# Thursday August 29th: Outer Measure

Today: 
1.2 in Stein, the Lebesgue Outer Measure

Some preliminary results about open sets:

**Theorem 1.3 (Stein):**
Every open subset of $\RR$ can be written as a countable union of disjoint open intervals. Moreover, this representation is unique.


**Theorem 1.4 (Stein):** 
As a partial analog of 1.3, every subset of $\RR^n$ can be written as a countable union of *almost disjoint* closed cubes.

**Definition:**
$A, B$ are *almost disjoint* iff $A^\circ \intersect B^\circ = \emptyset$.

We'll now attempt to assign a preliminary notion of measure for all subsets of $\RR$ which extends the notion of volume.

**Definition (Outer Measure):**
If $E \subseteq \RR^n$, then the **Lebesgue outer measure** of $E$, denoted $m_*(E)$, is defined as follows:
$$
m_*(E) = \inf_{\union_{i\in\NN} Q_i \supseteq E, ~ Q_i \text{closed}} \sum_{i\in\NN} \abs{Q_j}
$$
where we take the infimum over all coverings of $E$ by countably many closed cubes.

*Remarks:*

- $m_*$ is well-defined for *all* subsets $E \subseteq \RR^n$.

- $m_*(E) \in [0, \infty]$

- For all $\varepsilon > 0$, there exists a covering $E \subseteq \union_{i\in \NN}$ such that 
$$
\sum_{i\in\NN} \abs{Q_i} \leq m_*(E) + \varepsilon
.$$

- We would not want to merely require coverings by _finite_ collections of closed cubes.
  (See challenge problem and *Jordan content* of sets)

*Examples:*
If $E$ is countable, then $m_*(E) = 0$.

This follows from the fact that any point is a closed cube with zero volume

**Proposition:**
If $E \subset \RR$, then $E$ is null $\iff m_*(E) = 0$.


$\implies:$ 
We can cover by open intervals with lengths summing to $<\varepsilon$, so just close them (which doesn't increase the length).

$\impliedby$: 
**(Easy exercise.)** 

Increase the length of the $n$th open interval by $\varepsilon / 2^n$.
$\qed$

*Example:*
If $Q$ is a closed cube, the $m_*(Q) = \abs{Q}$, the usual volume.

Since $Q \subseteq Q$, $Q$ covers itself and $m_*(Q) \leq \abs{Q}$. For the other direction, fix $\varepsilon > 0$; we will show $\abs{Q} \leq m_*(G) + \varepsilon$ for every $\varepsilon$.

Let $\theset{Q_j}$ be an arbitrary covering of $Q$ by closed cubes. Idea: enlarge the cubes a bit so they're open, and use compactness to get a finite subcover.

Let $S_j$ denote an open cube with the property that $Q_j \subseteq S_j$ and
$$
\abs{Q_j} \leq \abs{S_j} \leq (1+\varepsilon)\abs{Q_j}
$$
Since $Q$ is compact, there is a finite $N$ such that $E \subseteq \union_{j=1}^N S_j$, and the claim is that $\abs{Q} \leq \sum_{j=1}^N \abs{S_j}$ (Lemma 1.2, Stein).

> Recall 1-dimensional setting, we did the same thing to prove that $\RR$ was not null.

We then have
$$
\abs{Q} \leq \sum_{j=1}^N \abs{S_j} \leq (1+\varepsilon) \sum_{j=1}^N \abs{Q_j} \leq (1+\varepsilon) \sum_{j=1}^\infty \abs{Q_j},
$$

which is what we wanted to show.

*Exercise*: 
Let $Q$ be open, show that $m_*(Q) = \abs Q$.

**Proposition**:
$$
m_*(\RR^n) = \infty
.$$

This would follow if we could show that $\abs{Q} \leq m_*(\RR^n)$ for any $Q$, and we can take $Q$ to be arbitrarily large.
This is because any covering of $\RR^n$ is also a covering of $Q$.

## Properties of Outer Measure

1. **Monotonicity**: 
If $E_1 \subseteq E_2$ then $m_*(E_1) \leq m_*(E_2)$.

2. **Countable Subadditivity**: 
If $E = \union_{i\in\NN} E_i$ for any countable union, then
$$
m_*(\union E_i) \leq \sum m_*(E_i)
$$

3. If $E \subseteq \RR^n$ then $\forall \varepsilon > 0$ there exists an *open* set $G \supseteq E$ such that
$$
m_*(E) \leq m_*(G) \leq m_*(E) + \varepsilon.
$$

> Note: This does not imply every set is measurable! In fact,
$$
m_*(G) - m_*(E) \neq m(G\setminus E)
.$$ 
If we try to write $G = E \disjoint (G\setminus E)$, we only get an equality if there's a positive distance between these two sets! Otherwise, we only have subadditivity, and 
$$
m_*(G\setminus E) \geq m_*(G) - m_*(E)
.$$ 
But this is the wrong direction if we want to say something like 
$$
m_*(G) - m_*(E) \leq \varepsilon
.$$

4. **Almost Disjoint Additivity:** 
If $E_1, E_2 \subseteq \RR^n$ and

\begin{align*}
\mathrm{dist}(E_1, E_2) \definedas
\inf_{x\in E_1, y\in E_2} \abs{x - y} > 0 
\implies \\
m_*(E_1 \dot\union E_2) = m_*(E_1) + m_*(E_2)
.\end{align*}


5. If $E = \dot\disjoint_{j\in \NN} Q_j$ with the $Q_j$ *almost disjoint*, then $m_*(E) = \sum_j \abs{Q_j}$.

*Remark:* 
Property 4 does *not* hold in general if we merely assume that $E_1 \intersect E_2 = \emptyset$. 
It *will* be true if we restrict the collection of sets we consider to be "measurable", so any counterexample will have to involve a pathological set.

**Warning**: 
Any $E_j$ could have $m_*(E) = \infty$, so we have to be careful with our assumptions and how we work with inequalities, particularly when subtracting measures.

## Proofs of Properties of Outer Measure

### Property 1
Straightforward, since any covering of $E_2$ is also a covering of $E_1$.

We are thus taking infimums over *larger* collections of sets, so it can only get smaller.

### Property 2

If $m_*(E_j) = \infty$ for any $j$, this is vacuous, so assume $m_*(E_j) < \infty$ for every $j$.

Let $\varepsilon > 0$. 
For each $j$, there exists a covering $E_j \subseteq \union_k Q_{j, k}$ where 
$$
\sum_k \abs{Q_{j,k}} \leq m_*(E_j) + \varepsilon / 2^j
.$$

But now $E \subseteq \union_{j, k} Q_{j, k}$, so
$$
m_*(E) \leq \sum_{j, k} \abs{Q_{j, k}} = \sum_j \sum_k \abs{Q_{i, j}} \leq \sum_j (m_*(E_j) + \varepsilon/2^j) = \varepsilon + \sum_j m_*(E_j).
$$

### Property 3

> Idea: enlarge open sets in a summable way.

Let $\varepsilon > 0$. 
Then there exists a covering $E \subseteq \union_{j\in \NN} Q_j$ such that 
$$
\sum \abs{Q_j} \leq m_*(E) + \varepsilon/2
.$$ 

Let $S_j$ be an open cube such that $Q_j \subset S_j$ and 
$$
\abs{S_j} \leq \abs{Q_j} + \varepsilon/2^{j+1}
.$$

So define $G \definedas \union_j S_j$, which is open. 

Using subadditivity, we have
$$
m_*(G) \leq \sum_j \abs{S_j} = \sum_j \left( \abs{Q_j} + \varepsilon/2^{j+1} \right) \leq m_*(E) + \varepsilon.
$$

### Property 4

We Just need to show that 
$$
m_*(E_1 \union E_2) \leq m_*(E_1) + m_*(E_2)
,$$ 
since the reverse direction follows from (2). 

*Proof to follow in next section.*

> **Key idea:** by subdividing cubes, we can assume that no cube intersects both sets.

*Remark:* 
It is possible to construct closed disjoint subsets of $\RR^n$ such that the distance between them is still zero. 
Take $X = \NN$ and $Y = \theset{n + \frac 1 {2n} \suchthat n \in \NN}$.

> **Exercise (a good one!)**: 
> Show that if $F$ is closed and $K$ is **compact**, then $\mathrm{dist}(X, Y) > 0$.