> [!definition] > > Let $X$ be a [[Set|set]] and $\alg \subseteq \pow{X}$ be an [[Algebra|algebra]], then a [[Function|function]] $\mu_0: \alg \to [0, \infty]$ is a **premeasure** if > - $\mu_0(\emptyset) = 0$ > - If $\bracs{A_i}_{1}^{\infty}$ is a [[Sequence|sequence]] of disjoint sets in $\alg$ where $\bigcup_{i \in \nat}A_i \in \alg$, then $\mu_0\paren{\bigcup_{i \in \nat}A_i} = \sum_{i \in \nat}\mu_0(A_i)$. > > In other words, $\mu_0$ is a finitely-additive [[Measure Space|measure]] on $\alg$, with the occasional countable additivity when the union happens to also be in $\alg$. > [!theorem] > > Let $X$ be a set, $\alg \subseteq \pow{X}$ be an algebra, and $\mu_0$ be a premeasure on $\alg$, then $\mu_0$ induces an [[Outer Measure|outer measure]] > $ > \mu^*(A) = \inf\bracs{ > \sum_{i \in I}\mu_0(E_i): E_i \in \alg, \bigcup_{i \in I}E_i \supseteq A > } > $ > on $X$. > > *Proof*. Since $\mu_0$ defines a notion of measure on a family of sets $\ce = \alg$, the "outer approximation" theorem applies. $\mu^*$ acts as an "least upper bound" on the mass of $A$, taking the infimum of all possible upper bounds on its mass. > [!theorem] > > Let $X$ be a set, $\alg \subseteq \pow{X}$ be an algebra, $\mu_0$ be a premeasure on $\alg$, and $\mu^*$ be the induced outer measure, then > - $\mu^*|\alg = \mu_0$. > - Every set in $\alg$ is [[Carathéodory's Theorem|outer measurable]]. > > *Proof*. The first part essentially means that the elementary sets $\alg$ and the notion of measure $\mu_0$ is one that "makes sense". Since $\alg$ is an algebra and $\mu_0$ is a premeasure, there are no ways to chop any elementary sets into smaller elementary sets and somehow "lose" mass in the process. > > To begin, each elementary set $E \in \alg$ can simply be approximated by itself from the outside, giving $\mu^*(E) \le \mu_0(E)$. > > ![[extension_lemma_cut.png]] > > Let $\bracs{E_i}_{1}^{\infty}$ be an elementary cover of $E \in \alg$. Since each part of the cover is in $\alg$, we can remove any overlap. With $E$ also being in $\alg$, we can remove any "over-covering". So the only parts relevant to the outer measure is the tightest part of it, or simply its way of cutting $E$ into disjoint pieces. > > Now, let $\bracs{E_i}_1^{\infty}$ be a sequence of sets such that $\bigcup_{i \in \nat}E_i \supseteq E$. Take $F_1 = E_1$ and $F_k = E \cap \paren{E_{k} \setminus \bigcup_{i = 1}^{k - 1}E_i}$, then $\bracs{F_i}_1^\infty$ is a *disjoint* sequence of sets in $\alg$ where $\bigcup_{i \in \nat}F_i = E$. However, as $\mu_0$ is a premeasure and each $F_i \in \alg$, the mass of $E$ is the combination of the masses of each $F_i$ in the first place, so there is no mass lost with this partition: > $ > \mu_0(E) = \sum_{i \in \nat}\mu_0(F_i) \le \sum_{i \in \nat}\mu_0(E_i) \Rightarrow \mu^*(E) \ge \mu_0(E) > $ > Therefore $\mu^*(E) = \mu_0(E)$ for any $E \in \alg$, and $\mu^*|\alg = \mu_0$. > > ![[premeasure_split.png]] > > A set is outer-measurable if the distribution of mass along its borders is approachable by $\mu^*$. Since for any elementary set $E \in \alg$, $\mu^*$ uses $E$ itself to approach it, the distribution of mass along the border is already precisely measured. $E$ itself can be used to cut the approximating sets cleanly, which then separates the approximation into two parts with no ambiguity along its border. > > To see this, let $A \in \pow{X}$, $\pb{E} \in \alg$ and, $\bracs{E_i}_{1}^{\infty}$ be an elementary cover of $A$. Then we can take > $ > \pb{I_i} = \pb{E \cap E_i} \quad \po{O_i} = \po{E^c \cap E_i} > $ > then $\bracs{\pb{I_i}}_1^\infty$ is an elementary cover of $\pb{E \cap A}$ and $\bracs{\po{O_i}}_1^{\infty}$ is an elementary cover of $\po{E^c \cap A}$, and together they cover $A$. > $ > (\pb{E \cap A}) \cup (\po{E^c \cap A}) = \paren{\bigcup_{i \in \nat}\pb{E \cap E_i}} \cup \paren{\bigcup_{i \in \nat}\po{E^c \cap A}} > $ > Since $\bracs{\pb{I_i}}_{1}^{\infty}$ and $\bracs{\po{O_i}}_1^\infty$ are elementary covers, we have > $ > \begin{align*} > \mu^*(\pb{E \cap A}) &\le \sum_{i \in \nat}\mu_0(\pb{I_i}) \\ > \mu^*(\po{E^c \cap A}) &\le \sum_{i \in \nat}\mu_0(\po{O_i}) > \end{align*} > $ > Moreover, $E_i = \pb{I_i} \cup \po{O_i}$ is a disjoint union, so > $ > \mu_0(E_i) = \mu_0(\pb{I_i}) + \mu_0(\po{O_{i}}) > $ > and we have > $ > \begin{align*} > \mu^*(\pb{E \cap A}) + \mu^*(\po{E^c \cap A}) > &\le \sum_{i \in \nat}\mu_0(\pb{I_i}) + \sum_{i \in \nat}\mu_0(\po{O_i}) \\ > &= \sum_{i \in \nat}\mu_0(E_i) > \end{align*} > $ > meaning that $\mu^*(\pb{E \cap A}) + \mu^*(\po{E^c \cap A})$ is a lower bound for > $ > \bracs{\sum_{j \in J}\mu_0(E_j): |J| = |\nat|, > E_j \in \alg, \bigcup_{j \in J}\mu_0(E_j) \supseteq A > } > $ > and > $ > \mu^*(\pb{E \cap A}) + \mu^*(\po{E^c \cap A}) \le \mu^*(A) > $ > Therefore > $ > \mu^*(\pb{E \cap A}) + \mu^*(\po{E^c \cap A}) = \mu^*(A) > $ > and $E \in \cm$ is $\mu^*$ measurable.