$\require{physics}\newcommand{\cbrt}[1]{\sqrt[3]{#1}}\newcommand{\sgn}{\text{sgn}}\newcommand{\ii}[1]{\textit{#1}}\newcommand{\eps}{\varepsilon}\newcommand{\EE}{\mathbb E}\newcommand{\PP}{\mathbb P}\newcommand{\Var}{\mathrm{Var}}\newcommand{\Cov}{\mathrm{Cov}}\newcommand{\pperp}{\perp\kern-6pt\perp}\newcommand{\LL}{\mathcal{L}}\newcommand{\pa}{\partial}\newcommand{\AAA}{\mathscr{A}}\newcommand{\BBB}{\mathscr{B}}\newcommand{\CCC}{\mathscr{C}}\newcommand{\DDD}{\mathscr{D}}\newcommand{\EEE}{\mathscr{E}}\newcommand{\FFF}{\mathscr{F}}\newcommand{\WFF}{\widetilde{\FFF}}\newcommand{\GGG}{\mathscr{G}}\newcommand{\HHH}{\mathscr{H}}\newcommand{\PPP}{\mathscr{P}}\newcommand{\Ff}{\mathcal{F}}\newcommand{\Gg}{\mathcal{G}}\newcommand{\Hh}{\mathbb{H}}\DeclareMathOperator{\ess}{ess}\newcommand{\CC}{\mathbb C}\newcommand{\FF}{\mathbb F}\newcommand{\NN}{\mathbb N}\newcommand{\QQ}{\mathbb Q}\newcommand{\RR}{\mathbb R}\newcommand{\ZZ}{\mathbb Z}\newcommand{\KK}{\mathbb K}\newcommand{\SSS}{\mathbb S}\newcommand{\II}{\mathbb I}\newcommand{\conj}[1]{\overline{#1}}\DeclareMathOperator{\cis}{cis}\newcommand{\abs}[1]{\left\lvert #1 \right\rvert}\newcommand{\norm}[1]{\left\lVert #1 \right\rVert}\newcommand{\floor}[1]{\left\lfloor #1 \right\rfloor}\newcommand{\ceil}[1]{\left\lceil #1 \right\rceil}\DeclareMathOperator*{\range}{range}\DeclareMathOperator*{\nul}{null}\DeclareMathOperator*{\Tr}{Tr}\DeclareMathOperator*{\tr}{Tr}\newcommand{\id}{1\!\!1}\newcommand{\Id}{1\!\!1}\newcommand{\der}{\ \mathrm {d}}\newcommand{\Zc}[1]{\ZZ / #1 \ZZ}\newcommand{\Zm}[1]{\left(\ZZ / #1 \ZZ\right)^\times}\DeclareMathOperator{\Hom}{Hom}\DeclareMathOperator{\End}{End}\newcommand{\GL}{\mathbb{GL}}\newcommand{\SL}{\mathbb{SL}}\newcommand{\SO}{\mathbb{SO}}\newcommand{\OO}{\mathbb{O}}\newcommand{\SU}{\mathbb{SU}}\newcommand{\U}{\mathbb{U}}\newcommand{\Spin}{\mathrm{Spin}}\newcommand{\Cl}{\mathrm{Cl}}\newcommand{\gr}{\mathrm{gr}}\newcommand{\gl}{\mathfrak{gl}}\newcommand{\sl}{\mathfrak{sl}}\newcommand{\so}{\mathfrak{so}}\newcommand{\su}{\mathfrak{su}}\newcommand{\sp}{\mathfrak{sp}}\newcommand{\uu}{\mathfrak{u}}\newcommand{\fg}{\mathfrak{g}}\newcommand{\hh}{\mathfrak{h}}\DeclareMathOperator{\Ad}{Ad}\DeclareMathOperator{\ad}{ad}\DeclareMathOperator{\Rad}{Rad}\DeclareMathOperator{\im}{im}\renewcommand{\BB}{\mathcal{B}}\newcommand{\HH}{\mathcal{H}}\DeclareMathOperator{\Lie}{Lie}\DeclareMathOperator{\Mat}{Mat}\DeclareMathOperator{\span}{span}\DeclareMathOperator{\proj}{proj}$ > [!definition] Measure Space > A ==**measure space**== is a triple $(\Omega, \AAA, \mu)$ where $\Omega$ is a set, $\AAA$ is a $\sigma$-[[measurable space|algebra of subsets]], and $\mu$ is a [[measure]]. > > In probability, $\Omega$ is commonly called the ==**state space**== or ==**outcome space**==, and elements of $\AAA$ are called ==**events**==. > [!problem] Basic Properties > > 1. If $A\subseteq B$ are measurable, then $\mu(A) \leq \mu(B)$, because $\mu(B) = \mu(A) + \mu(B\setminus A)$. > 2. We write $A_i \uparrow A$ if $A_1 \subseteq A_2 \subseteq \dots$ and $A = \bigcup_i A_i$. Then, > $ \mu(A) = \lim_i \mu(A_i).$ > 3. We write $A_i \downarrow A$ if $A_1 \supseteq A_2 \supseteq \dots$ and $A = \bigcap_i A_i$. If $\mu(A_1) < \infty$, then > $ \mu(A) = \lim_i \mu(A_i) $ > [!solution]- > > 1. $B = A \cup (B \setminus A)$, and the two sets are disjoint. > 2. $A = \bigcup_i (A_i \setminus A_{i-1})$, each of which is disjoint. This summation converges because its partial sums are increasing. > 3. Apply the previous result to $A_1 \setminus A_i$. > [!idea] Countable Additivity > The problem with uncountable sets is that they're a bit too large. Example: Let $S$ be an uncountable set of positive real numbers. Then, if you split into buckets $(\frac{1}{n + 1}, \frac{1}{n}]$, then there are only countably many buckets, thus some bucket has infinitely many elements of $S$. Conclusion: if $S$ is uncountable, there exist finite sets that have arbitrarily large sums. > [!idea] Almost Surely > A property $P$ holds ==**almost surely**== if it holds for all $\omega \in \Omega$ except for a set of measure zero. In measure theory, essentially all statements are only true almost surely; measure theory cannot distinguish between two sets that differ by a set of measure zero. > [!problem] Almost Sure Things > A ==**probability space**== is a measure space $(\Omega, \AAA, \mu)$ where $\mu(\Omega) = 1$. Show that the intersection of countably many measurable sets of measure $1$ has measure $1$. > [!solution]- > Let $A_1, A_2, \dots$ be measurable sets of measure $1$. Then, their complements $\conj{A}_1, \conj{A}_2, \dots$ have measure $0$. By countable additivity, the union of these complements has measure $0$. Thus, the intersection of the $A_i$ has measure $1$. The most famous measure is the [[Lebesgue Measure]] over $\RR$.