Let f be a continuous real-valued function on X whose support S has compact intersection with the saturation of every compact subset of X.
a) For every x∈X, the function ξ↦f(xξ) on H belongs to K(H); one sets
fχ(x)=∫Hf(xξ)χ(ξ)−1dβ(ξ).
b) The function fχ is continuous, is zero outside SH, and satisfies fχ(xξ)=χ(ξ)fχ(x).
c) If g is a continuous real-valued function on X and satisfies g(xξ)=χ(ξ)g(x), then (fg)χ=f1g (f1 being given by the formula (1) with χ replaced by the representation ξ↦1 of H in R+∗).
d) If η∈H, then (δ(η)f)χ=χ(η)ΔH(η)−1fχ.
Let x0∈X and let V be a compact neighborhood of x0 in X. The set of ξ∈H such that Vξ intersects S is also the set of ξ∈H such that Vξ intersects S∩VH, hence is compact in H since S∩VH is compact and H operates properly in X (GT, III, §4, No. 5, Th. 1); then Lemma 1 of §1, No. 1 proves a) and the continuity of fχ. The rest of b) is obvious. Finally, c) and d) result from the following calculations:
(fg)χ(x)=∫Hf(xξ)g(xξ)χ(ξ)−1dβ(ξ)=∫Hf(xξ)g(x)χ(ξ)χ(ξ)−1dβ(ξ)
=g(x)∫Hf(xξ)dβ(ξ)=g(x)f1(x)
(δ(η)f)χ(x)=∫Hf(xξη)χ(ξ)−1dβ(ξ)
=ΔH(η)−1∫Hf(xξ)χ(ξη−1)−1dβ(ξ)
=χ(η)ΔH(η)−1∫Hf(xξ)χ(ξ)−1dβ(ξ).
Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.