It suffices to treat the case that T is a subspace of a compact space X. Let V be an open subset of X such that $V \cap T = U$; denote by $V'$ an open set in X containing K such that $\overline{V'} \subset V$, by g a continuous function on X with values in I (resp. in $\mathbf{C}$) extending f and zero on $X - V$ (GT, IX, §4, No. 1, Prop. 1). The condition a) is satisfied by taking $U' = V' \cap T$, and b) by taking $f'$ to be the restriction of g to T.

(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 $t$.

Let us take up again the proof of the second part of Proposition 1, on setting $\theta = \eta - \eta'$. We can require the open set U to be integrable for $|\eta|$ and $|\eta'|$. The function $g$ is then integrable for these two measures, and the relation $\theta(g) = 0$ implies $a < 0$; therefore $|\theta|^*(f) = 0$ for every positive lower semi-continuous function $f$, whence finally $|\theta| = 0$, on taking $f = +\infty$.

By Prop. 15 of §1, No. 10, it suffices to show that if K is compact in T, and if $g$ is the extension to T by 0 of a function in $\mathscr{C}_+(K)$ between 0 and 1, then there exists a function $f \in \mathscr{C}_+^b(T)$, with support contained in a $\mu$-integrable open set, such that $\|f - g\|_p$ is arbitrarily small. Now, let $\varepsilon$ be a number $> 0$, U an open neighborhood of K such that $\mu^*(U - K) < \varepsilon$, V an open neighborhood of K such that $\overline{V} \subset U$, and $f$ a function with values in $[0, 1]$, continuous, equal to $g$ on K and to 0 outside V (Lemma). The function $|f - g|^p$ is then bounded above by $\varphi_{U-K}$; therefore $\|f - g\|_p \leq \varepsilon^{1/p}$, which establishes the proposition.

Remark 3). — There is an analogous statement for functions with values in a Banach space F: the subspace $\mathscr{H} \otimes F$ of $\mathscr{C}^b(T; F)$ is dense in $\mathscr{L}_F^p(\mu)$.

Necessity is obvious. To establish sufficiency, let us take up again the proof of the preceding proposition, on taking $p = 1$ and $\mu = |\theta|$; the notations being the same, the relation $\mu^*(|f - g|) \leq \varepsilon$ and the inequality $\theta(f) \geq 0$ imply $\theta_K(g_K) = \theta(g) \geq -\varepsilon$; since $g_K$ is an arbitrary element of $\mathscr{C}(K)$ between 0 and 1, the measure $\theta_K$ is positive; the compact set K being arbitrary, this means that $\theta$ is positive.

(M) *For every number $\varepsilon > 0$, there exists a compact subset K of T such that the relations $g \in \mathscr{C}^b(T; \mathbf{C}), |g| \leq 1, g_K = 0$ imply $|I(g)| \leq \varepsilon$.

The measure $\theta$ is then unique.*

Uniqueness follows from Prop. 2 of No. 1. Let us show that the condition (M) is necessary. Let $\theta$ be a bounded complex measure; let K be a compact set such that $|\theta|^*(T - K) \leq \varepsilon$ (§ 1, No. 2, Remark 3). The hypotheses $|g| \leq 1, g_K = 0$ imply $|g| \leq \varphi_{\mathbf{C}_K}$, therefore $|\theta(g)| \leq |\theta|^*(\varphi_{\mathbf{C}_K}) \leq \varepsilon$.

Let us pass to the proof of sufficiency. Let X be the Stone–Čech compactification of T (GT, IX, § 1, Exer. 7; or TG, IX, § 1, No. 6). For every function $f \in \mathscr{C}(X; \mathbf{C})$, set $\nu(f) = I(f_T)$; we define in this way a continuous linear form on $\mathscr{C}(X; \mathbf{C})$, that is, a complex measure on the compact space X. Let $\varepsilon$ be a number $> 0$, K a compact set satisfying (M); the function $\varphi_{\mathbf{C}_K}$ being lower semi-continuous and positive on X, the formula (2) gives us the following relations, where $\mathscr{G}$ denotes the set of functions $g \in \mathscr{C}(X; \mathbf{C})$ such that $|g| \leq \varphi_{\mathbf{C}_K}$:

Let $(K_n)_{n \geq 1}$ be a sequence of compact subsets of T, such that each $K_n$ satisfies (M) for $\varepsilon = 1/n$, and let $S = \bigcup K_n$; S is a Borel set in X, contained in T, and $|\nu|^*(X - T) \leq |\nu|^*(X - S) \leq |\nu|^*(X - K_n) \leq 1/n$ for all n, so that T is $\nu$-measurable and $\nu$ is concentrated on T. Let $f$ be a bounded continuous function on T; since X is the Stone–Čech compactification of T, $f$ may be extended by continuity to a function $g \in \mathscr{C}(X; \mathbf{C})$. Now let $\mu$ be the measure induced by $\nu$ on $T$; one has $\mu(f) = \nu(f^0)$. (1) Since $\nu$ is concentrated on $T$, the functions $f^0$ and $g$ are equal $\nu$-almost everywhere, therefore $\mu(f) = \nu(g) = I(g_T) = I(f)$, which completes the proof.

a) Let f be a lower semi-continuous numerical function $\geqslant 0$ defined on T; then the function $\mu \mapsto |\mu|^*(f)$ is lower semi-continuous on $\mathscr{M}^b(T)$.

b) Let f be an upper semi-continuous bounded function defined on T; then the function $\mu \mapsto \mu(f)$ is upper semi-continuous on $\mathscr{M}_+^b(T)$.

For, one sees by Prop. 1 b) of No. 1 that $\mu \mapsto |\mu|^*(f)$ is the upper envelope of a family of functions of the form $\mu \mapsto |\mu(g)|$ with $g \in \mathscr{C}^b(T)$, hence continuous for the tight topology. This establishes a). To prove b), it suffices to choose a constant upper bound C for f, and to write $\mu(f) = \mu(C) - \mu(C - f)$; the function $\mu \mapsto \mu(C)$ is continuous, and the function $\mu \mapsto \mu(C - f)$ is lower semi-continuous on $\mathscr{M}_+^b(T)$ by the foregoing.

For every $t \in T$, set $f'(t) = \liminf_{s \to t} f(s)$, $f''(t) = \limsup_{s \to t} f(s)$. Obviously $f' \leqslant f \leqslant f''$, with equality at every point of T where f is continuous (hence $\mu$-almost everywhere). On the other hand, $f'$ is lower semi-continuous, $f''$ is upper semi-continuous and bounded (GT, IV, §6, No. 2, Prop. 4). We therefore have the following relations by Prop. 6,

One concludes by observing that $\mu^*(f') = \mu^*(f'')$, because $f'$ and $f''$ are equal locally $\mu$-almost everywhere.

We denote again by i the mapping $\mu \mapsto i(\mu)$ of $\mathscr{M}_+^b(T)$ into $\mathscr{M}_+^b(X)$; i is injective (§2, No. 4, Prop. 8) and maps $\mathscr{M}_+^b(T)$ into W (§2, No. 3, Prop. 7). If $\lambda \in W$, then $\lambda = i(\lambda_T)$ (§2, No. 3, Prop. 7 b)). Consequently, i is a bijection of $\mathscr{M}_+^b(T)$ onto W, and the inverse bijection of i is the mapping $r : \lambda \mapsto \lambda_T$ on $W$. On the other hand, $i$ is continuous: for, if $f \in \mathscr{C}^b(X)$, then $\langle i(\mu), f \rangle = \langle \mu, f \circ i \rangle$, and $f \circ i$ belongs to $\mathscr{C}^b(T)$. Thus, everything comes down to showing that, for every measure $\mu \in W$ and every function $f \in \mathscr{C}^b_+(T)$, one has

or again

Let $f^\infty$ be the function on $X$ that coincides with $f$ on $T$ and with $+\infty$ on $X - T$, and let $f'$ and $f''$ be, respectively, the upper semi-continuous regularization of $f^0$ and the lower semi-continuous regularization of $f^\infty$ (GT, IV, §6, No. 2). The relations

immediately imply that $f'$ and $f''$ both coincide with $f$ and $f^0$ on $T$. Prop. 6 then yields

But one can replace $f'$ and $f''$ by $f^0$ in these two formulas, since the measures $\lambda$ and $\mu$ are carried by $T$; we have thus obtained the desired relation.

The statement of Prop. 8 is only valid for positive measures: the mapping $\mu \mapsto i(\mu)$ of $\mathscr{M}^b(T)$ into $\mathscr{M}^b(X)$ is injective and continuous, but is not in general a homeomorphism of $\mathscr{M}^b(T)$ onto its image. For example, take $X = \mathbf{R}$, $T = \mathbf{R} - \{0\}$; the measures $\lambda_t = \varepsilon_t - \varepsilon_{-t}$ ($t > 0$) converge tightly to 0 in $X$ as $t$ tends to 0, but do not converge tightly to 0 in $T$ (the characteristic function of $]0, +\infty[$ belongs to $\mathscr{C}^b(T)$) (cf. however the Cor. of Th. 1 of No. 5).

The condition is obviously necessary. To show that it is sufficient, let us denote by $X$ the Alexandroff compactification of $T$ (GT, I, §9, No. 8) and by $i$ the canonical injection of $T$ into $X$. By Prop. 8, everything comes down to showing that $\lambda \mapsto i(\lambda)$ converges tightly to $i(\mu)$ in $\mathscr{M}^b(X)$ with respect to $\mathfrak{F}$. Since $\mu(1) < +\infty$, there exists a set $A \in \mathfrak{F}$ such that the total masses of the measures in $A$ are bounded by a number $M$; it therefore suffices to verify that

for functions $g \in \mathscr{C}^b(X)$ forming a total set in $\mathscr{C}^b(X)$. Now, this equality is satisfied when $g$ has compact support in $T$, because of the vague convergence of $\mathfrak{F}$ to $\mu$, and also when $g$ is a constant function on $X$, from the fact that $\lim_{\lambda, \mathfrak{F}} \lambda(1) = \mu(1)$. Since the functions of the preceding two types form a total set in $\mathscr{C}^b(X)$ (Ch. III, §1, No. 2, Prop. 3), this completes the proof.

We begin by treating the case that $T$ is Polish and compact. The set $U$ of positive measures with mass $\leq 1$ is then compact (Ch. III, §1, No. 9, Cor. 2 of Prop. 15), and the topology induced on $U$ by the tight topology (which here coincides with the vague topology) is also induced by the topology of pointwise convergence on a total subset of $\mathscr{C}(T)$ (loc. cit., No. 10, Prop. 17). Now, there exists in $\mathscr{C}(T)$ a countable total set (GT, X, §3, No. 3, Th. 1); consequently, $U$ is a metrizable compact space. The set $V$ of positive measures of mass $< 1$ is open in $U$, hence is a Polish locally compact space. Now, the mapping $\mu \mapsto \frac{1}{1 + \mu(1)} \mu$ of $\mathscr{M}_+^b(T)$ onto $V$ is a homeomorphism, the mapping $\lambda \mapsto \frac{1}{1 - \lambda(1)} \lambda$ being the inverse homeomorphism.

Let us pass to the case that $T$ is Polish; we can suppose that $T$ is the intersection of a decreasing sequence $(G_n)$ of open sets in a metrizable compact space $X$ (GT, IX, §6, No. 1, Cor. 1 of Th. 1); the space $\mathscr{M}_+^b(T)$ is then homeomorphic to the subspace $W$ of $\mathscr{M}_+^b(X)$ consisting of the measures concentrated on $T$ (No. 3, Prop. 8), and it will suffice to show that $W$ is the intersection of a sequence of open sets in the Polish space $\mathscr{M}_+^b(X)$ (GT, loc. cit., Th. 1). Now, let $W_n$ be the set of measures $\mu \in \mathscr{M}_+^b(X)$ concentrated on $G_n$; the mapping $h_n : \mu \mapsto \mu^*(X - G_n)$ on $\mathscr{M}_+^b(X)$ is upper semi-continuous (No. 3, Prop. 6), and the set $A_k^n$ of measures $\mu \in \mathscr{M}_+^b(X)$ such that $h_n(\mu) < 1/k$ is therefore open for every $k \geq 1$ and every $n \in \mathbf{N}$. The proof is completed by observing that $W = \bigcap_n W_n = \bigcap_{n,k} A_k^n$.

For, let $\hat{T}$ be the completion of T for a metric defining the topology of T; the space $\hat{T}$ is Polish, and $\mathscr{M}_+^b(T)$ is homeomorphic to the subspace of the Polish space $\mathscr{M}_+^b(\hat{T})$ consisting of the measures concentrated on T (No. 3, Prop. 8). But every subspace of a Polish space is metrizable of countable type (GT, IX, §2, No. 8).

For, consider a Polish space P and a continuous mapping $f$ of P onto T (GT, IX, §6, No. 2, Def. 2). Let $\tilde{f}$ be the continuous mapping $\mu \mapsto f(\mu)$ of $\mathscr{M}_+^b(P)$ into $\mathscr{M}_+^b(T)$; the space $\mathscr{M}_+^b(P)$ is Polish by Prop. 10, and $\tilde{f}$ is surjective (§2, No. 4, Prop. 9); the space $\mathscr{M}_+^b(T)$ is therefore Souslin. Similarly, if T is Lusin, then $f$ may be assumed to be injective (GT, loc. cit., No. 4, Prop. 12); then $\tilde{f}$ is injective (§2, No. 4, Prop. 8), and so $\mathscr{M}_+^b(T)$ is Lusin (GT, loc. cit., No. 4, Prop. 12).

Let T be a completely regular Souslin space (recall that for this, it suffices that T be Souslin and regular (TG, App. 1, Cor. of Prop. 2)), and let H be a compact subset of $\mathscr{M}_+^b(T)$; then H is compact and Souslin, hence metrizable, for the tight topology (loc. cit., App. 1, Cor. 2 of Prop. 3).

For, the functions $\mu \mapsto |\mu|^*(1)$, $\mu \mapsto |\mu|^*(T - K_\varepsilon)$ are lower semi-continuous on $\mathscr{M}^b(T)$ by Prop. 6 of No. 3.

The interest of Prokhorov’s condition comes from the following theorem, whose converse will be studied later on (Th. 2).

Theorem 1 (Prokhorov). — Let T be a completely regular space, and let H be a subset of $\mathscr{M}^b(T)$ that satisfies Prokhorov’s condition; then H is relatively compact in $\mathscr{M}^b(T)$ for the tight topology.

We can suppose that T is a subspace of a compact space X; let i be the canonical injection of T into X. We can on the other hand suppose that H is closed in $\mathscr{M}^b(T)$, by Prop. 11. It will then suffice to show that every ultrafilter $\mathfrak{U}$ on H converges in $\mathscr{M}^b(T)$.

We shall begin with the case that $H \subset \mathscr{M}_+^b(T)$. The total masses of the measures $\mu \in H$ being bounded by hypothesis, $i(\mu)$ converges vaguely with respect to $\mathfrak{U}$, in $\mathscr{M}_+(X)$, to a measure $\nu \in \mathscr{M}_+(X)$ (Ch. III, §1, No. 9, Cor. 2 of Prop. 15); by Prop. 8 of No. 3, everything comes down to proving that $\nu$ is concentrated on T. Now, let $\varepsilon$ be a number $> 0$, and let $K_\varepsilon$ be a compact subset of T satisfying the formula (6). Since $X - K_\varepsilon$ is open in X, we have, by Prop. 6 of No. 3 applied in X, the inequalities

since $\varepsilon > 0$ is arbitrary, the theorem is established in this special case.

Let us pass to the general case; for every measure $\mu$ on T, set

since $\mu = a_1(\mu) - a_2(\mu) + ia_3(\mu) - ia_4(\mu)$, it will suffice to show that the mappings $a_j$ ($j = 1, 2, 3, 4$) converge tightly with respect to $\mathfrak{U}$. But the set $H_j$ of measures $a_j(\mu)$, where $\mu$ runs over H, satisfies Prokhorov’s condition by virtue of the relation $|a_j(\mu)| \leq |\mu|$, and is contained in $\mathscr{M}_+^b(T)$; it is therefore relatively compact in $\mathscr{M}_+^b(T)$ by the special case, and the theorem then follows at once.

It suffices to treat the case that H is closed (Prop. 11), hence compact; the conclusion then follows from the fact that $\mu \mapsto i(\mu)$ is continuous and injective.

Recall that this result is also valid for an arbitrary subset of $\mathscr{M}_+^b(T)$ (No. 3, Prop. 8).

We may restrict ourselves to the case that H is closed, hence compact. The total masses of the measures $\mu \in H$ are obviously bounded, because the mapping $\mu \mapsto \mu(1)$ is continuous, and everything comes down to proving the assertion b) of Def. 2.

Suppose first that T is locally compact. Let $\varepsilon$ be a number > 0. Let us associate to every measure $\mu \in H$ a compact set $K_\mu$ in T such that $\mu^\bullet(T - K_\varepsilon) < \varepsilon$, then a relatively compact open neighborhood $U_\mu$ of $K_\mu$. The function $\lambda \mapsto \lambda^\bullet(T - U_\mu)$ being upper semi-continuous on $\mathscr{M}_+^b(T)$ (No. 3, Prop. 6), the set $V^\mu$ of measures $\lambda \in H$ such that $\lambda^\bullet(T - U_\mu) < \varepsilon$ is a neighborhood of $\mu$ in H. Therefore there exists a finite subset $H'$ of H such that the sets $V^\mu$ ($\mu \in H'$) cover H. Denoting by K the compact set $\bigcup_{\mu \in H'} \overline{U}_\mu$, we have $\lambda^\bullet(T - K) < \varepsilon$ for all $\lambda \in H$.

Suppose next that T is Polish. We do not restrict the generality by assuming that T is the intersection of a decreasing sequence $(T_p)_{p \geq 1}$ of open subsets of a compact space X (GT, IX, §6, No. 1, Cor. 1 of Th. 1). Let $i_p$ be the injection of T into $T_p$, and let $H_p$ be the set of measures of the form $i_p(\lambda)$ for $\lambda \in H$; since $H_p$ is compact in $\mathscr{M}_+^b(T_p)$, it follows that there exists a compact set $K_p \subset T_p$ such that $\nu^\bullet(T_p - K_p) \leq \varepsilon 2^{-p}$ for every measure $\nu \in H_p$, by the preceding result applied to the locally compact space $T_p$. Therefore also $\nu^\bullet(T - (T \cap K_p)) \leq \varepsilon 2^{-p}$, and finally $\lambda^\bullet(T - (T \cap K_p)) \leq \varepsilon 2^{-p}$ for every measure $\lambda \in H$. Now set $K = \bigcap_p K_p$; the set K is compact and is contained in T, and, for every measure $\lambda \in H$, we have $\lambda^\bullet(T - K) \leq \sum_p \lambda^\bullet(T - (T \cap K_p)) \leq \sum_p \varepsilon 2^{-p} = \varepsilon$. Prokhorov’s condition is thus verified.

Let us show that the condition in the statement is necessary. Let $\theta$ be a bounded complex measure on T, $\varepsilon$ a number $> 0$, and K a compact subset of T such that $|\theta|^*(T - K) < \varepsilon$. Let $f \in B$; we denote by U the neighborhood of $f$ in B for the topology of compact convergence, formed by the functions $g \in B$ such that $\sup_{x \in K} |g(x) - f(x)| \leq \varepsilon$. Then, for every $g \in U$,

because $|g - f|$ is bounded above by $\varepsilon$ on K and by 2 on $T - K$.

Conversely, consider a linear form I on $\mathscr{C}^b(T; \mathbf{C})$ whose restriction to B is continuous for the topology of compact convergence. Then, for every number $\varepsilon > 0$, there exist a number $a > 0$ and a compact subset K of T such that the relations $f \in B, \sup_{x \in K} |f(x)| \leq a$ imply $|I(f)| \leq \varepsilon$. Prop. 5 of No. 2 then implies the existence of a unique bounded complex measure $\theta$ such that $I(f) = \theta(f)$ for all $f \in \mathscr{C}^b(T; \mathbf{C})$.

Let $\mu \in \mathscr{M}_+^b(T), f \in H$, and let M be a real number such that $||\mu|| < M$, and $|g| \leq M$ for all $g \in H$. Let $\varepsilon$ be a number $> 0$ and choose a compact subset K of T such that $\mu^*(T - K) < \varepsilon$, then a relatively compact open neighborhood S of K. The set U of measures $\lambda \in \mathscr{M}_+^b(T)$ satisfying the inequalities

is then a neighborhood of $\mu$ in $\mathscr{M}_+^b(T)$ (No. 3, Prop. 6). In addition, let V be the neighborhood of $f$ in H consisting of the functions $g \in H$ such that

Let $\lambda \in U$ and $g \in V$; since the function $|g - f|$ is bounded above by $\varepsilon$ in $S$, and by $2M$ in $T - S$, we have

from which one deduces This proves the continuity of the mapping $(\lambda, g) \mapsto \lambda(g)$ at the point $(\mu, f)$ of $\mathscr{M}_+^b(T) \times H$.