— a) Let f be a function on X/H', with values in a Banach space or in R, integrable for μ/β′. There exists a (μ/β)-negligible subset N of X/H having the following property: if x∈X is such that π(x)∈/N, then the function f∘ωx on H/H', that is, the function ξ˙↦f(π′(xξ)), is integrable for β/β′. The integral ∫H/H′f(π′(xξ))d(β/β′)(ξ˙) depends only on x˙=π(x), and is a (μ/β)-integrable function of x˙; and
∫X/H′fd(μ/β′)=∫X/Hd(μ/β)(x˙)∫H/H′f(π′(xξ))d(β/β′)(ξ˙).
b) Let f be a function ≥0 on X/H′, measurable for μ/β′ and zero outside a countable union of (μ/β′)-integrable sets. Then
π(x)↦∫H/H′∗f(π′(xξ))d(β/β′)(ξ˙)
is (μ/β)-measurable, and
∫X/H′∗fd(μ/β′)=∫X/H∗d(μ/β)(x˙)∫H/H′∗f(π′(xξ))d(β/β′)(ξ˙).
c) Let f be a function on X/H′ with values in a Banach space or in R, measurable for μ/β′ and zero outside a countable union of (μ/β′)-integrable sets. Then, for f to be (μ/β′)-integrable, it is sufficient that
∫X/H∗d(μ/β)(x˙)∫H/H′∗∣f(π′(xξ))∣d(β/β′)(ξ˙)<+∞.
Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.
Nothing.