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I'm currently reading an article and the author defines the following objects. Let $\mathbb{Z}_{n}$ be the cyclic group of integers mod $n$, for some $n \ge 1$ and define $$\mathcal{G} := \bigoplus_{k=0}^{\infty}\mathbb{Z}_{n}$$ Furthermore, define the subgroups: $$\mathcal{G}_{k} :=\{x \in \mathcal{G}, \hspace{0.1cm} \mbox{$x_{i}=0$, for all $i \ge k$}\}$$ if $k\ge 1$ and $\mathcal{G}_{0}:=\{0\}$. Now, if $L\ge 2$, we introduce a notion of norm: $$|x| :=\begin{cases} \displaystyle 0 \quad \mbox{if $x=0$}\\ \displaystyle L^{p} \quad \mbox{where $p=\inf \{k,\hspace{0.1cm} x\in \mathcal{G}_{k}\}$ if $x \neq 0$} \end{cases} $$ Now, the author defines, in his words "$dx$ to be a Haar measure which is also the counting measure on $\mathcal{G}$". I'm not familiar with the concept of a Haar measure on a group, but I've read that it should be defined on some locally compact Hausdorff topological group. Now, how do I conclude that $\mathcal{G}$ is such a group? Does $|\cdot|$ define a topology on $\mathcal{G}$? How can I define a topology on $\mathcal{G}$ with the above informations? I imagine $|\cdot|$ defines a topology in some way, but I don't know how. Besides, does this topology fulfill all properties (Hausdorff, locally compact) to define the Haar measure?

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    $\begingroup$ The Haar measure is the unique (up to a multiple) measure invariant under left translations. On $\mathbb{Z}_k$ the counting measure is clearly invariant under addition. So you can take the “direct sum” of those measures for $\mathcal{G}$. $\endgroup$
    – meiji163
    Nov 25, 2019 at 20:03
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    $\begingroup$ The topology on your group is just the discrete topology (and the norm is also discrete). $\endgroup$
    – YCor
    Nov 25, 2019 at 21:38
  • $\begingroup$ Means I have to take the set of all subsets of $\mathcal{G}$ as my $\sigma$-algebra? $\endgroup$
    – IamWill
    Nov 25, 2019 at 21:40

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