# Tuesday September 3rd

## Lebesgue Measurability
Recall the definition of the Lebesgue measure:

**Definition:**
For any $E \subseteq \RR^n$, we define
$$
\mu(E) = \inf \theset{\sum \abs {Q_i} \suchthat E \subset \union Q_i, Q_i ~\text{a closed cube}}
$$

This satisfies properties (1) through (5).

> Note we don't have finite additivity for the outer measure.

**Definition:**
A set $E$ is said to be *measurable* iff
$$
\forall\varepsilon > 0, ~\exists \text{ an open }G \supseteq E  \suchthat\quad m_*(G\setminus E) < \varepsilon.
$$

**Observations**:

- If $E$ is open, $E$ is measurable

- If $m_*(E) = 0$, then $E$ is measurable. (Quite a special property!)

- If $E$ is closed, $E$ is measurable. (Needs a proof.)


**Theorem 1:**
The collection $\mathcal M$ of all measurable sets is a *$\sigma\dash$algebra*, i.e. $\mathcal M$ is closed under

- Countable unions

- Complements

- Countable intersections

**Theorem 2:**
The Lebesgue measure (on measurable sets) is countably additive, i.e. if $\theset{E_i}_{i\in\NN}$ is a countable collection of *pairwise disjoint* measurable sets, then
$$
m(\disjoint E_i) = \sum m(E_i).
$$

## Lebesgue Measurable Sets Form a Sigma Algebra

*Proof of Theorem 1:*

*Part 1:*
Let $E = \union_{i\in \NN} E_i$ with each $E_i$ measurable; we want to show $E$ is measurable. 

Given $\varepsilon > 0$, for each $j$ choose $G_j \supseteq E_j$ such that 
$$
m_*(G_j \setminus E_j) < \varepsilon /2^j
.$$ 
Then let $G = \union G_j$, which is open and $G \supseteq E$ and $G\setminus E = \union G_j\setminus E_j$. 

Using monotonicity, and then subadditivity, we have
$$
m_*(G\setminus E) =  m_*(\union G_j \setminus E_j) \leq \sum m_*(G_j - E_j) \leq \sum \varepsilon /2 = \varepsilon.
$$


*Part 2:*
Let $E \in \mathcal M$. 
Then for all $k\in \NN$, there is an open $G_k \supseteq E$ with 
$$
m_*(G_k\setminus E) \leq \frac 1 k
.$$

> Lemma to prove later: $G_k^c$ is closed and measurable.

By (1), the set $S \definedas \union G_k^c$ is measurable, and $S \subseteq E^c$, since $E^c = S \union (E^c \setminus S)$. 
So we just need to show that $E^c \setminus S = E^c \intersect S^c$ is measurable.

But where does $S$ live? 
Since 
$$
E^c \setminus S \subset G_k \setminus E = G_k \intersect E^c \text{ for every } k
,$$
we have 
$$
m_*(E^c \setminus S) \leq m_*(G_k\setminus E) < \frac 1 k \text{ for all } k
,$$ 
which says that $m_*(E^c\setminus S) = 0$ and thus $E^c \setminus S$ is measurable.

$\qed$

> Think further about why outer measure zero sets should be Lebesgue measurable!

**Next time:**

- Closed sets are measurable, 

- Proof of theorem 2, 

- Characterizations of measurability.