Muhammad Haris Rao
Definition: Given a set $X$, an outer measure on $X$ is a function $\mu : \mathcal{P}(X) \longrightarrow [0, \infty]$ such that
Outer measures can be easily constructed using the following method:
Theorem: Let $X$ be a set and $\mathcal{A} \subseteq \mathcal{P}(X)$ a collection of subsets which covers all of $X$ and contains $\emptyset$, and $\mu_0 : \mathcal{A} \longrightarrow [0, \infty]$ a set function with $\mu_0(\emptyset) = 0$. Then the function $\mu : \mathcal{P}(X) \longrightarrow [0, \infty]$ defined by \begin{align*} \mu \left( A \right) &= \text{inf} \left\{ \sum_{n \ge 1} \mu_0 \left( A_n \right) \mid A_n \in \mathcal{A} \text{ and } A \subseteq \bigcup_{n \ge 1} A_n \right\} \end{align*} is an outer measure on $X$.
Proof. It is clear that $\mu$ is well-defined since $\mathcal{A}$ covers all of $X$, and it is clear that $\mu(\emptyset) = 0$. Monotonicity follows from the fact that if $A \subseteq B$, then \begin{align*} \left\{ \sum_{n \ge 1} C_n \mid C_n \in \mathcal{A} \text{ and } A \subseteq \bigcup_{n \ge 1} C_n \right\} \supseteq \left\{ \sum_{n \ge 1} C_n \mid C_n \in \mathcal{A} \text{ and } B \subseteq \bigcup_{n \ge 1} C_n \right\} \end{align*} Taking infimums on both sides to evaluate $\mu(A), \mu(B)$ introduces the desired inequality. It only remains to prove countable subadditivity. So let $\{ A_n \}_{n \ge 1} \subseteq \mathcal{P}(X)$, and let $\varepsilon > 0$ be arbitrary. By standard properties of the infimum, there exists for each $n \ge 1$ a sequence $\{ B_k^{(n)} \}_{k \ge 1} \subseteq \mathcal{A}$ with $A_n \subseteq \bigcup_{k \ge 1} B_k^{(n)}$ such that \begin{align*} \sum_{k \ge 1} \mu_0 \left( B_k^{(n)} \right) < \mu \left( A_n \right) + \frac{\varepsilon}{2^n} \end{align*} Then the countable collection $\{ B_k^{(n)} \}_{n, k \ge 1}$ covers $A$ and \begin{align*} \sum_{n, k \ge 1} \mu_0 \left( B_k^{(n)} \right) &= \sum_{n \ge 1} \sum_{k \ge 1} \mu_0 \left( B_k^{(n)} \right) \le \sum_{n \ge 1} \left( \mu\left( A_n \right) + \frac{\varepsilon}{2^n} \right) = \sum_{n \ge 1} \mu \left( A_n \right) + \varepsilon \end{align*} The re-indexing of the infinite sums are valid as all summands are non-negative. The above inequality is true for arbitrary $\varepsilon > 0$ so it follows that \begin{align*} \sum_{n, k \ge 1} \mu_0 \left( B_k^{(n)} \right) \le \sum_{n \ge 1} \mu \left( A_n \right) \end{align*} Thus, we have \begin{align*} \mu\left( \bigcup_{n \ge 1} A_n \right) &= \mu\left( \bigcup_{n, k \ge 1} B_k^{(n)} \right) = \text{inf} \left\{ \sum_{\ell \ge 1} C_\ell \mid C_\ell \in \mathcal{A} \text{ and } \bigcup_{n, k \ge 1} B_k^{(n)} \subseteq \bigcup_{\ell \ge 1} C_\ell \right\} \le \sum_{n, k \ge 1} B_k^{(n)} \le \sum_{n \ge 1} \mu \left( A_n \right) \end{align*} which proves countable subadditivity.
In the statement above, $\mu$ is called the outer measure induced by the set function $\mu_0$. The construction above works for any collection of subsets which contains the empty set and covers $X$ and any non-negative set function on this collection of subsets which vanishes on the empty set. Usually, what we really want is a way to construct measures, and the outer measure constructed above can fail very badly in being a measure.
For example, let $\mathcal{A} \subseteq \mathcal{P}(\mathbb{R})$ be $\mathcal{A} = \{\emptyset, \{ 0 \}, \mathbb{R} \}$, and define $\mu_0 : \mathcal{A} \longrightarrow [0, \infty]$ by $\mu_0(\emptyset) = \mu_0(\{0\}) = 0$ and $\mu_0\left( \mathbb{R} \right) = 1$. From this information, there is an induced outer measure $\mu : \mathcal{P}(\mathbb{R}) \longrightarrow [0, \infty]$ defined on $A \subseteq \mathbb{R}$ by $$ \mu(A) = \begin{cases} 0 &\text{ , if $A \subseteq \{ 0 \}$} \\ 1 &\text{ , otherwise} \end{cases} $$ Very obviously, this is not a measure on the measurable space $\left( \mathbb{R}, \mathcal{P}(\mathbb{R}) \right)$. For example, it fails continuity: Even though $(-1/n, 1/n)$ converge to $\{ 0 \}$ as $n \to \infty$, $$ \lim_{n \to \infty} \mu((-1/n, 1/n)) \ne \mu(\{ 0 \}) $$ However, restricting $\mu$ to the trivial $\sigma$-algebra $\{ \emptyset, \mathbb{R} \} \subseteq \mathcal{P}(\mathbb{R})$ does give a measure space.
The theorem below furthers this idea by showing how to restrict the outer measure in a way that we obtain a valid measure.
Theorem(Caratheodery Extension Theorem): Let $\mathcal{A} \subseteq \mathcal{P}(X)$ be a collection of subsets covering $X$ and such that $\emptyset \in \mathcal{A}$, and $\mu_0 : \mathcal{A} \longrightarrow [0, \infty]$ a set function on $\mathcal{A}$ with $\mu_0(\emptyset) = 0$. Let $\mu : \mathcal{P}(X) \longrightarrow [0, \infty]$ the outer measure induced by $\mu_0$. The collection of subsets \begin{align*} \Sigma &= \left\{ A \in \mathcal{P}(X) \mid \mu(B) = \mu(B \cap A) + \mu(B \cap A^c) \text{ for all $B \in \mathcal{P}(X)$} \right\} \end{align*} is a $\sigma$-algebra, and $\mu$ is a measure when restricted to $\Sigma$. Moreover, if $\mathcal{A}$ is an algebra of sets and $\mu_0$ a premeasure, then $\mathcal{A} \subseteq \Sigma$ and $\mu(A) = \mu_0(A)$ for all $A \in \mathcal{A}$.
Proof. The subsets in $\Sigma$ are called the Caratheodery measurable subsets. It is obvious that $\emptyset \in \Sigma$, and the fact that $\Sigma$ is closed under complementation is also clear. We need to prove that it is closed under countable unions. First, we prove this for finite unions. Suppose $A, B \in \Sigma$. Then, for an arbitrary $C \in \mathcal{P}(X)$ we have \begin{align*} \mu(C) &= \mu(C \cap A) + \mu(C \cap A^c) = \mu\left(C \cap A \cap B \right) + \mu\left(C \cap A \cap B^c \right) + \mu\left(C \cap A^c \cap B \right) + \mu\left(C \cap A^c \cap B^c \right) \end{align*} where the second equality follows from applying the condition on $B$ being Caratheodery measurable to $\mu(C \cap A)$ and $\mu(C \cap A^c)$. By subadditivity of $\mu$, \begin{align*} \mu \left( C \cap (A \cap B) \right) + \mu \left( C \cap (A^c \cap B) \right) + \mu \left( C \cap (A \cap B^c) \right) \ge \mu \left(C \cap (A \cup B) \right) \end{align*} Hence, we have \begin{align*} \mu(C) &\ge \mu \left(C \cap (A \cup B) \right) + \mu\left(C \cap A^c \cap B^c \right) = \mu \left(C \cap (A \cup B) \right) + \mu\left( C \cap (A \cup B)^c \right) \end{align*} The reverse inequality follows from subadditivity so then, \begin{align*} \mu\left( C \right) &= \mu \left(C \cap (A \cup B) \right) + \mu\left( C \cap (A \cup B)^c \right) \end{align*} This means that $A \cup B \in \Sigma$ as well so $\Sigma$ is closed under finite unions. Thus, it is an algebra of sets. Further, if $A, B \in \Sigma$ are disjoint then we have for any $C \in \mathcal{P}(X)$ that \begin{align*} \mu \left( C \cap (A \cup B) \right) &= \mu \left( \left(C \cap (A \cup B)\right) \cap A \right) + \mu \left( \left( C \cap (A \cup B) \right) \cap A^c \right) = \mu\left( C \cap A \right) + \mu \left( C \cap B \right) \end{align*} Induction on this shows that if $\{ A_n \}_{n \ge 1}$ is a sequence of disjoint elements from $\Sigma$ then for any $C \in \mathcal{P}(X)$ and $n \ge 1$, \begin{align*} \mu \left( C \cap \bigcup_{k = 1}^n A_k \right) = \sum_{k = 1}^n \mu \left( C \cap A_k \right) \end{align*} Thus, using the fact that $\bigcup_{k = 1}^n A_k \in \Sigma$ we obtain \begin{align*} \mu\left( C \right) &= \mu\left( C \cap \bigcup_{k = 1}^n A_k \right) + \mu\left( C \cap \left( \bigcup_{k = 1}^n A_k \right)^c \right) \\ &= \sum_{k = 1}^n \mu\left( C \cap A_k \right) + \mu\left( C \cap \left( \bigcup_{k = 1}^n A_k \right)^c \right) \\ &\ge \sum_{k = 1}^n \mu\left( C \cap A_k \right) + \mu\left( C \cap \left( \bigcup_{k \ge 1} A_k \right)^c \right) \end{align*} Taking $n \to \infty$ and then applying subadditivity, \begin{align*} \mu\left( C \right) &\ge \sum_{k \ge 1} \mu\left( C \cap A_k \right) + \mu\left( C \cap \left( \bigcup_{k \ge 1} A_k \right)^c \right) \ge \mu \left( C \cap \bigcup_{k \ge 1} A_k \right) + \mu\left( C \cap \left( \bigcup_{k \ge 1} A_k \right)^c \right) \end{align*} This proves that countable disjoint unions of elements from $\Sigma$ are in $\Sigma$ as well. Since $\Sigma$ was closed under arbitrary finite unions and complements, it is possible to 'disjointify' any countable union into a countable union of disjoint sets from $\Sigma$. Specifically, if $\{ A_n \}_{n \ge 1} \subseteq \Sigma$ is any not necessarily disjoint sequence of sets from $\Sigma$, we can write their union as a disjoint union of elements from the algebra of sets $\Sigma$ as \begin{align*} \bigcup_{n \ge 1} A_n &= \bigcup_{n \ge 1} \left( A_n \bigg\backslash \bigcup_{k < n} A_k \right) \end{align*} Thus, any countable union of sets from $\Sigma$ remains in $\Sigma$ which proves that $\Sigma$ is a $\sigma$-algebra. Now to show that $\mu$ is a measure when restricted to $\Sigma$. Let $\{ A_n \}_{n \ge 1}$ be a sequence of disjoint elements from $\Sigma$. Then, applying monotonicity and the fact that the $\{ A_n \}_{n \ge 1}$ are disjoint \begin{align*} \mu \left( \bigcup_{k \ge 1} A_n \right) \ge \mu \left( \bigcup_{k = 1}^n A_k \right) = \sum_{k = 1}^n \mu\left( A_k \right) \end{align*} Taking $n \to \infty$ and then using the fact that $\mu$ is sub-additive yields the equality \begin{align*} \mu \left( \bigcup_{k \ge 1} A_n \right) &= \sum_{k \ge 1} \mu \left( A_k \right) \end{align*} Now to show that in the case where $\mathcal{A}$ is an algebra and $\mu_0$ a premeasure, $\mathcal{A} \subseteq \Sigma$ and the extension $\mu$ restricts to $\mu_0$ on $\mathcal{A}$. So fix $A \in \mathcal{A}$, and let $C \in \mathcal{P}(X)$. If $\varepsilon > 0$ then there is $\{ A_n \}_{n \ge 1} \subseteq \mathcal{A}$ covering $C$ such that \begin{align*} \mu \left( C \right) + \varepsilon > \sum_{n \ge 1} \mu_0\left( A_n \right) \end{align*} Then, we will have \begin{align*} \mu\left( A \cap C \right) + \mu \left( A^c \cap C \right) &\le \sum_{n \ge 1} \mu_0 \left( A \cap A_n \right) + \sum_{n \ge 1} \mu_0\left( A^c \cap A_n \right) \\ &= \sum_{n \ge 1} \mu_0 \left( \left( A \cap A_n \right) \cup \left( A^c \cap A_n \right) \right) \\ &= \sum_{n \ge 1} \mu_0 \left( A_n \right) \\ &< \mu \left( C \right) + \varepsilon \end{align*} where the first equality follows from $\mu_0$ being a pre-measure on $\mathcal{A}$. It follows from this that \begin{align*} \mu\left( A \cap C \right) + \mu \left( A \cap C \right) &\le \mu \left( C \right) \end{align*} and hence $A \in \Sigma$. This proves $\mathcal{A} \subseteq \Sigma$. Finally, to prove that $\mu$ restricts to $\mu_0$ on $\mathcal{A}$. Since any $A \in \mathcal{A}$ covers itself, $\mu(A) \le \mu_0(A)$. Now let $\varepsilon > 0$ and $\{ A_n \}_{n \ge 1} \subseteq \mathcal{A}$ covering $A$ be such that \begin{align*} \mu(A) > \sum_{n \ge 1} \mu_0(A_n) - \varepsilon \end{align*} Now make them disjoint by setting \begin{align*} A_n^* = A_n \bigg\backslash \bigcup_{k < n} A_k \end{align*} so that $A$ is covered by $\{ A_n^* \}_{n \ge 1}$. The we have the equality \begin{align*} A = \bigcup_{n \ge 1} (A_n^* \cap A) \end{align*} Since $\mathcal{A}$ is an algebra, we have also $\{ A_n^* \cap A \}_{n \ge 1} \subseteq \mathcal{A}$. Since $\mu_0$ is a premeasure, \begin{align*} \mu(A) > \sum_{n \ge 1} \mu_0(A_n) - \varepsilon \ge \sum_{n \ge 1} \mu_0(A_n^*) - \varepsilon \ge \sum_{n \ge 1} \mu_0(A_n^* \cap A) - \varepsilon = \mu_0\left( \bigcup_{n \ge 1} (A_n^* \cap A )\right) - \varepsilon = \mu_0(A) - \varepsilon \end{align*} This holds for any $\varepsilon > 0$ so we can conclude that This proves $\mu_0(A) \ge \mu(A)$ for all $A \in \mathcal{A}$. Combined with the other inequality, it follows that we have equality $\mu_0(A) = \mu(A)$ so $\mu_0 = \mu \mid_{\mathcal{A}}$. This completes the proofs for all the claims.