Corollary 1

— Suppose that $f$ is $\nu$-integrable; then $f$ is $\lambda_\alpha$-integrable for every $\alpha \in A$, and formula (6) holds. Conversely, if the set $A$ is finite and $f$ is $\lambda_\alpha$-integrable for every $\alpha \in A$, then the function $f$ is $\nu$-integrable.

If $f$ is $\nu$-integrable, then $f$ is essentially $\nu$-integrable and $\nu$-moderated (§ 1, No. 3, Cor. of Prop. 9); $f$ is therefore essentially $\lambda_\alpha$-integrable and $\lambda_\alpha$-moderated, hence $\lambda_\alpha$-integrable, for every $\alpha \in A$. Conversely, if $A$ is finite and if $f$ is $\lambda_\alpha$-integrable for all $\alpha \in A$, then $f$ is essentially $\nu$-integrable by Prop. 3, and it suffices to verify that $\nu^*(|f|) < +\infty$; this follows at once from the relation $\nu^* = \sum_{\alpha \in A} \lambda_\alpha^*$ (Ch. IV, §1, No. 3, Prop. 15).