Proposition 14

For a measure $\mu$ on a locally compact space $X$ to be a discrete measure (§ 1, No. 3, Example I), it is necessary and sufficient that $\operatorname{Supp}(\mu)$ be a discrete closed subspace of $X$.

Let $\mu$ be a discrete measure on $X$, defined by the masses $h(x) \neq 0$ placed at the points $x$ of a discrete closed subspace $N$ of $X$; let us show that $\operatorname{Supp}(\mu) = N$. For every $a \in N$ and every neighborhood $V$ of $a$, there exists a function $f \in \mathscr{K}(X; \mathbf{C})$ with support contained in $V$, equal to 1 at the point $a$ and to 0 at the other points of $N$, whence $\mu(f) = h(a) \neq 0$. On the other hand if $b \notin N$ then there exists a neighborhood $W$ of $b$ not intersecting $N$; for every function $g \in \mathscr{K}(X; \mathbf{C})$ with support contained in $W$, we therefore have $\mu(g) = 0$, which proves that $b \notin \operatorname{Supp}(\mu)$.

Conversely, let $\mu$ be a measure such that $\operatorname{Supp}(\mu)$ is a discrete closed subspace $N$ of $X$. For every $a \in N$, there exists an open neighborhood $V_a$ of $a$ that contains no point of $N$ distinct from $a$; the restriction of $\mu$ to $V_a$ is therefore a point measure with support $\{a\}$ (No. 2, Prop. 5), hence (No. 4, Prop. 12) is of the form $h(a)\varepsilon_a$, where $h(a) \neq 0$. Setting $h(x) = 0$ at the points of $\mathbf{C}\mathbf{N}$, and denoting by $\nu$ the measure defined by the masses $h(x)$, the principle of localization shows that $\nu = \mu$.

We thus see that on a discrete space $X$, every measure is discrete.