# Tuesday September 10th: 

## A Brief Introduction to (Actual) Measure Theory

**Definition:**
Instead of just $\RR^n$, just consider a set $X$ with a $\sigma\dash$algebra $\mathcal A$.

Then the pair $(X, \mathcal A)$ is a *measurable* space, i.e. it is ready to be equipped with a measure.

A *measure* space is a measurable space with a *measure*.

**Definition:**
A set function $\mu: \mathcal A \to [0, \infty]$ satisfying

- $\mu(\emptyset) = 0$
- $\mu$ is countably additive,

is said to be a *measure*.

> What we've done so far is construct something we've called "the Lebesgue measure", then verified that it was actually a measure. 
We constructed a set function on $\RR^n$ called $m$ (the outer measure), then restricted attention to a class of sets $\mathcal M$, and produced a measure space $(\RR^n, \mathcal M, m)$.

Note that countable additivity implies monotonicity and subadditivity, which were what we originally discovered about $m_*$.

## Continuity of Measure
**Theorem (Fundamental Theorem of Measure Theory):**

Let $\theset{E_i} \subseteq \mathcal M$.

- If $E_i \nearrow E$, so $E_1 \subseteq E_2 \subseteq \cdots$ and $\union E_i = E$, then $\mu(E) = \lim \mu(E_i)$ (continuity from below).
- If $E_i \searrow E$, so $E_1 \supseteq E_2 \supseteq \cdots$ and $\intersect E_i = E$, then $\mu(E) = \lim \mu(E_i)$ as long as $\mu(E_1) < \infty$i (continuity from above).

*Exercise:*
Show that $\mu(E_1) < \infty$ is necessary in the 2nd result above.

*Proof:*

Part (1):

![Image](figures/2019-09-05-11:20.png)\

Let 

- $A_1 = E_1$
- $A_2 = E_2 \setminus E_1$,
- $\cdots A_j = E_j \setminus E_{j-1}$.


Then $\theset{A_j}$ are disjoint, and $E = \disjoint A_j$, so

\begin{align*}
m(E) &= \sum_j m(A_j) \\
&= \lim_{k\to\infty} \sum _{j=1}^k \mu (A_j) \\
&= \lim_{k\to\infty} \mu(\union_{j=1}^k A_j) \\
&= \lim_{k\to\infty} \mu (E_k). \quad \qed
\end{align*}

Part (2):

![Image](figures/2019-09-05-11:25.png)\

Let $A_i = E_j\setminus E_{j+1}$, so $\theset{A_i}$ are disjoint.

Then (**important!!**) $E_1 = \union A_i \union E$, which are **disjoint and measurable**. Then,

\begin{align*}
\mu(E_1) &= \sum \mu(A_i) + \mu(E) \\
&= \lim_{k\to\infty} \sum_{i=1}^{k-1} \mu(A_j) \\
&= \lim_{k\to\infty} \sum_{i=1}^{k-1} \mu(E_i) - \mu(E_{i+1}) + \mu(E),\quad\text{(which is telescoping)} \\
&= \lim_{k\to\infty} (\mu(E_1) - \mu(E_k)) + \mu(E) \\
\implies \mu(E_1) &= \mu(E_1) - \lim_k \mu(E_k) + \mu(E) \quad\text{since } \mu(E_1) < \infty \\
\implies \mu(E) &= \lim_k \mu(E_k) \hfill
.\end{align*}

$\qed$

Recall that if $E\subseteq \RR^n$, then

\begin{align*}
m_*(E) &= \inf\theset{m_*(G) \suchthat E\subset G~\text{open}} \\
&\iff  \\ 
\forall \varepsilon > 0,~~ \exists G \supseteq E &\suchthat m_*(G) \leq m_*(E) + \varepsilon
.\end{align*}

> Note: this says that the measure is *regular*.

## Approximation by Compact Sets

**Theorem:**
If $E\subseteq \RR^n$ is measurable, then

\begin{align*}
m(E) = \sup&\theset{m(K) \suchthat K\subseteq E~\text{and $K$ is compact}} \\
\iff & \\
\forall \varepsilon ~\exists ~\text{compact}~ K\subseteq E  \suchthat& m(K) \geq m(E) - \varepsilon
.\end{align*}

*Proof:*

*Case 1:* 
Suppose $E$ is bounded.

Let $\varepsilon > 0$, then by the closed characterization of measurability, we have $m(E\setminus F) < \varepsilon$ for some closed set $F\subseteq E$. 

Since $E$ is bounded, $m(E) < \infty$, and so

\begin{align*}
m(E\setminus F) &< \varepsilon  \implies \\
m(E) - m(F) &< \varepsilon \implies \\
m(F) &> m(E) - \varepsilon
.\end{align*}

where since $F$ is a closed and bounded set in $\RR^n$, $F$ is compact.

*Case 2:*
Suppose $E$ is unbounded.
Write 
$$
E_j = E \intersect_{j=1}^\infty B(j, 0)
,$$ 
so $E_j \nearrow E$.

Using continuity from below, we have $m(E) = \lim_j m(E_j)$.

Suppose $m(E) < \infty$.
Then for some $j$, 
$$
m(E_j) \geq m(E) - \varepsilon
.$$

But $E_j$ is bounded.

By case 1, there is a compact $K \subseteq E_j$ where 
$$
m(K) \geq m(E_j) - \varepsilon
.$$

But then $m(K) \geq m(E) - 2\varepsilon$.

Suppose now that $m(E) = \infty$.

For any $M > 0$, we can find an $E_j$ such that 
$$m(E_j) > M.$$ 

Since $E_j$ is bounded, by case 1 we get a compact $K \subseteq E_j \subseteq E$ with $m(K) > M$.

$\qed$

> Qual alert: very similar arguments are often used on the quals.


## Caratheodory Characterization

A subset $E\subseteq \RR^n$ is measurable iff for all $A\subseteq \RR^n$,
$$
m_*(A) = m_*(E\intersect A) + m_*(E \intersect A^c).
$$

![Image](figures/2019-09-05-12:01.png)\

Note that this can be interpreted in terms of *inner measures*, in which we're approximating $E$ with cubes from the inside. If we also think of this in terms of probability spaces, we could interpret the RHS as saying that the probability of an event either happening or *not* happening should be 1.

> Note: $G_\delta$ sets are measurable.

Note that the $\leq$ case is satisfied automatically by subadditivity, and the $\geq$ case comes from if $A\subseteq V$ then

\begin{align*}
m(A) \geq \cdots \text{Exercise!}
.\end{align*}


**Theorem:**
Let $E\subseteq \RR^n$ be measurable. Then

1. For all $h\in \RR^n$, then $E+h$ is measurable, and $\mu(E+h) = \mu(E)$.
  
2. For every $x\in \RR$, the set $cE$ is measurable and $\mu(cE) = \abs{c}^n \mu(E)$.

We can say more, and determine measures of all linear transformations of a set in terms of the determinant. 
Note that because we're working with cubes in the outer measure, so the only content here is that these new sets are actually measurable.

If $E$ is measurable, $E = H \union Z$ where $H \in F_\sigma$ and $\mu(Z) = 0$. 
But then $E+h = (H+h) \union (Z + h)$, but $H+h$ is still $F_\sigma$ because shifts of closed sets are still closed. 
Moreover, $\mu(Z+h) = \mu(Z) = 0$, so were done.

> Moral: it suffices to check things on Borel sets.