§ 4. Integrable functions and sets

1. Extension of the integral

It follows from the definition of the space $\mathscr{L}_F^p$ that the subspace $\mathscr{K}_F$ of continuous functions with compact support is dense in $\mathscr{L}_F^p$ (§ 3, No. 4, Def. 2). Every continuous (for the topology of convergence in mean of order $p$ ) linear function, defined on $\mathscr{K}_F$ and taking its values in a complete Hausdorff topological vector space $G$ , can therefore be extended by continuity in a unique manner, to a continuous linear function defined on $\mathscr{L}_F^p$ with values in $G$ (GT, II, § 3, No. 6, Th. 2 and III, § 3, No. 1, Prop. 3).

Now, for every continuous function $f$ with compact support, with values in the Banach space $F$ , we have defined (in Ch. III, § 3, No. 1) the integral $\mu(f) = \int f \, d\mu$ with respect to $\mu$ , which is an element of $F$ , and we have proved (Ch. III, § 3, No. 2, Prop. 6) the inequality

\left| \int f \, d\mu \right| \leqslant \int |f| \, d|\mu| = N_1(f).

This inequality proves that the linear mapping $f \mapsto \int f \, d\mu$ of $\mathscr{K}_F$ into $F$ is continuous for the topology of convergence in mean in $\mathscr{K}_F$ . It can therefore be extended by continuity to the entire space $\mathscr{L}_F^1$ , and we may make the following definition:

Definition 1

The functions belonging to $\mathscr{L}_F^1(X, \mu)$ are said to be integrable with respect to the measure $\mu$ * (or, again, that they are $\mu$ -integrable). *The integral (with respect to $\mu$ ) of the integrable function $f$ is by definition the value at $f$ of the extension by continuity to $\mathscr{L}_F^1$ of the linear mapping $g \mapsto \int g\, d\mu$ of $\mathscr{H}_F$ into $F$ ; it is again denoted $\mu(f)$ or $\int f\, d\mu$ , or $\int f(x)\, d\mu(x)$ or $\int f\mu$ , or $\int f(x)\mu(x)$ .

Example

— Let $X$ be a discrete space, $\mu$ a measure on $X$ , and set $\alpha(x) = \mu(\varphi_{\{x\}})$ for every $x \in X$ . The functions in $\mathscr{F}_F^1$ are then integrable, in other words $\mathscr{L}_F^1 = \mathscr{F}_F^1$ ; moreover, for every function $f \in \mathscr{L}_F^1$ ,

\int f\, d\mu = \sum_{x \in X} \alpha(x)f(x).

For, let $f \in \mathscr{F}_F^1$ ; we have $|\mu|^*(|f|) = \sum_{x \in X} |\alpha(x)| \cdot |f(x)| < +\infty$ (§ 1, No. 3, Example); for every $\varepsilon > 0$ , there exists a finite subset $M$ of $X$ such that

\sum_{x \in X - M} |\alpha(x)| \cdot |f(x)| \leq \varepsilon.

The function $g$ equal to $f$ at the points $x \in M$ where $|f|$ is finite, and to 0 elsewhere, belongs to $\mathscr{H}(X; F)$ and, by the conventions that have been made,

|\mu|^*(|f - g|) \leq \sum_{x \in X - M} |\alpha(x)| \cdot |f(x)| \leq \varepsilon,

which proves that $f \in \mathscr{L}_F^1$ . On the other hand,

\left| \mu(g) - \sum_{x \in X} \alpha(x)f(x) \right| \leq \sum_{x \in X - M} |\alpha(x)| \cdot |f(x)| \leq \varepsilon,

whence the second assertion.

In other words, the $\mu$ -integrable functions $f$ are those for which the family $(\alpha(x)f(x))_{x \in X}$ is absolutely summable (GT, IX, §3, No. 6), and the integral $\int f\, d\mu$ is the sum of this family.

Since $\mu(f)$ is continuous on $\mathscr{L}_F^1$ by definition, and since it takes its values in a Hausdorff space, we have $\mu(f) = 0$ for every function that belongs to the closure of 0 in $\mathscr{L}_F^1$ , that is, is negligible; if $f$ and $g$ are two equivalent integrable functions, then $\mu(f) = \mu(g)$ . In other words, the value of $\mu(f)$ depends only on the class $\tilde{f}$ of the integrable function $f$ ; it is again denoted $\mu(\tilde{f})$ , and the function $\tilde{f} \mapsto \mu(\tilde{f})$ is a continuous linear mapping of $L^1_F$ into $F$ . If a function $f$ , with values in $F$ and defined almost everywhere in $X$ , is equivalent to an integrable function, we again say that $f$ is integrable and we write $\int f\, d\mu = \mu(\tilde{f})$ ; one defines similarly an integrable function with values in $\overline{\mathbf{R}}$ , defined and finite almost everywhere, as well as its integral.

2. Properties of the integral

Proposition 1

For every positive $\mu$ -integrable numerical function $f$ ,

\int f\, d|\mu| = \int^* f\, d|\mu| = N_1(f) \geqslant 0.

For, $\int f\, d|\mu|$ and $N_1(f)$ are continuous on $\mathscr{L}^1$ and are equal for every continuous function $f \geqslant 0$ with compact support; on the other hand, every function $f \geqslant 0$ in $\mathscr{L}^1$ is the limit (in the sense of convergence in mean) of a sequence of continuous functions $\geqslant 0$ with compact support (§ 3, No. 5, Prop. 11); whence the proposition.

Corollary 1

For every integrable function $f \in \mathscr{L}^1_F$ , $|f|$ is integrable and

\int |f|\, d|\mu| = \int^* |f|\, d|\mu| = N_1(f).

We shall make frequent use of Prop. 1 and its Cor. 1, on replacing $\int^* f\, d|\mu|$ or $N_1(f)$ by $\int f\, d|\mu|$ when dealing with an integrable function $\geqslant 0$ . For example, for two integrable functions $f, g$ to be equivalent, it is necessary and sufficient that $\int |f - g|\, d|\mu| = 0$ .

We recall that, for a function $f$ to belong to $\mathscr{L}^p_F$ , it is necessary and sufficient that the function $|f|^{p-1} \cdot f$ belong to $\mathscr{L}^1_F$ (§ 3, No. 8, Cor. 1 of Th. 7), that is, that it be integrable; this is the reason for the terminology 'p-th power integrable function'. Moreover:

Corollary 2

For every function $f \in \mathscr{L}^p_F$ , the numerical function $|f|^p$ is integrable and

N_p(f) = \left( \int |f|^p\, d|\mu| \right)^{1/p}.

This follows at once from the fact that $|f|$ belongs to $\mathscr{L}^p$ (§ 3, No. 5, Prop. 11) and formula (2).

Proposition 2

— For every integrable function $f$ ,

\left| \int f \, d\mu \right| \leq \int |f| \, d|\mu|.

This follows at once from the inequality (1) by passage to the limit, on taking into account (3) and the continuity of $N_1(f)$ on $\mathscr{L}_F^1$ .

Theorem 1

— Let $F$ and $G$ be two Banach spaces, $u$ a continuous linear mapping of $F$ into $G$ . For every integrable function $f$ with values in $F$ , $u \circ f$ is integrable and

\int u(f(x)) \, d\mu(x) = u \left( \int f(x) \, d\mu(x) \right).

We already know that $u \circ f$ is integrable (§ 3, No. 5, Th. 4); the relation (6), being valid for every $f \in \mathscr{H}_F$ , extends to every integrable function $f$ by the principle of extension of identities: for, $f \mapsto u \circ f$ is continuous for the topology of convergence in mean, as follows from the inequality $N_1(u \circ f) \leq \|u\| \cdot N_1(f)$ .

Corollary 1

— Let $a'$ be a continuous linear form on $F$ . If $f$ is an integrable function with values in $F$ , then the numerical function $\langle f, a' \rangle$ is integrable and

\int \langle f(x), a' \rangle \, d\mu(x) = \left\langle \int f(x) \, d\mu(x), a' \right\rangle.

We shall see in Ch. VI, § 1, Exers. 7, 11 and 12 that there can exist functions $f$ , with values in an infinite-dimensional Banach space $F$ , such that $\langle f, a' \rangle$ is integrable for every continuous linear form $a'$ on $F$ , without $f$ being integrable.

Corollary 2

— If the $a_k$ ( $1 \leq k \leq n$ ) are vectors in $F$ and the $f_k$ ( $1 \leq k \leq n$ ) are integrable numerical functions, then the function $f = \sum_{k=1}^n a_k f_k$ is integrable and

\int \left( \sum_{k=1}^n a_k f_k \right) d\mu = \sum_{k=1}^n a_k \int f_k \, d\mu.

3. Passage to the limit in integrals

Proposition 3

— Let $\mathscr{B}$ be a filter base on $\mathscr{L}_F^1$ . Assume that there exists a compact set $K \subset X$ such that, for every set $M \in \mathscr{B}$ , all of the functions $f \in M$ have their support in $K$ . Under these conditions, if $\mathscr{B}$ converges uniformly on $X$ to $f_0$ , then the function $f_0$ is integrable and

\int f_0 \, d\mu = \lim_{\mathscr{B}} \int f \, d\mu .

For, $\mathscr{B}$ converges in mean to $f_0$ (§ 3, No. 3, Prop. 4).

Proposition 4

— Let $(f_n)$ be an increasing (resp. decreasing) sequence of integrable numerical functions. For the upper (resp. lower) envelope $f$ of the sequence to be integrable, it is necessary and sufficient that $\sup_n \int f_n \, d|\mu| < +\infty$ (resp. $\inf_n \int f_n \, d|\mu| > -\infty$ ), in which case

\int f \, d\mu = \lim_{n \to \infty} \int f_n \, d\mu .

We limit ourselves to considering an increasing sequence. The sequence $g_n = f_n + f_1^-$ is increasing and consists of integrable functions $\geqslant 0$ ; since its upper envelope is $g = f + f_1^-$ , the proposition follows from Th. 5 of § 3, No. 6.

Theorem 2

— Let $A$ be a set of indices, filtered by a filter $\mathfrak{F}$ with a countable base. Let $(f_\alpha)_{\alpha \in A}$ be a family of integrable functions that, with respect to the filter $\mathfrak{F}$ , converge pointwise almost everywhere to a function $f$ ; if there exists a numerical function $g \geqslant 0$ such that $\int^* g \, d|\mu| < +\infty$ and such that $|f_\alpha(x)| \leqslant g(x)$ almost everywhere in $X$ for each $\alpha \in A$ , then the function $f$ is integrable and

\int f \, d\mu = \lim_{\mathfrak{F}} \int f_\alpha \, d\mu .

The theorem follows from Lebesgue’s theorem (§ 3, No. 7, Cor. of Th. 6) since, under the conditions of the statement, $f_\alpha$ converges in mean to $f$ with respect to $\mathfrak{F}$ .

Corollary 1

— Let $\Omega$ be a topological space, $t_0$ a point of $\Omega$ admitting a countable fundamental system of neighborhoods, $f$ a mapping of $X \times \Omega$ into $F$ having the following properties: a) for every $t \in \Omega$ , the function $x \mapsto f(x, t)$ is integrable; b) for every $x \in X$ , the function $t \mapsto f(x, t)$ is continuous at $t_0$ ;

c) there exist a neighborhood U of t_0 and a numerical function g \geqslant 0 defined on X, such that $\int^* g d|\mu| < +\infty$ and $|\mathbf{f}(x,t)| \leqslant g(x)$ for all $x \in X$ and $t \in U$ .

Under these conditions, the mapping $t \mapsto \int \mathbf{f}(x,t) d\mu(x)$ of $\Omega$ into F is continuous at the point $t_0$ .

Corollary 2

— Let $(\mathbf{f}_n)$ be a sequence of integrable functions such that the series with general term $\mathbf{f}_n(x)$ converges almost everywhere; if there exists a function $g \geqslant 0$ such that $\int^* g d|\mu| < +\infty$ and such that, for every integer $n$ , $\left| \sum_{k=1}^n \mathbf{f}_k(x) \right| \leqslant g(x)$ almost everywhere, then the sum $\mathbf{f}(x)$ (defined almost everywhere) of the series with general term $\mathbf{f}_n(x)$ is integrable and

\int \mathbf{f} d\mu = \sum_{n=1}^\infty \int \mathbf{f}_n d\mu

('term-by-term integration of a series').

4. Characterizations of integrable numerical functions

Proposition 5

— For a numerical function $f \geqslant 0$ (finite or not), lower semi-continuous on X, to be integrable, it is necessary and sufficient that $\int^* f d|\mu| < +\infty$ .

It all comes down to proving that the condition is sufficient. The definition of $|\mu|^*(f)$ (§ 1, No. 1, Def. 1) proves that, for every $\varepsilon > 0$ , there exists a continuous function $g \geqslant 0$ , with compact support, such that $g \leqslant f$ and $|\mu|^*(f) \leqslant |\mu|(g) + \varepsilon$ . But $f - g$ is lower semi-continuous and $\geqslant 0$ , therefore (§ 1, No. 1, Th. 2)

|\mu|^*(f) = |\mu|(g) + |\mu|^*(f - g),

in other words $N_1(f - g) = |\mu|^*(f - g) = |\mu|^*(f) - |\mu|(g) \leqslant \varepsilon$ , which proves that $f$ is integrable (§ 3, No. 4, Prop. 7).

Corollary 1

— For a finite numerical function $f \geqslant 0$ , upper semi-continuous on X, to be integrable, it is necessary and sufficient that $\int^* f d|\mu| < +\infty$ .

For, if $|\mu|^*(f) < +\infty$ , then there exists a lower semi-continuous function $h$ such that $f \leqslant h$ and $|\mu|^*(h) < +\infty$ ; $h - f$ is everywhere defined and lower semi-continuous, and $|\mu|^*(h - f) \leqslant |\mu|^*(h) < +\infty$ ; therefore $h - f$ is integrable, and since $f(x) = h(x) - (h(x) - f(x))$ almost everywhere, $f$ is integrable.

Corollary 2

— Let $\mathbf{H}$ be a nonempty set, directed for the relation $\leqslant$ (resp. $\geqslant$ ), of lower (resp. upper) semi-continuous and integrable numerical functions; if

\sup_{f \in \mathbf{H}} \int f \, d|\mu| < +\infty \quad \text{(resp. } \inf_{f \in \mathbf{H}} \int f \, d|\mu| > -\infty \text{)},

then the upper (resp. lower) envelope $g$ of $\mathbf{H}$ is integrable,

\int g \, d\mu = \lim_{f \in \mathbf{H}} \int f \, d\mu,

and $\int g \, d|\mu| = \sup_{f \in \mathbf{H}} \int f \, d|\mu|$ (resp. $\int g \, d|\mu| = \inf_{f \in \mathbf{H}} \int f \, d|\mu|$ ).

We may limit ourselves to the case of lower semi-continuous functions; the functions $f^+$ (resp. $f^-$ ), as $f$ runs over $\mathbf{H}$ , then form a directed set for $\leqslant$ (resp. $\geqslant$ ) of lower (resp. upper) semi-continuous functions $\geqslant 0$ ; the upper (resp. lower) envelope of the $f^+$ (resp. $f^-$ ), for $f \in \mathbf{H}$ , is equal to $g^+$ (resp. $g^-$ ). On the other hand, one can replace $\mathbf{H}$ by one of its sections (which is cofinal to it), consisting of the $f \in \mathbf{H}$ that are $\geqslant f_0$ , for some function $f_0 \in \mathbf{H}$ ; then $\int f^+ \, d|\mu| \leqslant \int f \, d|\mu| + \int f_0^- \, d|\mu|$ ; we thus see that we are reduced to proving the two assertions of the corollary when $\mathbf{H}$ consists of positive functions. If $\mathbf{H}$ is directed for $\leqslant$ and consists of lower semi-continuous functions $\geqslant 0$ , then we know (§ 1, No. 1, Th. 1) that

|\mu|^*(g) = \sup_{f \in \mathbf{H}} |\mu|^*(f) = \sup_{f \in \mathbf{H}} \int f \, d|\mu| < +\infty,

therefore $g$ , which is lower semi-continuous, is integrable by Prop. 5; we have $\int g \, d|\mu| = \lim_{f \in \mathbf{H}} \int f \, d|\mu|$ and, since $f \leqslant g$ , $\lim_{f \in \mathbf{H}} N_1(g - f) = 0$ , which shows that $f$ converges in mean to $g$ with respect to $\mathbf{H}$ , and thus proves the corollary in this case. If $\mathbf{H}$ is directed for $\geqslant$ and consists of upper semi-continuous integrable functions $f$ such that $0 \leqslant f \leqslant f_1$ with $f_1 \in \mathbf{H}$ , then there exists a lower semi-continuous integrable function $h$ such that $f_1 \leqslant h$ ; we may write $f = h - f'$ , where $f'(x) = h(x) - f(x)$ when $f(x) < +\infty$ , and $f'(x) = 0$ otherwise. It is clear that the $f'$ form a directed set, for $\leqslant$ , of lower semi-continuous integrable functions $\geqslant 0$ , with

\int f' \, d|\mu| \leqslant \int h \, d|\mu| < +\infty;

we can apply to them what has been proved above; if $g'$ is the upper envelope of the $f'$ , then $h$ and $g'$ are finite almost everywhere, therefore $h - g'$ is defined almost everywhere and is equal to $g$ almost everywhere; from this, the conclusions of the corollary follow at once in this case.

Corollary 3

— Let $f$ be a bounded numerical function, upper semi-continuous on $X$ and with compact support. Then, the mapping $\mu \mapsto \int f\, d\mu$ is upper semi-continuous on $\mathscr{M}_+(X)$ for the vague topology.

If $h$ is a function in $\mathscr{K}_+(X)$ such that $|f| \leq h$ (Ch. III, §1, No. 2, Lemma 1) then $0 \leq f + h \leq 2h$ , and since $f + h$ is upper semi-continuous, it follows from Cor. 1 that $f$ is $\mu$ -integrable for every measure $\mu$ on $X$ . Moreover, $\mu(f) = \mu(h) - \mu(h - f)$ and $h - f$ is a lower semi-continuous function $\geq 0$ . Since the mapping $\mu \mapsto \mu(h - f)$ is lower semi-continuous on $\mathscr{M}_+(X)$ for the vague topology (§1, No. 1, Prop. 4), this proves the corollary.

Theorem 3

— For a numerical function $f \geq 0$ to be integrable, it is necessary and sufficient that, for every $\varepsilon > 0$ , there exist an upper semi-continuous function $g \geq 0$ , with finite values and compact support, and a lower semi-continuous integrable function $h$ , such that $g \leq f \leq h$ and $\int (h - g)\, d|\mu| \leq \varepsilon$ .

The condition is sufficient by a general criterion for integrability (§3, No. 4, Prop. 8), Prop. 5 and its Corollary 1. Let us show that the condition is necessary. If $f \geq 0$ is integrable then, for every $\varepsilon > 0$ , there exists a function $u \geq 0$ , continuous and with compact support, such that $N_1(f - u) \leq \varepsilon/4$ . By the definition of $N_1$ , this implies that there exists a lower semi-continuous function $v \geq 0$ such that $|\mu|^*(v) \leq \varepsilon/2$ and $|f - u| \leq v$ . Thus, $-v(x) \leq f(x) - u(x) \leq v(x)$ for all $x \in X$ , and since $u(x)$ is everywhere finite, it follows that $(u(x) - v(x))^+ \leq f(x) \leq u(x) + v(x)$ for all $x \in X$ . The functions $g = (u - v)^+$ and $h = u + v$ meet the requirements.

Corollary

— For every integrable (resp. integrable and $\geq 0$ ) numerical function $f$ , there exist an increasing sequence $(g_n)$ of upper semi-continuous functions that are integrable (resp. integrable, with finite values $\geq 0$ , and with compact support), and a decreasing sequence $(h_n)$ of lower semi-continuous integrable functions, such that: $1^\circ$ $g_n(x) \leq f(x) \leq h_n(x)$ for all $x \in X$ and every integer $n$ ; $2^\circ$ $f(x)$ is equal almost everywhere to the lower envelope $h$ of the sequence $(h_n)$ and to the upper envelope $g$ of the sequence $(g_n)$ ; $3^\circ$ $\int f\, d\mu = \lim_{n \to \infty} \int g_n\, d\mu = \lim_{n \to \infty} \int h_n\, d\mu$ .

Suppose first that $f \geq 0$ . By Th. 3 there exist, for every $n$ , a lower semi-continuous integrable function $v_n$ , and an upper semi-continuous function $u_n \geq 0$ with finite values and compact support, such that

u_n \leq f \leq v_n \quad \text{and} \quad \int (v_n - u_n)\, d|\mu| \leq 1/n;

setting $g_n = \sup(u_1, u_2, \ldots, u_n)$ and $h_n = \inf(v_1, v_2, \ldots, v_n)$ , the sequences $(g_n)$ and $(h_n)$ meet the requirements. For, since $g \leq f$ , $g$ is integrable by Prop. 4 of No. 3, and since

\int (f - g_n) d|\mu| \leq \int (v_n - u_n) d|\mu| \leq 1/n

we have

\int (f - g) d|\mu| = \lim_{n \to \infty} \int (f - g_n) d|\mu| = 0

(No. 3, Prop. 4), which proves that

f

and

g

are equivalent. One argues similarly for the sequence

(h_n)

If $f$ is not positive, we can apply the foregoing to $f^+$ and $f^-$ , thus there are two increasing sequences $(g'_n), (g''_n)$ of upper semi-continuous integrable functions, and two decreasing sequences $(h'_n), (h''_n)$ of lower semi-continuous integrable functions, such that: $1^\circ \ g'_n \leq f^+ \leq h'_n,\ g''_n \leq -f^- \leq h''_n;$ $2^\circ \ f^+$ (resp. $-f^-$ ) is equal almost everywhere to the upper envelope of the $g'_n$ and to the lower envelope of the $h'_n$ (resp. to the upper envelope of the $g''_n$ and to the lower envelope of the $h''_n$ ); and $3^\circ$ :

\int f^+ d\mu = \lim_{n \to \infty} \int g'_n d\mu = \lim_{n \to \infty} \int h'_n d\mu,

-\int f^- d\mu = \lim_{n \to \infty} \int g''_n d\mu = \lim_{n \to \infty} \int h''_n d\mu.

Moreover, we can suppose that the

g'_n

and the

h''_n

are everywhere finite; it is then clear that the sequences

g_n = g'_n + g''_n

and

h_n = h'_n + h''_n

meet the requirements.

Example

For every positive measure $\mu$ on $\mathbf{R}$ , every step function with compact support is $\mu$ -integrable; for, the characteristic function of an open (resp. closed) interval is lower (resp. upper) semi-continuous, and every step function is a linear combination of such characteristic functions. It follows that if $\mathbf{f}$ is a regulated function on $\mathbf{R}$ with compact support (FRV, II, §1, No. 3), then $\mathbf{f}$ is integrable, because it is the uniform limit of a sequence of step functions $\mathbf{g}_n$ with support contained in a fixed compact set (No. 3, Prop. 3); moreover, $\int \mathbf{f}\, d\mu = \lim_{n \to \infty} \int \mathbf{g}_n\, d\mu$ .

If, in particular, $\mu$ is taken to be Lebesgue measure, one sees that for every regulated function $\mathbf{f}$ with compact support, the integral $\int \mathbf{f}\, d\mu$ is equal to the integral $\int_{-\infty}^{+\infty} \mathbf{f}(x)\, dx$ defined in FRV, II, §2, No. 1.

Remark 1

Let $\mathbf{f}$ be a regulated function on $\mathbf{R}$ that is integrable with respect to Lebesgue measure $\mu$ ; then $|\mathbf{f}|$ is also integrable (No. 2, Cor. 1 of Prop. 1), and, setting $I_n = [-n, n]$ , $|\mathbf{f}|$ is the upper envelope of the increasing sequence of regulated functions $|\mathbf{f}| \varphi_{I_n}$ , therefore $\int |\mathbf{f}|\, d\mu = \lim_{n \to \infty} \int_{-n}^{n} |\mathbf{f}(x)|\, dx$ by Th. 2 of No. 3; thus, the integral $\int_{-\infty}^{+\infty} f(x) dx$ is absolutely convergent (FRV, II, §2, No. 3). Moreover, $\int f\, d\mu = \int_{-\infty}^{+\infty} f(x) dx$ by Th. 2 of No. 3. Conversely, suppose that $\int_{-\infty}^{+\infty} f(x) dx$ is absolutely convergent; again, by Th. 2 of No. 3, $\int f\, d\mu = \int_{-\infty}^{+\infty} f(x) dx$ . Note that if the integral $\int_{-\infty}^{+\infty} f(x) dx$ is convergent but not absolutely convergent; then $f$ is not integrable with respect to Lebesgue measure.

Remark 2

Applied to Lebesgue measure and to regulated functions, Prop. 3 of No. 3 yields anew the theorem on passage to the limit for integrals of regulated functions on a compact interval (FRV, II, §3, No. 1, Prop. 1); for sequences (or for filters with a countable base) of regulated functions, Th. 2 of No. 3 greatly improves on this proposition since, for uniformly bounded regulated functions on a compact interval, it substitutes pointwise convergence for uniform convergence (cf. §5, No. 4, Th. 2). However, as regards the passage to the limit for absolutely convergent integrals of regulated functions on a noncompact interval, we note that the conditions of Th. 2 of No. 3 imply that the integrals under consideration are uniformly convergent (in the sense defined in FRV, II, §3, No. 2), and thus do not improve on the conditions for convergence given in Book IV (loc. cit.) except as concerns the convergence of the functions $f_\alpha$ on every compact interval. Finally, the conditions for passing to the limit given for non absolutely convergent integrals of regulated functions remain outside the theory developed in this chapter.

5. Integrable sets

Definition 2

A subset* $A$ of a locally compact space $X$ is said to be integrable with respect to a measure $\mu$ on $X$ (or, that it is $\mu$ *-integrable)* if the characteristic function $\varphi_A$ of $A$ is integrable. The finite number $\mu(A) = \int \varphi_A\, d\mu$ is called the measure of $A$ .

For every integrable set $A$ , $|\mu|(A) = |\mu|^*(A)$ (No. 2, Prop. 1); for a set to be negligible, it is necessary and sufficient that it be of measure zero with respect to $|\mu|$ .

Proposition 6

The union of a finite family* $(A_i)_{1 \leq i \leq n}$ of integrable sets is integrable, and

|\mu| \left( \bigcup_{i=1}^n A_i \right) \leq \sum_{i=1}^n |\mu|(A_i).

If, moreover, the $A_i$ are pairwise disjoint, then

\mu \left( \bigcup_{i=1}^n A_i \right) = \sum_{i=1}^n \mu(A_i).

For, if $A = \bigcup_{i=1}^n A_i$ then $\varphi_A = \sup \varphi_{A_i}$ , therefore (§ 3, No. 5, Cor. of Prop. 12) if the $A_i$ are integrable then so is $A$ ; the relation (13) is a special case of the analogous relation for outer measures (§ 1, No. 4, Prop. 18), on taking account of the relation $|\mu|(A) = |\mu|^*(A)$ ; finally, if the $A_i$ are pairwise disjoint, then $\varphi_A = \sum_{i=1}^n \varphi_{A_i}$ , whence (14).

Proposition 7

1° *If A and B are two integrable sets such that $B \subset A$ , then the set $C = A - B$ is integrable and

\mu(C) = \mu(A) - \mu(B).

2° The intersection of a countable family of integrable sets is integrable.

The first part follows from the fact that $\varphi_C = \varphi_A - \varphi_B$ . On the other hand, if $(A_n)$ is a sequence of integrable sets and $A$ is its intersection, then $\varphi_A = \inf_n \varphi_{A_n}$ , therefore $A$ is integrable (No. 3, Prop. 4).

Corollary

If $(A_n)$ is a decreasing sequence of integrable sets, then* $\mu(\bigcap_n A_n) = \lim_{n \to \infty} \mu(A_n)$ .

For, if $A = \bigcap_n A_n$ then $\varphi_A$ is the lower envelope of the decreasing sequence $(\varphi_{A_n})$ (No. 3, Prop. 4).

Proposition 8

Let $(A_n)$ be an increasing sequence of integrable sets; for the union $A = \bigcup_n A_n$ to be integrable, it is necessary and sufficient that* $\sup_n |\mu|(A_n) < +\infty$ , *in which case,

\mu(A) = \lim_{n \to \infty} \mu(A_n).

For, the $\varphi_{A_n}$ form an increasing sequence of integrable functions, and $\varphi_A = \sup_n \varphi_{A_n}$ ; thus, the proposition follows from Prop. 4 of No. 3.

Corollary

Let $(A_n)$ be a sequence of integrable sets such that* $\sum_{n=1}^\infty |\mu|(A_n) < +\infty$ ; the union $A = \bigcup_n A_n$ *is integrable, and

|\mu| \left( \bigcup_n A_n \right) \leq \sum_{n=1}^\infty |\mu|(A_n).

For, $\varphi_A = \sup_n \varphi_{A_n}$ and

|\mu|^*(A) \leq \sum_{n=1}^\infty |\mu|^*(A_n) = \sum_{n=1}^\infty |\mu|(A_n) < +\infty

(§1, No. 4, Prop. 18); therefore A is integrable (§3, No. 6, Cor. 2 of Th. 5) and, since $|\mu|(A) = |\mu|^*(A)$ , we indeed have (17).

Proposition 9

— Let $(A_n)$ be a sequence of pairwise disjoint integrable sets such that $\sum_{n=1}^\infty |\mu|(A_n) < +\infty$ ; then

\mu\left(\bigcup_n A_n\right) = \sum_{n=1}^\infty \mu(A_n).

For, if $A = \bigcup_n A_n$ then $\varphi_A = \sum_{n=1}^\infty \varphi_{A_n}$ , and the proposition follows from (17) and Cor. 2 of Th. 2 of No. 3.

The relation (18) is also expressed by saying that the measure $\mu$ is completely additive in the set of integrable subsets of $X$ .

6. Criteria for the integrability of a set

Proposition 10

— For an open (resp. closed) set $A$ in $X$ to be integrable, it is necessary and sufficient that $|\mu|^*(A) < +\infty$ .

Since $\varphi_A$ is then lower (resp. upper) semi-continuous, the proposition follows from Prop. 5 of No. 4 and its Corollary 1.

Corollary 1

— Every compact set is integrable; every relatively compact open set is integrable.

Corollary 2

— For every positive measure $\mu$ on $X$ , $A \mapsto \mu^*(A)$ is a capacity on $X$ (cf. GT, IX, §6, No. 9, Example).

Example

For Lebesgue measure $\mu$ on $\mathbf{R}$ , it follows from Prop. 10 that every bounded open interval $[a, b[$ is integrable and has measure $b - a$ (\§1, No. 2, Prop. 9). Since every set reducing to a point is negligible for Lebesgue measure, it follows that all of the intervals with endpoints $a$ and $b$ have the same measure $b - a$ .

Proposition 11

— Let $\mathscr{G}$ be a set, directed for the relation $\subset$ , of integrable open sets in $X$ ; for $A = \bigcup_{G \in \mathscr{G}} G$ to be integrable, it is necessary and sufficient that $\sup_{G \in \mathscr{G}} |\mu|(G) < +\infty$ , in which case $\mu(A) = \lim_\mathscr{G} \mu(G)$ and $|\mu|(A) = \sup_{G \in \mathscr{G}} |\mu|(G)$ .

For, one knows that $|\mu|^*(A) = \sup_{G \in \mathscr{G}} |\mu|(G)$ (\§1, No. 2, Prop. 7); the proposition therefore follows from Prop. 10.

Corollary

— Let $\mathfrak{F}$ be a set, directed for the relation $\supset$ , of integrable closed sets in $X$ ; then the closed set $B = \bigcap_{H \in \mathfrak{F}} H$ is integrable, and one has

\mu(B) = \lim_{\mathfrak{F}} \mu(H) \quad \text{and} \quad |\mu|(B) = \inf_{H \in \mathfrak{F}} |\mu|(H).

For, let

H_0

be a set in

\mathfrak{F}

; since

H_0

is integrable, it is contained in an integrable open set

U

(§ 1, No. 4, Prop. 19); the open sets

U \cap \mathbf{C}H

form a set directed for the relation

\subset

, are contained in

U

, and have union

U \cap \mathbf{C}B

; we are thus reduced to Prop. 11.

Theorem 4

— For a set $A$ to be integrable, it is necessary and sufficient that, for every $\varepsilon > 0$ , there exist an integrable open set $G$ and a compact set $K$ , such that $K \subset A \subset G$ and

|\mu|(G - K) = |\mu|(G) - |\mu|(K) \leq \varepsilon.

a) The condition is sufficient, because it means that $\varphi_K \leq \varphi_A \leq \varphi_G$ and $\int (\varphi_G - \varphi_K) d|\mu| \leq \varepsilon$ ; since $\varphi_G$ and $\varphi_K$ are integrable, so is $\varphi_A$ (§ 3, No. 4, Prop. 8).

b) The condition is necessary. If $A$ is integrable, there exists an open set $G \supset A$ such that $|\mu|^*(G)$ is arbitrarily close to $|\mu|^*(A) = |\mu|(A)$ (§ 1, No. 4, Prop. 19); thus, it all comes down to proving that for every $\varepsilon > 0$ , there exists a compact set $K \subset A$ such that $|\mu|(A) - |\mu|(K) \leq \varepsilon$ . Since $\varphi_A$ is integrable, there exists a function $f \geq 0$ , upper semi-continuous and with compact support $S$ , such that $f \leq \varphi_A$ and $\int (\varphi_A - f) d|\mu| \leq \varepsilon/2$ (No. 4, Th. 3). Let $\delta > 0$ be an arbitrary number and let $K$ be the set of points $x \in X$ such that $f(x) \geq \delta$ ; $K$ is closed and is contained in $S$ , hence is compact, and since $f \leq \varphi_A$ we have $K \subset A$ . The set $B = A - K$ is integrable, and $f \leq \varphi_K + \delta \varphi_B$ , whence

\int f d|\mu| \leq |\mu|(K) + \delta \cdot |\mu|(B) \leq |\mu|(K) + \delta \cdot |\mu|(A),

and finally

|\mu|(A) \leq \int f d|\mu| + \frac{\varepsilon}{2} \leq |\mu|(K) + \delta \cdot |\mu|(A) + \frac{\varepsilon}{2},

which completes the proof, since

\delta

is arbitrary.

Corollary 1

— For a set $A$ to be integrable, it is necessary and sufficient that, for every $\varepsilon > 0$ , there exist a compact set $K \subset A$ such that $|\mu|^*(A - K) \leq \varepsilon$ . The measure $|\mu|(A)$ is then the supremum of the set of measures $|\mu|(K)$ of the compact sets $K \subset A$ .

The condition is necessary, because if $G$ and $K$ satisfy the conditions of Theorem 4, then $|\mu|^*(A - K) \leq |\mu|^*(G - K) \leq \varepsilon$ .

The condition is sufficient, because it says that, for the topology of convergence in mean, $\varphi_A$ is in the closure of the set of integrable functions $\varphi_K$ ( $K$ an arbitrary compact subset of $A$ ).

Corollary 2

— For every integrable set $A$ , there exist:

$1^\circ$ a set $A_1 \supset A$ , a countable intersection of integrable open sets, such that $A_1 - A$ is negligible;

$2^\circ$ a set $A_2 \subset A$ , a countable union of pairwise disjoint compact sets, such that $A - A_2$ is negligible.

$1^\circ$ For every integer $n$ , there exists an integrable open set $G_n \supset A$ such that $|\mu|(G_n) - |\mu|(A) \leq 1/n$ ; if $A_1$ is the intersection of the $G_n$ , then $|\mu|(A_1) = |\mu|(A)$ (No. 5, Cor. of Prop. 7), therefore $A_1 - A$ is negligible.

$2^\circ$ Let us define compact sets $K_n$ inductively as follows: $K_1 \subset A$ and $|\mu|(A - K_1) \leq 1$ ; $K_n \subset A - \bigcup_{i=1}^{n-1} K_i$ and $|\mu|(A \cap \mathbf{C}(\bigcup_{i=1}^{n-1} K_i) \cap \mathbf{C}K_n) \leq 1/n$ for $n > 1$ (Th. 4); if $A_2$ is the union of the $K_n$ , then $|\mu|(A_2) = |\mu|(A)$ (No. 5, Prop. 8), therefore $A - A_2$ is negligible.

Corollary 3

— Every set of finite outer measure is contained in the union of a negligible set and a countable family of pairwise disjoint compact sets the sum of whose measures is finite.

It suffices to apply Corollary 2 to an integrable open set containing the given set.

Corollary 4

— For every open set $U$ in $X$ , $|\mu|^*(U)$ is the supremum of the measures $|\mu|(K)$ of the compact sets $K \subset U$ .

If $|\mu|^*(U) < +\infty$ , this is immediate from Th. 4. The following argument covers also the case that $|\mu|^*(U) = +\infty$ . Since $X$ is locally compact and $U$ is open, $\varphi_U$ is the upper envelope of the set $H$ of functions $f \in \mathscr{K}_+$ such that $f \leq \varphi_U$ and $\operatorname{Supp}(f) \subset U$ (cf. the proof of §1, No. 1, Lemma), and, since $H$ is directed for $\leq$ , we have $|\mu|^*(U) = \sup_{f \in H} |\mu|(f)$ by §1, No. 1, Th. 1; the corollary is then immediate from the fact that if $f \in H$ and $K = \operatorname{Supp}(f)$ , then $f \leq \varphi_K \leq \varphi_U$ .

Note that $|\mu|^*(U)$ is also the supremum of the measures $|\mu|(G)$ of the relatively compact open sets such that $\overline{G} \subset U$ . For, if $K$ is a compact set contained in $U$ then, for every $x \in K$ , there exists a relatively compact open neighborhood $V$ of $x$ such that $\overline{V} \subset U$ . On covering $K$ by a finite number of these neighborhoods, their union $G$ is a relatively compact open set such that $\overline{G} \subset U$ and $K \subset G$ , whence $|\mu|(K) \leq |\mu|(G) \leq |\mu|^*(U)$ .

7. Characterization of bounded measures

Proposition 12

— For a measure $\mu$ on a locally compact space $X$ to be bounded (Ch. III, §1, No. 8), it is necessary and sufficient that $X$ be an integrable set with respect to $\mu$ (or, what comes to the same thing, that every finite constant function be integrable); in this case,

\| \mu \| = | \mu |(X) = \int d|\mu|.

For, we have seen that $|\mu|^*(X) = \| \mu \|$ (§ 1, No. 2); the proposition therefore follows from Prop. 10 of No. 6.

For every bounded measure $\mu$ , we again say that $\mu(X)$ is the total mass of $\mu$ .

It follows from Th. 4 of No. 6 that if $\mu$ is a bounded measure then, for every $\varepsilon > 0$ , there exists a compact set $K$ such that $|\mu|(CK) \leq \varepsilon$ .

Proposition 13

— Let $\mu$ be a bounded measure on $X$ . Let $\mathscr{B}$ be a filter base on $\mathscr{L}_F^p$ having the following properties: $1^\circ$ there exists a set $M \in \mathscr{B}$ such that the functions $f \in M$ are uniformly bounded on $X$ ; $2^\circ$ $\mathscr{B}$ converges uniformly on every compact subset of $X$ to a function $f_0$ . Under these conditions, $f_0$ belongs to $\mathscr{L}_F^p$ and $\mathscr{B}$ converges in mean of order $p$ to $f_0$ .

We note first of all that if $|f(x)| \leq a$ for every $x \in X$ and every function $f \in M$ , then also $|f_0(x)| \leq a$ for every $x \in X$ . This being so, for every $\varepsilon > 0$ there exist a compact set $K$ such that $|\mu|(CK) \leq \varepsilon^p$ and a set $N \in \mathscr{B}$ such that, for every function $f \in N$ , $|f(x) - f_0(x)| \leq \varepsilon(|\mu|(K))^{-1/p}$ for all $x \in K$ . Now, we may write

f - f_0 = (f - f_0)\varphi_K + (f - f_0)\varphi_{CK};

it follows from the foregoing that if $f \in M \cap N$ then $N_p((f - f_0)\varphi_K) \leq \varepsilon$ and $N_p((f - f_0)\varphi_{CK}) \leq 2a\varepsilon$ , whence $N_p(f - f_0) \leq (2a + 1)\varepsilon$ , which proves the proposition.

Corollary

— For a bounded measure $\mu$ on $X$ , every bounded continuous mapping $f$ of $X$ into $F$ belongs to each of the $\mathscr{L}_F^p$ ( $1 \leq p < +\infty$ ).

For every compact subset $K$ of $X$ , let $M_K$ be the set of mappings of $X$ into $F$ of the form $hf$ , where $h$ is a continuous mapping of $X$ into $[0, 1]$ equal to 1 on $K$ and with compact support. It is clear that the sets $M_K$ form a filter base $\mathscr{B}$ on $\mathscr{L}_F^p$ , that the functions belonging to $M_K$ are uniformly bounded, and that $\mathscr{B}$ converges uniformly to $f$ on every compact subset of $X$ , whence the corollary.

In particular, the function $f$ is integrable and its integral $\int f\, d\mu$ is the limit with respect to $\mathscr{B}$ of the integrals $\int h f\, d\mu$ .

We will obtain anew the Cor. of Prop. 13 as a consequence of a general criterion for integrability in §5, No. 6.

In the notations of Ch. III, §1, No. 2, $|f| \leq \|f\| \cdot 1$ for every function $f \in \mathscr{C}^b(X; F)$ , whence, by the formulas (3) and (4) of No. 2,

N_p(f) \leq \|f\| \cdot N_p(1) = \|f\| \cdot \|\mu\|^{1/p}.

In particular, for $p = 1$ , formula (5) of No. 2 yields

\left| \int f\, d\mu \right| \leq \|f\| \cdot \|\mu\|,

consequently the mapping $f \mapsto \int f\, d\mu$ is continuous on the Banach space $\mathscr{C}^b(X; F)$ ; its restriction to the closure $\mathscr{C}^0(X; F)$ of $\mathscr{K}(X; F)$ in $\mathscr{C}^b(X; F)$ , that is, to the space of continuous functions tending to 0 at the point at infinity (Ch. III, §1, No. 2, Prop. 3), is therefore the extension by continuity of the integral to $\mathscr{C}^0(X; F)$ .

8. Integration with respect to a measure with compact support

Let $\mu$ be a measure on $X$ whose support $S = \operatorname{Supp}(\mu)$ is compact; the open set $X - S$ is negligible (§2, No. 2, Prop. 5). For every function $f$ with values in a vector space $F$ or in $\overline{\mathbf{R}}$ , the functions $f$ and $f \varphi_S$ are therefore equivalent (§2, No. 4); for $f$ to be $\mu$ -integrable (when $F$ is a Banach space), it is therefore necessary and sufficient that $f \varphi_S$ be so, in which case (No. 1)

\int f\, d\mu = \int f \varphi_S\, d\mu.

If, moreover, $f$ is bounded on $S$ , it follows from (20) that

\left| \int f\, d\mu \right| \leq \|\mu\| \cdot \sup_{x \in S} |f(x)|.

In particular, if $f$ is continuous on $X$ then $f$ is $\mu$ -integrable, since $fh \in \mathscr{K}(X; F)$ for every function $h \in \mathscr{K}(X; \mathbf{R})$ equal to 1 on $S$ (Ch. III, §1, No. 2, Lemma 1). More precisely:

Proposition 14

— Let X be a locally compact space, F a Banach space not reduced to 0; equip the space $\mathscr{C}(X; F)$ of all continuous mappings of X into F with the topology of compact convergence. For a measure $\mu$ on X to be such that the linear mapping $f \mapsto \int f d\mu$ of $\mathscr{K}(X; F)$ into F is extendible to a continuous linear mapping of $\mathscr{C}(X; F)$ into F, it is necessary and sufficient that $\operatorname{Supp}(\mu)$ be compact; such an extension is unique and coincides with the integral defined in No. 1.

We have just seen that if $\mu$ has compact support, then the integral $\int f d\mu$ is defined for every function $f \in \mathscr{C}(X; F)$ and that the mapping $f \mapsto \int f d\mu$ of $\mathscr{C}(X; F)$ into F is continuous for the topology of compact convergence. Conversely, suppose that $f \mapsto \int f d\mu$ is continuous in $\mathscr{K}(X; F)$ for the topology of compact convergence. Then, there is a compact set $K \subset X$ and a number $a > 0$ such that $|\mu(f)| \leq a \cdot \sup_{x \in K} |f(x)|$ for every function $f \in \mathscr{K}(X; F)$ ; in particular, if the support of $g \in \mathscr{K}(X; F)$ does not intersect K, then $\mu(g) = 0$ . Taking $g = h a$ , where $a \neq 0$ is a vector in F and $h \in \mathscr{K}(X; \mathbf{C})$ , we see that $\mu(h) = 0$ for every function $h \in \mathscr{K}(X; \mathbf{C})$ whose support does not intersect K, which proves that $\operatorname{Supp}(\mu) \subset K$ . Finally, the uniqueness of the extension follows from the fact that $\mathscr{K}(X; F)$ is dense in $\mathscr{C}(X; F)$ for the topology of compact convergence (Ch. III, §1, No. 2, Prop. 4).

Prop. 14 permits identifying a measure on X with compact support with its continuous extension to $\mathscr{C}(X; \mathbf{C})$ . The set of measures on X with compact support may therefore be identified with the dual $\mathscr{C}'(X; \mathbf{C})$ of the Hausdorff locally convex space $\mathscr{C}(X; \mathbf{C})$ . Recall that $\mathscr{C}(X; \mathbf{C})$ is complete (GT, X, §1, No. 6, Cor. 3 of Th. 2), but it is not necessarily barreled (Exer. 17). However, if X is countable at infinity, hence is the union of an increasing sequence of compact sets $K_n$ such that $K_n \subset \overset{\circ}{K}_{n+1}$ , then the topology of $\mathscr{C}(X; \mathbf{C})$ can be defined by the countable family of semi-norms $p_n(f) = \sup_{x \in K_n} |f(x)|$ , therefore $\mathscr{C}(X; \mathbf{C})$ is a Fréchet space in this case. Consequently, for every covering $\mathscr{G}$ of $\mathscr{C}(X; \mathbf{C})$ by bounded sets, the space $\mathscr{C}'(X; \mathbf{C})$ is then quasi-complete for the $\mathscr{G}$ -topology (TVS, III, §4, No. 2, Cor. 4 of Th. 1).

We shall consider above all on $\mathscr{C}'(X; \mathbf{C})$ the topology of compact convergence (the topology of uniform convergence on the compact subsets of $\mathscr{C}(X; \mathbf{C})$ ). Recall that the relatively compact subsets H of $\mathscr{C}(X; \mathbf{C})$ are characterized by the following properties (GT, X, §2, No. 5, Cor. 3 of Th. 2):

$1^\circ$ H is equicontinuous;

$2^\circ$ for every $x \in X$ , the set $H(x)$ of the $f(x)$ , where $f$ runs over H, is bounded in $\mathbf{C}$ .

Proposition 15

— Let X be a locally compact space and, for every $x \in X$ , let $\varepsilon_x$ be the Dirac measure at the point x. The mapping $x \mapsto \varepsilon_x$ of X into $\mathscr{C}'(X; \mathbf{C})$ is continuous for the topology of compact convergence on $\mathscr{C}'(X; \mathbf{C})$ .

Consider a neighborhood of $\varepsilon_{x_0}$ in $\mathscr{C}'(X; \mathbf{C})$ for this topology, which we can suppose to be defined by taking a number $\delta > 0$ , a compact subset H of $\mathscr{C}(X; \mathbf{C})$ , and considering the set of measures $\mu$ on X with compact support such that $|\mu(f) - \varepsilon_{x_0}(f)| \leq \delta$ for every function $f \in H$ . Since H is equicontinuous, there exists a neighborhood U of $x_0$ in X such that the relation $f \in H$ implies $|f(x) - f(x_0)| \leq \delta$ for all $x \in U$ , which may also be written $|\varepsilon_x(f) - \varepsilon_{x_0}(f)| \leq \delta$ and proves the proposition.(*)

Proposition 16

— Let K be a compact subset of X, L the vector space of measures $\mu$ on X with support contained in K. On L, the topologies induced by the topology $\mathscr{T}$ of compact convergence on $\mathscr{C}'(X; \mathbf{C})$ and the topology $\mathscr{T}'$ of strictly compact convergence on $\mathscr{M}(X; \mathbf{C})$ (Ch. III, §1, No. 10) coincide.

It is clear that on L, the topology induced by $\mathscr{T}$ is finer than the topology induced by $\mathscr{T}'$ . Conversely, let H be a compact subset of $\mathscr{C}(X; \mathbf{C})$ , h a function in $\mathscr{K}(X; \mathbf{C})$ equal to 1 on K. It is clear that the set H' of functions $fh$ , where $f$ runs over H, is strictly compact in $\mathscr{K}(X; \mathbf{C})$ , and, for every measure $\mu \in L$ , $\mu(f) = \mu(fh)$ for every function $f \in H$ , whence the conclusion.

Corollary 1

— For every compact subset K of X and every number $a > 0$ , the set B of measures $\mu$ on X such that $\operatorname{Supp}(\mu) \subset K$ and $\| \mu \| \leq a$ is an equicontinuous subset of $\mathscr{C}'(X; \mathbf{C})$ that is compact for the topology $\mathscr{T}$ of compact convergence.

For, let H be a subset of $\mathscr{C}(X; \mathbf{C})$ consisting of functions that are uniformly bounded on K; there exists a number $c > 0$ such that $|\mu(f)| \leq c \cdot \| \mu \| \leq ac$ for every function $f \in H$ and every measure $\mu \in B$ , by virtue of (22); therefore $B \subset acH^\circ$ in the dual $\mathscr{C}'(X; \mathbf{C})$ of $\mathscr{C}(X; \mathbf{C})$ , which proves the equicontinuity of B; the fact that B is compact for $\mathscr{T}$ follows from the fact that, on B, $\mathscr{T}$ and the vague topology induce the same topology (Prop. 16 and Ch. III, §1, No. 10, Prop. 17) and the fact that B is vaguely compact (Ch. III, §1, No. 9, Cor. 2 of Prop. 15 and §2, No. 2, Prop. 6).

Corollary 2

— Every measure with compact support (resp. every positive measure with compact support) $\mu$ is in the closure in $\mathscr{C}'(X; \mathbf{C})$ , for the topology $\mathscr{T}$ of compact convergence, of the set of measures (resp.

(*) In fact, the mapping $x \mapsto \varepsilon_x$ is a homeomorphism of X into $\mathscr{C}'(X; \mathbf{C})$ (Ch. VI, §1, No. 6, Remark 1).

positive measures) whose support is finite and contained in Supp(\mu) and whose norm is equal to $\| \mu \|$ .

For, on the set B of measures $\nu$ such that $\mathrm{Supp}(\nu) \subset \mathrm{Supp}(\mu)$ and $\| \nu \| \leq \| \mu \|$ , the topology induced by the vague topology is identical to the topology induced by $\mathscr{T}$ , and the corollary therefore follows from Ch. III, §2, No. 4, Cors. 2 and 3 of Th. 1.

9. Clans and additive set functions

Definition 3

A nonempty set $\Phi$ of subsets of a set $A$ is said to be a clan if there exists an algebra $\mathscr{A}$ (over $\mathbf{R}$ ) consisting of real-valued functions defined on $A$ , such that the relations $M \in \Phi$ and $\varphi_M \in \mathscr{A}$ are equivalent.

Example

If $\mu$ is a measure on a locally compact space $X$ then the linear combinations, with real coefficients, of the characteristic functions of integrable sets form an algebra $\mathscr{A}$ , because, for any two integrable sets $M, N$ , the function $\varphi_M \varphi_N = \varphi_{M \cap N}$ is integrable (No. 5, Prop. 7); it then follows from Defs. 2 and 3 that the set of integrable subsets of $X$ is a clan.

Proposition 17

In order that a nonempty set $\Phi$ of subsets of a set $A$ be a clan, it is necessary and sufficient that it satisfy the following condition:

(CL) For every pair of sets $M, N$ belonging to $\Phi$ , the sets $M \cup N$ and $M \cap \mathbf{C}N$ belong to $\Phi$ .¹

The condition is necessary, by virtue of the relations

\varphi_{M \cup N} = \varphi_M + \varphi_N - \varphi_M \varphi_N , \quad \varphi_{M \cap \mathbf{C}N} = \varphi_M - \varphi_M \varphi_N .

To show that it is sufficient, we first observe that it implies that, for any two sets $M, N$ in $\Phi$ , $M \cap N$ belongs to $\Phi$ since $M \cap N = M \cap \mathbf{C}(M \cap \mathbf{C}N)$ . Let $\mathscr{E}(\Phi)$ be the set of linear combinations, with real coefficients, of the characteristic functions of the sets of $\Phi$ . Since $\varphi_M \varphi_N = \varphi_{M \cap N}$ , $\mathscr{E}(\Phi)$ is an algebra. Everything comes down to showing that if $M$ is a subset of $A$ such that $\varphi_M = \sum_i c_i \varphi_{M_i}$ , where the $M_i$ belong to $\Phi$ , then $M \in \Phi$ . This will result from the following lemma:

Lemma

Let $\Phi$ be a nonempty set of subsets of $A$ satisfying the axiom (CL). Given a finite family $(M_i)_{1 \leq i \leq n}$ of sets in $\Phi$ , there exists a

¹ A clan $\Phi$ of subsets of a set $A$ is also known as a ring (or Boolean ring) of sets; if, moreover, $A \in \Phi$ , then $\Phi$ is called an algebra (or Boolean algebra) of sets (cf. GT, I, §6, Exer. 20 and II, §4, Exer. 12). A Boolean algebra closed under countable unions is called a tribe, or $\sigma$ -*algebra* (GT, IX, §6, No. 3, Def. 3).

finite family $(N_j)_{1 \leq j \leq m}$ of pairwise disjoint sets in $\Phi$ such that each $M_i$ is the union of a certain number of the $N_j$ .

For, consider the $2^n - 1$ sets of the form $\bigcap_{i=1}^n P_i$ , where $P_i = M_i$ for certain indices $i$ , $P_i = \mathbf{C} M_i$ for the others, at least one of the $P_i$ being equal to $M_i$ . Let $(N_j)_{1 \leq j \leq m}$ be the sequence of these sets arranged in some order; they are pairwise disjoint and belong to $\Phi$ ; on the other hand, every set $M_k$ is the union of the sets $N_j = \bigcap_{i=1}^n P_i$ corresponding to the families $(P_i)$ such that $P_k = M_k$ , which proves the lemma.

The lemma established, every function of the form $\sum_{i=1}^n c_i \varphi_{M_i}$ , where $M_i \in \Phi$ , may be written in the form $\sum_{j=1}^m d_j \varphi_{N_j}$ , where the $N_j$ belong to $\Phi$ and are pairwise disjoint; if $\varphi_M = \sum_{j=1}^m d_j \varphi_{N_j}$ then necessarily $d_j = 0$ or $d_j = 1$ for each index $j$ , thus $M$ is the union of a certain number of the $N_j$ , and therefore belongs to $\Phi$ .

Every clan $\Phi$ of subsets of $A$ contains the empty subset $\varnothing$ of $A$ ; for, there exists at least one subset $M \in \Phi$ , therefore $M - M = \varnothing$ belongs to $\Phi$ . Note also that the set of subsets of $A$ consisting of the single subset $\varnothing$ is a clan.

Definition 4

Given a clan $\Phi$ of subsets of a set $A$ , and a Banach space $F$ , one calls step function*² over the sets of $\Phi$ (or $\Phi$ -step function), with values in $F$ , every function of the form $\sum_i a_i \varphi_{M_i}$ , *where the $a_i$ belong to $F$ , and the $M_i$ to $\Phi$ .

It is clear that the set $\mathscr{E}_F(\Phi)$ of $\Phi$ -step functions with values in $F$ is a vector space over $\mathbf{R}$ or $\mathbf{C}$ . We have just seen in Prop. 17 that the set $\mathscr{E}(\Phi)$ of real-valued $\Phi$ -step functions is an algebra over $\mathbf{R}$ ; it is also the linear subspace of $\mathbf{R}^A$ generated by the characteristic functions of the sets of $\Phi$ .

By the Lemma, every function in $\mathscr{E}_F(\Phi)$ may be written $f = \sum_j c_j \varphi_{N_j}$ , where the $N_j \in \Phi$ are pairwise disjoint; from this it follows that $|f| = \sum_j |c_j| \varphi_{N_j}$ belongs to $\mathscr{E}(\Phi)$ . In particular, $\mathscr{E}(\Phi)$ is a Riesz space, since the upper envelope of two functions in $\mathscr{E}(\Phi)$ belongs to $\mathscr{E}(\Phi)$ .

Remark

— It is easily seen that Def. 4 is equivalent to the following: a $\Phi$ -step function with values in $F$ is a function $f$ that takes on only a finite number of values and which is such that, for every $a \neq 0$ in $F$ , the set $f(a)^{-1}$ belongs to $\Phi$ .

² Fonction étagée, whence the notation $\mathscr{E}(\Phi)$ in what follows.

Definition 5

— A real-valued function $\lambda$ defined on a clan $\Phi$ of subsets of a set $A$ is said to be additive if, for every pair $M, N$ of disjoint sets belonging to $\Phi$ , $\lambda(M \cup N) = \lambda(M) + \lambda(N)$ .

It follows in particular from this definition that $\lambda(\varnothing) = 0$ .

Proposition 18

— Let $\lambda$ be an additive set function defined on a clan $\Phi$ . There exists one and only one linear form (again denoted $\lambda$ ) on the vector space $\mathscr{E}(\Phi)$ of real-valued $\Phi$ *-step functions*, such that $\lambda(\varphi_M) = \lambda(M)$ for every set $M \in \Phi$ ; if, moreover, $\lambda(M) \geqslant 0$ for every $M \in \Phi$ , then $\lambda$ is a positive linear form on $\mathscr{E}(\Phi)$ .

The uniqueness of the linear form $\lambda$ is clear, since the characteristic functions of the sets in $\Phi$ generate the vector space $\mathscr{E}(\Phi)$ . To prove the existence of $\lambda$ , it suffices to prove that the relation $\sum_i c_i \varphi_{M_i} = 0$ , where the $M_i$ are nonempty sets belonging to $\Phi$ , implies $\sum_i c_i \lambda(M_i) = 0$ .

Now, by the Lemma there exists a finite family $(N_j)$ of pairwise disjoint nonempty sets in $\Phi$ such that, for every index $i$ , $\varphi_{M_i} = \sum_j a_{ij} \varphi_{N_j}$ with $a_{ij} = 0$ or $a_{ij} = 1$ . The relation $\sum_i c_i \varphi_{M_i} = 0$ , which may be written $\sum_j \left( \sum_i c_i a_{ij} \right) \varphi_{N_j} = 0$ , therefore implies that $\sum_i c_i a_{ij} = 0$ for every index $j$ . By Def. 5, we then have

\sum_i c_i \lambda(M_i) = \sum_j \left( \sum_i c_i a_{ij} \right) \lambda(N_j) = 0,

which proves the existence of $\lambda$ . Finally, suppose that $\lambda(M) \geqslant 0$ for every $M \in \Phi$ ; for every function $f \in \mathscr{E}(\Phi)$ , one can write $f = \sum_i c_i \varphi_{M_i}$ , where the $M_i \in \Phi$ are pairwise disjoint; if $f \geqslant 0$ , it follows that $c_i \geqslant 0$ for every index $i$ such that $M_i$ is nonempty, whence $\lambda(f) = \sum_i c_i \lambda(M_i) \geqslant 0$ .

10. Approximation of continuous functions by step functions

Proposition 19

— Let $X$ be a locally compact space, $\Phi$ a clan of subsets of $X$ , containing the set of compact subsets of $X$ . For every continuous mapping $f$ of $X$ into a Banach space $F$ (resp. every continuous, real-valued function $f \geqslant 0$ on $X$ ) with compact support $K$ , there exists a sequence $(g_n)$ of functions in $\mathscr{E}_F(\Phi)$ with support contained in $K$ (resp. a sequence $(g_n)$ of functions in $\mathscr{E}(\Phi)$ such that $0 \leqslant g_n \leqslant f$ for every $n$ ) that converges uniformly to $f$ *(resp. $f$ ).

Since $f$ is uniformly continuous on $K$ , one can cover $K$ by a finite number of compact sets $M_i$ ( $1 \leq i \leq m$ ) such that the oscillation of $f$ on each $M_i$ is $\leq 1/n$ . Since the $M_i$ and $K$ belong to $\Phi$ , there exists a partition of $K$ into sets $N_j \in \Phi$ such that each of the sets $M_i \cap K$ is the union of a certain number of the $N_j$ (No. 9, Lemma). Let $a_j$ be an element of $F$ such that $|f(x) - a_j| \leq 1/n$ on $N_j$ . Setting $g_n = \sum_j a_j \varphi_{N_j}$ , we have $|f - g_n| \leq 1/n$ , whence the proposition in this case. One argues similarly for a continuous real-valued function $f$ , on taking $a_j = \inf_{x \in N_j} f(x)$ and $g_n = \sum_j a_j \varphi_{N_j}$ .

Corollary 1

Let $\mu$ be a measure on $X$ ; the space $\mathscr{E}_F(\Phi)$ is dense in each of the spaces $\mathscr{L}_F^p$ ( $1 \leq p < +\infty$ ).

For, it follows from Prop. 19 and the criterion for convergence in mean for uniform limits of functions with compact support (§ 3, No. 3, Prop. 4) that $\mathscr{E}_F(\Phi)$ is dense, for the topology of convergence in mean of order $p$ , in the closure of the space $\mathscr{K}_F$ of continuous functions with compact support, whence the corollary.

Corollary 2

For every closed subset $S$ of $X$ , every function $f \in \mathscr{K}(X, S; \mathbf{C})$ is the uniform limit of linear combinations $\sum_i \lambda_i \varphi_{K_i}$ , where the $\lambda_i$ belong to $\mathbf{C}$ and the $K_i$ are compact subsets of $S$ .

The set $\mathscr{A}$ of such linear combinations is a $\mathbf{C}$ -algebra. Let $\Phi$ be the set of subsets $M$ of $X$ such that $\varphi_M \in \mathscr{A}$ ; $\Phi$ is thus a clan all of whose elements are subsets of $S$ , containing the compact subsets of $S$ , and $\mathscr{E}_C(\Phi) \subset \mathscr{A}$ . It then suffices to apply Prop. 19 to the locally compact space $S$ and the clan $\Phi$ .

Corollary 3

If $\mu$ and $\nu$ are two measures on $X$ such that $\mu(K) = \nu(K)$ for every compact subset $K$ of $X$ , then $\mu = \nu$ .

For, it follows from Cor. 2 and the definition of a measure that, for every compact subset $S$ of $X$ , $\mu$ and $\nu$ take on the same values in $\mathscr{K}(X, S; \mathbf{C})$ .

11. Extension of a measure defined on a family of sets

Let $\Phi$ be a nonempty set of subsets of a locally compact space $X$ . Given a real-valued function $M \mapsto \alpha(M)$ , defined and $\geq 0$ on $\Phi$ , we propose to seek conditions under which there exists a positive measure $\mu$ on $X$ such that the sets of $\Phi$ are $\mu$ -integrable and $\mu(M) = \alpha(M)$ for all $M \in \Phi$ . We shall limit ourselves to considering the case that the set $\Phi$ satisfies the following conditions:

(PC_I) The union and intersection of two sets of $\Phi$ belong to $\Phi$ .

(PC_{II}) For every pair consisting of a compact set K and an open set U in X such that $K \subset U$ , there exists a set $M \in \Phi$ such that $K \subset M \subset U$ .

Note that the condition (PC_{II}) implies that $\varnothing \in \Phi$ , by taking $K = U = \varnothing$ . However, the set $\Phi$ is not necessarily a clan; for example, the set of all compact subsets of X satisfies the conditions (PC_I) and (PC_{II}), but in general is not a clan, because if M and N are compact, the same is not in general true of $M \cap \mathbf{C}N$ .

We shall assume in addition that the function $\alpha$ defined on $\Phi$ satisfies the following conditions (obviously necessary for the problem to have a solution):

(PM_I) The relation $M \subset N$ implies $\alpha(M) \leq \alpha(N)$ . (PM_{II}) For any M and N in $\Phi$ , $\alpha(M \cup N) \leq \alpha(M) + \alpha(N)$ . (PM_{III}) The relation $M \cap N = \varnothing$ implies $\alpha(M \cup N) = \alpha(M) + \alpha(N)$ .

On taking $N = \varnothing$ in the condition (PM_{III}), one deduces that $\alpha(\varnothing) = 0$ ; the condition (PM_I) then shows that $\alpha(M) \geq 0$ for every $M \in \Phi$ .

Theorem 5

Let $\Phi$ be a set of subsets of a locally compact space X, satisfying (PC_I) and (PC_{II}), and let $\alpha$ be a real-valued function, defined on $\Phi$ , satisfying the conditions (PM_I), (PM_{II}) and (PM_{III}). In order that there exist a positive measure $\mu$ on X such that the sets of $\Phi$ are $\mu$ -integrable and $\mu(M) = \alpha(M)$ for all $M \in \Phi$ , it is necessary and sufficient that $\alpha$ satisfy in addition the following condition:

(PM_{IV}) For every $\varepsilon > 0$ and every $M \in \Phi$ , there exist a compact set $K \subset M$ and an open set $U \supset M$ such that, for every $N \in \Phi$ satisfying the relation $K \subset N \subset U$ , one has $|\alpha(N) - \alpha(M)| \leq \varepsilon$ .

Moreover, if the condition (PM{IV}) is satisfied, then the measure $\mu$ is unique; for every compact set $K$ , $\mu(K) = \inf_{M \in \Phi, M \supset K} \alpha(M)$ ; for every open set $U$ , $\mu^*(U) = \sup_{M \in \Phi, M \subset U} \alpha(M)$ .*

Note that the condition (PM_{IV}) is equivalent to the conjunction of the following two:

(PM'_{IV}) For every $\varepsilon > 0$ and every $M \in \Phi$ , there exists an open set $U \supset M$ such that, for every $N \in \Phi$ contained in $U$ , $\alpha(N) \leq \alpha(M) + \varepsilon$ . (PM''_{IV}) For every $\varepsilon > 0$ and every $M \in \Phi$ , there exists a compact set $K \subset M$ such that, for every $N \in \Phi$ containing $K$ , $\alpha(N) \geq \alpha(M) - \varepsilon$ .

For, it is obvious that (PM'_{IV}) and (PM''_{IV}) imply (PM_{IV}). Conversely, let us show for example that (PM_{IV}) implies (PM'_{IV}): let K be a compact set and U an open set such that $K \subset M \subset U$ and $|\alpha(P) - \alpha(M)| \leq \varepsilon$ for every $P \in \Phi$ satisfying $K \subset P \subset U$ . Then, if $N \in \Phi$ and $N \subset U$ , $M \cup N$ belongs to $\Phi$ and $K \subset M \cup N \subset U$ , whence $\alpha(M \cup N) \leq \alpha(M) + \varepsilon$ and a fortiori $\alpha(N) \leq \alpha(M) + \varepsilon$ .

When the set $\Phi$ , satisfying (PC_I) and (PC_{II}), consists of compact sets, then the condition (PM''_{IV}) is trivially verified, and (PM_{IV}) is then equivalent to (PM'_{IV}).

The condition (PM_{IV}) is necessary: this follows at once from Th. 4 of No. 6 on the 'approximation' of an integrable set by a compact set and an open set. To prove the other assertions of the theorem, we proceed in several steps.

1° Definition of a topology on $\mathfrak{P}(X)$ .

For every pair $(K, U)$ consisting of a compact set $K$ and an open set $U$ in $X$ , we denote by $I(K, U)$ the set of subsets $M \subset X$ such that $K \subset M \subset U$ ; in order that $I(K, U)$ be nonempty, it is necessary and sufficient that $K \subset U$ . If $(K', U')$ is a second pair, formed by a compact set $K'$ and an open set $U'$ , we have

I(K, U) \cap I(K', U') = I(K \cup K', U \cap U').

Let

T

be the topology on

\mathfrak{P}(X)

generated by the set of subsets

I(K, U)

K

runs over the set of compact subsets of

X

, and

U

over the set of open subsets of

X

; by the foregoing, the

I(K, U)

form a base for the topology

T

(GT, I, §1, No. 3).

We observe that the definition of $T$ implies that, in $\mathfrak{P}(X)$ , the set of compact subsets of $X$ is dense. The condition (PC_{II}) expresses that $\Phi$ is dense in $\mathfrak{P}(X)$ , and condition (PM_{IV}) expresses that the function $\alpha$ is continuous on $\Phi$ for the topology induced by $T$ . Finally, Th. 4 of No. 6 expresses that the function $M \mapsto \mu(M)$ is continuous on the clan of $\mu$ -integrable sets, for the topology induced by $T$ .

2° Uniqueness of $\mu$ .

We denote by $\overline{\Phi}$ the set of subsets $M \subset X$ such that $\alpha(N)$ tends to a finite limit as $N$ tends to $M$ (for the topology $T$ ) while remaining in $\Phi$ ; we may then extend $\alpha$ in only one way to a continuous mapping $\overline{\alpha}$ of $\overline{\Phi}$ into $\mathbf{R}$ (GT, I, §8, No. 5, Th. 1). If there exists a measure $\mu$ meeting the requirements, the above remarks prove that the clan $\Psi$ of $\mu$ -integrable sets is contained in $\overline{\Phi}$ and that $\mu(M) = \overline{\alpha}(M)$ for every $M \in \Psi$ ; this relation holds in particular for every compact subset $M$ of $X$ , which proves the uniqueness of $\mu$ (No. 10, Cor. 3 of Prop. 19).

3° Extension of $\alpha$ to the compact sets.

Without assuming the existence of $\mu$ , we are now going to study the set $\overline{\Phi}$ and the extension $\overline{\alpha}$ of $\alpha$ to $\overline{\Phi}$ . We first show that every compact set $K$ belongs to $\overline{\Phi}$ and that $\overline{\alpha}(K) = \inf_{P \in \Phi, P \supset K} \alpha(P)$ . Set $a = \inf_{P \in \Phi, P \supset K} \alpha(P)$ ; for every $\varepsilon > 0$ , there exists an $M \in \Phi$ such that $K \subset M$ and $\alpha(M) \leq a + \varepsilon$ . By (PM'_{IV}), there exists an open set $U \supset M$ such that, for every $N \in \Phi$ contained in $U$ , we have $\alpha(N) \leq \alpha(M) + \varepsilon \leq a + 2\varepsilon$ ; for every $N \in \Phi$ such that $K \subset N \subset U$ , we therefore have $a \leq \alpha(N) \leq a + 2\varepsilon$ , which, by the definitions, shows that $K \in \overline{\Phi}$ and $\overline{\alpha}(K) = a$ .

This result proves at once that if $K_1$ and $K_2$ are two compact sets such that $K_1 \subset K_2$ , then $\overline{\alpha}(K_1) \leq \overline{\alpha}(K_2)$ . If $K_1$ and $K_2$ are any two compact sets, we have $\overline{\alpha}(K_1 \cup K_2) \leq \overline{\alpha}(K_1) + \overline{\alpha}(K_2)$ by (PM_{II}). We shall see, moreover, that if $K_1$ and $K_2$ are disjoint then $\overline{\alpha}(K_1 \cup K_2) = \overline{\alpha}(K_1) + \overline{\alpha}(K_2)$ . For, there then exist two disjoint open sets $U_1, U_2$ such that $K_1 \subset U_1$ , $K_2 \subset U_2$ (GT, II, §4, Prop. 4). Therefore, by (PC_{II}), there also exist two sets $M_1 \in \Phi$ , $M_2 \in \Phi$ such that $K_1 \subset M_1 \subset U_1$ and $K_2 \subset M_2 \subset U_2$ . Now let $P$ be any set of $\Phi$ containing $K_1 \cup K_2$ ; the union of the two sets $P \cap M_1$ and $P \cap M_2$ belongs to $\Phi$ by (PC_{I}), and since these two sets are disjoint, application of (PM_I) and (PM_{III}) yields

\alpha(P) \geq \alpha(P \cap M_1) + \alpha(P \cap M_2) \geq \overline{\alpha}(K_1) + \overline{\alpha}(K_2),

which establishes our assertion.

4° Extension of $\alpha$ to the open sets.

We shall now see that, for an open set $U$ to belong to $\overline{\Phi}$ , it is necessary and sufficient that, as $K$ runs over the set of compact subsets of $U$ , the supremum of the numbers $\overline{\alpha}(K)$ is finite; moreover, $\overline{\alpha}(U)$ is then equal to this supremum.

For, let $U$ be an open set belonging to $\overline{\Phi}$ ; for every $\varepsilon > 0$ there exists a compact set $K \subset U$ such that, for every set $M \in \Phi$ satisfying $K \subset M \subset U$ , one has $|\overline{\alpha}(U) - \alpha(M)| \leq \varepsilon$ , whence $|\overline{\alpha}(U) - \overline{\alpha}(K)| \leq \varepsilon$ ; on the other hand, if $K'$ is any compact set contained in $U$ , then $K \subset K \cup K' \subset U$ , whence $|\overline{\alpha}(U) - \overline{\alpha}(K \cup K')| \leq \varepsilon$ and so $\overline{\alpha}(U) \geq \overline{\alpha}(K \cup K') - \varepsilon \geq \overline{\alpha}(K') - \varepsilon$ ; $\overline{\alpha}(U)$ is therefore indeed equal to the supremum of the numbers $\overline{\alpha}(K)$ as $K$ runs over the set of compact subsets of $U$ .

Conversely, let $U$ be an open set such that $b = \sup_{K \subset U} \overline{\alpha}(K) < +\infty$ ( $K$ running over the set of compact subsets of $U$ ), and let us show that $U \in \overline{\Phi}$ . For every $\varepsilon > 0$ , there exists a compact set $K \subset U$ such that $b - \varepsilon \leq \overline{\alpha}(K) \leq b$ ; by (PM''_{IV}), for every set $M \in \Phi$ such that $K \subset M \subset U$ , there exists a compact set $K' \subset M$ such that

\alpha(M) \leq \overline{\alpha}(K') + \varepsilon \leq b + \varepsilon;

therefore $b - \varepsilon \leq \alpha(M) \leq b + \varepsilon$ , which proves that $U \in \overline{\Phi}$ .

From this characterization of the open sets $U \in \overline{\Phi}$ , and of $\overline{\alpha}(U)$ , it follows first of all that if $U_1$ and $U_2$ are two open sets such that $U_1 \subset U_2$ and $U_2 \in \overline{\Phi}$ , then $U_1 \in \overline{\Phi}$ and $\overline{\alpha}(U_1) \leq \overline{\alpha}(U_2)$ . On the other hand, if $U_1$ and $U_2$ are two open sets belonging to $\overline{\Phi}$ , then the same is true of $U_1 \cup U_2$ , and $\overline{\alpha}(U_1 \cup U_2) \leq \overline{\alpha}(U_1) + \overline{\alpha}(U_2)$ . For, let $K$ be any compact set contained in $U_1 \cup U_2$ ; for every point $x \in K$ , there exists a compact neighborhood of $x$ contained in either $U_1$ or $U_2$ ; one can therefore cover $K$ by a finite number of these neighborhoods; if $K_1$ (resp. $K_2$ ) is the union of those that are contained in $U_1$ (resp. $U_2$ ), then $K \subset K_1 \cup K_2$ , whence

\overline{\alpha}(K) \leq \overline{\alpha}(K_1 \cup K_2) \leq \overline{\alpha}(K_1) + \overline{\alpha}(K_2) \leq \overline{\alpha}(U_1) + \overline{\alpha}(U_2),

which establishes the asserted property.

5° Properties of $\overline{\Phi}$ and $\overline{\alpha}$ .

The definition of $\overline{\Phi}$ and $\overline{\alpha}$ can now be transformed as follows (taking into account (PC_{II})): in order that $M \in \overline{\Phi}$ , it is necessary and sufficient that, for every $\varepsilon > 0$ , there exist a compact set $K$ and an open set $U \in \overline{\Phi}$ such that $K \subset M \subset U$ and $\overline{\alpha}(U) - \overline{\alpha}(K) \leq \varepsilon$ ; $\overline{\alpha}(M)$ is, moreover, the infimum of the $\overline{\alpha}(U)$ for the open sets $U \in \overline{\Phi}$ containing $M$ , and the supremum of the $\overline{\alpha}(K)$ for the compact sets $K \subset M$ .

From this, we shall first deduce that if $M_1, M_2$ and $M_1 \cup M_2$ belong to $\overline{\Phi}$ , then $\overline{\alpha}(M_1 \cup M_2) \leq \overline{\alpha}(M_1) + \overline{\alpha}(M_2)$ . Indeed, if $U_1$ and $U_2$ are two open sets of $\overline{\Phi}$ containing $M_1$ and $M_2$ , respectively, and such that $\overline{\alpha}(U_1) \leq \overline{\alpha}(M_1) + \varepsilon$ and $\overline{\alpha}(U_2) \leq \overline{\alpha}(M_2) + \varepsilon$ , then $U_1 \cup U_2$ belongs to $\overline{\Phi}$ , contains $M_1 \cup M_2$ , and consequently

\overline{\alpha}(M_1 \cup M_2) \leq \overline{\alpha}(U_1 \cup U_2) \leq \overline{\alpha}(U_1) + \overline{\alpha}(U_2) \leq \overline{\alpha}(M_1) + \overline{\alpha}(M_2) + 2\varepsilon,

whence our assertion.

Next, let us show that if $K$ is a compact set and $U$ is an open set of $\overline{\Phi}$ such that $K \subset U$ , then $\overline{\alpha}(U - K) = \overline{\alpha}(U) - \overline{\alpha}(K)$ . By the foregoing, we have $\overline{\alpha}(U) \leq \overline{\alpha}(K) + \overline{\alpha}(U - K)$ . On the other hand, for every compact set $K' \subset U - K$ ,

\overline{\alpha}(K \cup K') = \overline{\alpha}(K) + \overline{\alpha}(K') \leq \overline{\alpha}(U);

since $U - K$ is open and belongs to $\overline{\Phi}$ , $\overline{\alpha}(U - K)$ is the supremum of the $\overline{\alpha}(K')$ , which shows that $\overline{\alpha}(K) + \overline{\alpha}(U - K) \leq \overline{\alpha}(U)$ .

The definition of $\overline{\Phi}$ may therefore now be expressed in the following way: in order that $M \in \overline{\Phi}$ , it is necessary and sufficient that, for every $\varepsilon > 0$ , there exist a compact set $K$ and an open set $U \in \overline{\Phi}$ such that $K \subset M \subset U$ and $\overline{\alpha}(U - K) \leq \varepsilon$ .

We are now in a position to prove that $\overline{\Phi}$ is a clan and $\overline{\alpha}$ an additive set function on $\overline{\Phi}$ . We first show that if $M$ and $N$ belong to $\overline{\Phi}$ then so do $M \cap \mathbf{C}N$ and $M \cup N$ . By hypothesis, for every $\varepsilon > 0$ there exist two compact sets $K, K'$ and two open sets $U, U'$ of $\overline{\Phi}$ such that

K \subset M \subset U,\quad K' \subset N \subset U',\quad \overline{\alpha}(U - K) \leq \varepsilon,\quad \overline{\alpha}(U' - K') \leq \varepsilon.

The set $K'' = K \cap \mathbf{C}U'$ is compact, the set $U'' = U \cap \mathbf{C}K'$ is open and belongs to $\overline{\Phi}$ , and $K'' \subset M \cap \mathbf{C}N \subset U''$ ; on the other hand, $U'' - K''$ is contained in the union of $U \cap \mathbf{C}K$ and $U' \cap \mathbf{C}K'$ , whence $\overline{\alpha}(U'' - K'') \leq 2\varepsilon$ , which proves that $M \cap \mathbf{C}N \in \overline{\Phi}$ . Similarly, $U_1 = U \cup U'$ is open and belongs to $\overline{\Phi}$ , $K_1 = K \cup K'$ is compact, and $K_1 \subset M \cup N \subset U_1$ ; on the other hand, $U_1 - K_1$ is contained in the union of $U - K$ and $U' - K'$ , whence again $\overline{\alpha}(U_1 - K_1) \leq 2\varepsilon$ , and $M \cup N$ belongs to $\overline{\Phi}$ . Finally, if $M$ and $N$ are disjoint, then

\overline{\alpha}(K_1) = \overline{\alpha}(K) + \overline{\alpha}(K') \geq \overline{\alpha}(M) + \overline{\alpha}(N) - 2\varepsilon,

consequently $\overline{\alpha}(M \cup N) \geq \overline{\alpha}(M) + \overline{\alpha}(N) - 2\varepsilon$ ; since $\varepsilon$ is arbitrary, we have $\overline{\alpha}(M \cup N) = \overline{\alpha}(M) + \overline{\alpha}(N)$ .

6° Existence of the measure $\mu$ .

By Prop. 18 of No. 9, there exists one and only one positive linear form $\beta$ on the vector space $\mathscr{E}(\overline{\Phi})$ of $\overline{\Phi}$ -step functions, such that $\beta(\varphi_M) = \overline{\alpha}(M)$ for all $M \in \overline{\Phi}$ . For every compact subset $K$ of $X$ , let us denote by $\mathscr{G}(K)$ the space of uniform limits of functions of $\mathscr{E}(\overline{\Phi})$ whose support is contained in $K$ . Since $\beta$ is positive, $|\beta(f)| \leq \overline{\alpha}(K) \cdot \|f\|$ for every function $f \in \mathscr{E}(\overline{\Phi})$ whose support is contained in $K$ ; the restriction of $\beta$ to the space of these functions is a continuous linear form for the topology of uniform convergence; it may therefore be extended to a positive continuous linear form $\overline{\beta}_K$ on $\mathscr{G}(K)$ . Moreover, if $K \subset K_1$ , then the restriction of $\overline{\beta}_{K_1}$ to $\mathscr{G}(K)$ is identical to $\overline{\beta}_K$ , therefore there exists a positive linear form $\overline{\beta}$ on the union $\mathscr{G}$ of the $\mathscr{G}(K)$ , that extends each of the forms $\overline{\beta}_K$ .

Now, since every compact set belongs to $\overline{\Phi}$ , the space $\mathscr{H}$ of continuous real-valued functions with compact support is a subspace of $\mathscr{G}$ (No. 10, Prop. 19); the restriction to $\mathscr{H}$ of the positive linear form $\overline{\beta}$ is therefore a positive measure $\mu$ . Let us show that for every compact set $K$ , $\mu(K) = \overline{\alpha}(K)$ . For every $\varepsilon > 0$ , there exists an open set $U \in \overline{\Phi}$ such that $K \subset U$ , $\mu(U) \leq \mu(K) + \varepsilon$ and $\overline{\alpha}(U) \leq \overline{\alpha}(K) + \varepsilon$ . Let $f$ be a continuous mapping of $X$ into $[0, 1]$ whose support is contained in $U$ and such that $f(x) = 1$ on $K$ (Ch. III, §1, No. 2, Lemma 1). Then $\mu(K) \leq \mu(f) \leq \mu(U) \leq \mu(K) + \varepsilon$ , and, on the other hand,

\overline{\alpha}(K) = \beta(\varphi_K) \leq \overline{\beta}(f) \leq \beta(\varphi_U) = \overline{\alpha}(U) \leq \overline{\alpha}(K) + \varepsilon;

since $\mu(f) = \overline{\beta}(f)$ , we see that $|\mu(K) - \overline{\alpha}(K)| \leq \varepsilon$ , and since $\varepsilon$ is arbitrary, $\mu(K) = \overline{\alpha}(K)$ .

The characterization of the open sets belonging to $\overline{\Phi}$ , combined with Cor. 4 of Th. 4 of No. 6, then shows that the open sets belonging to $\overline{\Phi}$ are none other than the $\mu$ -integrable open sets, and that, for such a set $U$ , we have $\mu(U) = \overline{\alpha}(U)$ . Th. 4 of No. 6 and the characterization of the sets of $\overline{\Phi}$ given in 5° then show that the $\mu$ -integrable sets are the sets of $\overline{\Phi}$ and that, for such a set $M$ , $\mu(M) = \overline{\alpha}(M)$ . Finally, the fact that $\mu^*(U) = \sup_{M \in \Phi, M \subset U} \alpha(M)$ for every open set $U$ follows at once from (PC $_{II}$ ) and Cor. 4 of Th. 4 of No. 6.

Theorem 5 is thus completely proved.

Corollary

— *Let X be a locally compact space with a countable base, $\Psi$ the set of Borel sets of X, $\beta$ a mapping of $\Psi$ into $[0, +\infty]$ satisfying the following conditions:

(i) If $(B_1, B_2, \ldots)$ is a sequence of pairwise disjoint Borel sets of X, then $\beta(B_1 \cup B_2 \cup \ldots) = \beta(B_1) + \beta(B_2) + \ldots$ .

(ii) If B is a compact subset of X, then $\beta(B) < +\infty$ .

Then, there exists one and only one positive measure $\mu$ on X such that $\beta(B) = \mu^*(B)$ for all $B \in \Psi$ .

Let $\Phi$ be the set of compact subsets of $X$ and let $\alpha$ be the restriction of $\beta$ to $\Phi$ . The conditions (PC $_I$ ), (PC $_{II}$ ), (PM $_I$ ), (PM $_{II}$ ), (PM $_{III}$ ) and (PM $_{IV}'$ ) are then satisfied. Let $K$ be a compact subset of $X$ , and $\varepsilon > 0$ . Then $K$ is the intersection of a decreasing sequence $(U_1, U_2, \ldots)$ of relatively compact open sets of $X$ (GT, IX, §2, No. 5, Prop. 7). We have $\sum_{n=1}^{\infty} \beta(U_n - U_{n+1}) = \beta(U_1 - K) < +\infty$ , therefore

\beta(U_n) - \beta(K) = \beta(U_n - K) = \sum_{p=n}^{\infty} \beta(U_p - U_{p+1})

tends to 0 as $n$ tends to $\infty$ . This proves that the condition $(\mathrm{PM}'_{\mathrm{IV}})$ is satisfied. By Th. 5, there exists a positive measure $\mu$ on $X$ such that $\mu(K) = \alpha(K)$ for every compact subset $K$ of $X$ . Since every open set $U$ of $X$ is the union of an increasing sequence of compact subsets, we have $\mu^*(U) = \beta(U)$ . Let $L$ be a compact subset of $X$ . By Prop. 7 of No. 5, the $\mu$ -integrable subsets of $L$ form a tribe of subsets of $L$ . Therefore, if $B$ is an element of $\Psi$ contained in $L$ , then $B$ is $\mu$ -integrable; for every $\varepsilon > 0$ , there then exist a compact set $K$ and an open set $U$ in $X$ such that $K \subset B \subset U$ and $\mu^*(U) - \mu(K) \leq \varepsilon$ (No. 6, Th. 4). Since $\beta(U) = \mu^*(U)$ and $\beta(K) = \mu(K)$ , we see that $|\mu^*(B) - \beta(B)| \leq 2\varepsilon$ . Therefore $\beta(B) = \mu^*(B)$ . Finally, every Borel set $C$ of $X$ is the union of a sequence of pairwise disjoint, relatively compact Borel sets, whence $\beta(C) = \mu^*(C)$ . The uniqueness of $\mu$ follows at once from Th. 5.

Exercises

See the exercises for § 4.