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Let $G$ be an LCA group. We say $\tilde \chi$ is a multiplicative character if it is a continuous function : $\tilde \chi : G \to S^1 $ where $ S^1 : = \{ z \in \mathbb C : |z| =1 \}$ is the unit circle, which is also a homomorphism, i.e., $ \tilde\chi (x +y ) = \tilde\chi (x) \tilde\chi (y) $ for all $x,y \in G$.

Now let $\mu$ be a non-trivial Haar measure on $G$, and let $ \chi : G \to S^1 $ be merely a measurable function such that $ \chi ( x +y ) = \chi (x ) \chi (y)$ for almost every $x,y \in G$. Show that $ \chi $ is equal almost everywhere to a multiplicative character $\tilde \chi$ of $G$.

The hint says the function $ \tau_x \chi (y) : = \chi ( y -x) = \chi (-x) \chi (y)$ for almost every $y$ and almost every $x$. And $ \tau_x \chi $depends continuously on $x$ in the local $L^1$ topology. I don't have any idea about this one and can't even get start with the hint. Any help is appreciated.

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Hi, could you tell me where you have met this fact? I asked an essentially same question recently, and I want to understand this stuff better. – ougao Aug 12 '14 at 18:52

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