Corollary 3

— Let $\mu$ be a bounded measure on $X$; if $f \in \mathscr{K}(X; \mathbf{C})$ is such that $|f(x)| \leq a$ on $\operatorname{Supp}(\mu)$, then $|\mu(f)| \leq a \|\mu\|$.

For, $\operatorname{Supp}(|\mu|) = \operatorname{Supp}(\mu)$, and if $h$ is a continuous mapping of $X$ into $[0, 1]$, equal to 1 on $\operatorname{Supp}(f)$ and with compact support, then $|f(x)| \leq ah(x)$ on $\operatorname{Supp}(\mu)$, therefore

by Corollary 2; the conclusion then follows from formula (13) of § 1, No. 6.