Bourbaki

Proposition 5

Tag 098M, permanent. Cite it as [Bourbaki, Tag 098M].

Integration, chapter VI, § 1. Integration of vector-valued functions

— Let μ\mu be a bounded positive measure on TT, SS a μ\mu-measurable set carrying μ\mu (Ch. V, §5, No. 7, Def. 4), ff a scalarly μ\mu-integrable () function with values in FF. Let DD be the closed convex envelope of f(S)f(S) in the space F′F' equipped with the topology σ(F′,F′)\sigma(F', F'). Then ∫f dμ∈μ(T)D\int f\, d\mu \in \mu(T)D.

Since DD is the intersection of the closed half-spaces containing f(S)f(S) (TVS, II, §5, No. 3, Cor. 1 of Prop. 4), it suffices to prove that the relation ⟨f(t),z′⟩≤a\langle f(t), z' \rangle \leq a for all t∈St \in S (where z′∈F′,a∈Rz' \in F', a \in \mathbf{R}) implies ⟨z′,∫f dμ⟩≤a⋅μ(T)\langle z', \int f\, d\mu \rangle \leq a \cdot \mu(T); but since ∫f dμ=∫Sf dμ\int f\, d\mu = \int_S f\, d\mu, this follows from Ch. IV, §4, No. 2, Cor. 1 of Th. 1.

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

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