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I am trying to prove this result:

If $f$ is a non-constant function defined on $[a, b]$ such that $f$ is differentiable on $[a, b]$ with a bounded derivative and $f'(x) = 0$ for all $x$ in a dense subset of $[a, b]$ then $f'$ is not Riemann integrable on $[a, b]$.

The conclusion deals with $f'$ and I don't seem to have enough details on $f'$ except that it is bounded and vanishes in a dense subset of $[a, b]$ (meaning every open sub-interval of $[a, b]$ contains points where derivative $f'$ vanishes). Perhaps I am missing some implication of vanishing of $f'$ on a dense set.

Any hints or a solution based on elementary techniques (i.e. not involving measure theory) will be highly appreciated.

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  • $\begingroup$ Have a look at math.stackexchange.com/questions/1799792/… Applying the fundamental theorem of calculus I believe with the above we arrive at a contradiction to the fact that $f$ is non-constant. $\endgroup$ – Kayle of the Creeks May 28 '16 at 9:59
  • $\begingroup$ @KayleoftheCreeks: Got it! I was trying to think more about discontinuities of $f'$ (on which there was no info) and did not think of deriving a contradiction. Thanks a lot. $\endgroup$ – Paramanand Singh May 28 '16 at 10:07
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The comments to this question point to this question which shows that if a function is Riemann integrable over an interval and vanishes on a dense subset then its integral is $0$. This happens because in this case given any partition of the interval we can choose the tags so that the Riemann sum becomes $0$.

Now the answer to the current question is obvious. Since $f$ is non-constant on $[a, b]$ there is a point $c \in (a, b]$ for which $f(c) \neq f(a)$. If $f'$ is integrable over $[a, b]$ then it is also integrable over $[a, c]$ and then $$0 \neq f(c) - f(a) = \int_{a}^{c}f'(x)\,dx$$ Moreover since $f'$ vanishes in a dense subset of $[a, b]$ it also vanishes on a dense subset of $[a, c]$ and hence the above integral is $0$. This contradiction shows that $f'$ can not be Riemann integrable on $[a, b]$.

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