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.

A nonmeasurable subset of a null line shows that the product of complete measures need not be complete

Statement refuted

Assuming the Axiom of Countable Choice, the product of two complete measure spaces is always complete.

Counterexample

technique · direct

Let $N \subseteq R$ be a non-Lebesgue-measurable set, and consider $E : = {0} \times N \subseteq R^{2} .$

Facts & Assumptions

Given: The Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )), a non-Lebesgue-measurable set $N \subseteq R$ , and the set $E = {0} \times N$ .

[L1]

Every section of a product-measurable set is measurable. (Every section of a product-measurable set is measurable)

[L2]

Assuming countable choice, Lebesgue measure is sigma-finite. (Lebesgue measure is sigma-finite, and every metrically bounded subset of $R^{n}$ has finite outer measure)

[L3]

For sigma-finite factors, the product measure satisfies the rectangle formula. (For sigma-finite factors, the product measure exists, has the rectangle formula, is sigma-finite, and is unique)

[L4]

Assuming countable choice, every singleton is Lebesgue null. (Every at most countable subset of $R^{n}$ is Lebesgue null; in particular $λ_{1} (Q) = 0$ )

[L5]

Assuming countable choice, Euclidean Lebesgue measure is complete. (Assuming countable choice, $L (R^{n})$ is a sigma-algebra containing every elementary set and $λ_{n}$ is a complete measure extending elementary volume)

Verification

1.1

By [L2], both factors are sigma-finite. For each integer $k \geq 1$ , [L3] and [L4] give $(λ \times λ) ({0} \times [- k, k]) = λ ({0}) λ ([- k, k]) = 0.$ Hence ${0} \times R = ⋃_{k \geq 1} {0} \times [- k, k]$ is product-measurable and product-null, and $E$ is a subset of a product-null set.

L2L3L4algebra

2.1

If $E$ belonged to $L (R) \otimes L (R)$ , then its horizontal section $E_{0}$ would be $N$ , which is not Lebesgue measurable. Thus $E$ is not product-measurable, even though step 1.1 places it inside a product-null set. Since [L5] makes both factor spaces complete, their product measure space is not complete.

step 1.1L1L5∎

Depends on

Used by

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

Dependency tree · two levels

54 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)