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I know that $f(x)$ is bounded by the extreme value theorem, because $f(x)$ is defined on a closed bounded interval. Now, how can I prove that its derivative is integrable knowing that it is continuous? Or, is there a counterexample?

Mik
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2 Answers2

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Any continuous function on a closed and bounded interval $[a,b]$ is integrable. Since, $f$ is given to be continuously differentiable, its derivative is continuous on the domain, and hence integrable.

If you only want to assume that $f'$ is continuous on $(a,b)$ then it is not so simple. Consider the function $f : [0,1] \to \mathbb{R}$ defined by $$ f(x) = \begin{cases} 0, & x=0;\\ x^2 \sin (1/x^2), & x \neq 0. \end{cases} $$ It is differentiable everywhere, and its derivative is continuous on $(0,1)$, but the derivative is unbounded as $x \to 0^+$. You can take a look at the answers to this question for more examples of bounded functions with unbounded derivatives: Can the graph of a bounded function ever have an unbounded derivative?

  • So, if a function that is defined on [a,b] is continuously differentiable, then its derivative is continuous on [a,b] and not on (a,b)? – Mik Oct 08 '18 at 06:31
  • @Kim it depends on your definition. The definition I have seen is that $f : [a.b] \to \mathbb{R}$ is said to be differentiable if there is an open set $U \subset [a,b]$ such that $f$ can be extended to a continuously differentiable function on $U$. –  Oct 08 '18 at 07:58
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    This isn't correct. The derivative can become unbounded near $a,b$ and we need care to conclude the integral makes sense. – Wojowu Oct 08 '18 at 09:12
  • We need OP's definition of continuously differentiable to know if this is correct. – Zach Boyd Oct 08 '18 at 10:03
  • @Wojowu quite right, this is really sloppy of me. I am reverting my answer to its previous form. –  Oct 08 '18 at 11:28
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    @ZachBoyd quite right, this is really sloppy of me. I am reverting my answer to its previous form. –  Oct 08 '18 at 11:28
  • @Kim Correction to my first comment: I meant to say "$f : [a,b] \to \mathbb{R}$ is said to be continuously differentiable if $\dots$" –  Oct 08 '18 at 12:39
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If, according to your question, $f'$ is continuous on a bounded interval (i.e. a compact set), it makes no doubt it is Riemann-integrable (and hence also Lebesgue integrable):

A bounded function on a compact interval $[a, b]$ is Riemann integrable if and only if it is continuous almost everywhere (the set of its points of discontinuity has measure zero, in the sense of Lebesgue measure).

See: Riemann integral

Picaud Vincent
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  • @Brahadeesh I did not realize I had so much typos in my answer, thanks for the corrections. – Picaud Vincent Oct 08 '18 at 04:43
  • As in the other answer, we need to be careful about whether f' is continuous on (a,b) or [a,b] since there are different ways of handling the endpoints, and it can change the answer. – Zach Boyd Oct 08 '18 at 10:05
  • @ZachBoyd Yes sure, here I assumed that $f'$ was continuous on $[a,b]$ as it was not explicitly stated otherwise in the question – Picaud Vincent Oct 08 '18 at 16:31
  • @PicaudVincent Interesting. So if I am supposed to prove some result using a certain set of assumptions, I can make additional assumptions as long as it is not explicitly stated otherwise? Solving my exercise sheets will become much easier now :) – Filippo Nov 11 '21 at 07:17