Bourbaki

Exercise 16, § 6

General Topology, IX, p. 122

Let XX be a metrizable space, ff a mapping from the set R(E)\mathscr{R}(E) of compact subsets of XX into R‾\overline{\mathbf{R}}, which satisfies condition (CA_1) (for two compact subsets of XX) and is such that, for every decreasing sequence (Kn)(K_n) of compact subsets of XX, one has

f(⋂nKn)=inf⁡f(Kn). f(\bigcap_n K_n) = \inf f(K_n).
For every subset AA of XX, f∗(A)f_*(A) denotes the upper bound of the f(K)f(K) for all compact subsets KK of AA, and f∗(A)f^*(A) the lower bound of the f∗(U)f_*(U) for all open sets UU containing AA. We say that AA is admissible for ff if f∗(A)=f∗(A)f_*(A) = f^*(A). a) Prove that for every compact subset KK of XX and every ε>0\varepsilon > 0, there exists an open set U⊃KU \supset K such that for every compact subset LL of XX such that K⊂L⊂UK \subset L \subset U, one has
f(L)≤f(K)+ε f(L) \leq f(K) + \varepsilon
(reason by contradiction). Deduce that every compact subset and every open subset of XX is admissible for ff. b) Suppose that ff satisfies the following condition: (AL) For every pair of compact sets K,K′K, K' of XX,
f(K∪K′)+f(K∩K′)≤f(K)+f(K′). f(K \cup K') + f(K \cap K') \leq f(K) + f(K').
Prove that if (Ki)(K_i) is a finite family of compact sets such that f(⋃iKi)<+∞f\left( \bigcup_i K_i \right) < +\infty, (Ki′)(K'_i) a second finite family of compact sets having the same set of indices, such that Ki′⊂KiK'_i \subset K_i and f(Ki′)>−∞f(K'_i) > -\infty for every ii, one has
f(⋃iKi)−f(⋃iKi′)≤∑i(f(Ki)−f(Ki′)). f\left( \bigcup_i K_i \right) - f\left( \bigcup_i K'_i \right) \leq \sum_i (f(K_i) - f(K'_i)).

EXERCISES

Let (Ai),(Bi)(A_i), (B_i) be two finite families of subsets of XX, having the same set of indices, such that Bi⊂AiB_i \subset A_i for every ii, and such that f∗(⋃iAi)<+∞f^*\left( \bigcup_i A_i \right) < +\infty and f∗(Bi)>−∞f^*(B_i) > -\infty for every ii. Prove that if ff satisfies (AL), one has

f∗(⋃iAi)−f∗(⋃iBi)≤∑i(f∗(Ai)−f∗(Bi)). f^*\left( \bigcup_i A_i \right) - f^*\left( \bigcup_i B_i \right) \leq \sum_i (f^*(A_i) - f^*(B_i)).

(Reduce to a set of indices with two elements. First consider the case where the AiA_i and BiB_i are open, and use the following lemma: if U1,U2U_1, U_2 are two open sets, KK a compact set such that K⊂U1∪U2K \subset U_1 \cup U_2, there exist two compact sets K1⊂U1,K2⊂U2K_1 \subset U_1, K_2 \subset U_2 such that K⊂K1∪K2K \subset K_1 \cup K_2.)

c) Suppose that ff satisfies condition (AL). Prove that for every increasing sequence (An)(A_n) of subsets of XX such that

f∗(An)>−∞, f^*(A_n) > -\infty,

one has f∗(⋃nAn)=sup⁡nf∗(An)f^*\left( \bigcup_n A_n \right) = \sup_n f^*(A_n) (use bb). Deduce that if ff does not take the value −∞-\infty in R(X)\mathscr{R}(X), then f∗f^* is a right-continuous capacity on XX.

d) Suppose that ff satisfies condition (AL). Prove that every union of a sequence (An)(A_n) of admissible sets such that f∗(An)>−∞f^*(A_n) > -\infty, is admissible. (Use bb and cc).

Solution

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Machine translation of the English, by gpt-5-6, gpt-5-mini. Not checked by a person. That model is a cut down version of the one the rest of this language was written by.