In particular, if f is scalarly essentially $\mu$-integrable, then it remains scalarly essentially $\mu$-integrable when the topology of F is replaced by a coarser topology.

This is an immediate consequence of the formula $\langle \mathbf{z}', gf \rangle = g \langle \mathbf{z}', f \rangle$, for all $\mathbf{z}' \in F'$, and the formula $\int h d(g \cdot \mu) = \int hg d\mu$ for every essentially $\mu$-integrable scalar function $h$ (Ch. V, §5, No. 3, Th. 1).

A great many propositions about essentially integrable numerical functions may be transposed word for word into propositions about scalarly essentially integrable vector-valued functions. Among the more important, we call attention to the conditions for a function to be essentially integrable with respect to a measure defined by a density (Ch. V, §5, No. 3, Th. 1), or with respect to the image of a measure (Ch. V, §6, No. 2, Th. 1), or with respect to an induced measure (Ch. V, §7, No. 1, Th. 1), or with respect to the sum of a summable family of positive measures (Ch. V, §2, No. 2, Props. 1 and 3 and Cor. 3 of Prop. 1). These transpositions are left to the reader.

However, to obtain statements corresponding to the theorems on 'double integrals' (Ch. V, §3, No. 3, Th. 1 and §8, No. 4, Th. 1 (Lebesgue–Fubini theorem)), it is necessary to strengthen the hypotheses (cf. Exer. 1); on applying the previously cited theorems to each of the functions $\langle \mathbf{z}', f \rangle$, where $\mathbf{z}' \in F'$, one thus obtains the following propositions:

$^3$It suffices to assume that the family is scalarly essentially $\mu$-integrable, thus the proposition holds for either of the two interpretations of ' $\mu$-adequate'. $^4$Equality as elements of ${F'}^*$.

Since $D$ is the intersection of the closed half-spaces containing $f(S)$ (TVS, II, §5, No. 3, Cor. 1 of Prop. 4), it suffices to prove that the relation $\langle f(t), z' \rangle \leq a$ for all $t \in S$ (where $z' \in F', a \in \mathbf{R}$) implies $\langle z', \int f\, d\mu \rangle \leq a \cdot \mu(T)$; but since $\int f\, d\mu = \int_S f\, d\mu$, this follows from Ch. IV, §4, No. 2, Cor. 1 of Th. 1.

For every $z' \in F'$, $\langle z', f \rangle$ is $\mu$-measurable and bounded in $S$, hence integrable, which proves that $f$ is scalarly integrable. Moreover, since $A$ is compact in $F_\sigma$, it is closed in $F'$*, and the closed convex envelope of $f(S)$ in $F'$* is contained in $A$, whence the corollary.

Let $D$ be the set of $z \in F$ such that $q(z) \leq 1$; $D$ is closed, convex, and contains $0$, therefore $D = D^{oo}$ (TVS, II, §6, No. 3, Cor. 3 of Th. 1). It therefore suffices to prove that for every $z' \in D^\circ$, one has $|\langle z', \int f\, d\mu \rangle| \leq \int^\bullet (q \circ f)\, d\mu$; but this follows at once from the fact that $|\langle z', f(t) \rangle| \leq q(f(t))$ for every $t \in T$.

Note that the numerical function $q \circ f$ need not be $\mu$-measurable (Exer. 12).

(*) Recall that for a bounded positive measure $\mu$, the concepts of $\mu$-integrable function and essentially $\mu$-integrable function are the same (Ch. V, §1, No. 3, Cor. of Prop. 9).

5 The original notation for upper essential integral is $\overline{\int^*}$, changed to $\int^\bullet$ in the second edition of Ch. V.

in a weakly compact, balanced, convex subset of F. Then $\int f d\mu$ belongs to the bidual $F''$ of F.

For every compact subset K of T,

the Cor. of Prop. 5 can be applied to the bounded measure $\varphi_K \cdot \mu$ and the function $f \varphi_K$, consequently $\int f \varphi_K d\mu \in F$. For every $z' \in F'$, $\langle z', f \rangle$ is essentially $\mu$-integrable, consequently (Ch. V, §1, No. 3, Prop. 10)

the limit being taken with respect to the increasing directed set of compact subsets of T. One concludes that, with respect to this set, $\int f \varphi_K d\mu$ converges to $\int f d\mu$ for the topology $\sigma({F'}^*, F')$. Now,

which proves that the set of elements $\int f \varphi_K d\mu$ is a bounded subset of $F_\sigma$, hence also of F (TVS, IV, §1, No. 1, Prop. 1). Proposition 7 is therefore a consequence of the following lemma:

For, a bounded subset of F is contained in the polar (in $F''$) of a neighborhood of 0 in the strong dual $F'$ of F, hence is relatively compact in $F''$ for $\sigma(F'', F')$ (TVS, III, §3, No. 5, Prop. 7 and No. 4, Cor. 2 of Prop. 4); since $\sigma(F'', F')$ is induced by $\sigma({F'}^*, F')$, the lemma is proved.

For, every bounded subset of F is relatively weakly compact (TVS, IV, §2, No. 2, Th. 1), and $F = F''$.

Since S is $\mu$-integrable, there exists a partition of S formed by a $\mu$-negligible set N and a sequence $(K_n)$ of compact subsets such that the restriction of f to each $K_n$ is continuous (Ch. IV, §4, No. 6, Cor. 3 of Th. 4 and §5, No. 1, Def. 1); $f(K_n)$ is therefore a compact subset of $F$. The closed, balanced convex envelope $B_n$ of $f(K_n)$ is then pre-compact (TVS, II, §4, No. 1, Prop. 3) and is contained in the complete subset $B$ of $F$, therefore it is compact, and a fortiori weakly compact. Consequently (Cor. of Prop. 5) $f \varphi_{K_n}$ is scalarly $\mu$-integrable, and

For every continuous semi-norm $p$ on $F$, it follows that

since $B$ is bounded and since the series with general term $\mu(K_n)$ is convergent and has sum $\mu(T)$, one sees that the sequence with general term $s_n = z_1 + z_2 + \cdots + z_n$ is a Cauchy sequence in the complete subset $\mu(T)B$ of $F$. This sequence therefore converges to an element $s$ of $\mu(T)B$; since one can suppose that $f(t) = 0$ on $T - S$, Lebesgue’s theorem applied to each of the functions $\langle z', f \rangle$ ($z' \in F'$) proves that $s = \int f d\mu$.

a) For every $x \in G$, $\int (U(t) \cdot x) d\mu(t) \in H$.

b) For every equicontinuous subset $B'$ of $H'$, the set of linear forms $u_{y'} : x \mapsto \int \langle U(t) \cdot x, y' \rangle d\mu(t)$, where $y'$ runs over $B'$, is equicontinuous.

The conditions a) and b) are necessary. For, since for every $x \in G$, the mapping $\tilde{x} : V \mapsto V \cdot x$ of $\mathscr{L}_s(G; H)$ into $H$ is linear, one sees (No. 1,

Prop. 1) that $\widetilde{\mathbf{x}} \circ U : t \mapsto U(t) \cdot \mathbf{x}$ is scalarly essentially $\mu$-integrable and that

where $S = \int U\, d\mu \in \mathscr{L}_s(G; H)$. This proves a). Moreover, (1) may also be written

in other words ${}^t S \cdot \mathbf{y}' = u_{\mathbf{y}'}$. Since $S$ is continuous, ${}^t S$ transforms every equicontinuous subset of $H'$ into an equicontinuous subset of $G'$, whence b).

Conversely, suppose a) and b) verified. By virtue of a), the formula (1) defines a linear mapping $S$ of $G$ into $H$, and, for every $\mathbf{y}' \in H'$, this mapping satisfies (2) (No. 1, Prop. 1); but then, condition (b) says that $S$ is continuous (TVS, II, §6, No. 4, Props. 5 and 6, and III, §3, No. 5, Prop. 7), therefore $S \in \mathscr{L}_s(G; H)$. Finally, formula (2) proves that $S = \int U\, d\mu$.

$1^\circ$ The measure $\mu$ is bounded, and if $S$ is its support, then $U(S)$ is an equicontinuous subset of $\mathscr{L}(G; H)$.

$2^\circ$ The condition a) of Prop. 9 is satisfied, the space $G$ is barreled, and, for every compact subset $K$ of $T$, $U(K)$ is a bounded subset of $\mathscr{L}_s(G; H)$.

Let us first place ourselves in case $1^\circ$. We can restrict ourselves to the case that $S = T$ (Ch. V, §7, No. 1, Th. 1). Then, for every equicontinuous subset $B'$ of $H'$, there exists an equicontinuous, convex, balanced and weakly closed subset $A' \subset G'$ such that ${}^t U(t) \cdot \mathbf{y}' \in A'$ for all $\mathbf{y}' \in B'$ and all $t \in T$ (TVS, II, §6, No. 4, Prop. 6). Since $U$ is scalarly $\mu$-integrable, the mapping $t \mapsto {}^t U(t) \cdot \mathbf{y}'$ of $T$ into the dual $G'$ of $G$ equipped with $\sigma(G', G)$, is scalarly $\mu$-integrable, and one may write

Since $A'$ is convex and compact for $\sigma(G', G)$, the Cor. of Prop. 5 of No. 2 shows that $u_{\mathbf{y}'} \in \mu(T)A'$ for every $\mathbf{y}' \in B'$, which proves our assertion.

Let us now place ourselves in case $2^\circ$. For every $\mathbf{y}' \in H'$ and every compact subset $K$ of $T$, set

an element of the algebraic dual $G^*$ of $G$. Since $G$ is barreled, every bounded subset of $\mathscr{L}_s(G; H)$ is equicontinuous (TVS, III, §4, No. 2, Th. 1); the first part of the argument, applied to the function $\varphi_K U$ and the bounded measure $\varphi_K \cdot \mu$, shows that $u_{K, y'} \in G'$. Moreover, for the topology $\sigma(G^*, G)$, one has $u_{y'} = \lim_{K} = u_{K, y'}$, the limit being taken with respect to the increasing directed set of compact subsets of $T$ (Ch. V, §1, No. 3, Prop. 10). To verify condition b) of Prop. 9, it suffices, by Prop. 9, to prove that the linear mapping $S$ of $G$ into $H$ defined by (1) is continuous; moreover, $G$ is barreled, therefore it will suffice to prove that $S$ is continuous when $G$ and $H$ are equipped with their weakened topologies (TVS, IV, §1, No. 3, Prop. 7); finally, by virtue of (2), one is reduced to showing that $u_{y'} \in G'$ for every $y' \in H'$. Since $u_{y'}$ is in the closure, for $\sigma(G^*, G)$, of the set $M'$ of the $u_{K, y'}$, where $K$ runs over the set of compact subsets of $T$, it suffices to prove that $M'$ is equicontinuous (TVS, III, §3, No. 4, Prop. 4); and since $G$ is barreled, it comes to the same to say that for every $x \in G$, the set of $\langle x, u_{K, y'} \rangle$ is bounded (TVS, III, §4, No. 2, Th. 1). But this follows at once from the relations

a) $H$ is quasi-complete, $\mu$ is bounded and, if $S$ is its support, $U$ is $\mu$-measurable and $U(S)$ is equicontinuous.

b) $H$ is semi-reflexive, $\mu$ is bounded and, if $S$ is its support, $U$ is scalarly $\mu$-measurable and $U(S)$ is equicontinuous.

c) $H$ is semi-reflexive, $G$ is barreled, $U$ is scalarly essentially $\mu$-integrable, and, for every compact subset $K$ of $T$, $U(K)$ is bounded.

The fact that $U$ is scalarly essentially integrable is obvious in all three cases; by virtue of Prop. 9 and its corollary, it suffices in each case to verify condition a) of Prop. 9. Now, this condition follows from Prop. 8 of No. 2 in the first case, and from the Cor. of Prop. 7 of No. 2 in the other two cases.

Let V be a barrel in F, q is gauge, which is a semi-norm on F; let H be the Hausdorff space associated with the space F equipped with the topology defined by this single semi-norm. The completion $\widehat{H}$ of H is a Banach space; let $\pi$ be the canonical mapping of F into $\widehat{H}$; we are going to show that $\pi$ is continuous (for the original topology on F); this will establish the proposition, because V, the inverse image under $\pi$ of the unit ball of $\widehat{H}$, will then be a neighborhood of 0 in F. To establish the continuity of $\pi$ it will suffice, by virtue of (GDF), to show that the graph of $\pi$ is closed in $F \times \widehat{H}$; in other words, we must see that if $\mathscr{F}$ is a filter on F, convergent to $x \in F$, and its image $\pi(\mathscr{F})$ converges to $y \in \widehat{H}$, then $y = \pi(x)$. Now, every element $x'$ of the polar $V^\circ$ of V in $F'$ may be extended in only one way to a continuous linear form on $\widehat{H}$ (again denoted $x'$), and the set of these forms is the unit ball of the dual of $\widehat{H}$; it therefore suffices to show that $\langle y, x' \rangle = \langle \pi(x), x' \rangle$ for every $x' \in V^\circ$. But this follows from the relations

Recall that the dual of $F'_s$ is F (TVS, II, §6, No. 2, Prop. 3). For every $z \in F$, the numerical function $\langle z, f \rangle$ is therefore essentially $\mu$-integrable; let $\theta(z)$ be its class in $L^1(\mu)$. To show that $\int f d\mu \in F'$, one must establish that the linear form $z \mapsto \langle z, \int f d\mu \rangle$ is continuous on F; in fact, we are going to prove the following stronger result:

By virtue of the (GDF) property, it suffices to show that for every sequence $(z_n)$ of elements of F converging to $z$, such that $(\theta(z_n))$ converges to $u \in L^p(\mu)$, one has $u = \theta(z)$. Now, replacing if necessary the sequence $(z_n)$ by a subsequence, one can suppose that the sequence of functions $\langle z_n, f \rangle$ converges locally almost everywhere to a function $h \in \overline{\mathscr{L}}^p(\mu)$, with class $u$ in $L^p(\mu)$ (Ch. IV, §3, No. 4, Th. 3 and Ch. V, §1, No. 3). Since, by hypothesis, for every $t \in T$ the sequence $(\langle z_n, f(t) \rangle)$ converges to $\langle z, f(t) \rangle$, we have $h(t) = \langle z, f(t) \rangle$ locally almost everywhere, consequently $u = \theta(z)$.

The space $F$ is in duality with the tensor product $\bigotimes_{i=1}^n G_i$, and the topology of pointwise convergence on $F$ is none other than the topology $\sigma(F, \bigotimes_{i=1}^n G_i)$. The algebraic dual ${F'}^*$ is therefore the space of all multilinear forms on $\prod_{i=1}^n G_i$. Let $z = (z_1, \ldots, z_n)$ be an element of $\prod_{i=1}^n G_i$; for every multilinear form $u \in {F'}^*$, the mapping $x \mapsto u(z_1, \ldots, z_{i-1}, x, z_{i+1}, \ldots, z_n)$ is a linear form on $G_i$, which we shall denote by $\lambda_i(z)(u)$; we thus obtain a linear mapping $\lambda_i(z)$ of ${F'}^*$ into the algebraic dual $G_i^*$ of $G_i$, continuous for the topologies $\sigma({F'}^*, \bigotimes_{i=1}^n G_i)$ and $\sigma(G_i^*, G_i)$. To say that $u \in F$ means that for every index $i$ and every $z \in \prod_{i=1}^n G_i$, one has $\lambda_i(z)(u) \in G_i'$. Now, by Prop. 1 of No. 1, the mapping $\lambda_i(z)$ of $F$ is a scalarly essentially $\mu$-integrable mapping of $T$ into $G_i'$ equipped with the topology $\sigma(G_i', G_i)$, and

Since $G$ is barreled (Prop. 11), $\mathscr{L}(G; H) = \mathscr{L}(G_\sigma; H_\sigma)$ (TVS, IV, §1, No. 3, Prop. 7); moreover, one can replace $F = \mathscr{L}_s(G; H)$ by the space $\mathscr{L}_s(G_\sigma; H_\sigma)$, since the two spaces have the same dual $G \otimes H'$ (TVS, II, §6, No. 2, Prop. 3, and the first paragraph of No. 3 above). If, for every $u \in \mathscr{L}(G; H) = \mathscr{L}(G_\sigma; H_\sigma)$, one sets $\tilde{u}(x, y') = \langle u(x, y') \rangle$ (for $x \in G$, y' \in H' ), the linear mapping $u \mapsto \tilde{u}$ is a bijection of F onto the space $F_1$ of separately continuous bilinear forms on $G_\sigma \times H'_s$, where $H'_s$ denotes the dual $H'$ equipped with the weak topology $\sigma(H', H)$ (App., No. 1); moreover, this mapping is an isomorphism of $\mathscr{L}_s(G_\sigma; H_\sigma)$ onto $F_1$ equipped with the topology of pointwise convergence (loc. cit.). But since by hypothesis H is the dual of $H'_b$, $F_1$ is also the space of separately continuous bilinear forms on $G \times H'_b$. Cor. 2 therefore follows from Cor. 1.

Note that Cor. 2 is applicable in particular when G is Banach space and H is a reflexive Banach space.

For, F may be regarded as a subspace of a countable product $\prod_n E_n$ of Banach spaces (TVS, II, §1, No. 3, Prop. 3), and one can suppose that $\mathrm{pr}_n(F)$ is dense in $E_n$, which is therefore separable. For every n, the mapping $\mathrm{pr}_n \circ f$ is scalarly $\mu$-measurable, hence $\mu$-measurable (Ch. IV, §5, No. 5, Cor. 2 of Prop. 10), consequently f is $\mu$-measurable (Ch. IV, §5, No. 3, Th. 1).

Suppose first that F is metrizable and separable, and let D be a countable dense set in F. Let $(V_n)$ be a decreasing fundamental sequence of balanced, convex, open neighborhoods of 0 in F; the polar sets $V_n^\circ$ are equicontinuous and their union is all of $F'$. Let $T_n = f(V_n^\circ)$; the sequence $(T_n)$ is increasing and $T = \bigcup T_n$; let us show that each of the $T_n$ is $\mu$-*measurable*. Indeed, $D \cap V_n^n$ is dense in $V_n$; for every $y \in D \cap V_n$, let $S_y$ be the set of $t \in T$ such that $|\langle y, f(t) \rangle| \leq 1$; the hypothesis implies that each of the $S_y$ is measurable, and $T_n$ is the intersection of the countable family of the $S_y$ ($y \in D \cap V_n$). This being so, for every compact subset K of T and every $\varepsilon > 0$, there exists an integer n such that $\mu(K - (K \cap T_n)) \leq \frac{\varepsilon}{4}$, then a compact subset $K_1$ of $K \cap T_n$ such that $\mu((K \cap T_n) - K_1) \leq \frac{\varepsilon}{4}$; finally, there exists a compact subset $K_2$ of $K_1$ such that $\mu(K_1 - K_2) \leq \frac{\varepsilon}{2}$ and such that the restrictions to $K_2$ of all of the functions $\langle y, f \rangle$, where $y \in D$, are continuous (Ch. IV, §5, No. 1, Prop. 2). Since the set $f(K_2) \subset f(T_n) \subset V_n^\circ$ is equicontinuous, the topology induced by $\sigma(F', F)$ on $f(K_2)$ is identical to the topology of pointwise convergence in $D$ (GT, X, §2, No. 4, Th. 1); consequently, the restriction of $f$ to $K_2$ is continuous, whence our assertion in the first case.