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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?

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  • $\begingroup$ @Rahul no, there doesn't. That's equivalent to the question the OP is inquiring about. $\endgroup$ Apr 16 '17 at 22:23
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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...

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