Proposition 13

Let $\mu$ be a measure on a locally compact space $X$. For a point $x_0$ to belong to $\mathrm{Supp}(\mu)$, it is necessary and sufficient that the point measure $\varepsilon_{x_0}$ be in the vague closure of the set of measures $g \cdot \mu$, where $g$ runs over the set of continuous functions with compact support such that $\| g \cdot \mu \| \leq 1$.

The condition is obviously sufficient by virtue of Prop. 6 of No. 2. To see that it is necessary, suppose $x_0 \in \mathrm{Supp}(\mu)$; consider a finite number of functions $f_k$ ($1 \leq k \leq n$) in $\mathscr{K}(X; \mathbf{C})$, and an arbitrary number $\delta > 0$; we are to prove that there exists a function $g \in \mathscr{K}(X; \mathbf{C})$ such that $\| g \cdot \mu \| \leq 1$ and such that

for $1 \leq k \leq n$. Let $U$ be a relatively compact open neighborhood of $x_0$ such that the oscillation of each of the $f_k$ ($1 \leq k \leq n$) on $U$ is $\leq \delta / 2$. By hypothesis, since $x_0 \in \mathrm{Supp}(\mu)$, there exists a function $g_0 \in \mathscr{K}(X; \mathbf{C})$ whose support is contained in $U$ and such that $\mu(g_0) \neq 0$; the measure $\nu = g_0 \cdot \mu$ is not zero, since for every function $f \in \mathscr{K}(X; \mathbf{C})$ equal to 1 on $U$, $\nu(f) = \mu(g_0) \neq 0$. Moreover, $\nu$ is bounded (No. 3, Prop. 11);

multiplying $g_0$ by a scalar, we can suppose that $\| \nu \| = 1$. This being so, setting $\alpha_k = f_k(x_0)$ we can write, for $1 \leq k \leq n$ and for every function $h \in \mathscr{K}(X; \mathbf{C})$,

Since $\nu$ has its support in $U$, we may identify it with its restriction to $U$; the hypothesis $\| \nu \| = 1$ then implies that there exists a function $h \in \mathscr{K}(X; \mathbf{C})$, with support contained in $U$, such that $\| h \| \leq 1$ and such that $| \alpha_k (1 - \nu(h)) | \leq \delta / 2$ for $1 \leq k \leq n$. The definition of $U$ moreover shows that $| (\alpha_k - f_k(x)) h(x) | \leq \delta / 2$ for all $x \in U$; since $\| \nu \| = 1$ and $\operatorname{Supp}(\nu) \subset U$ we therefore have $| \nu((\alpha_k - f_k)h) | \leq \delta / 2$ and so, setting $g = g_0 h$,

This proves the proposition, since $\| g \cdot \mu \| = \| (g_0 h) \cdot \mu \| \leq \| g_0 \cdot \mu \| = 1$.