Alphabeta Math
LemmaStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-07
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.

Riemann-integrable half-space extensions of chart coefficients

Statement

Let n1, Hn={xRn:xn0}, and let U be relatively open in Hn. If f is smooth on U with compact support KU, set f~=f on U and f~=0 on RnU. Then f~ is bounded, compactly supported, smooth away from {xn=0}, and Riemann integrable. Its Euclidean integral is independent of the bounding rectangle and of any auxiliary smooth extension. For an interior chart URn, the zero extension is smooth everywhere.

Facts & Assumptions

[F1]

Compact support of a differential form: Let M be a smooth manifold, possibly with boundary, and k0. For ωΩk(M) define suppω={pM:ωp0}M,Ωck(M)={ωΩk(M):suppω is compact}. The closure and compactness are in M, including its genuine boundary. Zero is the intrinsic zero of each exterior-power fiber, so this definition is independent of trivialization. The zero form has empty support.

[F2]

Smooth functions and tensor fields extend locally across the boundary: Every smooth function or tensor field on a manifold with boundary extends smoothly across each boundary point to some neighbourhood in its double; the extension is not canonical.

[F3]

Smooth extension from a closed neighbourhood: Let C be a closed subset of a smooth manifold M, let UM be open with CU, and let f:UR be smooth. Then there exists a smooth function F:MR such that F=f on an open neighbourhood of C and supp(F)U.

[F4]

Lebesgue's criterion in Rm: a bounded function on a closed nondegenerate rectangle is Riemann integrable iff its discontinuity set is null: A bounded real function on a closed nondegenerate rectangle in Rm, m1, is Riemann integrable if and only if its discontinuity set is null.

[F5]

Measure zero and content zero in Rm by countable and finite cube covers: Fix m1. A closed cube is a rectangle j<m[aj,aj+] with 0; its volume is m. A set ERm is null when, for every ε>0, it is covered by a sequence of closed cubes whose nonnegative volume series converges with sum at most ε. It has content zero when such a cover can be finite. The series and finite sums are def-series and def-finite-sum, and their nonnegative bounds use thm-nonnegative-series-bounded-partial-sums and lem-finite-sum-laws. Both properties pass to subsets. Padding a finite cover with degenerate zero-volume cubes proves that content zero implies null. This terminology defines only cover-nullity; it does not define a measure on arbitrary sets.

[F6]

For every ε>0 in a complete ordered field there is a natural n1 with 1/n<ε: Let F be a complete ordered field (def-complete-ordered-field) and let εF with ε>0. Then there is a natural number n1 such that 1n1F<ε, where n1F is the canonical natural of F (thm-of-archimedean) and 1/(n1F) is its multiplicative inverse (def-field). As is standard we abbreviate n1F to n and write the conclusion 1/n<ε. This is the reciprocal form of the Archimedean property. thm-of-archimedean on its own delivers only the assertion that the canonical naturals are cofinal, x<n1F; the form actually used in analysis, that the reciprocals of the naturals get below every positive bound, is the statement above, and it is recorded separately so that no proof has to reconstruct the inversion step in passing.

[F7]

The Riemann integral of a compactly supported function is independent of its bounding rectangle: Let n1 and let f:RnR have compact support. If f is Riemann integrable on one closed rectangle whose interior contains its support, then it is integrable on every such rectangle, and all the resulting integrals are equal. This includes the empty-support case.

Proof

Given: The objects and hypotheses in the statement above.

1.1

By compact support, f vanishes on UK and is bounded on K. A point at an artificial edge of U lies outside the closed Euclidean compact set K; a neighborhood missing K has zero extended coefficient. Inside U the coefficient is smooth up to the genuine face. Thus f~ is smooth off that face and supported in K. This includes f=0.

F1F2
1.2

Auxiliary extensions can be constructed near K: choose finitely many extension neighborhoods, smooth Euclidean bump functions supported there and positive on smaller neighborhoods covering K, and divide by their sum near K. The weighted extensions agree with f on the half-space near K. Cut off on a smaller neighborhood of K to obtain a compactly supported smooth Euclidean function there. The cutoff is one near K; its restriction to the half-space, extended by zero at artificial edges, is f. Such cutoffs follow by applying the closed-neighborhood extension lemma to the constant function one and, if needed, composing with a smooth nonnegative function.

F2F3
1.3

Choose R>0 so K(R,R)n. Partition the first n1 coordinates of [R,R]n1 into at most (2R/δ+2)n1 cells of side at most δ. Center a closed cube of side 2δ on each face cell. They cover the face in the bounding cube and have total volume at most 2n(2R+2δ)n1δ. This tends to zero; reciprocal integers give arbitrarily small δ. For n=1 this is a single interval of length 2δ.

F5F6
2.1

The discontinuities of f~ in [R,R]n lie in that content-zero, hence null, face. Boundedness and the null-discontinuity criterion imply Riemann integrability. The criterion is used in its sufficient direction only.

F4step 1.1step 1.3
3.1

The compact-support integral lemma makes the value independent of any larger bounding rectangle. Every auxiliary extension after restriction to Hn gives the same zero-extended function, hence the same integral. With an interior chart there is no genuine face, so the artificial-edge argument proves smoothness everywhere.

F7step 1.1step 1.2step 2.1

Depends on

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