real analysis - Is there a function which is "midpoint linear" but not linear?

Let a function $f:\mathbb{R}\rightarrow\mathbb{R}$ be midpoint linear if for all $x,y\in\mathbb{R}$, $f(\frac{x+y}{2})=\frac{f(x)+f(y)}{2}$. If a midpoint linear function $f$ is continuous, then I believe you can show through a density argument that $f$ is also linear. However, if we drop the assumption of continuity, I wonder if we can construct (or at least describe) a counterexample that is midpoint linear function but is not in fact linear. Because midpoint convexity does not imply convexity without the additional assumption of continuity, I believe that such a counterexample should exist, and that it will likely be through a Vitali set sort of argument that involves taking a basis for $\mathbb{R}$ over $\mathbb{Q}$. I am not sure of the details though.

Answer

I think you're exactly right: take a Hamel basis for $\Bbb R$ over $\Bbb Q$, and then you can define $f$ arbitrarily on each basis element and extend it linearly to all of $\Bbb R$.

For example, if $\pi$ is a member of your Hamel basis, then every real number $x$ can be written uniquely as $x = r\pi + {}$(a rational linear combination of finitely many other basis elements), where $r$ is rational (possibly $0$). Then you can define $f(x) = r$. This function is midpoint-linear (indeed, rational-linear) but not linear: it takes the value $0$ on a whole lot of real numbers, including some between $\pi$ and $2\pi$ for instance.