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Baby Rudin's Fundamental Theorem of Calculus (Theorem 6.21), in my professor's words,states:

Let $f: [a,b] \to \mathbb{R}$ be a Riemann integrable function. If $F: [a,b] \to \mathbb{R}$ is an antiderivative of $f$, then $\int_a^b \! f(x) \, \mathrm{d}x = F(b)-F(a)$.

During the proof, one of my peers asked if the hypothesis that $f$ is Riemann integrable was needed since we have right after that the derivative of $F$ is little $f$. That is, does the second hypothesis imply the first? $F$ is differentiable, so it's continuous on $[a,b]$, and furthermore bounded. Does this then imply that $f$ is also continuous and bounded? If it does, that mean we can exclude the first hypothesis, or is necessary?

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  • $\begingroup$ Well, you need to know that the integral exists. $\endgroup$ – copper.hat Dec 4 '13 at 20:21
  • $\begingroup$ True. So we cannot just say "$F$, on a compact interval, has a derivative called $f$, and therefore, the integral from $a$ to $b$ exists" correct? $\endgroup$ – user70551 Dec 4 '13 at 20:34
  • $\begingroup$ See mathoverflow.net/q/6711/31729. $\endgroup$ – copper.hat Dec 4 '13 at 20:39
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The derivative of a bounded differentiable function isn't necessarily bounded or continuous. A standard example is to let $1 < \alpha < 2$ and define $f(x) = x^\alpha \sin \frac 1x$ if $x \not= 0$, and $f(0) = 0$. In this case $f'(0) = 0$ but $f'$ is unbounded in every neighborhood of $0$.

This doesn't provide a counterexample to the fundamental theorem, though. An example of a differentiable function whose derivative is not Riemann integrable is Volterra's function.

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