Corollary 2

— Every bounded measure $\mu$ on $X$ is in the vague closure of the set of measures whose support is finite and contained in that of $\mu$ and whose norm is equal to $\| \mu \|$.

We can suppose that $\mu \neq 0$. Let $V$ be an open neighborhood of 0 for the vague topology; for every $\varepsilon$ such that $0 < \varepsilon < 1$, there exists, by virtue of Cor. 1, a measure $\nu_0$ whose support is finite and contained in $\mathrm{Supp}(\mu)$ and for which $\nu_0 - \mu \in V$ and $\| \mu \| \geq \| \nu_0 \| \geq (1 - \varepsilon) \| \mu \|$. Setting $\nu = (\| \mu \| / \| \nu_0 \|) \nu_0$, we have $\| \nu \| = \| \mu \|$ and $\| \nu - \nu_0 \| \leq \| \mu \|$; for $\varepsilon$ sufficiently small we therefore have $\nu - \mu \in V + V$, whence the conclusion.