Sigma-Finiteness as a Hypothesis in Product Integration

Why Sigma-Finiteness Appears in Product Integration

Tonelli’s theorem allows nonnegative measurable functions to be integrated in either order, while Fubini’s theorem allows the same exchange for absolutely integrable functions. In their standard product-measure forms, both results assume that the two factor measures are sigma-finite. This is not a technical condition to omit merely because a function is nonnegative: with uncountable counting measure, a measurable diagonal gives different iterated integrals.

Recall that a measure is finite if its value on the whole space is finite. A measure is sigma-finite if its space is a countable union of measurable sets of finite measure. Every finite measure is sigma-finite, but a sigma-finite measure need not be finite. In what follows, product sigma-algebras are denoted by \(\mathcal{M}\otimes\mathcal{N}\), and functions in Tonelli’s theorem take values in \([0,\infty]\).

Definition: A measure space \((X,\mathcal{M},\mu)\) is sigma-finite if there are sets \(X_i\in\mathcal{M}\), \(i\ge1\), such that \(X=\bigcup_{i\ge1}X_i\) and \(\mu(X_i)<\infty\) for every \(i\). The sets \(X_i\) are called a finite-measure exhaustion. They need not be disjoint.

What the Hypothesis Does

For sigma-finite measure spaces \((X,\mathcal{M},\mu)\) and \((Y,\mathcal{N},\nu)\), the standard product-measure theorem constructs a measure \(\mu\times\nu\) on \(\mathcal{M}\otimes\mathcal{N}\) satisfying \((\mu\times\nu)(A\times B)=\mu(A)\nu(B)\) for measurable rectangles \(A\times B\). The construction uses the rectangle prescription as a premeasure and the Carathéodory extension theorem. Sigma-finiteness is what makes this prescription a premeasure on the rectangle algebra and ensures that its extension is unique. The product measure is sigma-finite: if \(X=\bigcup_iX_i\) and \(Y=\bigcup_jY_j\) are finite-measure exhaustions, then the countable family \(X_i\times Y_j\) covers \(X\times Y\), and each such rectangle has finite product measure.

Uniqueness can be understood by reducing to finite pieces. On a rectangle of finite product measure, finite measures that agree on measurable rectangles agree throughout the product sigma-algebra. This follows from the \(\pi\)-\(\lambda\) theorem: rectangles form a \(\pi\)-system, and the sets on which two finite measures of equal total mass agree form a \(\lambda\)-system. Countably many finite-measure pieces then determine the measure on the whole product. The extension and \(\pi\)-\(\lambda\) results used here are standard measure-theoretic results; the point is that finite-measure localization is available because of sigma-finiteness.

Theorem (Tonelli and Fubini, sigma-finite form): Let \((X,\mathcal{M},\mu)\) and \((Y,\mathcal{N},\nu)\) be sigma-finite measure spaces, and let \(\mu\times\nu\) be their product measure on \(\mathcal{M}\otimes\mathcal{N}\). If \(f:X\times Y\to[0,\infty]\) is product-measurable, Tonelli’s theorem gives \(\int f\,d(\mu\times\nu)=\int_X\int_Y f(x,y)\,d\nu(y)\,d\mu(x)=\int_Y\int_X f(x,y)\,d\mu(x)\,d\nu(y)\), with infinity allowed. If \(f\) is real-valued or complex-valued and \(\int |f|\,d(\mu\times\nu)<\infty\), Fubini’s theorem gives the same equality, with the iterated integrals defined almost everywhere and finite.

The factor measures must be sigma-finite for these standard statements. Product measurability and nonnegativity do not replace that hypothesis. Nor does sigma-finiteness mean that the whole product space has finite measure: it only provides countably many pieces on which the measures are finite.

Worked Example: A Sigma-Finite Product with Counting Measure

Let \(X=\mathbb{N}=\{1,2,\ldots\}\) with counting measure \(\kappa\), and let \(Y=[0,1]\) with Lebesgue measure \(m\). Both are sigma-finite: \(\mathbb{N}=\bigcup_{j\ge1}\{1,\ldots,j\}\), and \(m(Y)=1\). On \(\mathbb{N}\times[0,1]\), set \(f(n,t)=2^{-n}\). This is a nonnegative measurable function.

For each \(n\), its section integral is \(\int_0^1f(n,t)\,dm(t)=2^{-n}\cdot1=2^{-n}\). Thus the iterated integral that sums over \(n\) is

\int_{\mathbb{N}}\int_{[0,1]}f(n,t)\,dm(t)\,d\kappa(n) =\sum_{n=1}^{\infty}2^{-n}=1.

For each fixed \(t\), the integral over \(\mathbb{N}\) is \(\sum_{n=1}^{\infty}2^{-n}=1\), so integrating this constant over \([0,1]\) also gives \(1\). Tonelli applies because both factor measures are sigma-finite and \(f\ge0\). Check: direct summation of the geometric series gives \(\sum_{n=1}^{\infty}2^{-n}=(1/2)/(1-1/2)=1\), and the other order gives \(1\cdot m([0,1])=1\).

Uncountable Counting Measure: A Diagonal Counterexample

Let \(X=Y=[0,1]\), let \(\kappa\) be counting measure on the Borel sigma-algebra \(\mathcal{B}_{[0,1]}\), and let \(m\) be Lebesgue measure on \([0,1]\). Counting measure assigns to a set its cardinality if it is finite and \(+\infty\) if it is infinite. It is not sigma-finite here: every set of finite counting measure is finite, and a countable union of finite sets is countable, so such a union cannot cover \([0,1]\). The measure \(m\), in contrast, is finite.

Consider the diagonal \(D=\{(x,y)\in[0,1]^2:x=y\}\). It is Borel measurable. For any \(E\in\mathcal{B}_{[0,1]}\otimes\mathcal{B}_{[0,1]}\), its section \(E_x=\{y:(x,y)\in E\}\) is Borel measurable. Define a measure on the product sigma-algebra by \(\lambda(E)=\sum_{x\in[0,1]}m(E_x)\), where an uncountable sum of nonnegative numbers means the supremum of the sums over finite subsets of indices. This is countably additive: for disjoint measurable \(E_n\), the sections \((E_n)_x\) are disjoint, so countable additivity of \(m\), followed by interchange of nonnegative sums, gives \(\lambda(\bigcup_nE_n)=\sum_n\lambda(E_n)\). For a rectangle \(A\times B\), the formula gives \(\lambda(A\times B)=\sum_{x\in A}m(B)\), agreeing with the rectangle product prescription, with the convention \(0\cdot\infty=0\).

Worked Example: The Diagonal and Unequal Iterated Integrals

Take \(f=\chi_D\), which is nonnegative and product-measurable. For each fixed \(x\), the section \(D_x=\{y:(x,y)\in D\}\) is the singleton \(\{x\}\). A singleton has Lebesgue measure zero, so \(\int_Y f(x,y)\,dm(y)=m(\{x\})=0\) for every \(x\). Integrating this zero function with respect to \(\kappa\) gives \(0\).

In the other order, for each fixed \(y\), the section \(\{x:(x,y)\in D\}\) is the singleton \(\{y\}\). Its counting measure is \(1\), since it contains exactly one point. Hence the inner integral with respect to \(\kappa\) equals \(1\) for every \(y\), and the outer integral is \(\int_{[0,1]}1\,dm=1\). Meanwhile, the product measure just defined satisfies \(\lambda(D)=\sum_{x\in[0,1]}m(\{x\})=0\). Therefore

\int_X\int_Y\chi_D(x,y)\,dm(y)\,d\kappa(x)=0, \qquad \int_Y\int_X\chi_D(x,y)\,d\kappa(x)\,dm(y)=1, \qquad \int_{X\times Y}\chi_D\,d\lambda=0.

Check: in the first order every inner integral is \(0\), so its counting integral is \(0\). In the second order every inner integral is \(1\), and the outer interval has Lebesgue measure \(1\), giving \(1\). The product integral is \(\lambda(D)=0\) by the defining section-sum formula. Thus the values disagree exactly as computed. Tonelli’s theorem cannot be invoked: the counting measure \(\kappa\) is not sigma-finite.

Why the Diagonal Does Not Contradict the Theorem

The diagonal example identifies a failed hypothesis, not a failure of Tonelli’s theorem. Tonelli’s conclusion is guaranteed under its stated sigma-finiteness assumptions; here one factor lacks a finite-measure exhaustion. The example also shows why the phrase “the product measure” needs care outside the standard setting: without sigma-finiteness, a rectangle prescription need not have a unique extension, and product-measure conclusions cannot be inferred from the usual theorem.

Worked Example: The Same Diagonal with Finite Factors

Now equip both copies of \([0,1]\) with Lebesgue measure. These measures are finite and hence sigma-finite. For the same diagonal indicator \(\chi_D\), each section is a singleton and has Lebesgue measure zero. Both iterated integrals are therefore \(\int_{[0,1]}0\,dm=0\). The diagonal has two-dimensional Lebesgue measure zero as well: for every \(\varepsilon>0\), it is covered by the \(n\) squares \([j/n,(j+1)/n]\times[j/n,(j+1)/n]\), \(j=0,\ldots,n-1\), whose total area is \(n(1/n)^2=1/n\); choosing \(n\) with \(1/n<\varepsilon\) shows its outer measure is zero. Thus all three integrals equal zero, in agreement with Tonelli. The contrast is the factor measure, not the diagonal function.

A Reliable Hypothesis Check

A common incorrect argument is: “The function is nonnegative and measurable, so Tonelli gives equality of the iterated integrals.” Applied to \(\chi_D\) with \(\kappa\) and \(m\), this would assert \(0=1\). The correction is to check the theorem’s hypotheses before using its conclusion: the function is product-measurable and nonnegative, but \(\kappa\) is not sigma-finite. Nonnegativity satisfies one hypothesis; it does not compensate for the missing measure-space hypothesis.

For Fubini, the same discipline applies. In addition to sigma-finiteness and product measurability, the function must be absolutely integrable with respect to the product measure. For Tonelli, the function must be nonnegative and product-measurable, and the factor measures must be sigma-finite in the standard theorem. Keeping these conditions distinct prevents an invalid exchange of integrals, especially when one factor is an uncountable counting measure.

Check Your Understanding

For each question, identify the measure hypothesis as well as the relevant function property.

Why is counting measure on \([0,1]\) not sigma-finite?
For the diagonal indicator with counting measure in the \(x\)-variable and Lebesgue measure in the \(y\)-variable, compute each iterated integral and the product integral defined by \(\lambda(E)=\sum_xm(E_x)\).
Which hypotheses of the standard Tonelli theorem hold for the diagonal example, and which one fails?
Explain why a finite-measure exhaustion gives a countable cover of the product by finite-product-measure rectangles.
If both factors on \([0,1]\) carry Lebesgue measure, what are the two iterated integrals of the diagonal indicator, and why is Tonelli applicable?