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

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Total-variation function of an absolutely continuous function

Statement

Assume the Axioms of Countable Choice and Dependent Choice. If $F \in A C [a, b]$ , then $V_{F} (x) = \int_{a}^{x} ∣ F^{'} (t) ∣ d t and V_{F}^{'} = ∣ F^{'} ∣ a.e.$

Facts & Assumptions

Given: Countable choice, dependent choice, $F \in A C [a, b]$ , and its total-variation function $V_{F}$ .

Proof

technique · direct

1.1

The sharp FTC gives $F (v) - F (u) = \int_{u}^{v} F^{'}$ , so every partition sum is at most $\int_{a}^{x} ∣ F^{'} ∣$ .

givenalgebra

2.1

Approximate the sign of $F^{'}$ by a finite step function and use its sign-change endpoints as a partition; the corresponding partition sum approaches $\int_{a}^{x} ∣ F^{'} ∣$ . Thus the supremum defining Total-variation function on a compact interval equals that integral.

step 1.1choose

3.1

Since $∣ F^{'} ∣ \in L^{1} [a, b]$ , The indefinite integral of an $L^{1}$ function is absolutely continuous makes the right side an absolutely continuous function, and The indefinite integral of an $L^{1}$ function is differentiable almost everywhere gives its derivative $∣ F^{'} ∣$ almost everywhere. Step 2.1 identifies that function with $V_{F}$ .

step 2.1∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

20 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

R. K. Srivastava, MA550 Measure Theory Lecture Notes, Corollary 4.38 (standard reference, not scraped)