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this is probably a very common question, but i couldn't find the answer on my books.
is every darboux function the derivative of a function? even the nowhere continuous ones?

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up vote 2 down vote accepted

No. Unfortunately, there is no such simple criterion for a derivative. At least none is known.

Ref: Bruckner, Andrew, Differentiation of real functions. Second edition. CRM Monograph Series, 5. American Mathematical Society, Providence, RI, 1994. xii+195 pp. ISBN: 0-8218-6990-6, MR1274044 (94m:26001)

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thanks, that's what i feared. i'll get the Bruckner asap. i think it's necessary to assume that the function is at least $L^1$. – alberto.bosia Feb 19 '12 at 12:08

Let $\phi(x)=x\chi_{[-1,1]}(x)+\text{sgn}(x)\chi_{[-1,1]^c}(x)$, $f$ be the Conway base 13 function, then $\phi\circ f$ is a Darboux function and is nowhere continuous. Moreover, $\phi\circ f$ is Borel measurable and bounded (in particular, it is $L^1$).

But $\phi\circ f$ is not a derivative because any derivative is a limit of continuous functions, hence its discontinuity points form a set of first category (in particular cannot be the whole interval). See this post.

In sum, $\phi\circ f$ is a bounded Borel measurable Darboux function but is not a derivative.

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