(i) The restriction to $\mathscr{K}(X \times Y, K \times L; \mathbf{C})$ of the canonical bijection

(S, R, §4, No. 14) is an isometry of the Banach space $\mathscr{K}(X \times Y, K \times L; \mathbf{C})$ onto the Banach space $\mathscr{K}(X, K; \mathscr{K}(Y, L; \mathbf{C}))$.

(ii) The vector space $\mathscr{K}(X, K; \mathbf{C}) \otimes_{\mathbf{C}} \mathscr{K}(Y, L; \mathbf{C})$, identified canonically with a subspace of $\mathscr{K}(X \times Y, K \times L; \mathbf{C})$ (A, II, §7, No. 7, comments following the Cor. of Prop. 15), is dense in this Banach space.

It is immediate that the image under $\omega$ of $\mathscr{K}(X \times Y, K \times L; \mathbf{C})$ is contained in $\mathscr{K}(X, K; \mathscr{K}(Y, L; \mathbf{C}))$. Conversely, if $u$ is a continuous mapping of $X$ into $\mathscr{K}(Y, L; \mathbf{C})$, with support contained in $K$, then the mapping $(x, y) \mapsto u(x)(y)$ of $X \times Y$ into $\mathbf{C}$ is continuous and has support contained in $K \times L$, therefore the restriction of $\omega$ to $\mathscr{K}(X \times Y, K \times L; \mathbf{C})$ is a bijection of this space onto $\mathscr{K}(X, K; \mathscr{K}(Y, L; \mathbf{C}))$; the fact that this restriction is a Banach space isometry follows from the relation

This proves (i); on the other hand the image under $\omega$, of

identified with a subspace of $\mathscr{K}(X \times Y, K \times L; \mathbf{C})$, is again the space $\mathscr{K}(X, K; \mathbf{C}) \otimes_{\mathbf{C}} \mathscr{K}(Y, L; \mathbf{C})$ but this time identified canonically with a space of mappings of $X$ into $\mathscr{K}(Y, L; \mathbf{C})$ (A, II, §7, No. 7, Cor. of Prop. 15); but this subspace of $\mathscr{K}(X, K; \mathscr{K}(Y, L; \mathbf{C}))$ is known to be dense in the latter space (§ 1, No. 2, Prop. 5), thus the conclusion of (ii) follows from the fact that the restriction of $\omega$ is a topological isomorphism.

Having proved the lemma, we now observe that every compact subset of $X \times Y$ is contained in a product $K \times L$, where $K$ (resp. $L$) is a compact subset of $X$ (resp. $Y$). It therefore follows from Lemma 1, (ii) that the subspace $\mathscr{K}(X; \mathbf{C}) \otimes_{\mathbf{C}} \mathscr{K}(Y; \mathbf{C})$ is dense in $\mathscr{K}(X \times Y; \mathbf{C})$; since the relation (1) may also be written as $\nu(g \otimes h) = \lambda(g)\mu(h)$ for $g \in \mathscr{K}(X; \mathbf{C})$, $h \in \mathscr{K}(Y; \mathbf{C})$, the uniqueness of $\nu$ follows at once. To prove the existence of $\nu$, we shall make use of the following lemma:

belongs to $\mathscr{K}(Y, L; \mathbf{C})$.

For every function $u \in \mathscr{K}(X; \mathscr{K}(Y, L; \mathbf{C}))$, the integral $\int u(x) d\lambda(x)$ belongs to $\mathscr{K}(Y, L; \mathbf{C})$ since the latter is a Banach space (§ 3, No. 3, Cor. 1 of Prop. 7); but for $u = \omega(f)$ and for every $y \in Y$,

whence the lemma.

Consider, then, for every function $f \in \mathscr{K}(X \times Y; \mathbf{C})$, the number $\nu(f) = \mu(\int f(x, y) d\lambda(x))$ (which we shall also write $\int d\mu(y) \int f(x, y) d\lambda(x)$ by an abuse of notation); if K (resp. L) is a compact subset of X (resp. Y), then there exists a number $a_K$ (resp. $b_L$) such that, for every function $g \in \mathscr{K}(X, K; \mathbf{C})$ (resp. $h \in \mathscr{K}(Y, L; \mathbf{C})$), we have $|\lambda(g)| \leq a_K \|g\|$ (resp. $|\mu(h)| \leq b_L \|h\|$). It follows that for every function $f \in \mathscr{K}(X \times Y, K \times L; \mathbf{C})$,

for every $y \in Y$, whence $|\nu(f)| \leq a_K b_L \|f\|$. The linear form $\nu$ on $\mathscr{K}(X \times Y; \mathbf{C})$ is thus a measure on $X \times Y$ that obviously satisfies (1), which completes the proof of Th. 1.

Because of the last formula, the integral of $f$ with respect to the product measure $\nu$ is usually denoted $\iint f\, d\lambda\, d\mu$, or $\iint f\, d\mu\, d\lambda$, or $\iint f\, \lambda\mu$, or $\iint f\, \mu\lambda$, or $\iint f(x, y)\, d\lambda(x)\, d\mu(y)$, or $\iint f(x, y)\, d\mu(y)\, d\lambda(x)$, or $\iint f(x, y)\, \lambda(x)\mu(y)$, or $\iint f(x, y)\, \mu(y)\lambda(x)$; it is said to be the double integral of $f$ with respect to $\lambda$ and $\mu$. With this notation, the formula (3) may be written

Formula (3) shows that if $\lambda$ and $\mu$ are real (resp. positive) measures, then the product measure $\nu$ is real (resp. positive).

For every function $f \in \mathscr{K}(X \times Y; \mathbf{C})$, we have, by virtue of formula (3) of No. 1,

which proves formula (6).

We first observe that the relation $\lambda \otimes \mu = 0$ implies that one of the measures $\lambda, \mu$ is zero (A, II, §7, No. 7, Prop. 16, (ii)). On the other hand, if U (resp. V) is an open set in X (resp. Y), then the restriction of $\lambda \otimes \mu$ to the product $U \times V$ is the product of the restrictions of $\lambda$ to U and of $\mu$ to V, as follows from Th. 1 of No. 1 and the definition of the restriction of a measure to an open set (§ 2, No. 1). It therefore follows that, for the restriction of $\lambda \otimes \mu$ to $U \times V$ to be zero, it is necessary and sufficient that either the restriction of $\lambda$ to U or the restriction of $\mu$ to V be zero, which proves the proposition, on taking into account the definition of the support of a measure (§ 2, No. 2).

Let $f \in \mathscr{K}_+(X \times Y),\ g \in \mathscr{K}(X \times Y; \mathbf{C})$ be such that $|g| \leq f$; we have (§ 1, No. 6, formula (13))

It follows that $\langle f, |\lambda \otimes \mu| \rangle \leq \langle f, |\lambda| \otimes |\mu| \rangle$, and so

On the other hand, let $u \in \mathscr{K}_+(X),\ v \in \mathscr{K}_+(Y)$. For every $\varepsilon > 0$, there exist $u_1 \in \mathscr{K}(X; \mathbf{C}),\ v_1 \in \mathscr{K}(Y; \mathbf{C})$ such that $|u_1| \leq u,\ |v_1| \leq v$ and

(§ 1, No. 6). It follows that $|u_1 \otimes v_1| \leq u \otimes v$ and

Since $\varepsilon$ is arbitrary, we infer that

Taking (8) into account, we see that

Since every function in $\mathscr{K}(X; \mathbf{C})$ (resp. $\mathscr{K}(Y; \mathbf{C})$) is a linear combination of functions in $\mathscr{K}_+(X)$ (resp. $\mathscr{K}_+(Y)$), the preceding formula remains true for $u \in \mathscr{K}(X; \mathbf{C})$ and $v \in \mathscr{K}(Y; \mathbf{C})$; the proposition therefore follows from the fact that $\mathscr{K}(X; \mathbf{C}) \otimes_{\mathbf{C}} \mathscr{K}(Y; \mathbf{C})$ is dense in $\mathscr{K}(X \times Y; \mathbf{C})$.

For, by virtue of Prop. 3,

The argument for $(\lambda \otimes \mu)^-$ is similar.

with the convention that the second member is to be replaced by 0 whenever one of the factors is 0 and the other is $+\infty$. In particular, if $\lambda$ and $\mu$ are bounded then $\lambda \otimes \mu$ is bounded.

By the above Proposition 3, and §1, No. 8, Cor. 1 of Prop. 10, we may limit ourselves to the case that $\lambda$ and $\mu$ are positive measures. If $\lambda = 0$ or $\mu = 0$, the result is trivial; let us therefore assume that $\lambda \neq 0$ and $\mu \neq 0$. Let us first prove that $\|\lambda \otimes \mu\| \leq \|\lambda\| \cdot \|\mu\|$. We may suppose $\lambda$ and $\mu$ to be bounded. For every $f \in \mathscr{K}_+(X \times Y)$,

and

for every $x \in X$, therefore

which proves our assertion. On the other hand, let

be two real numbers $\geqslant 0$. There exist $g \in \mathscr{K}_+(X) , \ h \in \mathscr{K}_+(Y)$ such that

Then $g \otimes h \in \mathscr{K}_+(X \times Y) , \ \| g \otimes h \| \leqslant 1$ and $\langle g \otimes h , \lambda \otimes \mu \rangle \geqslant \alpha \beta ;$ therefore $\| \lambda \otimes \mu \| \geqslant \alpha \beta$ and finally $\| \lambda \otimes \mu \| \geqslant \| \lambda \| \cdot \| \mu \|$, which completes the proof.

For every function $f \in \mathscr{K}(X \times Y; \mathbf{C})$, we know that the function $h(y) = \int f(x, y) d\lambda_0(x)$ belongs to $\mathscr{K}(Y; \mathbf{C})$ (No. 1, Lemma 2), and $\langle f , \lambda_0 \otimes \mu \rangle = \langle h , \mu \rangle$, whence the proposition.

Let $K \subset X , \ L \subset Y$ be two compact sets, $A$ a compact subset of $\mathscr{K}(X \times Y, K \times L; \mathbf{C})$, and $B$ a vaguely bounded and closed subset of $\mathscr{M}(X; \mathbf{C})$; it is known that $B$ is vaguely compact (§ 1, No. 9, Prop. 15), hence also compact for the topology of strictly compact convergence (§ 1, No. 10, Prop. 17). On the other hand, the Banach space $\mathscr{K}(X \times Y, K \times L; \mathbf{C})$ is isometric to $\mathscr{K}(X, K; \mathscr{K}(Y, L; \mathbf{C}))$ (No. 1, Lemma 1); the mapping $\varphi$ of $\mathscr{K}(X, K; \mathscr{K}(Y, L; \mathbf{C})) \times \mathscr{M}(X; \mathbf{C})$ into $\mathscr{K}(Y, L; \mathbf{C})$, such that $\varphi(g, \lambda)$ is the function $h$ defined by $h(y) = \int g(x, y) d\lambda(x)$, is separately continuous by virtue of § 3, No. 4, Props. 8 and 9. Since $\mathscr{K}(X, K; \mathscr{K}(Y, L; \mathbf{C}))$ is barreled, it follows that the mapping $\varphi$ is hypocontinuous relative to the vaguely bounded subsets of $\mathscr{M}(X; \mathbf{C})$ (TVS, III, § 5, No. 3, Prop. 6); the restriction of this mapping to $A \times B$ is therefore continuous (loc. cit., Prop. 4). The image $C$ of $A \times B$ under this mapping is consequently a compact subset of the Banach space $\mathscr{K}(Y, L; \mathbf{C})$. Now, $C$ is none other than the set of functions $h(y) = \int f(x, y) d\lambda(x)$ as $f$ runs over $A$ and $\lambda$ runs over $B$; by virtue of formula (3) of No. 1, the conditions $\lambda \in B$ and $\mu \in C^\circ$ therefore imply $\lambda \otimes \mu \in A^\circ$. In view of the definition of the topology of strictly compact convergence, this proves the proposition (TVS, III, § 5, No. 3, Def. 2).

The conclusion of Prop. 6 is no longer valid when the topology of strictly compact convergence is replaced by the vague topology (Exer. 2 c)). However, if B (resp. C) is a vaguely bounded subset of $\mathscr{M}(X; \mathbf{C})$ (resp. $\mathscr{M}(Y; \mathbf{C})$), then the image of $B \times C$ under the mapping $(\lambda, \mu) \mapsto \lambda \otimes \mu$ is vaguely bounded in $\mathscr{M}(X \times Y; \mathbf{C})$ and therefore the restriction of this mapping to $B \times C$ is vaguely continuous, by virtue of Prop. 6, of §1, No. 10, Prop. 17, and of Prop. 4 of TVS, III, §5, No. 3 (cf. Exer. 3).

For, these two measures coincide, by (12), for every function in $\bigotimes_{i=1}^n \mathscr{K}(X_i; \mathbf{C})$.

The integral of a function $f \in \mathscr{K}(X; \mathbf{C})$ with respect to the product measure is denoted

or or or also or with $n$ signs $\int$; it is said to be a multiple integral of order $n$, or an $n$-tuple integral. By virtue of the associativity of the product of measures and the theorem on inverting the order of integration (No. 1, Th. 2), we have, for every permutation $\sigma$ of $[1, n]$, (14)

The integral notation and formula (14) may be extended in an obvious way to functions $f \in \mathscr{K}(X; E)$ with values in a Hausdorff locally convex space $E$, such that $f(X)$ is contained in a complete convex subset of $E$. We leave to the reader the task of generalizing to the product of any finite number of measures the results of Nos. 2 and 3 concerning the product of two measures.

In particular, one calls Lebesgue measure on $\mathbf{R}^n$ the product of $n$ measures identical to the Lebesgue measure on $\mathbf{R}$; the integral of a function $f \in \mathscr{K}(\mathbf{R}^n; E)$, satisfying the preceding condition, is denoted

and is equal to Lebesgue measure on $\mathbf{R}^n$ is invariant under every translation.

(iv) *If the $p_{\alpha \beta}$ are surjective, then every inverse system $(\mu_\alpha)$ of positive measures has an inverse limit $\mu$, which is a positive measure on $X$, and $\| \mu \| = \| \mu_\alpha \|$ for all $\alpha$.

(i) We first prove the following lemma:

We note first of all that if $g = g_\beta \circ p_\beta$ and $h = h_\gamma \circ p_\gamma$, where $g_\beta \in \mathscr{C}(X_\beta; \mathbf{C})$ and $h_\gamma \in \mathscr{C}(X_\gamma; \mathbf{C})$, then there exists an $\alpha \in I$ such that $\alpha \geq \beta$ and $\alpha \geq \gamma$, and therefore $p_\beta = p_{\beta \alpha} \circ p_\alpha$, $p_\gamma = p_{\gamma \alpha} \circ p_\alpha$, which shows that

belongs to F; one sees similarly that $gh \in F$; F is thus a C-subalgebra of $\mathscr{C}(X; \mathbf{C})$, which contains the constants and is such that the relation $f \in F$ implies $\overline{f} \in F$. Finally, if $x \neq y$ are two points of X, there exists an $\alpha \in I$ such that $p_\alpha(x) \neq p_\alpha(y)$, therefore there is a function $f_\alpha \in \mathscr{C}(X_\alpha; \mathbf{C})$ such that $f_\alpha(p_\alpha(x)) \neq f_\alpha(p_\alpha(y))$. The conclusion therefore follows from the Stone–Weierstrass theorem (GT, X, §4, No. 4, Prop. 7).

The lemma having been established, let $\mu, \mu'$ be two measures on X such that $(p_\alpha)_*(\mu) = (p_\alpha)_*(\mu')$ for all $\alpha \in I$; this means that, for every $\alpha \in I$ and every function $f_\alpha \in \mathscr{C}(X_\alpha; \mathbf{C})$, we have

that is, $\langle f_\alpha \circ p_\alpha, \mu \rangle = \langle f_\alpha \circ p_\alpha, \mu' \rangle$; in other words, $\mu$ and $\mu'$ coincide on the subspace F of $\mathscr{C}(X; \mathbf{C})$, which is dense by Lemma 3, therefore $\mu = \mu'$, which proves (i).

(ii) The relation (15) applied to $p_\alpha$ shows that if $\mu$ is the inverse limit of the inverse system $(\mu_\alpha)$, necessarily

for all $\alpha \in I$.

(iii) Suppose the $p_{\alpha \beta}$ are surjective; one knows that the same is then true of the $p_\alpha$ (GT, I, §9, No. 6, Prop. 8). Consider an inverse system of measures $(\mu_\alpha)$ and let us first show that there exists a linear form $\lambda$ on F (in the notations of Lemma 3) such that, for every $\alpha \in I$ and every $f_\alpha \in \mathscr{C}(X_\alpha; \mathbf{C})$, $\lambda(f_\alpha \circ p_\alpha) = \mu_\alpha(f_\alpha)$. To that end, let $\beta, \gamma$ be two indices in I, and $f_\beta \in \mathscr{C}(X_\beta; \mathbf{C})$, $f_\gamma \in \mathscr{C}(X_\gamma; \mathbf{C})$ two functions such that $f_\beta \circ p_\beta = f_\gamma \circ p_\gamma$; then there exists an index $\alpha \in I$ such that $\alpha \geq \beta$ and $\alpha \geq \gamma$, therefore

since $p_\alpha$ is surjective, this implies $f_\beta \circ p_{\beta \alpha} = f_\gamma \circ p_{\gamma \alpha}$, therefore

but by definition the last relation may also be written $\mu_\beta(f_\beta) = \mu_\gamma(f_\gamma)$, whence our assertion.

This being so, suppose that there exists a finite number $a > 0$ such that $\| \mu_\alpha \| \leq a$ for all $\alpha \in I$; then, for every function $f_\alpha \in \mathscr{C}(X_\alpha; \mathbf{C})$,

since $p_\alpha$ is surjective. This shows that the linear form $\lambda$ is continuous on $F$, and it follows from Lemma 3 that $\lambda$ may be extended to a measure $\mu$ on $X$ such that $(p_\alpha)_*(\mu) = \mu_\alpha$ for all $\alpha \in I$, which proves (iii).

(iv) To prove the existence of $\mu$ it suffices, by (iii), to verify that the family of norms $(\| \mu_\alpha \|)$ is bounded. But $\| \mu_\alpha \| = \mu_\alpha(1)$ and, when $\alpha \leq \beta$, the relation $\mu_\alpha = (p_{\alpha \beta})_*(\mu_\beta)$ implies that $\mu_\alpha(1) = \mu_\beta(1)$; since $I$ is directed, the total masses of all the measures $\mu_\alpha$ are therefore equal, whence our assertion. Moreover, the subspace $F$ obviously satisfies the property (P) of §1, No. 7, Prop. 9, therefore the inverse limit measure $\mu$ of $(\mu_\alpha)$ is positive. Finally, the relation $\mu_\alpha = (p_\alpha)_*(\mu)$ shows as above that $\mu(1) = \mu_\alpha(1)$.