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CorollaryStatement: AI-adaptedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-13
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Assuming Countable Choice, the integral function of a Riemann-integrable function is Lipschitz and differentiable almost everywhere, with derivative equal to the integrand there

Statement

Assume the Axiom of Countable Choice. Let a<b, let f:[a,b]R be Riemann integrable, and let

F(x):=axf.

Then F is Lipschitz on [a,b]. If D is the set of discontinuities of f, then D has measure zero and

F(x)=f(x)(x[a,b]D).

Thus the set of points where F fails to exist or differs from f is contained in a set of measure zero. At a and b, the displayed derivative is the relative one-sided derivative on [a,b].

Facts & Assumptions

Given: Reals a<b, an integrable f:[a,b]R, its integral function F, and Countable Choice.

[L1]

An integrable f is bounded, and if fK then its integral function satisfies F(y)F(x)Kyx (The integral function of a bounded integrable f is Lipschitz, hence uniformly continuous).

[L2]

Assuming Countable Choice, a bounded function is Riemann integrable if and only if its discontinuity set has measure zero (Lebesgue's criterion for Riemann integrability: a bounded f on [a,b] is Riemann integrable if and only if its set of discontinuities has measure zero).

[L3]

If f is integrable and continuous at c[a,b], then its integral function is differentiable at c and F(c)=f(c), with relative endpoint derivatives (The first fundamental theorem: if f is integrable on [a,b] and continuous at c, then F(c)=f(c); in particular a continuous f has F as a primitive).

Proof

technique · direct
1.1

Integrability supplies a bound fK, so F is Lipschitz by [L1].

givenL1
1.2

The discontinuity set D has measure zero by the forward implication of [L2], whose stated proof is the sole use of Countable Choice.

givenL2
2.1

Every x[a,b]D is a continuity point of f, so [L3] gives F(x)=f(x) there, including the relative derivative at an endpoint outside D.

step 1.2L3
3.1

Hence every point at which the derivative conclusion fails lies in D, and that containing set has measure zero.

step 1.2step 2.1

Depends on

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