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 $F, G \in A C [a, b]$ , $\int_{a}^{b} F G^{'} + \int_{a}^{b} F^{'} G = F (b) G (b) - F (a) G (a) .$

Facts & Assumptions

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

Proof

technique · direct

1.1givenalgebra

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

2.1step 1.1

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

3.1step 2.1algebra∎

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

Depends on

Dependency tree · 2 levels

Sources

Donald L. Cohn, Measure Theory, 2nd ed., Corollary 6.3.9