# Derivative and Antiderivative: Suppose $F: [a,b]\rightarrow \mathbb{R}$ is differentiable on $[a,b]$.

Suppose $F: [a,b]\rightarrow \mathbb{R}$ is differentiable on $[a,b]$. Does it follow that $\int_{a}^{b} F' = F(b)-F(a)$?

This is an interesting question. Don't we have to assume that $F'(x)$ is always equal to $f(x)$ for this to work? In other words, does $F(x)$ being differentiable always guarantee that $F'(x)$ is integrable?

Thanks!

• It's possible for $F'$ to be unbounded, hence not Riemann integrable, e.g., $F(x) = x^{2} \sin(1/x^{2})$ for $x \neq 0$, $F(0) = 0$. More interestingly, it's possible for $F'$ to exist everywhere, but to be discontinuous on a set of positive Lebesgue measure, see How discontinuous can a derivative be?. – Andrew D. Hwang Apr 29 '17 at 21:11
• To expand on Andrew's answer, $F'$ may even be bounded and not Riemann integrable. Volterra's function is a standard example. – MathematicsStudent1122 Apr 29 '17 at 21:13
• Note, however, that this deficiency only exists with the Riemann integral. If $F:[a,b] \to \mathbb{R}$ is differentiable and $F'$ is bounded, then $F'$ is both Lebesgue integrable and Henstock–Kurzweil integrable. – MathematicsStudent1122 Apr 29 '17 at 21:20
• You ask whether one must assume something is equal to $f(x)$ but you have not defined anything called $f(x). \qquad$ – Michael Hardy Apr 29 '17 at 21:26
• @Math_QED That's fairly common notation beyond first year calculus. – MathematicsStudent1122 Apr 29 '17 at 21:40

There are several questions in the post that are intertwined, and I'll try to list a bunch of definitions/conclusions below in the hope to answer your question (all integrations below are Lebesgue integration):

Firstly, if $F$ is an antiderivative of $f$, then the derivative of $F$ is $f$ a.e.

Suppose $f:\mathbb R \mapsto \mathbb R$ is integrable and $a \in \mathbb R$. Define $F(x) = \int_a^xf(y)\,dy$ as the antiderivative of $f$. Then $F$ is differentiable almost everywhere and $F'(x)=f(x)$ a.e.

Secondly however, if $f$ is the derivative of $F$ a.e., $F$ is not necessarily the antiderivative of $f$. One good-enough relationship is based on functions of bounded variation. (But notice that we need more condition in order to guarantee the equality)

Functions of bounded variation are differentiable. In particular, if $F$ is increasing, then $F'$ exists a.e., and $\int_a^bF'(x)\,dx \le F(b) - F(a)$; and any function with bounded variation can be written as the difference of two increasing functions.

Notice that here the equality is not guaranteed.

Example: let $F$ be the Cantor-Lebesgue function, then it is differentiable with $F'(x)= 0$ a.e., and thus $$1=F(1)-F(0) \color{red}> 0 = \int_0^1F'(x) \, dx$$

Finally, here is one stricter condition under which the equality is achieved.

If $F$ is absolutely continuous, then $F'$ exists a.e. and $\int_a^bF'(x)\,dx = F(b) - F(a)$.

One could consider absolutely continuous as a special case of bounded variation, because it could be proved that if $f$ is absolutely continuous, then it is of bounded variation.