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I have two questions that i'm curious about.

  1. If $f$ is differentiable real function on its domain, then $f'$ is Riemann integrable.

  2. If $g$ is a real function with intermediate value property, then $g$ is Riemann integrable.

Thank you in advance.

Katlus
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3 Answers3

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Hints:

1) The function $f$ defined by $$f(x)=\cases{ x^2\sin(1/x^2),&$x\ne0$ \cr 0,&$x=0$}$$ is differentiable on $[-1,1]$; but its derivative is unbounded on $[-1,1]$.

2) Derivatives enjoy the Intermediate Value Property (by Darboux's Theorem).

David Mitra
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  • Note that one can verify that $f'$, with $f$ is as above, has the Intermediate Value Property directly, rather than appealing to Darboux. – David Mitra Dec 12 '12 at 14:30
  • I am sorry, but on (1) the derivative is not integrable? Thank you. https://www.wolframalpha.com/input/?i=integrate_0%5E1+%28+2+x+sin%281%2Fx%5E2%29+-+%282+cos%281%2Fx%5E2%29%29%2Fx%29 – Quiet_waters May 03 '21 at 15:20
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    @Quiet_waters WA is computing an improper integral. The derivative here is not Riemann integrable (such functions are bounded). – David Mitra May 03 '21 at 16:21
  • Thank you so much. Maybe you could help me on this question. Thank you! https://math.stackexchange.com/questions/4125797/seeking-an-example-of-continuous-function-that-has-no-integrable-derivative – Quiet_waters May 03 '21 at 16:48
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  1. A Riemann integrable function $f$ on an interval $[a,b]$ must be bounded on that interval. So if you take $$f(x)=\cases{x^{\frac{3}{2}} \sin(\frac{1}{x}),&$x\ne0$ \cr 0,&$x=0$}$$ on $[0,1]$, you can check that this is a continuous and differentiable function but with an unbounded derivative and so it is not integrable. You can even construct an example of a differentiable function whose derivative is bounded but is still not Riemann integrable - see Volterra's function.
  2. Again, you take a function such as $$f(x)=\cases{\frac{1}{x} \sin(\frac{1}{x}),&$x\ne0$ \cr 0,&$x=0$}$$ on $[0,1]$. This is a discontinuous, unbounded function that satisfies the intermediate value property, but not Riemann integrable. A bounded example is given by the derivative of Volterra's function.
user26857
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levap
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  • I am sorry, but in (1) the derivative is not integrable? Thank you in advance https://www.wolframalpha.com/input/?i=integrate_0%5E1+%7B%28-2+cos%281%2Fx%29+%2B+3+x+sin%281%2Fx%29%29%2F%282+sqrt%28x%29%29%7D. – Quiet_waters May 03 '21 at 15:25
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    @Quiet_waters: The derivative of $(1)$ is not bounded so the derivative is not Riemann integrable. Wolfram alpha computes the integral as an improper integral which is something different. – levap May 03 '21 at 15:31
  • Thank you so much. Maybe you could help me on this question. Thank you! https://math.stackexchange.com/questions/4125797/seeking-an-example-of-continuous-function-that-has-no-integrable-derivative – Quiet_waters May 03 '21 at 16:48
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From the two examples provided above, it turns out that the desired function $f$ is chosen in such a way that $f'$ is unbounded. However, turning a function $f$ into unbounded $f'$ is somewhat not interesting. If you really want to look for an example of $f$ such that $f'$ is defined and bounded everywhere, the Volterra function is what you are looking for. This function is defined by making use of the Smith–Volterra–Cantor set and "copies" of the function defined by $$\left\{ \begin{aligned} & x^2 \sin ⁡ \frac 1x & & \text{for } x\neq 0 \\ & 0 & & \text{for } x=0. \end{aligned} \right.$$ The first three steps in the construction of Volterra's function are as in the picture below.

enter image description here

QA Ngô
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