# CH3 exercise

## 8

Let $\mathscr{C}$ be a $\sigma$-algebra containing the closed sets of $\mathbb{R}^n$ ;

$\mathscr{B}$ be a $\sigma$-algebra containing the open sets of $\mathbb{R}^n$.

For $\mathcal{C}\in\mathscr{C}$, by the properties of $\sigma$-algebra, $\mathcal{C}^c\in \mathscr{C}$.

Since $\mathcal{C}$ is closed set, $\mathcal{C}^c$ is open set. Furthermore, $\mathscr{B}$ contains open sets, then we have $\mathscr{C}\subset\mathscr{B}$.

## 9

Given an $\epsilon>0$ there exists an $N_0\in\mathbb{N}$ such that
$$
\displaystyle \Sigma_{k=N_0}^{\infty}|E_k|_e < \epsilon.
$$
Moreover,
$$
\displaystyle \limsup_kE_k = \displaystyle \cap_{k>0}\displaystyle \cup_{j\geq k}E_j \subset \cup_{j\geq N_0}E_j.
$$
We have
$$
|\cup_{j\geq N_0}E_j|_{e}<\epsilon
$$
Therefore,
$$
\displaystyle |\limsup_k E_k|_e\leq|\cup_{j\geq N_0}E_j|_{e}<\epsilon
$$
We conclude that $\limsup_kE_k$ is measure zero set.

## 10

$$
E_1\cup E_2 = E_1\cup (E_2-E_1)
$$

Since $E_1$ and $E_2-E_1$ are disjoint, $|E_1\cup E_2| = |E_1|+|E_2-E_1|$. ---(1)

Furthermore,
$$
E_2 = (E_2-E_1)\cup(E_1\cap E_2).
$$
We have $|E_2| = |E_2-E_1|+|E_1\cap E_2|$. ---(2)

Combine (1) and (2), we have the following result:
$$
|E_1\cup E_2| = |E_1|+|E_2|-|E_1\cap E_2|
$$

## 11

$\left(\Rightarrow\right)$

Since $E$ is measurable, given $\epsilon<0$ there exists $G$ be open set such that $E\subset G$ and $|G-E|_e<\epsilon$.

Moreover, $G$ can be written as union of nonoverlapping intervals, that is, $G = \displaystyle \cup_k I_k$.

By the hypothesis, $|E|_e<\infty$ we have $\displaystyle \Sigma_{k}|I_k|_e<\infty$.

For $\epsilon<0$, there exists an $N_0\in \mathbb{N}$ such that $\displaystyle \Sigma_{k>N_0}|I_k|<\epsilon$.

Finally, Since $E = G-(G-E)$.
$$
G = \cup_{k\leq N_0}I_k \cup_{k>N_0}I_k;\\
E = \left(\cup_{k\leq N_0}I_k \cup \cup_{k>N_0}I_k\right)-\left(G-E\right).
$$
$N_1 = \cup_{k>N_0}I_k, N_2 = (G-E)$, with $|N_1|_e , |N_2|_e< \epsilon$

$\left(\Leftarrow\right)$

Given $\epsilon >0$. $E = (S\cup N_1)-N_2$ for $S$ be the union of finite nonoverlapping intervals, with $|N_1|_e,|N_2|_e <\epsilon$

Since $S$ is measurable(for intervals are measurable), there exists an open set $G_1$, with $E\subset G_1$ and $|G_1-E|_e<\epsilon$.

On the other hand, $|N_1|_e<\epsilon$, there exists an open set $G_2$, with $N_1\subset G_2$ and $|N_1|_e+\epsilon>|G_2|_e$ .

We have $E\subset G_1\cup G_2$.
$$
|G_1\cup G_2-E_1|_e\leq |G_1-S|_e+|N_2|_e+|G_2|_e+|N_2|_e\\
\leq \epsilon+\epsilon+2\epsilon+\epsilon = 5\epsilon
$$
This shows that $E$ is measurable.
## 12
First, for $I$ be a given measure zero interval, $G$ be any open set.
$G \times I$ is measurable and measure zero.

$\left(Proof\right)$
$\because I$ is measurable
given any $\epsilon > 0$ then $\exists H$ be open set such that $|H-I|_e<\epsilon$
$\because H$ is open $H = \cup_{k=1}^{\infty} J_k$ for $J_k$'s be nonoverlapping closed interval.
## 13

(i)
For any closed set $F\subset E$ and any union of intervals $\{I_k\}$ which is a cover of $E$, we have $F\subset E\subset \bigcup\limits_{k} I_k$.

Then $|F|\leq |\bigcup\limits_{k} I_k|_e\leq \sum\limits_{k}|I_k|_e$.

Since $|E|_i=\sup\limits_{F}|F|_e$ and $|E|_e=\inf\limits_{I_k}\sum\limits_k|I_k|_e$, we have $|F|\leq |E|_i\leq |E|_e\leq \sum\limits_k|I_k|_e$.

Hence, $|E|_i\leq |E|_e$.

(ii)
$(\Rightarrow)$
Suppose that $E$ is measurable.

By Lemma 3.22, given $\epsilon > 0$, there exists $F\subset E$ such that $|E-F|<\epsilon$ and $|F|<+\infty$.

By Cor 3.25, $|E-F|=|E|_e-|F|<\epsilon$.

Therefore,we have $|E|_e-\epsilon<|F|\leq|E|_i$, and that implies $|E|_e\leq |E|_i$.

By (i),we have $|E|_e\geq |E|_i$.

Hence $|E|_e= |E|_i$.

$(\Leftarrow)$
Suppose $|E|_e= |E|_i$.

Given $\epsilon > 0$, there exists a closed set $F$ in an open set $E$ such that $$
|F|+\frac{\epsilon}{4}\geq|E|_i=|E|_e\geq |G|-\frac{\epsilon}{4}.
$$

So we have $|G|-|F|\leq \frac{\epsilon}{2}< \epsilon.$

Note that $|G-F|=|G|-|F|$ by Cor3.25 since $F$ and $G$ are measurable , $F\subset G$ and $|F|<+\infty.$

Hence , $|G-E|_e\leq |G-F|=|G|-|F|<\epsilon$ since $G-E\subset G-F$ and Cor3.25.

This shows that $E$ is measurable.