Corollary

*Let $T$ be a completely regular space, $X$ a topological space, and $f$ a complex-valued function defined on $T \times X$, continuous and bounded. For every bounded measure $\mu$ on $T$, let $F_\mu$ be the function on $X$ defined by $F_\mu(x) = \int_T f(t, x) d\mu(t)$ for all $x \in X$.

a) The function $F_\mu$ is continuous and bounded for every bounded measure $\mu$.

b) Suppose that $T$ is locally compact. The mapping $\mu \mapsto F_\mu$ of $\mathscr{M}_+^b(T)$ into $C^b(X; \mathbf{C})$ is then continuous, if $\mathscr{M}_+^b(T)$ is equipped with the tight topology, and $C^b(X; \mathbf{C})$ with the topology of compact convergence.

For every $x \in X$, denote by $f_x$ the continuous and bounded function $t \mapsto f(t, x)$ on $T$; the mapping $x \mapsto f_x$ of $X$ into $C^b(T; \mathbf{C})$ has bounded image, and it is continuous if $C^b(T; \mathbf{C})$ is equipped with the topology of compact convergence (GT, X, §3, No. 4, Th. 3). Since $F_\mu(x) = \mu(f_x)$, the function $F_\mu$ is continuous by Prop. 12. Suppose $T$ is locally compact; Prop. 13 shows that the mapping $(\mu, x) \mapsto F_\mu(x)$ of $\mathscr{M}_+^b(T) \times X$ into $\mathbf{C}$ is continuous; the assertion b) follows from this (loc. cit.).