Proposition 8

— Let $\mu$ be a measure on a locally compact space $X$. For every function $f \in \mathscr{K}(X; \mathbf{C})$ that is zero on $\operatorname{Supp}(\mu)$, $\mu(f) = 0$.

Set $K = \operatorname{Supp}(f)$, $S = \operatorname{Supp}(\mu)$. Given a number $\varepsilon > 0$, let $V$ be the set of $x \in X$ such that $|f(x)| < \varepsilon$; $V$ is an open set containing $S$ by hypothesis, therefore $CS$ is a neighborhood of the compact set $CV$. It follows that there exists a continuous mapping $h$ of $X$ into $[0, 1]$, equal to 1 on $CV$ and with support contained in $CS$ (§ 1, No. 2, Lemma 1). Since the support of $fh$ does not intersect $S$, $\mu(fh) = 0$. On the other hand, $f = fh$ on $K \cap CV$, and $|fh| \leq |f|$ on $X$, therefore $|f - fh| \leq 2\varepsilon$ on $X$, by the choice of $V$. Observe finally that there exists a number $M_K$ such that $|\mu(g)| \leq M_K \|g\|$ for every function $g \in \mathscr{K}(X; \mathbf{C})$ whose support is contained in $K$; since the support of $f - fh$ is contained in $K$, $|\mu(f - fh)| \leq 2M_K \varepsilon$ and consequently $|\mu(f)| = |\mu(f - fh)| \leq 2M_K \varepsilon$; since $\varepsilon$ is arbitrary, $\mu(f) = 0$.