Proposition 6

— Let X be a locally compact space, ($K_\alpha$) a family of compact subsets of X whose interiors $\overset{\circ}{K}_\alpha$ form an open covering of X. For a linear form $\mu$ on $\mathscr{K}(X; \mathbf{C})$ to be a measure on X, it is necessary and sufficient that, for every $\alpha$, there exist a number $M_\alpha$ such that

for every function $f \in \mathscr{K}(X, K_\alpha; \mathbf{C})$.

The condition being obviously necessary, it suffices to prove that (5) implies (4) for every compact subset $K$ of $X$. Now, $K$ is covered by a finite number of open sets $\overset{\circ}{K}_{\alpha_i}$ ($1 \leq i \leq n$); applying Lemma 1 of No. 2 to $K$ and to the $\overset{\circ}{K}_{\alpha_i}$, there exist continuous functions $g_i \geq 0$ on $X$ such that $\operatorname{Supp}(g_i) \subset K_{\alpha_i}$, $0 \leq \sum_{i=1}^n g_i(x) \leq 1$ for all $x \in X$ and $\sum_{i=1}^n g_i(x) = 1$ for $x \in K$. For every function $f \in \mathscr{K}(X, K; \mathbf{C})$, we can therefore write $f = \sum_{i=1}^n f g_i$ and we have $f g_i \in \mathscr{K}(X, K_{\alpha_i}; \mathbf{C})$ and $\| f g_i \| \leq \| f \|$; if $M_K = \sum_{i=1}^n M_{\alpha_i}$, we then have the relation (4).

We denote by $\mathscr{M}(X; \mathbf{C})$, or simply $\mathscr{M}(X)$ if no confusion can result, the vector space of measures on $X$, in other words, the dual of $\mathscr{K}(X; \mathbf{C})$.

One knows that for every set $\mathscr{S}$ of bounded subsets of $\mathscr{K}(X;\mathbf{C})$, there is defined on $\mathscr{M}(X;\mathbf{C})$ the $\mathscr{S}$-topology, which is locally convex (TVS, III, §3, No. 1, Cor. of Prop. 1). We denote the topological vector space, obtained by equipping $\mathscr{M}(X;\mathbf{C})$ with the $\mathscr{S}$-topology, by $\mathscr{M}_{\mathscr{S}}(X;\mathbf{C})$ or $\mathscr{M}_{\mathscr{S}}(X)$.