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If $f=x^2\sin(x^{-2})$, then $f'$ exists everywhere (including $x=0$) but $f'$ is not Lebesgue integrable on $[0,1]$ (precisely because of the singularity at $x=0).$ I'm trying to find a function $f$ such that $f'$ exists everywhere but $f'$ is not Lebesgue integrable on any interval. Perhaps someone can recall a standard example?

Aubrey
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    Something like taking an enumeration $(q_i)$ of $\mathbb{Q} \cap [0,1]$ then setting $$g(x) = \sum_{n=1}^\infty 2^{-i} f(x - q_i)$$ may work, but I'm not sure whether it's differentiable or not (the usual method using local uniform convergence of partial sums of derivatives clearly won't work; and I don't think it would for anything unbounded, which is neccasary!). – Matt Rigby Sep 22 '14 at 12:50
  • @Aubrey, I am sorry, why $f'$ is not Lebesgue 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 17:07
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    @Quiet_waters That's to do with the definition of Lebesgue integrability for signed functions. In this case, $ \int _ { [ 0 , 1 ] } | f ' | \ \mathrm d \mu = + \infty $, hence non-integrability of $ f ' $. – Mohsen Shahriari Sep 15 '21 at 07:26

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There are no such functions. Since $f$ is continuous, the derivative $$ f'(x) = \lim_{n\to\infty} n(f(x+1/n)-f(x)) $$ is a pointwise limit of continuous functions, i.e., a function of Baire class $1$. Functions of Baire class $1$ have many points of continuity. Every point of continuity has a neighborhood in which the function is bounded; since it's also a Borel function, it is Lebesgue integrable in such neighborhood.

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