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Does there exists any diffentiable function $f$ such that $f'$ is discontinuous exactly on $\Bbb{Q} $ and continuous on $\Bbb{R}\setminus \Bbb{Q}$ ?


Since $\Bbb{Q}$ is $F_{\sigma}$ , we can produce a function which is discontinuous only on $\Bbb{Q}$.

For an example we can pick Thomae's function.But Thomae's function has no primitive. Because if thomae's function $f$ has a primitive $F$ then $F'=f $ . Since $F'$ is Darboux function, image of $F'=f$ must contains an intervals and this is not possible as Thomae's function doesn't attain irrational values.

By choosing a particular example, we can conclude the impossibility of existence of such function.


If $f'$ is Darboux function and belongs to Baire class $1$ then $f'$ has a primitive $f$ .

Hence our goal is to create a Darboux function $f'$ of Baire class $1$ which is continuous on $\Bbb{Q}$ and discontinuous on $\Bbb{R}\setminus \Bbb{Q}$.

How to produce such function?

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    $\begingroup$ Dave Renfro's comprehensive answer to How discontinuous can a derivative be? is probably of interest. $\endgroup$ Jun 15, 2022 at 1:36
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    $\begingroup$ Just saw this question now. Near the beginning of the answer @Andrew D. Hwang cites is: More precisely, a subset $D$ of $\mathbb R$ can be the discontinuity set for some derivative if and only if $D$ is an $F_{\sigma}$ first category (i.e. an $F_{\sigma}$ meager) subset of $\mathbb R.$ The rationals are such a set -- $F_{\sigma}$ because a countable union of singleton sets each of which is a closed set; first category because a countable union of singleton sets each of which is a nowhere dense set -- so the answer is YES. Maybe for an answer someone can give an explicit example. $\endgroup$ Jul 18, 2022 at 16:08
  • $\begingroup$ Since $\Bbb{Q}$ is $G_{\delta}$ , we can produce a function which is discontinuous only on $\Bbb{Q}$. --- This is not correctly worded, and probably should be revised to one of the following two versions: "Since $\Bbb{Q}$ is $F_{\sigma}$ , we can produce a function which is discontinuous only on $\Bbb{Q}$." OR "Since $\Bbb{R}\setminus \Bbb{Q}$ is $G_{\delta}$ , we can produce a function which is continuous only on $\Bbb{R}\setminus \Bbb{Q}$." $\endgroup$ Jul 18, 2022 at 16:17
  • $\begingroup$ Sorry $\Bbb{Q}$ is an $F_{\sigma}$ set. I have made a mistake there. I will fix immediately. $\endgroup$ Jul 18, 2022 at 16:20
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    $\begingroup$ @Andreas: That does not work because $f$ has jump discontinuities at the rational numbers, so it can not be the derivative of some function $F$ (it does not have the intermediate value property). $\endgroup$
    – Martin R
    Jul 18, 2022 at 17:51

1 Answer 1

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The following result is quoted in How discontinuous can a derivative be?, with references to several proofs:

A subset $D$ of $\mathbb R$ can be the discontinuity set for some derivative if and only if $D$ is an $F_{\sigma}$ first category (i.e. an $F_{\sigma}$ meager) subset of $\mathbb R.$

One of the references which is also online available is

$\Bbb Q$ surely is an $F_{\sigma}$ first category set, so there does exist a differentiable function $f$ on $\Bbb R$ such that $f'$ is discontinuous exactly on $\Bbb Q$. Inspecting the proof from the Bruckner/Leonard article, one can construct such a function for this set:

Let $h: \Bbb R \to \Bbb R$ be a function with the following properties:

  • $h$ is differentiable everywhere.
  • $h'$ is continuous everywhere except at $x=0$.
  • $h$ and $h'$ are bounded.

(For example, $h(x) = x^2 \sin(1/x)$ for $x\ne 0$, $h(0) = 0$.)

Now let $(q_n)$ be an enumeration of the rational numbers, and define $f: \Bbb R \to \Bbb R$ as $$ f(x) = \sum_{n=1}^\infty 3^{-n} h(x-q_n) \, . $$

Both $\sum_{n=1}^\infty 3^{-n} h(x-q_n) $ and $\sum_{n=1}^\infty 3^{-n} h'(x-q_n) $ converge uniformly on $\Bbb R$, so that $f$ is differentiable with $$ f'(x) = \sum_{n=1}^\infty 3^{-n} h'(x-q_n) \, . $$

$f'$ is continuous at every irrational point $x_0$ because all functions $3^{-n} h'(x-q_n)$ are continuous at $x_0$ and the series for $f'$ converges uniformly.

And $f'$ is discontinuous at every $q_m \in \Bbb Q$ because in $$ f'(x) = 3^{-m} h'(x-q_m) + \sum_{n \ne m} 3^{-n} h'(x-q_n) \, . $$ the first term on the right is discontinuous at $q_m$, whereas the second term is continuous at $q_m$, again as a uniformly convergent series of functions which are all continuous at $q_m$.

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  • $\begingroup$ Very interesting.... $\endgroup$ Jul 18, 2022 at 16:50
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    $\begingroup$ @Lost in Space: For those interested, here is a freely available copy of the Bruckner/Leonard paper. $\endgroup$ Jul 18, 2022 at 19:50

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