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TheoremStatement: Literature-sourcedProof: AI-generatedPipeline-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-06
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Integration by parts for absolutely continuous functions

Statement

Assume the Axioms of Countable Choice and Dependent Choice. For F,G∈AC[a,b], ∫abFG′+∫abF′G=F(b)G(b)−F(a)G(a).

Facts & Assumptions

Given: Countable choice, dependent choice, and absolutely continuous functions F,G on [a,b].

Proof

technique · direct
1.1

The product lemma makes FG AC. At the common full-measure set where F′ and G′ exist, the ordinary product rule gives (FG)′=F′G+FG′. The sharp FTC makes F′,G′ integrable, and continuity on the compact interval bounds F,G, so both products are integrable.

givenalgebra
2.1

Apply Fundamental theorem of calculus for absolutely continuous functions to FG and integrate the displayed derivative.

step 1.1
3.1

Rearranging gives the stated identity, with identical zero sides on [a,a].

step 2.1algebra∎

Depends on

Used by

Dependency tree · two levels

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Sources