# Almost everywhere measurable homomorphism

Let $G$ be a locally compact abelian group (where the group operation is written as $+$) which is $\sigma$-compact and $\mu$ a nontrivial Haar measure on $G$. Define $f: G \rightarrow \mathbb{C}$ such that $|f(x)| = 1$ for all $x \in G$ and $f(x + y) = f(x)f(y)$ for almost every $x, y \in G$. Why must $f$ be almost everywhere equal to a continuous function $g: G \rightarrow \mathbb{C}$ satisfying $g(x + y) = g(x)g(y)$ for all $x, y \in G$?

I was thinking perhaps Lusin's Theorem might work: Let $A$ be a subset of $G$ of finite measure (such a set exists since $\mu$ is a nontrivial Haar measure). Then Lusin's Theorem gives: for every $\epsilon > 0$, there is a closed set $E$ with $\mu(A \backslash E) < \epsilon$ such that $f$ restricted to $E$ is continuous. By Steinhaus' theorem, since $G$ is locally compact, $E + E$ and $E - E$ have nonempty interior. Are there alternative approaches?

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Could you give any reference for this result? –  ougao Aug 11 '14 at 3:24