# Math4121 Lecture 29

## Continue on Measure Theory

### Lebesgue Measure

Caratheodory's criterion:

$S$ is Lebesgue measurable if for all $A\subset S$,

$$
m_e(X) = m_e(X\cap S) + m_e(X\cap S^c)
$$

Let $\mathfrak{M}$ be the collection of all Lebesgue measurable sets.

1. $\phi\in\mathfrak{M}$
2. $\mathfrak{M}$ is closed under countable unions (proved last lecture)
3. $\mathfrak{M}$ is closed under complementation ($\mathfrak{M}$ is a $\sigma$-algebra) (goal today)

> Desired properties of a measure:
>
> 1. $m(I)=\ell(I)$ for all intervals $I$
> 2. If $\{S_n\}_{n=1}^{\infty}$ is a set of pairwise disjoint Lebesgue measurable sets, then
>
> $$ m\left(\bigcup_{n=1}^{\infty}S_n\right) = \sum_{n=1}^{\infty}m(S_n)$$
> 3. If $R\subset S$, then $m(S\setminus R) = m(S) - m(R)$

Recall the Borel $\sigma$-algebra $\mathcal{B}$ was the smallest $\sigma$-algebra containing closed intervals. Therefore $\mathcal{B}\subset\mathfrak{M}$.

Towards proving $\mathfrak{M}$ is closed under countable unions:

#### Theorem 5.9 (Finite union/intersection of Lebesgue measurable sets is Lebesgue measurable)

Any finite union/intersection of Lebesgue measurable sets is Lebesgue measurable.

<details>
<summary>Proof</summary>

Suppose $S_1, S_2$ is a measurable, and we need to show that $S_1\cup S_2$ is measurable. Given $X$, need to show that

![Finite union cut](https://notenextra.trance-0.com/Math4121/Finite_union_cut.png)

$$
m_e(X) = m_e(X_1\cup X_2\cup X_3)+ m_e(X_4)
$$

Since $S_1$ measurable, $m_e(X_1\cup X_2\cup X_3)=m_e(X_3)+m_e(X_1\cup X_2)$.

Since $S_2$ measurable, $m_e(X_3\cup X_4)=m_e(X_3)+m_e(X_4)$.

Therefore,

$$
\begin{aligned}
m_e(X) &= m_e(X_1\cup X_2\cup X_3) + m_e(X_4) \\
&= m_e(X_1\cup X_2) + m_e(X_3)+m_e(X_4) \\
&= m_e(X_1\cup X_2) + m_e(X_3\cup X_4) \\
&= m_e(X)
\end{aligned}
$$

by measurability of $S_1$ again.

</details>

#### Theorem 5.10 (Countable union/intersection of Lebesgue measurable sets is Lebesgue measurable)

Any countable union/intersection of Lebesgue measurable sets is Lebesgue measurable.

<details>
<summary>Proof</summary>

Let $\{S_j\}_{j=1}^{\infty}\subset\mathfrak{M}$. Definte $T_j=\bigcup_{k=1}^{j}S_k$ such that $T_{j-1}\subset T_j$ for all $j$.

And $U_1=T_1$, $U_j=T_j\setminus T_{j-1}$ for $j\geq 2$.

Then $\bigcup_{j=1}^{\infty}S_j=\bigcup_{j=1}^{\infty}T_j=\bigcup_{j=1}^{\infty}U_j$. Notice that $\{U_j\}_{j=1}^{\infty}$ are pairwise disjoint, and $\{T_j\}_{j=1}^{\infty}$ are monotone.

Let $X$ have finite outer measure. Since $U_n$ is measurable,

$$
\begin{aligned}
m_e(X\cap T_n) &= m_e(X\cap T_n\cap U_n)+ m_e(X\cap T_n\cap U_n^c) \\
&= m_e(X\cap U_n)+ m_e(X\cap T_{n-1}) \\
&= \sum_{j=1}^{n}m_e(X\cap U_j)
\end{aligned}
$$

Since $T_n$ is measurable and $T_n\subset S$, $S^c\subset T_n^c$. $m_e(X\cap T_n^c)\geq m_e(X\cap S^c)$.

Therefore,

$$
m_e(X)=m_e(X\cap T_n)+m_e(X\cap T_n^c)\\
\geq \sum_{j=1}^{n}m_e(X\cap U_j)+m_e(X\cap S^c)
$$

Take the limit as $n\to\infty$,

$$
\begin{aligned}
m_e(X) &\geq \sum_{j=1}^{\infty}m_e(X\cap U_j)+m_e(X\cap S^c) \\
&= m_e(\bigcup_{j=1}^{\infty}(X\cap U_j))+m_e(X\cap S^c) \\
&= m_e(X\cap S)+m_e(X\cap S^c) \\
&\geq m_e(X)
\end{aligned}
$$

Therefore, $m_e(X\cap S)=m_e(X)$.

Therefore, $S$ is measurable.

</details>

#### Corollary from the proof

Every open or closed set is Lebesgue measurable.

(Every open set is a countable union of disjoint open intervals)