ExampleConstruction: AI-adaptedVerification: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-09

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.

Euclidean volume from chart gluing

Example

On $R^{n}$ for $n \geq 1$ , the standard density $∣ d x^{1} \dots d x^{n} ∣$ induces Borel Lebesgue measure. Its completion is ordinary Lebesgue measure. For a box with side lengths $b_{j} - a_{j}$ , its mass is $\prod_{j} (b_{j} - a_{j})$ .

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 $λ_{0} (R^{0}) = 1$ . Euclidean identity-chart instance and box calculation.

[F1]

Intrinsic density measure and its chart restriction: The measure of a Borel chart subset is the coordinate coefficient integral.

[F2]

$L (R^{n})$ is exactly the completion of the restriction of $λ_{n}$ to the Borel sets: Under countable choice, completing Borel Lebesgue measure gives the Lebesgue sigma-algebra and Lebesgue measure.

[F3]

A box in $R^{n}$ with parameters $a_{i} \leq b_{i}$ is Lebesgue measurable of measure $\prod_{i < n} (b_{i} - a_{i})$ , whichever of its faces are included: The measure of any coordinate box is the product of its side lengths.

Verification

1.1

Take the identity chart on $R^{n}$ and partition $φ = 1$ . The coefficient is one, hence for each Borel $E$ , $μ_{r} (E) = \int_{E} 1 d λ_{n} = λ_{n} (E)$ . In particular $μ_{r} (\prod_{j} [a_{j}, b_{j}]) = \prod_{j} (b_{j} - a_{j})$ ; for the unit cube the result is one, and if a side has length zero the result is zero.

F1F3

2.1

The equality on Borel sets identifies the completed domain and measure with those in the Lebesgue completion theorem. Thus the completion is $(R^{n}, L (R^{n}), λ_{n})$ . Empty sets have zero measure in both domains; the case $n = 1$ is the usual interval-length formula.

F2step 1.1∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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Sources

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