proof of Choquet’s capacitability theorem

Let (X,) be a paved space such that is closed under finite unions and finite intersectionsMathworldPlanetmath, and let I be an -capacity ( We prove the capacitability theorem, which states that all -analytic ( sets are (,I)-capacitable ( The idea is to deduce it from the following special case.


With the above notation, every set in Fσδ is (F,I)-capacitable.

Recall that σδ is the collectionMathworldPlanetmath of countableMathworldPlanetmath intersections of countable unions in and, since countable unions and intersections of analytic sets are analytic, all such sets are analytic. According to the capacitability theorem they should then be capacitable, and the lemma is indeed a special case.

Supposing that the lemma is true, then the capacitability theorem can be deduced as follows. For an -analytic setMathworldPlanetmath AX, there is a compact paved space ( (K,𝒦) and S(×𝒦)σδ such that A=πX(S), where πX is the projection map from X×K to X. Letting 𝒢 be the closure under finite unions and finite intersections of the paving ×𝒦, then the compositionMathworldPlanetmathPlanetmath IπX is a 𝒢-capacity, and the projection of any (𝒢,IπX)-capacitable set onto X is itself (,I)-capacitable (see extending a capacity to a Cartesian product ( In particular, S𝒢σδ so, by the lemma, is (𝒢,IπX)-capacitable. Therefore, A=πX(S) is (,I)-capacitable. It only remains to prove the lemma.

Proof of lemma.

If Sσδ then there exists Sm,n such that


For any positive integers m1,m2,,mk let us write


In particular, S()=S and, I(S())=I(S). For any ϵ>0 and k suppose that we have chosen positive integers m1,,mk-1 such that I(S(m1,,mk-1))>I(S)-ϵ. Since I is a capacity and S(m1,,mk) increases to S(m1,,mk-1) as mk increases to infinityMathworldPlanetmathPlanetmath,


as mk tends to infinity. So, by choosing mk large enough, we have


Then, by inductionMathworldPlanetmath, we can find an infiniteMathworldPlanetmath sequence m1,m2, such that this inequality holds for every k. Setting


then AS. Furthermore, Ak contains S(m1,,mk) and decreases to A as k tends to infinity. As I is an -capacity this gives


So S is (,I)-capacitable, as required. ∎

Title proof of Choquet’s capacitability theorem
Canonical name ProofOfChoquetsCapacitabilityTheorem
Date of creation 2013-03-22 18:47:51
Last modified on 2013-03-22 18:47:51
Owner gel (22282)
Last modified by gel (22282)
Numerical id 4
Author gel (22282)
Entry type Proof
Classification msc 28A05
Classification msc 28A12