CounterexampleConstruction: AI-adaptedVerification: AI-generatedverified 2026-09-23 (gpt-6-sol)

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A completed-product measurable set can have a nonmeasurable exceptional section

Statement refuted

Assuming the Axiom of Choice, the assertion that every section of a completed-product measurable function is measurable is false.

Counterexample

technique · direct

Apply AC to the nonempty equivalence classes of $x \sim y ⟺ x - y \in Q$ on $[0, 1]$ , selecting one point from each class. This gives a Vitali set $N \subseteq [0, 1]$ as in Vitali set on $[0, 1]$ , and Assuming the Axiom of Choice, a Vitali set is not Lebesgue measurable proves that $N$ is not Lebesgue measurable. Let $E : = {0} \times N \subseteq R^{2} .$ Put $f : = 1_{E}$ .

Facts & Assumptions

Given: The function $f = 1_{E}$ above.

[L1]

The set $E = {0} \times N$ is contained in a planar null set, so it becomes measurable after completing the product measure. (A nonmeasurable subset of a null line shows that the product of complete measures need not be complete)

[A1]

The Axiom of Choice supplies the Vitali selector in the construction above; the same AC premise covers the Lebesgue-measure contracts used by [L1] and the Vitali nonmeasurability theorem.

Verification

1.1A1L1

By [A1] and [L1], the indicator $f$ is measurable for the completed product measure.

2.1step 1.1∎

The section at $0$ is $f_{0} = 1_{N}$ , whose support $N$ is not Lebesgue measurable. Hence $f_{0}$ is not measurable. So completed-product measurability gives section measurability only almost everywhere, not at every parameter.

Depends on

Used by

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Dependency tree · two levels

18 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

John K. Hunter, Measure Theory, Example 5.20 (standard reference, not scraped)