Counterexamples to Unjustified Order Switching

When Integration Order Is Not Automatic

Fubini’s theorem permits an order switch for an integrable function on a product measure space. Tonelli’s theorem permits order switching for a nonnegative measurable function, with infinite values allowed. Outside those settings, an iterated integral may fail to exist, may be infinite, or may depend on the order.

The distinction is important: a disagreement between two finite iterated integrals is not a harmless computational anomaly. It demonstrates that a hypothesis needed for Fubini’s theorem is absent. Conversely, equal infinite iterated integrals do not establish integrability or produce a finite value for the integral.

Definition: Let \((X,\mathcal A,\mu)\) and \((Y,\mathcal B,\nu)\) be measure spaces, and let \(f\) be measurable on their product. Its iterated integrals are obtained by first integrating in one coordinate and then integrating the resulting section integral in the other coordinate, whenever these operations are defined. Fubini’s theorem applies when \(\int_{X\times Y}|f|\,d(\mu\times\nu)<\infty\); Tonelli’s theorem applies when \(f\geq 0\), even if the resulting integral is infinite.

In the examples below, the spaces are either the positive integers with counting measure or the open unit interval with Lebesgue measure. These are sigma-finite measure spaces, so the usual Tonelli and Fubini theorems apply under their respective hypotheses.

A Diagnostic Consequence of Fubini’s Theorem

A useful way to test a claimed order switch is to work backwards from Fubini’s theorem. If a measurable function on a sigma-finite product space has finite iterated integrals in both orders, but those values differ, it cannot be absolutely integrable.

Theorem: Suppose \((X,\mathcal A,\mu)\) and \((Y,\mathcal B,\nu)\) are sigma-finite measure spaces, and \(f\) is measurable on \(X\times Y\). If both iterated integrals are finite and unequal, then \(\int_{X\times Y}|f|\,d(\mu\times\nu)=\infty\).

Proof. Suppose instead that \(\int_{X\times Y}|f|\,d(\mu\times\nu)<\infty\). Then \(f\) is integrable. Fubini’s theorem implies that the section integrals are defined almost everywhere, are integrable in the outer variable, and both orders of integration equal the same product-space integral:

\int_X\left(\int_Y f(x,y)\,d\nu(y)\right)d\mu(x) = \int_{X\times Y} f\,d(\mu\times\nu) = \int_Y\left(\int_X f(x,y)\,d\mu(x)\right)d\nu(y).

This contradicts the assumption that the two finite iterated integrals are unequal. Therefore \(f\) is not absolutely integrable, so its integral of \(|f|\) is infinite. \(\square\)

This result is a diagnostic, not a replacement for checking the hypotheses directly. A disagreement tells us that absolute integrability fails, but it does not by itself identify whether the positive part, the negative part, or both have infinite integral. The examples make those details visible.

Worked Example: A Matrix with Different Iterated Sums

Worked Example: A Counting-Measure Order Switch

Take \(X=Y=\mathbb N\), each with counting measure. Define the measurable function \(a:\mathbb N\times\mathbb N\to\mathbb R\) by \(a(m,n)=1\) when \(n=m\), \(a(m,n)=-1\) when \(n=m+1\), and \(a(m,n)=0\) otherwise. Each row has just two nonzero entries. Thus for every \(m\in\mathbb N\),

\sum_{n=1}^{\infty}a(m,n)=1+(-1)=0, \qquad \sum_{m=1}^{\infty}\left(\sum_{n=1}^{\infty}a(m,n)\right)=0.

Now sum down each column. The first column has the single nonzero entry \(a(1,1)=1\). For every \(n\geq 2\), column \(n\) has \(a(n,n)=1\) and \(a(n-1,n)=-1\), with all other entries zero. Therefore,

\sum_{m=1}^{\infty}a(m,1)=1, \qquad \sum_{m=1}^{\infty}a(m,n)=1+(-1)=0 \quad (n\geq2),

and consequently

\sum_{n=1}^{\infty}\left(\sum_{m=1}^{\infty}a(m,n)\right)=1.

The two iterated sums are finite but different: summing rows first gives \(0\), while summing columns first gives \(1\). The failure is exactly the lack of absolute summability. There is one entry \(1\) and one entry \(-1\) in every row, so

\sum_{m=1}^{\infty}\sum_{n=1}^{\infty}|a(m,n)| = \sum_{m=1}^{\infty}2 = \infty.

In fact, the positive and negative parts each have infinite sum: there are infinitely many entries equal to \(1\) and infinitely many equal to \(-1\). Thus the signed sum over the whole product is not a finite integral to which Fubini applies. This example shows why checking that each individual row and column has a finite sum is not enough.

Worked Example: Different Finite Integrals on the Unit Square

Worked Example: A Signed Function with a Singular Corner

On \((0,1)^2\) with two-dimensional Lebesgue measure, define \(f(x,y)=(x-y)/(x+y)^3\). The function is continuous on this open square, but it becomes unbounded near the excluded corner \((0,0)\). Fix \(x\in(0,1)\), substitute \(u=x+y\), and note that \(x-y=2x-u\). An antiderivative in \(y\) is \(-x/(x+y)^2+1/(x+y)\). Hence

\begin{aligned} \int_0^1 f(x,y)\,dy &=\left[-\frac{x}{(x+y)^2}+\frac{1}{x+y}\right]_{y=0}^{y=1}\\ &=-\frac{x}{(x+1)^2}+\frac{1}{x+1} -\left(-\frac{x}{x^2}+\frac{1}{x}\right)\\ &=\frac{1}{(x+1)^2}. \end{aligned}

The lower-endpoint expression is zero because \(-x/x^2+1/x=0\). Integrating the resulting section integral gives

\int_0^1\left(\int_0^1 f(x,y)\,dy\right)dx = \int_0^1\frac{dx}{(x+1)^2} = \left[-\frac{1}{x+1}\right]_0^1 = \frac12.

For the reverse order, \(f(y,x)=-f(x,y)\). Applying the same calculation with the variables interchanged therefore gives \(\int_0^1 f(x,y)\,dx=-1/(y+1)^2\), and so

\int_0^1\left(\int_0^1 f(x,y)\,dx\right)dy = -\int_0^1\frac{dy}{(y+1)^2} = -\frac12.

These finite iterated integrals disagree. The theorem above already implies that \(f\) is not absolutely integrable; we can verify the stronger fact that its positive and negative parts both have infinite integral. In the region \(0<y<x/2\), we have \(x-y>x/2\) and \(x+y<3x/2\), and hence

f(x,y)>\frac{x/2}{(3x/2)^3}=\frac{4}{27x^2}.

Integrating this lower bound over that region yields

\int_0^1\int_0^{x/2} f(x,y)\,dy\,dx \geq \frac{2}{27}\int_0^1\frac{dx}{x} = \infty.

Thus the positive part has infinite integral. By exchanging \(x\) and \(y\), the region \(0<x<y/2\) gives the same divergence for the negative part. Since both parts have infinite integral, the signed product-space integral is undefined as a finite Lebesgue integral. The two computed iterated integrals do not contradict Fubini: its absolute-integrability hypothesis fails.

What the example establishes: Finite section integrals, even when they can be evaluated explicitly for every fixed coordinate, do not guarantee that the function is integrable on the product space. The behavior of the outer integral and the total absolute mass still matter.

Worked Example: Tonelli Allows an Infinite Answer

Worked Example: A Nonnegative Function with Infinite Integral

Consider \(g(x,y)=1/(x+y)^2\) on \((0,1)^2\). It is measurable and nonnegative. For each fixed \(x\in(0,1)\), the section integral is finite:

\int_0^1 g(x,y)\,dy = \left[-\frac{1}{x+y}\right]_{y=0}^{y=1} = \frac{1}{x}-\frac{1}{x+1}.

But the outer integral diverges. For \(0<x<1\), \(1/(x+1)<1\), so the section integral is greater than \(1/x-1\); its integral near zero is infinite. More explicitly, for \(0<x<1/2\), \(1/(x+1)<2/3\) and \(1/x\geq2\), giving \(1/x-1/(x+1)\geq 2/(3x)\). Therefore

\int_0^1\left(\int_0^1 g(x,y)\,dy\right)dx = \infty.

The function is symmetric in \(x\) and \(y\), so the reverse iterated integral is also infinite. Tonelli’s theorem guarantees equality of these nonnegative integrals, with value \(+\infty\). It does not say that \(g\) is integrable: indeed, its integral is not finite. There is no signed cancellation here, but there is still no finite value to use in Fubini’s theorem.

What Each Failure Tells You

The examples illustrate two different ways an unjustified argument can go wrong.

Unequal finite answers: The matrix and signed unit-square examples have different finite iterated values. The diagnostic theorem shows that absolute integrability fails. In the continuous example, both positive and negative parts have infinite integral.
Equal infinite answers: The nonnegative example has the same infinite value in both orders. Tonelli permits this equality, but it does not turn the integral into a finite one.
Failure to define an inner integral: In other signed examples, a section integral may itself have both positive and negative parts infinite. Then it is undefined as a Lebesgue integral, so the corresponding iterated integral cannot be treated as an ordinary finite quantity.

A common pitfall is to calculate one order, obtain a plausible finite number, and then assume the other order must give the same answer. The right check is not whether the algebra looks symmetric or whether individual sections can be integrated. Check that the function is measurable and that the relevant theorem’s hypotheses hold. For a signed Fubini argument, the central condition is \(\int |f|<\infty\); for a nonnegative Tonelli argument, an infinite result is allowed, but should be reported as infinite.

Identify the setting.
Specify the product measure and whether the integrand is nonnegative or signed.

Check the applicable hypothesis.
For Fubini, verify absolute integrability; for Tonelli, verify measurability and nonnegativity.

Interpret the outcome correctly.
A finite discrepancy rules out absolute integrability, while equal infinite values do not establish finite integrability.

Check Your Understanding

Use the examples and criteria above to explain what an attempted order switch does and does not establish.

For the matrix in the first example, calculate the row sums and column sums. Where does absolute summability fail?
Why do the two finite iterated integrals for the signed unit-square function not contradict Fubini’s theorem?
Which regions demonstrate that the positive and negative parts of that function each have infinite integral?
What does Tonelli’s theorem conclude for the nonnegative function \(1/(x+y)^2\), and what does it not conclude?
If two finite iterated integrals are unequal on a sigma-finite product space, what can you conclude about the integral of the absolute value?