TheoremStatement: Literature-sourcedProof: AI-generatedPipeline-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-06

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Absolutely continuous functions have integrable derivatives

Statement

Assume the Axiom of Countable Choice. If $F \in A C [a, b]$ , then $F^{'}$ exists almost everywhere on $(a, b)$ and belongs to $L^{1} [a, b]$ .

Facts & Assumptions

Given: Countable choice and an absolutely continuous real function $F$ on $[a, b]$ .

Proof

technique · direct

1.1

The hierarchy $C^{1}$ implies Lipschitz, Lipschitz implies absolutely continuous, and absolutely continuous implies continuous and bounded variation makes $F$ continuous and BV. By Jordan decomposition for functions of bounded variation, write $F = P - N$ with $P, N$ nondecreasing.

given

2.1

The monotone derivative theorem For a nondecreasing function, the derivative is measurable and integrable and its integral is bounded by the total increase gives derivatives $P^{'}, N^{'}$ almost everywhere and $P^{'}, N^{'} \in L^{1}$ . Hence $F^{'} = P^{'} - N^{'}$ exists almost everywhere and is integrable.

step 1.1algebra

3.1

Degenerate intervals have no interior derivative assertion and the zero function in $L^{1}$ , as required.

step 2.1∎

Depends on

Used by

Fundamental theorem of calculus for absolutely continuous functions Theorem

Dependency tree · two levels

33 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

Donald L. Cohn, Measure Theory, 2nd ed., Lemma 6.3.7 (standard reference, not scraped)