proof of Choquet’s capacitability theorem

proof of Choquet’s capacitability theorem

Let

(

X

,

ℱ

)

be a

paved space

such that

ℱ

is

closed under

finite unions and finite

intersections

, and let

I

be an

ℱ

-

capacity

(

http://planetmath.org/ChoquetCapacity

). We prove the

capacitability theorem

, which states that all

ℱ

-

analytic

(

http://planetmath.org/AnalyticSet2

) sets are

(

ℱ

,

I

)

-capacitable (

http://planetmath.org/ChoquetCapacity

). The idea is to deduce it from the following special case.

Lemma

.

With the above notation, every set in

F

σ

⁢

δ

is

(

F

,

I

)

-capacitable.

Recall that

ℱ

σ

⁢

δ

is the

collection

of

countable

intersections of countable unions in

ℱ

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

ℱ

-

analytic set

A

⊆

X

, there is a compact paved space (

http://planetmath.org/PavedSpace

)

(

K

,

𝒦

)

and

S

∈

(

ℱ

×

𝒦

)

σ

⁢

δ

such that

A

=

π

X

⁢

(

S

)

, where

π

X

is the

projection map

from

X

×

K

to

X

. Letting

𝒢

be the

closure

under finite unions and finite intersections of the paving

ℱ

×

𝒦

, then the

composition

I

∘

π

X

is a

𝒢

-capacity, and the projection of any

(

𝒢

,

I

∘

π

X

)

-capacitable set onto

X

is itself

(

ℱ

,

I

)

-capacitable (see extending a capacity to a

Cartesian product

(

http://planetmath.org/ExtendingACapacityToACartesianProduct

)). In particular,

S

∈

𝒢

σ

⁢

δ

so, by the lemma, is

(

𝒢

,

I

∘

π

X

)

-capacitable. Therefore,

A

=

π

X

⁢

(

S

)

is

(

ℱ

,

I

)

-capacitable. It only remains to prove the lemma.

Proof of lemma.

If

S

∈

ℱ

σ

⁢

δ

then there exists

S

m

,

n

∈

ℱ

such that

S

=

⋂

n

⋃

m

S

m

,

n

.

For any positive integers

m

1

,

m

2

,

…

,

m

k

let us write

S

⁢

(

m

1

,

m

2

,

…

,

m

k

)

≡

(

⋂

n

≤

k

⋃

m

≤

m

n

S

m

,

n

)

⁢

⋂

(

⋂

n

>

k

⋃

m

S

m

,

n

)

.

In particular,

S

⁢

(

)

=

S

and,

I

⁢

(

S

⁢

(

)

)

=

I

⁢

(

S

)

. For any

ϵ

>

0

and

k

∈

ℕ

suppose that we have chosen positive integers

m

1

,

…

,

m

k

-

1

such that

I

⁢

(

S

⁢

(

m

1

,

…

,

m

k

-

1

)

)

>

I

⁢

(

S

)

-

ϵ

. Since

I

is a capacity and

S

⁢

(

m

1

,

…

,

m

k

)

increases to

S

⁢

(

m

1

,

…

,

m

k

-

1

)

as

m

k

increases to

infinity

,

I

⁢

(

S

⁢

(

m

1

,

…

,

m

k

)

)

→

I

⁢

(

S

⁢

(

m

1

,

…

,

m

k

-

1

)

)

as

m

k

tends to infinity. So, by choosing

m

k

large enough, we have

I

⁢

(

S

⁢

(

m

1

,

…

,

m

k

)

)

>

I

⁢

(

S

)

-

ϵ

.

Then, by

induction

, we can find an

infinite

sequence

m

1

,

m

2

,

…

such that this inequality holds for every

k

.
Setting

A

k

≡

⋂

n

≤

k

⋃

m

≤

m

n

S

m

,

n

∈

ℱ

,

A

≡

⋂

n

⋃

m

≤

m

n

S

m

,

n

=

⋂

k

A

k

∈

ℱ

δ

,

then

A

⊆

S

. Furthermore,

A

k

contains

S

⁢

(

m

1

,

…

,

m

k

)

and decreases to

A

as

k

tends to infinity. As

I

is an

ℱ

-capacity this gives

I

⁢

(

A

)

=

lim

k

→

∞

⁡

I

⁢

(

A

k

)

≥

lim

k

→

∞

⁡

I

⁢

(

S

⁢

(

m

1

,

…

,

m

k

)

)

≥

I

⁢

(

S

)

-

ϵ

.

So

S

is

(

ℱ

,

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

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