proof of Carathéodory’s lemma

proof of Carathéodory’s lemma

A set

S

⊆

X

is

μ

-

measurable

if and only if

μ

⁢

(

E

)

≥

μ

⁢

(

E

∩

S

)

+

μ

⁢

(

E

∩

S

c

)

(1)

for every

E

⊆

X

.
As this

inequality

is clearly satisfied if

S

=

∅

and is unchanged when

S

is replaced by

S

c

, then

𝒜

contains the

empty set

and is

closed under

taking

complements

of sets.
To show that

𝒜

is a

σ

-algebra, it only remains to show that it is closed under taking

countable

unions of sets. Choose any sets

A

,

B

∈

𝒜

and

E

⊆

X

. Then,

μ

⁢

(

E

)

≥

μ

⁢

(

E

∩

A

)

+

μ

⁢

(

E

∩

A

c

)

≥

μ

⁢

(

E

∩

A

)

+

μ

⁢

(

E

∩

A

c

∩

B

)

+

μ

⁢

(

E

∩

A

c

∩

B

c

)

≥

μ

⁢

(

E

∩

(

A

∪

B

)

)

+

μ

⁢

(

E

∩

A

c

∩

B

c

)

The first two inequalities here follow from applying (

1

) with

A

and then

B

in place of

S

, and the third uses the subadditivity of

μ

together with

A

∪

(

A

c

∩

B

)

=

A

∪

B

. So (

1

) is satisfied with

A

∪

B

in place of

S

, showing that

𝒜

is closed under taking pairwise unions and is therefore an

algebra of sets

on

X

. If

A

,

B

are

disjoint sets

in

𝒜

then replacing

E

by

E

∩

(

A

∪

B

)

and

S

by

A

in (

1

) gives

μ

⁢

(

E

∩

(

A

∪

B

)

)

≥

μ

⁢

(

E

∩

A

)

+

μ

⁢

(

E

∩

B

)

. As the reverse inequality follows from subadditivity of

μ

, this implies that

μ

⁢

(

E

∩

(

A

∪

B

)

)

=

μ

⁢

(

E

∩

A

)

+

μ

⁢

(

E

∩

B

)

.

So, the map

A

↦

μ

⁢

(

E

∩

A

)

is an additive set function on

𝒜

. In particular, taking

E

=

X

shows that

μ

is

additive

on

𝒜

.

Now choose a

sequence

A

i

∈

𝒜

, and set

B

i

≡

⋃

j

=

1

i

A

j

which are in the algebra

𝒜

. To prove that

𝒜

is a

σ

-algebra it needs to be shown that

A

≡

⋃

i

A

i

=

⋃

i

B

i

is itself in

𝒜

.
First, as

B

i

∈

𝒜

and

A

c

⊆

B

i

c

,

μ

⁢

(

E

)

≥

μ

⁢

(

E

∩

B

i

)

+

μ

⁢

(

E

∩

B

i

c

)

≥

μ

⁢

(

E

∩

B

i

)

+

μ

⁢

(

E

∩

A

c

)

.

As

C

i

≡

B

i

∖

B

i

-

1

are

pairwise disjoint

sets in

𝒜

satisfying

⋃

j

=

1

i

C

j

=

B

i

the additivity of

C

↦

μ

⁢

(

E

∩

C

)

on

𝒜

gives

μ

⁢

(

E

)

≥

∑

j

=

1

i

μ

⁢

(

E

∩

C

j

)

+

μ

⁢

(

E

∩

A

c

)

.

So, letting

i

increase to

infinity

, the subadditivity of

μ

applied to

⋃

j

(

E

∩

C

j

)

=

E

∩

A

gives

μ

⁢

(

E

)

≥

∑

j

μ

⁢

(

E

∩

C

j

)

+

μ

⁢

(

E

∩

A

c

)

≥

μ

⁢

(

E

∩

A

)

+

μ

⁢

(

E

∩

A

c

)

.

This shows that

A

is

μ

-measurable and so

𝒜

is a

σ

-algebra.

It only remains to show that the

restriction

of

μ

to

𝒜

is a

measure

, for which it needs to be shown that

μ

is countably additive on

𝒜

. So, choose any pairwise disjoint sequence

A

i

∈

𝒜

and set

A

=

⋃

i

A

i

. The following inequality

∑

j

=

1

i

μ

⁢

(

A

j

)

=

μ

⁢

(

⋃

j

=

1

i

A

j

)

≤

μ

⁢

(

A

)

≤

∑

j

μ

⁢

(

A

j

)

follows from the additivity of

μ

on

𝒜

, the requirement that

μ

is

increasing

and from the

countable subadditivity

of

μ

. Letting

i

increase to infinity gives

μ

⁢

(

A

)

=

∑

j

μ

⁢

(

A

j

)

and

μ

is indeed countably additive on

𝒜

.

Title

proof of Carathéodory’s lemma

Canonical name

ProofOfCaratheodorysLemma

Date of creation

2013-03-22 18:33:25

Last modified on

2013-03-22 18:33:25

Owner

gel (22282)

Last modified by

gel (22282)

Numerical id

5

Author

gel (22282)

Entry type

Proof

Classification

msc 28A12

Related topic

CaratheodorysLemma

Related topic

OuterMeasure

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