Is a function absolutely continuous if and only if its derivative is in $L^1$?

Question

So I was reading about the Fundamental Theorem of Calculus for Lebesgue Integrals. It said $F$ is absolutely continuous on $[a,b]$ iff $F'$ exists $a.e.$, $F'\in L^1$ and $$F(x)-F(a)=\int_a^x F'dm.$$ Well, but the Fundamental Theorem of Calculus for Riemann Integrals we know that if $F$ is differentiable everywhere, then $$F(x)-F(a)=\int_a^xF'(x)dx.$$ Given that the Riemann Integral, when it exists, equals the Lebesgue integral for $L^1$ functions, would this mean that if $F$ is not absolutely continuous, would I be correct in concluding that $F'\not\in L^1.$ In particular, at least one of $$\int_a^b (F')^+dm,$$ $$\int_a^x(F')^-dm$$ was infinite.

Answer 1

Yes and no. What is true is that $F$ is absolutely continuous if and only if it has a derivative almost everywhere, this derivative is in $L^1$, and $F(x)-F(a)=\int_a^xF'(x)dx$. However, the Heaviside step function also has a derivative almost everywhere (except at $0$), which is $0$, and hence in $L^1$, but it is not absolutely, or even just, continuous. Neither is the Cantor's staircase, which is even continuous with $F'=0$ a.e. So $\int_a^b (F')^\pm\,dm$ need not be infinite for absolute continuity of $F$ to fail.

Almost everywhere is just a wrong notion of the derivative for such functions. One can define the derivative in a more general, distributional, sense. In that sense the derivative of the Heaviside function is the $\delta$-function, and the derivative of the Cantor's staircase is a singular continuous measure. If we take $F'$ in this sense it will be true that it is in $L^1$ if and only if $F$ is absolutely continuous.

Answer 2

Correction

If a differentiable function $f$ is continuously differentiable except a measure-zero set of points on a closed interval, then Newton-Leibniz formula holds for $f$.
There is a function called Volterra's function differentiable everywhere, having a bounded derivative but its derivative is not Riemann integrable.

Answer

If you assume the differentibility of $f$ on the whole closed interval, then the answer is positive.

In fact, we have the following conclusion.

If $f$ is a continuous function of differentibility except countable points on closed interval $[a,b]$, then following conditions are equivalent:
(1) Newton-Leibniz formula holds for $f$ and $f'$ on every subinterval of $[a,b]$ in the sense of Lebesgue integral.
(2) $f$ is of absolute continuity.
(3) $f$ is of bounded variation.
(4) $f'$ is Lebesgue integrable.

Reference: J. Yeh, Real Analysis Theory of Measure and Integration, pp. 272-274.

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Puzzlement

If we denote the set of absolutely continuous functions by $AC[a,b]$, the set of functions of bounded variation by $BV[a,b]$, the set of continuous functions which are differentiable except countable points on $[a,b]$ by $ D^{ec}[a,b] $, and the set of functions a.e. differentiable whose derivatives are included in $L^1[a,b]$ by $L^{1,1}[a,b]$.

It is clear that $BV[a,b]\cap D^{ec}[a,b]=L^{1,1}[a,b]\cap D^{ec}[a,b]=AC[a,b]\cap D^{ec}[a,b]$.
I wonder if $AC[a,b]\subseteq D^{ec}[a,b]$. If true, we have $L^{1,1}[a,b]\cap D^{ec}[a,b]=AC[a,b]$, that is amazing. Or could we construct an absolutely continuous function undifferentiable on an uncountable set?