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\newtheorem{defn}[thm]{Definition}
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\title{Caratheodory's extension theorem}
\author{DBW}

\begin{document}
\maketitle

These notes are meant as introductory notes on Caratheodory's extension theorem. The presentation is not completely my own work; the presentation heavily relies on the presentation of Noel Vaillant on {\tt http://www.probability.net/WEBcaratheodory.pdf}. To make the line of arguments as clear as possible, the starting point is the notion of a ring on a topological space and not the notion of a semi-ring.

\section{Elementary definitions and properties}

We fix a topological space $\Omega$. The power set of $\Omega$ is denoted $\mathcal{P}(\Omega)$ and consists of all subsets of $\Omega$.

\begin{defn}
A ring on on $\Omega$ is a subset $\R$ of $\PO$, such that

$(i)$ $\emptyset \in \R$

$(ii)$ $A,B\in \R ~~ \Rightarrow ~~ A\cup B \in \R$

$(iii)$ $A,B\in \R ~~ \Rightarrow ~~ A\setminus B\in \R$
\end{defn}

\begin{defn}
A $\sigma$-algebra on $\Omega$ is a subset $\Sigma$ of $\PO$ such that

$(i)$ $\emptyset \in \Sigma$

$(ii)$ $(A_n)_{n\in \mathbb{N}} \in \Sigma ~~ \Rightarrow ~~ \cup_{n} A_n\in \Sigma$

$(iii)$ $A\in \Sigma ~~\Rightarrow ~~ A^c\in \Sigma$
\end{defn}

Since $A\cap B = A\setminus (A\setminus B)$ it follows that any ring on $\Omega$ is closed under finite intersections; hence any ring is also a semi-ring. Since $\cap_n A_n = (\cup_n A^c)^c$ it follows that any $\sigma$-algebra is closed under arbitrary intersections. And from $A\setminus B = A\cap B^c$ we deduce that any $\sigma$-algebra is also a ring.

If $(R_i)_{i\in I}$ is a set of rings on $\Omega$ then it is clear that $\cap_I R_i$ is also a ring on $\Omega$. Let $S$ be any subset of $\PO$, then we call the intersection of all rings on $\Omega$ containing $S$ the ring generated by $S$.

\begin{defn}
Let $\A$ be a subset of $\PO$. A measure on $\A$ is a map $\mu:\A\to [0,+\inf ]$ such that

$(i)$ $\mu(\emptyset)=0$

$(ii)$ If $A_n\in \A$ are disjoint and $A = \uplus_n A_n \in \A ~~\Rightarrow ~~\mu(A) =\sum_n \mu(A_n)$.
\end{defn}

If $\A$ is a $\sigma$-algebra, we don't need to assume that in addition $\uplus_n A_n\in \A$. By taking all but finitely many $A_n$ to be the empty set one sees that $\mu( A_1\uplus\cdots \uplus A_N) = \mu(A_1)+\ldots+\mu(A_n)$. If $A\subset B$ then $A \uplus (B\setminus A) = B$ and hence $\mu(B) = \mu (A) +\mu(B\setminus A)\geq \mu(A)$.


\begin{defn}
We call an outer measure on $\Omega$ a map $\lambda:\PO\to [0,+\infty]$ with

$(i)$ $\lambda (\emptyset) = 0$

$(ii)$ $A\subset B ~~ \Rightarrow ~~\lambda(A)\leq \lambda(B)$

$(iii)$ $(A_n)_{n\in \mathbb{N}}\in \PO\,,~ \lambda(\cup_n A_n) \leq \sum_n \lambda(A_n)$
\end{defn}

By taking all but finitely many $A_n$ to be the empty set one sees that an outer measure is subadditive; $\lambda(A\cup B) \leq \lambda(A)+\lambda(B)$.


\section{The interplay between $\sigma$-algebras and (outer) measures}

Let $\lambda$ be an outer measure on $\Omega$. We define $\Sigma_\lambda$ to be the set of all subsets $A\subset \Omega$ such that for any $X\subset \Omega$ we have
$$
\lambda(X) = \lambda(X\cap A)+\lambda(X\cap A^c)\,.
$$
In other words, $\Sigma_\lambda$ consists of all subsets $A\subset \Omega$ that cut $\Omega$ in two in a good way. Clearly, $\Omega\in \Sigma_\lambda$ and by the very form of the definition of $\Sigma_\lambda$, we have $A\in \Sigma_\lambda \Leftrightarrow A^c\in \Sigma_\lambda$. We can now present the following proposition, whose proof is a bit tedious, but which contains loads of information on the exact interplay.

\begin{prop}
Let $\lambda$ be an outer measure on $\Omega$ and let $\Sigma_\lambda$ be as defined above. Then $\Sigma_\lambda$ is a $\sigma$-algebra on $\Omega$.
\end{prop}

\begin{proof}
After the preliminary remarks preceding the proposition, it only remains to show that $\Sigma_\lambda$ is closed under countable unions. We will first prove that $\Sigma_\lambda$ is closed under finite intersections and unions.

Let $A,B\in \Sigma_\lambda$ and let $X$ be any subset of $\Omega$. We have $X\cap A^c = X\cap (A\cap B)^c\cap A^c$ since $(A\cap B)^c \supset A^c$. On the other hand we have $(A\cap B)^c = A^c\cup B^c$ and hence $X\cap (A\cap B)^c\cap A = (X\cap A\cap B^c)\cup (X\cap A\cap A^c) = X\cap A\cap B^c $. Therefore we have $\lambda(X\cap (A\cap B)^c) = \lambda(X\cap (A\cap B)^c \cap A) + \lambda(X\cap (A\cap B)^c \cap A^c = \lambda(X\cap A^c)+\lambda(X\cap A\cap B^c)$. Now adding $\lambda(X\cap A\cap B)$ and using that $\lambda(X\cap A) = \lambda(X\cap A\cap B)+\lambda(X\cap A\cap B^c)$ one obtains $\lambda(X\cap A\cap B)+\lambda(X\cap (A\cap B)^c) = \lambda(X)$. Hence $A\cap B\in \Sigma_\lambda$.

Since $A\cup B = (A^c\cap B^c)^c$ and $A\setminus B = A\cap B^c$ we see that $\Sigma_\lambda$ is closed under finite unions and the set-theoretic difference. Thus $\Sigma_\lambda$ is a ring on $\Omega$.

If $A,B\in \Sigma_\lambda$ are disjoint and $X\subset \Omega$ then $\lambda(X\cap (A\uplus B)) = \lambda(X) - \lambda(X\cap A^c\cap B^c) = \lambda(X) - \lambda(X\cap A^c) + \lambda(X \cap A^c\cap B) = \lambda(X\cap A)+\lambda(X\cap B)$ as $A^c\cap B = B$. Using induction we obtain $\lambda(X\cap \biguplus_{n=1}^{N}A_n) = \sum_{n=1}^{N} \lambda(X\cap A_n)$ whenever $A_n$ are in $\Sigma_\lambda$ and pairwise disjoint.

Now we fix a sequence $A_n$ in $\Sigma_\lambda$ which are pairwise disjoint and we denote the union $\cup_n A_n$ by $A$. Furthermore, we fix an arbitrary $X\in \Omega$ and an arbitrary large integer $N$.

Since $X\cap A^c \subset X\cap (\biguplus_{n=1}^{N}A_n)^c$ and $\Sigma_\lambda$ is closed under finite unions we have
$\lambda(X\cap A^c) + \lambda(X\cap (\biguplus_{n=1}^{N}A_n)) \leq \lambda(X\cap (\biguplus_{n=1}^{N}A_n)^c)+\sum_n\lambda(X\cap A_n) = \lambda(X)$. But $N$ is arbitrary in this equation and we can safely let it go to $\infty$ and obtain
\begin{equation}\label{eqn1}
\lambda(X\cap A^c) + \sum_n\lambda(X\cap A_n) \leq \lambda(X)\,.
\end{equation}

On the other hand we have $\lambda(X) \leq \lambda(X\cap A^c) + \lambda(X\cap A)$, which again by the definition of an outer measure is less or equal $\lambda(X\cap A^c) + \sum_n \lambda(X\cap A_n)$. Hence using (\ref{eqn1}) we obtain
\[
\lambda(X) \leq \lambda(X\cap A^c) + \lambda(X\cap A) \leq \lambda(X\cap A^c) + \sum_n \lambda(X\cap A_n) \leq \lambda(X)\,.
\]
It follows that we must equality. From this we conclude that $\Sigma_\lambda$ is indeed closed under countable unions and, by taking $X=A$, that $\lambda(A) = \sum_n \lambda(A_n)$. Therefore the restriction of $\lambda$ to $\Sigma_\lambda$ is a measure on $\Sigma_\lambda$.

\end{proof}

We will call $\Sigma_\lambda$ the $\sigma$-algebra related to $\lambda$.

Now we come to a critical step; we want to associate an outer measure $\lambda_\mu$ to a given measure $\lambda$ on some ring $\R$. Of course, we want the restriction of the outer measure $\lambda_\mu$ to the ring to coincide with the measure $\mu$. 

Let $\R$ be a ring on $\Omega$ and let $\mu$ be a measure on $\R$. If $X\subset \Omega$ is any subset we can cover $X$ with sets from $\R$ to approximate $X$ inside $\R$ - we call an $\R$-cover of $X$ a countable subset $(A_n)_n$ of $\R$ with $X\subset \cup_n A_n$. This leads to the following definition; for any $X\subset \Omega$ we $\lambda_\mu(X)$ to be the infimum of all sums $\sum_n \mu(A_n)$ where $(A_n)_{n\in \mathbb{N}}$ is any countable cover of $X$ with $A_n$ in $\R$. We need to check that this is an outer measure.

\begin{prop}
The map $\lambda_\mu:\PO\to [0,+\infty]$ defined in the above paragraph defines an outer measure on $\Omega$.
\end{prop}

\begin{proof}
Since $\emptyset\in\R$ we have $\lambda_\mu(\emptyset) =0$. If $X\subset Y$ are two subsets of $\Omega$, then any cover of $Y$ with sets from $\R$ also covers $X$ and hence $\lambda_\mu(X)\leq \lambda_\mu(Y)$.

Now let $X_n$ be any sequence of subsets of subsets. By the definition of the infimum we can find for each $\epsilon>0$ and for each $n$ an $\R$-cover $(A_{n,m})_m$ of $X_n$ such that $\sum_m \mu(A_{n,m}) < \lambda_\mu(X)+\tfrac{\epsilon}{2^n}$. The sets $A_{n,m}$ form a countable cover of $X = \cup_n X_n$; we can for example set $B_1=A_{1,1}$, $B_2 = A_{2,1}$, $B_3 = A_{1,2}$, $B_4 = A_{3,1}$ and so on, similar to Cantor's proof of the countability of $\mathbb{Q}$. But then $\lambda_\mu(X)  = \lambda_\mu(\cup_n X_n) \leq \sum_{n,m} \mu(A_{n,m}) < \sum_n (\lambda_\mu(X_n) + \tfrac{\epsilon}{2^n}) = \sum_{n} \lambda_\mu(X_{n})+\epsilon$. But $\epsilon$ was arbitrary and hence $\lambda_\mu(\cup_n X_n)\leq \sum_n \lambda_\mu(X_n)$.

\end{proof}


\begin{prop}\label{prop123}
The restriction of $\lambda_\mu$ to $\R$ is $\mu$.
\end{prop}

\begin{proof}

For any $A\in \R$ the set $A$ itself forms a cover and hence $\lambda_\mu(A)\leq \mu(A)$. 

On the other hand, let $(A_n)$ be an $\R$-cover of $A$. We define $B_1 = A_1\cap A$ and $B_{n+1} = (A_n\cap A)\setminus \cup_{k\geq n}(A_k \cap A)$ for $n\geq 1$. Then clearly $B_n\in \R$, the $B_n$ are disjoint, $\uplus_n B_n = A$ and $\mu(B_n)\leq \mu(A_n)$. Since $\mu$ is a measure on $\R$ we have $\mu(A) = \sum_n \mu(B_n)$ which is less than or equal to $\sum_n\mu(A_n)$. Since this holds for any $\R$-cover of $A$ we have $\mu(A) \leq \lambda_\mu(A)$. Therefore equality holds and the proposition is proved.

\end{proof}

We will call $\lambda_\mu$ the outer measure associated to $\mu$.


So we now have two constructions; given a ring and a measure on it we can construct an outer measure. Given an outer measure we can construct a $\sigma$-algebra such that the restriction of the outer measure to the $\sigma$-algebra is a measure on the $\sigma$-algebra. So it seems feasible that we can construct a measure on a $\sigma$-algebra starting from a ring with a measure on it. That this really works and that all things work out nicely is the content of Caratheodory's theorem.

\section{Caratheodory's theorem: Statement and Proof}


\begin{lem}\label{lem1}
Let $\R$ be a ring on $\Omega$ and let $\mu$ be a measure on $\R$. Let $\lambda$ be the outer measure associated to $\mu$. Let $\Sigma$ be the $\sigma$-algebra related to $\lambda$. Then $\R \in \Sigma$.
\end{lem}

\begin{proof}
Let $A$ be an element of $\R$ and let $X$ be any subset of $\Omega$.
Since $\lambda$ is an outer measure on $\Omega$ we have $\lambda(X) = \lambda( (X\cap A)\cup (X\cap A^c)) \leq \lambda(X\cap A)+\lambda(X\cap A^c)$.

Now let $(A_n)_{n\in \mathbb{N}}$ be any $\R$-cover of $X$. Then the $A_n\cap A$ form an $\R$-cover of $X\cap A$ and the $A_n\cap A^c$ form an $\R$-cover of $X\cap A^c$. Hence we have that $\lambda(X\cap A)+\lambda(X\cap A^c) \leq \sum_n \mu(A_n\cap A) +\sum_n (A_n\cap A^c) = \sum_n \mu(A_n)$, where the last step follows from the fact that $\mu$ is a measure and hence $\mu(C\uplus D) = \mu(C)+\mu(D)$. Since the inequality holds for any $\R$-cover of $X$ we need $\lambda(X\cap A)+\lambda(X\cap A^c) \leq \lambda(X)$. We thus need equality; for any $X\subset \Omega$ we have $\lambda(X) = \lambda(X\cap A)+\lambda(X\cap A^c)$, or in other words $A\in \Sigma$ and since $A$ was an arbitrary element of $\R$ the lemma is proved.
\end{proof}


\begin{rem}
Any ring generates a $\sigma$-algebra; one simply enlarges the ring with countable unions. Or, the $\sigma$-algeba generated by the ring $\R$ is the intersection of all $\sigma$-algebras that contain $\R$. Therefore the above lemma shows that the $\sigma$-algebra generated by $\R$ is contained in $\Sigma$.
\end{rem}

Now we come to Caratheodory's theorem:

\begin{thm}
Let $\R$ be a ring on $\Omega$ and let $\mu$ be a measure on $\R$. Then there exists a measure $\mu'$ on the $\sigma$-algebra generated by $\R$ such that the restriction of $\mu'$ to $\R$ coincides with $\mu$.
\end{thm}

\begin{proof}

Let $\lambda$ be the outer measure on $\Omega$ associated to $\mu$. Let $\Sigma$ be the $\sigma$-algebra associated to $\lambda$. Then by lemma \ref{lem1} the $\sigma$-algebra generated by $\R$ is contained in $\Sigma$. Hence $\lambda$ restricts to a measure on the $\sigma$-algebra generated by $\R$. By proposition \ref{prop123} this restriction of $\lambda$ to $\R$ coincides with $\mu$.

\end{proof}

\begin{exa}
Let $\Omega$ be the real line. Then the open intervals generate a $\sigma$-algebra $\Sigma$. For any open interval $(a,b)$ with $a<b$ we can put $\mu((a,b)) = b-a$. Then there exists a measure $\mu'$ on $\Sigma$ such that $\mu'((a,b)) = b-a$. Indeed, for countable unions of disjoint intervals we can define $\mu(\cup_n (a_n,b_n)) = \sum_n (b_n-a_n)$. Hence $\mu$ does give rise to a measure on the ring generated by all intervals.
\end{exa}


\end{document} 