False statementConstruction: AI-adaptedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (gpt-5.6-terra)audited 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.

FALSE: the product of two complete measure spaces is complete

Statement

If $(X, A, μ)$ and $(Y, B, ν)$ are complete measure spaces, then the product measure space $(X \times Y, A \otimes B, μ \times ν)$ is complete.

Facts & Assumptions

Given: Lebesgue measure $λ$ on $R$ , a non-Lebesgue-measurable set $N \subseteq R$ , the set $E : = {0} \times N \subseteq R^{2}$ , and the line $Z : = {0} \times R$ .

[L1]

The completed product measure is generally a genuine completion of the uncompleted product measure. (The Euclidean Lebesgue measure is the completion of the product of the factor Lebesgue measures)

[L2]

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

[L3]

On measurable rectangles, the product measure satisfies $(λ \times λ) (A \times B) = λ (A) λ (B)$ . (For sigma-finite factors, the product measure exists, has the rectangle formula, is sigma-finite, and is unique)

Refutation

technique · direct

1.1

For each $n \geq 1$ , the rectangle $Z_{n} : = {0} \times [- n, n]$ satisfies $(λ \times λ) (Z_{n}) = λ ({0}) λ ([- n, n]) = 0$ by [L3], so $Z = ⋃_{n \geq 1} Z_{n}$ is $(λ \times λ)$ -null.

L3algebra

2.1

Because $E \subseteq Z$ , [L1] makes $E$ Lebesgue measurable in $R^{2}$ .

L1step 1.1

3.1

If $E$ belonged to $L (R) \otimes L (R)$ , its horizontal section at $0$ would be $E_{0} = N$ , which is not Lebesgue measurable, contradicting [L2]. Thus the product sigma-algebra misses a subset of a product-null set, so the product of two complete Lebesgue spaces need not be complete. This is exactly why [L1] needs an actual completion step.

L1L2step 2.1∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

22 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)