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\date{26 March 2018}
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\begin{document}

\begin{frame}{Lebesgue outer measure}
Define the notation $E+a=\{e+a \colon e \in E\}$.

The Lebesgue outer measure of $A$ is defined to be the following number:
$$m^*(A) = \inf \left\{ \displaystyle\sum_k b_k-a_k \colon A \subseteq \bigcup_k (a_k,b_k) \right\}.$$
\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 $m^* \colon \mathscr{P}(X) \rightarrow [0,\infty]$ (note: we allow $\infty$!) is an outer measure on $X$ 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}{Properties of outer measure}
\textbf{Proposition}: The union of a list of zero sets (with respect to outer measure $m^*$) is itself a zero set.  \\
\textbf{Proposition}: (translation invariance) Let $E \subset \mathbb{R}$ and $a \in \mathbb{R}$. Then $m^*(E+a)=m^*(E)$.
\end{frame}

\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 measures}
Let $M$ be a set and let $\omega \colon \mathscr{P}(M) \rightarrow [0,\infty)$ be an outer measure. We say that $E \subset M$ 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: $m(E)$.
\end{frame}

\begin{frame}{Properties of measurable sets}
The following theorem is the statement that measurable sets form a ``$\sigma$-algebra". \\
\textbf{Theorem}: The collection $\mathcal{M}$ of measurable sets with respect to any outer measure on any set $M$ is a $\sigma$-algebra: i.e. the complement of a measurable set is measurable, the union and intersection of a list of measurable sets is measurable, and every set with outer measure zero is measurable.

\textbf{Theorem}: If $E$ is measurable and $E=E_1 \cup E_2$ where $E_1 \cap E_2=\emptyset$, then the outer measure is ``countably additive" meaning
$$m(E) = m(E_1)+m(E_2).$$
\end{frame}

\begin{frame}{A nonmeasurable set}
Let $x, y \in [0,1)$. Define their ``sum modulo $1$" $\oplus$ by 
$$x \oplus y = \left\{ \begin{array}{ll}
x+y, & \quad x+y < 1 \\
x+y-1, & \quad x+y \geq 1.
\end{array} \right.$$

\textbf{Quiz}: What is $0.5 \oplus 0.2$? What is $0.9 \oplus 0.9$?
\end{frame}

\begin{frame}{A nonmeasurable set}
\textbf{Lemma}: Let $E \subset [0,1)$ be a measurable set. Then for each $y \in [0,1)$, the set $E \oplus y$ is measurable and $m(E \oplus y)=m(E)$. \\
\textit{Proof}: Let $E_1=E \cap [0,1-y)$ and $E_2=E \cap [1-y,1)$. Then $E_1$ and $E_2$ are disjoint measurable sets whose union is $E$, hence by countable additivity,
$$m(E) = m(E_1) + m(E_2).$$
\onslide<2->{Since $y \in [0,1)$ we observe that $E_1 \oplus y = E_1+y \subset [y,1)$ and by translation invariance, we must conclude that $E_1 \oplus y$ is measurable and $m(E_1 \oplus y) = m(E_1)$. \onslide<3->{We can also conclude that $E_2 \oplus y = E_2 + (y-1) \subset [0,y)$ and by translation invariance we must conclude that $E_2 \oplus y$ is measurable and $m(E_2 \oplus y)=m(E_2)$. \onslide<4->{But since $E \oplus y = (E_1 \oplus y) \bigcup (E_2 \oplus y)$ and are disjoint, we may compute
$$m(E \oplus y) \stackrel{ctb.add.}{=} m(E_1 \oplus y) + m(E \oplus y)\stackrel{above}{=} m(E_1)+m(E_2) \stackrel{ctb.add.}{=} m(E). \blacksquare$$}}}
\end{frame}

\begin{frame}{A nonmeasurable set}
\textbf{Theorem}: A nonmeasurable set exists. \\
\textit{Proof}: Define an equivalence relation: if $x-y \in \mathbb{Q}$ then we say that $x$ is equivalent to $y$ and write $x \sim y$. \\
\textbf{Quiz}: This defines an equivalence relation. \\
This creates a set of equivalence classes: two elements from the same class always differ by a rational while two elements from different classes differ by an irratonal.
\end{frame}

\begin{frame}{An aside about the axiom of choice}
If $\mathscr{X}$ is a set containing one set $X$, then we may pick an $x \in X$ no problem. If $\mathscr{X}=\{X_1,X_2,\ldots,X_n\}$, then we can prove via induction that there is always a way to pick an $x_1 \in X_1, x_2 \in X_2, \ldots, x_n \in X_n$ and form the set $\{x_1,\ldots,x_n\}$.\\
\vspace{0.1in}
\onslide<2->{The axiom of choice is an axiom of set theory that generalizes this concept. It says: if $\mathscr{X}$ is any collection of sets, then there is a set $P$ with the property that $P$ contains one element of every $X \in \mathscr{X}$.\\
\vspace{0.1in}
\onslide<3->{It is ``controversial" because it (or something like it) is required to prove nonintuitive things like the exitence of nonmeasurable sets and the so-called Banach-Tarski paradox. \\
\vspace{0.1in}
\onslide<4->{On the other hand, if you reject it, then you lose ``nice" things like every vector space having a basis. Also rejecting the axiom of choice allows us to have a function $f \colon \mathbb{R} \rightarrow \mathbb{R}$ so that if $x_n \rightarrow x$ then $f(x_n) \rightarrow f(x)$ (``sequential continuity") but $f$ is not continuous in the $\epsilon$-$\delta$ sense.}}}
\end{frame}

\begin{frame}{A nonmeasurable set}
By the axiom of choice, we know there is a set $P$ which contains exactly one element from each equivalence class. Let $(r_i)$ be \textbf{the} (unordered, of course) list of rational numbers in $[0,1)$ (note: this is possible because $\mathbb{Q}$ is a ``countable" set) and arrange for $r_0=0$.  \\
\vspace{0.1in}
\onslide<2->{Define the sets $(P_i)$ by $P_i=P\oplus r_i$. We know then that $P_0=P$. If $x \in P_i \cap P_j$, then there are elements $p_i, p_j \in P$ so that $x=p_i\oplus r_i=p_j \oplus r_j$. \onslide<3->{It is easy to show by cases that $p_i-p_j=r_j-r_i$ is a rational number, and so $p_i \sim p_j$. But since $P$ has one element from each equivalence class, we must conclude that $i=j$. \onslide<4->{Hence if $i \neq j$, then $P_i \cap P_j = \emptyset$, meaning that the sequence $(P_i)$ is a sequence of disjoint sets.}}}
\end{frame}

\begin{frame}{A nonmeasurable set}
Each $x \in [0,1)$ is in some equivalence class and hence is equivalent to some element of $P$ (since $P$ contains one thing from each class). If $x$ differs from an element in $P$ by a rational $r_i$, then $x \in P_i$. \onslide<2->{Consequently: $\displaystyle\bigcup P_i = [0,1)$. By Lemma, since each $P_i$ is a translation modulo $1$ of $P$, each $P_i$ is measurable if $P$ is. \onslide<3->{\textbf{Assume that $P$ is measurable.} Then,
$$m([0,1)) = m \left( \displaystyle\bigcup_{i=1}^{\infty} P_i \right) =\displaystyle\sum_{i=1}^{\infty} m(P_i) = \displaystyle\sum_{i=1}^{\infty} m(P).$$
We know $m([0,1))=1$, BUT the sum on the right is zero or infinite (why?). This is a contradiction, hence $P$ cannot be measurable.}}
\end{frame}

\begin{frame}{Measurable functions}
\textbf{Definition}: Let $X \subset \mathbb{R}$. A function $f \colon X \rightarrow [-\infty,\infty]$ is called measurable if $X$ is measurable and for all $\alpha \in \mathbb{R}$, the set $\{x \colon f(x) \geq \alpha\}$ is measurable. \\
\textbf{Proposition}: If $c$ is a constant and $f,g$ are measurable functions, then $f+c, cf, f+g, g-f,$ and $fg$ are measurable. \\
\textbf{Proposition}: Continuous functions are measurable, step functions are measurable, restrictions of measurable functions to measurable domains are measurable.  \\
\end{frame}

\begin{frame}{Littlewood's 3 principles}
\begin{enumerate}[1.]
\item Every measurable set is nearly a finite union of intervals. 
\item Every measurable function is nearly continuous.
\item Every convergence sequence of measurable functions is nearly uniformly convergent.
\end{enumerate}

\end{frame}

\begin{frame}{Lebesgue integration -- ``simple functions"}
We saw earlier: Riemann and Darboux integration. Now we build up the measure theoretic integral (in turbo speed). 

Let $E \subset \mathbb{R}$ and define the characteristic function $\chi_E \colon \mathbb{R} \rightarrow \{0,1\}$ by
$$\chi_E(x) = \left\{ \begin{array}{ll}
1, & \quad x \in E \\ 
0, & \quad x \not\in E
\end{array} \right.$$
A linear combination of characteristic functions is called a simple function: 
$$\phi(x) = \displaystyle\sum_{i=1}^n a_i \chi_{E_i}(x).$$
Let $A_i = \{x \colon \phi(x)=a_i\}$. If $\phi \equiv 0$ outside a set of finite measure, then we write
$$\displaystyle\int \phi(x) \mathrm{d}x = \displaystyle\sum_{i=1}^n a_i m(A_i).$$
\end{frame}

\begin{frame}{Lebesgue integration}
When we want to restrict integration to a certain set $E$, we use the following notation:
$$\displaystyle\int_E \psi(x) \mathrm{d}x = \displaystyle\int \psi(x) \chi_E(x) \mathrm{d}x$$
You can show with some work the following properties: \\
\textbf{Proposition}: If $\phi$ and $\psi$ are simple functions and $\alpha, \beta \in \mathbb{R}$, then
$$\displaystyle\int \alpha \phi(x) + \beta \psi(x) \mathrm{d}x = \alpha \displaystyle\int \phi(x) \mathrm{d}x + \beta \displaystyle\int \psi(x) \mathrm{d}x.$$
\textbf{Proposition}: In order for the following equality to hold for a measurable set $E$ with $m(E)$ finite:
$$\displaystyle\inf_{f \leq \psi} \displaystyle\int_E \psi(x) \mathrm{d}x = \displaystyle\sup_{f \geq \phi} \displaystyle\int_E \phi(x) \mathrm{d}x,$$
it is necessary and sufficient for $f$ to be measurable.
\end{frame}

\begin{frame}{Lebesgue integral (for bounded functions)}
If $f$ is a \textbf{bounded} measurable function defined on a measurable set $E$ with $m(E)$ finite, the Lebesgue integral of $f$ over $E$ is
$$\displaystyle\int_E f(x) \mathrm{d}x = \displaystyle\inf_{\psi \geq f} \displaystyle\int_E \psi(x) \mathrm{d}x,$$
where $\psi$ is a simple function. \\
\textbf{Proposition}: Let $f$ be a bounded function defined on $[a,b]$. if $f$ is Riemann-integrable on $[a,b]$, then it is measurable and 
$$\mathrm{Riemann} \displaystyle\int_a^b f(x) \mathrm{d}x = \mathrm{Lebesgue}\displaystyle\int_a^b f(x) \mathrm{d}x.$$
\end{frame}

\begin{frame}{Lebesgue integral for non-negative functions (a pattern!)}
We essentially repeat the process: if $f$ is nonnegative measurable function defined on a measurable set $E$, then for bounded functions $h$,
$$\displaystyle\int_E f = \sup_{h \leq f} \displaystyle\int_E h.$$
\textbf{Fatou's Lemma}: If $(f_n)$ is a sequence of nonnegative measurable functions and $f_n \rightarrow f$ almost everywhere on a set $E$, then
$$\displaystyle\int_E f \leq \displaystyle\liminf_{n \rightarrow \infty} \displaystyle\int_E f_n.$$
\textbf{Monotone convergence}: If $(f_n)$ is an increasing sequence of nonnegative measurable functions and $f_n \rightarrow f$ almost everywhere, then 
$$\displaystyle\int f = \displaystyle\lim_{n \rightarrow \infty} \displaystyle\int f_n.$$
\end{frame}

\begin{frame}{The general Lebesgue integral (FINALLY)} 
Let $f^+$ denote the positive part of a function $f$, i.e.
$$f^+(x) = \max \{f(x),0\}.$$
Similarly let $f^-$ be the negative part:
$$f^-(x) = \min \{-f(x),0\}.$$
Fact: if $f$ is measurable, then so are $f^+$ and $f^-$ and we have
$$f=f^+ - f^-.$$
A measurable function $f$ over a measurable set $E$ is called (Lebesgue) integrable over $E$ if $f^+$ and $f^-$ are both integrable over $E$, i.e.
$$\displaystyle\int_E f = \displaystyle\int_E f^+ - \displaystyle\int_E f^-.$$
\end{frame}
\end{document}