Bourbaki

Lemma 1

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

Integration, chapter VII, Appendix 2.

Let X, B be two locally compact spaces, π a mapping of X into B, ν a positive measure on B. Let b ↦ λb (b ∈ B) be a ν-adequate family of positive measures on X such that, for every b ∈ B, the measure λ_b is concentrated on π‾(b)\overline{\pi}(b). Set μ=∫λbdν(b)\mu = \int \lambda_b d\nu(b), and assume that the mapping π is μ-measurable.

a) If N ⊂ B is locally ν-negligible, then π‾(N)\overline{\pi}(N) is locally μ-negligible. b) If f is a ν-measurable function on B (with values in a topological space), then f∘πf \circ \pi is μ-measurable on X.

Let K be a compact subset of X. We are to prove that π‾(N)∩K\overline{\pi}(N) \cap K is μ-negligible and that the restriction of f∘πf \circ \pi to K is μ-measurable. Now, K is the union of a μ-negligible set and a sequence of compact subsets KnK_n such that π∣Kn\pi|K_n is continuous. It suffices to show that π‾(N)∩Kn\overline{\pi}(N) \cap K_n is μ-negligible and that the restriction of f∘πf \circ \pi to KnK_n is μ-measurable. We may therefore assume henceforth that π∣K\pi|K is continuous. Then π(K)=K′\pi(K) = K' is compact. Since π‾(N)∩K=π‾(N∩K′)∩K\overline{\pi}(N) \cap K = \overline{\pi}(N \cap K') \cap K and N∩K′N \cap K' is ν-negligible, we may assume henceforth that N is ν-negligible. Then N is contained in a ν-negligible set N′N' that is a countable intersection of open sets (Ch. IV, §4, No. 6, Cor. 2 of Th. 4). Since π\pi is μ-measurable, π‾(N′)\overline{\pi}(N') is μ-measurable (Ch. IV, §5, No. 5, Prop. 7), therefore π‾(N′)∩K\overline{\pi}(N') \cap K is μ-integrable, and (Ch. V, §3, No. 4, Th. 1)

μ(π‾(N′)∩K)=∫Bλb(π‾(N′)∩K)dν(b). \mu(\overline{\pi}(N') \cap K) = \int_B \lambda_b (\overline{\pi}(N') \cap K) d\nu(b).

*Cf. Ch. V, §6, No. 6, Cor. 1 of Prop. 10; the second edition of Ch. V was not available when Ch. VII was published.

Now, if b∉N′b \notin N', π−1(N′)∩K\pi^{-1}(N') \cap K is λb\lambda_b-negligible since λb\lambda_b is concentrated on π−1(b)\pi^{-1}(b) by hypothesis. Therefore μ(π−1(N′)∩K)=0\mu(\pi^{-1}(N') \cap K) = 0. A fortiori, π−1(N)∩K\pi^{-1}(N) \cap K is μ\mu-negligible and a) has indeed been proved. On the other hand, there exists a partition of K′K' formed by a ν\nu-negligible set MM and a sequence (Kn′)(K'_n) of compact sets such that f∣Kn′f|_{K'_n} is continuous. Then the restriction of f∘πf \circ \pi to each set π−1(K)∩Kn′\pi^{-1}(K) \cap K'_n is continuous, and π−1(M)∩K\pi^{-1}(M) \cap K is μ\mu-negligible by a), therefore the restriction of f∘πf \circ \pi to KK is indeed μ\mu-measurable.

Exercises

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

Cited by

Nothing.

Cites

Transcribed from the printed volume. Not translated.