Proposition 1

Let g be a function defined locally almost everywhere in T (for the positive measure $\mu$), with values in a Banach space F (resp. in $\overline{\mathbf{R}}$). The following properties are equivalent:

a) For every point $t \in T$, there exists a neighborhood V of t such that the function $g \varphi_V$ is $\mu$-integrable.

b) The function g is $\mu$-measurable and, for every compact set $K \subset T$, $\int^* |g|\varphi_K\,d\mu < +\infty$.

c) For every numerical function $h \in \mathscr{K}(T)$, $gh$ is $\mu$-integrable.

Let us show that a) implies b); for, the function g is measurable by the principle of localization (Ch. IV, §5, No. 2, Prop. 4). On the other hand, for every $t \in K$ there exists, by hypothesis, a neighborhood $V_t$ of $t$ in $T$ such that $g \varphi_{V_t}$ is integrable; one can therefore cover $K$ by a finite number of neighborhoods $V_i$ ($1 \leq i \leq n$) such that the functions $g \varphi_{V_i}$ are integrable. Since $|g| \varphi_K \leq \sum_{i=1}^n |g| \varphi_{V_i}$, one has $\int^* |g| \varphi_K \, d\mu < +\infty$.

Secondly, b) implies c), because $gh$ is then measurable, and if $L$ is the compact support of $h$ then $|gh| \leq \|h\| \cdot |g| \varphi_L$, therefore $\int^* |gh| \, d\mu < +\infty$ by hypothesis; $gh$ is therefore integrable by the criterion for integrability (Ch. IV, §5, No. 6, Th. 5).

Finally, c) implies a). Indeed, for every $t \in T$ let $V$ be a compact neighborhood of $t$. There exists a continuous mapping $h$ of $T$ into $[0,1]$, equal to 1 on $V$ and with compact support (Ch. III, §1, No. 2, Lemma 1); by hypothesis $gh$ is integrable, therefore so is $g \varphi_V = (gh) \varphi_V$ (Ch. IV, §5, No. 6, Cor. 3 of Th. 5).