Proposition 6

— Let $(g_\alpha)_{\alpha \in A}$ be a locally countable family of locally $\mu$-integrable positive numerical functions defined on T. In order that the function $g = \sum_{\alpha \in A} g_\alpha$ be locally $\mu$-integrable, it is necessary and sufficient that the family of measures $(g_\alpha \cdot \mu)_{\alpha \in A}$ be summable, in which case

It is clear that $g$ is $\mu$-measurable (Ch. IV, §5, No. 2, Prop. 4 and No. 4, Cor. 1 of Th. 2). For $g$ to be locally $\mu$-integrable, it is therefore necessary and sufficient that $\mu^\bullet(gf)$ be finite for every $f \in \mathscr{H}_+(T)$. Now, since the set of $\alpha \in A$ such that $g_\alpha f \neq 0$ is countable, we have $\mu^\bullet(gf) = \sum_{\alpha \in A} \mu^\bullet(g_\alpha f)$ (§1, No. 1, Cor. of Prop. 2). Set $\nu_\alpha = g_\alpha \cdot \mu$; the condition $\mu^\bullet(gf) < +\infty$ is equivalent to the condition $\sum_{\alpha \in A} \nu_\alpha(f) < +\infty$: in other words, $g$ is locally $\mu$-integrable if and only if the family $(\nu_\alpha)$ is summable. Denoting the sum of this family by $\nu$, the preceding calculation yields the equality $\nu(f) = \mu^\bullet(gf)$, which is equivalent to (8).