CorollaryStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-21

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The indefinite Henstock–Kurzweil integral of a derivative is a primitive

Statement

Let $F$ be differentiable on $[a, b]$ , $a < b$ , put $f = F^{'}$ , and define $G (x) = \int_{a}^{x} f$ . Then $G$ is a primitive of $f$ on $[a, b]$ : $G^{'} (x) = f (x)$ at every point, including the domain-relative endpoints.

Facts & Assumptions

Given: The differentiable $F$ , its derivative $f$ , and the integral function $G$ .

[L1]

Every derivative is Henstock–Kurzweil integrable and its integral is the endpoint increment (Every derivative is Henstock–Kurzweil integrable and satisfies Newton–Leibniz).

[L2]

On a degenerate interval, the Henstock–Kurzweil integral is $0$ (The Henstock–Kurzweil integral on a compact interval).

[L4]

A function is differentiable at $c$ when its domain-relative difference quotient has a limit there (The derivative $f^{'} (c) = \lim_{x \to c} \frac{f (x) - f (c)}{x - c}$ of $f : A \to R$ at a point $c \in A$ that is a limit point of $A$ , and differentiability on a set).

Proof

technique · direct

1.1

For $x > a$ , restriction to $[a, x]$ leaves the domain-relative difference quotients of $F$ unchanged at every limit point of that interval, so [L4] makes the restriction differentiable with derivative $f$ . Apply [L1] to obtain $G (x) = F (x) - F (a)$ ; at $x = a$ the same identity follows from [L2].

givenL1L2L4

2.1

The constant $F (a)$ cancels from every domain-relative difference quotient in step 1.1, so [L4] gives $G^{'} = F^{'} = f$ throughout $[a, b]$ , endpoints included.

step 1.1L4algebra∎

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Sources

Alessandro Fonda, The Kurzweil-Henstock Integral for Undergraduates, Ch. 1, Cor. 1.15 (standard reference, not scraped)
Andrew Bruckner, Judith Bruckner and Brian Thomson, Real Analysis, Section 1.21 (standard reference, not scraped)