> [!definitionb] Definition > > ![[caratheodory_measurable.png|500]] > > For any set $E \subseteq X$, use $A$ to *cut* $E$ into two pieces, one inside and one outside of $A$. > > If outer-measuring the pieces individually yields the same measure as outer-measuring the entire set, then the "borders" of $A$ is "compatible" with the outer measure: the distribution of mass created by the cut can be approached by the outer measure. > > If the outer measure uses an elementary collection of sets with measure to approximate the measure of a given set, then the distribution of mass around the "borders" of a measurable set should be able to be approached by those elementary sets. > > We formalise the idea as follows: Let $X$ be a [[Set|set]] and $\mu^*: \pow{X} \to [0, \infty]$ be an [[Outer Measure|outer measure]] on $X$. Then a set $A \subseteq X$ is **$\mu^*$-measurable** if > $ > \mu^*(E) = \mu^*(E \cap A) + \mu^*(E \cap A^c) \quad \forall E \in \pow{X} > $ > [!theoremb] Theorem > > If $\mu^*$ is an [[Outer Measure|outer measure]] on $X$, then the collection $\cm$ of $\mu^*$ measurable sets is a [[Sigma Algebra|sigma algebra]], and the restriction of $\mu^*$ to $\cm$ is a [[Complete Measure|complete]] [[Measure Space|measure]]. > > *Proof*. A set is $\mu^*$ measurable if the distribution of mass around its *border* can be approached by $\mu^*$. For any $A \in \cm$, the *cuts* made by $A$ is the same as the cuts made by its complement: > $ > \begin{align*} > A \in \cm \Rightarrow \mu^*(E) &= \mu^*(E \cap A) + \mu^*(E \cap A^c) &\forall E \in \pow{X} \\ > &= \mu^*(E \cap A^c) + \mu^*(E \cap A^{cc}) \\ > &\Rightarrow A^c \in \cm > \end{align*} > $ > > It doesn't matter which side of the border is filled, so $\cm$ is closed under complements. > > ![[caratheodory_borders.png|500]] > > Now, let $A, B \in \cm$, then the border of $A \cup B$ is just a combination of the old ones. Cutting a set with $A \cup B$ is equivalent to using some part of $As border, and part of $Bs. Since each part of the new border is approachable by $\mu^*$, the entire border is also approachable. > > ![[caratheodory_finite_union.png]] > > To see this, let $\pb A, \po B \in \cm$ and $E \in \pow{X}$. First cut $E$ with $\pb A$, which > $ > \mu^*(E) = \mu^*(\pb{E \cap A}) + \mu^*(\pp{E \cap A^c}) > $ > cleanly distributes the mass into two pieces. Then cut each piece with $\po B$: > $ > \begin{align*} > \mu^*(\pb{E \cap A}) &= \mu^*(\pg{E \cap A \cap B}) + \mu^*(\pb{E \cap A \cap B^c}) \\ > \mu^*(\pp{E \cap A^c}) &= \mu^*(\po{E \cap A^c \cap B}) + \mu^*(\pp{E \cap A^c \cap B^c}) > \end{align*} > $ > which cleanly distributes the mass into four pieces in total, yielding > $ > \begin{align*} > \mu^*(E) &= \mu^*(\pg{E \cap A \cap B}) + \mu^*(\pb{E \cap A \cap B^c}) \\ > &+ \mu^*(\po{E \cap A^c \cap B}) + \mu^*(\pp{E \cap A^c \cap B^c}) > \end{align*} > $ > Since the cuts made by $\pb A$ and $\po B$ are clean, we can "re-stitch" parts of them together along those cuts as follows. > > First mend the blue and green piece together with the cut made by $\po B$. > $ > \begin{align*} > \pb A \cup \po B \supseteq \pb A &\Rightarrow [\ps{E \cap (A \cup B)}] \cap \pb A = \pb{E \cap A}\\ > \mu^*([\ps{E \cap (A \cup B)}] \cap \pb A) &=\mu^*(\pb{E \cap A}) \\&= \mu^*(\pg{E \cap A \cap B}) + \mu^*(\pb{E \cap A \cap B^c}) > \end{align*} > $ > "Reformat" $\po{E \cap A^c \cap B}$ like so to isolate the cut made by $\pb A$: > $ > \begin{align*} > [\ps{E \cap (A \cup B)}] \cap A^c&= E \cap [(A \cup B) \cap A^c] \\ > &= E \cap [(A \cap A^c) \cup (B \cap A^c)] \\ > &= E \cap B \cap A^c \\ > \mu^*([\ps{E \cap (A \cup B)}] \cap A^c) &= \mu^*(E \cap B \cap A^c) > \end{align*} > $ > Then combine the two pieces like so by mending the cut made by $\pb A$: > $ > \begin{align*} > \mu^*(\ps{E \cap (A \cup B)}) &= \mu^*([\ps{E \cap (A \cup B)}] \cap \pb A) \\ > &+ \mu^*([\ps{E \cap (A \cup B)]}] \cap A^c) \\ > &= \mu^*(\pg{E \cap A \cap B}) + \mu^*(\pb{E \cap A \cap B^c}) \\ > &+ \mu^*(\po{E \cap A^c \cap B}) > \end{align*} > $ > We "reformat" $\pp{E \cap A^c \cap B^c}$ to combine the cuts made by $A$ and $B$: > $ > \pp{E \cap A^c \cap B^c} = E \setminus (A \cup B) = \pp{E \cap (A \cup B)^c} > $ > And finally, we combine $\ps{E \cap (A \cup B)}$ and $\pp{E \cap (A \cup B)^c}$ to find that > $ > \begin{align*} > &\mu^*(\ps{E \cap (A \cup B)}) + \mu^*(\pp{E \cap (A \cup B)^c}) \\ > &= \mu^*(\pg{E \cap A \cap B}) + \mu^*(\pb{E \cap A \cap B^c}) +\mu^*(\po{E \cap A^c \cap B}) \\ > &+ \mu^*(\pp{E \cap A^c \cap B^c}) \\ > &= \mu^*(E) > \end{align*} > $ > the cuts made by $\pb A \cup \po B$ distribute mass in a way that can be approached by the outer measure, meaning that $\pb A \cup \po B \in \cm$. $\cm$ is closed under finite union, and is an [[Algebra|algebra]]. > > ![[caratheodory_finite_additive.png|400]] > > Since the borders of $\mu^*$-measurable sets cleanly distributes mass along them, there is no ambiguity when the borders are close to each other, as they "pull" the approached mass towards the inside of their sets. When two $\mu^*$ measurable sets share a border, that border also cleanly splits the mass between the two. > > To see this, let $\pb A, \po B \in \cm, \pb A \cap \po B$ and consider $\mu^*(\pb A \cup \po B)$. We use $\pb A$ to cut $\pb A \cup \po B$ back to $\pb A$ and $\po B$ to obtain > $ > \begin{align*} > \mu^*(\pb A \cup \po B) &= \mu^*((\pb A \cup \po B) \cap \pb A) + \mu^*((\pb A \cup \po B) \cap A^c) \\ > &= \mu^*(\pb A) + \mu^*(\po B) > \end{align*} > $ > Since we only used $\pb{A}$ to make the cut, this works even if $\po{B} \not\in \cm$. So $\mu^*$ is finitely additive on $\cm$. > > ![[caratheodory_limit.png]] > > Now, since $\mu^*$ works by approaching the mass of sets from the outside, all sets in $\cm$ are "limit points" of $\mu^*s approximations. As we have all the "limit points", $\cm$ should be "closed", and for any sequence of sets in $\cm$, their "limit" can also be approached by $\mu^*$. > > To see this, let $\bracs{\pb{A_j}}_{1}^{\infty} \subseteq \cm$ be a sequence of pairwise disjoint sets. > $ > \begin{align*} > \mu^*\paren{\bigcup_{i \in \nat}\pb{A_i}} &\ge \mu^*\paren{\pb{\bigcup_{i \in I}A_i}} \quad\forall I \subset \nat, |I| < |\nat| \\ > &= \sum_{i \in I}\mu^*(\pb{A_i}) \quad\forall I \subset \nat, |I| < |\nat| \\ > \mu^*\paren{\bigcup_{i \in \nat}\pb{A_i}}&\ge \sup \bracs{\sum_{i \in I}\mu^*(\pb{A_i}): I \subset \nat, |I| < |\nat|} \\ > \mu^*\paren{\bigcup_{i \in \nat}\pb{A_i}} &= \sum_{i \in \nat}\mu^*(\pb{A_i}) > \end{align*} > $ > The expanding borders of $\pb{\bigcup_{i \in I}{A_i}}$ approaches $\bigcup_{i \in \nat}\pb{A_i}$, allowing $\mu^*$ to also approach the union. So $\mu^*$ is countably additive for sets in $\cm$. > > ![[caratheodory_multicut.png|300]] > > Now, let $E \in \pow{X}$, $A = \bigcup_{i \in \nat}\pb{A_i}$, and $I \subset \nat, |I| < |\nat|$. Then we can cut $E$ with $\pb{\bigcup_{i \in I}A_i}$. > $ > \begin{align*} > \mu^*(E) &= \mu^*\paren{E \cap \pb{\bigcup_{i \in I}A_i}} \\ > &+ \mu^*\paren{E \cap \pb{\paren{\bigcup_{i \in I}A_i}^c}} > \end{align*} > $ > We can cut the left part with each $\pb{A_i}$ as > $ > \begin{align*} > &\mu^*\paren{E \cap \pb{\bigcup_{i \in I}A_i}} \\ > &= \mu^*\paren{E \cap \pb{\bigcup_{i \in I}A_i} \cap \pb{A_{i_1}}} + \mu^*\paren{E \cap \pb{\bigcup_{i \in I}A_i} \cap \pb{A_{i_1}^c}} \\ > &= \mu^*(E \cap \pb{A_{i_1}}) + \mu^*\paren{E \cap \pb{\bigcup_{i \in I, i \ne i_1}A_i}} \\ > &\vdots \\ > &= \sum_{i \in I}\mu^*(E \cap \pb{A_i}) > \end{align*} > $ > into $|I|$ disjoint pieces, preserving the total mass with the clean cuts of each $\pb{A_i}$. > > ![[caratheodory_approach.png|500]] > > We now relax the borders of the complement and let $E \cap \pb{\bigcup_{i \in I}A_i}$ approach $E \cap \bigcup_{i \in \nat}\pb{A_i}$ from below. > $ > \begin{align*} > \mu^*(E) &= \mu^*\paren{E \cap \pb{\bigcup_{i \in I}A_i}} + \mu^*\paren{E \cap \pb{\paren{\bigcup_{i \in I}A_i}^c}} \\ > &\ge \mu^*\paren{E \cap \pb{\bigcup_{i \in I}A_i}} + \mu^*\paren{E \cap \paren{\bigcup_{i \in \nat}\pb{A_i}}^c} \\ > &= \sum_{i \in I}\mu^*(E \cap \pb{A_i}) + \mu^*\paren{E \cap \paren{\bigcup_{i \in \nat}\pb{A_i}}^c} > \end{align*} > $ > Mending the cuts made by each $\pb{A_i}$ yields a subset of the entire union. Since the cuts of each $\pb{A_i}$ cleanly distributes the mass among them, there is no double-counting and the mass can never exceed that of the union. > $ > \begin{align*} > \sum_{i \in I}\mu^*(E \cap \pb{A_i}) &= \mu^*\paren{E \cap \bigcup_{i \in I}\pb{A_i}} \le \mu^*\paren{E \cap \bigcup_{i \in \nat}\pb{A_i}} \\ > \sum_{i \in \nat}\mu^*(E \cap \pb{A_i}) &\le \mu^*\paren{E \cap \bigcup_{i \in \nat}\pb{A_i}} > \end{align*} > $ > And in this process, we also cover the entirety of the union, so the measured mass can also be no less than the mass of the union. > $ > \sum_{i \in \nat}\mu^*(E \cap \pb{A_i}) = \mu^*\paren{E \cap \bigcup_{i \in \nat}\pb{A_i}} > $ > which yields > $ > \begin{align*} > \mu^*(E) &\ge \sum_{i \in \nat}\mu^*(E \cap \pb{A_i}) + \mu^*\paren{E \cap \paren{\bigcup_{i \in \nat}\pb{A_i}}^c} \\ > &= \mu^*\paren{E \cap \bigcup_{i \in \nat}\pb{A_i}} + \mu^*\paren{E \cap \paren{\bigcup_{i \in \nat}\pb{A_i}}^c} \\ > &\ge \mu^*(E) > \end{align*} > $ > and > $ > \mu^*(E) = \mu^*\paren{E \cap \pb{\bigcup_{i \in \nat}A_i}} + \mu^*\paren{E \cap \pb{\paren{\bigcup_{i \in \nat}}^c}} > $ > As the distribution of mass around border of the union is approached by finite unions of $\bracs{\pb{A_i}}$, it ended up also being one approachable by $\mu^*$. So $\pb{\bigcup_{i \in \nat}A_i} \in \cm$ and $\cm$ is a $\sigma$-algebra, and $\mu^*$ is a measure on $\cm$. > > ![[caratheodory_complete.png]] > > Finally, let $\pb{N} \in \cm: \mu^*(\pb{N}) = 0$, then the distribution of mass around its borders can be approached by $\mu^*$. However, the lack of mass inside the borders of $\pb{N}$ means that there is no mass to "leak" into $\pb{N}$. Any subsets of $\pb{N}$ *simply do not have any mass to distribute*, they just push all of the mass outside. > > To see this, let $\pb{N} \in \cm$, $F \subset \pb{N}$ and $E \in \pow{X}$, then we first cut $E$ with $\pb{N}$ > $ > \mu^*(E) = \mu^*(E \cap \pb{N}) + \mu^*(E \cap \pb{N^c}) > $ > and find that $\mu^*(E \cap \pb{N}) \le \mu^*(\pb{N}) = 0$. Now use $F$ to cut $E \cap \pb{N}$ to obtain > $ > \mu^*(E \cap \pb{N}) = \mu^*(E \cap \pb{N} \cap F) + \mu^*(E \cap \pb{N} \cap F^c) = 0 > $ > as there is no mass to distribute. We mend the cut made by $\pb{N}$ to obtain > $ > \begin{align*} > \mu^*(E) &= \mu^*(E \cap \pb{N}) + \mu^*(E \cap \pb{N^c}) \\ > &= \mu^*(E \cap \pb{N} \cap F) + \mu^*(E \cap \pb{N} \cap F^c) + \mu^*(E \cap \pb{N^c}) \\ > &= \mu^*(E \cap F) + \mu^*(E \cap \pb{N} \cap F^c) + \mu^*(E \cap \pb{N^c} \cap F^c) \\ > &= \mu^*(E \cap \pb{F}) + \mu^*(E \cap \pb{F^c}) > \end{align*} > $ > and $\pb{F} \in \cm$ is $\mu^*$ measurable. Since all subsets of null sets are measurable, $\mu^*|_\cm$ is a complete measure. >