Proposition 1

Tag 0972, permanent. Cite it as [Bourbaki, Tag 0972].

Integration, chapter V, § 8. Products of measures

— Let $(\mu_\alpha)_{\alpha \in A}$ (resp. $(\mu'_\beta)_{\beta \in B}$ ) be a summable family of positive measures on T (resp. on T'), with sum denoted by $\mu$ (resp. by $\mu'$ ). The family $(\mu_\alpha \otimes \mu'_\beta)_{(\alpha, \beta) \in A \times B}$ is then summable on $T \times T'$ , and

\mu \times \mu' = \sum_{(\alpha, \beta) \in A \times B} \mu_\alpha \otimes \mu'_\beta.

These properties are obvious when A and B are finite. It follows that if $A'$ (resp. $B'$ ) is a finite subset of A (resp. of B), then

\sum_{(\alpha, \beta) \in A' \times B'} \mu_\alpha \otimes \mu'_\beta \leq \mu \otimes \mu'.

The family $(\mu_\alpha \otimes \mu'_\beta)$ is therefore summable. To show that the two members of (4) are equal, it suffices to prove that the second member satisfies the characteristic property of product measures (Ch. III, §4, No. 1, Th. 1), which is shown by the following calculation.

Let $f$ be an element of $\mathscr{K}_+(T)$ , $f'$ an element of $\mathscr{K}_+(T')$ ; recall that $f \otimes f'$ denotes the function $(t, t') \mapsto f(t)f'(t')$ on $T \times T'$ , which belongs to $\mathscr{K}_+(T \times T')$ (A, II, §7, No. 7). Then, by the definition of product measures,

\sum_{(\alpha, \beta) \in A \times B} \langle \mu_\alpha \otimes \mu'_\beta, f \otimes f' \rangle = \sum_{(\alpha, \beta) \in A \times B} (\langle \mu_\alpha, f \rangle \langle \mu'_\beta, f' \rangle) = \left( \sum_{\alpha \in A} \langle \mu_\alpha, f \rangle \right) \left( \sum_{\beta \in B} \langle \mu'_\beta, f' \rangle \right) = \langle \mu, f \rangle \langle \mu', f' \rangle = \langle \mu \otimes \mu', f \otimes f' \rangle.

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

Cited by

Proposition 7, § 8

Cites

III, §4, No. 1, Th. 1