Show that $f(x+y)=f(x)+f(y)$ implies $f$ continuous $\Leftrightarrow$ $f$ measurable [duplicate]

One implication is trivial. If a function is continuous, then it is measurable. The converse is more tricky.

You can find a very nice proof in the following document. Another proof can be found considering the function $F(x)=\int_0^x f(t)dt$, which is well defined since $F$ is measurable.

Another approach is the following: prove that a discontinuous solution for the functional equation is not bounded on any open interval. It can be shown that for a discontinuous solution the image of any interval is dense in $\Bbb{R}$, and therefore we have problems with the measurability.

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.

Here is a solution using convolution that complements the ones above. We see that $f(x)=f(1)x$ for rational $x$ by first checking the integers, then numbers of the form $1/n$. The hard part is to take care of the irrational numbers. We may suppose without loss of generality that $f(x)=0$ on the rationals (because subtracting off $f(1)x$ doesn't affect the property in the definition).

Fix some $j\in C_c^\infty(\mathbb R)$. By the defining property of $f(x)$, we have the following expression for the convolution: $$(e^{2\pi if(x)}\star j)(x)=e^{2\pi i f(x)}\int j(y)e^{-2\pi i f(y)}\, dy.$$

In particular, the integral is a constant, so $e^{2\pi i f(x)}$ is continuous, since the convolution is continuous (as $j$ is smooth and compactly supported -- this is a standard theorem).

We must be careful, because we cannot conclude from this that $f(x)$ is continuous without further work. The logarithm function is multi-valued, and it is possible for $f(x)$ to 'skip' by multiples of $1$.

However, we can conclude that $f$ must be integer valued. From this we conclude that it is zero everywhere: given $x\in R$, apply the defining property $n$ times to $x/n$ to conclude that $n|f(x)$ for every $n$, which implies $f(x)=0$.

Note that we use the measurability hypothesis to make sure the integral defining the convolution is well-defined. We use a complex exponential instead of $e^f$ (which would make the conclusion easier) because the latter invites concerns about integrability when defining the convolution.