# Is a measurable solution continuous?

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 $$\mathbb R/\mathbb Z$$.

• 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. Nov 14, 2018 at 5:26
• @FedorPetrov In the torus case, the question is still: whether $f$ is equal to a continuous function almost everywhere or not. Nov 14, 2018 at 5:51
• I get this, I just mean that the answers for torus and line are the same (I do not know what is the answer) Nov 14, 2018 at 6:40
• Sorry, why do you add words on "a.e."? Do you have an example of such discontinuous measurable function? Nov 14, 2018 at 12:37
• @Ilya_Bogdanov: the characteristic function of the rationals. Nov 14, 2018 at 14:06

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.
• @IlyaBogdanov I do not know. Also I do not know if $q$ is allowed to be any real number. Nov 15, 2018 at 10:22