The number a>0 such that φ−1(μ)=aμ is called the modulus of the automorphism φ and is denoted modGφ or simply modφ.
If f is a μ-integrable function on G, then
∫f(φ−1(x))dμ(x)=(modφ)∫f(x)dμ(x).
If A is a μ-integrable subset of G, then
μ(φ(A))=(modφ)μ(A).
In particular, for s∈G, let is be the inner automorphism x↦s−1xs. Then is−1=δ(s)γ(s), therefore
is−1(μ)=δ(s)μ=ΔG(s)μ,
consequently
modis=ΔG(s).
If G is either discrete or compact, then its normalized Haar measure is transformed into itself by every automorphism φ 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.