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

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The indefinite integral of an $L^{1}$ function is absolutely continuous

Statement

For $f \in L^{1} [a, b]$ , the function $I_{f} (x) = \int_{a}^{x} f$ is absolutely continuous.

Facts & Assumptions

Given: $f \in L^{1} [a, b]$ and its indefinite integral $I_{f}$ .

Proof

technique · direct

1.1givenchoose

Given $ε > 0$ , apply Absolute continuity of the integral to $∣ f ∣$ and obtain $δ > 0$ .

2.1step 1.1algebra

For disjoint intervals $[u_{j}, v_{j}]$ of total length below $δ$ , their union $E$ has measure below $δ$ , and $\sum_{j} ∣ I_{f} (v_{j}) - I_{f} (u_{j}) ∣ \leq \int_{E} ∣ f ∣ < ε$ .

3.1step 2.1∎

This is the defining finite-family condition for absolute continuity; the empty family and $a = b$ give zero sums.

Depends on

Dependency tree · 2 levels

Sources

Christopher Heil, Absolute Continuity and the Banach--Zaretsky Theorem, Lemmas 14 and 22

The indefinite integral of an L1L1 function is absolutely continuous