Bourbaki

Exercise 3, § 1

Integration, V

Bourbaki marks this one as harder than the rest.

a) Let (fα)α∈A(f_\alpha)_{\alpha \in A} be a family of numerical functions ⩾0\geqslant 0, directed for the relation ⩽\leqslant, such that the mapping t↦(fα(t))t \mapsto (f_\alpha(t)) of TT into RA\mathbf{R}^A is μ\mu-measurable. Let ff be the upper envelope of the family (fα)(f_\alpha). Then ff is μ\mu-measurable and

∫∙f dμ=sup⁡α∈A∫∙fα dμ. \int^\bullet f \, d\mu = \sup_{\alpha \in A} \int^\bullet f_\alpha \, d\mu .

(1)

b) Let (fα)α∈A(f_\alpha)_{\alpha \in A} be a family of numerical functions ⩾0\geqslant 0 defined on TT and having the following property: for every compact subset KK of TT and every ε>0\varepsilon > 0, there exists a compact set K′⊂KK' \subset K such that μ(K−K′)⩽ε\mu(K - K') \leqslant \varepsilon and such that the restrictions to K′K' of all the functions fαf_\alpha are lower semi-continuous. Let ff be the upper envelope of the family (fα)(f_\alpha). Show that ff is measurable and satisfies the relation (1).

Give an example of a mapping t↦(fα(t))t \mapsto (f_\alpha(t)), satisfying the above condition, that is not μ\mu-measurable.

Solution

No solution has been written yet.

Transcribed from the printed volume. Not translated.