proof of Choquet’s capacitability theorem
Let be a paved space such that is closed under finite unions and finite intersections, and let be an -capacity (http://planetmath.org/ChoquetCapacity). We prove the capacitability theorem, which states that all -analytic (http://planetmath.org/AnalyticSet2) sets are -capacitable (http://planetmath.org/ChoquetCapacity). The idea is to deduce it from the following special case.
With the above notation, every set in is -capacitable.
Recall that is the collection of countable 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 set , there is a compact paved space (http://planetmath.org/PavedSpace) and such that , where is the projection map from to . Letting be the closure under finite unions and finite intersections of the paving , then the composition is a -capacity, and the projection of any -capacitable set onto is itself -capacitable (see extending a capacity to a Cartesian product (http://planetmath.org/ExtendingACapacityToACartesianProduct)). In particular, so, by the lemma, is -capacitable. Therefore, is -capacitable. It only remains to prove the lemma.
Proof of lemma.
If then there exists such that
For any positive integers let us write
In particular, and, . For any and suppose that we have chosen positive integers such that . Since is a capacity and increases to as increases to infinity,
as tends to infinity. So, by choosing large enough, we have
then . Furthermore, contains and decreases to as tends to infinity. As is an -capacity this gives
So is -capacitable, as required. ∎
|Title||proof of Choquet’s capacitability theorem|
|Date of creation||2013-03-22 18:47:51|
|Last modified on||2013-03-22 18:47:51|
|Last modified by||gel (22282)|