Theorem 2

— Let T be a topological space, and let I be a function defined on $\mathscr{B}(T)$, with values in $\overline{\mathbf{R}}_+$. In order that there exist a measure $\mu$ on T such that $\mu^*(A) = I(A)$ for every $A \in \mathscr{B}(T)$, it is necessary and sufficient that I be countably additive, locally bounded and inner regular. The measure $\mu$ is then unique.

These three conditions are necessary: for, the mapping $A \mapsto \mu^*(A)$ on $\mathscr{B}(T)$ is countably additive (§ 1, No. 5, Cor. of Prop. 4), locally bounded by the definition of measures (§ 1, No. 2, Def. 5), and inner regular by Remark 3 of § 1, No. 2.

We pass to existence. It is clear that the restriction of I to $\mathscr{K}(T)$ satisfies conditions 1), 2), 3) and 5) of the statement of Th. 1; let us show that 4) is satisfied as well. Let K be a compact subset of T, the intersection of a decreasing directed family $(K_\alpha)_{\alpha \in A}$ of compact sets, and let $\varepsilon$ be a number $> 0$; I being locally bounded, there exists an open (hence Borel) neighborhood V of K such that $I(V) < +\infty$, and then there exists an index $\alpha$ such that $K_\alpha \subset V$; changing notation if necessary, we can suppose that $K_\alpha \subset V$ for all $\alpha \in A$. By the inner regularity of I, there exists a compact set $L \subset V - K$ such that $I(L) \geq I(V - K) - \varepsilon$; since L does not intersect K, there exists an index $\alpha$ such that $L \cap K_\alpha = \varnothing$, and one then has $I(V - K_\alpha) \geq I(L) \geq I(V - K) - \varepsilon$. Since $K_\alpha \subset V$, it follows that $I(K_\alpha) \leq I(K) + \varepsilon$ and the condition 4) is verified.

By Th. 1, there exists a measure $\mu$ such that $\mu^\bullet(K) = I(K)$ for all $K \in \mathscr{K}(T)$. The inner regularity of the set functions $\mu^\bullet$ and $I$ on $\mathscr{B}(T)$ then implies that $\mu^\bullet(A) = I(A)$ for all $A \in \mathscr{B}(T)$, and existence is proved. The uniqueness of $\mu$ follows from the uniqueness assertion of Th. 1.