\documentclass{book}
\input{../preamble.tex}


\title{MATH4408 Notes}

\begin{document}
\maketitle
\chapter{Sets}
\section{Review of sets}
We always work within a universal set $R$ and deal exclusively with subset of $R$. De Morgan's Laws state that
\[\overline{\bigcup _\alpha A_\alpha} = \bigcap _\alpha \overline{A}_\alpha
\mt{and}
\overline{\bigcap _\alpha A_\alpha} = \bigcup _\alpha \overline{A}_\alpha.
\]


\begin{definition}[Ring]
A set $\mc R$ is called a \textbf{ring} if is is non-empty, and $A\cup B$ and $A\setminus B$ are in $\mc R$ whenever $A, B\in \mc R$.
\end{definition}

\begin{definition}[Algebra]
    A set $\mc R$ is an \textbf{algebra} if it is a ring and contains a universal set $R$.
\end{definition}

\begin{remark}
    $A\Delta B=(A\setminus B)\cup (B\setminus A)$ and $A\cap B = (A\cup B)\setminus(A\Delta B)$ lie in $\mc R$. If $\mc R$ is an algebra then $\bar A \in \mc R$.
\end{remark}

\begin{definition}[$\sigma$-ring]
    A ring $\mc R$ is a $\sigma$-ring if
    \[\bigcup_{n=1}^\infty A_n \in \mc R \quad\text{whenever}\quad (A_n)_{n=1}^\infty\subset \mc R.\]
\end{definition}

\begin{definition}[$\sigma$-algebra]
    A $\sigma$-algebra is a $\sigma$-ring containing the universal set $R$.
\end{definition}


\begin{example}
    \begin{enumerate}
        \item $\{\emptyset, R\}$ is a $\sigma$-algebra
        \item $2^R$ is a $\sigma$-algebra
        \item Open interval in $\bb R$ do not even form a ring.
    \end{enumerate}
\end{example}

\begin{theorem}[Ring generated by a collection]
    If $\ms U$ is a non-empty collection of subsets of $R$, then there exists a unique ring $\mc R(\ms U)$ containing $\ms U$ and contained in every other ring that contains $\ms U$.
\end{theorem}
\begin{proof}
    Let $\Sigma$ be the class of all rings that contain $\ms U$. Since $2^R \in \Sigma$, the class is non-empty. Define 
    \[\mc R(\ms U) = \bigcap_{r\in \Sigma}r.\]
    By construction it is contained within all rings that contain $\ms U$. It is easy to see that it is a ring.
\end{proof}

\begin{definition}[Ring generated by a collection]
    If $\mc R$ is a ring containing $\ms U$ and is contained by all rings which contain $\ms U$, then it is called the \textbf{ring generated by} $\ms U$, denoted $\mc {R}(\ms U)$.
\end{definition}

Intuitively, we obtain $\mc R(\ms U)$ by throwing in the differences and the unions of sets in $\ms U$. Analogous theorems hold for rings, algebra, $\sigma$-rings and $\sigma$-algebras.

\begin{example}
    Let $\ms U = \{[a, b) \mid a < b\}$, $R = \bb R$. The ring generated by $\ms U$ is 
    \[\mc R(\ms U) = \{[a_1, b_1)\cup ...\cup [a_n, b_n)\mid a_i < b_i\}\].
    This is not an algebra. The $\sigma$-algebra generated by $\ms U$ is huge and hard to describe. It contains singletons, $\bb Q$, and $\bb R \setminus \bb Q$, and much more.
\end{example}

\chapter{Measure}
\begin{definition}[Measure]
    Let $\mc R$ be an algebra of sets. A function $\mu: \mc R\to [0, \infty]$ is a \textbf{measure} if:
    \begin{enumerate}
    \item $\mu(\emptyset)=0$
    \item $\sigma$\textbf{-additivity.} If $(A_n)_{n=1}^{\infty}\subset \mc R$ are such that $A_i\cap A_j=\emptyset$ for $i\neq j$, and $\bigcup_{n=1}^\infty A_n \in \mc R$, then 
    \[\mu \left(\bigcup_{n=1}^\infty A_n\right) = \sum_{n=1}^\infty \mu (A_n).\]
    \end{enumerate}
\end{definition}

\begin{remark}
    \begin{enumerate}
        \item $\sigma$-additivity implies additivity.
        \item $\mu (\emptyset) = \mu (\emptyset\cup \emptyset) = 2\mu (\emptyset)$, hence $\mu (\emptyset) = 0$ provided $\mu (R) < \infty$.
        \item If $\mu (R) = 1$ then $\mu$ is a \textbf{probability}.
    \end{enumerate}
\end{remark}

\begin{proposition}[Properties of measure]
    \begin{enumerate}
        \item \textbf{Monotonicity}. If $A\subseteq B$ then $\mu (A)\leq \mu (B)$. Indeed, $$\mu(B) = \mu(A\cap B)+\mu (B\setminus A)\geq \mu (A).$$
        \item If $A\setminus B$ then $\mu (B\setminus A) = \mu (B) - \mu (A)$.
        \item \textbf{Semi-additivity}. If $(A_n)_{n=1}^\infty \subset \mc R$ and $\cup _{n=1}^\infty A_n \in \mc R$ then 
        \[\mu \left(\bigcup_{n=1}^\infty A_n\right)\leq \sum_{n=1}^\infty \mu (A_n).\]
    \end{enumerate}
\end{proposition}

\begin{lemma}
    There exists $(B_n)_{n=1}^\infty \subset \mc R$ such that 
    \begin{enumerate}
        \item $B_i \subset A_i $ for all $i$,
        \item $B_i \cap B_j = \emptyset$ for $i\neq j$,
        \item $\bigcup_{i=1}^\infty = \bigcup_{n=1}^\infty A_i$.
    \end{enumerate}
\end{lemma}
\begin{proof}
    Take $B_1 = A_1$, $B_2 = A_2 \setminus A_1$, $B_3 = A_3 \setminus (A_1\cup A_2)$, $B_n = A_n \setminus\bigcup_{i=1}^\infty A_i$.
\end{proof}

\begin{proof}[Proof (of semi-additivity).]
    By monotonicity, $\mu (B_i) \leq \mu (A_i)$. Therefore, 
    \[\mu \left(\bigcup_{n=1}^\infty A_n \right) = \mu \left(\bigcup_{n=1}^\infty B_n \right) = \sum_{n=1}^\infty \mu (B_n) \leq \sum_{n=1}^\infty \mu (A_n).\]
\end{proof}

\begin{example}
    Let $R = [0, 1]$ and $\mc R$ be the algebra generated by $[a, b)$ with $0\leq a < b \leq 1$. Define $\mu ([a,b) = b-a$ and extend to $\mc R$ by additivity. This is measure. The proof is non-trivial and will be given later.
\end{example}

\begin{remark}
    In probability, $R$ is the space of elementary events, subsets of $R$ are events, and $\mu$ is the probability. 

    The idea of independence does not have an analogue in measure theory.
\end{remark}

\section{Outer Measure}
For now on, assume $\mu (R) < \infty$. We will later generalise our result to $\sigma$-finite measures.

\begin{definition}[Outer measure]
    Let $\mu$ be a measure on an algebra $\mc R\subset 2^R$. The \textbf{outer measure} associated with $\mu$ is the function $\mu^\star: 2^R \to \bb R$ given by  
    \[\mu^\star (A) = \inf \sum_{i=1}^{n (\infty)} \mu(E_i)\]
    where the infimum is taken over all possible coverings of $A$ by sets $E_1, ...,E_n,...$ from $\mc R$. The coverings may or may not be finite.
\end{definition}

\begin{theorem}
    If $A \in \mc R$ then $\mu ^\star (A) = \mu (A)$. 
\end{theorem}
\begin{proof}
    Since $A\in \mc R$ and it covers itself, we have $\mu ^\star (A)\leq \mu (A).$ Choose $\epsilon> 0$. There exists a cover $(E_i)_{i=1}^\infty\subseteq \mc R$ such that 
    \[\mu^\star (A) > \left(\sum_{n=1}^\infty \mu (E_i)\right) - \epsilon.\]
    Now, $A = A \cap \left(\bigcup _{i=1}^\infty E_i\right) = \bigcup_{i=1}^\infty (A\cap E_i)$ so 
    \[\mu (A) \leq \sum_{i=1}^\infty \mu (A\cap E_i) \leq \sum_{i=1}^\infty \mu (E_i) \leq \mu^\star (A)+\epsilon.\]
    As $\epsilon \to 0$, we must have $\mu(A)\leq \mu^\star (A)$.
\end{proof}

\begin{remark}
    Some observations:
    \begin{enumerate}
        \item $\mu ^\star (A)\geq 0$ by definition.
        \item $\mu ^\star(\emptyset) = 0$ since $\emptyset\in \mc R$.
        \item If $A\subseteq B$ then $\mu^\star (A)\leq \mu^\star (B)$ because every cover of $B$ also covers $A$.
    \end{enumerate}
\end{remark}

\begin{counterexample}
Outer measure may fail to be additive. Indeed, take $R = \bb R$, $\mc R = \{\emptyset, \bb R\}$, with $\mu (\bb R) = 1.$ We have $\mu^\star (\{0\}) = 1 = \mu^\star (\{1\})$, but 
\[\mu^\star (\{0, 1\}) = 1\neq \mu^\star (\{0\}\cup \{1\}).\]
\end{counterexample}

\begin{theorem}
    The outer measure $\mu^\star$ is $\sigma$-semi-additive, i.e. given $(A_n)_{n=1}^\infty\subset 2^R$ we have 
    \[\mu^\star \left(\bigcup_{n=1}^\infty A_n \right)\leq \sum_{n=1}^\infty \mu^\star (A_n).\]
\end{theorem}

\begin{proof}
    For each $n$ and $\epsilon> 0$, there exists a cover $(E_{in})\si[i]\subset \mc R$ such that $A_n \subset \bigcup\si E_{in}$ and
    \[\mu^\star(A_n) > \sum\si \mu (E_{in}) - \frac{\epsilon}{2^n}.\]
    Clearly, 
    \[\bigcup\si A_n \subset \bigcup\si E_{in}\]
    and 
    \begin{align*}
        \mu^\star \left(\bigcup \si A_n\right) &< \sum_{i,n=1}^\infty \mu (E_{in}) \\
        &\leq \sum_{n=1}^\infty \mu^\star (A_n)+\frac{\epsilon}{2^n}
    \end{align*}
\end{proof}

\section{Measurable Sets}

\begin{definition}[Measureable]
    A set $A\subset R$ is \textbf{measurable} if 
    \begin{equation}
        \mu^\star = \mu^\star(A\cap E)+\mu^\star(\bar A \cap E) \label{eq:measurable}
    \end{equation}
    for all $E\subset R$.
\end{definition}

Recall that measure $\mu$ is defined on the algebra $\mc R$. Consider the restriction $\mu^\star|_{\tilde {\mc R}}$ where $\tilde{ \mc R}$ is the collection of all measurable sets. Denote this restriction by $\tilde \mu$.

\begin{theorem}
    The class $\tilde {\mc R}$ is a $\sigma$-algebra containing $\mc R$. The restriction $\tilde \mu$ is a measure on $\tilde {\mc R}$
\end{theorem}

\begin{lemma}
    $\tilde {\mc R}$ is an algebra.
\end{lemma}

\begin{proof}
    Given $A_1, A_2\in \tilde{\mc R}$, let us prove that $A_1\cup A_2\in \tilde{\mc R}$. Take $E\subset \mc R$, and we prove \eqref{eq:measurable} for $A = A_1\cup A_2$. Since $A_2$ is measurable, 
    \[\mu^\star (E\cap A_1) = \mu^\star (E\cap A_1\cap A_2)+\mu^\star(E\cap A_2 \cap\bar A_2).\]
    Also, 
    \[\mu^\star (E\cap \bar A_1) = \mu^\star (E\cap \bar A_1 \cap A_2)+\mu^\star (E\cap \bar A_1\cap \bar A_2).\]
    Adding, we find that 
    \begin{align}
        \mu(E) = \mu^\star (E\cap A_1\cap A_2) +\mu^\star (E\cap A_2\cap \bar A_2)\nonumber \\ \qquad \qquad+\mu^\star (E\cap \bar A_1\cap A_2)+ \mu^\star (E\cap \bar A_1\cap \bar A_2). \label{eq:twostar}
    \end{align}

    This holds for all $E$. Now, pick $E' \subset R$ and apply \eqref{eq:twostar} to $E = E'\cap (A_1\cup A_2)$. We get
    \begin{align}
        \mu^\star (E'\cap (A_1\cup A_2)) &= \mu^\star (E'\cap A_1\cap A_2)\nonumber \\ \qquad\qquad &+\mu^\star (E'\cap A_1\cap \bar A_2)+\mu^\star (E'\cap\bar A_1\cap A_2)+\mu^\star(\emptyset). \label{eq:threestar}
    \end{align}
    Combine this with \eqref{eq:twostar} for $E=E'$ and conclude that 
    \begin{align*}
        \mu^\star (E') &= \mu^\star (E'\cap (A_1\cup A_2))+\mu^\star (E'\cap (\bar A_1\cap \bar A_2))\\
        &= \mu^\star (E'\cap (A_1\cup A_2))+\mu^\star (E'\cap (\overline{A_1\cup A_2})).
    \end{align*}
    Thus, $A_1\cup A_2$ is measurable.

    For the difference, observe that \eqref{eq:measurable} does not change if we replace $A$ with its complement. If a collection of sets contains unions and complements then it contains differences. Hence, $\tilde {\mc R}$ is an algebra.
\end{proof}

\begin{remark} \label{rmk:additive}
    Applying \eqref{eq:threestar} to non-intersecting $A_1$ and $A_2$ gives 
    \begin{align*}
        \mu^\star (E'\cap (A_1\cup A_2)) &= \mu^\star (E'\cap A_1\cap \bar A_2) + \mu^\star (E'\cap\bar  A_1\cap A_2) \\
        &= \mu^\star (E'\cap A_1) + \mu^\star (E'\cap A_2).
    \end{align*}
    For $E' = R$, this means that $\mu^\star$ is additive on $\tilde{\mc R}$.
\end{remark}

\begin{lemma}
    $\tilde{\mc R}$ is a $\sigma$-algebra.
\end{lemma}
\begin{proof}
    Take $(A_i)\si[i]\subset \tilde{\mc R}$, we show that $\cup \si [i] \in \tilde{R}$. We need 
    \[\mu^\star (E) = \mu^\star \left(E \cap \left(\cup\si [i] A_i\right)\right) + \mu^\star \left(E \cap \left(\overline{\cup\si [i] A_i}\right)\right).\]
    Without loss of generality, assume that $A_i \cap A_j = \emptyset$ for $i\neq j$. Using $\sigma$-semiadditivity, it is clear that 
    \[\mu^\star (E) \leq \mu^\star \left(E \cap \left(\cup\si [i] A_i\right)\right) + \mu^\star \left(E \cap \left(\overline{\cup\si [i] A_i}\right)\right).\]
    Using Lemma 1, $\cup_{i=1}^nA_i \in \tilde{\mc R}$ for all $n \in \bb N$, so 
    \[\mu^\star (E) = \mu^\star \left(E \cap \left(\cup_{i=1}^n A_i\right)\right) + \mu^\star \left(E \cap \left(\overline{\cup_{i=1}^n A_i}\right)\right).\]
    Remark \ref{rmk:additive} establishes countable additivity, so we have that 
    \[\mu^\star (E\cap (\cup_{i=1}^n A_i )) = \sum_{i=1}^n \mu^\star(E \cap A_i).\]
    Also, $E\cap \overline{\cup _{i=1}^n A_i}\subset E\cap \overline{\cup\si[i]A_i}$, so 
    \[\mu^\star (E) \geq \sum_{i=1}^n \mu^\star (E\cap A_i )+\mu^\star (E\cap \overline{\cup \si [i]A_n})\]
    Finally, take $n\to \infty$ and note that by $\sigma$-semiadditivity,
    \[\sum\si [i] \mu^\star(E\cap A_i )\geq \mu^\star (E\cap (\cup \si [i]A_i)).\]
\end{proof}

\begin{lemma}
    $\mu^\star$ is $\sigma$-additive on $\tilde{\mc R}$. Therefore, $\tilde\mu = \mu^\star |_{\tilde{\mc R}}$ is a measure on $\tilde{\mc R}$.
\end{lemma}
\begin{proof}
    Take $(A_i)\si \subset \tilde{\mc R}$ such that $A_i \cap A_j = \emptyset$ for $i \neq j$. We have 
    \[\mu^\star \left(\bigcup \si[i]A_i\right) \geq \mu^\star \left(\bigcup_{i=1}^n A_i\right) = \sum_{i=1}^n \mu^\star (A_i).\]
    Now let $n\to \infty$, so 
    $\mu^\star \left(\bigcup \si[i]A_i\right)  m\geq \sum_{i=1}^\infty \mu^\star (A_i)$
    and the opposite inequality holds by $\sigma$-semiadditivity.
\end{proof}

\begin{lemma}
    The algebra $\mc R$ is contained in the $\sigma$-algebra $\tilde{\mc R}$.
\end{lemma}

\begin{proof}
    We need to show that 
    \[\mu^\star (E) = \mu^\star (E\cap A)+\mu^\star (E\cap \bar A)\]
    when $E\subset R$ and $A\in \mc R$.
\end{proof}
The inequality in the $\leq$ direction follows from semi-additivity. Let $(E_i)\si[i]\subset \mc R$ such that $E\subset \cup \si[i]E_i$ and 
\[\sum\si [i] \mu(E_i)< \mu^\star (E) +\epsilon.\]
Clearly, 
\begin{align*}
    E_i &= E_i \cap (A\cup \bar A)  \\ 
    &= (E_i \cap A)\cup (E_i\cap \bar A)
\end{align*}
so
\begin{align*}
    \mu^\star (E) +\epsilon > \sum\si[i] \mu(E_i\cap A)+\mu(E_i\cap \bar A).
\end{align*}
Observe that 
\[E\cap A \subset \left(\bigcup\si[i]E_i \right)\cap A = \bigcup\si [i] E_i\cap A\]
and 
\[E\cap \bar A \subset \bigcup \si [i] E_i \cap \bar A.\]
It follows that 
\[\mu^\star (E\cap A)\leq \sum\si [i]\mu^\star(E_i\cap A)\]
and 
\[\mu^\star (E\cap\bar A)\leq \sum\si [i]\mu^\star(E_i\cap\bar A)\]
so we obtain 
\[\mu^\star (E)+\epsilon> \mu^\star (E\cap A)+\mu^\star (E\cap \bar A).\]
Letting $\epsilon\to 0$, we obtain the desired result.

\begin{corollary}
    Every measure on $\mc R$ can be extended to a $\sigma$-algebra containing $\mc R$. One such extension is $\tilde \mu = \mu^\star_{\tilde{\mc R}}$. We can always extend $\mu$ to the $\sigma$-algebra $\mc R_\sigma$ generated by $\mc R$.

    An extension to $\mc R_\sigma$ is $\tilde \mu |_{\mc R_\sigma}$, called the \textbf{minimal extension} of $\mu$.
\end{corollary}

\subsection{Properties of measure}
\begin{definition}[Complete]
    A measure $\mu$ on an algebra $\mc R$ is called \textbf{complete} if every subset of a set $A\in \mc R$ such that $\mu (A)=0$ lies in $\mc R$.
\end{definition}

\begin{theorem}
    If $\mu$ is a measure on an algebra $\mc R$ with outer measure $\mu^\star$, and $\mu^\star(A) = 0$ for some $A\subset R$ then $A$ is measurable.
\end{theorem}
\begin{proof}
    We need 
    \[\mu^\star (E)\geq \mu^\star (E\cap A)+\mu^\star (E\cap \bar A)\]
    for $E\subset R$. We know 
    \[\mu^\star (E\cap A)\leq \mu^\star (A) = 0\]
    and $\mu^\star (E\cap \bar A)\leq \mu^\star (E)$ by monotonicity.
\end{proof}

\begin{corollary}
    The extension $\tilde \mu = \mu^\star |_{\tilde {\mc R}}$ is complete.
\end{corollary}
\begin{proof}
    If $A\in \tilde{\mc R}$ and $\tilde\mu (A) = 0$ then $A$ is measurable, hence $A\in \tilde{\mc R}$.
\end{proof}
\end{document}