CounterexampleConstruction: AI-adaptedVerification: AI-generatedprecheck 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.

A set can have measurable horizontal and vertical sections and still fail to be product-measurable

Statement refuted

If $E \subseteq X \times Y$ has measurable horizontal and vertical sections for every parameter, then $E$ is product-measurable.

Counterexample

technique · direct

Assume the Axiom of Countable Choice. Let $X$ be the set of countable ordinals, let $M$ be the sigma-algebra of countable and cocountable subsets of $X$ , and define $E : = {(x, y) \in X \times X : y < x} .$

Facts & Assumptions

Given: The set $E \subseteq X \times X$ above.

[L1]

A sigma-algebra is closed under complements and countable unions (Sigma-algebras), and under countable choice a countable union of countable sets is countable (The Axiom of Countable Choice ( ${A C}_{ω}$ ), Countable unions of at most countable sets, assuming ${A C}_{ω}$ ). Hence the countable-cocountable family on an uncountable set is a sigma-algebra.

[L2]

For sigma-finite measures, the two iterated section-measure integrals of a product-measurable set agree. (For sigma-finite measures, the two section-measure integrals of a measurable set agree)

[A1]

Define $ν$ on $M$ by $ν (A) = 0$ for countable $A$ and $ν (A) = 1$ for cocountable $A$ . The same countable-union argument as in [L1] shows that $ν$ is a finite measure on $(X, M)$ .

Verification

1.1givenL1

For each $x \in X$ , the section $E_{x} = {y : y < x}$ is countable by the choice of $X$ , hence measurable for $M$ . For each $y \in X$ , the section $E^{y} = {x : y < x}$ has countable complement ${x : x \leq y}$ , hence is cocountable and measurable.

2.1A1L2step 1.1∎

Suppose for contradiction that $E$ were product-measurable for $M \otimes M$ . Since $ν (X) = 1$ , the measure $ν$ is finite and hence sigma-finite, so [L2] would give $\int_{X} ν (E_{x}) d ν = \int_{X} ν (E^{y}) d ν .$ But step 1.1 makes $ν (E_{x}) = 0$ for every $x$ and $ν (E^{y}) = 1$ for every $y$ , so the two sides are $0$ and $1$ , a contradiction. Therefore $E$ is not product-measurable, even though all of its sections are measurable. Thus the displayed implication is false.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

19 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 47 (standard reference, not scraped)