Bourbaki

Example

Tag 09A4, permanent. Cite it as [Bourbaki, Tag 09A4].

Integration, chapter VI, § 2. Vectorial measures . "

— 1) The identity mapping of K(T)\mathscr{K}(T) is a vectorial measure on T with values in K(T)\mathscr{K}(T).

2) Let H be a complex Hilbert space, L(H)\mathscr{L}(H) the normed algebra of continuous endomorphisms of H. Let A be a subalgebra of L(H)\mathscr{L}(H), commutative, closed, self-adjoint and containing the identity; one shows that there exist a compact space X and an isomorphism of the normed algebra A onto the algebra KC(X)=CC(X)\mathscr{K}_\mathbf{C}(X) = \mathscr{C}_\mathbf{C}(X) of continuous complex functions on X, equipped with the norm ∥f∥=sup⁡x∈X∣f(x)∣\|f\| = \sup_{x \in X} |f(x)|. The inverse isomorphism, restricted to K(X)\mathscr{K}(X), is a vectorial measure m on X, with values in L(H)\mathscr{L}(H), such that m(fg)=m(f)m(g)m(fg) = m(f)m(g)._

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.