Integrability of derivatives - MathOverflow most recent 30 from http://mathoverflow.net2013-05-24T05:23:49Zhttp://mathoverflow.net/feeds/question/6711http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/6711/integrability-of-derivativesIntegrability of derivativesMark Meckes2009-11-24T18:37:36Z2010-03-05T20:03:45Z
<p>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?</p>
<p>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.</p>
<p>Note that derivatives of everywhere differentiable functions cannot be arbitrarily badly behaved. For example, they satisfy the conclusion of the intermediate value theorem.</p>
http://mathoverflow.net/questions/6711/integrability-of-derivatives/6713#6713Answer by Gerald Edgar for Integrability of derivativesGerald Edgar2009-11-24T18:49:34Z2010-03-05T11:03:44Z<p>Open interval $(a,b)$ easy ... make $f'$ unbounded, say $f(x) = \sqrt{x}$ on $(0,1)$.</p>
<p>Requiring differentiability even at the endpoint, the counterexample must be more elaborate. But still an unbounded function is not Riemann integrable, so take some $x^a \sin^b x$.</p>
<p>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.</p>
http://mathoverflow.net/questions/6711/integrability-of-derivatives/6716#6716Answer by Pete L. Clark for Integrability of derivativesPete L. Clark2009-11-24T19:06:02Z2009-11-24T19:06:02Z<p>I believe this answers the question:</p>
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<p>MR0425042 (54 #13000)
Goffman, Casper
A bounded derivative which is not Riemann integrable.
Amer. Math. Monthly 84 (1977), no. 3, 205--206.</p>
<p>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. </p>
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<p>The paper is available here:</p>
<p><a href="http://www.math.uga.edu/~pete/Goffman77.pdf" rel="nofollow">http://www.math.uga.edu/~pete/Goffman77.pdf</a></p>
http://mathoverflow.net/questions/6711/integrability-of-derivatives/6717#6717Answer by Gian Maria Dall'Ara for Integrability of derivativesGian Maria Dall'Ara2009-11-24T19:06:41Z2009-11-24T19:06:41Z<p>$f(x) = x Sin(1/x)$ on $(0,1)$ should work. Or with $x^2$ replacing $x$ if you want differentiability at the boundary.</p>
http://mathoverflow.net/questions/6711/integrability-of-derivatives/17172#17172Answer by HenrikRüping for Integrability of derivativesHenrikRüping2010-03-05T11:48:41Z2010-03-05T20:03:45Z<p>I remember, that there was an example of such a function in the book <a href="http://books.google.de/books?id=cDAMh5n4lkkC&printsec=frontcover&dq=counterexamples+in+analysis&cd=1#v=onepage&q=&f=false" rel="nofollow">Counterexamples in Analysis</a>. Just wanted to mention it for the sake of completeness.</p>