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Suppose that $f : \mathbb{R} \rightarrow \mathbb{R}$ is differentiable but $f \notin C^1 ( \mathbb{R} )$ . It means that $f'$ exist but it is not continuous.

Question 1 Is function $f'$ locally integrable. I.e. does there exist for every $a , b \in \mathbb{R}$

$$ \int_{a}^{b} f'(x) dx $$

I think, I should ask about existence of Lebesgue integral.

Question 2 If it exist, does the Newton-Leibniz formula holds?

$$ \int_{a}^{b} f'(x) dx = f(b) - f(a) $$

Comment. I am asking because I wanted to prove Cauchy's integral theorem using Stokes' theorem. One told me that I am not allowed to use Stokes' theorem if derivatives are not continuous.. So I wonder whether it is important. The simplest case of Stokes' theorem is Newton-Leibniz formula.

quinque
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3 Answers3

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Question 1. 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 $x=0$).(Taken from here).

Question 2. By this thread, $f$ is absolutely continuous on each segment $[a,b]$. Then the positive answer follows from the fundamental theorem of calculus for Lebesgue integral (see, for instance, [?, Theorem 6.10], [Ba], [Be, Theorem 2]).

References

[Ba] Diómedes Bárcenas, The Fundamental Theorem of Calculus for Lebesgue Integral, Divulgaciones Matemáticas 8:1 (2000), 75-85.

[Ba] Stephen Becker, Absolutely continuous functions, Radon-Nikodym Derivative, APPM 5450 Spring 2016 Applied Analysis 2.

[?] “Section 6.5. Integrating Derivatives: Differentiating Indefinite Integrals”.

Alex Ravsky
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    It is very interesting. So it seems like we do not have a proper notion of integral which would work for all derivatives of differentiable functions. This is mysterious since the operation of derivating is injective up to constant (hence, formally such integral can be defined as inverse operator). In your example $x^2 \sin(x^{-2})$, one just has to take integral in the sense of principal value. – quinque Jun 18 '19 at 10:21
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    @quinque I guess that the problem is that we consider the set of all derivatives of differentiable functions as a (proper) subspace of some natural function space, which does not have natural integral operator. – Alex Ravsky Jun 18 '19 at 10:53
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If f is a continuous function of differentibility except countable points on closed interval [a,b], then following conditions are equivalent:

(1) Newton-Leibniz formula holds for f and f' on every subinterval of [a,b] in the sense of Lebesgue integral.

(2) f is of absolute continuity.

(3) f is of bounded variation.

(4) f' is Lebesgue integrable.

Remark: (3)=>(2) is the most curious outcome under the assumption that f is differentiable except countable points and (4) is the most practicable.

yshykc
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Q1: Since $f$ is differentiable on $(a,b)$ then $f(x):=\int_a^x{f'(t)dt}$, where $f'$ is absolutely continuous on $(a,b)$ (we can guarantee this case by Lebesgue theorem), so the answer is Yes $f$ is locally integrable in both senses Riemann and Lebesgue.

Q2: No cannot apply this. Since $f$ is not continuous on $(a,b)$. You may refer to the fundamental theorem of integral calculus.

Mohammad W. Alomari
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  • As I understand, the answer on the second question is "no". But to be sure I would like to see an example of such failure. Or at least an argument why such a function do exist. – quinque Jun 25 '15 at 04:11
  • You can apply the Newton-Leibniz rule under the weaker condition of Riemann integrability: https://math.stackexchange.com/users/72031/paramanand-singh. So as long as $f'$ is Riemann integrable, we can say that $\int_a^b f'dx = f(b) - f(a)$, even if $f'$ is not continuous. – ryan221b May 05 '19 at 13:17
  • On Q1: The function $f′$ cannot be absolutely continuous on each $(a,b)$, because, according to the question condition, it is even not continuous. – Alex Ravsky May 07 '19 at 00:17
  • On Q2: $f$ is continuous on $[a,b]$, because $f$ is differentiable. – Alex Ravsky May 07 '19 at 01:20