CounterexampleConstruction: AI-adaptedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-29

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

The function (x^2-y^2)/(x^2+y^2)^2 shows that Fubini's integrability hypothesis is not decorative

Statement refuted

Fubini's theorem remains valid if one deletes the assumption $f \in L^{1} (μ \times ν)$ .

Counterexample

technique · direct

On $(0, 1)^{2}$ , let $f (x, y) : = \frac{x^{2} - y^{2}}{(x^{2} + y^{2})^{2}} .$

Facts & Assumptions

Given: The function $f$ above.

[L1]

The principal inverse tangent satisfies $(\arctan u)^{'} = \frac{1}{1 + u^{2}}$ and $\arctan u = \int_{0}^{u} \frac{d t}{1 + t^{2}} .$ (Principal arctangent: derivative, integral, power series, and the Gregory–Leibniz series)

[A1]

On $(0, 1)^{2}$ one has $x^{2} + y^{2} > 0$ , so the rational functions written below are well-defined and differentiable.

Verification

1.1

Direct differentiation gives $f (x, y) = \partial_{y} (\frac{y}{x^{2} + y^{2}}) = - \partial_{x} (\frac{x}{x^{2} + y^{2}}) .$

A1algebra

2.1

Integrating the first identity of step 1.1 in $y$ from $0$ to $1$ gives $\int_{0}^{1} f (x, y) d y = \frac{1}{1 + x^{2}} .$ Integrating in $x$ and applying [L1] yields $\int_{0}^{1} (\int_{0}^{1} f (x, y) d y) d x = \int_{0}^{1} \frac{d x}{1 + x^{2}} = \frac{π}{4} .$

step 1.1L1

3.1

Repeating the same calculation with the second identity of step 1.1 gives $\int_{0}^{1} (\int_{0}^{1} f (x, y) d x) d y = - \int_{0}^{1} \frac{d y}{1 + y^{2}} = - \frac{π}{4} .$ Therefore the iterated integrals exist and are unequal, so the conclusion of Fubini fails. In particular $f \notin L^{1} ((0, 1)^{2})$ , because otherwise Fubini's theorem for L^1 functions on a sigma-finite product would force them to agree.

step 1.1step 2.1L1∎

Depends on

Used by

Equal iterated integrals still do not imply product integrability Counterexample

Dependency tree · two levels

25 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)
Terence Tao, An Introduction to Measure Theory, Exercise 1.7.23 (standard reference, not scraped)