Let $f: \mathbb R\to \mathbb R$ be a Borel measurable function. Suppose that for each $q\in \mathbb Q$, the function $f(q+x)-f(x)$ is continuous on $\mathbb R$. Is it true that there is a continuous function $g: \mathbb R\to \mathbb R$ such that $f(x)=g(x)$ for Lebesgue almost every $x\in \mathbb R$?

If the answer to the above question is negative, how about assuming in addition that $f(x+m)=f(x)$ for all $m\in \mathbb Z$. Namely, we ask the same question for a function defined on $\mathbb R/\mathbb Z$.

  • $\begingroup$ Additional assumption does change the answer: if $f(x+1)-f(x)=g(x)$ is continuous, we may easily find a continuous $h(x)$ such that $h(x+1)-h(x)=g(x)$, then $f-h$ is 1-periodic. $\endgroup$ Nov 14, 2018 at 5:26
  • $\begingroup$ @FedorPetrov In the torus case, the question is still: whether $f$ is equal to a continuous function almost everywhere or not. $\endgroup$
    – ronggang
    Nov 14, 2018 at 5:51
  • $\begingroup$ I get this, I just mean that the answers for torus and line are the same (I do not know what is the answer) $\endgroup$ Nov 14, 2018 at 6:40
  • $\begingroup$ Sorry, why do you add words on "a.e."? Do you have an example of such discontinuous measurable function? $\endgroup$ Nov 14, 2018 at 12:37
  • $\begingroup$ @Ilya_Bogdanov: the characteristic function of the rationals. $\endgroup$
    – Nik Weaver
    Nov 14, 2018 at 14:06

1 Answer 1


Consider the 1-periodic function $f$ with Fourier series $$\sum n^{-1}\cos (2\pi n! x).$$ Note that it satisfies your property, since all but finitely many summands are $h$-periodic for any rational $h$. On the other hand, if it were a Fourier series of a continuous function $F$, its partial sums would be Cesàro convergent to the values of $F$, but for $x=0$ the sum of series is infinite (both Cesàro or usual).

Well, strictly speaking the above function $f$ is defined only on a set of full measure (not pointwise) and $f(x+h)-f(x)$ is equivalent to a continuous function, but not genuine pointwise continuous. Is it ok for you? If not, you may carefully correct it on a set of measure 0. For example, define $f(x)$ as a sum of above series when it converges (by Carleson's theorem the series converges almost everywhere to the initial function). After that it remains to define the values of $f$ on the exceptional set. This exceptional set has measure 0 and is invariant under shifts by rational numbers, we must force the difference $f(x+h)-f(x)$ take the necessary values $f_N(x+h)-f_N(x)$, where $N$ is chosen so that $N!h$ is integer, and $f_N$ denotes the corresponding finite sum. This is possible since these values agree in a natural sense.

  • $\begingroup$ Hm. What if the function is bounded? $\endgroup$ Nov 15, 2018 at 6:02
  • $\begingroup$ @IlyaBogdanov I do not know. Also I do not know if $q$ is allowed to be any real number. $\endgroup$ Nov 15, 2018 at 10:22
  • 1
    $\begingroup$ If not, you may carefully correct it on a set of measure 0 Are you sure that the correction described below that line gives you a Borel measurable function? I do not see it immediately :-( $\endgroup$
    – fedja
    Nov 15, 2018 at 16:28
  • $\begingroup$ @fedja ops, indeed. I was thinking about Lebesgue measurability (which I am always thinking about). $\endgroup$ Nov 15, 2018 at 18:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.