If $f=F'$ and $|f|$ is Riemann integrable, how to show that $f$ is Riemann integrable? Let $f:[a,b]\to \mathbb R $. Suppose that there exists $F$ such that $F'(x)=f(x)$ and that $|f|$ is Riemann integrable.
How to show $f$ is Riemann integrable?
 A: $f$ is continuous everywhere $|f|$ is.  This is not true in general for real valued functions, but it is true for derivatives.
Suppose that $|f|$ is continuous at $c$.  So $\lim\limits_{x\to c}|f(x)|=|f(c)|$.  If $f(c)=0$, then it follows that $\lim\limits_{x\to c}f(x)=0$.  Suppose that $f(c)\neq 0$, and suppose, to reach a contradiction, that $f$ is not continuous at $c$.  Because $|f|$ is continuous at $c$, this implies that $f$ takes on both positive and negative values in every neighborhood of $c$.  Let $\delta>0$ be such that $|x-c|<\delta$ implies $||f(x)|-|f(c)||<|f(c)|$.  Then $|x-c|<\delta$ implies that $f(x)\neq 0$.  Since $f$ takes on both positive and negative values on $\{x:|x-c|<\delta\}$, this implies that $f$ does not have the intermediate value property.  This violates Darboux's theorem. It was the assumption that $f$ is not continuous at $c$ that led to this contradiction, so $f$ is in fact continuous at $c$.
A bounded function on $[a,b]$ is Riemann integrable if and only if the set of points where it is discontinuous has measure zero.  This holds for $|f|$, hence also for $f$.
