Lemma 1

For $1 \leq i \leq n$, let $X_i, Y_i$ be two locally compact spaces, $\mu_i$ a measure on $X_i$, and $\varphi_i$ a continuous mapping of $X_i$ into $Y_i$. Let $X = \prod_i X_i$, $Y = \prod_i Y_i$, $\mu = \bigotimes_i \mu_i$, and $\varphi$ the mapping of $X$ into $Y$ that is the product of the $\varphi_i$. If $\varphi$ is $\mu$-proper and $\mu_i \neq 0$ for each $i$, then the $\varphi_i$ are $\mu_i$-proper and $\varphi(\mu) = \bigotimes_i \varphi_i(\mu_i)$.

We can suppose that the $\mu_i$ are positive and $n = 2$. Let $f_1 \in \mathscr{K}_+(Y_1)$. Since $\mu_2 \neq 0$, there exists an $f_2 \in \mathscr{K}_+(Y_2)$ such that $f_2 \circ \varphi_2$ is not $\mu_2$-negligible. The function $(x_1, x_2) \mapsto f_1(\varphi_1(x_1)) f_2(\varphi_2(x_2))$ is essentially $\mu$-integrable and continuous, hence $\mu$-integrable. Therefore there exists an $x_2 \in X_2$ such that $f_2(\varphi_2(x_2)) \neq 0$ and such that the function $x_1 \mapsto f_1(\varphi_1(x_1)) f_2(\varphi_2(x_2))$ is $\mu_1$-integrable. Therefore $f_1 \circ \varphi_1$ is $\mu_1$-integrable,

²The lemma follows by induction on part b) of the cited Prop. 11, which is the case $n = 2$; the corresponding result in the first edition of Ch. V (§8, No. 3, Prop. 7) did not include the result of part b).

which proves that $\varphi_1$ is $\mu_1$-proper. One argues similarly for $\varphi_2$. Then $\varphi(\mu) = \bigotimes_i \varphi_i(\mu_i)$ by Prop. 11 of Ch. V, §8, No. 5.

The following lemma completes Prop. 4 of Ch. V, §6, No. 3.\footnote{The assertion of this lemma is in fact part b) of the cited Prop. 4, a part that was not included in the first edition of Ch. V.}