---- > [!definition] Definition. ([[measure]]) > Suppose $X$ is a set and $\Sigma$ is a [[σ-algebra]] on $X$. A **(positive) measure** on $(X, \Sigma)$ is a function $\mu:\Sigma \to [0, \infty]$ such that $\mu(\emptyset)=0$ and countable additivity holds: $\mu(\bigsqcup_{k=1}^{\infty}E_{k})=\sum_{k=1}^{\infty}\mu(E_{k})$ for every disjoint sequence $E_{1},E_{2},\dots$ of sets in $\Sigma$. > A **measure space** is a triple $(X, \Sigma, \mu)$ where $X$ is a set, $\Sigma$ is a $\sigma$-algebra on $X$, and $\mu$ is a measure on $(X,\Sigma)$. > An **atom** is a [[σ-algebra|measurable set]] of positive measure that has no subsets of positive measure. > [!basicexample] > > - The **counting measure** on a set $X$ is the measure $\mu$ defined on the [[σ-algebra|discrete σ-algebra]] (or, by restriction, to any other) by setting $\mu(E)=n$ if $E$ is a finite set of $n$ elements and $\mu(E)=\infty$ if $E$ is not a finite set. > - Given a set $X$, $\sigma$-algebra $\Sigma$ on $X$, and $c \in X$, the **[[Dirac measure]] $\delta_{c}$ on $(X,\Sigma)$** is the [[probability|probability measure]] $\delta_{c}(E)=\begin{cases} > 1 & c \in E \\ > 0 & c \not \in E. > \end{cases}$ > - If $w:X \to [0, \infty]$ is a 'weighting function' on the points of $X$, we get a measure $\mu$ on $(X, \Sigma)$ as $\mu(E)=\sum_{x \in E}w(x),$ > where the sum here is defined as the [[supremum]] of all finite subsums $\sum_{x \in D} w(x)$ as $D$ ranges over all finite subsets of $E$. > [!basicnonexample] > Let $2^{\mathbb{R}}$ denote the [[σ-algebra|discrete σ-algebra]] on $\mathbb{R}$, i.e. the $\sigma$-algebra consisting of all subsets of $\mathbb{R}$. The [[Lebesgue outer measure|Lebesgue]] [[outer measure]] $|\cdot|:2^{\mathbb{R}} \to [0, \infty]$ is *not* a measure on $(\mathbb{R}, 2^{\mathbb{R}})$ (it is not additive). However, $|\cdot|$ *is* a measure on $(\mathbb{R}, \mathcal{B}(\mathbb{R}))$, where $\mathcal{B}(\mathbb{R})$ denotes the [[Borel set|Borel σ-algebra]] on $\mathbb{R}$, as we shall soon see. ^nonexample > [!basicproperties] > Suppose $(X, \Sigma, \mu)$ is a [[measure|measure space]]. > > *(Monotonicity)* Let $D,E \in \Sigma$ such that $D \subset E$. Then $\mu(D) \leq \mu(E)$. > > *(Commutes with set difference).* Let $D,E \in \Sigma$ such that $D \subset E$. Then $\mu(E-D)=\mu(E)-\mu(D)$ provided that $\mu(D)<\infty$.[^1] > > The above two points both follow from the identity > $\mu(E)=\mu((E-D) \sqcup D)=\mu(E-D)+\mu(D).$ > > > [^1]: The hypothesis that $\mu(D)<\infty$ is needed to avoid undefined expressions of the form $\infty-\infty$. > > *(Countable subadditivity)* > Let $E_{1},E_{2},\dots \in \Sigma$. Then > $\mu\left(\bigcup_{k=1}^{\infty}E_{k}\right) \leq \sum_{k=1}^{\infty}\mu(E_{k}).$ > > > > [!proof]- Proof. > > Let $D_{1}=\emptyset$ and $D_{k}=E_{1} \cup \dots \cup E_{k-1}$ for $k \geq 2$. Then $E_{1}-D_{1}, E_{2}-D_{2}, E_{3}-D_{3},\dots$ > > is a disjoint sequence of subsets of $X$ whose union equals $\bigcup_{k=1}^{\infty}E_{k}$. Now > > $\mu(\bigcup_{k=1}^{\infty}E_{k})=\mu(\bigsqcup_{k=1}^{\infty} E_{k}-D_{k})=\sum_{k=1}^{\infty}\mu(E_{k}-D_{k})=\sum_{k=1}^{\infty} \mu(E_{k})-\mu(D_{k})$ > > which is less than or equal to $\sum_{k=1}^{\infty}\mu(E_{k})$. > > > *(Increasing union - continuity from below)* Let $E_{1} \subset E_{2} \subset \cdots$ be an increasing sequence of sets in $\Sigma$. Then $\mu(\bigcup_{k=1}^{\infty}E_{k})=\lim_{k \to \infty} \mu(E_{k}).$ > > > [!proof]- Proof. > > If $\mu(E_{k})=\infty$ for some $k \in \mathbb{N}$ then the required equation holds because both sides equal $\infty$. So assume $\mu(E_{k}) < \infty$ for all $k$. > > > > For convenience, let $E_{0}=\emptyset$. Then $\bigcup_{k=1}^{\infty}E_{k}=\bigsqcup_{j=1}^{\infty}(E_{j} - E_{j-1}),$ > > thus $\begin{align} > > \mu(\bigcup_{k=1}^{\infty}E_{k})&= \mu(\bigsqcup_{j=1}^{\infty}(E_{j} - E_{j-1})) \\ > > &= \sum_{j=1}^{\infty} \mu(E_{j}) - \mu(E_{j-1}) \\ > > &= \lim_{k \to \infty} \underbrace{ \sum_{j=1}^{k} \mu(E_{j})- \mu(E_{j-1}) }_{ \text{telescopes} } \\ > > &= \lim_{k \to \infty} \mu(E_{k}). > > \end{align}$ > > > > > *(Decreasing intersection - continuity from above)* Suppose $E_{1} \supset E_{2} \supset \cdots$ is a decreasing sequence of sets in $\Sigma$ with $\mu(E_{1})<\infty$. Then $\mu(\bigcap_{k=1}^{\infty}E_{k})=\lim_{k \to \infty}\mu (E_{k}).$ > > > [!proof]- Proof. > > One of [[De Morgan's Laws]] tells us that $E_{1} - \bigcap_{k=1}^{\infty}E_{k}=\bigcup_{k=1}^{\infty}(E_{1} - E_{k})$ > > Now $E_{1} - E_{1} \subset E_{1}-E_{2} \subset E_{1}-E_{3} \subset \cdots$ is an increasing sequence of sets in $\Sigma$. Applying the result above for increasing unions, we get $\underbrace{ \mu(E_{1}-\bigcap_{k=1}^{\infty}E_{k}) }_{ \mu(E_{1}) - \mu(\bigcap_{k=1}^{\infty} E_{k}) }=\underbrace{ \lim_{k \to \infty}\mu(E_{1} - E_{k}) }_{ \mu (E_{1}) - \lim_{k \to \infty}\mu(E_{k}) }.$ > > Cancelling $\mu(E_{1})$ on both sides, the result follows. > > > *(Inclusion-Exclusion)* Suppose $D,E \in \Sigma$ with $\mu(D \cap E)<\infty$. Then $\mu(D \cap E)=\mu(D)+\mu(E)-\mu(D \cap E).$ > > We have $D \cup E= \big( D - (D \cap E) \big) \sqcup \big( E- (D \cap E) \big) \sqcup (D \cap E).$ > $\begin{align} > \mu(D \cup E)&= \mu\big( D - (D \cap E) \big) + \mu\big( E- (D \cap E) \big) + \mu(D \cap E) \\ > &= \mu(D) - \mu(D \cap E) + \mu(E) \cancel{ - \mu(D \cap E) + \mu(D \cap E) } > \end{align}$ ---- #### ---- #### 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 > ```