Bourbaki

Exercise 13, § 5

Integration, IV, p. 129

Bourbaki marks this one as harder than the rest.

Let μ\mu be a positive measure on XX. If AA is measurable, show that for every subset BB of XX,

μ∗(B)=μ∗(B∩A)+μ∗(B∩CA) \mu^*(B) = \mu^*(B \cap A) + \mu^*(B \cap \mathbf{C}_A)

(if μ∗(B)<+∞\mu^*(B) < +\infty, consider an integrable set B1B_1 such that B⊂B1B \subset B_1 and μ∗(B)=μ(B1)\mu^*(B) = \mu(B_1) (§4§ 4, Exer. 7 b))). Conversely, show that if AA satisfies this condition, then AA is measurable (cf. §4§ 4, Exer. 6 e)).

Solution

No solution has been written yet.

Transcribed from the printed volume. Not translated.