Bourbaki

Exercise 7, § 5

Integration, V, p. 100

Bourbaki marks this one as harder than the rest.

a) Let μ\mu be a positive measure on TT, and let AA be a μ\mu-integrable set such that μ(A)>0\mu(A) > 0. Show that if, for every μ\mu-integrable set B⊂AB \subset A, either μ(B)=0\mu(B) = 0 or μ(B)=μ(A)\mu(B) = \mu(A), then there exists a point a∈Aa \in A such that μ({a})=μ(A)\mu(\{a\}) = \mu(A). (Consider the intersection of the compact sets K⊂AK \subset A such that μ(K)=μ(A)\mu(K) = \mu(A); show that it is nonempty, has measure equal to μ(A)\mu(A), and reduces to a single point.)

b) Suppose that μ\mu is a diffuse measure. For every μ\mu-integrable set AA such that μ(A)>0\mu(A) > 0, and every ε>0\varepsilon > 0, show that there exists a μ\mu-integrable subset BB of AA such that 0<μ(B)≤ε0 < \mu(B) \leq \varepsilon (observe, with the help of aa), that there exists a μ\mu-integrable subset CC of AA such that 0<μ(C)≤12μ(A)0 < \mu(C) \leq \frac{1}{2} \mu(A)). From this, deduce that when XX runs over the set of μ\mu-integrable subsets of AA, the set of values of μ(X)\mu(X) is the closed interval [0,μ(A)][0, \mu(A)] (for every number β\beta such that 0<β<μ(A)0 < \beta < \mu(A), let γ\gamma be the supremum of the measures of the measurable subsets XX of AA such that μ(X)≤β\mu(X) \leq \beta; first show that γ=β\gamma = \beta, using the preceding result, then prove that there exists an increasing sequence (Xn)(X_n) of measurable subsets of AA such that lim⁡n→∞μ(Xn)=β\lim_{n \to \infty} \mu(X_n) = \beta).

Solution

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Transcribed from the printed volume. Not translated.