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Let $ f:[a,b]\to\mathbb{R} $ be an integrable function with a primitive function.

Define:

$ F\left(x\right)=\intop_{a}^{x}f\left(t\right)dt. $

Prove/disprove:

$F$ is differentiable and it follows that $ F'=f $

So, I'm pretty sure that this is false. Otherwise we probably wouldn't mention continious function in the fundamental theorem of calculus.

But I'm having trouble finding a counter example. Obviously we do not want a continuous function as a counter example, so we are looking for a non continuous function, but the discontinuities of the function should be essential discontinuity, because otherwise it wouldn't follow Darbo's theorem and $ f $ wouldn't has an antiderviative.

Also, we still want $ f $ to be bounded, because otherwise it wouldn't be integrable.

So I guess it should be some trigonometric function that "explodes" around its discontinuities points.

I'd like some help finding one. Thanks in advance.

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FreeZe
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  • What exactly should "primitive function" indicate here? Usually it would mean that there is a function $g$ such that $g'(t) = f(t)$ holds for all $t$. – Daniel Fischer Aug 16 '20 at 13:49
  • Im not familier with thia function, but from a quick google: isnt it continious? – FreeZe Aug 16 '20 at 13:49
  • @Danirl Ficher Indeed that's what I ment when I wrote "primitive function". Though, if $ f $ is not continious I think it does not necesseraly correct that F is also an antideriviative of $ f(t) $ – FreeZe Aug 16 '20 at 13:58
  • In that case, Theorem 7.21 of Rudin's Real And Complex Analysis shows that $F'(t) = f(t)$ for all $t$. This deals with the Lebesgue integral, but I think assuming that $f$ is Riemann integrable allows only very small simplifications in the proof. – Daniel Fischer Aug 16 '20 at 14:05
  • Im sorry Im not familier at all with Lebesgue integral (its not in our course syllabus, we'll learn it in the next calculus course). Do you claim that this is actually correct? – FreeZe Aug 16 '20 at 14:07
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    Yes. If $g \colon [a,b] \to \mathbb{R}$ is differentiable at every point of $[a,b]$, and $f(t) = g'(t)$ is integrable on $[a,b]$, then $$g(x) - g(a) = \int_a^x f(t),dt$$ holds for all $x \in [a,b]$. Thus $F$ differs from $g$ by a constant, showing that $F$ is differentiable and $F'(t) = g'(t) = f(t)$ on $[a,b]$. – Daniel Fischer Aug 16 '20 at 14:17
  • @DanielFischer I guess you used here Newton leibniz theorem. I know that if $ g $ is an antideriviative of $ f $ then $ \intop_{a}^{b}f\left(t\right)dt=g\left(b\right)-g\left(a\right) $. But is it still hold's when $ x $ is'nt constant and $ f $ does not necessary continious? – FreeZe Aug 16 '20 at 14:34
  • I don't know which version of the FTC is called Newton Leibniz theorem. (I don't really see either of the two considering pathological functions.) What I'm saying is that it is sufficient that a) $f$ is a derivative (has a primitive) and b) $f$ is integrable to deduce that the indefinite integral of $f$ is a primitive of $f$. We don't need continuity for that. But the proof is much harder than the continuous case, which explains that the FTC as stated in earlier courses refers to continuous $f$. – Daniel Fischer Aug 16 '20 at 14:42
  • See this answer. – Paramanand Singh Aug 20 '20 at 10:34

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