From Nonnegative Integration to Signed Integration
Tonelli’s theorem handles nonnegative functions: it permits the two orders of integration to be compared even when an integral is infinite. For a real-valued function, positive and negative contributions must instead be kept separate. If the integral of the absolute value is finite, Tonelli controls both contributions in each order, so their difference is finite as well. This is the mechanism behind the real-valued Fubini identity.
Throughout, let \((X,\mathcal{M},\mu)\) and \((Y,\mathcal{N},\nu)\) be sigma-finite measure spaces, and use their product measure \(\mu\times\nu\) on \(\mathcal{M}\otimes\mathcal{N}\). No completeness assumption is required. Let \(f:X\times Y\to\mathbb{R}\) be measurable. Its sections are \(f_x(y)=f(x,y)\) and \(f^y(x)=f(x,y)\). Sections of a product-measurable function are measurable; we also use Tonelli’s theorem for nonnegative product-measurable functions.
The definition of the signed integral requires both nonnegative integrals on the right to be finite. The absolute-integrability assumption guarantees this: \(0\le f^+,f^-\le |f|\). The same control must also hold after taking sections. The following result makes that step explicit. We use Tonelli’s theorem in its sigma-finite form: for any nonnegative measurable \(h\) on \(X\times Y\), both section-integral functions are measurable and \(\int h\,d(\mu\times\nu)=\int_X\int_Y h(x,y)\,d\nu(y)\,d\mu(x)=\int_Y\int_X h(x,y)\,d\mu(x)\,d\nu(y)\), with infinite values allowed.
Proof. Apply Tonelli’s theorem to the nonnegative functions \(f^+\) and \(f^-\). Define \(A(x)=\int_Y f^+(x,y)\,d\nu(y)\) and \(B(x)=\int_Y f^-(x,y)\,d\nu(y)\). These are measurable extended nonnegative functions, and Tonelli gives \(\int_X A\,d\mu=\int_{X\times Y}f^+\,d(\mu\times\nu)\) and \(\int_X B\,d\mu=\int_{X\times Y}f^-\,d(\mu\times\nu)\). Both right-hand sides are finite, since \(f^+,f^-\le |f|\). Consequently \(A\) and \(B\) are finite \(\mu\)-almost everywhere: if, for example, \(A=\infty\) on a set of positive measure, its nonnegative integral would be infinite. The measurable set \(N=\{x:A(x)+B(x)=\infty\}\) is therefore null.
For \(x\notin N\), \(f_x\) is integrable because \(\int_Y|f_x|\,d\nu=A(x)+B(x)<\infty\). Its signed integral is \(A(x)-B(x)\). Define \(g(x)=A(x)-B(x)\) on \(X\setminus N\), and \(g(x)=0\) on \(N\). This is a measurable representative of the section integral, defined on all of \(X\). Since \(A\) and \(B\) are integrable, linearity of the integral for integrable functions gives \(g\in L^1(\mu)\) and \(\int_Xg\,d\mu=\int_XA\,d\mu-\int_XB\,d\mu\). Changing the values on the measurable null set \(N\) does not alter this integral.
Substituting the two Tonelli identities yields \(\int_Xg\,d\mu=\int f^+\,d(\mu\times\nu)-\int f^-\,d(\mu\times\nu)=\int f\,d(\mu\times\nu)\). Both terms are finite, so this subtraction is valid. Applying the same argument with \(X,\mu\) and \(Y,\nu\) interchanged gives an integrable representative \(h(y)\) of the other section integral for \(\nu\)-almost every \(y\), and \(\int_Yh\,d\nu=\int f\,d(\mu\times\nu)\). Thus both iterated integrals are finite and equal to the product integral. \(\square\)
Why the Difference Is Legitimate
The proof does not apply Tonelli directly to a signed function. Instead, it applies Tonelli separately to \(f^+\) and \(f^-\), and only subtracts after showing the resulting integrals are finite. In particular, the hypothesis \(\int|f|<\infty\) implies \(\int f^+<\infty\) and \(\int f^-<\infty\), both on the product space and, through Tonelli, for almost every section. This rules out the undefined subtraction \(+\infty-(+\infty)\).
The section conclusion is an almost-everywhere statement, not necessarily a statement about every fiber. On a null set of outer variables, a section may fail to be integrable. Defining the section-integral function to be zero there gives a measurable representative without changing its outer integral. This distinction matters even when the product integral is zero.
Worked Example: A Signed Function on the Unit Square
Take \(X=Y=[0,1]\) with Lebesgue measure \(m\), and let \(f(x,y)=x-y\). Compute the positive and negative integrals, the absolute integral, and the iterated signed integral.
The positive part is nonzero when \(y<x\), and there \(f^+(x,y)=x-y\). For fixed \(x\), \(\int_0^1 f^+(x,y)\,dy=\int_0^x(x-y)\,dy=x^2/2\). Therefore \(\int f^+\,d(m\times m)=\int_0^1x^2/2\,dx=1/6\). The negative part is nonzero when \(y>x\), and there \(f^-(x,y)=y-x\). Thus \(\int_0^1 f^-(x,y)\,dy=\int_x^1(y-x)\,dy=(1-x)^2/2\), giving \(\int f^-\,d(m\times m)=\int_0^1(1-x)^2/2\,dx=1/6\).
It follows that \(\int|f|\,d(m\times m)=1/6+1/6=1/3<\infty\), while \(\int f\,d(m\times m)=1/6-1/6=0\). For each \(x\), the signed section integral is \(\int_0^1(x-y)\,dy=x-1/2\), and hence \(\int_0^1(x-1/2)\,dx=1/2-1/2=0\). Check: the absolute section integral is \(x^2/2+(1-x)^2/2\); its outer integral is \((1/2)(1/3+1/3)=1/3\), agreeing with the product absolute integral. The opposite order gives the same signed value, by the theorem.
Worked Example: An Absolutely Summable Signed Array
Let \(X=Y=\mathbb{N}=\{1,2,\ldots\}\) with counting measure, and set \(a_{mn}=(-1)^{m+n}2^{-m-n}\). Counting measure is sigma-finite, and integration is summation. Compute the absolute sum, the positive and negative sums, and an iterated sum.
The absolute sum factors into two geometric series: \(\sum_{m,n\ge1}|a_{mn}|=(\sum_{m\ge1}2^{-m})(\sum_{n\ge1}2^{-n})=1\cdot1=1\). The sum over odd positive indices is \(\sum_{k\ge0}2^{-(2k+1)}=(1/2)/(1-1/4)=2/3\); the sum over even positive indices is \(\sum_{k\ge1}2^{-2k}=(1/4)/(1-1/4)=1/3\). The terms are positive when \(m,n\) have the same parity. Thus \(\sum a_{mn}^+=(2/3)^2+(1/3)^2=4/9+1/9=5/9\). They are negative when the parities differ, so \(\sum a_{mn}^-=2(2/3)(1/3)=4/9\). Their difference is \(5/9-4/9=1/9\).
For fixed \(m\), the inner signed sum is \(\sum_{n\ge1}a_{mn}=(-1)^m2^{-m}\sum_{n\ge1}(-1)^n2^{-n}=(-1)^{m+1}/(3\cdot2^m)\), since the last geometric series is \(-1/3\). Summing over \(m\) gives \(\sum_{m\ge1}(-1)^{m+1}/(3\cdot2^m)=(-1/3)(-1/3)=1/9\). Check: the signed sum is also the product of the two one-dimensional signed sums, \((-1/3)(-1/3)=1/9\); the positive and negative sums differ by exactly \(1/9\). The absolute sum \(1\) verifies the required finiteness.
Worked Example: An Exceptional Fiber on a Null Set
Let \(X=Y=\mathbb{R}\) with Borel sigma-algebras and Lebesgue measures, and define \(f(x,y)=\chi_{\{0\}\times\mathbb{R}}(x,y)\). This is product-measurable because \(\{0\}\times\mathbb{R}=\{0\}\times\mathbb{R}\) is a measurable rectangle. Compute the product integral and the section integrals.
The set \(\{0\}\times\mathbb{R}\) has product measure \(m(\{0\})m(\mathbb{R})=0\cdot\infty\) only as an unhelpful formal expression, so instead exhaust the second factor by \([-k,k]\). Each rectangle \(\{0\}\times[-k,k]\) has product measure \(m(\{0\})m([-k,k])=0\cdot 2k=0\), and their countable union is \(\{0\}\times\mathbb{R}\). Thus \(\int|f|\,d(m\times m)=0\).
For \(x\ne0\), \(f_x\) is the zero function and its integral is \(0\). At \(x=0\), \(f_0(y)=1\) for every \(y\), so \(\int_{\mathbb{R}}f_0\,dm=\infty\). Therefore the inner section integral is infinite at one outer point but zero everywhere else; its outer integral is zero. Check: the exceptional set \(\{0\}\) has measure zero, and the product integral is also zero. This confirms that absolute integrability guarantees integrable sections almost everywhere, not at every point.
A Common Mistake
A tempting but incorrect inference is: “If \(f\) is product-integrable, then every section \(f_x\) is integrable.” In the null-fiber example, this would assert that the constant-one function on \(\mathbb{R}\) has a finite integral, contradicting \(\int_{\mathbb{R}}1\,dm=\infty\). The correct conclusion is that \(f_x\) is integrable for \(\mu\)-almost every \(x\). The exceptional fiber lies above a null set, and its infinite section integral does not change the outer integral of the nonnegative section-integral function.
Another error is to write \(\int f=\int f^+-\int f^-\) when both terms are infinite. That expression is undefined, not zero. The proof above uses the finiteness of \(\int|f|\) to show each term is finite before subtracting. Without this condition, the argument does not establish Fubini’s identity.
This positive-negative decomposition is useful whenever an integral identity is first available for nonnegative functions. It isolates exactly what must be checked before passing to a real-valued integrand: the total positive and negative contributions must both be finite, and the section integrals must be defined outside at most a null set. The complete Fubini theorem develops the full result from these ingredients.
Check Your Understanding
Use the positive and negative parts, and state which space and measure each integral refers to.
- For a real-valued measurable \(f\), write the pointwise identities relating \(f\), \(f^+\), \(f^-\), and \(|f|\).
- Why does \(\int_{X\times Y}|f|\,d(\mu\times\nu)<\infty\) imply that both \(\int f^+\,d(\mu\times\nu)\) and \(\int f^-\,d(\mu\times\nu)\) are finite?
- In the unit-square example, compute \(\int f^+\,d(m\times m)\) and \(\int f^-\,d(m\times m)\) from their triangular regions.
- What does the theorem guarantee about section integrability, and why is “for every \(x\)” too strong?
- Why is the expression \(+\infty-(+\infty)\) not a valid definition of a signed integral?