Is there such a function that is continuous at every irrational, discontinuous at every rational, and is also of bounded variation on $[0,1]$? A candidate is Thomae’s Function,

$$f(x) = \begin{cases} \frac{1}{x} & x = \frac{p}{q}, \text{and gcd}(p,q) = 1, q > 0 \\ 0 & x \text{is irrational} \end{cases},$$

but it can be shown to be not of bounded variation since we end up with a series $\sum_k^m \frac{1}{k}$.


2 Answers 2


Let $F(x)=\sum_{r_n \leq x} \frac 1 {2^{n}}$ where $(r_n)$ is an ennumeration of rationals in $[0,1]$. (The sum is over all n such that $r_n \leq x$). Then $F$ has all the desired properties.


The answer by @KaviRamaMurthy is elegant and ingenious since his $\ F\ $ is monotone (increasing) -- thus, while the construction of $\ F\ $ is a bit involved, the proof is instant.

The following is perhaps more straightforward(?); I'll make it general so that it's even easier to see what's going on (altogether not too much):

Let $\ A\subseteq X\subseteq\mathbb R,\ $ where $\ X\setminus A\ $ is dense in $\ X,\ $ and let $\ g:A\rightarrow\mathbb R\setminus\{0\}\ $ be such that

$$ s\ :=\ \sum_{a\in A}\,|g(a)|\ <\ \infty $$

(it follows that $\ A\ $ is countable). Define $\ G:X\rightarrow\mathbb R\ $ as the extension of $\ g\ $ by $\ 0,\ $ i.e. $\ \forall_{x\in X\setminus A}\, G(x)=0.\ $

It follows that $\ G\ $ is continuous at every $\ x\in X\setminus A,\ $ and that the total variation is $\ \le\ 2\cdot s$.

Since $\ X\setminus A\ $ is dense in $\ X,\ $ function $\ G\ $ is discontinuous at every $\ a\in A.$


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .