Ex3-7

**Chapter 3 Question 7**

Lemma 3.15: If

{

I

k

}

# k
1

N

{\displaystyle \{ I\_{k} \}\_{k=1}^{N}}

 is a finite collection of nonoverlapping intervals in

R

n

{\displaystyle \mathbf R^n}

, then

⋃

# k
1

N

I

k

{\displaystyle \bigcup\_{k=1}^{N} I\_{k}}

 is measurable and

μ
(

⋃

# k
1

N

I

k

# k
1

N

μ
(

I

k

)

{\displaystyle \mu(\bigcup\_{k=1}^{N} I\_{k}) = \sum\_{k=1}^{N} \mu(I\_{k}) }

.

Proof: Corollary 3.13 tells us that if

I

{\displaystyle I}

 is an interval, then

I

{\displaystyle I}

 is measurable and

μ
(
I
# )
v
(
I
)

{\displaystyle \mu(I) = v(I)}

. Since a finite collection of sets is clearly countable, and each set

I
k

{\displaystyle I\_k}

 is measurable, Theorem 3.12 tells us that the union

⋃

# k
1

N

I

k

{\displaystyle \bigcup\_{k=1}^{N} I\_{k}}

 is measurable. We are left to show that the equality

μ
(

⋃

# k
1

N

I

k

# k
1

N

μ
(

I

k

)

{\displaystyle \mu(\bigcup\_{k=1}^{N} I\_{k}) = \sum\_{k=1}^{N} \mu(I\_{k}) }

 holds.

Note that since

⋃

# k
1

N

I

k

{\displaystyle \bigcup\_{k=1}^{N} I\_{k}}

 is measurable, its Lebesgue measure must agree with its outer measure. Thus it suffices to calculate the outer measure of

⋃

# k
1

N

I

k

{\displaystyle \bigcup\_{k=1}^{N} I\_{k}}

. By Theorem 3.4, we have the bound

⋃

# k
1

N

I

k

)

e

≤

∑

# k
1

N

I

k

e

.

Note also that

I

k

e

=
v
(

I

k

)

{\displaystyle |I\_{k}|\_{e} = v(I\_{k})}

 for each

k

{\displaystyle k}

, so we can rewrite the above as

⋃

# k
1

N

I

k

)

e

≤

∑

# k
1

N

v
(

I

k

)

{\displaystyle |\bigcup\_{k=1}^{N} I\_{k})|\_{e} \leq \sum\_{k=1}^{N} v(I\_{k})}

Thus we are reduced to the problem of showing the other direction, namely

⋃

# k
1

N

I

k

)

e

≥

∑

# k
1

N

v
(

I

k

)

{\displaystyle |\bigcup\_{k=1}^{N} I\_{k})|\_{e} \geq \sum\_{k=1}^{N} v(I\_{k})}

We show the above by using compactness. Let

# S

⋃

l

J
l

{\displaystyle S=\bigcup\_{l}J\_l}

 be a covering of

⋃

# k
1

N

I

k

{\displaystyle \bigcup\_{k=1}^{N} I\_{k}}

 by intervals. Given an arbitrary

ϵ
>
0

{\displaystyle \epsilon > 0}

, let

J

l

∗

{\displaystyle J^{\*}\_{l}}

 contain

J

l

{\displaystyle J\_{l}}

 in its interior such that

v
(

J

l

∗

)
≤
(
1
ϵ
)
v
(

J

l

)

{\displaystyle v(J^{\*}\_{l}) \leq (1 + \epsilon) v(J\_{l})}

 (\*)

Then we have

I

k

⊂

⋃

l

(

J
l

∗

)

∘

{\displaystyle I\_{k} \subset \bigcup\_{l}(J^{\*}\_l)^{\circ}}

 for every

k

{\displaystyle k}

. Since

⋃

# k
1

N

I

k

{\displaystyle \bigcup\_{k=1}^{N} I\_{k}}

 is clearly compact (it is closed and bounded) in the Euclidean topology, by Heine-Borel we can extract a finite subcover

⋃

# l
1

M

J
l

∗

{\displaystyle \bigcup\_{l=1}^{M}J^{\*}\_l}

 of

⋃

# k
1

N

I

k

{\displaystyle \bigcup\_{k=1}^{N} I\_{k}}

.

By a basic fact about the combinatorics of intervals in

R

n

{\displaystyle \mathbf R^n}

 which is intuitive but tedious to prove, and because

{

J
l
∗

∩

I
k

}

1
≤
l
≤
M

{\displaystyle \{J\_l^\*\cap I\_k\}\_{1\leq l\leq M}}

 is a cover for

I
k

{\displaystyle I\_k}

, we have

v
(

I
k

)
≤

∑

# l
1

M

v
(

J
l

∗

∩

I
k

)

{\displaystyle v(I\_k) \leq \sum\_{l=1}^{M}v(J^{\*}\_l\cap I\_k)}

Summing the inequalities over

k

{\displaystyle k}

, we get

∑

# k
1

N

v
(

I
k

)
≤

∑

# k
1

N

∑

# l
1

M

v
(

J
l

∗

∩

I
k

# l
1

M

∑

# k
1

N

v
(

J
l
∗

∩

I
k

)
≤

∑

# l
1

M

v
(

J
l
∗

)

{\displaystyle \sum\_{k=1}^{N} v(I\_k) \leq \sum\_{k=1}^N\sum\_{l=1}^{M}v(J^{\*}\_l\cap I\_k) = \sum\_{l=1}^M \sum\_{k=1}^N v(J^\*\_l \cap I\_k)\leq \sum\_{l=1}^M v(J\_l^\*)}

since the collection

{

I
k

}

1
≤
k
≤
N

{\displaystyle \{I\_k\}\_{1\leq k\leq N}}

 is nonoverlapping.

Using (\*), we then have

∑

k

N

v
(

I
k

)
≤

∑

# l
1

M

v
(

J
l

∗

)
≤
(
1
ϵ
)

∑

# l
1

M

v
(

J
l

)

{\displaystyle \sum\_{k}^{N} v(I\_k) \leq \sum\_{l=1}^{M}v(J^{\*}\_l) \leq (1 + \epsilon) \sum\_{l=1}^{M}v(J\_l)}

Since

ϵ
>
0

{\displaystyle \epsilon > 0}

 was chosen arbitrarily, we conclude

∑

k

N

v
(

I
k

)
≤

∑

# l
1

M

v
(

J
l

)

{\displaystyle \sum\_{k}^{N} v(I\_k) \leq \sum\_{l=1}^{M}v(J\_l)}

Since we also have

∑

# l
1

M

v
(

J
l

)
≤
σ
(
S
)

{\displaystyle \sum\_{l=1}^{M}v(J\_l) \leq \sigma(S)}

, we can conclude that

∑

k

N

v
(

I
k

)
≤
σ
(
S
)

{\displaystyle \sum\_{k}^{N} v(I\_k) \leq \sigma(S)}

and so by taking infimums on both sides, taken over all possible interval coverings, we arrive at

∑

k

N

v
(

I
k

)
≤

⋃

k

I

k

)

e

{\displaystyle \sum\_{k}^{N} v(I\_k) \leq |\bigcup\_{k} I\_{k})|\_{e}}

as desired.

Comment: I used

μ

{\displaystyle \mu}

 throughout to denote the Lebesgue measure in Euclidean space, which differs from the book. I suppose somebody who is bothered by this can help me change it to the notation used in the book :).