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I have no idea about this problem (even with hint) in showing that a multiplicative Haar measurable function is equal to a multiplicative character almost everywhere. It is Exercise 11 in Tao's blog. There is already a post about this problem but no one answered it, so I ask it again here.

Exercise 11. Let $G$ be an LCA group with non-trivial Haar measure $\mu$, and let $\chi: G \rightarrow S^{1}$ be a measurable function such that $\chi(x) \chi(y)=\chi(x+y)$ for almost every $x, y \in G$. Show that $\chi$ is equal almost everywhere to a multiplicative character $\tilde{\chi}$ of $G$. (Hint: on the one hand, $\tau_{x} \chi=\chi(-x) \chi$ a.e. for almost every $x$. On the other hand, $\tau_{x} \chi$ depends continuously on $x$ in, say, the local $L^{1}$ topology.)

Some notations: $\tau_x$ means the translation of a function by a vector $x$, i.e., $\tau_xf = f(\cdot - x)$. A multiplicative character $\tilde{\chi}$ is a continuous group homomorphism from an (additive) group $G$ to the multiplicative group $S^1$ (satisfying $\tilde{\chi}(x+y) = \tilde{\chi}(x)\tilde{\chi}(y)$).

Tao has also given some comments: "The hint asks for you to exploit the continuity of the map $x \mapsto \tau_x \chi$ from $G$ to $L^1_{loc}(G)$, which can be done by approximating $\tau_x$ locally by a continuous function in $L^1$." But I still don't know how to construct the needed multiplicative character $\tilde{\chi}$.

Any help is appreciated!

ScienceAge
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  • Seems like you already know $\chi$ is "almost" a multiplicative character, so the idea is to show that you can fill in the parts where $\chi$ is not a multiplicative character in such a way that the homomorphism property holds. The hint is trying to guide you towards using continuity to fill in the gaps. – User203940 Jun 21 '22 at 22:04
  • @User203940 I can feel the purpose of the hint but indeed I cannot figure out how I can argue to fill in the gaps using continuity as you said. – ScienceAge Jun 21 '22 at 22:12
  • Here's my vague idea (note: these details are probably not right, but I think you can maybe fix them). I think for every $x \in G$, you can find a continuous function $f_x$ which agrees almost everywhere with $\tau_x \chi$ and this varies continuously with $x$. Notice that $\tau_x f_0 = f_x$ almost everywhere, and since they are continuous they are equal. Furthermore, $\tau_x f_0(y) = \chi(-x) \chi(y) = f_0(-x) f_0(y)$ almost everywhere, so by continuity they are equal, hence $f_0$ is a character. – User203940 Jun 22 '22 at 16:00

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