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I understand that the Volterra function $V$ satisfies the properties that $V$ is everywhere differentiable, the derivative $V'$ is everywhere bounded, but $V'$ is not Riemann integrable. I've read on how to construct $V$ and I understand that since $V$ is discontinuous on a set that has positive Lesbesgue measure it can't be Riemann integrable, but what is the intuition here beyond abstract, analytic results? If I wanted to explain this to someone with a calc 2 understanding of Riemann integrability, what could I tell them?

YungJohn_Nash
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  • Maybe make use of the intuitive/pictural idea that for "nice" graphs for $a \leq x \leq b$ (such as $y=x^2$ and $y = xe^{3x} - \sin x)$ the area of a tube containing the graph goes to zero as the tube radius approaches zero (that area is bounded above by a constant times its length and radius), and the tube is a geometric representation of the difference between upper and lower Riemann sums. When the curves have infinite length you use separate tubes for the "infinitely wiggly parts", and attempt to decrease tube lengths sufficiently fast while still covering the curve (continued) – Dave L. Renfro Jan 17 '23 at 08:52
  • so as to have the sum of the tube areas approach zero, which is possible if there is only one localized "infinitely wiggly part", or they are isolated, but when the "infinitely wiggly parts" start clustering together too much (but even this isn't too much), then it becomes impossible to do this (e.g. see Mark McClure's answer to Discontinuous derivative). – Dave L. Renfro Jan 17 '23 at 08:52

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Suppose f is a differentiable function. It can be shown that the set of points of discontinuity of f' must be a meagre, F-$\sigma$ set. A theorem from Lebesgue says that a function is Riemann integrable if and only if its set of points of discontinuity is set of measure zero. The twist? Meagre sets can have non-zero Lebesgue measure, and a famous example of these are provided by the so-called "fat Cantor sets" (or less colorfully, the Cantor-Smith-Volterra sets). Volterra constructed a function whose derivative was discontinuous on such a set, and hence could not be Riemann integrable.

Evan T Young
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