Alphabeta Math
TheoremStatement: AI-adaptedProof: AI-adaptedprecheck passaudited 2026-08-01
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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.

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The graph of a continuous function on a closed nondegenerate rectangle in Rm has content zero in Rm+1

Statement

Let m≥1, let Q⊆Rm be a closed nondegenerate rectangle, and let f:Q→R be continuous. Its graph has content zero in Rm+1.

Proof

technique · constructive
1.1L1L2givenchooseconstruct

Given ε>0, choose a uniform coordinate grid with cell widths at most δ, where uniform continuity makes the oscillation of f on each cell below a vertical amount η. Since Q is nondegenerate, the grid may be chosen so that the number Nδ of cells satisfies Nδδm≤CQ for a constant depending only on Q.

2.1step 1.1L2given

One horizontal cube footprint of side δ covers each domain cell. Above it, stack (m+1)-cubes of side δ across the graph's vertical range. Integer part: for every real x there is exactly one integer m with m≤x<m+1 bounds their number by η/δ+2, so all stacks together have volume at most Nδ(ηδm+2δm+1)≤CQ(η+2δ).

3.1step 2.1givenchoose

Summing over the finitely many domain cells gives total covering volume at most a rectangle-dependent constant times η+δ. Choose η and then δ to make this below ε.

4.1step 3.1discharge-construct∎

This finite cube cover proves content zero in the sense of Measure zero and content zero in Rm by countable and finite cube covers.

Depends on

Dependency tree · 2 levels

Sources