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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 $\mathbb R/\mathbb Z$.

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  • $\begingroup$ 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. $\endgroup$ – Fedor Petrov Nov 14 '18 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 '18 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) $\endgroup$ – Fedor Petrov Nov 14 '18 at 6:40
  • $\begingroup$ Sorry, why do you add words on "a.e."? Do you have an example of such discontinuous measurable function? $\endgroup$ – Ilya Bogdanov Nov 14 '18 at 12:37
  • $\begingroup$ @Ilya_Bogdanov: the characteristic function of the rationals. $\endgroup$ – Nik Weaver Nov 14 '18 at 14:06
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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.

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  • $\begingroup$ Hm. What if the function is bounded? $\endgroup$ – Ilya Bogdanov Nov 15 '18 at 6:02
  • $\begingroup$ @IlyaBogdanov I do not know. Also I do not know if $q$ is allowed to be any real number. $\endgroup$ – Fedor Petrov Nov 15 '18 at 10:22
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    $\begingroup$ 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 '18 at 16:28
  • $\begingroup$ @fedja ops, indeed. I was thinking about Lebesgue measurability (which I am always thinking about). $\endgroup$ – Fedor Petrov Nov 15 '18 at 18:30

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