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?
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?
$f$ 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 $f$.