The sharp fundamental theorem of calculus (absolute continuity)

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

1. $F$ is absolutely continuous on $[a,b]$.
2. $F$ is differentiable almost everywhere on $[a,b]$, $F^{\mathrm{\prime }}\in L^1[a,b]$, and

$F(x)=F(a)+{\int }_a^x{F}^{\mathrm{\prime }}d\lambda \text{for every }x\in [a,b]\mathrm{.}$

3. There exists $g\in L^1[a,b]$ with $F(x)=F(a)+{\int }_a^{x}gd\lambda$ for every $x\in [a,b]$; and then $g=F^{\mathrm{\prime }}$ almost everywhere.

In particular, for $F\in AC[a,b]$ the Newton-Leibniz formula $F(b)-F(a)={\int }_a^b{F}^{\mathrm{\prime }}d\lambda$ holds, and $AC[a,b]$ is exactly the class of functions for which it holds in this sense.

The identity must be required at every $x$, not only at $x=b$. The endpoint identity alone does not characterise absolute continuity. Let $c$ be the Cantor function and set $F(x)=c(2x)$ on $[0,1\mathrm{/}2]$ and $F(x)=c(2-2x)$ on $[1\mathrm{/}2,1]$. Then $F$ is continuous, $F^{\mathrm{\prime }}=0$ almost everywhere, $F^{\mathrm{\prime }}\in L^1$, and

${\int }_0^1{F}^{\mathrm{\prime }}d\lambda =0=F(1)-F(0),$

yet $F$ is not absolutely continuous, since it is not constant while carrying all its variation on a null set. The identity fails at $x=1\mathrm{/}2$, where the left side is $1$ and the right side is $0$.

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 $g=F^{\mathrm{\prime }}$ almost everywhere is the Lebesgue differentiation theorem (Lebesgue differentiation theorem for $L^1$ functions ‡). The hard direction, from 1 to 2, uses that $F\in AC$ is of bounded variation, hence differentiable almost everywhere by Lebesgue's differentiation theorem for monotone functions ‡, that $F^{\mathrm{\prime }}\in L^1$ with ${\int }_a^x{F}^{\mathrm{\prime }}\le F(x)-F(a)$ for increasing $F$, and then a Vitali covering argument to upgrade the inequality to equality using the $\delta$ 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 $f$ is Riemann integrable on $[a,b]$ and $F$ is any antiderivative of $f$ on $[a,b]$, then ${\int }_a^{b}f=F(b)-F(a)$; and if $f$ is continuous then $x\mapsto {\int }_a^{x}f$ 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: $f$ is Lebesgue integrable iff $f$ and $\mathrm{\mid }f\mathrm{\mid }$ are both HK integrable ‡ records the integral for which the Newton-Leibniz formula holds for every everywhere-differentiable $F$, with no integrability hypothesis on $F^{\mathrm{\prime }}$ whatsoever.