Ex3-12

> **Question 3.12**
>
>
> If E1 and E2 are measurable subsets of
>
>
>
>
> R
> 1
>
>
>
> {\displaystyle R^1}
>
>
>
> **Solution:**
>
>
>
>
> **Proof:**
>
>
> Let
>
>
>
> ϵ
> >
> 0
>
>
> {\displaystyle \epsilon > 0}
>
> . Since |Z| = 0, there exists intervals
>
>
>
> {
>
> I
> k
>
> }
>
>
> {\displaystyle \{I\_k\}}
>
> such that
>
>
>
> Z
> ⊆
> ∪
>
> I
> k
>
>
>
> {\displaystyle Z\subseteq\cup I\_k}
>
>  and
>
>
>
> ∪
>
>
>
> I
> k
>
>
>
> <
> ϵ
>
>
> {\displaystyle \cup |I\_k|<\epsilon}
>
> .
>
>
> Write
>
>
>
>
> A
> n
>
>
>
> {\displaystyle A\_n}
>
>  = A ∩ [−n, n]. Then
>
>
>
> A
> =
> ∪
>
> A
> n
>
>
>
> {\displaystyle A=\cup A\_n}
>
> .n. Note that
>
>
>
>
>
>
>
> A
> n
>
> ×
> Z
> ⊆
> [
> −
> n
> ,
> n
> ]
> ×
> ∪
>
> I
> k
>
> =
> ∪
> (
> [
> −
> n
> ,
> n
> ]
> ×
>
> I
> k
>
> )
>
>
> {\displaystyle A\_n \times Z\subseteq [-n,n] \times \cup I\_k = \cup( [-n,n]\times I\_k)}
>
>
>
>
> so
>
>
>
>
>
>
>
> A
>
> n
>
>
> ×
> Z
>
>
>
> e
>
>
> ≤
>
> ∑
>
> k
> =
> 1
>
>
> ∞
>
>
> 2
> n
>
>
> I
>
> k
>
>
>
> =
> 2
> n
> ϵ
>
>
>
> ,
>
>
> Since
>
>
>
> ϵ
> >
> 0
>
>
> {\displaystyle \epsilon > 0}
>
>  is arbitrary,
>
>
>
>
>
>
> A
>
> n
>
>
> ×
> Z
>
>
> =
> 0
>
>
> {\displaystyle \left|A\_{n} \times Z\right|=0}
>
>  for each  n . So
>
>
>
>
>
>
>
>
> A
> ×
> Z
>
>
>
>
> e
>
>
> =
>
>
>
>
> ⋃
>
> n
> =
> 1
>
>
> ∞
>
>
>
> (
>
>
> A
>
> n
>
>
> ×
> Z
>
> )
>
>
>
>
> e
>
>
> ≤
>
> ∑
>
> n
> =
> 1
>
>
> ∞
>
>
>
>
>
>
> A
>
> n
>
>
> ×
> Z
>
>
>
> e
>
>
> =
> 0
>
>
> {\displaystyle
>
> }
>
>
>
>
>
>
>
>
>
> A
> ×
> Z
>
>
> =
> 0
>
>
> {\displaystyle |A \times Z|=0 }
>
>   as desired.
>
>
> Now, since
>
>
>
>
> E
> 1
>
>
>
> {\displaystyle E\_1 }
>
> is measurable,
>
>
>
>
> E
>
> 1
>
>
> =
>
> H
>
> 1
>
>
> ∪
>
> Z
>
> 1
>
>
>
>
> {\displaystyle E\_{1}=H\_{1} \cup Z\_{1}}
>
> , where
>
>
>
>
> H
> 1
>
>
>
> {\displaystyle H\_1}
>
>
>
>
>
> Z
> 1
>
>
>
> {\displaystyle Z\_1}
>
> | = 0. Similarly, write
>
>
>
>
> E
>
> 2
>
>
> =
>
> H
>
> 2
>
>
> ∪
>
> Z
>
> 2
>
>
>
>
> {\displaystyle E\_{2}=H\_{2} \cup Z\_{2}}
>
> where
>
>
>
>
> H
> 2
>
>
>
> {\displaystyle H\_2}
>
>
>
>
>
> Z
> 2
>
>
>
> {\displaystyle Z\_2}
>
> | = 0.
>
>
> Then
>
>
>
>
>
>
>
> E
>
> 1
>
>
> ×
>
> E
>
> 2
>
>
> =
>
> (
>
>
> H
>
> 1
>
>
> ×
>
> H
>
> 2
>
>
>
> )
>
> ∪
>
> (
>
>
> H
>
> 1
>
>
> ×
>
> Z
>
> 2
>
>
>
> )
>
> ∪
>
> (
>
>
> Z
>
> 1
>
>
> ×
>
> H
>
> 2
>
>
>
> )
>
> ∪
>
> (
>
>
> Z
>
> 1
>
>
> ×
>
> Z
>
> 2
>
>
>
> )
>
> .
>
>
> {\displaystyle E\_{1} \times E\_{2}=\left(H\_{1} \times H\_{2}\right) \cup\left(H\_{1} \times Z\_{2}\right) \cup\left(Z\_{1} \times H\_{2}\right) \cup\left(Z\_{1} \times Z\_{2}\right) .}
>
>
>
>
> Note that
>
>
>
>
> H
> 1
>
> ×
>
> H
> 2
>
>
>
> {\displaystyle H\_1\times H\_2}
>
>  is of type Fσ, while the other terms have measure zero by the previous lemma. Thus
>
>
>
>
> E
> 1
>
> ×
>
> E
> 2
>
>
>
> {\displaystyle E\_1\times E\_2}
>
>  is measurable.
>
>
> **Case 1)** Suppose
>
>
>
>
>
>
> E
> 1
>
>
>
>
>
> {\displaystyle |E\_1|}
>
>  and
>
>
>
>
>
>
> E
> 2
>
>
>
>
>
> {\displaystyle |E\_2|}
>
>  are both finite.
>
>
> Since
>
>
>
>
> E
> 1
>
>
>
> {\displaystyle E\_1 }
>
> ,
>
>
>
>
> E
> 2
>
>
>
> {\displaystyle E\_2}
>
> are measurable, for each k ∈ N there are open sets
>
>
>
>
> S
> k
>
>
>
> {\displaystyle S\_k}
>
>  ⊇
>
>
>
>
> E
> 1
>
>
>
> {\displaystyle E\_1 }
>
> ,
>
>
>
>
> T
> k
>
>
>
> {\displaystyle T\_k}
>
>  ⊇
>
>
>
>
> E
> 2
>
>
>
> {\displaystyle E\_2}
>
>
>
>
>
> S
> k
>
>
>
> {\displaystyle S\_k}
>
>  \
>
>
>
>
> E
> 1
>
>
>
> {\displaystyle E\_1 }
>
>
>
>
>
> T
> k
>
>
>
> {\displaystyle T\_k}
>
>  \
>
>
>
>
> E
> 2
>
>
>
> {\displaystyle E\_2}
>
> | < 1/k. We may assume
>
>
>
>
>
>
>
> S
>
> k
> +
> 1
>
>
>
>
> {\displaystyle S\_{k+1}}
>
>  ⊆
>
>
>
>
> S
> k
>
>
>
> {\displaystyle S\_k}
>
> ,
>
>
>
>
> T
>
> k
> +
> 1
>
>
>
>
> {\displaystyle T\_{k+1}}
>
>  ⊆
>
>
>
>
> T
> k
>
>
>
> {\displaystyle T\_k}
>
>  (if
>
>
>
>
> S
>
> k
> +
> 1
>
>
> ⊈
>
> S
>
> k
>
>
>
>
> {\displaystyle S\_{k+1} \nsubseteq S\_{k}}
>
>  , then define
>
>
>
>
> S
>
> k
> +
> 1
>
>
> ′
>
>
> =
>
> S
>
> k
> +
> 1
>
>
> ∩
>
> S
>
> k
>
>
>
>
> {\displaystyle S\_{k+1}^{\prime}=S\_{k+1} \cap S\_{k}}
>
> instead).
>
>
> Since
>
>
>
>
> S
> k
>
>
>
> {\displaystyle S\_k}
>
>  is open,
>
>
>
>
> S
> k
>
>
>
> {\displaystyle S\_k}
>
>  =
>
>
>
>
> S
>
> k
>
>
> =
>
> ⋃
>
> k
> ∈
>
> N
>
>
>
>
> I
>
> i
>
>
>
>
> {\displaystyle S\_{k}=\bigcup\_{k \in \mathbb{N}} I\_{i}}
>
>  for some nonoverlapping closed intervals.
>
>
> Similarly,
>
>
>
>
> T
>
> k
>
>
> =
>
> ⋃
>
> j
> ∈
>
> N
>
>
>
>
> J
>
> j
>
>
>
>
> {\displaystyle T\_{k}=\bigcup\_{j \in \mathbb{N}} J\_{j}}
>
> for some nonoverlapping closed intervals.
>
>
> So
>
>
>
>
>
>
>
>
>
>
>
>
> S
>
> k
>
>
> ×
>
> T
>
> k
>
>
>
>
>
>
>
> =
>
>
>
> ⋃
>
> (
> i
> ,
> j
> )
> ∈
>
> N
>
> ×
>
> N
>
>
>
>
> (
>
>
> I
>
> i
>
>
> ×
>
> J
>
> j
>
>
>
> )
>
>
>
>
>
>
>
>
>
> =
>
> ∑
>
> i
> ,
> j
> ∈
>
> N
>
>
>
>
>
>
> I
>
> i
>
>
> ×
>
> J
>
> j
>
>
>
>
>
>
>
>
>
>
> =
>
> ∑
>
> i
> ,
> j
> ∈
>
> N
>
>
>
>
>
> I
>
> i
>
>
>
>
>
> J
>
> j
>
>
>
>
>
>
>
>
>
> =
>
> (
>
>
> ∑
>
> i
> ∈
>
> N
>
>
>
>
>
> I
>
> i
>
>
>
>
> )
>
>
> (
>
>
> ∑
>
> j
> ∈
>
> N
>
>
>
>
>
> J
>
> j
>
>
>
>
> )
>
>
>
>
>
>
>
> =
>
>
> S
>
> k
>
>
>
>
>
> T
>
> k
>
>
>
> .
>
>
>
>
>
> {\displaystyle \begin{aligned}
> &=\sum\_{i, j \in \mathbb{N}}\left|I\_{i} \times J\_{j}\right| \\
> \end{aligned}}
>
>
>
>
> Write
>
>
>
> S
> =
>
> ⋂
>
> k
> =
> 1
>
>
> ∞
>
>
>
> S
>
> k
>
>
> ,
> T
> =
>
> ⋂
>
> k
> =
> 1
>
>
> ∞
>
>
>
> T
>
> k
>
>
>
>
> {\displaystyle S=\bigcap\_{k=1}^{\infty} S\_{k}, T=\bigcap\_{k=1}^{\infty} T\_{k}}
>
> .Then
>
>
>
>
>
> S
> ∖
>
> E
>
> 1
>
>
>
>
> =
>
>
> T
> ∖
>
> E
>
> 2
>
>
>
>
> =
> 0
>
>
>
> .Hence,
>
>
>
>
>
>
>
>
>
>
>
>
> E
>
> 1
>
>
> ×
>
> E
>
> 2
>
>
>
>
>
>
>
> =
>
>
> S
> ×
> T
>
>
>
>
>
>
>
>
> =
>
> lim
>
> k
> =
> ∞
>
>
>
>
>
> S
>
> k
>
>
> ×
>
> T
>
> k
>
>
>
>
>
>
>
>
>
>
> =
>
> lim
>
> k
> →
> ∞
>
>
>
>
> S
>
> k
>
>
>
>
>
> T
>
> k
>
>
>
>
>
>
>
>
>
> =
>
>
> E
>
> 1
>
>
>
>
>
> E
>
> 2
>
>
>
> ,
>
>
>
>
>
> {\displaystyle \begin{aligned}
> &=\lim \_{k=\infty}\left|S\_{k} \times T\_{k}\right| \\
> \end{aligned}}
>
>
>
>
> where the second equality follows by MCT for measure, since Sk×Tk & S×T
>
>
>
>
> S
>
> k
>
>
> ×
>
> T
>
> k
>
>
> ↘
> S
> ×
> T
>
>
> {\displaystyle S\_{k} \times T\_{k} \searrow S \times T}
>
>
>
>
> and
>
>
>
>
>
>
> S
>
> k
>
>
> ×
>
> T
>
> k
>
>
>
>
> <
> ∞
>
>
> {\displaystyle \left|S\_{k} \times T\_{k}\right|<\infty}
>
>
>
>
>
> E
> 1
>
>
>
> {\displaystyle E\_1 }
>
>
>
>
>
> E
> 2
>
>
>
> {\displaystyle E\_2}
>
> | are both finite. The last equality also follows by MCT for measure.
>
>
>
>
>
>
> E
> 1
>
>
>
> {\displaystyle E\_1 }
>
>
>
>
>
> E
> 2
>
>
>
> {\displaystyle E\_2}
>
> | are infinite
>
>
> If
>
>
>
>
>
> E
>
> 1
>
>
>
> =
> ∞
>
>
> {\displaystyle \left|E\_{1}\right|=\infty}
>
> and
>
>
>
>
>
> E
>
> 2
>
>
>
> >
> 0
>
>
> {\displaystyle \left|E\_{2}\right|>0}
>
>  , then write
>
>
>
>
> E
>
> 1
>
>
> n
>
>
> =
>
> E
>
> 1
>
>
> ∩
> [
> −
> n
> ,
> n
> ]
>
>
> {\displaystyle E\_{1}^{n}=E\_{1} \cap[-n, n] }
>
> .
>
>
>
>
>
>
>
>
>
>
>
>
> E
>
> 1
>
>
> ×
>
> E
>
> 2
>
>
>
>
>
>
>
> =
>
> lim
>
> n
> →
> ∞
>
>
>
>
>
> E
>
> 1
>
>
> n
>
>
> ×
>
> E
>
> 2
>
>
>
>
>
>
>
>
>
>
> =
>
> lim
>
> n
> →
> ∞
>
>
>
>
> E
>
> 1
>
>
> n
>
>
>
>
>
> E
>
> 2
>
>
>
>
>
>
>
>
>
> =
>
>
> E
>
> 1
>
>
>
>
>
> E
>
> 2
>
>
>
>
>
>
>
>
>
> =
> ∞
>
>
>
>
>
> {\displaystyle
>
> \begin{aligned}
> &=\infty
> \end{aligned}
> }
>
>
>
>
>
>
>
>
> where the first equality follows by MCT for measure, since
>
> E
>
> 1
>
>
> n
>
>
> ×
>
> E
>
> 2
>
>
> ↗
>
> E
>
> 1
>
>
> ×
>
> E
>
> 2
>
>
>
>
> {\displaystyle \text{where the first equality follows by MCT for measure, since} E\_{1}^{n} \times E\_{2} \nearrow E\_{1} \times E\_{2}}
>
> .
>
>
>
>
>
>
>  If
>
>
> E
>
> 1
>
>
>
> =
> ∞
>  and
>
>
> E
>
> 2
>
>
>
> =
> 0
> ,
>
>
>
> E
>
> 1
>
>
> ×
>
> E
>
> 2
>
>
>
>
> =
> 0
>  by our first lemma.
>
>
>
>
>
>
> (proved)
>
>
>