Ex3-11

**Solution to Chapter 3, Exercise 11**

*Question: Prove Theorem 3.29.*

**Theorem 3.29** *Suppose that

E

e

<
∞

{\displaystyle |E|\_{e}<+\infty }

. Then

E

{\displaystyle E}

is measurable if and only if given

ϵ
>
0

{\displaystyle \epsilon >0}

,

# E
(
S
∪

N

1

)
−

N

2

{\displaystyle E=(S\cup N\_{1})-N\_{2}}

, where

S

{\displaystyle S}

is a finite union of nonoverlapping intervals and

N

1

e

,

N

2

e

<
ϵ

.*

First, suppose that

E

{\displaystyle E}

is measurable. Given *ϵ
>
0

{\displaystyle \epsilon >0}*, we can find an open set

G

{\displaystyle G}

so that we have

G
−
E

e

<
ϵ

{\displaystyle |G-E|\_{e}<\epsilon }

.

Since

G

{\displaystyle G}

is open, by Theorem 1.11

G

{\displaystyle G}

can be written as a countable union of nonoverlapping intervals

{

I

k

}

# k
1

∞

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

.

# k
1

∞

I

k

{\displaystyle G=\bigcup \_{k=1}^{\infty }I\_{k}}

Since

E

e

<
∞

{\displaystyle |E|\_{e}<\infty }

, we can choose the nonoverlapping intervals so that

G

e

=

⋃

# k
1

∞

I

k

e

=

∑

# k
1

∞

I

k

e

<
∞

Then since

∑

# k
1

∞

I

k

e

{\displaystyle \sum \_{k=1}^{\infty }|I\_{k}|\_{e}}

is finite, there exists an

N
∈

N

{\displaystyle N\in \mathbb {N} }

such that

∑

# k
1

∞

I

k

e

−

∑

# k
1

N

I

k

e

<
ϵ

∑

# k
N
1

∞

I

k

e

=

∑

# k
N
1

∞

I

k

e

<
ϵ

Let

S

=

⋃

# k
1

N

I

k

N

1

=

⋃

# k
N
1

∞

I

k

N

2

=
G
−
E

{\displaystyle {\begin{aligned}S&=\bigcup \_{k=1}^{N}I\_{k}\\N\_{1}&=\bigcup \_{k=N+1}^{\infty }I\_{k}\\N\_{2}&=G-E\end{aligned}}}

Then

N

1

e

=

∑

# k
N
1

∞

I

k

e

<
ϵ

N

2

e

=

G
−
E

e

<
ϵ

and since

E
⊂
G

{\displaystyle E\subset G}

,

# E
G
−

N

2

=
(
S
∪

N

1

)
−

N

2

.

{\displaystyle E=G-N\_{2}=(S\cup N\_{1})-N\_{2}.}

Conversely, given any

ϵ
>
0

{\displaystyle \epsilon >0}

. There exists

N

1

,

N

2

{\displaystyle N\_{1},N\_{2}}

and

S

{\displaystyle S}

 such that

# E
(
S
∪

N

1

)
−

N

2

{\displaystyle E=(S\cup N\_{1})-N\_{2}}

, where

S

{\displaystyle S}

is a finite union of nonoverlapping intervals and

N

1

e

,

N

2

e

<

ϵ
4

.

We know that

S

{\displaystyle S}

is measurable, then there exists an open set

G

{\displaystyle G}

such that

S
⊂
G

{\displaystyle S\subset G}

and

G
−
S

<

ϵ
4

{\displaystyle |G-S|<{\frac {\epsilon }{4}}}

.

By Theorem 3.6, we can find an open set

G

1

{\displaystyle G\_{1}}

such that

N

1

⊂

G

1

{\displaystyle N\_{1}\subset G\_{1}}

and

G

1

e

≤

N

1

e

ϵ
4

<

ϵ
2

.

Since

S
⊂
G

{\displaystyle S\subset G}

and

N

1

⊂

G

1

{\displaystyle N\_{1}\subset G\_{1}}

, we have

S
∪

N

1

⊂
G
∪

G

1

{\displaystyle S\cup N\_{1}\subset G\cup G\_{1}}

.

Then

# E
(
S
∪

N

1

)
−

N

2

⊂
S
∪

N

1

⊂
G
∪

G

1

{\displaystyle E=(S\cup N\_{1})-N\_{2}\subset S\cup N\_{1}\subset G\cup G\_{1}}

. Also, we have

(
G
∪

G

1

)
−
E

e

=

(
G
−
E
)
∪
(

G

1

−
E
)

e

≤

G
−
E

e

G

1

−
E

e

≤

G
−
E

e

G

1

e

≤

G
−
E

e

ϵ
2

since

G

1

−
E
⊆

G

1

{\displaystyle G\_{1}-E\subseteq G\_{1}}

. Furthermore,

G
−
E

=
G
−
(
(
S
∪

N

1

)
−

N

2

)

=
G
−
(
(
S
∪

N

1

)
∩

N

2

C

)

=
G
∩
(
(
S
∪

N

1

)
∩

N

2

C

)

C

=
G
∩
(
(
S
∪

N

1

)

C

∪

N

2

)

=
(
G
∩
(
S
∪

N

1

)

C

)
)
∪
(
G
∩

N

2

)

=
(
G
∩

S

C

∩

N

1

C

)
∪
(
G
∩

N

2

)

⊆
(
G
∩

S

C

)
∪

N

2

=
(
G
−
S
)
∪

N

2

{\displaystyle {\begin{aligned}G-E&=G-((S\cup N\_{1})-N\_{2})\\&=G-((S\cup N\_{1})\cap N\_{2}^{C})\\&=G\cap ((S\cup N\_{1})\cap N\_{2}^{C})^{C}\\&=G\cap ((S\cup N\_{1})^{C}\cup N\_{2})\\&=(G\cap (S\cup N\_{1})^{C}))\cup (G\cap N\_{2})\\&=(G\cap S^{C}\cap N\_{1}^{C})\cup (G\cap N\_{2})\\&\subseteq (G\cap S^{C})\cup N\_{2}\\&=(G-S)\cup N\_{2}\end{aligned}}}

We used the fact that

A
−
# B
A
∩

B

C

{\displaystyle A-B=A\cap B^{C}}

. Then we have

G
−
E

e

≤

G
−
S

e

N

2

e

<

ϵ
4

ϵ
4

Therefore,

(
G
∪

G

1

)
−
E

e

<

ϵ
4

ϵ
4

ϵ
2

=
ϵ

{\displaystyle |(G\cup G\_{1})-E|\_{e}<{\frac {\epsilon }{4}}+{\frac {\epsilon }{4}}+{\frac {\epsilon }{2}}=\epsilon }

Hence, we have found an open set

G
′

=
G
∪

G

1

{\displaystyle G'=G\cup G\_{1}}

 such that

E
⊂

G
′

{\displaystyle E\subset G'}

and

G
′

−
E

e

<
ϵ

{\displaystyle |G'-E|\_{e}<\epsilon }

.