Alphabeta Math
TheoremStatement: AI-adaptedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-01
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's criterion on a nondegenerate rectangle in Rm\mathbb{R}^m: integrability is equivalent to arbitrarily small Darboux gaps

Statement

A bounded f:QRf:Q\to\mathbb R on a nondegenerate rectangle is Riemann integrable if and only if, for every ε>0\varepsilon>0, some grid PP satisfies U(f,P)L(f,P)<εU(f,P)-L(f,P)<\varepsilon.

Facts & Assumptions

Proof

technique · direct
1.1

If the two integrals equal II, choose PP_- with L(f,P)>Iε/2L(f,P_-)>I-\varepsilon/2 and P+P_+ with U(f,P+)<I+ε/2U(f,P_+)<I+\varepsilon/2. A common refinement PP has gap below ε\varepsilon.

L1L2L3
1.2

Conversely, a common refinement shows every lower sum is at most every upper sum, so for every PP, 0QfQfU(f,P)L(f,P)0\le\overline{\int_Q}f-\underline{\int_Q}f\le U(f,P)-L(f,P). Arbitrarily small gaps force the integral difference to be 00.

L1L3given
2.1

Thus the conditions are equivalent.

step 1.1step 1.2

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 59 results over 17 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources