# Math4121 Lecture 27

## Lebesgue Measure

### Outer Measure

$$
m_e(S)=\inf\left\{\sum_{n=1}^\infty \ell(I_n): S\subset \bigcup_{n=1}^\infty I_n\right\}
$$

where $I_j$ is an open interval

**Properties:**

1. $m_e(I)=\ell(I)$
2. Countably sub-additive: $m_e\left(\bigcup_{n=1}^\infty S_n\right)\leq \sum_{n=1}^\infty m_e(S_n)$ (Prove today)
3. does not respect complementation (Build in to Borel measure)

Why does Jordan content respect complementation?

$(\text{Finite union of intervals })^C=\text{another finite union of intervals}$

We know this failed for countable unions.

Example:

$$
\bigcup_{n=1}^\infty \left(q_n-\frac{\epsilon}{2^n},q_n+\frac{\epsilon}{2^n}\right)
$$

Where $q_n$ is dense.

### Inner Measure

Say $S\subset I$

$$
m_i(S)=m(I)-m_e(I\setminus S)
$$

where $m(I)=\ell(I)$

Say $S$ is (Lebesgue) measurable if $m_i(S)=m_e(S)$, call this value $m(S)=m_e(S)=m_i(S)$ the (Lebesgue) measure of $S$.

#### Corollary of measurability of subsets

If $S$ is measurable, and $S\subset T$, then

$$
m(S)=m_e(S)=m(I)-m_e(I\setminus S)
$$

$$
m(I\setminus S)=m(I)-m(S)
$$

$I\setminus S$ is Lebesgue measurable and $m(I)=m(S)+m(I\setminus S)$

#### Proposition 5.8 (Countable additivity over measurable sets)

If $S_n$ are measurable, then

$$
m_e\left(\bigcup_{n=1}^\infty S_n\right)\leq\sum_{n=1}^\infty m(S_n)
$$

<details>
<summary>Proof</summary>

Let $\epsilon>0$ and for each $j$, let $\{I_{i,j}\}_{i=1}^\infty$ be a cover of $S_j$ s.t.

$$
\sum_{i=1}^\infty \ell(I_{i,j})<m(S_j)+\frac{\epsilon}{2^j}
$$

Then $\bigcup_{j=1}^\infty \bigcup_{i=1}^\infty I_{i,j}$ is a countable cover of $\bigcup_{j=1}^\infty S_j$ and 

$$
m_e\left(\bigcup_{j=1}^\infty S_j\right)\leq \sum_{j=1}^\infty \sum_{i=1}^\infty \ell(I_{i,j})<\sum_{j=1}^\infty \left(m_e(S_j)+\frac{\epsilon}{2^j}\right)=\sum_{j=1}^\infty m_e(S_j)+\epsilon
$$

Since $\epsilon$ is arbitrary, we have

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

</details>

#### Corollary: inner measure is always less than or equal to outer measure

$$
m_i(S)\leq m_e(S)
$$

<details>
<summary>Proof</summary>

$$
m_i(S)=m(I)-m_e(I\setminus S)\leq m(I)-m_i(I\setminus S)=m_e(S)
$$

</details>

### Caratheodory's Criterion

#### Lemma 5.7 (Local additivity)

If $\{I_j\}_{j=1}^\infty$ are pairwise disjoint open intervals, then

$$
m_e\left(S\cap \left(\bigcup_{j=1}^\infty I_j\right)\right)=m_e\left(\bigcup_{j=1}^\infty (S\cap I_j)\right)=\sum_{j=1}^\infty m_e(S\cap I_j)
$$

<details>
<summary>Proof</summary>

For each $j$, let $\{J_i\}_{i=1}^\infty$ be a cover of $S\cap \left(\bigcup_{j=1}^\infty I_j\right)$ such that $\sum_{i=1}^\infty \ell(J_i)<c_e(S\cap \left(\bigcup_{j=1}^\infty I_j\right))+\epsilon$. Since $\{I_j\}_{j=1}^\infty$ are pairwise disjoint, so is $\{J_i\cap I_j\}_{j=1}^\infty$ for each $i$.

$$
\sum_{j=1}^\infty m_e(J_i\cap I_j)=m_e(J_i)
$$

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

Since $\epsilon$ is arbitrary, we have

$$
m_e\left(S\cap \left(\bigcup_{j=1}^\infty I_j\right)\right)\leq \sum_{j=1}^\infty m_e(S\cap I_j)
$$

</details>

#### Theorem 5.6 (Caratheodory's Criterion)

A set $S$ is measurable if and only if for every set $X\in \mathbb{R}$ of finite outer measure,

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

Lebesgue: $X=I$ and $S\subset I$ we can cut any set by a measurable set to get a measurable set. (no matter how big the set is)