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\date{26 March 2018}
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\begin{frame}{Measure theory (Ch. 6)}
\textbf{Fundamental question of measure theory}: in what way can we assign a numerical value to ``measure" the volume (or length or area or...) of a set? \\
\textbf{Easy question}: What is the length of the interval $(a,b)$? \\ \onslide<2->{ \textbf{Answer}: $b-a$ \\
This leads to a natural definition of what is called an ``outer measure". \\
\textbf{Definition}: The length of $(a,b)$ is $b-a$. Let $A \subset \mathbb{R}$. The Lebesgue outer measure of $A$ is defined to be the following number:
$$m^*(A) = \inf \left\{ \displaystyle\sum_k b_k-a_k \colon A \subseteq \bigcup_k (a_k,b_k) \right\}.$$
We call the union in this definition a ``covering" of $A$.
}
\end{frame}

\begin{frame}{Outer measure}
\underline{Axioms of Outer Measure}
Recall the power set $\mathscr{P}(X)$ of a set $X$ is the set of all subsets. We say that a function $m^* \colon \mathscr{P}(\mathbb{R}) \rightarrow [0,\infty]$ (note: we allow $\infty$!) is an outer measure on $\mathbb{R}$ provided it satisfies the following three axioms.
\begin{enumerate}[a.)]
\item The outer measure of the empty set is zero.
\item (\textbf{monotonicity}) If $A \subseteq B$, then $m^* A \leq m^* B$.
\item (\textbf{countable subadditivity}) If $A = \bigcup_{n=1}^{\infty} A_n$, then $m^* A \leq \displaystyle\sum_{n=1}^{\infty} m^* A_n$.
\end{enumerate}
\end{frame}

\begin{frame}{Lebesgue outer measure is an outer measure}
We must verify axioms a.), b.), and c.). \\
\onslide<2->{
\textbf{Axiom a.)} holds because the empty set is a subset of \textbf{all} intervals, so since we take inf, we must get zero. \\
\onslide<3->{
\textbf{Axiom b.)} holds because any set of intervals ``covering" $B$ will also ``cover" $A$, but not vice versa. \\
\onslide<4->{
\textbf{Axiom c.)} holds because if $\epsilon > 0$, then for each $n$, there is a covering $\{I_{nk}\}$ of $A_n$ so that
$$\displaystyle\sum_{k=1}^{\infty} m^* I_{nk} < m^* A_n + \dfrac{\epsilon}{2^n}.$$
Then summing over $n$ yields
$$\displaystyle\sum_{n=1}^{\infty} \displaystyle\sum_{k=1}^{\infty} m^* I_{nk} \leq \displaystyle\sum_{n=1}^{\infty} \left(m^*(A_n) + \dfrac{\epsilon}{2^n} \right) = \epsilon + \displaystyle\sum_{n=1}^{\infty} m^*(A_n).$$
}
}
}
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\begin{frame}{Lebesgue outer measure is an outer measure}
We have $\displaystyle\sum_{n=1}^{\infty} \displaystyle\sum_{k=1}^{\infty} m^* I_{nk} \leq \epsilon + \displaystyle\sum_{n=1}^{\infty} m^*(A_n).$
Taking the infimum of this over all possible coverings yields
$$m^*(A) \leq \displaystyle\sum_{n=1}^{\infty} m^*(A_n),$$
proving Axiom c.)
\end{frame}

\begin{frame}{Measure in the plane}
The area of a rectangle $R=(a,b) \times (c,d)$ is $m^*(R)=(b-a)(d-c)$. This idea also yields an outer measure on $\mathbb{R}^2$. \\
\vspace{0.1in}
If $X$ is a set equipped with an outer measure $m^*$, then we say that a set $Y \subset X$ is a zero set (with respect to $m^*$) provided that $m^*(Y)=0$. 
\end{frame}

\begin{frame}{Properties of outer measure}
\textbf{Proposition}: The union of a list of zero set (with respect to outer measure $m^*$) is itself a zero set.  \\
\onslide<2->{\textbf{Proof}: Let $Z_1, Z_2, \ldots$ be a list of zero sets and consider the set $Z = \bigcup_{n=1}^{\infty} Z_n$. Since each $Z_n$ is a zero set, $m^*(Z_n)=0$. Thus, by countable subadditivity of the outer measure,
$$m^*(Z) = m^* \left( \displaystyle\bigcup_{n=1}^{\infty} Z_n \right) \leq \displaystyle\sum_{n=0}^{\infty} m^*(Z_n) = 0. \blacksquare$$}
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\begin{frame}
Define the notation $E+a=\{e+a \colon e \in E\}$. \\
\textbf{Quiz question}: What is $[0,2] \cup [5,6] + 2$? \\
\onslide<2->{\textbf{Proposition}: (translation invariance) Let $E \subset \mathbb{R}$ and $a \in \mathbb{R}$. Then $m^*(E+a)=m^*(E)$. \\
\onslide<3->{\textbf{Proof}: Suppose that $(a_n,b_n)$ is a sequence of intervals that covers $E$. Then $(a_n+a,b_n+a)$ is a sequence of intervals that covers $E+a$. Then,
$$m^*(E+a) \stackrel{sub.add.}{\leq} \displaystyle\sum_{n=1}^{\infty} (b_n+a)-(a_n+a) = \displaystyle\sum_{n=1}^{\infty} b_n-a_n.$$
From this we conclude by taking infimums that $m^*(E+a) \leq m^*(E).$
Similarly, if $(a_n',b_n')$ is a sequence of intervals that covers $E+a$, then $(a_n'-a,b_n'-a)$ is a sequence of intervals that covers $E$. Thus
$$m^*(E) \stackrel{sub.add.}{\leq} \displaystyle\sum_{n=1}^{\infty} (b_n'-a)-(a_n'-a) = \displaystyle\sum_{n=1}^{\infty} b_n'-a_n'.$$
Taking infimums yields $m^*(E) \leq m^*(E+a)$. This completes the proof that $m^*(E)=m^*(E+a)$. $\blacksquare$
}}
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\begin{frame}{Measurabilitiy}
Outer measure is not quite strong enough for what we want. Any ``good" measure should not merely be countably subadditive, but rather be ``countably additive", i.e. if $A_1, A_2, \ldots$ are mutually disjoint sets, then
$$\mathrm{measure} \left( A_1 \cup A_2 \cup A_3 \cup \ldots \right) = \displaystyle\sum_{k=1}^{\infty} \mathrm{measure}(A_k).$$
While $\mathscr{P}(X)$ has an outer measure for any set $X$, it is (\textbf{surprisingly!}) not true that all sets have a measure on $\mathscr{P}(X)$.
\end{frame}

\begin{frame}{Abstract outer measure}
Let $M$ be any set and let $\mathscr{P}(M)$ denote its power set (book uses $2^M$ for this). An abstract outer measure on $M$ is a function $\omega \colon \mathscr{P}(M) \rightarrow [0,\infty]$ that satisfies the three axioms: $\omega(\emptyset)=0$, monotonicity, and countable subadditivity. \\
\textbf{Quiz}: Prove that if $X=\{x_1,\ldots,x_n\}$ is a finite set, then the function $\omega \colon \mathscr{P}(X) \rightarrow [0,\infty]$ defined by $\omega(A)=\mathrm{cardinality}(A)$ defines an outer measure on $X$.
\end{frame}

\begin{frame}{Abstract outer measures}
We say that a set $E \subset M$ is measurable with respect to $\omega$ provided that $\forall X \subset M$, 
$$\omega(X) = \omega (X \cap E) + \omega( X \cap E^c).$$

We use $\mathcal{M}$ to denote the set of Lebesgue measurable subsets of $\mathbb{R}$. If $E$ is measurable, then its Lebesgue measure is defined to be $m^*(E)$, and we often drop the $*$ to denote it: $m(E)$.

A $\sigma$-algebra is a collection of sets that includes the empty set, is closed under complement, and closed under ``countable" unions.

\textbf{Theorem}: The collection $\mathcal{M}$ of measurable sets with respect to any outer measure on any set $M$ is a $\sigma$-algebra and the outer measure restricted to this $\sigma$-algebra is countably additive. In particular, Lebesgue measure is countably additive.
\end{frame}
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