular, \(\nu_1\) and \(\nu_2\) agree on the generating rectangles but disagree on a Borel set. The counting measure on the first factor is not sigma-finite, so the usual uniqueness theorem for product measures does not apply.

The example does not say that agreement on rectangles is never useful without sigma-finiteness, or that every non-sigma-finite product has multiple extensions. It shows that sigma-finiteness is a substantive hypothesis in a general uniqueness result: removing it allows failure, even for familiar spaces and measures.

It also highlights a common pitfall. When one factor has infinite measure and another set has measure zero, writing the rectangle value as \(\infty\cdot 0\) does not specify a value. Here both measures assign zero to rectangles with an infinite first side and a null second side, but that agreement does not resolve their disagreement on the diagonal. A sound argument must keep rectangle values, their possible ambiguities, and the uniqueness hypotheses distinct.

Takeaway: Sigma-finiteness is not needed to define every useful measure, but it is essential to the standard guarantee that values on measurable rectangles determine a unique measure on the product sigma-algebra.

Check Your Understanding

Use the construction above to test whether you can distinguish rectangle agreement from uniqueness on the whole product sigma-algebra.

Why can the uncountable interval not be covered by countably many sets of finite counting measure?
What values do \(\nu_1\) and \(\nu_2\) assign to \(A\times B\) when \(A\) is finite and \(B\) has positive Lebesgue measure?
Why must the rectangle case \(\mu(A)=\infty\) and \(\lambda(B)=0\) be handled without multiplying extended real numbers?
Calculate \(\nu_1(D)\) and \(\nu_2(D)\) from the horizontal and vertical sections of the diagonal.
Which hypothesis of the usual uniqueness theorem is missing in this example?