# Integrability of derivatives

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.

• Mar 30, 2022 at 14:13
• @MartinSleziak: In fact the only answer on the second post links right back to this one. Mar 30, 2022 at 16:00

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:

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. It can be found in Chapter 8 (Sets and Measure on the Real Axis), Example 35 (A bounded function possessing a primitive on a closed interval but failing to be Riemann-integrable there.)

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.

• But can you make a bounded example? Nov 24, 2009 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. Nov 24, 2009 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. Mar 5, 2010 at 11:06

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

The function $$0\neq x\mapsto x^2\cos\frac1{x^2}$$, $$0=x\mapsto0$$, is everywhere differentiable with a derivative that is not Riemann integrable, not Lebesegue integrable, but Denjoy--Perron--Luzin--Henstock--Kurzweil integrable. See Wikipedia & Granath.