Bourbaki

Definition 4

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

Integration, chapter VII, § 1. Construction of a Haar measure

The number a>0a > 0 such that φ−1(μ)=aμ\varphi^{-1}(\mu) = a \mu is called the modulus of the automorphism φ\varphi and is denoted mod⁡Gφ\operatorname{mod}_G \varphi or simply mod⁡φ\operatorname{mod} \varphi.

If ff is a μ\mu-integrable function on G, then

∫f(φ−1(x))dμ(x)=(mod⁡φ)∫f(x)dμ(x). \int f(\varphi^{-1}(x)) d\mu(x) = (\operatorname{mod} \varphi) \int f(x) d\mu(x).

If A is a μ\mu-integrable subset of G, then

μ(φ(A))=(mod⁡φ)μ(A). \mu(\varphi(A)) = (\operatorname{mod} \varphi) \mu(A).

In particular, for s∈Gs \in G, let isi_s be the inner automorphism x↦s−1xsx \mapsto s^{-1} x s. Then is−1=δ(s)γ(s)i_s^{-1} = \delta(s) \gamma(s), therefore

is−1(μ)=δ(s)μ=ΔG(s)μ, i_s^{-1}(\mu) = \delta(s) \mu = \Delta_G(s) \mu,

consequently

mod⁡is=ΔG(s). \operatorname{mod} i_s = \Delta_G(s).

If G is either discrete or compact, then its normalized Haar measure is transformed into itself by every automorphism φ\varphi of G, as one sees immediately by transport of structure. Thus an automorphism of a discrete or compact group has modulus 1.

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.