Proposition 11

Let $\pi$ (resp. $\pi'$) be a mapping of $T$ (resp. $T'$) into a locally compact space $T_1$ (resp. $T'_1$).

a) If $\pi$ (resp. $\pi'$) is $\mu$-proper (resp. $\mu'$-proper), then the mapping $\pi \times \pi'$ is $(\mu \otimes \mu')$-proper and $(\pi \times \pi')(\mu \otimes \mu') = \pi(\mu) \otimes \pi'(\mu')$.

b) Conversely, if $\pi \times \pi'$ is $(\mu \otimes \mu')$-proper and $\mu' \neq 0$, then $\pi$ is $\mu$-proper.

a) For, $\pi \times \pi'$ is $(\mu \times \mu')$-measurable by Cor. 1 of Prop. 3 of No. 2. On the other hand, if $K$ (resp. $K'$) is a compact subset of $T_1$ (resp. $T'_1$), then $\pi^{-1}(K)$ and ${\pi'}^{-1}(K')$ are essentially integrable for $\mu$ and $\mu'$, respectively, therefore $\pi^{-1}(K) \times {\pi'}^{-1}(K')$ is essentially integrable for $\mu \otimes \mu'$ (Cor. 2 of Prop. 8). This proves that $\pi \times \pi'$ is $(\mu \times \mu')$-proper. Now let $\mu_1 = \pi(\mu)$, $\mu'_1 = \pi'(\mu')$, $\nu_1 = (\pi \times \pi')(\mu \otimes \mu')$; for $f \in \mathscr{K}(T_1)$ and $f' \in \mathscr{K}(T'_1)$, we have

(Cor. 2 of Prop. 8), which proves that $\nu_1 = \mu_1 \otimes \mu'_1$ (Ch. III, § 4, No. 1, Th. 1).

b) Now suppose that $\pi \times \pi'$ is $\mu \otimes \mu'$-proper and that $\mu' \neq 0$. Let $\mu_1$ be a measure $\leq \mu$ with compact support. The function $\pi \times \pi'$ being measurable for $\mu_1 \otimes \mu'$, the mapping $t \mapsto (\pi(t), \pi'(t'))$ is $\mu$-measurable except for $t'$ forming a locally $\mu'$-negligible set (No. 2, Prop. 2). Since $\mu' \neq 0$, it follows that $\pi$ is $\mu_1$-measurable, and finally that $\pi$ is $\mu$-measurable

(§2, No. 3, Prop. 4 and §2, No. 2, Prop. 2). It remains to show that $\mu^\bullet(f \circ \pi) < +\infty$ for every function $f \in \mathscr{K}_+(\mathrm{T}_1)$. If $\mu$ is zero, this property is obvious. If $\mu$ is not zero, then neither is $\mu \otimes \mu'$, consequently $(\pi \times \pi')(\mu \otimes \mu') \neq 0$ (§6, No. 2, Cor. 1 of Prop. 2). By Lemma 1 of Ch. III, §4, No. 1, there exist two functions $g \in \mathscr{K}_+(\mathrm{T}_1)$, $g' \in \mathscr{K}_+(\mathrm{T}'_1)$ such that

This expression being equal to $\langle \mu \otimes \mu', (g \circ \pi) \otimes (g' \circ \pi') \rangle$ by the definition of image measures, Prop. 8 implies that ${\mu''}^\bullet(g' \circ \pi') \neq 0$. We then have, by Prop. 8 and by Prop. 2 of §6, No. 2,

The first integral in the first member is therefore finite, which completes the proof.

This result extends at once to the product of two complex measures (apply the statement to their absolute values). The same is true for the following proposition.