ExampleConstruction: AI-adaptedVerification: AI-adaptedPipeline-generatedverified 2026-09-26 (gpt-6-sol)

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.

Metric balls need no curvature comparison for measurability

Example

Assume countable choice. On $(0, 1)$ with Euclidean distance and density $x^{- 1} ∣ d x ∣$ , the ball $B (1 / 2, 1 / 4)$ has volume $\log 3$ , whereas $B (1 / 2, 1) = (0, 1)$ has infinite volume. Both are Borel and positive in volume.

Facts & Assumptions

Given: Assume ${A C}_{ω}$ . Manifolds are Hausdorff, second countable and smooth, with boundary allowed; $n = 0$ is allowed unless excluded. Densities are pointwise Borel, $0 \cdot \infty = 0$ , and $λ_{0} (R^{0}) = 1$ . Two explicit metric-ball volumes, showing the role of relative compactness.

[F1]

Positive open-set and metric-ball volume: Topology-compatible positive-radius balls are Borel and positive in volume; compact closure implies finite volume.

[F2]

Weighted interval volume: The Example and Verification compute $μ ((a, b)) = \log (b / a)$ and $μ ((0, 1)) = \infty$ .

Verification

1.1F1F2

The inequality $∣ x - 1 / 2 ∣ < 1 / 4$ with $x \in (0, 1)$ is equivalent to $1 / 4 < x < 3 / 4$ . The closure $[1 / 4, 3 / 4]$ is compact inside M, and the weighted interval computation gives $μ (B (1 / 2, 1 / 4)) = \log ((3 / 4) / (1 / 4)) = \log 3$ . The ball is open Borel and has positive finite measure.

2.1F1F2step 1.1∎

Every $x \in (0, 1)$ satisfies $∣ x - 1 / 2 ∣ < 1 / 2 < 1$ , so $B (1 / 2, 1) = (0, 1)$ . Its mass is infinite by the interval example. Its closure in M is all of M, which is not compact: the open cover ${(1 / N, 1) : N \geq 2}$ of M has no finite subcover. Thus the finite-volume hypothesis on the closure is absent in precisely this example. Both radii are strictly positive; neither ball is empty.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

14 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

Folland, Real Analysis, second edition, §11.4 pp.361–363; Theorems 2.14–2.15 pp.50–51 (standard reference, not scraped)