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Is there a (preferably simple) example of a function $f:(a,b)\to \mathbb{R}$ which is everywhere differentiable, such that $f'$ is not Riemann integrable?

I ask for pedagogical reasons. Results in basic real analysis relating a function and its derivative can generally be proved via the mean value theorem or the fundamental theorem of calculus. Proofs via FTC are often simpler to come up with and explain: you just integrate the hypothesis to get the conclusion. But doing this requires $f'$ (or something) to be integrable; textbooks taking such an approach typically stipulate that $f'$ is continuous. Proofs via MVT can avoid such unnecessary assumptions but may require more creativity. So I'd like an example to show that the extra work does actually pay off.

Note that derivatives of everywhere differentiable functions cannot be arbitrarily badly behaved. For example, they satisfy the conclusion of the intermediate value theorem.

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up vote 25 down vote accepted

I believe this answers the question:

MR0425042 (54 #13000) Goffman, Casper A bounded derivative which is not Riemann integrable. Amer. Math. Monthly 84 (1977), no. 3, 205--206.

In 1881 Volterra constructed a bounded derivative on $[0,1]$ which is not Riemann integrable. Since that time, a number of authors have constructed other such examples. These examples are generally relatively complicated and/or involve nonelementary techniques. The present author provides a simple example of such a derivative $f$ and uses only elementary techniques to show that $f$ has the desired properties.

The paper is available here:

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For whatever reason, your link doesn't seem to work for me if I include the 'www'. Removing it fixed the problem. – User Aug 29 '15 at 2:05

$f(x) = x Sin(1/x)$ on $(0,1)$ should work. Or with $x^2$ replacing $x$ if you want differentiability at the boundary.

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Open interval $(a,b)$ easy ... make $f'$ unbounded, say $f(x) = \sqrt{x}$ on $(0,1)$.

Requiring differentiability even at the endpoint, the counterexample must be more elaborate. But still an unbounded function is not Riemann integrable, so take some $x^a \sin^b x$.

Even allowing improper Riemann integrals or Lebesgue integral is not enough to avoid the hypothesis that $f'$ is integrable. The Henstock-Kurzweil integral is needed to recover $f$ from $f'$ which exists everywhere on $[a,b]$ in general.

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But can you make a bounded example? – Greg Kuperberg Nov 24 '09 at 18:51
Yes, I forgot to say I want f' to be bounded, so you can't work around the issue by restricting the domain. – Mark Meckes Nov 24 '09 at 19:44
Yes, there are examples with f' bounded (see Goffman paper) ... but then you cannot get by with just finitely many discontinuities in f' ... If f' is bounded and continuous except at finitely many points, then f' is integrable. – Gerald Edgar Mar 5 '10 at 11:06

I remember, that there was an example of such a function in the book Counterexamples in Analysis. Just wanted to mention it for the sake of completeness.

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