Additivity + Measurability $\implies$ Continuity

Using Lusin's theorem, there exists a compact $K \subset [0,1]$ such that $\mu(K)>2/3$ and $f$ continuous on $K$. Let $\epsilon>0$. In fact, $f$ is uniformly continuous on $K$, so there exists $\delta>0$ such that $|x-y|<\delta$ implies $|f(x)-f(y)|<\epsilon$; without loss of generality, suppose $\delta<1/3$.

Let $h<\delta$. Notice that the intersection between $K$ and $K-h$ is nonempty; otherwise, $1+h= \mu([-h,1]) \geq \mu(K \cup K-h)= \mu(K)+\mu(K-h)=2\mu(K)>4/3$, so $h>1/3$ whereas $h<\delta<1/3$ by assumption.

Let $x_0 \in K \cap (K-h)$. We have $|f(x_0+h)-f(x_0)|<\epsilon$, hence $|f(h)|<\epsilon$ because $f$ is additive. You deduce that $f$ is continuous at $0$.

Finally, it is straightforward to conclude that $f$ is continuous on $\mathbb{R}$.

Edit: For a simple proof of Lusin's theorem, see for example: Marcus B. Feldman, A proof of Lusin's theorem, The American Mathematical Monthly Vol. 88, No. 3 (Mar., 1981) (pp. 191-192).