Lemma 1

Assume that T is compact, and set l = I(T). For every $A \subset T$, set

and let $\Phi$ be the set of $A \subset T$ such that $J(A) + J(\mathbf{C}A) = l$. The set $\Phi$ is then a clan that contains $\mathscr{K}(T)$, and the function $J$ on $\Phi$ is increasing and additive.

It is clear that $J$ is an increasing set function, extending $I$, and that $J(A) + J(\mathbf{C}A) \leq l$ for every $A \subset T$.

Let $K$ and $S$ be two compact sets in $T$; we are first going to show that

On considering the restrictions of $I$ to $\mathscr{K}(S)$ and of $J$ to $\mathfrak{P}(S)$, one reduces immediately to the case that $S = T$. Since $T$ is normal, $K$ is the intersection of the decreasing directed family of its compact neighborhoods, and condition 4) implies the existence, for every $\varepsilon > 0$, of a compact neighborhood $H$ of $K$ such that $I(H) \leq I(K) + \varepsilon$. Let $L$ be the closure of $T - H$; $L$ is compact, $L \cap K = \varnothing$ and $H \cup L = T$, therefore $l = I(H \cup L) \leq I(H) + I(L) \leq I(K) + I(L) + \varepsilon$ (condition 2)), whence the relation $J(K) + J(\mathbf{C}K) \geq I(K) + I(L) \geq l - \varepsilon$. Since $\varepsilon$ is arbitrary, $J(K) + J(\mathbf{C}K) = l$. This proves the formula (2), as well as the inclusion $\mathscr{K}(T) \subset \Phi$.

Let us now prove that $\Phi$ is a clan. Since $\Phi$ is obviously stable under passage to the complement, it suffices to show that if $A_1$ and $A_2$ denote elements of $\Phi$, then $A_1 \cup A_2 \in \Phi$, or again that

Denote by $\varepsilon$ a number $> 0$, and, for $i = 1, 2$, let $K_i$ be a compact set contained in $A_i$, and $L_i$ a compact set contained in $\mathbf{C}A_i$, such that

Set $M_1 = K_1 \cup L_1$; the relations $l = J(M_1) + J(\mathbf{C}M_1)$ and

imply that $J(\mathbf{C}M_1) \leq 2\varepsilon$. Then if $S$ is a compact subset of $T$, the relation (2) (applied to $K = M_1$) implies $J(S) \leq J(M_1 \cap S) + 2\varepsilon$, whence

Let us add the inequalities obtained by making $S = K_2$ and $S = L_2$, and take into account the inequality $J(K_2) + J(L_2) \geq l - 2\varepsilon$ and the fact that $K_1 \cap K_2$, $L_1 \cap K_2$ and $K_1 \cap L_2$ are three disjoint compact sets contained in $A_1 \cup A_2$. Denoting by $C$ the union of these three compact sets, it follows that

whence immediately the desired formula (3), in view of the arbitrariness of $\varepsilon$. This having been established, the preceding inequalities imply that $J(C) \geq J(A_1 \cup A_2) - 6\varepsilon$; if $A_1$ and $A_2$ are disjoint, then $C$ is the union of $K_1 \cap L_2 \subset A_1$ and $K_2 \cap L_1 \subset A_2$, from which one deduces that $J(A_1 \cup A_2) \leq J(A_1) + J(A_2)$. The reverse inequality being obvious, $J$ is indeed additive on $\Phi$, and the lemma is established.

Let us complete the proof of the theorem for the case that $T$ is compact. Let $\mathscr{E}(\Phi)$ be the vector space of $\Phi$-step functions on $T$, equipped with the topology of uniform convergence (Ch. IV, §4, No. 9, Def. 4); we shall again denote by $J$ the positive linear form on $\mathscr{E}(\Phi)$ associated with the additive function $J$ (loc. cit., Prop. 18). Since $J(T) = l$, $J$ is continuous and has norm $l$. Let then $\mathscr{H}$ be the closure of $\mathscr{E}(\Phi)$ for the topology of uniform convergence; one verifies at once that $J$ may be extended by continuity to a positive linear form on $\mathscr{H}$, again denoted $J$. Since $\mathscr{H}$ contains $\mathscr{C}(T)$ (loc. cit., No. 10, Prop. 19) the restriction of $J$ to $\mathscr{C}(T)$ is a positive measure $\mu$. It remains to show that $\mu^\bullet(K) = I(K)$ for every compact subset $K$ of $T$. Now, we have $\mu^\bullet(K) = \inf_{f \in S_K} \mu^\bullet(f)$, where $S_K$ denotes the set of elements of $\mathscr{C}(T)$ that are $\geq \varphi_K$ (§ 1, No. 6, Prop. 5). Since $J(f) = \mu^\bullet(f)$ for $f \in \mathscr{C}(T)$, it clearly suffices to show that $J(K) \geq \inf_{f \in S_K} J(f)$. As in the proof of Lemma 1, let $H$ be a compact neighborhood of $K$ such that $J(H) \leq J(K) + \varepsilon$, and let $f$ be a continuous function on $T$, between 0 and 1, equal to 1 on $K$ and to 0 outside $H$ (GT, IX, §4, No. 1, Prop. 1). Then

$\varepsilon$ being arbitrary, the desired inequality is proved, and the theorem is thus established when $T$ is compact.

Let us now pass to the general case. For every compact set $L$ in $T$, let $I_L$ be the restriction of $I$ to $\mathfrak{K}(L)$. By the special case just treated, there exists a measure $\mu_L$ on $L$, unique, such that $\mu_L(K) = I_L(K)$ for every compact set $K \subset L$. Let then $L'$ be a compact set contained in $L$; we have $(\mu_L)_{L'}^\bullet(K) = \mu_{L'}^\bullet(K) = \mu_{L'}^\bullet(K)$ for every compact set $K \subset L'$, therefore $\mu_{L'} = (\mu_L)_{L'}$; the mapping $\mu : L \mapsto \mu_L$ is a premeasure. The condition 5) expresses that $\mu$ is a measure, and the relation $I(K) = \mu^*(K)$ for all compact $K \subset T$ is obvious.