CounterexampleConstruction: AI-generatedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (gpt-5.6-terra)audited 2026-08-29

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Equal iterated integrals still do not imply product integrability

Statement refuted

If both iterated integrals of a function exist and are equal, then the function must belong to $L^{1} (μ \times ν)$ .

Counterexample

technique · direct

Let $f (x, y) : = \frac{x^{2} - y^{2}}{(x^{2} + y^{2})^{2}}$ on $(0, 1)^{2}$ , and define $g$ on $((0, 1) \times {0, 1}) \times (0, 1)$ by $g ((x, 0), y) : = f (x, y), g ((x, 1), y) : = f (y, x) .$ Give $(0, 1) \times {0, 1}$ the product of Lebesgue measure with counting measure on ${0, 1}$ , and give $(0, 1)$ Lebesgue measure.

Facts & Assumptions

Given: The function $g$ above.

[L1]

The function $f$ from The function (x^2-y^2)/(x^2+y^2)^2 shows that Fubini's integrability hypothesis is not decorative has two existing iterated integrals equal to $π / 4$ and $- π / 4$ , and it is not in $L^{1}$ .

Verification

1.1

The first copy of $g$ contributes the two iterated values of $f$ , while the second copy contributes the same values with the order reversed. Therefore both iterated integrals of $g$ exist and are equal to $π / 4 + (- π / 4) = 0.$

2.1

The absolute integral of $g$ is the sum of the absolute integrals of the two copies, so it is still infinite because each copy carries the non- $L^{1}$ singularity of [L1]. Thus equal iterated integrals do not imply $L^{1}$ -integrability.

L1step 1.1∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

8 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Gerald B. Folland, Real Analysis, 2nd ed., Exercise 55(a) (standard reference, not scraped)