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\title{MATH4408 Notes}

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\maketitle

\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{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{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}

\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{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{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}

\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}

\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{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}

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