TITLE: Show that $f(x+y)=f(x)+f(y)$ implies $f$ continuous $\Leftrightarrow$ $f$ measurable
QUESTION [18 upvotes]: Let $f:\mathbb R \rightarrow \mathbb R$, and for every $x,y\in \mathbb R$ we have $f(x+y)=f(x)+f(y)$.
Show that $f$ measurable $\Leftrightarrow f$ continuous.

REPLY [8 votes]: Let $f: \mathbb{R} \to \mathbb{R}$ be a measurable group homomorphism. We are going to show $f$ is continuous. For any $\delta > 0$, there exists an interval $I = (a, b)$ of  length $b - a = \delta$ such that the set   $E: = f^{-1}(I)$   has positive measure. Now, using that $f$ is a group homomorphism, we can see that $$f^{-1} ((-\delta, \delta)) = f^{-1}( I - I) = E - E$$ (where for a set $S \subseteq \mathbb R$, we denote $S - S:= \{x - y | x, y \in S\}$). By the Steinhous theorem, $E - E$ contains an open neighborhood at the origin. Thus, for some $\varepsilon > 0$, we have 
$$|x| < \varepsilon \implies x \in E - E \implies |f(x)| < \delta.$$ Thus $f$ is continuous at $0$, and since $f$ is a group homomorphism, this proves that $f$ is continuous everywhere. 
Note that the continuity of $f$ implies that $f$ is actually linear! Thus, any measurable group homomorphism $f: \mathbb{R} \to \mathbb{R}$ is linear.