Proposition 10

— Let $g$ (resp. $g'$) be a complex function (or a function with values in $\overline{\mathbf{R}}$) defined on $T$ (resp. $T'$). a) If $g$ (resp. $g'$) is locally integrable for $\mu$ (resp. $\mu'$), then the function $g \otimes g': (t, t') \mapsto g(t)g'(t')$ is locally integrable for $\nu = \mu \otimes \mu'$, and

b) Conversely, if $g \otimes g'$ is locally $\nu$-integrable, and if $g'$ is not locally $\mu'$-negligible, then $g$ is locally $\mu$-integrable.

a) Let $K$ and $K'$ be two compact subsets of $T$ and $T'$, respectively; Cor. 2 of Prop. 8 shows that the function $(t, t') \mapsto g(t)g'(t')\varphi_{K \times K'}(t, t')$, equal to $(g\varphi_K) \otimes (g'\varphi_{K'})$, is $\nu$-integrable. Consequently, $g \otimes g'$ is locally $\nu$-integrable. One then verifies immediately that the second member of (15) satisfies the characteristic property of product measures (Ch. III, § 4, No. 1, Th. 1).

b) Now suppose that $g \otimes g'$ is locally $\nu$-integrable, and that $g'$ is not locally $\mu'$-negligible. Let $\mu_1$ be a positive measure with compact support such that $\mu_1 \leq \mu$; $g \otimes g'$ being $(\mu_1 \otimes \mu')$-measurable, $t \mapsto g(t)g'(t')$ is $\mu_1$-measurable except for a locally $\mu'$-negligible set of values of $t'$ (Prop. 2).

Since $g'$ is not zero locally $\mu'$-almost everywhere, from this we deduce that that $g$ is $\mu_1$-measurable, then $\mu$-measurable on decomposing $\mu$ into a sum of a family of measures with compact support (§ 2, No. 3, Prop. 4 and § 2, No. 2, Prop. 2). Having established this point, we may reduce to the case that $g$ and $g'$ are $\geqslant 0$, on replacing $g$ and $g'$ by their absolute values if necessary. Let $K$ be any compact subset of $T$, and let $K'$ be a compact subset of $T'$ such that $\int g' \varphi_{K'} d\mu' \neq 0$. By Prop. 8,

The first factor of the first member is therefore finite, and this completes the proof.

This proposition extends to complex measures, thanks to Prop. 3 of Ch. III, § 4, No. 2.