CounterexampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (deepseek-v4-pro + claude-opus-5[1m])audited 2026-08-24

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 three-point outer measure has nonmeasurable subsets despite passing the whole-space split

Statement refuted

To check Carathéodory measurability of $E$ , it is enough to test the defining split only with $A = X$ .

Facts & Assumptions

Given: The set $X = {1, 2, 3}$ and $μ^{*} (\emptyset) = 0$ , $μ^{*} (X) = 2$ , and $μ^{*} (A) = 1$ for every other nonempty $A \subseteq X$ .

[F1]

An outer measure on a set $X$ is a function $μ^{*} : P (X) \to [0, + \infty]$ that vanishes at the empty set, is monotone, and is countably subadditive. (Outer measures)

[F2]

A set $E \subseteq X$ is Carathéodory measurable for $μ^{*}$ when $μ^{*} (A) = μ^{*} (A \cap E) + μ^{*} (A ∖ E)$ for every $A \subseteq X$ . (Carathéodory measurable sets)

Counterexample

technique · direct

1.1F1algebra

Normalization and monotonicity follow by exhausting subset sizes. For subadditivity, a nonempty proper union has value $1$ and some member has value at least $1$ ; a union equal to $X$ either has an $X$ member of cost $2$ or requires at least two nonempty proper members of total cost at least $2$ . Thus [F1] holds.

2.1step 1.1F2choosealgebra∎

Every $E$ passes the whole-space equation: for nonempty proper $E$ , the two pieces have value $1$ and sum to $2 = μ^{*} (X)$ . But choose $x \in E$ and $y \notin E$ and test [F2] with $A = {x, y}$ ; then $μ^{*} (A) = 1$ while the two singleton pieces sum to $2$ . Thus every nonempty proper set fails, and only $\emptyset, X$ are measurable.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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