---- > [!definition] Definition. ([[Lebesgue outer measure]]) > The default example of an [[outer measure]] is the **(Lebesgue) outer measure** on $\mathbb{R}$, given by $\begin{align} |\cdot|: 2^{\mathbb{R}} &\to [0, \infty] \\ A & \mapsto \inf \left\{ \sum_{i=1}^{\infty} \ell(I_{k}) : I_{1},I_{2},\dots \text{form an open cover of }A \text{ by intervals} \right\}. \end{align}$Observe that the notion of [[measure zero]] studied in undergraduate real analysis refers to *outer* measure zero. ^definition > [!basicnonexample] Nonadditivity. > > > $|\cdot|$ is not an earnest measure, because it is not additive. See the counter example below using a [[Vitali set]]. > > > > [!proposition] > > There exist disjoint subsets $A$ and $B$ of $\mathbb{R}$ such that $|A \cup B| \neq |A|+|B|$. > ^proposition > > > [!proof] > > Define an [[equivalence relation]] $\sim$ on $[-1,1]$ as $a \sim c \iff a- c \in \mathbb{Q}.$ > > We denote the [[equivalence class]] of $a \in [-1,1]$ under $\sim$ by $\tilde{a}$. It contains all the elements of $[-1,1]$ which differ from $a$ by a rational number. Note (by general [[equivalence class]] properties) that if $a,b \in [-1,1]$ with $\tilde{a} \cap \tilde{b} \neq \emptyset$, then $\tilde{a}=\tilde{b}$. Also note $[-1,1]=\sqcup_{a \in [-1,1]}\tilde{a}$ (equivalence classes $\iff$ [[partition]]). > > > > Let $V$ be a set that contains exactly one representative from each class $\tilde{a}$, i.e. a set that contains exactly one element from each element of $[-1,1] / {\sim}$ (this is using the [[Axiom of Choice]]). Enumerate the collection $[-2,2] \cap \mathbb{Q}$ of rational numbers as $r_{1},r_{2},\dots$. Then $[-1,1] \subset \bigcup_{k=1}^{\infty}(r_{k}+V).$ > > Applying monotonicity and countable subadditivity of outer measure to this inclusion yields $\underbrace{ |[-1,1]| }_{ =2 } \leq \sum_{k=1}^{\infty}|r_{k}+V|.$ > > Translation invariance of $|\cdot|$ now gives $2 \leq \sum_{k=1}^{\infty} |V|.$ > > This tells us that $|V|>0$. > > > > Note that the translates $r_{k}+V$ are disjoint. Let $n \in \mathbb{N}$. Clearly $\bigcup_{k=1}^{n}(r_{k}+V) \subset [-3,3]$, since $V \subset [-1,1]$ and each $r_{k} \in [-2,2]$. Hence $|\bigsqcup_{k=1}^{n}(r_{k}+V)| \leq 6.$ > > However $\sum_{k=1}^{n} |r_{k}+V|=\sum_{k=1}^{n} | V|=n | V|.$ > > > > If we choose $N$ large enough so that $N | V|>6$, we obtain $|\bigsqcup_{k=1}^{n} (r_{k}+V)| < \sum_{k=1}^{n} |r_{k}+V|.$ > > > > > [!justification] > Note that $|\cdot|$ is clearly monotone: $|A| \leq |B|$ when $A \subset B$ since any open cover of $B$ is also an open cover of $A$. Let us show that countable subadditivity holds. Suppose $A_{1},A_{2},\dots, \subset \mathbb{R}$. If $|A_{k}|=\infty$ for some $k$, then it certainly holds. Thus assume $|A_{k}|<\infty$ for all $k$. We have to show $|\bigcup_{k=1}^{\infty}A_{k}| \leq \sum_{k=1}^{\infty} |A_{k}|$. Fix $\varepsilon>0$. For each $k$, let $I_{1,k},I_{2,k}$ be a sequence of open intervals [[cover|covering]] $A_{k}$ such that $ \sum_{j=1}^{\infty}\ell(I_{j,k}) - |A_{k}| \leq \frac{\varepsilon}{2^{k}}.$ Now rearrange and sum over $k$ thus: $\sum_{k=1}^{\infty} \sum_{j=1}^{\infty} \ell(I_{j,k}) \leq \underbrace{ \sum_{k=1}^{\infty} \frac{\varepsilon}{2^{k}} }_{ =\varepsilon } + \sum_{k=1}^{\infty} |A_{k}|=\varepsilon+ \sum_{k=1}^{\infty} | A_{k}| $ Since the doubly-indexed collection of open intervals $\{ I_{j,k}: j,k \in \mathbb{N} \}$ can be rearranged into a sequence of intervals covering $\bigcup_{k=1}^{\infty} A_{k}$, we are done. ^justification > [!basicproperties] > > > > > > - Finite sets have outer measure $0$. > > [!proof]- > > > Indeed, suppose $A =\{ a_{1},\dots,a_{m} \}$. Fix $\varepsilon>0$. [[cover|Cover]] $A$ by the [[open interval|open intervals]] $I_{i}=\left( a_{i}-\frac{\varepsilon }{4m}, a_{i}+\frac{ \varepsilon }{4m} \right)$, $i \in [m]$, so that $\ell(I_{i})=\frac{\varepsilon}{2m}$. Then $\sum_{i=1}^{m}\ell(I_{i})=\frac{\varepsilon}{2}< \varepsilon$. > ^proof > > - More generally, countable subsets of $\mathbb{R}$ have outer measure zero. > > [!proof]- > > > Indeed, suppose $A=\{ a_{1},a_{2},\dots \} \subset \mathbb{R}$. Fix $\varepsilon>0$. Put $I_{i}:=\left( a_{i}-\frac{\varepsilon}{2^{i+1}}, a_{i}+ \frac{\varepsilon}{2^{i+1}} \right)$, so that $\ell(I_{i})=\frac{\varepsilon}{2^{i}}$. Then $\sum_{i=1}^{\infty} \ell(I_{i})=\varepsilon \sum_{i=1}^{\infty}\left( \frac{1}{2} \right)^{i}=\varepsilon$, where we are using that the [[geometric series]] $\sum_{i=0}^{\infty} (\frac{1}{2})^{i}$ [[sequence|converges]] to $\frac{1}{1 - \frac{1}{2}}=2$ and so $\sum_{i=1}^{\infty} (\frac{1}{2})^{i}=1$. > > ^proof > > - The outer measure $|\cdot|$ on $\mathbb{R}$ is [[group-invariant function|translation invariant]], since the length function $\ell(\cdot)$ is: for all $t \in \mathbb{R}$, $A \subset \mathbb{R}$, $|t+A|=|A|$. > > - If $a<b$, then the outer measure of the [[closed interval]] $[a,b]$ is $b-a$ (this is not obvious!) > > > [!proof]- > > Indeed, for any $\varepsilon>0$, $(a-\varepsilon, b+\varepsilon)$ covers $[a,b]$ and so $|[a,b]| \leq b+\varepsilon - (a-\varepsilon)= b-a+2\varepsilon$. Since $\varepsilon$ was arbitrary: $|[a,b]| \leq b-a.$ > > Proving the reverse inequality $b-a \leq |[a,b]|$ should not be obvious! Doing so requires that the [[complete|completeness]] of $\mathbb{R}$ be used in some form. > > > > If we knew that $|(a,b)|=b-a$, then the reverse inequality would follow from monotonicity. Instead, we will prove it by other means and then deduce $|(a,b)|=b-a$ as a corollary. Suppose $I_{1},I_{2},\dots$ is any sequence of open intervals covering $[a,b]$. [[closed intervals are compact|Since]] $[a,b]$ is [[compact]] (that assertion requires [[complete|completeness]] of $\mathbb{R}$ to prove!), we can in fact assume $[a,b] \subset I_{1} \cup \dots \cup I_{n}$ > > for some $n \in \mathbb{N}$. We will then prove by induction on $n$ that the inclusion above implies $\sum_{k=1}^{n} \ell(I_{k}) \geq b-a.$ > > This will then imply $\sum_{k=1}^{\infty}\ell(I_{k}) \geq \sum_{k=1}^{n}\ell(I_{k}) \geq b-a$, completing the proof. If $n=1$ then the result clearly holds. Suppose $n>1$. Suppose $I_{1},\dots,I_{n},I_{n+1}$ are open intervals such that $[a,b] \subset I_{1} \cup \dots \cup I_{n} \cup I_{n+1}.$ > > Now, $b$ is in at least one of the intervals $I_{1},\dots,I_{n},I_{n+1}$, WLOG $b \in I_{n+1}:=(c,d)$. If $c \leq a$, then $\ell(I_{n+1}) \geq b-a$ and there is nothing to prove. Thus we can assume $a < c <b < d$. > > > > > > ```handdrawn-ink > > { > > "versionAtEmbed": "0.1.19", > > "filepath": "Ink/Drawing/2025.8.12 - 22.10pm.drawing" > > } > > ``` > > This means $[a,c] \subset I_{1} \cup \dots \cup I_{n}$, to which we may apply the induction hypothesis to obtain $\sum_{k=1}^{n} \ell(I_{n}) \geq c-a$. Thus $\begin{align} > > \sum_{k=1}^{n+1} \ell(I_{k}) &\geq (c-a) + \ell(I_{n+1}) \\ > > &= (c-a) + (d-c) \\ > > &= d-a \\ > > &\geq b-a, > > \end{align}$ > > completing the proof. > > ---- #### ---- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```