Library / Integration / Chapter VI / § 1
Proposition 1
Tag 098I, permanent. Cite it as [Bourbaki, Tag 098I].
Integration, chapter VI, § 1. Integration of vector-valued functions
— If f is a mapping of T into F that is scalarly essentially μ \mu μ u ∘ f u \circ f u ∘ f μ \mu μ
∫ ( u ∘ f ) d μ = u ( ∫ f d μ ) .
\int (u \circ f)\, d\mu = u \left( \int f\, d\mu \right).
∫ ( u ∘ f ) d μ = u ( ∫ f d μ ) . For every
z ′ ∈ G ′ \mathbf{z}' \in G' z ′ ∈ G ′ ,
⟨ z ′ , u ∘ f ⟩ = ⟨ t u ( z ′ ) , f ⟩ \langle \mathbf{z}', u \circ f \rangle = \langle ^t u(\mathbf{z}'), f \rangle ⟨ z ′ , u ∘ f ⟩ = ⟨ t u ( z ′ ) , f ⟩ , whence the first assertion; the second follows from the formula
⟨ z ′ , ∫ ( u ∘ f ) d μ ⟩ = ∫ ⟨ z ′ , u ∘ f ⟩ d μ = ⟨ t u ( z ′ ) , ∫ f d μ ⟩ = ⟨ z ′ , u ( ∫ f d μ ) ⟩ .
\left\langle \mathbf{z}', \int (u \circ f)\, d\mu \right\rangle = \int \langle \mathbf{z}', u \circ f \rangle\, d\mu = \left\langle ^t u(\mathbf{z}'), \int f\, d\mu \right\rangle = \left\langle \mathbf{z}', u \left( \int f\, d\mu \right) \right\rangle.
⟨ z ′ , ∫ ( u ∘ f ) d μ ⟩ = ∫ ⟨ z ′ , u ∘ f ⟩ d μ = ⟨ t u ( z ′ ) , ∫ f d μ ⟩ = ⟨ z ′ , u ( ∫ f d μ ) ⟩ . In particular, if f is scalarly essentially μ \mu μ μ \mu μ
Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.
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Transcribed from the printed volume. Not translated.