Definition 3

A measure on a locally compact space X is said to be bounded if it is continuous on $\mathscr{K}(X; \mathbf{C})$ for the topology of uniform convergence.

It comes to the same to say that there exists a finite number $M \geq 0$ such that, for every function $f \in \mathscr{K}(X; \mathbf{C})$,

(where $\|f\|$ is defined by formula (3) of No. 2).

To say that $\mu$ is a bounded measure thus signifies that $\mu$ belongs to the dual of the space $\mathscr{K}(X; \mathbf{C})$ normed by $\|f\|$; we shall denote this dual by $\mathscr{M}^1(X; \mathbf{C})$ (or simply $\mathscr{M}^1(X)$ when no confusion can result). We know that $\mathscr{M}^1(X; \mathbf{C})$ is equipped with a norm, $\|\mu\|$ being the smallest of the numbers $M \geq 0$ for which the inequality (19) holds for every function $f \in \mathscr{K}(X; \mathbf{C})$, or again,

Equipped with this norm, $\mathscr{M}^1(X; \mathbf{C})$ is known to be a Banach space (TVS, III, §3, No. 8, Cor. 2 of Prop. 12).

The definition of $\|\mu\|$ by the formula (20) may be extended to every measure $\mu$ on X and, by an abuse of language, $\|\mu\|$ is again said to be the norm of $\mu$; for $\mu$ to be bounded, it is necessary and sufficient that $\|\mu\|$ be finite.

If X is compact, then every measure on X is bounded.