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\begin{document}
\title{Notes}
\date{January 23, 2015}
\maketitle
$E$ is measurable if $\dots$

irrational numbers have a measure because rationals have measure 0 and the complement of a measurable set is 0

\section*{today}
explain why any open interval is measurable

if $E_1$ and $E_2$ are disjoint and measurable then $m*(A\cap(E_1\cup E_2)=m*(A\cap E_1)+m*(A\cap E_2)$ for any $A$

$E_1,E_2,E_3,\dots,E_n$ disjoint and measurable then $m*(A\cap(\bigcup\limits_{i=1}^nE_i))=\sum\limits_{i=1}^n{m*(A\cap E_i)}$

\subsection*{theorem}
if $\{E_i\}_{i=1}^\infty$ are measurable, countable then
\begin{enumerate}
\item
$\bigcup\limits_{i=1}^\infty$ measurable
\item
$\bigcap\limits_{i=1}^\infty$ measurable
\item
$m*(\bigcup\limits_{i=1}^\infty E_i)\le\sum\limits_{i=1}^\infty{m*(E_i)}$
\item
if $E_i\cap E_j=\emptyset$ for all $i,j$

$m*(\bigcup\limits_{i=1}^\infty E_i)=\sum\limits_{i=1}^\infty{m*(E_i)}$
\end{enumerate}
\subsubsection*{proof}
{\bfseries case 1}

$E_i\cap E_j=\emptyset, i\ne j$

$F_n=\bigcup\limits_{i=1}^n E_i$

we know that $F_n$ is measurable so $m*(A\cap F_n)=\sum\limits_{i=1}^n{m*(A\cap E_i)}$

$m*(A)=m*(A\cap F_n)+m*(A\cap F_n^C)$


for all $n$
\begin{align*}
  m*(A)&=m*(A\cap F_n)+m*(A\cap F_n^C)\\
  &\ge m*(A\cap F_n)+m*(A\cap(\bigcup_{i=1}^\infty E_i)^C)\\
\intertext{$m*(A\cap F_n^C)$ contains $m*(A\cap(\bigcup_{i=1}^\infty E_i)^C)$}
  &= \sum\limits_{i=1}^n{m*(A\cap E_i)}+m*(A\cap(\bigcup_{i=1}^\infty E_i)^C)\\
  m*(A)&\ge \sum\limits_{i=1}^n{m*(A\cap E_i)}+m*(A\cap(\bigcup_{i=1}^\infty E_i)^C)\\
  &\ge m*(A\cap \bigcup_{i=1}^\infty E_i)+m*(A\cap(\bigcup_{i=1}^\infty E_i)^C)
\end{align*}

{\bfseries case 2} not disjoint

$E_1=\Omega_1$ and $E_2\setminus\Omega_1=\Omega_2$ and $E_3\setminus(\Omega_1\cup\Omega_2)=\Omega_3$

$\{E_i\}_{i=1}^\infty\to \{\Omega_i\}_{i=1}^\infty$ and $\Omega_i\cap\Omega_j=\emptyset\forall i\ne j$
\begin{enumerate}
\item
$\bigcup\limits_{i=1}^\infty E_i=\bigcup\limits_{i=1}^\infty \Omega_i$
\item
$\Omega_i$ is measurable ($A\setminus B=A\cap B^C$)
\item
use case one to get 
$\bigcup\limits_{i=1}^\infty \Omega_i$ is measurable and imply that
$\bigcup\limits_{i=1}^\infty E_i$ is measurable
\end{enumerate}


\begin{enumerate}
\item
union of countable measurable sets is measurable
\item
$\{E_i\}_{i=1}^\infty$ with $E_i$ measurable

$\bigcap_{i=1}^\infty E_i=\left(\bigcup\limits_{i=1}^\infty(E_i)^C\right)^C$
\item
we already knew that $m*(\bigcup\limits_{i=1}^\infty E_i\le \sum\limits_{i=1}^\infty{m*(E_i)}$
\item
$\{E_i\}_{i=1}^\infty$ disjoint and measurable

$m*(\bigcup_{i=1}\infty E_i$

\begin{enumerate}
\item
$m*(\bigcup\limits_{i=1}^\infty E_i\le \sum\limits_{i=1}^\infty{m*(E_i)}$

$m*(\bigcup\limits_{i=1}^n E_i= \sum\limits_{i=1}^n{m*(E_i)}$

$m*(\bigcup_{i=1}^\infty E_i)\ge m*(\bigcup\limits_{i=1}^n E_i= \sum\limits_{i=1}^n{m*(E_i)}$

and then

$m*(\bigcup_{i=1}^\infty E_i)\ge = \sum\limits_{i=1}^\infty{m*(E_i)}$

\end{enumerate}
\end{enumerate}

\subsection*{theorem}
any open set is measurable
\subsubsection*{proof}
based on homework we know that $(a,\infty)$ is measurable

\begin{enumerate}
\item
need to show that $[b,\infty)$ is measurable for any $b$

$[b,\infty)=\bigcap_{n=1}^\infty(b-\frac{1}{n},\infty)$
\item
$(-\infty,c)$ is measurable because it's the complement of $[c,\infty)$
\item
$(a,d)$ is measurable for any $a<d$

$(a,d)=(a,\infty)\cap(-\infty,d)$
\item
any open interval $\mathbb{O}$ is measurable

$\mathbb{O}=\bigcup_{i=1}^\infty(a_i,b_i)$
Lindlöfs theorem

or any closed set is measurable
\end{enumerate}

cantor set is the intersection of a countable number of closed sets

\subsection*{proposition}
if $E_i\supseteq E_{i+1}$ in $\{E_i\}_{i=1}^\infty$ and $m*(E_1)$ is finite then $m*(\bigcap_{i=1}^\infty E_i)=\lim_{n\to\infty}m*(E_n)$ 

$\{n,n+1)\}_{n=1}^\infty$ will not work, why?

$\bigcap_{i=i}^\infty E_i\subseteq E_n$ for all $n$. $m*(\bigcap_{i=1}^\infty E_i\le m*(E_n)$ for all $n$
\end{document}
