proof that the outer (Lebesgue) measure of an interval is its length

proof that the outer (Lebesgue) measure of an interval is its length

We begin with the case in which we have a

bounded interval

, say

[

a

,

b

]

. Since the

open interval

(

a

-

ε

,

b

+

ε

)

contains

[

a

,

b

]

for each positive number

ε

, we have

m

*

⁢

[

a

,

b

]

≤

b

-

a

+

2

⁢

ε

. But since this is true for each positive

ε

, we must have

m

*

⁢

[

a

,

b

]

≤

b

-

a

. Thus we only have to show that

m

*

⁢

[

a

,

b

]

≥

b

-

a

; for this it suffices to show that if

{

I

n

}

is a

countable

open cover

by

intervals

of

[

a

,

b

]

, then

∑

l

⁢

(

I

n

)

≥

b

-

a

.

By the

Heine-Borel theorem

, any

collection

of open intervals

[

a

,

b

]

contains a finite subcollection that also cover

[

a

,

b

]

and since the sum of the lengths of the finite subcollection is no

greater

than the sum of the original one, it suffices to prove the

inequality

for finite collections

{

I

n

}

that cover

[

a

,

b

]

. Since

a

is

contained

in

⋃

I

n

, there must be one of the

I

n

’s that contains

a

. Let this be the interval

(

a

1

,

b

1

)

. We then have

a

1

<

a

<

b

1

. If

b

1

≤

b

, then

b

1

∈

[

a

,

b

]

, and since

b

1

∉

(

a

1

,

b

1

)

, there must be an interval

(

a

2

,

b

2

)

in the collection

{

I

n

}

such that

b

1

∈

(

a

2

,

b

2

)

, that is

a

2

<

b

1

<

b

2

. Continuing in this fashion, we obtain a

sequence

(

a

1

,

b

1

)

,

…

,

(

a

k

,

b

k

)

from the collection

{

I

n

}

such that

a

i

<

b

i

-

1

<

b

i

. Since

{

I

n

}

is a finite collection our process must terminate with some interval

(

a

k

,

b

k

)

. But it terminates only if

b

∈

(

a

k

,

b

k

)

, that is if

a

k

<

b

<

b

k

. Thus

∑

l

⁢

(

I

n

)

≥

∑

l

⁢

(

a

i

,

b

i

)

=

(

b

k

-

a

k

)

+

(

b

k

-

1

-

a

k

-

1

)

+

…

+

(

b

1

-

a

1

)

=

b

k

-

(

a

k

-

b

k

-

1

)

-

(

a

k

-

1

-

b

k

-

2

)

-

…

-

(

a

2

-

b

1

)

-

a

1

>

b

k

-

a

1

,

since

a

i

<

b

i

-

1

. But

b

k

>

b

and

a

1

<

a

and so we have

b

k

-

a

1

>

b

-

a

, whence

∑

l

⁢

(

I

n

)

>

b

-

a

. This shows that

m

*

⁢

[

a

,

b

]

=

b

-

a

.

If

I

is any finite interval, then given

ε

>

0

, there is a

closed interval

J

⊂

I

such that

l

⁢

(

J

)

>

l

⁢

(

I

)

-

ε

. Hence

l

⁢

(

I

)

-

ε

<

l

⁢

(

J

)

=

m

*

⁢

J

≤

m

*

⁢

I

≤

m

*

⁢

I

¯

=

l

⁢

(

I

¯

)

=

l

⁢

(

I

)

,

where by

I

¯

we the topological closure of

I

.
Thus for each

ε

>

0

, we have

l

⁢

(

I

)

-

ε

<

m

*

⁢

I

≤

l

⁢

(

I

)

,
and so

m

*

⁢

I

=

l

⁢

(

I

)

.

If now

I

is an

unbounded

interval, then given any

real number

Δ

, there is a closed interval

J

⊂

I

with

l

⁢

(

J

)

=

Δ

. Hence

m

*

⁢

I

≥

m

*

⁢

J

=

l

⁢

(

J

)

=

Δ

. Since

m

*

⁢

I

≥

Δ

for each

Δ

, it follows

m

*

⁢

I

=

∞

=

l

⁢

(

I

)

.

References

Royden, H. L.

Real analysis

. Third edition

. Macmillan Publishing Company, New York, 1988.

Title

proof that the outer (Lebesgue) measure of an interval is its length

Canonical name

ProofThatTheOuterLebesgueMeasureOfAnIntervalIsItsLength

Date of creation

2013-03-22 14:47:04

Last modified on

2013-03-22 14:47:04

Owner

Simone (5904)

Last modified by

Simone (5904)

Numerical id

6

Author

Simone (5904)

Entry type

Proof

Classification

msc 28A12

Related topic

LebesgueOuterMeasure

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