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Conjecture: Let $f: \mathbb R \to \mathbb R$ be an everywhere differentiable function and assume that $f'(x) \in \mathbb Z$ almost everywhere. Then is $f$ necessarily an affine function?

Can you give me a proof or a counter-example ? I thought of the devil's staircase, but this is not differentiable everywhere.

Pascal
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    Use Darboux's theorem for the derivative. https://teachingcalculus.com/2014/08/18/darbouxs-theorem/ to deduce that the derivative is constant. – uniquesolution Apr 04 '17 at 12:11
  • Do you assume that $f $ is everywhere or almost everywhere differentiable? – PhoemueX Apr 04 '17 at 12:16
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    @uniquesolution But the derivative is only assumed to be integral a.e., so the application of Darboux, if it works at all, isn't all that straightforward. Or did I miss something? – Harald Hanche-Olsen Apr 04 '17 at 14:16
  • The application of Darboux is very simple. Suppose $f'(a)=z_1$ and $f'(b)=z_2$. Then, between a and b, $f'$ has to take every value between $z_1$ and $z_2$, both integer and non-integer. If it can't be an integer, then $z_1=z_2$ and the derivative everywhere is that constant value. – Paul Apr 05 '17 at 01:31
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    @Paul Why can't $f'$ attain all noninteger values in $[z_1,z_2]$ on a set of measure zero? – user7530 Apr 05 '17 at 01:46
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    I found this which can answer my question https://mathoverflow.net/questions/266377/how-quickly-can-the-derivative-of-an-everywhere-differentiable-function-change-s – Pascal Apr 05 '17 at 07:16
  • Perhaps I shall return to this question$,\ldots\qquad$ – Michael Hardy Aug 14 '17 at 20:08

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Yes, $f'$ must be constant. This has been answered in the comments; the result in https://mathoverflow.net/questions/266377/how-quickly-can-the-derivative-of-an-everywhere-differentiable-function-change-s gives a kind of positive measure Darboux theorem. This is a special case of How irregular can $f'$ be beyond Darboux's Theorem?

Dap
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