> [!definition] > > Let $X$ be a [[Set|set]] and $\cm$ be a [[Sigma Algebra|sigma algebra]] over $X$. A **measure** on $\cm$ is a [[Function|function]] $\mu: \cm \to [0, \infty]$ satisfying > - $\mu(\bracs{}) = 0$ > - If $\bracs{E_j}_{1}^{\infty}$ is a [[Cardinality|countable]] [[Sequence|sequence]] of disjoint sets in $\cm$, then $\mu\paren{\bigcup_{j = 1}^{\infty}E_j} = \sum_{j = 1}^{\infty}\mu(E_j)$. > > If $\mu$ does not satisfy *countable additivity*, but does satisfy finite additivity, then $\mu$ is a **finitely additive** measure. > [!definition] > > Let $X$ be a set and $\cm$ be a sigma algebra over $X$, then $(X, \cm)$ is a *measurable* space. If $\mu$ is a measure on $\cm$, then $(X, \cm, \mu)$ is a **measure space**. > [!definition] Properties > > | Property | Definition | > | ---- | ---- | > | Finite | $\mu(X) < \infty$ | > | $\sigma$-Finite | $X = \bigcup_{i \in I}E_i$, $\mu(E_i) < \infty \forall i \in I$ | > | $\sigma$-Finite for $\mu$ | $E = \bigcup_{i \in I}E_i, \mu(E_i) < \infty \forall i \in I$ | > | Semifinite | $\forall E \in \cm, \mu(E) = \infty, \exists F \in \cm, F \subset E: \mu(F) < \infty$ | > [!theorem] > > Let $(X, \cm, \mu)$ be a measure space. If $\mu$ is $\sigma$-finite, then $\mu$ is semifinite. > > *Proof*. Let $E \in \cm$ such that $\mu(E) = \infty$. Since $X = \bigcup_{i \in I}E_i$ where $\mu(E_i) < \infty \forall i \in I$, $E \cap E_i \subseteq E, E_i$ and has finite measure. > [!theorem] > > Let $(X, \cm, \mu)$ be a measure space. Then > 1. **Monotonicity:** $\forall E, F \in \cm, E \subseteq F \Rightarrow \mu(E) \le \mu(F)$. > 2. **Subadditivity**: For any sequence $\bracs{E_i} \subset \cm$, $\mu\paren{\bigcup_{i \in I}E_i} \le \sum_{i \in I}\mu(E_i)$. > 3. **[[Continuity]] From Below:** For any sequence $\bracs{E_i} \subset \cm$ where $E_k \subseteq E_{k + 1}$, then $\mu\paren{\bigcup_{i \in I}E_i} = \lim_{i \to \infty}\mu(E_i)$. > 4. **Continuity From Above:** For any sequence $\bracs{E_i} \subset \cm$ where $E_k \supseteq E_{k + 1}$ and $\mu(E_1) < \infty$, then $\mu(\bigcap_{i \in I}E_i) = \lim_{i \to \infty}E_i$. > > *Proof*. If $E \subseteq F$, then $E \cup (F \setminus E) = F$ and $\mu(F) = \mu(E) + \mu(F \setminus E)$, $\mu(F) \ge \mu(E)$. > > If we take $F_k = \bigcup_{i = 1}^{k}E_i \setminus \bigcup_{i = 1}^{k - 1}E_i$, then $\bigcup_{i = 1}^{k}E_i = \bigcup_{i = 1}^{k}F_i$ for all $k \in \nat$, and $\bracs{F_k}$ is a disjoint sequence of sets. Therefore > $ > \mu\paren{\bigcup_{i = 1}^{\infty}E_i} = \mu\paren{\bigcup_{i = 1}^{\infty}F_i} = \sum_{i = 1}^{\infty}\mu(F_i) \le \sum_{i = 1}^{\infty}\mu(E_i) > $ > since $F_k \subseteq E_k$. > > Let $E_0 = \emptyset$ and $F_k = E_{k} \setminus E_{k - 1}$, then similar to earlier, the $F_i$s are disjoint and $\bigcup_{i = 1}^{\infty}F_i = \bigcup_{i = 1}^{\infty}E_i$. Since $\bracs{E_i}$ is an ascending sequence, > $ > \bigcup_{i = 1}^{k}F_k = \bigcup_{i = 1}^{k}E_i \setminus E_{i - 1} = E_k > $ > as well. This means that > $ > \begin{align*} > \mu\paren{\bigcup_{i = 1}^{\infty}E_i} &= \mu\paren{\bigcup_{i = 1}^{\infty}F_i} \\ > &= \sum_{i = 1}^{\infty}\mu(F_i) \\ > &= \limv{n}\sum_{i = 1}^{n}\mu(F_i) \\ > &= \limv{n}\sum_{i = 1}^{n}[\mu(E_i) - \mu(E_{i - 1})] \\ > &= \limv{n}[\mu(E_n) - \mu(\emptyset)] \\ > &= \limv{n}\mu(E_n) > \end{align*} > $ > > Finally, let $\bracs{E_i}$ be a shrinking sequence of sets with $\mu(E_1)<\infty$. Take $F_k = E_1 \setminus E_k$, then $F_k \cup E_k = E_1$ is a disjoint union, $\bigcup_{i = 1}^{k}F_i \cup \bigcap_{i = 1}^{k}E_i$ is also a disjoint union, and > $ > \begin{align*} > \mu\paren{\bigcap_{i = 1}^{\infty}E_i} = \mu\paren{E_1 \setminus \bigcup_{i = 1}^{\infty}F_i} &= \mu (E_1) - \mu\paren{\bigcup_{i = 1}^{\infty}F_i} \\ > &= \mu(E_1) - \lim_{i \to \infty}\mu\paren{F_i} \\ > &= \lim_{i \to \infty}[\mu(E_1) - \mu(F_i)] \\ > &= \lim_{i \to \infty}\mu\paren{E_1 \setminus F_i} \\ > &= \lim_{i \to \infty}\mu(E_i) > \end{align*} > $ > [!theorem] > > Let $(X, \cm, \mu)$ be a measure space and $\bracs{E_j}$ be a [[Sequence|sequence]] of sets in $\cm$, then the measure of the [[Limit Superior and Inferior|limit inferior]] is less than the limit inferior of the measure > $ > \mu(\liminf E_j) \le \lim\inf \mu(E_j) > $ > and the measure of the limit superior is greater than the limit superior of the measure > $ > \mu\paren{\bigcup_{j = 1}^{\infty}E_j} < \infty \Rightarrow \mu(\limsup E_j) \ge \limsup \mu(E_j) \quad > $ > > *Proof*. First we note that > $ > \bigcap_{j = k}^\infty E_j \subseteq \bigcap_{j = k + 1}^\infty E_j \quad \forall k \in \nat > $ > and that > $ > \bigcap_{j = k}^\infty E_j \subseteq E_l \quad \forall l \ge k > $ > This gives > $ > \mu\paren{\bigcap_{j = k}^{\infty}E_j} \le \mu(E_l)\ \forall l \ge k \Rightarrow \mu\paren{\bigcap_{j = k}^{\infty}E_j} \le \inf \bracs{\mu(E_l): l \ge k} > $ > and we have > $ > \begin{align*} > \mu\paren{\bigcup_{k = 1}^{n}\bigcap_{j = k}^{\infty}E_j} \le \mu\paren{\bigcap_{j = n}^{\infty}E_j} \le \inf\bracs{\mu(E_j): j \ge n} > \end{align*} > $ > By continuity from below, > $ > \begin{align*} > \mu\paren{\bigcup_{k = 1}^{\infty}\bigcap_{j = k}^\infty E_j} = \lim_{n \to \infty}\mu\paren{\bigcup_{k = 1}^n \bigcap_{j = k}^{\infty}E_j} &\le \lim_{n \to \infty}\inf\bracs{\mu(E_j): j \ge n} \\ > &= \liminf \mu(E_j) > \end{align*} > $ > We now note that > $ > \bigcup_{j = k}E_j \supseteq \bigcup_{j = k + 1}E_j \quad \forall k \in \nat > $ > and that > $ > \bigcup_{j = k}E_j \supseteq E_l \quad \forall l \ge k > $ > This gives > $ > \mu\paren{\bigcup_{j = k}^{\infty}E_j} \ge \mu(E_l)\ \forall l \ge k \Rightarrow \mu\paren{\bigcup_{j = k}^{\infty}E_j} \ge \sup\bracs{\mu(E_l): l \ge k} > $ > and by continuity from above (which we can use since $\mu\paren{\bigcup_{j = 1}^{\infty}E_j} < \infty$) > $ > \begin{align*} > \mu\paren{\bigcap_{k = 1}^{\infty}\bigcup_{j = k}^{\infty}E_k} = \lim_{n \to \infty}\mu\paren{\bigcap_{k = 1}^{n}\bigcup_{j = k}E_j} &\ge \lim_{n \to \infty} \sup \bracs{\mu(E_j): j \ge n} \\ > &= \limsup \mu(E_j) > \end{align*} > $ > [!theorem] > > Let $(X, \cm, \mu)$ be a measure space and $E, F \in \cm$, then > $ > \mu(E) + \mu(F) = \mu(E \cup F) + \mu(E \cap F) > $ > *Proof*. We break $E, F$ into pieces $E \setminus F, E \cap F, F \setminus E$, then > $ > \begin{align*} > \mu(E) + \mu(F) &= \mu(E \setminus F) + \mu(E \cap F) + \mu(F \setminus E) + \mu(E \cap F)\\ > &= \mu(E \cup F) + \mu(E \cap F) > \end{align*} > $ ### Common Measures > [!definition] Counting Measure > > Let $S$ be a finite set and take the Boolean algebra to be the [[Power Set|power set]] $\mathcal{P}(S)$. For each $A \in \mathcal{P}(S)$, define $m(A) = \#(A)$ as the **counting measure**. > [!definition] Dirac Measure at a > > Let $S = \real$ and $\Sigma(S)$ be the set of all [[Broken Line|broken lines]] $\mathcal{I}(\real)$. Fix a number $a \in \real$, and define a the **Dirac measure at a** $\delta_a$ by: > $ > \begin{align*} > \delta_a(J) = 1 &\quad \text{if}\ a \in J, \\ > \delta_{a}(J) = 0 &\quad \text{if}\ a \not\in J > \end{align*} > $ > for all broken lines $J$. > > Note that only one broken line $J$ in a set of disjoint broken lines can have a Dirac measure of 1.