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seen Dec 16 at 19:17

Oct
16
awarded  Yearling
Sep
10
comment Extending a measurable map $f: G \to \mathbb{R}/\mathbb{Z}$ to a continuous group homomorphism
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.
Sep
10
comment Extending a measurable map $f: G \to \mathbb{R}/\mathbb{Z}$ to a continuous group homomorphism
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.
Sep
10
comment Extending a measurable map $f: G \to \mathbb{R}/\mathbb{Z}$ to a continuous group homomorphism
.. more details?I suspect we would run into trouble of some suttle issues that Kleppner faced.
Sep
10
comment Extending a measurable map $f: G \to \mathbb{R}/\mathbb{Z}$ to a continuous group homomorphism
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..
Sep
9
comment Extending a measurable map $f: G \to \mathbb{R}/\mathbb{Z}$ to a continuous group homomorphism
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$?
Sep
9
revised Extending a measurable map $f: G \to \mathbb{R}/\mathbb{Z}$ to a continuous group homomorphism
added 37 characters in body
Sep
8
answered Extending a measurable map $f: G \to \mathbb{R}/\mathbb{Z}$ to a continuous group homomorphism
Sep
6
comment Extending a measurable map $f: G \to \mathbb{R}/\mathbb{Z}$ to a continuous group homomorphism
both are true. Is it related to my above comment?
Sep
6
comment Extending a measurable map $f: G \to \mathbb{R}/\mathbb{Z}$ to a continuous group homomorphism
"...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.
Sep
6
revised Extending a measurable map $f: G \to \mathbb{R}/\mathbb{Z}$ to a continuous group homomorphism
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Sep
6
comment Extending a measurable map $f: G \to \mathbb{R}/\mathbb{Z}$ to a continuous group homomorphism
$Z\subset G\times G$ is not necessarily of product type.
Sep
4
revised Extending a measurable map $f: G \to \mathbb{R}/\mathbb{Z}$ to a continuous group homomorphism
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Aug
20
awarded  Disciplined
Aug
12
comment a measurable function on a LCA group coincide with an mulitplicative character almost everywhere
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.
Aug
12
asked Extending a measurable map $f: G \to \mathbb{R}/\mathbb{Z}$ to a continuous group homomorphism
Aug
11
comment Existence of proper I.C.C. subgroup
my fault, I should mention that $G$ is inner amenable group if there exists a finite additive probability measure on the power set of $G$ which is invariant under conjugate action of $G$.(Recall for amenable group, we require that measure is invariant under left action of $G$).
Aug
11
comment induction exercise doubt
Hint: When do induction, you can not change the assumption, in your case, the assumption, $x_1x_2... x_{n+1}=1$ is the assumption, but you could translate your assumption into different form. Try set $x_i'=x_i for 1\leq i\leq n-1, x_n'=x_nx_{n+1}$.
Aug
11
comment What is the closed form for this sequence, powers of $4$?
try oeis.org, but it seems does not work...
Aug
11
comment Proof of Cauchy's Lemma in the case that G is abelian
Do reduction on order of $G$. As you did, take $U$, when $r>0$, focus on $U$, when $r=0$, consider $G/U$.