# Wednesday August 28th

## Outer Measure

**Definition (Lebesgue Outer Measure):**
For any $E \subseteq \RR^n$ define
$$
m_*(E) = \inf \sum \abs{Q_i}
$$

where the infimum is taken cover all countable coverings of $E$ by closed cubes $Q_i$.

*Proof of property (4):*
Since we have property (2), we just need to show 
$$
m_*(E_1 \union E_2) \geq m_*(E_1) + m_*(E_2)
.$$ 

Choose $\delta$ such that $0 < \delta < \mathrm{dist}(E_1, E_2)$ and let $\varepsilon > 0$.

Then there exists a covering of $E_1 \union E_2$ such that
$$
m_*(E_1 \union E_2) \leq \sum \abs{Q_i} \leq m_*(E_1 \union E_2) + \varepsilon.
$$

We can assume (possibly after subdividing) that $\mathrm{diam}(Q_i) < \delta$.
Then each $Q_i$ can intersect at most one of $E_1, E_2$.

Let 
\begin{align*}
J_1 &= \theset{j \suchthat Q_j \intersect E_1 \neq \emptyset} \\
J_2 &= \theset{j \suchthat Q_j \intersect E_2 \neq \emptyset}
.\end{align*}

Note that $J_1, J_2$ are disjoint, and we have
\begin{align*}
E_1 \subseteq \union_{j\in J_1} Q_j \\
E_2 \subseteq \union_{j\in J_2} Q_j
.\end{align*} 

But then 
$$m_*(E_1) + M_*(E_2) \leq \sum_{j\in J_1} \abs{Q_j} + \sum_{j\in J_2} \abs{Q_j}
$$ by definition, since $m_*$ is an infimum. 

But this is less than summing over *all* $j$, which is the term appearing in the cover we choose above.
$\qed$

## Covering by Cubes

*Proof of property (5):*

> Qual/Exam problem alert (DZG)

Let $\varepsilon > 0$, we will show
$$
\sum \abs{Q_j} \leq m_*(E) + \varepsilon
.$$

Start by shrinking the cubes. 
Choose 
$$
\tilde{Q_j} \leq \abs{Q_j} \leq \abs{\tilde{Q_j}} + \varepsilon /2^j
.$$
Then for any finite $N$, any collection of $N$ different $Q_j$s are disjoint.

> **Exercise**: If $K$ is compact, $F$ is closed, and $K \intersect F = \emptyset$, the $\mathrm{dist}(K, F) > 0$.

Note that although this is certainly true for the *entire* infinite collection, we take finitely many so we can get a $\delta$ that uniformly bounds the distance between any two from below.

But then
$$
m_*(\union_{j=1}^N \tilde{Q_j}) = \sum_{j=1}^N \abs{\tilde{Q_j}} \geq \sum_{j=1}^N \abs{\tilde{Q_j}} - \varepsilon
$$

and since $\union_{j=1}^N \tilde{Q_j} \subseteq E$, for all $N$ we have 
$$
\sum_{j=1}^N \abs{\tilde{Q_j}} -\varepsilon \geq m_*(E)
.$$ 

> (Missing details, finish/fill in.)


**Definition (Measurable):**

A set $E\subseteq \RR^n$ is (Lebesgue) measurable if
$$
\forall \varepsilon >0 \quad \exists G ~\text{open}~ \suchthat \quad m_*(G\setminus E) < \varepsilon.
$$

**Important Observations:**

- If $m_*(E) = 0$, the $E$ is automatically measurable.

- If $F\subseteq E$ and $m_*(E) = 0$, then $F$ is automatically measurable.