Extending a measurable map $f: G \to \mathbb{R}/\mathbb{Z}$ to a continuous group homomorphism

Question

Let $G$ be a compact abelian metrizable group (where the group operation is written as $+$) and $\mu$ is the Haar measure on $G$.

Suppose we have a measurable function $f: G \rightarrow \mathbb{T}\cong \mathbb{R}/\mathbb{Z}$ such that $f(x + y) = f(x)+f(y)$ for all $(x, y)\in Z$, where $Z\subset G\times G$ with $(\mu\times\mu)(Z)=1$.

Question:

Can we find a continuous, group homomorphism $\phi: G \rightarrow \mathbb{T}\cong \mathbb{R}/\mathbb{Z}$, i.e., $\phi$ lies in the dual group of $G$, such that $\phi(x)=f(x)$ almost everywhere?

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Remarks:

1, This is essentially a question asked here by someone else with some change. Since no answer appeared, I think it is OK to ask it again here. Note that $Z$ is not necessarily of product type.

2, It seems to be able to extend $f$ to a continuous, almost everywhere group homomorphism, but I do not see how to get a group homomorphism.(this claim seems Not true)

3, Any help, suggestions, references are appreciated, thanks in advance!

Answer 1

With someone else's help, I learnt that this is essentially proved in the following paper

A.Kleppner, Measurable homomorphisms of locally compact groups, Proc. Amer. Math. Soc. 106(1989), no. 2, 391-395.

But I have not checked the proof.

Answer 2

In order to produce a continuous homomorphism, a good method is to take somehow the average of f. For example, you can define $$F(x)= \frac{1}{\mu(G)} \int _G(f(x+y)+f(-y))d\mu(y)$$ By assumption, for a.e. $x,y\in G$ we have $f(x+y)+f(-y)=f(x)$. Try to show that $F(x)=f(x)$ for a.e. $x\in G$ and $F$ is continuous.