Proposition 4

— Let $\mu$ be a positive measure on a locally compact space $T$, and let $\mathfrak{K}$ be a $\mu$-dense set of compact subsets of $T$. There exists a summable family $(\mu_\alpha)_{\alpha \in A}$ of positive measures on $T$ such that $\mu = \sum_{\alpha \in A} \mu_\alpha$, and such that the supports of the measures $\mu_\alpha$ belong to $\mathfrak{K}$ and form a locally countable family of pairwise disjoint compact sets.

If the measure $\mu$ is moderated, the index set $A$ may be taken to be countable.

Consider a locally countable family $(K_\alpha)_{\alpha \in A}$ of pairwise disjoint elements of $\mathfrak{K}$ such that the set $N = T - \bigcup_{\alpha \in A} K_\alpha$ is locally $\mu$-negligible

(Ch. IV, §5, No. 9, Prop. 14). For every function $f \in \mathscr{K}(T)$, set

the linear form $\mu_\alpha$ on $\mathscr{K}(T)$ is positive, therefore is a positive measure, with support contained in $K_\alpha$. Since every compact set contained in an element of $\mathfrak{K}$ belongs to $\mathfrak{K}$, $\operatorname{Supp}(\mu_\alpha) \in \mathfrak{K}$ for all $\alpha \in A$. It remains only to show that the family $(\mu_\alpha)$ is summable and that its sum is equal to $\mu$, in other words that $\sum_{\alpha \in A} \mu_\alpha(f) = \mu(f)$ for every function $f \in \mathscr{K}_+(T)$.

Now, let $S$ be the (compact) support of $f$, and let $A'$ be the countable set formed by the $\alpha \in A$ such that $S \cap K_\alpha \neq \varnothing$. Since the set $N \cap S$ is $\mu$-negligible,

This completes the proof of the general case. If $\mu$ is moderated, then the set $T$ is $\mu$-moderated and so $T$ is the union of a sequence $(L_n)$ of compact sets and a negligible set (§ 1, No. 2, Prop. 5); let $A'$ be the countable set of $\alpha \in A$ such that $K_\alpha$ intersects one of the $L_n$. Then $\mu_\alpha = 0$ for $\alpha \notin A'$, and the last sentence of the statement follows immediately.