# Thursday September 5th

## Measurability of Closed Sets

**Recall:**
A set $E$ is *Lebesgue measurable* iff there exists an open set $G$ with $E \subseteq G$ and $m(G\setminus E) < \varepsilon$ for any $\varepsilon > 0$, and the set $\mathcal M$ of all measurable sets forms a $\sigma\dash$algebra.

*Fact:*
If $F, K \in \RR^n$ with $F$ closed, $K$ compact, and $F \intersect K = \emptyset$, then $\mathrm{Dist}(F, K) > 0$.

*Proof:*
Towards a contradiction, suppose the distance in zero. Idea: we'll use sequential compactness.

We can produce sequences $\theset{x_n} \subset F, \theset{y_n}\subset K$ such that $\abs{x_n - y_n} \to 0$.

Since $K$ is compact, it is *sequentially compact*, so there is a subsequence $\theset{y_{n_k}}$ with $y_{n_k} \to y \in K$.

Then
$$
\abs{x_{n_k} - y} \leq \abs{x_{n_k} - y_{n_k}} + \abs{y_{n_k} - y} \to 0.
$$
$\qed$


We used the following:

**Lemma:**
Closed sets are measurable.

*Proof:*

**Claim**: It suffices to prove this for *compact* sets.

Let $F$ be closed. 
Then write $F = \union_k (F \intersect B(k, 0))$. 
But $F\intersect B(k, 0)$ is closed and bounded, thus compact. 

So if we show compact sets are measurable, we've written $F$ as a countable union of measurable sets, which is thus measurable.

So suppose $K$ is compact, we want to show that $m_*(K) < \infty$. Given $\varepsilon > 0$, we can find an open set $G \supseteq K$ such that 
$$
m_*(G) < m_*(K) + \varepsilon
.$$ 

Now, *since $K$ is bounded*, the outer measure is not infinite, and so we have $m_*(G) - m_*(K) < \varepsilon$.

*Goal*: 
We now want to show 
$$
m_*(G\setminus K) \leq m_*(G) + m_*(K)
.$$

Since $G$ is open, $G\setminus K$ is open as well.
We can write any open set as the union of almost disjoint closed cubes, so we have
$$
G\setminus K = \union_j Q_j,\quad \theset{Q_j} \text{ a collection of almost disjoint cubes.}
$$

Now by property (5), we have 
$$
m_*(G\setminus K) = \sum_j \abs{Q_j}
.$$

Since any finite union of closed sets is closed, we have $K \intersect (\union_{j=1}^N Q_j) = \emptyset$.
But then $\mathrm{dist}(K, \union^N Q_j) > 0$. Using (1) and (4),

\begin{align*}
m_*(G) \geq m_*(K \union_{j=1}^N Q_j) &= m(K) \\
&= m(\union^N Q_j) \\
&= m_*(K) + \sum^N \abs{Q_j},
.\end{align*}


and since $K$ is bounded and thus of finite measure, we obtain
$$
\sum^\infty \abs{Q_j} \leq m_*(G) - m_*(K).
$$
$\qed$

## Characterizations of Measurability

**Theorem**:
A set $E\subseteq \RR^n$ is measurable iff

1. For any $\varepsilon,~\exists F\subseteq E$ with $F$ closed and $m_*(E\setminus F) < \varepsilon$.
2. There exist $F$ closed, $G$ open, $F \subseteq E \subseteq G$ with $m_*(G\setminus F) < \varepsilon$.

We know that $E$ is measurable iff $E^c$ is measurable, so we'll apply the definition to $E^c$. So we know
$$
\forall \varepsilon > 0,~ \exists \text{ open } G \supseteq E^c \suchthat m_*(G\setminus E^c) < \varepsilon .
$$

and so
$$
\forall \varepsilon > 0,~ \exists \text{ closed } G^c \subseteq E \suchthat m_*(E \setminus G^c) < \varepsilon .
$$

since $G\setminus E^c = G \intersect E = E \setminus G^c$.
So just take $F = G^c$ and we're done.

**Definition:** 
A $\sigma\dash$algebra is any collection of sets which is closed under complements and countable unions.

> Note that if we intersect $\sigma\dash$algebras, we still get a $\sigma\dash$algebra.

*Examples:*

- $\mathcal P(\RR^n)$

- $\mathcal M$, the collection of all (Lebesgue) measurable sets.

- $\mathcal B(\RR^n)$, the *Borel* subsets of $\RR^n$, i.e. the smallest $\sigma\dash$algebra containing the open sets.

There are inclusions
$$
\mathcal P(\RR^n) \supsetneq \mathcal M(\RR^n) \supsetneq \mathcal B(\RR^n).
$$

> Qual problem alert!

**Theorem:**
TFAE:

1. $E \subseteq \RR^n$ is measurable 
2. $E = H \union Z$ where $H \in F_\sigma$ and $Z$ is null 
3. $E = V\setminus Z'$ where $V\in G_\delta$ and $Z'$ is null.

*Proof:* 

$2,3 \implies 1$ 
**Exercise.** 
This is the easy direction.

$1 \implies 2,3$: 
For all $k \in \NN$, we can find $F_ k\subseteq E \subseteq G_k$ with $F_k$ closed, $G_k$ open, and
$$
m_*(G_k\setminus F_k) < \frac 1 k
.$$

So let $V = \intersect G_k$ and $H = \union F_k$. Then $H \subseteq E \subseteq V$. Note that $H$ is an $F_\sigma$ and $V$ is a $G_\delta$.

Moreover, $V\setminus H \subseteq G_k \setminus F_k$ for all $k$.
By subadditivity, 
$$
m_*(V\setminus H) \leq m_*(G_k \setminus F_k) \to 0.
$$

Now, $E = H \union(E \setminus H)$ where $E\setminus H \subseteq V\setminus H$ which is a null set. We also have $E = V \setminus (V \setminus E)$ where $V\setminus E \subseteq V\setminus H$, which is null.
$\qed$

**Recall**:
If $E$ is measurable, then we define its *Lebesgue measure* by $m(E) = m_*(E)$.

**Theorem 2:**
The Lebesgue measure is countably additive, i.e.
$$
E_i \intersect E_j = \emptyset \implies m(\union E_i) = \sum m(E_i).
$$

*Proof:*
Assume each $E_j$ is bounded, so that $m_*(E_j) < \infty$. 
Given $\varepsilon > 0$, for each $j$ we can find a *compact* $K_j$ such that
$$
m(E_j \setminus K_j) \leq \varepsilon / 2^j
.$$

Then for any finite $N$, since the $E_j$ are disjoint, then $\theset{K_i}_{i=1}^N$ are also disjoint, so
$$
m(\union^N K_j) = \sum m(K_j)
.$$

However, we have $m(E_k) - m(K_j) < \varepsilon / 2^j$, and so $m(K_j) > m(E_j) - \varepsilon / 2^j$. Then
$$
m(\union^N K_j) = \sum m(K_j) \geq \sum m(E_j) - \varepsilon / 2^j = \sum m(E_j) - \varepsilon.
$$

But since 
$$
\union^N K_j \subset E \definedas \union^\infty E_j
,$$

we have
\begin{align*}
m(E) \geq m(\union^N K_j) &\geq \sum^N m(E_j) \implies \\
\sum^n m(E_j) &\leq m(E) + \varepsilon \forall N \implies \\
\sum^\infty m(E_j) &\leq m(E) + \varepsilon 
&\to m(E)
.\end{align*}

So this shows the bounded case.

In general, let

\begin{align*}
A_1 = B(1, 0) \\
A_2 = B(2, 0) \setminus B(1, 0) \\
\vdots
\end{align*}

Then let $E_{i, j} = E_i \intersect A_j$, so $\union_i E_i = \union_{i, j} E_{i, j}$, where all of the $E_{i, j}$ are still disjoint but also now bounded.

Then
\begin{align*}
m(\union E_j) 
&= m(\disjoint_{i, j} E_{i, j}) 
&= \sum_j \sum_i m(E_{i, j}) 
&= \sum_j m(E_i)
,\end{align*}

where the last two equalities follow from the bounded case.
$\qed$