Bourbaki

Proposition 3

Tag 08NT, permanent. Cite it as [Bourbaki, Tag 08NT].

Integration, chapter III, § 3. Integrals of continuous vector-valued functions

— For every function g∈C(X;C)g \in \mathscr{C}(X;\mathbf{C}) and every function f∈K~(X;E)f \in \widetilde{\mathscr{K}}(X;E), the function gfgf belongs to K~(X;E)\widetilde{\mathscr{K}}(X;E) and

∫f d(g⋅μ)=∫fg dμ. \int f\, d(g \cdot \mu) = \int fg\, d\mu.

Indeed, for every z′∈E′z' \in E', ⟨gf,z′⟩=g⟨f,z′⟩\langle gf, z' \rangle = g \langle f, z' \rangle, whence the first assertion; moreover,

⟨∫f d(g⋅μ),z′⟩=∫⟨f,z′⟩ d(g⋅μ)=∫g⟨f,z′⟩ dμ=∫⟨gf,z′⟩ dμ=⟨∫gf dμ,z′⟩, \left\langle \int f\, d(g \cdot \mu), z' \right\rangle = \int \langle f, z' \rangle\, d(g \cdot \mu) = \int g \langle f, z' \rangle\, d\mu \\ = \int \langle gf, z' \rangle\, d\mu = \left\langle \int gf\, d\mu, z' \right\rangle,

whence (3).

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

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Cites

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Transcribed from the printed volume. Not translated.