Bourbaki

Proposition 2

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

Integration, IX, p. 24, § 2. Operations on measures

If f is a positive locally μ\mu-integrable function, then, for every function g∈F+(T)g \in \mathscr{F}_+(T), one has the relation

(f⋅μ)∙(g)=μ∙(fg). (f \cdot \mu)^\bullet(g) = \mu^\bullet(fg).

Indeed, for every compact set K in T,

(f⋅μ)K∙(gK)=(fK⋅μK)∙(gK)=μK∙(fKgK)=μK∙((fg)K), (f \cdot \mu)_K^\bullet(g_K) = (f_K \cdot \mu_K)^\bullet(g_K) = \mu_K^\bullet(f_K g_K) = \mu_K^\bullet((fg)_K),

on using the definition of f⋅μf \cdot \mu and Prop. 3 of Ch. V, §5, No. 3. Prop. 2 follows from this on passing to the supremum over K.

Now let x∈Tx \in T and let V be a neighborhood of x such that μ∙(fφV)<+∞\mu^\bullet(f \varphi_V) < +\infty (Def. 2); then (f⋅μ)∙(V)=μ∙(fφV)<+∞(f \cdot \mu)^\bullet(V) = \mu^\bullet(f \varphi_V) < +\infty, therefore f⋅μf \cdot \mu is a measure.

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

Cited by

Cites

Transcribed from the printed volume. Not translated.