proof of capacity generated by a measure

proof of capacity generated by a measure

For a

finite measure space

(

X

,

ℱ

,

μ

)

, define

μ

*

:

𝒫

⁢

(

X

)

→

ℝ

+

,

μ

*

⁢

(

S

)

=

inf

⁡

{

μ

⁢

(

A

)

:

A

∈

ℱ

,

A

⊇

S

}

.

We show that

μ

*

is an

ℱ

-

capacity

and that a subset

S

⊆

X

is

(

ℱ

,

μ

*

)

-capacitable (

http://planetmath.org/ChoquetCapacity

) if and only if it is in the

completion

(

http://planetmath.org/CompleteMeasure

) of

ℱ

with respect to

μ

.

Note, first of all, that

μ

*

⁢

(

S

)

=

μ

⁢

(

S

)

for any

S

∈

ℱ

.
That

μ

*

is

increasing

follows directly from the definition. If

A

n

∈

ℱ

is a

decreasing sequence

of sets then

A

=

⋂

n

A

n

is also in

ℱ

and, by

continuity

from above (

http://planetmath.org/PropertiesForMeasure

) for

measures

,

μ

*

⁢

(

A

n

)

=

μ

⁢

(

A

n

)

→

μ

⁢

(

A

)

=

μ

*

⁢

(

A

)

as

n

→

∞

.

Now suppose that

S

n

is an

increasing sequence

of subsets of

X

and set

S

=

⋃

n

S

n

. Then,

μ

*

⁢

(

S

n

)

≤

μ

*

⁢

(

S

)

for each

n

and, hence,

lim

n

→

∞

⁡

μ

*

⁢

(

S

n

)

≤

μ

*

⁢

(

S

)

.

To prove the reverse

inequality

, choose any

ϵ

>

0

and

sequence

A

n

∈

ℱ

with

S

n

⊆

A

n

and

μ

⁢

(

A

n

)

≤

μ

*

⁢

(

S

n

)

+

2

-

n

⁢

ϵ

.
Then,

A

m

∩

A

n

⊇

S

n

whenever

m

≥

n

and, therefore,

μ

⁢

(

A

n

∖

A

m

)

=

μ

⁢

(

A

n

)

-

μ

⁢

(

A

m

∩

A

n

)

≤

μ

⁢

(

A

n

)

-

μ

*

⁢

(

S

n

)

≤

2

-

n

⁢

ϵ

.

Additivity

of

μ

then gives

μ

⁢

(

⋃

m

≤

n

A

m

)

≤

μ

⁢

(

A

n

)

+

∑

m

=

1

n

-

1

μ

⁢

(

A

m

∖

A

n

)

≤

μ

*

⁢

(

S

n

)

+

ϵ

.

So, by continuity from below for measures,

μ

*

⁢

(

S

)

≤

μ

⁢

(

⋃

n

A

n

)

=

lim

n

→

∞

⁡

μ

⁢

(

⋃

m

≤

n

A

m

)

≤

lim

n

→

∞

⁡

μ

*

⁢

(

S

n

)

+

ϵ

.

Choosing

ϵ

arbitrarily small shows that

μ

*

⁢

(

S

n

)

→

μ

*

⁢

(

S

)

and, therefore,

μ

*

is indeed an

ℱ

-capacity.

Now suppose that

S

is in the completion of

ℱ

with respect to

μ

, so that there exists

A

,

B

∈

ℱ

with

A

⊆

S

⊆

B

and

μ

⁢

(

B

∖

A

)

=

0

. Then,

μ

*

⁢

(

A

)

=

μ

⁢

(

A

)

=

μ

⁢

(

B

)

≥

μ

*

⁢

(

S

)

and

S

is indeed

(

ℱ

,

μ

*

)

-capacitable.

Conversely

, let

S

be

(

ℱ

,

μ

*

)

-capacitable. Then, there exists

A

n

,

B

n

∈

ℱ

such that

A

n

⊆

S

⊆

B

n

and

μ

⁢

(

A

n

)

≥

μ

*

⁢

(

S

)

-

1

/

n

,

μ

⁢

(

B

n

)

≤

μ

*

⁢

(

S

)

+

1

/

n

.

Setting

A

=

⋃

n

A

n

and

B

=

⋂

n

B

n

gives

A

⊆

S

⊆

B

and

μ

⁢

(

A

)

≥

μ

*

⁢

(

S

)

≥

μ

⁢

(

B

)

.

So

μ

⁢

(

B

∖

A

)

=

μ

⁢

(

B

)

-

μ

⁢

(

A

)

=

0

, as required.

Title

proof of capacity

generated by

a measure

Canonical name

ProofOfCapacityGeneratedByAMeasure

Date of creation

2013-03-22 18:47:55

Last modified on

2013-03-22 18:47:55

Owner

gel (22282)

Last modified by

gel (22282)

Numerical id

5

Author

gel (22282)

Entry type

Proof

Classification

msc 28A12

Classification

msc 28A05

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