If $f : \mathbb{R} \rightarrow \mathbb{R}$ is a Lebesgue measurable function, such that all distinct $x,y \in \mathbb{R}$ we have that:
$$\frac{f(x)-f(y)}{x-y} \not\in \mathbb{Q}$$
Can we conclude that $f$ is a linear function?
This is If the slope of a secant is always irrational, is the function linear? with the assumption that $f$ is measurable added. There the answer was shown to be yes, if $f$ is continuous and no in general, by an axiom of choice argument. It reminded me of the old question (of interest to Steinhaus and his circle) about whether $f(x+y)=f(x)+f(y)$ implies $f$ is of the form $f(x)=kx$. If memory serves, the answer in this case is affirmative for measurable $f$ and no in general, by an axiom of choice argument.