# 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?

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I made some small edits - hope I did the question justice. – Gerry Myerson Dec 22 '11 at 5:04

$f$ 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 $f$.
Nice answer Jonas. Could you also comment about the unbounded case ? This is just a curiosity since more likely the question is about bounded $f$. – Leandro Dec 22 '11 at 6:22
@Leandro: Strictly speaking $|f|$ must be bounded in order to be Riemann integrable, so $f$ must be bounded given the problem as stated. If $|f|$ is unbounded, and the integrability is meant in some improper sense, then the same reasoning applies on subintervals where $f$ is bounded, and convergence of the improper Riemann integral of $|f|$ implies convergence of the improper Rieman integral of $f$. E.g., if $\lim_{y\to b}\int_a^y|f(x)|dx$ exists, and $(x_n)\nearrow b$, then you can show that $(\int_a^{x_n}f(x)dx)$ is a Cauchy sequence, then argue that $\lim_{y\to b}\int_a^yf(x)dx$ exists. – Jonas Meyer Dec 22 '11 at 6:35
@[commenter]: "Did you mean the first sentence to only apply in the case when there exists $F$ such that $F'=f$?" Yes, and that is what I intended the second sentence to convey. "Since you didn't use that fact...." Notice that later Darboux's theorem is invoked; derivatives have the intermediate value property. [This was a reply to comments that were deleted seconds before I posted. I guess that the commenter realized the answer their own concerns, but I'll leave this reply for now in case it helps. Perhaps I'll delete it sometime.] – Jonas Meyer Jan 21 '13 at 4:24