Bourbaki

Exercise 18, § 8

Integration, V, p. 120

Bourbaki marks this one as harder than the rest.

The notations being the same as in Exer. 17, let νι\nu_\iota be a positive measure on TιT_\iota, of total mass equal to 1, and let ν=⨂ι∈Iνι\nu = \bigotimes_{\iota \in I} \nu_\iota.

a) Show that if one of the measures νι\nu_\iota is alien to the measure μι\mu_\iota with the same index, then ν\nu is alien to μ\mu.

b) For every ι∈I\iota \in I, write νι=μι′+μι′′\nu_\iota = \mu'_\iota + \mu''_\iota, where μι′\mu'_\iota has base μι\mu_\iota, and μι′′\mu''_\iota is alien to μι\mu_\iota (§ 5, Th. 3); suppose that μι′≠0\mu'_\iota \neq 0 for all ι∈I\iota \in I. Show that, for ν\nu not to be alien to μ\mu, it is necessary and sufficient that μι′′=0\mu''_\iota = 0 except for a countable family of indices, and that the product μ′=⨂ι∈Iμι′\mu' = \bigotimes_{\iota \in I} \mu'_\iota be a measure ≠0\neq 0 with base μ\mu; μ′′=ν−μ′\mu'' = \nu - \mu' is then alien to μ\mu. (If μι′′≠0\mu''_\iota \neq 0 for an uncountable infinity of indices, show that there exists a number α\alpha such that 0<α<10 < \alpha < 1 and such that, for a countable infinity of indices ιn\iota_n, one has μιn′(Tιn)≤α\mu'_{\iota_n}(T_{\iota_n}) \leq \alpha; show then that μ\mu is alien to ν\nu by using a); prove the second part of the proposition by using Exer. 17.)

Solution

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Transcribed from the printed volume. Not translated.