Proposition 3

Tag 090N, permanent. Cite it as [Bourbaki, Tag 090N].

Integration, chapter V, § 2. Summable families of positive measures

— For a mapping $f$ of $X$ into a Banach space $F$ to be essentially $\nu$ -integrable, it is necessary and sufficient that $f$ be essentially $\lambda_\alpha$ -integrable for every $\alpha \in A$ and that

\sum_{\alpha \in A} \int^\bullet |f| \, d\lambda_\alpha < +\infty .

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

\int f \, d\nu = \sum_{\alpha \in A} \int f \, d\lambda_\alpha .

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

\sum_{\alpha \in A} \left| \int f \, d\lambda_\alpha \right| \leq \sum_{\alpha \in A} \int |f| \, d\lambda_\alpha = \nu(|f|)

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).

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

Cited by

Cites

Prop. 1