Darboux's theorem on additive functions bounded on an interval

Remarks

Set $c = f(1)$ and $g(x) = f(x) - cx$. Then $g$ is additive and $g(q) = 0$ for every rational $q$, so $g(x + q) = g(x)$ for all real $x$ and rational $q$. Every real $x$ has a rational $q$ with $x + q \in (a, b)$, and $g$ is bounded above on $(a, b)$, so $g$ is bounded above on all of $\R$. If $g(x_0) \ne 0$ for some $x_0$, then $g(n x_0) = n g(x_0)$ or $g(-n x_0) = -n g(x_0)$ grows without bound as $n \to \infty$, a contradiction. So $g = 0$ and $f(x) = cx$. The same holds if $f$ is bounded below on an interval, since then $-f$ is bounded above. Without any bound the conclusion fails: additive functions built from a Hamel basis are not linear.

Going deeper

For a student

This is the basic regularity criterion for Cauchy's functional equation. Continuity at a point, monotonicity on an interval and measurability all imply boundedness on some interval, so they all reduce to this result. For measurable functions this reduction goes through the Steinhaus theorem; see measurable solutions of Cauchy's equation. The same idea shows that the graph of a non-linear additive function is dense in the plane.