> [!quote] Introduction > > ![[outer_inner_measure.png]] > > To calculate the area of a specific region without a handy formula, we can cover it with shapes whose area we do know, from the inside and from the outside to approximate the actual area. With finer and finer approximations, we can take the area of the shape to be its outer and inner measure, if they are equal. > [!definition] > > ![[outer_measure.png|400]] > > An **outer measure** on a non-empty [[Set|set]] $X$ is a [[Function|function]] $\mu^*: \pow{X} \to [0, \infty]$ that satisfies: > - $\mu^*(\emptyset) = 0$ > - $F \subseteq E \Rightarrow \mu^*(F) \le \mu^*(E)$ > - $\mu^*\paren{\bigcup_{i \in I}E_i} \le \sum_{i \in I}\mu^*(A_i), |I| \le |\nat|$ > > Outer measures approximate sets from above. See [[Upper and Lower Approximations]]. > [!theorem] Outer Approximation > > ![[outer_approximation.png|500]] > > Let $\ce \subseteq \pow{X}$ be a family of sets with $\emptyset \in \ce$ and $X \in \ce$, and $\rho: \ce \to [0, \infty]$ be a *notion* of measure with $\rho(\emptyset) = 0$. Define for any $A \subseteq X$, > $ > \mu^*(A) = \inf\bracs{ > \sum_{i \in I}\rho(E_i): E_i \in \ce, A \subseteq \bigcup_{i \in I}E_i > } > $ > where $I$ is a countable index set. Then $\mu^*$ is an outer measure. > > *Proof*. Firstly, the set that we take the infimum from is non-empty, as for $E_1 = X, E_k = \emptyset \forall k > 1$ forms an elementary cover for any $A \in \pow{X}$. So when $\mu^*(A) = \infty$, it comes from our notion of measure $\rho$, and not a lack of an elementary cover for $A$. > > ![[outer_approximation_monotone.png|200]] > > Now, let $A \subseteq B$ with $A, B \in \pow{X}$, then any elementary cover of $B$ also covers $A$, giving > $ > \begin{align*} > \bracs{I: E_i \in \ce, A \subseteq \bigcup_{i \in I}E_i} &\supseteq \bracs{I: E_i \in \ce, B \subseteq \bigcup_{i \in I}E_i} \\ > \inf\bracs{\sum_{i \in I}\rho(E_i): A \subseteq \bigcup_{i \in I}E_i} &\le \inf\bracs{\sum_{i \in I}\rho(E_i): B \subseteq \bigcup_{i \in I}E_i} \\ > \mu^*(A) &\le \mu^*(B) > \end{align*} > $ > > ![[outer_approximation_subadditive.png]] > > Finally, let $\bracs{A_j}_1^{\infty} \subseteq \pow{X}$ be a countable sequence of sets, and $\bracs{E_i}_{i \in I_j} \subseteq \ce$ be an elementary cover of $A_j$, where $\bigcup_{i \in I_j}E_i \supseteq A_j$. This provides a way of writing $A$ as an elementary cover from the elementary covers of each $A_j$: > $ > A = \bigcup_{j \in \nat}A_j \subseteq \bigcup_{j \in \nat}\bigcup_{i \in I_j}E_i \quad E_i \in \ce > $ > Let $K = \bigcup_{j \in \nat}I_j$, then $K$ is countable, and > $ > K \in \bracs{I: A \subseteq \bigcup_{i \in I}E_i} > $ > $\bigcup_{k \in K}E_k$ is a countable cover of $A$ with elementary sets, meaning that > $ > \mu^*(A) \le \sum_{k \in K}\rho(E_k) \le \sum_{j \in \nat}\sum_{i \in I_j}\rho(E_i) > $ > for any choice of elementary covers for each $A_j$. > > ![[approach_from_above.png]] > > This allows us to construct a sequence as follows. Firstly, since each $\mu^*(A_j)$ is an infimum, we can approach each one of them as a sequence from above, where for each $x_{j, k}$ ($j$-th sequence, $k$-th entry) > $ > \exists I_{j, k}: x_{j, k} = \sum_{i \in I_{j, k}}\rho(E_i) > $ > and $\lim_{k \to \infty}x_{j, k} = \mu^*(A_j)$. This yields a countable family of sequences $\bracs{\bracs{x_k}_j}$ where each $\bracs{x_k}_j$ approaches $\mu^*(A_j)$ from above. > > Now, since for any sequence in the family, we can take entries that are arbitrarily close to the infimum > $ > \forall \varepsilon_n > 0, j \in \nat, \exists k_n \in \nat: x_{j, k_n} < \mu^*(A_j) + \varepsilon_n2^{-j} > $ > and collect them such that their sum is arbitrarily close to the sum of the infimums. Using this, we can set $\varepsilon_n = 1/n$ and form a new sequence: > $ > \begin{align*} > x_n = \sum_{j \in \nat}x_{j, k_n} &< \sum_{j \in \nat}\mu^*(A_j) + \varepsilon_n2^{-j} \\ > &= \sum_{j \in \nat}\mu^*(A_j) + 2\varepsilon_n > \end{align*} > $ > that approach $\sum_{j \in \nat}\mu^*(A_j)$ from above. Since each $x_n$ represents a countable elementary cover of $A$, each entry pushes $\mu^*(A)$ closer and closer to $\sum_{j \in \nat}\mu^*(A_j)$, making it such that $\mu^*(A) \le \sum_{j \in \nat}\mu^*(A_j)$ at the limit.