Definition. An outer measure on a set \(X\neq\varnothing\) is a function \(\mu^{*}:\mathcal{P}(X) \to [0,\infty]\) that satisfies
\(\mu^{*}(\varnothing)=0\),
\(\mu^{*}(A) \leq \mu^{*}(B)\) if \(A \subseteq B\),
\(\mu^{*}\left(\bigcup_{n \in \mathbb{N}}A_n\right) \leq \sum_{n \in \mathbb{N}}\mu^{*}(A_n)\).
In other words, an outer measure \(\mu^*:\mathcal P(X) \to [0,\infty]\) is a monotone and countably subadditive set function such that \(\mu^*(\varnothing)=0\).
The most common way to obtain outer measures is to start with a family \(\mathcal E\) of “elementary sets” on which a notion of measure is defined (such as rectangles in the plane) and then to approximate arbitrary sets “from the outside” by countable unions of members of \(\mathcal E\).
Proposition. Let \(\mathcal{E} \subseteq \mathcal{P}(X)\) and \(\rho:\mathcal{E} \to [0,\infty]\) be a set function such that \(\varnothing,X \in \mathcal{E}\) and \(\rho(\varnothing)=0\). For any \(A \subseteq X\), define \[\mu^{*}(A)=\inf\bigg\{\sum_{j=1}^{\infty}\rho(E_j):(E_j)_{j\in\mathbb N} \subseteq \mathcal{E} \text{ and } A \subseteq \bigcup_{j=1}^{\infty}E_j\bigg\}.\] Then \(\mu^{*}\) is an outer measure.
Proof. For any \(A \subseteq X\) there exists \((E_j)_{j=1}^{\infty}\) such that \(A \subseteq \bigcup_{j=1}^{\infty}E_j\) (take \(E_j=X\) for all \(j\in\mathbb N\)) so the definition of \(\mu^{*}\) makes sense.
Obviously \(\mu^{*}(\varnothing)=0\) (take \(E_j=\varnothing\) for all \(j\in\mathbb N\)).
We also have \(\mu^{*}(A) \leq \mu^{*}(B)\) for any \(A \subseteq B\). Indeed, if \((E_j)_{j\in\mathbb N} \subseteq \mathcal{E}\) and \(B \subseteq \bigcup_{j=1}^{\infty}E_j\) then \(A \subseteq \bigcup_{j=1}^{\infty}E_j\), so a covering \((E_j)_{j\in\mathbb N} \subseteq \mathcal{E}\) for \(B\) is also a covering for \(A\). Hence \(\mu^{*}(A) \leq \mu^{*}(B)\).
To prove the countable subadditivity, let \((A_j)_{j=1}^{\infty}\subseteq \mathcal P(X)\) and \(\varepsilon>0\). For each \(A_j\) there is \((E_{j, k})_{k=1}^{\infty}\) such that \[A_j \subseteq \bigcup_{k=1}^{\infty}E_{j, k} \quad \text{ and }\quad {\color{blue}\sum_{k=1}^{\infty}\rho(E_{j, k}) \leq \mu^{*}(A_j)+\varepsilon{2^{-j}}}.\]
But then if \(A=\bigcup_{j=1}^{\infty}A_j\) we have \(A\subseteq \bigcup_{j=1}^{\infty}\bigcup_{k=1}^{\infty}E_{j, k}\) and \[\mu^*(A)\le \sum_{j=1}^{\infty}\sum_{k=1}^{\infty}\rho(E_{j, k}). \qquad {\color{red}\text{Why? Justify this!}}\]
Consequently \[\mu^*(A)\le \sum_{j=1}^{\infty}\sum_{k=1}^{\infty}\rho(E_{j, k})\leq \sum_{j=1}^{\infty}{\color{blue}\big(\mu^{*}(A_j)+\varepsilon{2^{-j}}\big)}\le \sum_{j=1}^{\infty}\mu^{*}(A_j)+\varepsilon.\]
Since \(\varepsilon\) is arbitrary, we are done. $$\tag*{$\blacksquare$}$$
Carathéodory’s condition. If \(\mu^{*}\) is an outer measure on \(X\), a set \(A \subseteq X\) is called \(\mu^{*}\)-measurable if \[\mu^{*}(E)=\mu^{*}(E \cap A)+\mu^{*}(E \cap A^c) \quad \text{ for all } \quad E \subseteq X.\]
Remark.
By subadditivity of \(\mu^*\) we clearly have \[\mu^{*}(E) \leq \mu^{*}(E \cap A)+\mu^{*}(E \cap A^c).\]
To prove that \(A\subseteq X\) is \(\mu^{*}\)-measurable, it suffices to prove the reverse inequality \[\mu^{*}(E) \geq \mu^{*}(E \cap A)+\mu^{*}(E \cap A^c).\] for all \(E\subseteq X\) such that \(\mu^{*}(E)<\infty\).
Suppose that \(A\subseteq X=(a, b)\times(c, d)\subset\mathbb R^2\) and let \(\mathcal C\) be a family of open rectangles in \(X\) and \(\rho\) be the area function. Define \[\mu^{*}(A)=\inf\bigg\{\sum_{j=1}^{\infty}\rho(C_j):(C_j)_{j\in\mathbb N} \subseteq \mathcal{C} \text{ and } A \subseteq \bigcup_{j=1}^{\infty}C_j\bigg\},\] which is an outer measure that approximates the area of a bounded set \(A\) from the outside by using rectangles.
Since \(\mu^*(X)<\infty\) we can define an inner measure by setting \[\mu_*(A)=\mu^*(X)-\mu^*(A^c),\] which approximates the area of a bounded set \(A\) from the inside.
If \(\mu^{*}(A)=\mu_{*}(A)\) it is reasonable to think that their common value is the area of \(A\), which leads to the measurability condition \[\mu^{*}(A)=\mu_{*}(A)\quad {\color{blue} \Longleftrightarrow \quad \mu^*(X)=\mu^*(X\cap A)+\mu^*(X\cap A^c).}\]
Theorem. Let \(\mu^{*}\) be an outer measure on \(X\), and let \[\mathcal M(\mu^*) =\big\{A\subseteq X: \mu^{*}(E)=\mu^{*}(E \cap A)+\mu^{*}(E \cap A^c) \quad \text{ for all } \quad E \subseteq X\big\}.\] be the collection of \(\mu^{*}\)-measurable sets. Then
\(\mathcal M(\mu^*)\) is a \(\sigma\)-algebra, (Carathéodory’s \(\sigma\)-algebra).
\((X, \mathcal M(\mu^*), \mu^{*})\) is a complete measure space.
Proof. Step 1. We first show that \(\mathcal M(\mu^{*})\) is an algebra.
Observe that \(\mathcal M(\mu^{*})\) is closed under complements since the definition of \(\mu^{*}\)-measurability of \(A\) is symmetric in \(A\) and \(A^c\).
Next, if \(A,B \in \mathcal M(\mu^{*})\), then we will show that \[A \cup B \in \mathcal M(\mu^{*}).\]
Let \(A,B \in \mathcal M(\mu^*)\) be such that \(A \cap B=\varnothing\). Then \({\color{blue}B \subseteq A^c}\) and \[\begin{aligned} \mu^{*}(A \cup B)={\color{red}\mu^{*}((A \cup B) \cap A)}+{\color{blue}\mu^{*}((A \cup B) \cap A^c)} ={\color{red}\mu^{*}(A)}+{\color{blue}\mu^{*}(B)}. \end{aligned}\]
Step 3. We now show that \(\mathcal M(\mu^{*})\) is a \(\sigma\)-algebra.
Since \(\mathcal M(\mu^{*})\) is an algebra, it suffices to show that \(\mathcal M(\mu^{*})\) is closed under countable disjoint unions. Let \((A_j)_{j=1}^{\infty} \subseteq \mathcal M(\mu^{*})\) be such that \(A_i \cap A_j=\varnothing\) for \(i \neq j\), and define \[B_0=\varnothing, \ \ B_n=\bigcup_{j=1}^{n}A_j, \ \ B=\bigcup_{j=1}^{\infty}A_j.\] Then for any \(E \subseteq X\) we have \[\begin{aligned} \mu^{*}(E \cap B_n)&=\mu^{*}(E \cap B_n \cap A_n)+\mu^{*}(E \cap B_n \cap A_n^c)\\ &=\mu^{*}(E \cap A_n)+{\color{red}\mu^{*}(E \cap B_{n-1})}. \end{aligned}\]
Step 4. We now show that \(\mu^*\) is countably additive on \(\mathcal M(\mu^{*})\).
From the previous calculation we have \[\mu^{*}(E) = \sum_{j=0}^{\infty}\mu^{*}(E \cap A_j)+\mu^{*}(E \cap B^c).\] Taking \({\color{red}E=B=\bigcup_{j=1}^{\infty}A_j}\) we obtain \[\mu^{*}(B)= \sum_{j=1}^{\infty}\mu^{*}( A_j)\] as claimed.
Step 5. We finally show that \(\mu^{*}\) is a complete measure on \(\mathcal M(\mu^{*})\).Recall that a measure whose domain includes all subsets of null sets is called complete.
Take any \(A \subseteq X\) such that \(\mu^{*}(A)=0\). We will show \(A \in \mathcal M(\mu^{*})\). For any \(E \subseteq X\) we have \[\begin{aligned} \mu^{*}(E) \leq \mu^{*}(E \cap A)+\mu^{*}(E \cap A^c) \leq \mu^{*}(E \cap A^c) \leq \mu^{*}(E), \end{aligned}\] since \(\mu^{*}(E \cap A)=0\). Thus \[\mu^{*}(E) = \mu^{*}(E \cap A)+\mu^{*}(E \cap A^c)\] and \(A \in \mathcal M(\mu^{*})\). The proof of Carathéodory’s theorem is finished.$$\tag*{$\blacksquare$}$$
\(\mu_0(\varnothing)=0\),
if \((A_n)_{n\in \mathbb N}\) is a sequence of disjoint sets in \(\mathcal A\) such that \(\bigcup_{n=1}^{\infty}A_n \in \mathcal A\), then \[\mu_0\bigg(\bigcup_{n=1}^{\infty}A_n\bigg)=\sum_{n=1}^{\infty}\mu_0(A_n).\]
If \(\mu_0\) is a premeasure on \(\mathcal A \subseteq \mathcal{P}(X)\), it induces an outer measure on \(X\): \[\mu^{*}(A)=\inf\bigg\{\sum_{j=1}^{\infty}\mu_0(E_j): (E_j)_{j\in\mathbb N} \subseteq \mathcal A \text{ and } A \subseteq \bigcup_{j=1}^{\infty}E_j\bigg\}, \qquad A\subseteq \mathcal P(X).\]
Proposition. Let \(X\neq\varnothing\) be a set and let \(\mu_0\) be a premeasure on an algebra \(\mathcal A\subseteq \mathcal P(X)\) and let \(\mu^{*}\) be an outer measure defined as before \[\mu^{*}(A)=\inf\bigg\{\sum_{j=1}^{\infty}\mu_0(E_j): (E_j)_{j\in\mathbb N} \subseteq \mathcal A \text{ and } A \subseteq \bigcup_{j=1}^{\infty}E_j\bigg\}, \qquad A\subseteq \mathcal P(X).\] Then one has that
\(\mu_0(E)=\mu^{*}(E)\) for all \(E \in \mathcal A\),
\(\mathcal A\subseteq \mathcal M(\mu^*)\). In other words, every set in \(\mathcal A\) is \(\mu^{*}\)-measurable.
Proof of (a). Suppose \(E \in \mathcal A\), we will show that \(\mu_0(E)=\mu^{*}(E)\).
Taking \(E_1=E\) and \(E_2=E_3=\ldots=\varnothing\) we see that \(E \subseteq \bigcup_{j=1}^{\infty}E_j\) and consequently \(\mu^{*}(E) \leq \mu_0(E)\).
We now reverse this inequality and show that \(\mu_0(E) \leq \mu^{*}(E)\).
If \(E \subseteq \bigcup_{j=1}^{\infty}A_j\) with \(A_j \in \mathcal A\), then we let \[B_n=E \cap \bigg(A_n \setminus \bigcup_{j=1}^{n-1}A_j\bigg)\in\mathcal A.\] Then \(B_n \cap B_m =\ \varnothing\) if \(n \neq m\) and \(E\subseteq \bigcup_{j=1}^{\infty}B_j=\bigcup_{j=1}^{\infty}A_j\).. Now \[\begin{aligned} \mu_0(E)\le\sum_{j=1}^{\infty}\mu_0(B_j) \leq \sum_{j=1}^{\infty}\mu_0(A_j). \end{aligned}\] It follows that \(\mu_0(E) \leq \mu^{*}(E)\) as claimed.
Proof of (b). We now prove that \(\mathcal A\subseteq \mathcal M(\mu^*)\).
Let \(C \in \mathcal A\) and \(E \subseteq X\) it suffices to show that \[\mu^{*}(E) \geq \mu^{*}(E \cap C)+\mu^{*}(E \cap C^c).\] Let \(\varepsilon>0\), then there is a sequence \((B_n)_{n=1}^{\infty}\subseteq \mathcal A\) such that \[E \subseteq \bigcup_{n =1}^{\infty}B_n \quad \text{ and } \quad \sum_{n=1}^{\infty}\mu_0(B_n) \leq \mu^{*}(E)+\varepsilon.\] \(\mu_0\) is additive on \(A\), thus \[\begin{aligned} \mu^{*}(E)+\varepsilon \geq \sum_{j=1}^{\infty}\mu_0(B_j)= \sum_{j=1}^{\infty}(\mu_0(B_j \cap C)+\mu_0(B_j \cap C^c)). \end{aligned}\]
Since \(\mu_0 \geq \mu^{*}\) and \(B=\bigcup_{j=1}^{\infty}B_j\) we obtain \[\begin{aligned} \mu^{*}(E)+\varepsilon &\geq\sum_{j=1}^{\infty}(\mu_0(B_j \cap C)+\mu_0(B_j \cap C^c))\\& \geq \sum_{j=1}^{\infty}(\mu^{*}(B_j \cap C)+\mu^{*}(B_j \cap C^c))\\&\geq (\mu^{*}(B \cap C)+\mu^{*}(B \cap C^c)) \\&\geq (\mu^{*}(E \cap C)+\mu^{*}(E \cap C^c)) \end{aligned}\] by subadditivity of \(\mu^*\). Letting \(\varepsilon \to 0\) we conclude \[\begin{aligned} \mu^{*}(E) \geq \mu^{*}(E \cap C)+\mu^{*}(E \cap C^c), \end{aligned}\] thus \(C\) is \(\mu^{*}\)-measurable and \(\mathcal A\subseteq \mathcal M(\mu^*)\) as desired.$$\tag*{$\blacksquare$}$$
Theorem. Let \(X\neq\varnothing\) be a set, let \(\mathcal A \subseteq \mathcal{P}(X)\) be an algebra, \(\mu_0\) a premeasure on \(\mathcal A\), and \(\mathcal M= \sigma(\mathcal A)\). There exists a measure \(\mu\) on \(\mathcal M\) whose restriction to \(\mathcal A\) is \(\mu_0\) and \(\mu(E)=\mu^{*}(E)\) for all \(E \in \mathcal M\), where \(\mu^{*}\) is given by \[\mu^{*}(A)=\inf\bigg\{\sum_{j=1}^{\infty}\mu_0(E_j): (E_j)_{j\in\mathbb N} \subseteq \mathcal A \text{ and } A \subseteq \bigcup_{j=1}^{\infty}E_j\bigg\}, \qquad A\subseteq \mathcal P(X).\]
If \(\nu\) is another measure on \(\mathcal M\) that extends \(\mu_0\), then \(\nu(E) \leq \mu(E)\) for all \(E \in \mathcal M\), with equality when \(\mu(E)<\infty\).
If \(\mu_0\) is \(\sigma\)-finite, then \(\mu\) is the unique extension of \(\mu_0\) to a measure on \(\mathcal M\).
Proof. For the first part of the theorem note that:
\((X, \mathcal M(\mu^*), \mu^*)\) is a complete measure space, which follows from the Carathéodory theorem.
\(\mathcal A\subseteq \mathcal M(\mu^*)\) by the previous proposition and consequently we obtain that \(\mathcal M=\sigma(\mathcal A)\subseteq \mathcal M(\mu^*)\).
Thus it suffices to define \(\mu\) to be the restriction of \(\mu^*\) to \(\mathcal M=\sigma(\mathcal A)\). Then we see that \(\mu(E)=\mu_0(E)\) for all \(E\in \mathcal A\).
Proof of (i). Assume that \(\nu\) is another measure on \(\mathcal M=\sigma(\mathcal A)\) such that \(\nu(E)=\mu_0(E)\) for all \(E \in \mathcal A\). We show that
\(\nu(E) \leq \mu(E)\) for all \(E \in \sigma(\mathcal A)\);
\(\nu(E)=\mu(E)\) if \(E \in \sigma(\mathcal A)\) and \(\mu(E)<\infty\).
For (a): If \(E \in \mathcal M\) and \(E \subseteq \bigcup_{j \in \mathbb{N}}A_j\), where \((A_j)_{j\in\mathbb N} \in \mathcal A\) then \[\nu(E) \leq \sum_{j=1}^{\infty}\nu(A_j)=\sum_{j=1}^{\infty}\mu_0(A_j)\] and we conclude that \({\color{red}\nu(E) \leq \mu(E)}\).
For (b): If we set \(A=\bigcup_{j=1}^{\infty}A_j\), then we have \[\color{blue} \nu(A)=\lim_{n \to \infty}\nu\big(\bigcup_{j=1}^{n}A_j\big) =\lim_{n \to \infty}\mu_0\big(\bigcup_{j=1}^{n}A_j\big)=\lim_{n \to \infty}\mu\big(\bigcup_{j=1}^{n}A_j\big)=\mu(A).\]
If \(\mu(E)<\infty\), let \(\varepsilon>0\) and choose \(A_j\)’s from \(\mathcal A\) such that \[\mu(A) \leq \mu(E)+\varepsilon,\] hence \(\mu(A \setminus E)<\varepsilon\) and \[\mu(E) \leq {\color{blue}\mu(A) =\nu(A)}=\nu(E)+{\color{red}\nu(A \setminus E)} \leq \nu(E)+{\color{red}\mu(A \setminus E)} \leq \nu(E)+\varepsilon.\]
Letting \(\varepsilon \to 0\) we see \(\mu(E) \leq \nu(E)\) if \(E \in \mathcal M=\sigma(\mathcal A)\) and \(\mu(E)<\infty\) and we conclude in this case that \[\nu(E)=\mu(E).\]Proof of (ii). Finally suppose that \(X=\bigcup_{j=1}^{\infty}A_j\), with \(\mu_0(A_j)<\infty\) and \(A_i \cap A_j=\varnothing\) if \(i \neq j\). Then for any \(E \in \mathcal M\) we have \[\mu(E)=\sum_{j=1}^{\infty}\mu(E \cap A_j)=\sum_{j=1}^{\infty}\nu(E \cap A_j)=\nu(E).\] This competes the proof of Carathéodory’s extension theorem. $$\tag*{$\blacksquare$}$$