Let $f:[0,1] \to \mathbb R$ be a differentiable function such that $\sup_{x\in [0,1]} |f'(x)|$ is finite. Then since $f'(x)=\lim_{n\to \infty} \dfrac{f(x+1/n)-f(x)}{1/n}$, so $f'(x)$ is measurable and also the Lebesgue integral $\int_0^1|f'(x)|dx$ is finite, thus $f' \in L^1([0,1])$.

My question is, is it true that $\int_0^1 f'(x)=f(1)-f(0)$ ?

Note that fundamental theorem of calculus does not apply here since $f'(x)$ is not continuous (not even known to be Riemann integrable)

  • $\begingroup$ Use Barrow rule $\endgroup$ – Tito Eliatron Nov 4 '18 at 22:55
  • $\begingroup$ @TitoEliatron: what is that ? $\endgroup$ – user521337 Nov 4 '18 at 22:57
  • $\begingroup$ Fundamental Theorem of Calculus en.wikipedia.org/wiki/Fundamental_theorem_of_calculus#Corollary In Spanish it is known as Barrow's Rule. $\endgroup$ – Tito Eliatron Nov 4 '18 at 22:58
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    $\begingroup$ @TitoEliatron: see my question again $\endgroup$ – user521337 Nov 4 '18 at 22:59
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    $\begingroup$ @TitoEliatron: again ... we do not know that $f'$ is Riemann integrable ... we only know it is Lebesgue integrable ... $\endgroup$ – user521337 Nov 4 '18 at 23:02

Lebesgue integrability is enough. You need the following version of FTC:

Let ${[a,b]}$ be a compact interval of positive length, let ${F: [a,b] \rightarrow {\bf R}}$ be a differentiable function, such that ${F'}$ is absolutely integrable. Then the Lebesgue integral ${\int_{[a,b]} F'(x)\ dx}$ of ${F'}$ is equal to ${F(b) - F(a)}$.

This is a standard result in real analysis. See for instance this excellent set of lecture notes by Terry Tao.

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