Proposition 1

— Let T be a completely regular space. a) Let $\mu$ be a positive measure on T, and f a numerical function $\geq 0$ defined on T and lower semi-continuous (resp. upper semi-continuous ous, finite, with compact support). Then

(1) $\mu^\bullet(f) = \sup_{g \in I_f} \mu^\bullet(g)$ (resp. $\mu^\bullet(f) = \inf_{g \in S_f} \mu(g)$),

where $I_f$ (resp. $S_f$) denotes the set of bounded continuous functions $g$ such that $0 \leq g \leq f$ (resp. $g \geq f$).

b) Let $\theta$ be a complex measure on $T$, and $f$ a numerical function $\geq 0$ defined on $T$ and lower semi-continuous. Then

(2) $|\theta|^\bullet(f) = \sup_g |\theta(g)|$,

where $g$ runs over the set of bounded and $|\theta|$-integrable continuous complex functions such that $|g| \leq f$.

The first of the formulas (1) is obvious, because $I_f$ is an increasing directed set of continuous functions whose upper envelope is $f$, and one can apply Prop. 5 of §1, No. 6. The same proposition will imply the second formula, if we show that $S_f$ contains a $\mu$-integrable bounded continuous function. Thus, let $K$ be the support of $f$, and $M$ the supremum of $f$; since $K$ is compact, $M$ is finite (GT, IV, §6, No. 2, Th. 3). Let $U$ be an open set containing $K$ and such that $\mu^\bullet(U) < +\infty$; there exists (Lemma) a continuous function $g$ with values in $[0, M]$, equal to $M$ on $K$ and zero outside $U$; then $g \in S_f$ and $\mu^\bullet(g) \leq M \mu^\bullet(U) < +\infty$.

Let us pass to b). It clearly suffices to show that $|\theta|^\bullet(f) \leq \sup_g |\theta(g)|$.

Let $a$ and $b$ be two real numbers such that $a < b < |\theta|^\bullet(f)$. By (1), there exists a function $h \in \mathscr{C}_+^b(T)$ such that $h \leq f$ and $|\theta|^\bullet(h) > b$; denote by $M$ the supremum of $h$. By the definition of $|\theta|^\bullet$ (§1, No. 2, Def. 4), there exists a compact subset $K$ of $T$ such that $|\theta|_K^\bullet(h_K) > b$. There then exists a continuous complex function $j$ on $K$ such that $|j| \leq h_K$ and $|\theta_K(j)| > b$ (Ch. III, §1, No. 6). Let us choose an open set $U$ containing $K$ and such that $|\theta|^\bullet(U - K) \leq \frac{b - a}{M}$ (§1, No. 9, Props. 13 and 14); extend $j$ to a continuous complex function $k$ on $T$, zero outside $U$ (Lemma); for every $t \in T$, set

(3)

Clearly $|g| \leq h \leq f$, and $g = j$ on $K$, therefore $||\theta_K(j)| - |\theta(g)|| = ||\theta(j^0)| - |\theta(g)|| \leq |\theta|^\bullet(|j^0 - g|) \leq M \cdot |\theta|^\bullet(U - K) \leq b - a$, consequently $|\theta(g)| > a$. Let us show on the other hand that $g$ is a continuous function:

since $a$ is subject to the sole condition $a < |\theta|^*(f)$, this will imply that the second member of (2) is $\geqslant$ the first, whence the proposition. Now, let F (resp. $F'$) be the set of $t \in T$ such that $|k(t)| \leq h(t)$ (resp. $|k(t)| \geq h(t)$). These sets being closed, and their union being T, it will suffice to show that $g_F$ and $g_{F'}$ are continuous: now, this property is obvious for $g_F = k_F$, and it is so for $g_{F'}$ at the points where $k(t) \neq 0$; on the other hand, if $t \in F'$ is such that $k(t) = 0$, then also $h(t) = 0$, and the inequality $|g| \leq h$ implies that $g$ is continuous at the point $t$.