Suppose that all of the measures μλ are positive and that the family (μλ(1))λ∈L is multipliable in R+ (with product possibly 0). Then there exists one and only one measure μ on X such that, for every finite subset J of L and every function fJ∈C(XJ;C),
μ(fJ∘prJ)=μJ(fJ)λ∈L−J∏μλ(1).
Moreover, the measure μ is positive and its total mass is given by
μ(1)=λ∈L∏μλ(1).
Let F be the vector space consisting of the functions on X of the form fJ∘prJ, where J runs over the directed set of finite subsets of L, and fJ∈C(XJ;C); we again say that F is the space of continuous functions on X that depend only on a finite number of variables. Lemma 3 of No. 5 shows that F is dense in C(X;C), which proves the uniqueness assertion. If μλ0=0 for some λ0∈L then the measure μ=0 meets the requirements, since in the second member of (18) we have μJ(fJ)=0 if λ0∈J and ∏λ∈L−Jμλ(1)=0 if λ0∈/J. We can therefore suppose that μλ=0 for all λ∈J and, since the measures μλ are positive, this implies that μλ(1)=0 for all λ∈L. Let us then set μλ′=μλ/μλ(1) for every λ∈L, so that μλ′ is a positive measure on Xλ such that μλ′(1)=1. It then follows from Prop. 8 of No. 5 that there exists a positive measure μ′ on X of total mass 1, such that μ′(fJ∘prJ)=μJ′(fJ) for every finite subset J of L and every function fJ∈C(XJ;C). The positive measure
μ=(λ∈L∏μλ(1))μ′
then meets the requirements, since
μJ(fJ)=μJ′(fJ)⋅λ∈J∏μλ(1),
λ∈L∏μλ(1)=λ∈J∏μλ(1)⋅λ∈L−J∏μλ(1).
The measure μ defined in Prop. 9 is called the product measure of the family of positive measures (μλ)λ∈L and is denoted by ⨂λ∈Lμλ.
Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.