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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?

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!

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ougao
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2 Answers2

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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.

Dimitris
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  • $Z\subset G\times G$ is not necessarily of product type. – ougao Sep 06 '14 at 01:49
  • The measure of $G\times G \setminus Z$ is zero if and only if the measure of the slices of $G\times G\setminus Z$ is zero, by Fubini. – Dimitris Sep 06 '14 at 04:25
  • "...iff the measure of the slices of $G\times G\Z$ is zero a.e., by Fubini". I still do not see how to use the assumption that $f(x+y)=f(x)+f(y), \forall (x,y)\in Z" to get $f(x+y)+f(-y)=f(x)$ for a.e. x, a.e. y in G. – ougao Sep 06 '14 at 12:33
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    Isn't $\mu$ a probability measure? i.e. $\mu(G)=1$? – Dimitris Sep 06 '14 at 16:04
  • both are true. Is it related to my above comment? – ougao Sep 06 '14 at 19:39
  • $$\int_ {G} \int_G |f(x+y)+f(-y)-f(x)| d\mu(x) d\mu(y) = \int_G \int_G |f(x)+f(-y) -f(x-y)| d\mu(x)d\mu(y) = \int_{G\times G} |f(x)+f(-y) -f(x-y)| d (\mu\times \mu)= 0$$ by assumption that $f(x+y)=f(x)+f(y)$ for all $(x,y)\in Z$, i.e. for a.e. $(x,y)\in G\times G$, since $\mu(G\times G\setminus Z)=0$. Thus $f(x+y)+f(-y)-f(x)=0$ for a.e. $(x,y)\in G\times G$. – Dimitris Sep 09 '14 at 20:05
  • Yes, you can show that $f(x+y)+f(-y)-f(x)=0$ for a.e. $(x,y)\in G\times G$. But the point is that you claim it holds for a.e. $x\in G$ and a.e. $y\in G$, i.e., you claim that you can find $V, W$, both are conull subset in $G$ such that the above equality holds for $x\in V, y\in W$. Could you tell me why? How do you argue that from $Z$ you can get $ V, W$? – ougao Sep 09 '14 at 23:46
  • That's again using Fubini's theorem. $$\int_G \int_G |f(x)+f(-y) -f(x-y)| d\mu(x)d\mu(y) =0$$ implies that $\int_G |f(x)+f(-y) -f(x-y)| d\mu(x)=0$ for a.e. $y\in G$. For such a $y$ you have $f(x)+f(-y) -f(x-y)=0$ for a.e. $x\in G$. So in fact, when we say that for a.e. $x,y\in G$ the relation is true, we mean that for a.e. $y$ there exists $V_y\subset G$ of measure $1$ such that for all $x\in V_y$ we have the relation. There is no ambiguity, since all those statements are equivalent to saying for a.e. $(x,y)\in G\times G$ etc. – Dimitris Sep 10 '14 at 00:04
  • Here are a couple of things I want to say. 1, when I see $a.e., x,y\in G$ I would understand it as for some conull subsets $V,W\subset G$, $x\in V, y\in W$ instead of what you mentioned above. 2, suppose some statement P holds for $(x,y)\in Z$ where $Z\subset G\times G$ is conull set. In general, you CAN NOT find conull $V,W\subset G$ such that P holds for $x\in V,y\in W$, since in general $Z$ is complicated, say $Z=G\times G-diagonal$. 3, In your answer $F(x)=f(x) a.e., x\in G$ is true, but I do not see how to show $F$ is continuous, a group homomorphism. If you have time, could you provide.. – ougao Sep 10 '14 at 02:44
  • .. more details?I suspect we would run into trouble of some suttle issues that Kleppner faced. – ougao Sep 10 '14 at 02:47
  • Suppose $f$ is a group homomorphism, then by definition of your $F$, $F(x)=f(x)\forall x\in X$, but we only assume $f$ is measurable, so you can not say that $F$ is continuous. – ougao Sep 10 '14 at 02:52
  • The definition implies that $F(x)=f(x)$ for a.e. $x\in G$, not for all. To see this, take an integral $\int _G |F(x)-f(x)| d\mu$ and show that it is $0$. Continuity of $F$ can be shown by a density argument ( continuous functions are dense in $L^1$ functions etc. ) That's a standard way of proving continuity and you can find this kinds of proofs in all graduate analysis books, e.g. Folland's. – Dimitris Sep 10 '14 at 03:52
  • Assume $f$ is a genuine group homomorphism, I.e.,$f(x+y)=f(x)+f(y)$ holds for all $(x,y)\in G\times G$, so in your definition of $F$, $f(x+y)+f(-y)=f(x)$, then $F(x)=f(x)$ for all $x\in G$, so how do you know $F$, which is now $f$ is continuous? To be honest, whenever I left some comments, you kept ignoring what I am talking about, I do not know why. – ougao Sep 10 '14 at 12:23
  • @ougao A measurable homomorphism is indeed continuous, see https://www.jstor.org/stable/2044356, or just modify the proof using Steinhaus theorem for topological groups – YuiTo Cheng Apr 17 '23 at 16:37
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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.

ougao
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