Bourbaki

Proposition 10

Tag 09P6, permanent. Cite it as [Bourbaki, Tag 09P6].

Integration, IX, p. 32, § 2. Operations on measures

Let ν\nu be the product premeasure of λ\lambda and μ\mu; for every function f∈F+(S)f \in \mathscr{F}_+(S) and every function g∈F+(T)g \in \mathscr{F}_+(T),

ν∙(f⊗g)=λ∙(f)μ∙(g). \nu^\bullet(f \otimes g) = \lambda^\bullet(f)\mu^\bullet(g) .

The premeasure ν\nu is the only premeasure on S×TS \times T that satisfies (4).

As KK (resp. LL) runs over the set of compact subsets of SS (resp. of TT), we have

ν∙(f⊗g)=sup⁡K,LνK×L∙((f⊗g)K×L)=sup⁡K,L(λK⊗μL)∙(fK⊗gL)=sup⁡K,LλK∙(fK)⋅μL∙(gL)=(sup⁡KλK∙(fK))(sup⁡LμL∙(gL))=λ∙(f)μ∙(g) \begin{align*} \nu^\bullet(f \otimes g) &= \sup_{K,L} \nu^\bullet_{K \times L}((f \otimes g)_{K \times L}) = \sup_{K,L} (\lambda_K \otimes \mu_L)^\bullet(f_K \otimes g_L) \\ &= \sup_{K,L} \lambda_K^\bullet(f_K) \cdot \mu_L^\bullet(g_L) = (\sup_K \lambda_K^\bullet(f_K)) (\sup_L \mu_L^\bullet(g_L)) \\ &= \lambda^\bullet(f)\mu^\bullet(g) \end{align*}

by Prop. 8 of Ch. V, §8, No. 3.

Let η\eta be a second premeasure on S×TS \times T satisfying (4), and let KK and LL be compact subsets of SS and TT respectively, ff and gg elements of F+(K)\mathscr{F}_+(K) and F+(L)\mathscr{F}_+(L) respectively. One has the relation (f⊗g)0=f0⊗g0(f \otimes g)^0 = f^0 \otimes g^0 between the extensions by 0, therefore (§1, No. 2, Prop. 2)

ηK×L∙(f⊗g)=η∙((f⊗g)0)=η∙(f0⊗g0)=λ∙(f0)μ∙(g0)=λK∙(f)μL∙(g). \eta^\bullet_{K \times L}(f \otimes g) = \eta^\bullet((f \otimes g)^0) = \eta^\bullet(f^0 \otimes g^0) \\ = \lambda^\bullet(f^0)\mu^\bullet(g^0) = \lambda_K^\bullet(f)\mu_L^\bullet(g) .

In particular, if one takes f∈K+(K)f \in \mathscr{K}_+(K), g∈K+(L)g \in \mathscr{K}_+(L), one sees that ηK×L\eta_{K \times L} has the characteristic property of the product measure λK⊗μL\lambda_K \otimes \mu_L (Ch. III, §4, No. 1, Th. 1). Therefore ηK×L=νK×L\eta_{K \times L} = \nu_{K \times L}; since every compact subset of S×TS \times T is contained in a set of the form K×LK \times L, the transitivity of induced measures implies that η=ν\eta = \nu.

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.