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Let $f:[a,b]\longrightarrow \mathbb R$ a derivable function. Is it true that for all $t\in [a,b]$ we have that $$f(t)=f(a)+\int_a^t f'(x)dx \ \ ?$$

The thing is since $f'$ is not supposed continuous, there is no reason for me for $f'$ to be Riemann integrable. So my questions are the followings one :

  • Q1) In Riemann sense, is the formula correct (if we don't have other hypothesis on $f'$). If no, do you have a counter example ?

  • Q2) If we assume $f'$ Riemann integrable, is the formula correct (in Riemann sense). If no, do you have a counter example ?

  • Q3) In Lebesgue sense, is the formula correct (if we don't have other hypothesis on $f'$). If no, do you have a counter example ?

  • Q4) If we assume $f'$ Lebesgue integrable, is the formula correct (in Lebesgue sense). If no, do you have a counter example ?

Peter
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  • See for Q2: https://math.stackexchange.com/questions/1899567/cant-understand-second-fundamental-theorem-of-calculus/1900844#1900844 – Zeekless Jul 17 '18 at 16:50

1 Answers1

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Q1: No, because $f'$ need not be Riemann integrable. For example $f'$ need not be bounded.

Q2: Yes. This is in fact very simple - the elegance of the proof seems to me to be a good reason for the Riemann integral to be defined exactly the way it is. Say $a=t_0<\dots<t_n=b$. Apply the Mean Value Theorem to each subinterval: $$f(b)-f(a)=\sum(f(t_{j+1})-f(t_j))=\sum f'(\xi_j)(t_{j+1}-t_j),$$precisely a Riemann sum for $\int_a^b f'(t)\,dt$.

Q3: No, $f'$ need not be Lebesgue integrable. For example if $a=-1$, $b=1$ and $$f(t)=\begin{cases}t^2\sin(1/t^{100}),&(t\ne0), \\0,&(t=0).\end{cases}$$

Q4: Pretty sure the answer is yes. Almost certain this is a theorem in Rudin - my copy of Real and Complex Analysis is missing. Edit: Thanks to @ChrisJanjigian for confirming that yes, it's Theorem 7.21. Or see here for a proof based on the Vitali-Caratheodory theorem.

Note this is assuming that $f$ is differentiable, which is to say that $f'$ exists at every point. There are certainly examples where $f$ is differentiable almost everywhere, $f'\in L^1$, but $f$ is not absolutely continuous; for example the Cantor function satisfies $f'=0$ almost everywhere.

David C. Ullrich
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    Q4 is Theorem 7.21 in Rudin. – Chris Janjigian Jul 17 '18 at 16:56
  • @ChrisJanjigian Thanks - I knew it was in there somewhere. – David C. Ullrich Jul 17 '18 at 17:02
  • @Abdullahi_A_Ibrahim No, it's not that simple. – David C. Ullrich Jul 17 '18 at 17:04
  • @Abdullahi_A_Ibrahim Really? What's the simple argument that $f'\in L^1$ implies $f$ is absolutely continuous? – David C. Ullrich Jul 17 '18 at 17:10
  • @Abdullahi_A_Ibrahim Sorry, I have no idea what the argument you're suggesting is. Yes, it's fine for $|f'|\le M$. So we define $E_n$ as you did, correcting the typo - now what? (In fact it's clear that $E_n$ is measurable, so I don't see the relevance of that link...) – David C. Ullrich Jul 17 '18 at 17:41
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    I once tried to simply Rudin's proof of Theorem 7.21. You may check it on my blog posting. – Sangchul Lee Jul 17 '18 at 20:36
  • @SangchulLee Looks good to me. Also it's plausible - Vitali-Cartheodory is just deep enough to take care of that nagging "it can't be that easy" feeling. – David C. Ullrich Jul 17 '18 at 22:35
  • Just one minor point regarding Q1. The derivative $f'$ may be bounded and yet fail to be Riemann integrable. The Volterra function derivative is the famous counter-example. – Paramanand Singh Jul 18 '18 at 07:00
  • @SangchulLee Minor comment about that - can't post this there since I don't have a google account. The definition of $G$ has a term $-[f(x)-f(a)]$. On a small screen this looked like $-|f(x)-f(a)|$ at first - I couldn't follow the rest of the argument, finally realized I'd read it wrong. Possibly $-(f(x)-f(a))$ would be more clear? – David C. Ullrich Jul 19 '18 at 13:44