Proposition 5

Let $(\lambda_\alpha)_{\alpha \in A}$ be a family of positive measures on $T$, directed for the relation $\leqslant$, admitting in $\mathscr{M}(T)$ a supremum $\lambda$. For a positive numerical function $g$ to be locally $\lambda$-*integrable, it is necessary and sufficient that* $g$ be locally $\lambda_\alpha$-*integrable for every* $\alpha \in A$ and that the family $(g \cdot \lambda_\alpha)$ be bounded above in $\mathscr{M}(T)$; *in this case,

It is clear that the condition is necessary. Conversely, suppose that $g$ is locally integrable for each of the measures $\lambda_\alpha$ and that the family $(g \cdot \lambda_\alpha)_{\alpha \in A}$ is bounded above; denote its supremum by $\lambda'$. The function $g$ is then $\lambda$-measurable (§ 1, No. 4, Cor. 2 of Prop. 11); moreover, for every function $h \in \mathscr{K}_+(T)$, (§ 1, No. 4, Prop. 11). This implies first of all that the first member is finite for any $h$, so that $g$ is locally $\lambda$-integrable; the symbol $\int^\bullet$ may therefore be replaced by $\int$, and the formula may be written $\int h \, d(g \cdot \lambda) = \int h \, d\lambda'$. It follows that $g \cdot \lambda = \lambda'$, and this completes the proof.