Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passverified 2026-09-23 (gpt-6-sol)
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.

Two integrable functions are equal almost everywhere exactly when all of their indefinite integrals agree

Statement

Let f,g∈L1(μ). Then the following are equivalent:

  1. f=g almost everywhere;
  2. for every measurable A, ∫Af dμ=∫Ag dμ.

For integrable real or complex h, the notation in condition 2 means ∫Ah dμ:=∫hχA dμ; the product is integrable because ∣hχA∣≤∣h∣.

Facts & Assumptions

Given: Integrable functions f,g∈L1(μ).

[L1]

The integral over a null set vanishes for nonnegative integrands (A nonnegative integral over a null set vanishes).

[L2]

The Lebesgue integral is linear on L1(μ) (The Lebesgue integral is linear on L1(μ)).

[L3]

A nonnegative measurable function has integral 0 exactly when it vanishes almost everywhere (A nonnegative measurable function has integral 0 exactly when it vanishes almost everywhere).

[L4]

Real and imaginary parts of an integrable complex function are integrable (Integrable real and complex functions, and their integrals).

Proof

technique · direct
1.1L1L2L4given

Assume f=g almost everywhere, with exceptional null set N. For every measurable A, the positive and negative parts of the real and imaginary components of (f−g)χA are supported on N. By [L1] their nonnegative integrals vanish, and [L2] and [L4] give ∫A(f−g) dμ=0, hence ∫Af dμ=∫Ag dμ.

1.2L2L3L4given

Assume instead that ∫Af dμ=∫Ag dμ for every measurable A. Apply this to the real part u:=Re⁡(f−g) on A+:={u>0} and to −u on A−:={u<0}. In each case the corresponding nonnegative integral is 0, so [L3] gives u=0 almost everywhere. The same argument for v:=Im⁡(f−g) shows v=0 almost everywhere. Hence f=g almost everywhere.

2.1step 1.1step 1.2∎

Step 1.1 proves (1)⇒(2) and step 1.2 proves (2)⇒(1).

Depends on

Used by

…and 5 more results.

Dependency tree · two levels

14 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources