— Let T be a locally compact space, and H a bounded subset of the normed space Cb(T;C). The mapping (μ,f)↦μ(f) of M+b(T)×H into C is then continuous, when M+b(T) is equipped with the tight topology, and H with the topology of compact convergence.
Let μ∈M+b(T),f∈H, and let M be a real number such that ∣∣μ∣∣<M, and ∣g∣≤M for all g∈H. Let ε be a number >0 and choose a compact subset K of T such that μ∗(T−K)<ε, then a relatively compact open neighborhood S of K. The set U of measures λ∈M+b(T) satisfying the inequalities
λ∗(T)<M,λ∗(T−S)<ε,∣λ(f)−μ(f)∣<ε
is then a neighborhood of μ in M+b(T) (No. 3, Prop. 6). In addition, let V be the neighborhood of f in H consisting of the functions g∈H such that
x∈Ssup∣g(x)−f(x)∣<ε.
Let λ∈U and g∈V; since the function ∣g−f∣ is bounded above by ε in S, and by 2M in T−S, we have
∣λ(g)−λ(f)∣≤∫T∣g−f∣dλ≤ελ∗(S)+2Mλ∗(T−S)≤3Mε, from which one deduces
∣λ(g)−μ(f)∣≤∣λ(g)−λ(f)∣+∣λ(f)−μ(f)∣≤(3M+1)ε. This proves the continuity of the mapping
(λ,g)↦λ(g) at the point
(μ,f) of
M+b(T)×H.
Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.