Denoting this mapping by $\Phi$, it is immediate that $\Phi(\lambda z') = \lambda \Phi(z')$ for all $\lambda \in \mathbf{R}$. Everything comes down to showing that $\Phi(y' + z') = \Phi(y') + \Phi(z')$. Set $\mu = |y' \circ m| + |z' \circ m|$; by the Lebesgue–Nikodym theorem, one can then write $y' \circ m = g \cdot \mu$ and $z' \circ m = h \cdot \mu$, where $g$ and $h$ are two bounded and locally $\mu$-integrable numerical functions (Ch. V, §5, No. 5, Th. 2); moreover, $(y' \circ m)^+ = g^+ \cdot \mu$ and $(y' \circ m)^- = g^- \cdot \mu$, and the analogous relations hold with $y'$ replaced by $z'$ (resp. $y' + z'$) and $g$ by $h$ (resp. $g + h$). This being so, it is immediate that $f$ is essentially $\mu$-integrable (Ch. V, §2, No. 2, Cor. 1 of Prop. 3), and the relation to be proved reduces to $(g + h)^+ - (g + h)^- = (g^+ - g^-) + (h^+ - h^-)$, which is obvious.

The proposition is obvious, in view of the equality $y' \circ u \circ m = {}^t u(y') \circ m$ for all $y \in G'$.

In general, if $f \in \mathscr{L}(m)$, the integral $\int f dm$ belongs to ${F'}^*$ but not to $F$ (see the above Example). However:

Let $A$ be the set of $f \in \mathscr{L}(m)$ such that $\sup_{t \in T} |f(t)| \leq 1$, and let $B = A \cap \mathscr{K}(T)$; by hypothesis, $m(B)$ is weakly relatively compact in $F$, therefore it suffices to show that $m(A)$ is contained in the closure (in ${F'}^*$) of $m(B)$ for the topology $\sigma({F'}^*, F')$; since $m(B)$ is convex and balanced, it suffices to prove that the polar of $m(B)$ in $F'$ is contained in that of $m(A)$ (TVS, II, §6, No. 3, Th. 1). Now, for a linear form $z' \in F'$ to belong to $(m(B))^\circ$, it is necessary and sufficient that $|\langle z', m(g)\rangle| = |\int g d(z'\circ m)| \leq 1$ for every function $g \in B$, which signifies that the scalar measure $|z'\circ m|$ is bounded and of norm $\leq 1$ (Ch. III, §1, No. 8); but by (2) the latter condition implies that $|\langle z', m(f)\rangle| \leq 1$ for every function $f \in A$, whence $z' \in (m(A))^\circ$.

The first assertion may be deduced immediately from Prop. 3: if $f$ is bounded and has compact support, and if U is a relatively compact open neighborhood of the support of $f$, then the restriction of $m$ to the subspace $\mathscr{H}(U)$ is a measure $m_U$ on U that satisfies the conditions of Prop. 3, and $\int f dm_U = \int f dm$ (Ch. V, §7, No. 1, Th. 1), therefore $\int f dm \in F$.

Now let $f$ be any element of $\mathscr{L}(m)$; for every compact subset K of T and every integer $n > 0$, let $f_{n,K}$ be the numerical function on T defined as follows: if $t \notin K$, $f_{n,K}(t) = 0$; if $t \in K$ and $|f(t)| \leq n$, $f_{n,K}(t) = f(t)$; finally, if $t \in K$ and $|f(t)| > n$, $f_{n,K}(t) = nf(t)/|f(t)|$. It is clear that for every $t \in T$, $f(t)$ is the limit of $f_{n,K}(t)$ with respect to the product filter of the Fréchet filter by the section filter of the (increasing directed) ordered set of compact subsets of T. Since $|f_{n,K}| \leq |f|$, it follows from Lebesgue’s theorem and Prop. 10 of Ch. V, §1, No. 3, applied to each scalar measure $|z'\circ m|$, that $f_{n,K}$ converges to $f$ in $\mathscr{L}(m)$ with respect to the preceding filter. Consequently, the integral $\int f dm$ is in the closure in ${F'}^*$ (for the topology $\sigma({F'}^*, F')$) of the set M of $m(f_{n,K})$. But the first part of the corollary shows that $M \subset F$, and, on the other hand, for every $z' \in F'$ one has $|\langle z', m(f_{n,K})\rangle| \leq \int |f| d|z'\circ m|$, which shows that M is bounded in $F_\sigma$, hence also in F (TVS, IV, §1, No. 1, Prop. 1). Lemma 1 of §1, No. 2 therefore shows that $\int f dm \in F''$.

For, the dual $F'$ may be identified with $\prod_{i=1}^n F_i'$, in such a way that if $x = (x_i) \in F$, $z' = (z_i') \in F'$, then $\langle x, z' \rangle = \sum_{i=1}^n \langle x_i, z_i' \rangle$. If $|\langle x, z' \rangle| \leq q(x)$ for every $x \in F$, then in particular $|\langle x_i, z_i' \rangle| \leq q_i(x_i)$ for $1 \leq i \leq n$, and the converse is obvious, which shows that the set $A'_q$ is the product of the $A'_{q_i}$. Since by hypothesis $|z_i' \circ m_i| \leq q_i(m_i)$ for $z_i' \in A'_{q_i}$, it follows that

for every $z' \in A'_q$, which proves the proposition.

$^1$ Majorées.

For,

by virtue of (1) and the relation $|z' \circ m| \leq q(m)$ for $z' \in A'_q$.

We shall make use of the following auxiliary result. Let $(\mu_\iota)_{\iota \in I}$ be an increasing directed family of positive measures on T. Let us denote by $\mathscr{L}^1((\mu_\iota)_{\iota \in I})$ the vector space of finite numerical functions on T, essentially $\mu_\iota$-integrable for every $\iota \in I$, equipped with the topology defined by the semi-norms $f \mapsto \mu_\iota(|f|)$ ($\iota \in I$). Let $\mathscr{L}_0$ be the linear subspace of $\mathscr{L}^1((\mu_\iota)_{\iota \in I})$ generated by the products $g \varphi_K$, where g runs over the set of continuous finite numerical functions on T, and K over the set of compact subsets of T.

To prove a), we may restrict ourselves to the case of an element of the form $f = g \varphi_K$ ($g \in \mathscr{C}(T)$), K compact in T). It is immediate (by virtue of Urysohn’s theorem) that $f$ is in the closure of the set B of functions of the form $gh$, where h describes the set of continuous mappings of T into $[0,1]$ that are equal to 1 on K and to 0 on the complement of a fixed compact neighborhood H of K. Moreover, the set B is bounded, because $\mu_\iota(|gh|) \leq \mu_\iota(|f\varphi_H|)$ for every function h having the preceding properties.

Let us now prove b); we may restrict ourselves to the case of an element $f \geq 0$ of $\mathscr{L}^1((\mu_\iota)_\iota \in I)$. For every $\iota \in I$ and every $\varepsilon > 0$, there exists a compact subset $K(\iota, \varepsilon)$ of T such that the restriction of $f$ to $K(\iota, \varepsilon)$ is continuous and $|\mu_\iota(|f - f\varphi_{K(\iota, \varepsilon)}|)| \leq \varepsilon$. It is clear that $f$ is in the closure of the set C of $f\varphi_{K(\iota, \varepsilon)}$ (where $\iota \in I,\ \varepsilon > 0$). By virtue of Urysohn’s theorem, the set C in contained in $\mathscr{L}_0$; moreover, it is bounded, because $\mu_\kappa(f\varphi_{K(\iota, \varepsilon)}) \leq \mu_\kappa(f)$ for all $\iota \in I,\ \kappa \in I$ and $\varepsilon > 0$.

Let us now prove Prop. 6: for every function $g \in \mathscr{K}(T)$ and every continuous seminorm q on F, $q\left( \int g\,dm \right) \leq \int |g|\,d(q(m))$ (Prop. 5), which implies that the mapping $g \mapsto \int g\,dm$ of $\mathscr{K}(T)$ into F is continuous when $\mathscr{K}(T)$ is equipped with the topology induced by that of $\mathscr{L}^1((q(m))_{q \in Q})$ (Q the set of continuous semi-norms on F). Consequently, by the preceding lemma and Prop. 10 of TVS, III, §1, No. 6, this mapping may be extended by continuity, first to a continuous linear mapping $v_0$ of $\mathscr{L}_0$ into F, then to a continuous linear mapping $v$ of $\mathscr{L}^1((q(m))_{q \in Q})$ into F. Moreover, for every $z' \in F'$ the relation $\langle z', v(f) \rangle = \int f\,d(z' \circ m)$ holds, by the definition of $v$, for every $f \in \mathscr{K}(T)$; since $|z' \circ m| \leq q(m)$ for $q(z) = |\langle z', z \rangle|$, the mapping $f \mapsto \int f\,d(z' \circ m)$ is continuous on $\mathscr{L}^1((q(m))_{q \in Q})$, therefore again by continuity, the relation $\langle z', v(f) = \int f\,d(z' \circ m)$ holds for every function $f \in \mathscr{L}^1((q(m))_{q \in Q})$. It follows that $v(f) = \int f\,dm$, which completes the proof.

For (No. 1, Remark 3), it suffices to show that, setting $m(g) = \int gf\,d\mu$, $z' \circ m$ is a scalar measure for every $z' \in F'$. But since $z'(m(g)) = \int g\langle z', f \rangle\,d\mu$, one has $z' \circ m = \langle z', f \rangle \cdot \mu$, whence our assertion.

a) If, for every compact subset $K$ of $T$, the upper integral $\int_K^*(q \circ f)\,d\mu$ is finite, then $m$ is $q$-majorizable.

b) If $m$ is $q$-majorizable, then $q(m)$ has base $\mu$; if in addition $f$ is $\mu$-measurable as a mapping of $T$ into $F_\sigma$, then $q \circ f$ is locally $\mu$-integrable and $q(m) = (q \circ f) \cdot \mu$.

a) For every finite subset $J$ of $A'_q$, let us denote by $\lambda_J$ the supremum of the measures $|z' \circ m|$, where $z'$ runs over $J$; if $g_J = \sup_{z' \in J} |\langle z', f \rangle|$ then $\lambda_J = g_J \cdot \mu$ (Ch. V, §5, No. 2, Prop. 2). For every relatively compact open subset $U$ of $T$, let $\lambda_{J,U}$ be the restriction of $\lambda_J$ to $U$; let us first show that as $J$ runs over the directed set $\mathfrak{F}$ of finite subsets of $A'_q$, the family $(\lambda_{J,U})$ is bounded above in $\mathscr{M}(U)$. Indeed, for every function $h \geq 0$ in $K(U)$,

whence our assertion (Ch. II, §2, No. 2). Let $\nu_U$ be the supremum of this family of measures in $\mathscr{M}(U)$. If $U'$ is a second relatively compact open subset of $T$ such that $U \subset U'$, then $\nu_U$ is the restriction of $\nu_{U'}$ to $U$, as follows immediately from the expression of the supremum of an increasing directed set of measures (Ch. II, §2, No. 2) and the fact that $\lambda_{J,U}$ is the restriction to $U$ of $\lambda_{J,U'}$. Thus there is one and only one positive measure $\nu$ whose restriction to each $U$ is $\nu_U$ (Ch. III, §2, No. 1, Prop. 1), and it is clear that $\nu = q(m)$.

b) Since the measures $\lambda_J$ have base $\mu$, so does their supremum $q(m)$ (Ch. V, §5, No. 5, Th. 2). If $f$ is $\mu$-measurable for the topology $\sigma(F, F')$ on $F$, it follows at once from the definitions that the mapping $g : t \mapsto (g_J(t))_{J \in \mathfrak{F}}$ of $T$ into the product space $\mathbf{R}^\mathfrak{F}$ is $\mu$-measurable. The restriction of $q \circ f = \sup_{J \in \mathfrak{F}} g_J$ to every compact subset of $T$ on which $g$ is continuous is lower semi-continuous; consequently $q \circ f$ is $\mu$-measurable (Ch. IV, §5, No. 5, Cor. of Prop. 8 and No. 10, Prop. 15). Let $K$ be a compact subset of T; it admits a partition consisting of a $\mu$-negligible set and a sequence $(K_n)$ of compact sets on which $g$ is continuous. Then $\int_{K_n}^*(q \circ f)\, d\mu = \sup_J \int_{K_n} g_J\, d\mu \leq \int_{K_n} dq(m)$ for all $n$ (Ch. IV, §1, No. 1, Th. 1 and Ch. V, §7, No. 1, Prop. 1), whence $\int_K^*(q \circ f)\, d\mu = \sum_n \int_{K_n}^*(q \circ f)\, d\mu \leq \int_K dq(m)$. But this proves that $q \circ f$ is locally $\mu$-integrable and that $\lambda_J \leq (q \circ f) \cdot \mu \leq q(m)$ for all $J \in \mathscr{F}$; whence, by definition, $q(m) = (q \circ f) \cdot \mu$.

Let $(e_i)_{1 \leq i \leq n}$ and $(e'_i)_{1 \leq i \leq n}$ be dual bases of F and F' (A, II, §2, No. 6) with $|e'_i| = 1$ for all i. Then $|e'_i \circ m| \leq |m|$ for every index i, therefore (Ch. V, §5, No. 5, Th. 2) $e'_i \circ m = h_i \cdot |m|$, where $h_i$ is bounded and $|m|$-measurable. Setting $h = \sum_{i=1}^n h_i \cdot e_i$, we therefore have $m = h \cdot |m|$. If $m = f \cdot |m|$, Prop. 8 shows that $|m| = |f| \cdot |m|$, whence $|f(t)| = 1$ locally almost everywhere for $|m|$ (Ch. V, §5, No. 3, Cor. 2 of Prop. 3). The final assertion follows at once from Prop. 8.

By the hypothesis, there exists a countable dense subset $H$ of $G$ that is a linear subspace with respect to the field $\mathbf{Q}$ of rational numbers; let $(h_n)$ be a (countable) basis of $H$ over $\mathbf{Q}$. For every integer $n$, let $h'_n$ be an element of $\mathscr{L}^\infty$ such that $\pi(h'_n) = h_n$, and let $\rho'$ be the $\mathbf{Q}$-linear mapping of $H$ into $\mathscr{L}^\infty$ defined by $\rho'(h_n) = h'_n$; it is clear that $\pi \circ \rho'$ is the identity on $H$. Moreover, for every $h \in H$ one has $|\rho'(h)(t)| \leq N_\infty(h)$ except at the points $t$ of a locally negligible set $A(h)$. Let $A$ be the union of the $A(h)$ for $h \in H$, which is also locally negligible. For every $h \in H$, denote by $\rho(h)$ the function $h'' \in \mathscr{L}^\infty$ such that $h''(t) = \rho'(h)(t)$ if $t \notin A$, and $h''(t) = 0$ if $t \in A$. It is clear that $\rho$ is a $\mathbf{Q}$-linear mapping of $H$ into the subspace $\mathscr{B}$ of bounded functions in $\mathscr{L}^\infty$, such that $\pi \circ \rho$ is the identity on $H$ and such that $|\rho(h)(t)| \leq N_\infty(h)$ for all $h \in H$ and $t \in T$. Since $\mathscr{B}$ is a Banach space for the norm $\|f\| = \sup_{t \in T} |f(t)|$ (Ch. IV, §5, No. 4, Th. 2), $\rho$ may be extended to a continuous $\mathbf{R}$-linear mapping, again denoted $\rho$, of $G$ into $\mathscr{B}$, which is obviously a lifting of $G$.