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There are examples showing that functions with almost everywhere 0 derivative can be increasing. However in those examples, functions are not differentiable everywhere. In fact, invoking theorem 7.21 from Rudin's Real and Complex Analysis, I can deduce that if a function $f$ is differentiable everywhere and its derivative equals $0$ a.e., then $f\equiv constant$. However, I'm wondering if there is some easier proof of such statement, since the proof of theorem 7.21 is quite weird to me. Is there any other theory that I can use to prove the statement?

Chaos
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  • @Rahul no, there doesn't. That's equivalent to the question the OP is inquiring about. – Stella Biderman Apr 16 '17 at 22:23
  • Your title does not match the body. –  Aug 27 '24 at 10:23
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    What about $f(x)-f(0)=\displaystyle\int_0^x f'(t),dt=\displaystyle\int_0^x0,dt$ and the definition of the Riemannian integral? –  Aug 27 '24 at 10:24
  • @YvesDaoust (i) Nice to see you again! (ii) Your proof is elegant ... if we can establish $f'$ is bounded and therefore Riemann integrable. But offhand, establishing boundedness looks delicate. (My first thought was to invoke Darboux's theorem: $f'$ has the intermediate value property, so must be $\equiv0$. But on closer inspection I don't see an easy way to prove this, either, not least because a measure-zero set can have the cardinality of the reals.) – Andrew D. Hwang Aug 27 '24 at 12:43
  • @AndrewD.Hwang: hi Andrew. Sorry, I think I missed the "a. e." after "equals $0$". –  Aug 27 '24 at 12:47

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Well, there is a whole theory of level sets of derivatives of everywhere differentiable functions, based largely on the theory of Henstock-Kurzweil (a.k.a. Denjoy or gauge) integral.

A good starting point, with a lot of references, could be [D. Preiss, Level sets of derivatives, TAMS 272(1):161–184], available at http://www.ams.org/tran/1982-272-01/S0002-9947-1982-0656484-0/S0002-9947-1982-0656484-0.pdf

This is certainly not simpler, though...

Mateusz Kwaśnicki
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