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 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}$to a continuous group homomorphism add more comment Sep 6 comment Extending a measurable map$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}$to a continuous group homomorphism Add comment Aug 20 awarded Disciplined Aug 12 comment a measurable function on a LCA group coincide with an mulitplicative character almost everywhere Hi, could you tell me where you have met this fact? I asked an essentially same question recently, and I want to understand this stuff better. Aug 12 asked Extending a measurable map$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$G/U$. Aug 11 comment Existence of proper I.C.C. subgroup Thanks for your quick answer, do you know any I.C.C. non amenable inner amenable groups satisfy this property? Aug 11 accepted Existence of proper I.C.C. subgroup Aug 11 comment Existence of proper I.C.C. subgroup I do not understand why "g is contained in two separate cyclic subgroups". I think the following argument also works. Clearly, we can pick some$h_i$with$h_i\not\in \langle g\rangle$then$\langle h_i, g\rangle=T$so$T$is abelian, a contradition. Aug 11 comment Compact operators on$L^2(G)$as a reduced cross product of$C_0(G)$and$G$. Why$\mathbb{C}1 \otimes C^*(\pi(c_0(G)), \{\lambda_g : g \in G\}) \cong \mathcal{K}(\ell^2(G))$? And a first feeling is that$\lambda_g\$ is not a compact operator. Aug 11 asked Existence of proper I.C.C. subgroup Aug 11 comment Almost everywhere measurable homomorphism Could you give any reference for this result?