# Math4121 Lecture 28

## Continue from last lecture

### Lebesgue Measure

#### Outer Measure

$$
m_e(S) = \inf_{S \subseteq \bigcup_{j=1}^{\infty} I_j} \sum_{j=1}^{\infty} \ell(I_j)
$$

If $S\subseteq I$ is measurable, then $m_i(S)=m_e(I)-m_e(I\setminus S)$

#### Lebesgue criterion for measurability

$S\subseteq I$ is measurable if and only if $m_e(I)=m_e(S)+m_e(I\setminus S)$

#### Caratheodory's criteria

Lebesgue criterion holds if and only if for any $X$ of finite outer measure,

$$
m_e(X)=m_e(X\cap S)+m_e(X\setminus S)
$$

> **Local additivity**
>
> $\{I_j\}_{j=1}^{\infty}$ is a collection of disjoint intervals, then 
>
> $$m_e\left(S\cap \bigcup_{j=1}^{\infty} I_j\right) = \sum_{j=1}^{\infty} m_e(S\cap I_j)$$
> Proved on Friday

<details>
<summary>Proof</summary>

$\implies$ If Lebesgue criterion holds for $S$, then for any $X$ of finite outer measure,

$$
m_e(X)=m_e(X\cap S)+m_e(X\setminus S)
$$

First, we extend Lebesgue criterion to intervals $I$ that may not contain $S$. Then we can find $J,K$ intervals neighboring $I$ such that $S\subseteq \tilde{I}=J\cup I\cup K$.

By Lebesgue criterion,

$$
\begin{aligned}
m_e(\tilde{I})&=m_e(\tilde{I}\cap S)+m_e(\tilde{I}\setminus S)\\
&=m_e(S)+m_e(\tilde{I}\setminus S)\\
&=m_e(S^c\cap \tilde{I})+m_e(S\cap \tilde{I})\\
&=\sum_{L\in \{J,I,K\}}m_e(L\cap S^c)+m_e(L\cap S)\\
&\geq \sum_{L\in \{J,I,K\}}m_e(L)\\
&=m_e(\tilde{I})
\end{aligned}
$$

Therefore, $m_e(I)=m_e(S^c\cap I)+m_e(S\cap I)$.

Now, let $X$ has finite outer measure, let $\epsilon>0$, we can find $\{I_j\}_{j=1}^{\infty}$ covering $X$ and

$$
\sum_{j=1}^{\infty} \ell(I_j)<m_e(X)+\epsilon
$$

$$
\begin{aligned}
m_e(X)&\leq m_e(X\cap S)+m_e(S^c\cap X)\\
&\leq m_e\left(\bigcup_{j=1}^{\infty} I_j\cap S\right)+m_e\left(\bigcup_{j=1}^{\infty} I_j\cap S^c\right)\\
&\leq \sum_{j=1}^{\infty} m_e(I_j\cap S)+m_e(I_j\cap S^c)\\
&=\sum_{j=1}^{\infty} m_e(I_j)\\
&<m_e(X)+\epsilon
\end{aligned}
$$

</details>

### Revisit Borel's criterion

1. $m(I)=\ell(I)$
2. If $\{S_j\}_{j=1}^{\infty}$ is a sequence of disjoint measurable sets, then $m\left(\bigcup_{j=1}^{\infty} S_j\right)=\sum_{j=1}^{\infty} m(S_j)$
3. If $R\subseteq S$, then $m(S\setminus R)=m(S)-m(R)$

#### Theorem 5.8 (Countable additivity for Lebesgue measure)

If $\{S_j\}_{j=1}^{\infty}$ is a sequence of disjoint measurable sets, whose union $S=\bigcup_{j=1}^{\infty} S_j$,  has finite outer measure, then

$$
m_e(S)=\sum_{j=1}^{\infty} m_e(S_j)
$$

<details>
<summary>Proof</summary>

First we prove $m_e(\bigcup_{j=1}^{\infty} S_j)=\sum_{j=1}^{\infty} m(S_j)$ by induction.

$n=1$ is trivial.

Let $n>1$ and suppose the statement holds for $n-1$. Take $X=\bigcup_{j=1}^{n-1} S_j$, then $S_n\cap X=S_n, X\setminus S_n=\bigcup_{j=1}^{n-1} (S_j)$.

By Caratheodory's criteria,

$$
\begin{aligned}
m_e(X)&=m_e(S_n)+m_e(\bigcup_{j=1}^{n-1} S_j)\\
m_e(\bigcup_{j=1}^{n} S_j)&=m(S_n)+\sum_{j=1}^{n-1} m(S_j)\\
&=\sum_{j=1}^{n} m(S_j)
\end{aligned}
$$

Take the limit as $n\to\infty$, and justify this.

$\sum_{j=1}^{\infty} m(S_j)=m_e(\bigcup_{j=1}^{\infty} S_j)\leq m_e(S)$

Since $m_e(S)$ is finite and $m(S_j)$ is monotone, the limit exists.

Therefore, $\sum_{j=1}^{\infty} m(S_j)\leq m_e(S)\leq \sum_{j=1}^{\infty} m(S_j)$

So $S$ is measurable.

</details>

#### Proposition 5.9 (Preview)

Any finite union (and intersection) of measurable sets is measurable.

<details>
<summary>Proof</summary>

Let $S_1, S_2$ be measurable sets.

We prove by verifying the Caratheodory's criteria for $S_1\cup S_2$.

</details>