Bourbaki

Exercise 2, § 3

Integration, IX, p. 108

Let E be a set, Φ\Phi a clan of subsets of E, and V a fully lattice-ordered Riesz space. A mapping m of Φ\Phi into V is said to be positive if m(A)≥0m(A) \geq 0 for all A∈ΦA \in \Phi, that it is additive if m(A∪B)=m(A)+m(B)m(A \cup B) = m(A) + m(B) when the sets A∈ΦA \in \Phi and B∈ΦB \in \Phi are disjoint, and that it is relatively bounded if the set of elements m(B)m(B) for B∈Φ,B⊂AB \in \Phi, B \subset A is bounded above in V for any A∈ΦA \in \Phi.

a) Let m be a relatively bounded mapping of Φ\Phi into V. One defines for every A the elements m+(A),m−(A)m_+(A), m_-(A) and ∣m∣(A)|m|(A) of V as in Exer. 1 a). Show that ∣m∣(A)|m|(A) is the supremum of the set of elements of V of the form ∑i=1p∣m(Ai)∣\sum_{i=1}^{p} |m(A_i)| for all the finite partitions (Ai)1≤i≤p(A_i)_{1 \leq i \leq p} of A into sets belonging to Φ\Phi. From this, deduce that the mapping ∣m∣|m| of Φ\Phi into V is additive, then that m is the difference of the additive positive mappings m+m_+ and m−m_- of Φ\Phi into V.

b) Let m′m' and m′′m'' be two additive positive mappings of Φ\Phi into V and let m=m′−m′′m = m' - m''. Show that m is relatively bounded, and that m′≥m+m' \geq m_+ and m′′≥m−m'' \geq m_-.

c) Generalize Exer. 1 d).

Solution

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Transcribed from the printed volume. Not translated.