Remark

Tag 08OF, permanent. Cite it as [Bourbaki, Tag 08OF].

Integration, chapter III, § 4. Products of measures

Let $E$ be a Hausdorff locally convex space, and let $f$ be a mapping in $\mathscr{K}(X \times Y; E)$ such that $f(X \times Y)$ is contained in a complete convex subset $C$ of $E$ . Then, for every $y \in Y$ , the integral $h(y) = \int f(x, y)\, d\lambda(x)$ belongs to $E$ (§3, No. 3, Prop. 7); moreover, the function $h$ belongs to $\widetilde{\mathscr{K}}(Y; E)$ : indeed, for every $z' \in E'$ we have

\langle h(y), z' \rangle = \int \langle f(x, y), z' \rangle\, d\lambda(x),

therefore $y \mapsto \langle h(y), z' \rangle$ belongs to $\mathscr{K}(Y; C)$ by Lemma 2. The integral $\int h\, d\mu$ is therefore defined (and a priori belongs to ${E'}^*$ ); let us show that

\iint f(x, y)\, d\lambda(x)\, d\mu(y) = \int d\mu(y) \int f(x, y)\, d\lambda(x)

= \int d\lambda(x) \int f(x, y)\, d\mu(y),

thus generalizing the formula (4). Indeed, for every $z' \in E'$ we have

\left\langle \int \int f d\lambda d\mu, z' \right\rangle = \int \int \langle f, z' \rangle d\lambda d\mu = \int d\mu \int \langle f, z' \rangle d\lambda \\ = \int \left\langle \int f d\lambda, z' \right\rangle d\mu = \left\langle \int d\mu \int f d\lambda, z' \right\rangle

by (4), whence (5).

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

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