7

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?

mathematics2x2life
  • 13,322
  • Well, you need to know that the integral exists. – copper.hat Dec 04 '13 at 20:21
  • 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? –  Dec 04 '13 at 20:34
  • See http://mathoverflow.net/q/6711/31729. – copper.hat Dec 04 '13 at 20:39
  • You can see that differentiation is slightly more powerful by noting that the limit in differentiation is taken at each point whereas the limit of the Riemann sums is taken for the whole range. E.g. when a function oscillates with a frequency approaching infinity at one edge, the Riemann sum cannot fit into the spaces between the oscillations. If you want perfect reversibility you need an integral which fixes that discrepancy. The gauge integral does just that at the expense of a slightly more complicated limiting procedure. – tobi_s Jul 21 '21 at 02:52

1 Answers1

6

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.

Umberto P.
  • 52,165