Absolute Integrability and Integrable Sections

Why Absolute Integrability Controls Sections

For a function on a product space, knowing that its integral exists is not by itself enough to justify taking integrals along every section. Some individual sections can have infinite absolute integral, even when the function is integrable on the product. The useful conclusion is almost-everywhere: absolute integrability on the product forces almost every section to be integrable, and the resulting section-integral function is itself integrable.

Let \((X,\mathcal{M},\mu)\) and \((Y,\mathcal{N},\nu)\) be sigma-finite measure spaces, and equip \(X\times Y\) with the product sigma-algebra \(\mathcal{M}\otimes\mathcal{N}\) and product measure \(\mu\times\nu\). Neither factor is assumed complete. Throughout, \(f:X\times Y\to\mathbb{R}\) is a product-measurable, finite-valued real function. For \(x\in X\), its section is \(f_x:Y\to\mathbb{R}\), defined by \(f_x(y)=f(x,y)\). A real-valued measurable function is integrable when the integral of its absolute value is finite.

We use Tonelli’s theorem for nonnegative product-measurable functions: their section integrals are measurable as extended nonnegative functions, and their iterated integral equals their product-space integral, with infinity allowed. The sigma-finiteness hypotheses above ensure this version applies. The central move is to apply it to \(|f|\), not to interchange integrals of a signed function.

The Result for Integrable Sections

Proof. Write \(C=\int_X h\,d\mu<\infty\). For each positive integer \(n\), let \(E_n=\{x:h(x)>n\}\). The pointwise inequality \(n\chi_{E_n}\le h\) gives \(n\mu(E_n)\le C\), so \(\mu(E_n)\le C/n\). The set \(\{h=\infty\}\) is contained in every \(E_n\), and therefore its measure is at most \(C/n\) for every \(n\). Letting \(n\to\infty\) shows \(\mu(\{h=\infty\})=0\). \(\square\)

Proof. Set \(h(x,y)=|f(x,y)|\). This is a nonnegative product-measurable function. By Tonelli’s theorem, the extended-valued function \(H(x)=\int_Y |f_x(y)|\,d\nu(y)\) is measurable and satisfies

The lemma shows \(H(x)<\infty\) for \(\mu\)-almost every \(x\). By the definition of integrability, this means that \(f_x\) is integrable for almost every \(x\).

To construct the section-integral function carefully, write \(f=f^+-f^-\), where \(f^+=\max(f,0)\) and \(f^-=\max(-f,0)\). Define the extended nonnegative functions \(A(x)=\int_Y f_x^+\,d\nu\) and \(B(x)=\int_Y f_x^-\,d\nu\). Tonelli’s theorem implies that \(A\) and \(B\) are measurable. Since \(|f|=f^++f^-\), additivity of the nonnegative integral gives \(H=A+B\). Thus \(G=\{x:H(x)<\infty\}\) is measurable, and on \(G\) both \(A\) and \(B\) are finite.

Define \(g(x)=A(x)-B(x)\) for \(x\in G\), and \(g(x)=0\) for \(x\notin G\). On the measurable set \(G\), the difference is finite and measurable; defining it to be zero on the measurable complement therefore gives a measurable real-valued function on all of \(X\). For \(x\in G\), the section \(f_x\) is integrable and \(g(x)=\int_Y f_x\,d\nu\). Also, \(|A(x)-B(x)|\le A(x)+B(x)=H(x)\) on \(G\), while \(g=0\) off \(G\). Hence \(|g|\le H\) everywhere, so

This proves that \(g\) is integrable and establishes all the claims. \(\square\)

Worked Examples

Exceptional Sets, Measurability, and a Common Pitfall

The theorem does not say that every section is integrable. It says that the set of exceptions is measurable and null. It also gives a specific safe way to define \(g\) everywhere: use the section integral on the measurable good set \(G\), then set \(g=0\) on its measurable complement. The proof shows directly that this piecewise-defined function is measurable and integrable.

A common mistake is to assume that any modification on a null set preserves measurability. That is guaranteed on a complete measure space, but the theorem does not assume completeness. For a concrete counterexample, let \(X=\{a,b,c\}\), \(\mathcal{M}=\{\varnothing,\{a,b\},\{c\},X\}\), and give \(\{a,b\}\) measure \(0\) and \(\{c\}\) measure \(1\). The zero function is measurable. Changing it to \(1\) at \(a\) and leaving it \(0\) elsewhere produces a function whose set of values greater than \(1/2\) is \(\{a\}\), which is not in \(\mathcal{M}\). The modified function is not measurable, although the change occurred inside the null set \(\{a,b\}\). By contrast, setting a function equal to zero on the whole measurable null set \(\{a,b\}\) preserves measurability: this is a piecewise definition along a measurable set. On a complete measure space, arbitrary changes on a subset of a measurable null set are measurable; without completeness, they need not be.

Finally, the estimate \(\int_X|g|\,d\mu\le\int_{X\times Y}|f|\,d(\mu\times\nu)\) is an absolute-integrability bound. The argument here does not assert an equality between the integral of \(f\) on the product and an iterated signed integral. Establishing that equality requires the full Fubini theorem. What has been established is the essential preparation: almost all sections are integrable, and their section integrals can be assembled into a measurable integrable function.