Bourbaki

Lemma 2

Tag 09I3, permanent. Cite it as [Bourbaki, Tag 09I3].

Integration, chapter VIII, § 1. Convolution

*Let T, T', T'' be three locally compact spaces, μ\mu a measure on T, π\pi a μ\mu-measurable mapping of T into T', π′\pi' a continuous mapping of T' into T'', and π′′=π′∘π\pi'' = \pi' \circ \pi. If π′′\pi'' is μ\mu-proper, then π\pi is μ\mu-proper, π′\pi' is π(μ)\pi(\mu)-proper, and π′′(μ)=π′(π(μ))\pi''(\mu) = \pi'(\pi(\mu)).

Let K' be a compact subset of T'. Then K′′=π′(K′)K'' = \pi'(K') is compact, therefore π′′(K′′)\pi''(K'') is essentially μ\mu-integrable, therefore π−1(K′)⊂π′′(K′′)\pi^{-1}(K') \subset \pi''(K'') is essentially μ\mu-integrable, thus π\pi is μ\mu-proper. Then π′\pi' is π(μ)\pi(\mu)-proper and π′′(μ)=π′(π(μ))\pi''(\mu) = \pi'(\pi(\mu)) by Ch. V, §6, No. 3, Prop. 4.

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

Cited by

Cites

Transcribed from the printed volume. Not translated.