For every $x \in X$ such that $f(x) > 0$, and for every real number $a$ such that $0 < a < f(x)$, there exists by hypothesis a compact neighborhood V of $x$ such that $f(y) \geq a$ on V; on the other hand, there exists a function $g \in \mathscr{K}_+$, with support contained in V, equal to $a$ at the point $x$ and $\leq a$

(1) 'I' as in 'inferior'; the letter $\mathscr{L}$ (as in 'lower') is preempted for other function spaces, to be discussed in §3.

on V (GT, IX, §1, No. 5, Th. 2); therefore $0 \leq g \leq f$ and $g(x) \geq a$, which proves the lemma.

Let $f = \sup_{g \in H} g$. We shall first prove the theorem for the special case that the functions $g \in H$ and their upper envelope $f$ belong to $\mathscr{H}_+$. It then follows from Dini’s theorem (GT, X, §4, No. 1, Th. 1) that the section filter of $H$ converges uniformly to $f$ on every compact subset of $X$, and in particular on the support K of $f$. Since $0 \leq g \leq f$ for every function $g \in \mathbf{H}$, the support of every function in $\mathbf{H}$ is contained in $K$; but by definition $\mu$ is continuous on the vector space $\mathscr{K}(X, K; \mathbf{C})$ of continuous functions with support contained in $K$, for the topology of uniform convergence; whence the relation (2) in this case.

Let us pass to the general case. It is clear that $\mu^*(g) \leq \mu^*(f)$ for every function $g \in \mathbf{H}$. By Def. 1, it all comes down to showing that, for every function $\psi \in \mathscr{K}_+$ such that $\psi \leq f$,

For every function $g \in \mathbf{H}$, let $\Phi_g$ be the set of functions $\varphi \in \mathscr{K}_+$ such that $\varphi \leq g$, and let $\Phi$ be the union of the sets $\Phi_g$ as $g$ runs over $\mathbf{H}$; since $\mathbf{H}$ is directed, so is $\Phi$, and $f = \sup_{\varphi \in \Phi} \varphi$. Since $\psi \leq f$, $\psi$ is the upper envelope of the set of functions $\inf(\psi, \varphi)$ as $\varphi$ runs over $\Phi$; but since $\psi$ and the functions $\inf(\psi, \varphi)$ belong to $\mathscr{K}_+$, the first part of the proof shows that $\mu(\psi) = \sup_{\varphi \in \Phi} \mu(\inf(\psi, \varphi))$. Now, each $\varphi \in \Phi$ belongs to a set $\Phi_g$, therefore

from which it follows at once that $\mu(\psi) \leq \sup_{g \in \mathbf{H}} \mu^*(g)$. We have thus proved that $\mu^*(f) = \sup_{g \in \mathbf{H}} \mu^*(g)$; the relation $\mu^*(f) = \lim_{g \in \mathbf{H}} \mu^*(g)$ is then a consequence of the monotone limit theorem (GT, IV, §5, No. 2, Th. 2).

As $\varphi_1$ (resp. $\varphi_2$) runs over the set of functions in $\mathscr{K}_+$ such that $\varphi_1 \leq f_1$ (resp. $\varphi_2 \leq f_2$), the functions $\varphi_1 + \varphi_2$ form a directed set (for $\leq$) whose upper envelope is $f_1 + f_2$. Therefore, by Th. 1,

where $(\varphi_1, \varphi_2)$ runs over the set of pairs of functions in $\mathscr{K}_+$ such that $\varphi_1 \leq f_1$ and $\varphi_2 \leq f_2$; since

(GT, IV, §5, No. 7, Cor. 2 of Prop. 12), the theorem is proved.

For every finite subset $J$ of $I$, it follows from Th. 2 (by induction on the number of elements of $J$) that $\mu^*\left( \sum_{\iota \in J} f_\iota \right) = \sum_{\iota \in J} \mu^*(f_\iota)$; as $J$ runs over the set of finite subsets of $I$, the functions $g_J = \sum_{\iota \in J} f_\iota$ belong to $J_+$ and form a directed set for the relation $\leq$, whose upper envelope is the function $\sum_{\iota \in I} f_\iota$; the proposition therefore follows from Th. 1.

For, this mapping is by definition the upper envelope of the mappings $\mu \mapsto \mu(g)$, where $g$ runs over the set of functions in $\mathscr{K}_+$ such that $g \leq f$; and by definition of the vague topology, the mappings $\mu \mapsto \mu(g)$ are continuous on $\mathscr{M}(X)$.

For, there exists in this case a function $f \in \mathscr{K}_+$ such that $\varphi_G \leq f$ (Ch. III, §1, No. 2, Lemma 1), whence

An open set of finite outer measure is not always relatively compact (Exer. 3).

The functions φ_G form a directed set (for ≤) in $\mathscr{I}_+$ and their upper envelope is the characteristic function of the union of the sets G ∈ $\mathfrak{B}$; the proposition is thus a consequence of Th. 1.

Moreover, if the G_ι are pairwise disjoint then

For, if G = $\bigcup_{\iota \in I} G_\iota$ then φ_G = sup_{ι \in I} φ_{G_ι} ≤ $\sum_{\iota \in I} \varphi_{G_\iota}$; when the G_ι are pairwise disjoint, φ_G = $\sum_{\iota \in I} \varphi_{G_\iota}$; the proposition is therefore a consequence of Props. 2 and 3.

Note in particular that if $G$ is an open set in $\mathbf{R}$ such that $\mu^*(G) = 0$, then $G$ is empty.

For every function $h_1 \in \mathscr{I}_+$ such that $f_1 \leqslant h_1$ and every function $h_2 \in \mathscr{I}_+$ such that $f_2 \leqslant h_2$, we have, by Th. 2,

Props. 10, 11 and 12 express that $\mu^*$ is an increasing, positively homogeneous and convex function on the set of numerical functions $\geqslant 0$ defined on X (Ch. I, No. 1). Note that if $f_1$ and $f_2$ are any two positive functions, the two members of (9) are not necessarily equal (§ 4, Exer. 8 d)); in §5, No. 6 we shall give conditions under which equality holds.

Since each of the functions $f_n$ is less than or equal to $\sup_n f_n$, everything comes down to proving that $\mu^*(\sup_n f_n) \leqslant \sup_n \mu^*(f_n)$; this is obvious if the second member of the inequality is $+\infty$. In the contrary case, $\mu^*(f_n) < +\infty$ for all $n$; we are going to show that, for every $\varepsilon > 0$, there exists an increasing sequence $(g_n)$ of functions in $\mathscr{I}_+$ such that $f_n \leqslant g_n$ and $\mu^*(g_n) \leqslant \mu^*(f_n) + \varepsilon$. If $g$ is the upper envelope of the sequence $(g_n)$, we will then have $\mu^*(g) = \sup_n \mu^*(g_n)$ (No. 1, Th. 1), whence $\mu^*(g) \leqslant \sup \mu^*(f_n) + \varepsilon$; since $\sup f_n \leqslant g$ and $\varepsilon$ is arbitrary, the theorem will then have been proved.

By hypothesis, there exists a function $h_n \in \mathscr{J}_+$ such that $f_n \leq h_n$ and $\mu^*(f_n) \leq \mu^*(h_n) \leq \mu^*(f_n) + \varepsilon/2^n$; let us show that the functions $g_n = \sup(h_1, h_2, \ldots, h_n)$ meet the requirements. They belong to $\mathscr{J}_+$, form an increasing sequence, and satisfy $f_n \leq g_n$ for all $n$; we shall prove that

Let us argue by induction on $n$; the case $n = 1$ is trivial. On the other hand $g_{n+1} = \sup(g_n, h_{n+1})$, $g_n \geq f_n$ and $h_{n+1} \geq f_{n+1} \geq f_n$, whence $\inf(g_n, h_{n+1}) \geq f_n$; since

it follows from Th. 2 of No. 1 that

Q.E.D.

For, there exists an increasing sequence of functions in $\mathfrak{G}$ having the same upper envelope as $\mathfrak{F}$: if $(f_n)$ is the sequence of functions of $\mathfrak{G}$, arranged in any order, let $(f_{n_k})$ be a subsequence defined recursively by the conditions $n_1 = 1$, $f_{n_{k+1}} \geq \sup(f_{n_k}, f_k)$; it is clear that this subsequence has the indicated properties.

It suffices to apply the relation (10) to the increasing sequence of functions $g_n = \sum_{k=1}^n f_k$ while taking into account that, by (9),

In §5, No. 6, we will give conditions under which the two members of (12) are equal.

For every integer $n$, set $g_n = \inf_{p \geqslant 0} f_{n+p}$; the sequence $(g_n)$ is increasing and $\liminf_{n \to \infty} f_n = \sup_n g_n$, whence, by (10),

but since $g_n \leqslant f_{n+p}$ for $p \geqslant 0$, we have $\mu^*(g_n) \leqslant \mu^*(f_{n+p})$, whence $\mu^*(g_n) \leqslant \inf_{p \geqslant 0} \mu^*(f_{n+p})$ and finally

If $f_0$ is the constant function equal to $+\infty$, then $f_0$ is the upper envelope of all the functions of $\mathscr{K}_+$, and, since $\mu \neq 0$, we have $\mu^*(f_0) > 0$; but since $f_0 = \alpha f_0$ for every $\alpha > 0$, necessarily $\mu^*(f_0) = +\infty$ (Prop. 11). The inequality (13) then shows that $\mu^*(f_n)$ tends to $+\infty$ with $n$.

Moreover, the relation $\mu \leq \nu$ implies $\mu^* \leq \nu^*$.

Let us prove the relation (15). Set $\lambda = \mu + \nu$; thus $\lambda(f) = \mu(f) + \nu(f)$ for $f \in \mathscr{K}_+$; for $f \in \mathscr{J}_+$, the value of $\lambda^*(f)$ (resp. $\mu^*(f)$, $\nu^*(f)$) is the limit of $\lambda(g)$ (resp. $\mu(g)$, $\nu(g)$) as $g$ runs over the directed set (for $\leq$) of all $g \in \mathscr{K}_+$ such that $g \leq f$; therefore $\lambda^*(f) = \mu^*(f) + \nu^*(f)$. Finally, if $f$ is any function $\geq 0$ defined on $X$, then $\lambda^*(f)$ (resp. $\mu^*(f)$, $\nu^*(f)$) is the limit of $\lambda^*(h)$ (resp. $\mu^*(h)$, $\nu^*(h)$) as $h$ runs over the directed set (for $\geq$) of all functions $h \in \mathscr{J}_+$ such that $h \geq f$; again, by passage to the limit, we therefore have $\lambda^*(f) = \mu^*(f) + \nu^*(f)$, which proves (15). The relation (14) is established similarly. Finally, if $\mu \leq \nu$ one can write $\nu = \mu + (\nu - \mu)$, where $\nu - \mu \geq 0$, therefore $\nu^* = \mu^* + (\nu - \mu)^*$, which shows that $\mu^* \leq \nu^*$.

For, such a set is contained in a relatively compact open set (GT, I, §9, No. 7, Prop. 10), whose outer measure is finite (No. 2, Prop. 5).

These propositions are the translations of Props. 10 and 13 and of Th. 3 of No. 3 for characteristic functions of sets.

The proposition is obvious if $\mu^*(A) = +\infty$. In the contrary case, for every $\varepsilon$ such that $0 < \varepsilon < 1$, there exists a function $f \in \mathscr{J}_+$ such that $\varphi_A \leq f$ and $\mu^*(A) \leq \mu^*(f) \leq \mu^*(A) + \varepsilon$. Let G be the set of $x \in X$ such that $f(x) > 1 - \varepsilon$. Since $f$ is lower semi-continuous, G is open (GT, IV, §6, No. 2, Prop. 1) and contains A; on the other hand $f \geq (1 - \varepsilon)\varphi_G$, whence

since $\varepsilon$ is arbitrary, we see that $\mu^*(G)$ differs as little as we please from $\mu^*(A)$, whence the proposition.