Proposition 2

— Let $G$ be a locally compact group, $\mu$ a left or right Haar measure on $G$. For $G$ to be discrete, it is necessary and sufficient that $\mu(\{e\}) > 0$. For $G$ to be compact, it is necessary and sufficient that $\mu^*(G) < +\infty$.

The conditions are obviously necessary. Let us show their sufficiency. Let $V$ be a compact neighborhood of $e$. If $\mu(\{e\}) > 0$, then $V$ is a finite set since $\mu(V) < +\infty$; since $G$ is Hausdorff, it is therefore discrete. Suppose $\mu^*(G) < +\infty$ and $\mu$ is, for example, left-invariant. Consider the set $\mathscr{E}$ of finite subsets $\{s_1, \ldots, s_n\}$ of $G$ such that $s_i V \cap s_j V = \varnothing$ for $i \neq j$; then

therefore $n \leq \mu^*(G)/\mu(V)$. We may therefore choose in $\mathscr{E}$ a maximal element $\{s_1, \ldots, s_n\}$. Then, for every $s \in G$, there exists an $i$ such that $s V \cap s_i V \neq \varnothing$, hence such that $s \in s_i V V^{-1}$. Thus $G$ is the union of the compact sets $s_i V V^{-1}$, hence is compact.