135 lines
3.4 KiB
Markdown
135 lines
3.4 KiB
Markdown
# 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>
|