Alphabeta Math
RemarkSession-authored (Fable 5 assisted) sources checked 2026-07-26 not proved here
Recorded, not proved here. This statement is included so the library can refer to it honestly, with a citation to the literature. It is not proved anywhere in this library: the track that would prove it has not been developed here yet.

The sharp fundamental theorem of calculus (absolute continuity)

Statement

Let F:[a,b]RF : [a,b] \to \mathbb{R}. The following are equivalent.

  1. FF is absolutely continuous on [a,b][a,b].
  2. FF is differentiable almost everywhere on [a,b][a,b], FL1[a,b]F' \in L^{1}[a,b], and

F(x)=F(a)+axFdλfor every x[a,b].F(x) = F(a) + \int_a^x F' \, d\lambda \qquad \text{for every } x \in [a,b].

  1. There exists gL1[a,b]g \in L^{1}[a,b] with F(x)=F(a)+axgdλF(x) = F(a) + \int_a^x g \, d\lambda for every x[a,b]x \in [a,b]; and then g=Fg = F' almost everywhere.

In particular, for FAC[a,b]F \in AC[a,b] the Newton-Leibniz formula F(b)F(a)=abFdλF(b) - F(a) = \int_a^b F' \, d\lambda holds, and AC[a,b]AC[a,b] is exactly the class of functions for which it holds in this sense.

The identity must be required at every xx, not only at x=bx = b. The endpoint identity alone does not characterise absolute continuity. Let cc be the Cantor function and set F(x)=c(2x)F(x) = c(2x) on [0,1/2][0, 1/2] and F(x)=c(22x)F(x) = c(2 - 2x) on [1/2,1][1/2, 1]. Then FF is continuous, F=0F' = 0 almost everywhere, FL1F' \in L^1, and

01Fdλ=0=F(1)F(0),\int_0^1 F' \, d\lambda = 0 = F(1) - F(0),

yet FF is not absolutely continuous, since it is not constant while carrying all its variation on a null set. The identity fails at x=1/2x = 1/2, where the left side is 11 and the right side is 00.

Remarks

Not proved in this library. It is recorded with citations and used in no proof here.

What would prove it. The implication from 3 to 1 is absolute continuity of the indefinite Lebesgue integral, which follows from the dominated convergence theorem (Dominated convergence theorem ). The implication from 3 to g=Fg = F' almost everywhere is the Lebesgue differentiation theorem (Lebesgue differentiation theorem for L1L^1 functions ). The hard direction, from 1 to 2, uses that FACF \in AC is of bounded variation, hence differentiable almost everywhere by Lebesgue's differentiation theorem for monotone functions , that FL1F' \in L^{1} with axFF(x)F(a)\int_a^x F' \le F(x) - F(a) for increasing FF, and then a Vitali covering argument to upgrade the inequality to equality using the δ\delta from absolute continuity. Every step is measure-theoretic.

Which page it serves. This is the natural endpoint of the fundamental theorems of calculus page, and the reason that page's results are called the working FTC rather than the FTC. That page proves: if ff is Riemann integrable on [a,b][a,b] and FF is any antiderivative of ff on [a,b][a,b], then abf=F(b)F(a)\int_a^b f = F(b) - F(a); and if ff is continuous then xaxfx \mapsto \int_a^x f is an antiderivative. Both statements carry hypotheses that the theorem above deletes. The library states the sharp version here so that no reader concludes the working FTC is the last word, and so that the counterexamples on that page (a derivative that is not Riemann integrable, the Cantor function, Volterra's function) have a stated theorem to be counterexamples to.

What this page's other items add. Banach-Zarecki theorem characterises the same class without mentioning an integral at all, and Henstock-Kurzweil versus Lebesgue: ff is Lebesgue integrable iff ff and f|f| are both HK integrable records the integral for which the Newton-Leibniz formula holds for every everywhere-differentiable FF, with no integrability hypothesis on FF' whatsoever.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · next 3 levels

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Sources