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My question:

Is there a function $f: \mathbb{R} \rightarrow \mathbb{R}$ that nowhere continuous on its domain, but has an antiderivative?

If there is no such a function, is it true to conclude that: to have an antiderivative, $f$ is necessary to be continuous at least at one point on its domain?

Any comments/ inputs are highly appreciated. Thanks in advance.

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    Thanks for your feedbacks! Previously, I also thought the same as Christ, but later on I was in doubt, therefore I asked this question... I think the right answer is "no such function", as verified by Sam and Kahen below, using Baire's theorem2012-02-13

2 Answers 2

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There is no such function.

In fact, by an application of Baire's theorem one can show that, given a sequence of continuous functions $f_n:\mathbb R \to \mathbb R$, which converges pointwise, i.e. $f_n(x)\to f(x)$ for each $x\in \mathbb R$, the set of points where $f$ is continuous is a dense $G_\delta$-set.

Applied to your situation, we can consider the sequence $f_n(x) = \frac{f(x+1/n) - f(x)}{1/n}$ All of these functions are continuous and converge pointwise to f'(x). So f'(x) must be continuous on a dense $G_\delta$-set.

So you are right with your second statement: For a function $g$ to be the derivative of some other function, $g$ necessarily has to have at least one point of continuity.

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    Thanks for your answer and analysis. It makes things clearer...2012-02-13
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If you want your antiderivative to be differentiable everywhere, this is impossible since the derivative of a differentiable function $F: \mathbb R \to \mathbb R$ can be realized as a pointwise limit of continuous functions (use the "$\lim_{h\to0} \frac{f(x+h)-f(x)}h$"-definition of differentiability to construct a suitable sequence), and the pointwise limit of a sequence of continuous functions from $\mathbb R$ to $\mathbb R$ must be continuous on a comeagre set by the Baire-Osgood theorem.

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    [Proof of Baire-Osgood](http://www.proofwiki.org/wiki/Baire-Osgood_Theorem).2012-05-01