Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-11
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.

Comparison tests for improper integrals

Statement

Suppose 0fg0\le f\le g eventually toward a singular end. If the improper integral of gg converges there, then the integral of ff converges. If instead fg|f|\le g eventually and g\int g converges, then f\int f converges absolutely and hence converges.

The same assertions hold separately at ++\infty, at -\infty, and at either finite singular endpoint.

Facts & Assumptions

Given: The stated eventual pointwise bounds and local Riemann integrability.

[L2]

A nonnegative improper integral converges exactly when its truncation integrals are bounded (A nonnegative improper integral converges iff its truncated integrals are bounded).

[L3]

Absolute convergence implies convergence (Absolute convergence implies improper convergence).

[L4]

Finite initial pieces do not affect convergence (Improper convergence is independent of finite truncations and split points).

Proof

technique · direct
1.1

Discard the finite portion before the eventual inequality using [L4]. On every remaining compact truncation, [L1] gives 0fg0\le\int f\le\int g. Convergence of g\int g bounds the latter truncations, so [L2] gives convergence of f\int f.

L4L1L2
2.1

If fg|f|\le g, step 1.1 applied to f|f| proves absolute convergence; [L3] then proves convergence of ff. The argument depends only on the direction of truncation and therefore proves every endpoint form.

L3step 1.1

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 94 results over 18 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