Bourbaki

Lemma 4

Tag 09AT, permanent. Cite it as [Bourbaki, Tag 09AT].

Integration, VI, p. 29, § 2. Vectorial measures . "

Let F be a Hausdorff locally convex space, ff an element of LFs′∞\mathscr{L}_{F'_s}^\infty. For every numerical function g∈L‾1g \in \overline{\mathscr{L}}^1, the function gfgf is scalarly essentially μ\mu-integrable and ∫gf dμ∈F′\int gf\, d\mu \in F'.

For every z∈Fz \in F, ⟨z,f⟩\langle z, f \rangle belongs to L∞\mathscr{L}^\infty, whence the first assertion. One can suppose moreover, without modifying ∫gf dμ\int gf\, d\mu, that f(T)⊂V∘f(T) \subset V^\circ, where V is a balanced convex neighborhood of 0 in F. Then the relation z∈Vz \in V implies ∣⟨z,f(t)⟩∣≤1|\langle z, f(t) \rangle| \leq 1 for all t∈Tt \in T, whence ∣⟨z,∫gf dμ⟩∣=∣∫⟨z,f⟩g dμ∣≤N‾1(g)|\langle z, \int gf\, d\mu \rangle| = |\int \langle z, f \rangle g\, d\mu| \leq \overline{N}_1(g), which proves that ∫gf dμ∈F′\int gf\, d\mu \in F'.

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

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