Corollary 1

— Every bounded measure $\mu$ on $X$ is in the vague closure of the convex set $A$ of measures whose support is finite and is contained in that of $\mu$, and whose norm is $\leq \| \mu \|$. Moreover, if $\nu$ tends vaguely to $\mu$ while remaining in $A$, then $\| \nu \|$ tends to $\| \mu \|$.

To prove the first assertion, it suffices to establish that the measure $\mu$ belongs to the polar set of the polar set $A^\circ$ of $A$ in $\mathscr{K}(X; \mathbf{C})$ (TVS, II, §6, No. 3, Th. 1 and §8, No. 4); this means that for $f \in \mathscr{K}(X; \mathbf{C})$, the relations $|\langle f, \varepsilon_a \rangle| \leq 1/\| \mu \|$ for all $a \in \mathrm{Supp}(\mu)$ imply that $|\langle f, \mu \rangle| \leq 1$; but this is a consequence of Cor. 3 of Prop. 8 of No. 3.

To prove the second assertion, we note that

since the function $\nu \mapsto \| \nu \|$ is lower semi-continuous for the vague topology (§1, No. 9, Cor. 4 of Prop. 15), and the conclusion follows from the fact that $\| \nu \| \leq \| \mu \|$ for $\nu \in A$ by definition.