Why Absolute Integrability Is the Diagnostic
When a function depends on two variables, it is often useful to integrate one variable first and then the other. But the existence of a value from one order of integration is not, by itself, a general justification for switching the order.
The practical test is to examine the integral of the absolute value over the product space. If that integral is finite, then both orders are justified and give the same answer. If it is infinite, this criterion gives no such guarantee; one must not infer that switching order is valid merely because one iterated integral has a finite value.
This tutorial develops that diagnostic on sigma-finite measure spaces. The key tool is Tonelli’s theorem, which applies to the nonnegative function \(|f|\) even when its integral is infinite.
Absolute Integrability on a Product Space
Let \((X,\mathcal A,\mu)\) and \((Y,\mathcal B,\nu)\) be sigma-finite measure spaces, and let \(\mu\otimes\nu\) denote their product measure. Suppose \(f:X\times Y\to\mathbb R\) is measurable. For each \(x\in X\), its section is the function \(y\mapsto f(x,y)\); for each \(y\in Y\), its section is \(x\mapsto f(x,y)\).
The absolute value matters because positive and negative contributions can cancel in an iterated integral. A finite result after cancellation does not measure how much positive and negative mass was present separately. The integral of \(|f|\) measures both contributions without cancellation.
The Absolute-Integrability Criterion
The result below gives the central practical consequence of the test. In its statement, “almost every” refers to the relevant measure on the space of the outer variable.
Proof. Apply Tonelli’s theorem to the nonnegative measurable function \(|f|\). It gives $$ \int_{X\times Y}|f|\,d(\mu\otimes\nu) = \int_X\left(\int_Y |f(x,y)|\,\nu(dy)\right)\mu(dx) = \int_Y\left(\int_X |f(x,y)|\,\mu(dx)\right)\nu(dy). $$ Each of these quantities is finite by hypothesis. A nonnegative measurable function with finite integral is finite almost everywhere. Therefore \(\int_Y|f(x,y)|\,\nu(dy)<\infty\) for \(\mu\)-almost every \(x\), and \(\int_X|f(x,y)|\,\mu(dx)<\infty\) for \(\nu\)-almost every \(y\). This proves the assertions about the sections.
For the equality of integrals, write \(f=f^+-f^-\), where \(f^+=\max(f,0)\) and \(f^-=\max(-f,0)\). Both are nonnegative measurable functions and satisfy \(f^+\leq |f|\) and \(f^-\leq |f|\). Hence both have finite product integrals. Tonelli’s theorem applies to each of them, so their iterated integrals equal their product integrals in either order. Subtracting the finite equalities for \(f^+\) and \(f^-\) gives the stated equality for \(f\). The finiteness of the absolute integral ensures that these subtractions involve finite quantities.
The proof shows how the diagnostic works: Tonelli first establishes control of the absolute values of the sections; that control then permits positive and negative parts to be combined without subtracting infinities.
Worked Examples: Testing the Criterion
Worked Example: An Exponential on the Positive Quadrant
Let \(X=Y=[0,\infty)\) with Lebesgue measure, and set \(f(x,y)=e^{-(x+y)}\). Since \(f\geq 0\), its absolute value is \(f\). Compute the product integral by integrating in \(y\) first: $$ \int_0^\infty\left(\int_0^\infty e^{-(x+y)}\,dy\right)dx = \int_0^\infty e^{-x}\left(\int_0^\infty e^{-y}\,dy\right)dx = \int_0^\infty e^{-x}\,dx = 1. $$ Thus the integral of \(|f|\) is finite. Integrating in the other order gives $$ \int_0^\infty\left(\int_0^\infty e^{-(x+y)}\,dx\right)dy = \int_0^\infty e^{-y}\,dy = 1. $$ The finite absolute integral supplies the justification for both orders; the matching computations verify the value in this example.
Worked Example: A Power-Law Tail
On \([0,\infty)^2\), let \(f(x,y)=(1+x+y)^{-3}\). This function is nonnegative, so its absolute integral is its integral. For each fixed \(x\), $$ \int_0^\infty (1+x+y)^{-3}\,dy = \frac{1}{2}(1+x)^{-2}. $$ Consequently, $$ \int_0^\infty\left(\int_0^\infty (1+x+y)^{-3}\,dy\right)dx = \frac12\int_0^\infty(1+x)^{-2}\,dx = \frac12. $$ In the reverse order, the same calculation with \(x\) and \(y\) interchanged also gives \(1/2\). Thus \(\int\!\int |f|=1/2<\infty\), and the criterion certifies that both orders are valid. This example also illustrates why an unbounded domain does not by itself prevent absolute integrability: the decay of the integrand can make the total mass finite.
Worked Example: One Finite Iterated Integral Without Absolute Integrability
On \([0,1]^2\), define $$ f(x,y)=\frac{x-y}{(x+y)^3} $$ when \(x+y>0\), and assign \(f(0,0)=0\). For each fixed \(x>0\), integrate in \(y\). An antiderivative with respect to \(u=x+y\) is \(-x/u^2+1/u\), because \(x-y=2x-u\). Hence $$ \int_0^1\frac{x-y}{(x+y)^3}\,dy = \left[-\frac{x}{(x+y)^2}+\frac{1}{x+y}\right]_{y=0}^{y=1} = \frac{1}{(1+x)^2}. $$ The inner integral is finite for every \(x>0\), and its outer integral is $$ \int_0^1\frac{1}{(1+x)^2}\,dx = \left[-\frac{1}{1+x}\right]_{0}^{1} = \frac12. $$ The section at \(x=0\) need not affect this Lebesgue outer integral, since it is a null set of outer-variable values.
However, the integral of \(|f|\) over the square is infinite. For \(0<x<1/3\) and \(2x<y<3x\), the point lies in the square, \(y-x\geq x\), and \(x+y\leq4x\). Therefore $$ |f(x,y)|=\frac{y-x}{(x+y)^3}\geq\frac{x}{(4x)^3}=\frac{1}{64x^2}. $$ Integrating this lower bound over that region gives $$ \int_0^{1/3}\int_{2x}^{3x}|f(x,y)|\,dy\,dx \geq \int_0^{1/3}\frac{1}{64x^2}(3x-2x)\,dx = \frac{1}{64}\int_0^{1/3}\frac{dx}{x} = \infty. $$ Thus one order has a finite iterated integral even though the absolute integral is infinite. The absolute-integrability criterion does not apply. This calculation does not, on its own, determine the value of the other order; it demonstrates why one finite iterated integral is not a substitute for the absolute-integrability test.
A Useful Consequence on Finite Measure Spaces
On a finite product space, a simple bound can establish absolute integrability without evaluating the integral. If \(|f|\leq M\) everywhere for some finite constant \(M\), then $$ \int_{X\times Y}|f|\,d(\mu\otimes\nu) \leq M\,(\mu\otimes\nu)(X\times Y) = M\,\mu(X)\nu(Y)<\infty, $$ provided \(\mu(X)<\infty\) and \(\nu(Y)<\infty\). The theorem therefore allows integration in either order. On an infinite measure space, boundedness alone does not give this conclusion: the total measure may be infinite, so the bound need not yield a finite absolute integral.
How to Use the Test in Practice
For a measurable function on a product space, a reliable workflow is to examine its absolute value before trying to exchange orders.
Confirm that the function is measurable on the product space and that the product measures satisfy the hypotheses being used, such as sigma-finiteness for the theorem above.
Consider \(\int_{X\times Y}|f|\,d(\mu\otimes\nu)\). Tonelli’s theorem can compute or estimate this integral, including when it is infinite.
If the absolute integral is finite, the sections are integrable almost everywhere and both iterated integrals equal the product integral.
If the absolute integral is infinite, the criterion gives no permission to exchange orders. A separately computed finite iterated integral does not fill that gap.
Why the Distinction Matters
The important distinction is between an integral whose finiteness is guaranteed by control of \(|f|\), and an iterated calculation that happens to produce a finite result after cancellation. Absolute integrability controls the total size of both the positive and negative contributions. A finite value for one order alone does not provide that control.
In applications, this is why it is often effective to estimate \(|f|\) before performing a complicated calculation. If the estimate proves the absolute integral finite, the order can be chosen for convenience. If it does not, the failure of the test is not itself a proof that the orders disagree; it means the automatic justification is unavailable and more analysis is needed.
Check Your Understanding
Answer these questions using the absolute-integrability criterion and the examples above.
- What does finiteness of \(\int_{X\times Y}|f|\,d(\mu\otimes\nu)\) imply about section integrability and the two iterated integrals?
- Which theorem applies to \(|f|\) even when its product integral is infinite?
- Why can a bounded measurable function on a finite-measure product space be integrated in either order?
- In the power-law example, what is the value of the absolute integral?
- What does the singular-kernel example establish about the relationship between one finite iterated integral and absolute integrability?