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Let $f:[0,1]$ be a real, Lebesgue/Riemann integrable, Lipschitz continuous and differentiable almost everywhere function. Is $f'$ Lebesgue integrable?

On one hand, $f'$ can be undefined for some points on $[0,1]$. On the other hand, the set where $f'$ is not defined has measure zero.

I think there is no sense to say that $f'$ is Lebesgue integrable is $\lvert f'\rvert$ is not defined.

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  • @MattSamuel That question doesn't mention Lipschitz continuity. Are the examples there Lipschitz continuous? – Avi Steiner Mar 23 '20 at 20:16
  • $f'$ is measurable (why?) and bounded (why?) on the compact interval $[0,1]$, and hence Lebesgue integrable (why?). – PhoemueX Mar 24 '20 at 06:11
  • @Phoemuex Can you argue why $f'$ is bounded? – Celine Harumi Mar 28 '20 at 03:46
  • Write down the definition of the derivative (limit of difference quotient), and use that $f$ is Lipschitz. – PhoemueX Mar 28 '20 at 06:28
  • @PhoemueX How can you guarantee that $f'$ exists on $[0,1]$, under the conditions of my question? $f'$ would be bounded only if I assume that $f$ is differentiable (everywhere). – Celine Harumi Mar 30 '20 at 00:35
  • You assume that $f'$ exists almost everywhere. What u mean by bounded is that there is $C > 0$ satisfying $|f'(x)| \leq C$ at every point at which $f'(x)$ exists (and hence almost everywhere). Since you don't care about null sets when integrating, this is enough. – PhoemueX Mar 30 '20 at 05:01
  • @PhoemueX Oh, you meant bounded "almost everywhere". I agree. You concluded that $f'$ is Lebesgue integrable on $[0,1]$. So $\int_{[0,1]} \lvert f'\rvert d\mu=\int_A \lvert f'\rvert d\mu +\int_{A^c} \lvert f'\rvert d\mu$, where $A$ is the set in $[0,1]$ where $f'$ exists. So the first term is bounded. I see that $\mu(A^c)$ is zero, but how can you argue that $\int_{A^c} \lvert f'\rvert d\mu$ is bounded if $f'$ is not defined on this set? – Celine Harumi Mar 30 '20 at 13:40

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