## Outer Measure > [!definition|*] Outer Measure > The (Lebesgue) **outer measure** of a set $S \subseteq \mathbb{R}$ is the infimum of the length of its interval-covers: $m^{*}(S)= \inf Z_{S},\, \text{where }Z_{S}=\left\{\Lambda(\{ I_{i} \}):\bigcup_{i}I_{i} \supseteq S \right\}$If $Z_{S}$ is unbounded, set $m^{*}=\infty$. Outer measures preserve orders: if $A \subseteq B$, then $m^{*}(A) \le m^{*}(B)$. Outer measures are **countably subadditive**: given sequence of sets $(E_{n})$, $m^{*}(\cup_{n} E_{n}) \le \sum_{n} m^{*}(E_{n})$ > [!proof]- > If $\mathrm{RHS}=\infty$, the result is trivial, so assume otherwise. > > It's sufficient to show that for any $\epsilon > 0$, $m^{*}(\cup_{n} E_{n}) \le \sum_{n} m^{*}(E_{n})+\epsilon$ > For each $E_{n}$, find interval-cover $\{ I^{(n)}_{i} \}$ satisfying $\Lambda(\{ I_{i}^{(n)} \}) \le m^{*}(E_{n})+\epsilon 2^{-n}.$ > Then their union $J=\cup_{n}\{ I^{(n)}_{i} \}$ is a countable interval-cover of $\cup_{n}E_{n}$, with length $\Lambda(J)=\sum_{n}\Lambda(\{ I^{(n)}_{i} \})\le\left(\sum_{n}m^{*}(E_{n})\right)+\epsilon.$So $m^{*}(\cup_{n}E_{n}) \le \Lambda(J) \le \sum_{n} m^{*}(E_{n})+\epsilon$. ### Null Sets in $\mathbb{R}$ * A set $S \subseteq \mathbb{R}$ is **null** or a **null set** if has outer measure of $m^{\ast}(S)=0$. * Equivalently, it is contained by a sequence of intervals of arbitrarily small (total) length: $\forall \epsilon>0, \exists (I_{i}): \Lambda((I_{i})) < \epsilon,\, S \subseteq \bigcup_{i=1}^{\infty}I_{i}$ > [!theorem|*] Union of Null Sets is Null > Countable unions of null sets are also null: if $S_{1},S_{2},\dots$ are all null, then $T=\cup_{i=1}^{\infty}S_{i}$ is also null. > > > [!proof]- > > Let $\{ I_{k}^{(i)} \}$ be an interval cover of $S_{i}$ with total length $\Lambda(\{ I_{k}^{(i)} \})< \epsilon 2^{-i}$. > > Then their union $\cup_{i}\{ I^{(i)}_{k}\}$ is a countable interval cover of $T$. Its total length is $\sum_{i}\sum_{k}I^{(i)}_{k}<\sum_{i} \epsilon 2^{-i}=\epsilon$. > - Hence *singletons, finite sets, and countable sets are all null*. - Uncountable sets can also be null: for example the **Cantor set** is uncountable but null. > [!theorem|*] Union with Null Sets Preserve Measure > Union with a null set does not change the measure. > > > [!proof]- > > Say $A,E \subseteq \mathbb{R}$, and $E$ null. Then $m^{*}(A \cup E)\le m^{*}(A)+\underset{0}{\cancel{m^{*}(E)}}\le m^{*}(A \cup E)$where the first inequality is subadditivity, and the second is from preserving the order $A \subseteq A \cup E$. So $m^*(A)=m^{*}(A\cup E)$. ### Consistency of Outer Measure and Length > [!bigidea] Outer Measure agrees with our common sense definition of Length. If $I=[a,b]$ is a closed interval, $m^{*}(I)=l(I)$. > [!proof] > $[\le]$ Trivial: $\{ I \} \in Z_{I}$ covers $I$. > $[\ge]$ The approximation property gives a cover $\{I_{i} \}: \Lambda(\{ I_{i} \}) \le m^{*}(I) + \epsilon$. > > Extend each of $I_{i}$ by $\pm 2^{-i}\epsilon$ to get open intervals $J_{i}$, another cover of $I$ with $\Lambda(\{ J_{i} \}) \le m^{*}(I) + 3 \epsilon$. > Heine-Borel gives a finite subcover $\{ J_{j}=(c_{j},d_{j}) \}_{j=1}^{k}$. > Let $c=\min \{ c_{j} \},d=\max \{ d_{j} \}$ be the two endpoints of $\{ J_{j} \}_{j=1}^{k}$, so that $c<a<b<d$. Then $(b-a)<(d-c)<\Lambda(\{ J_{j} \}_{j=1}^{k}) \le \Lambda(\{ J_{i} \})\le m^{*}(I)+3\epsilon$Since $\epsilon$ can be arbitrarily small, $l(I)=(b-a) \le m^*(I)$. * In general, if $I$ is any interval, $m^{*}(I)=l(I)$. > Proof: if $I$ unbounded, the result is trivial. > If $I=[a,b)$, union with the null set $\{ b \}$ changes neither the length or the outer measure. So the previous result applies to $[a,b)\cup \{ b \}=[a,b]$. > Similarly for $(a,b]$, $(a,b)$. ## The Lebesgue Measure > [!bigidea] > - The outer measure fails for bad sets. > - The **Lebesgue measurable sets** are the good sets. > - The **Lebesgue measure** is the outer measure restricted to those good sets. The problem with the outer measure $m^{\ast}:\mathcal{P}(\mathbb{R})\to \mathbb{R}$ is that *it is not countably additive*, where there exists a countable sequence of disjoint sets $(A_{n})$ such that $m^{\ast}(\cup_{n}A_{n})\lneq \sum_{n}m^{\ast}(A_{n})$see $\text{Example }2.4$ on page 7 of the 2022-23 lecture notes. We need to restrict the measure to sets with better behavior--the Lebesgue measurable sets. A set $E$ is **Lebesgue-measurable** if for any $A \subseteq \mathbb{R}$, $m^{*}(A)=m^{*}(A \cap E)+m^{*}(A \cap E^{c})$The set of Lebesgue measurable sets is denoted $\mathcal{M}_{Leb}$. * To show measurability, since $\mathrm{LHS}\le\mathrm{RHS}$ is guaranteed by subadditivity, it's sufficient to show $\mathrm{ LHS} \ge \mathrm{RHS}$. * $\mathcal{M}_{Leb}$ is a $\sigma$-algebra in $\mathbb{R}$, meaning that: * (1) $\mathbb{R} \in \mathcal{M}$, * (2) closed to complements: if $E \in \mathcal{M}$, then $E^{c}\in \mathcal{M}$, * (3) closed to countable unions: if $(E_{n})_{n=1}^{\infty} \subseteq \mathcal{M}$, then $\cup_{n}E_{n } \in \mathcal{M}$. * Null sets and intervals are measurable. * Since all open sets are countable unions of open intervals, all open sets are measurable. > Both proved in Sheet 1. - When restricted to $\mathcal{M}_{Leb}$, the outer measure is denoted $m:\mathcal{M}_{Leb} \to \mathbb{R}$, the **Lebesgue measure**. ### Properties of the Lebesgue Measure *Big idea: the Lebesgue measure works!* - The Lebesgue measure is **countably additive**: if $(E_{n})_{n=1}^{\infty} \subseteq \mathcal{M}_{Leb}$ are pairwise disjoint, then $m(\cup_{n} E_{n}) = \sum_{n} m(E_{n})$ > Proof: by subadditivity, it's sufficient to show that $\mathrm{LHS} \ge \mathrm{RHS}$. Suppose for contradiction $\mathrm{LHS}<\mathrm{RHS}$. > Since the partial sums of $\mathrm{RHS}=\lim_{ N \to \infty }\sum_{n=1}^{N}m(E_{n})$ are non-decreasing, there must be an $N$ when the partial sum surpasses $\mathrm{LHS}$: $\begin{align*} \exists N:m(\cup_{n}E_{n})<\sum_{n=1}^{N}m(E_{n}) && (*) \end{align*}$Apply the definition of measurability of $E_{n}$ to get $\begin{align*} m(\cup_{n}E_{n})&= m\big( E_{1} \cap (\cup_{n}E_{n}) \big) + m\big( E_{1}^{C} \cap (\cup_{n}E_{n}) \big)\\ &= m(E_{1})+m(\cup_{n \ge 2}E_{n})\\ &= m(E_{1})+m(E_{2})+m(\cup_{n \ge 3}E_{n}) \\ &\dots\\ &= \sum_{n=1}^{N}m(E_{n})+m(\cup_{n \ge N+1}E_{n}) \end{align*}$but then plugging back into $(*)$ gives $m(\cup_{n \ge N+1}E_{n})< 0$, a contradiction. * For any $E \in \mathcal{M}_{Leb}$, it can be arbitrarily well-fit into some open set: $\forall \epsilon>0,\exists O \text{ open} \subseteq \mathbb{R}:\,E \subseteq O,\,m(O-E)<\epsilon$Note that it is not one-set-fits-all-$\epsilon$. > Proof: > <u>Finite measure</u>: assume $m(E)<\infty$. Find an interval cover $\{ I_{i} \}$ of $A$ with total length less than $m^{*}(A)+\epsilon$. > Extend each $I_{n}$ by $\epsilon / 2^n$ to get the open interval-cover $\{ J_{i} \}:\Lambda(\{ J_{i} \})<m(E)+2\epsilon$. Then take the union $O=\cup_{i}J_{i}$. > Apply the definition of $E \in \mathcal{M}_{Leb}$ to verify $m(O-E)<2\epsilon$. > > <u>Infinite measure</u>: assume $m(E)=\infty$, then define $E_{n}=E \cap [-n,n]$. Fit open sets $O_{n}\supset E_{n}:m(O_{n}-E_{n})<\epsilon 2^{-n}$. > Then take $O \equiv \cup_{n}O_{n} \supset E$, so that $O-E=\cup_{n}(O_{n}-E_{n})$. > By subadditivity $m(O-E) \le \sum_{n}m(O_{n}-E_{n})<\sum_{n}\epsilon 2^{-n}=\epsilon$. * If $(A_{n}) \subset \mathcal{M}_{Leb}$, and $A_{n} \subset A_{n+1}$, then $m(\bigcup_{n}A_{n})=\lim_{ n \to \infty }m(A_{n})$. * *This property is used in a number of proofs in A8 Probability.* > Proof: define $B_{n}=A_{n}-A_{n-1}$, then $\cup_{n}A_{n}=\cup_{n}B_{n}$, and apply countable additivity to the disjoint $(B_{n})$.