Alphabeta Math
False statementConstruction: AI-adaptedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (gpt-5.6-terra)audited 2026-08-29
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FALSE: every section of a completed-product measurable function is measurable

Statement

If f is measurable for the completed product μ×ν, then fx is measurable for every x and fy is measurable for every y.

Facts & Assumptions

Given: Lebesgue measure λ on R, a non-Lebesgue-measurable set NR, the set E:={0}×NR2, and the indicator function f:=1E.

[L1]

For completed products, section measurability is guaranteed only for almost every parameter. (Tonelli and Fubini for the completed product, with only almost-everywhere section measurability)

[L2]

The completed product sigma-algebra contains every subset of a (λ×λ)-null set. (The completed product measure)

[L3]

On measurable rectangles, the product measure satisfies (λ×λ)(A×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 n1, the rectangle Zn:={0}×[n,n] satisfies (λ×λ)(Zn)=λ({0})λ([n,n])=0 by [L3], so Z:={0}×R=n1Zn is (λ×λ)-null.

L3algebra
2.1

Because EZ, [L2] puts E in the completed product sigma-algebra, so f=1E is measurable for the completed product.

L2step 1.1
3.1

The section at 0 is f0=1N, which is not measurable because N is not Lebesgue measurable. Thus the displayed claim fails even though f is measurable for the completed product, and [L1] is sharp.

L1step 2.1

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

20 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