This follows at once from Remark 1, Prop. 11 of §1, No. 3 and Prop. 3 of §1, No. 1.

The inequality is obvious if $f$ is not $\nu$-moderated, because then $\nu^*(f) = +\infty$ ($§ 1$, No. 2, Prop. 7). If $f$ is $\nu$-moderated then $f$ is $\lambda_\alpha$-moderated for every $\alpha \in A$, because every $\nu$-integrable open set is $\lambda_\alpha$-integrable; the relation (4) then follows at once from (3) and from Prop. 7 of $§ 1$, No. 2.

It can happen that the two members of (4) are not equal, even when $A$ is countable and each of the $\lambda_\alpha$ is a point measure ($§ 1$, Exer. 4 a)).

This follows at once from Cor. 2 of Prop. 11 of $§ 1$.

The family $(\int f \, d\lambda_\alpha)_{\alpha \in A}$ is then absolutely summable in $F$, and

Indeed, for $f$ to be essentially $\nu$-integrable (resp. essentially $\lambda_\alpha$-integrable), it is necessary and sufficient that $f$ be measurable for the measure $\nu$ (resp. $\lambda_\alpha$) and that $\nu^\bullet(|f|) < +\infty$ (resp. $\lambda_\alpha^\bullet(|f|) < +\infty$), by virtue of Prop. 9 of $§ 1$, No. 3. The first part of the statement therefore follows at once from Props. 2 and 1. If $f$ is essentially $\nu$-integrable, the inequality

implies that the family $(\int f \, d\lambda_\alpha)$ is absolutely summable in $F$, and that the norm of the sum is less than or equal to the norm of $f$ in $\mathscr{L}_F^1(\nu)$. The set of $f \in \mathscr{L}_F^1(\nu)$ that satisfy (6) is thus a closed linear subspace $\mathscr{H}$ of $\mathscr{L}_F^1(\nu)$; now, this subspace is also dense in $\mathscr{L}_F^1(\nu)$, because it contains the functions of the form $f \cdot a$, where $a \in F$ and $f$ denotes a finite integrable positive function (Prop. 1). Therefore $\mathscr{H} = \mathscr{L}_F^1(\nu)$ and the proposition is established.

Prop. 3 can also be deduced from the general theorem on integration that will be proved in §3 (No. 3, Th. 1).

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).

If $f$ is measurable (resp. essentially integrable, integrable) for $\theta$, then $f$ is by definition measurable (resp. essentially integrable, integrable) for the measure $|\theta|$, hence also for the measures $\theta_i$, which are $\leq |\theta|$. Conversely, if $f$ is measurable (resp. essentially integrable, integrable) for the measures $\theta_i$, then Prop. 2 (resp. Prop. 3, Cor. 1 of Prop. 3) implies that $f$ is measurable (resp. essentially integrable, integrable) for the measure $\theta_1 + \theta_2 + \theta_3 + \theta_4$, which is $\geq |\theta|$.

If the measure $\mu$ is moderated, the index set $A$ may be taken to be countable.

Consider a locally countable family $(K_\alpha)_{\alpha \in A}$ of pairwise disjoint elements of $\mathfrak{K}$ such that the set $N = T - \bigcup_{\alpha \in A} K_\alpha$ is locally $\mu$-negligible

(Ch. IV, §5, No. 9, Prop. 14). For every function $f \in \mathscr{K}(T)$, set

the linear form $\mu_\alpha$ on $\mathscr{K}(T)$ is positive, therefore is a positive measure, with support contained in $K_\alpha$. Since every compact set contained in an element of $\mathfrak{K}$ belongs to $\mathfrak{K}$, $\operatorname{Supp}(\mu_\alpha) \in \mathfrak{K}$ for all $\alpha \in A$. It remains only to show that the family $(\mu_\alpha)$ is summable and that its sum is equal to $\mu$, in other words that $\sum_{\alpha \in A} \mu_\alpha(f) = \mu(f)$ for every function $f \in \mathscr{K}_+(T)$.

Now, let $S$ be the (compact) support of $f$, and let $A'$ be the countable set formed by the $\alpha \in A$ such that $S \cap K_\alpha \neq \varnothing$. Since the set $N \cap S$ is $\mu$-negligible,

This completes the proof of the general case. If $\mu$ is moderated, then the set $T$ is $\mu$-moderated and so $T$ is the union of a sequence $(L_n)$ of compact sets and a negligible set (§ 1, No. 2, Prop. 5); let $A'$ be the countable set of $\alpha \in A$ such that $K_\alpha$ intersects one of the $L_n$. Then $\mu_\alpha = 0$ for $\alpha \notin A'$, and the last sentence of the statement follows immediately.