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%  Lecture 2
%  Source: handwritten notes (Mathpix-converted) + Kashlak STAT 571
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\section[Lecture 2 -- Constructing \texorpdfstring{$\sigma$}{sigma}-Fields and Measures]{Lecture 2 \textemdash{} Constructing \texorpdfstring{$\sigma$}{sigma}-Fields and Measures}
\label{sec:lec02}

\emph{Motivating question.} If we declare the measure of a half-open interval $(a,b]$ to be $b-a$, what other subsets of $\R$ can we then measure? The collection $\Acal=\{(a,b]:a<b\}$ is not a $\sigma$-field (e.g.\ $(a,b]\cup(c,d]\notin\Acal$), so we need a systematic procedure for enlarging $\Acal$ and extending the length function to the resulting $\sigma$-field. This lecture builds the existence half of that procedure: \emph{Carath\'eodory's extension theorem}.

\subsection{Semirings, rings, and fields}

We work below the level of a $\sigma$-field, with progressively richer set systems on which a candidate measure can live.

\begin{definition}{Semiring}{semiring}
A collection $\Acal$ of subsets of $\Omega$ is a \emph{semiring} if
\begin{itemize}
  \item $\emptyset\in\Acal$,
  \item for all $A,B\in\Acal$, $A\cap B\in\Acal$, and
  \item $B\setminus A=\bigcup_{i=1}^{n}C_i$ for some $C_i\in\Acal$.
\end{itemize}
The set difference need not lie in $\Acal$, but it must be expressible as a finite union of members of $\Acal$.
\end{definition}

\begin{example}[Half-open intervals]
The collection $\Acal=\{(a,b]:-\infty<a<b<\infty\}\cup\{\emptyset\}$ on $\Omega=\R$ is a semiring: intersections of half-open intervals are half-open intervals, and a set difference $(a,b]\setminus(c,d]$ splits into at most two such intervals.
\end{example}

\begin{definition}{Ring}{ring}
A collection $\Acal$ of subsets of $\Omega$ is a \emph{ring} if $\emptyset\in\Acal$ and for all $A,B\in\Acal$, both
\[
B\setminus A\in\Acal \qquad\text{and}\qquad A\cup B\in\Acal.
\]
\end{definition}

\begin{definition}{Field}{field}
A ring $\Acal$ is a \emph{field} if $\Omega\in\Acal$.
\end{definition}

\begin{remark}
A field that is closed under \emph{countable} unions is a $\sigma$-field. Fields and $\sigma$-fields are sometimes called \emph{algebras} and \emph{$\sigma$-algebras}, respectively. Finite disjoint unions of half-open intervals form a ring, but not a $\sigma$-field.
\end{remark}

\subsection{Set functions and pre-measures}

\begin{definition}{Set function}{setfunction}
A \emph{set function} is a map $\mu:\Acal\to\R^{+}$ (not necessarily a measure). For $A,B\in\Acal$, we say
\begin{itemize}
  \item $\mu$ is \emph{increasing} if $A\subset B\Rightarrow\mu(A)\leq\mu(B)$;
  \item $\mu$ is \emph{additive} if $\mu(A\cup B)=\mu(A)+\mu(B)$ whenever $A,B$ are disjoint;
  \item $\mu$ is \emph{countably additive} if for any pairwise disjoint $\{A_i\}_{i=1}^{\infty}$ with $\bigcup_i A_i\in\Acal$,
  \[
    \mu\!\left(\bigcup_{i=1}^{\infty}A_i\right)=\sum_{i=1}^{\infty}\mu(A_i);
  \]
  \item $\mu$ is \emph{countably subadditive} if for any $\{A_i\}_{i=1}^{\infty}\subseteq\Acal$ with $\bigcup_i A_i\in\Acal$ (not necessarily disjoint),
  \[
    \mu\!\left(\bigcup_{i=1}^{\infty}A_i\right)\leq\sum_{i=1}^{\infty}\mu(A_i).
  \]
\end{itemize}
We call $\mu$ a \emph{pre-measure} when it is countably additive and $\mu(\emptyset)=0$.
\end{definition}

\subsection{Outer measure and \texorpdfstring{$\mu^{*}$}{mu*}-measurability}

A pre-measure on a ring extends to \emph{any} subset of $\Omega$ via approximation from above by countable covers.

\begin{definition}{Outer measure}{outermeasure}
Let $\mu$ be a pre-measure on a ring $\Acal$. The \emph{outer measure} induced by $\mu$ is
\[
  \mu^{*}(E)\;=\;\inf\sum_{i}\mu(A_i),\qquad E\subseteq\Omega,
\]
where the infimum runs over all finite or countable collections $\{A_i\}\subseteq\Acal$ with $E\subseteq\bigcup_i A_i$.
\end{definition}

The outer measure is defined on \emph{all} of $\Pcal(\Omega)$, but it is generally only countably \emph{sub}additive there. To recover countable additivity we restrict to a distinguished subclass.

\begin{definition}{$\mu^{*}$-measurable set}{caratheodorymeasurable}
A set $B\subseteq\Omega$ is \emph{$\mu^{*}$-measurable} (in the sense of Carath\'eodory) if
\[
  \mu^{*}(E\cap B)+\mu^{*}(E\cap B^{c})\;=\;\mu^{*}(E)\qquad\text{for every }E\subseteq\Omega.
\]
We denote by $\mathcal{M}$ the collection of all $\mu^{*}$-measurable sets.
\end{definition}

\begin{remark}
The condition splits an arbitrary ``test set'' $E$ cleanly across $B$ and $B^{c}$. Subadditivity of $\mu^{*}$ already gives ``$\geq$'', so the content of the definition is the reverse inequality.
\end{remark}

\subsection{Carath\'eodory's extension theorem}

\begin{theorem}{Carath\'eodory Extension}{caratheodory}
Let $\Acal$ be a ring on $\Omega$ and let $\mu$ be a pre-measure on $\Acal$. Then $\mu$ extends to a measure on $\sigma(\Acal)$. In fact, the outer measure $\mu^{*}$ is countably additive on the $\sigma$-field $\mathcal{M}$, and
\[
  \Acal\;\subseteq\;\sigma(\Acal)\;\subseteq\;\mathcal{M}\;\subseteq\;\Pcal(\Omega),
\]
with $\mu^{*}\big|_{\Acal}=\mu$.
\end{theorem}

\begin{remark}
The proof proceeds in five steps: (1) verify $\mu^{*}(\emptyset)=0$, monotonicity, and countable subadditivity of $\mu^{*}$; (2) show $\mu$ and $\mu^{*}$ agree on $\Acal$; (3) show $\Acal\subseteq\mathcal{M}$; (4) show $\mathcal{M}$ is a $\sigma$-field; (5) conclude $\mu^{*}$ is a measure on $\sigma(\Acal)\subseteq\mathcal{M}$. The ``correct'' extension is therefore the outer measure restricted to $\sigma(\Acal)$.
\end{remark}

\begin{example}[{Lebesgue measure on $(0,1]$}]
Let $\Acal$ be the collection of finite disjoint unions of half-open intervals $(a,b]\subseteq(0,1]$ together with $\emptyset$, and define $\lambda(\bigcup_{j=1}^{n}I_j)=\sum_{j=1}^{n}\lambda(I_j)$ where $\lambda(I_j)$ is the length of $I_j$. Then $\Acal$ is a ring and $\lambda$ is a pre-measure, so Carath\'eodory's theorem extends $\lambda$ to a measure on $\sigma(\Acal)$. The chain
\[
  \Acal\;\subset\;\sigma(\Acal)\;\subseteq\;\mathcal{M}\;\subset\;\Pcal((0,1])
\]
identifies $\sigma(\Acal)$ with the Borel $\sigma$-field $\Bcal$ on $(0,1]$, while $\mathcal{M}$ is the strictly larger Lebesgue $\sigma$-field. Subsets of $(0,1]$ that fail to be Lebesgue measurable do exist, but their construction is non-trivial.
\end{example}

\begin{remark}
Existence is now settled. The next lecture takes up \emph{uniqueness}: if two measures agree on a generating class $\Acal$, must they agree on $\sigma(\Acal)$? This is handled by Dynkin's $\pi$-$\lambda$ theorem.
\end{remark}