Bourbaki

Remark 2

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

Integration, VII, p. 42, § 2. Quotient of a space by a group; homogeneous spaces

If G is a Lie group, we shall see later that the function ρ\rho of Th. 2 can be chosen to be infinitely differentiable.*

Under the conditions of Th. 2, certain results of Nos. 3 and 4 may be specialized as follows (on taking into account Ch. V, §4, Th. 2 and Prop. 2 for passing from properties relative to μ\mu to properties relative to ρ⋅μ\rho \cdot \mu):

a) Let ff be a μ\mu-measurable function on G, with values in a topological space, constant outside a countable union of μ\mu-integrable sets; then, the set of x˙∈G/H\dot{x} \in G/H such that the function ξ↦f(xξ)\xi \mapsto f(x\xi) is not β\beta-measurable is locally λ\lambda-negligible.

b) Let ff be a μ\mu-measurable function ⩾0\geqslant 0 on G, zero outside a countable union of μ\mu-integrable sets. Then, the function

x˙↦∫H∗f(xξ)dβ(ξ) \dot{x} \mapsto \int_{H}^{*} f(x\xi) d\beta(\xi)
on G/HG/H is λ\lambda-measurable and
∫G∗f(x)ρ(x)dμ(x)=∫G/H∗dλ(x˙)∫H∗f(xξ)dβ(ξ)(x˙=π(x)). \int_{G}^{*} f(x)\rho(x) d\mu(x) = \int_{G/H}^{*} d\lambda(\dot{x}) \int_{H}^{*} f(x\xi) d\beta(\xi) \quad (\dot{x} = \pi(x)).

c) Let ff be a ρ⋅μ\rho \cdot \mu-integrable function on G, with values in a Banach space or in R‾\overline{\mathbf{R}}. Then, the set of x˙∈G/H\dot{x} \in G/H such that ξ↦f(xξ)\xi \mapsto f(x\xi) is not β\beta-integrable is λ\lambda-negligible; the function x˙↦∫Hf(xξ)dβ(ξ)\dot{x} \mapsto \int_{H} f(x\xi) d\beta(\xi) is λ\lambda-integrable, and

∫Gf(x)ρ(x)dμ(x)=∫G/Hdλ(x˙)∫Hf(xξ)dβ(ξ). \int_{G} f(x)\rho(x) d\mu(x) = \int_{G/H} d\lambda(\dot{x}) \int_{H} f(x\xi) d\beta(\xi).

d) There exists a continuous function h⩾0h \geqslant 0 on G, whose support has compact intersection with the saturation KH of every compact subset K of G, and such that ∫Hh(xξ)dβ(ξ)=1\int_{H} h(x\xi) d\beta(\xi) = 1 for every x∈Gx \in G. For a function g on G/H to be measurable (resp. locally integrable, essentially integrable, integrable) for λ\lambda, it is necessary and sufficient that h⋅(g∘π)h \cdot (g \circ \pi) be so for ρ⋅μ=λ♯\rho \cdot \mu = \lambda^{\sharp}; and, when g is essentially integrable for λ\lambda, one has

∫G/Hg(u)dλ(u)=∫Gh(x)g(π(x))ρ(x)dμ(x). \int_{G/H} g(u) d\lambda(u) = \int_{G} h(x)g(\pi(x)) \rho(x) d\mu(x).

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

Cited by

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Cites

Transcribed from the printed volume. Not translated.