A function $f:\Bbb R \to \Bbb R$ is additive and Lebesgue measurable. Prove that $f$ is continuous.
I know that on $\Bbb Q$, $f$ comes out to be linear. So, if $f$ is to be continuous then $f$ must be linear in $\Bbb R$. But, I'm stuck here.
If anyone can please help. Thanks in advance.
Answer
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).
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