$\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] Measurable Space and $\sigma$-algebra > A ==**measurable space**== is a set $\Omega$ together with a non-empty collection $\AAA$ of subsets of $\Omega$. > >$\AAA$ is a ==**$\sigma$-algebra**==, and the elements of $\AAA$ are called ==**measurable sets**==. The set $\AAA$ is required to be closed under complements and *countable* unions. If we replace \"countable\" with \"finite,\" we get an ==**algebra**== instead. >People sometimes also use the word "field" instead of "algebra." >^algebra > [!problem] Countable Intersections Show that $\sigma$-algebras are closed under countable intersections, and contain $\Omega$ and $\emptyset$. > [!solution]- Let $\Omega$ be a set, and $\AAA$ a $\sigma$-algebra of subsets of $\Omega$. Let $A_1, A_2, \dots \in \AAA$. Then, $\conj{A}_1, \conj{A}_2, \dots \in \AAA$, so $\bigcup \conj{A}_i \in \AAA$. Thus, $\conj{\bigcup \conj{A}_i} \in \AAA$, so $\bigcap A_i \in \AAA$. Since $\AAA$ is closed under countable unions, $\emptyset = \bigcup_{A\in \emptyset} \in \AAA$. (Alternatively, use the nonempty condition.) Since $\AAA$ is closed under complements, $\Omega = \conj{\emptyset} \in \AAA$. It is quite hard to write down measurable sets explicitly, so for any set $\Omega$ and family of subsets $\FFF$, the ==**$\sigma$-algebra generated by $\FFF$**== is the smallest $\sigma$-algebra containing $\FFF$. > [!idea]- Details Specifically, take the intersection of all $\sigma$-algebras containing $\FFF$; this is guaranteed to be a $\sigma$-algebra containing $\FFF$. To prove that $\mathcal{A}$ is a $\sigma$-algebra, we need to show three things: >1. $\Omega \in \mathcal{A}$: Since each of the $\sigma$-algebras containing $\FFF$ must contain the sample space $\Omega$, it follows that the intersection of all such $\sigma$-algebras also contains $\Omega$. Therefore, $\Omega \in \mathcal{A}$. >2. If $A \in \mathcal{A}$, then $A^c \in \mathcal{A}$: Let $A \in \mathcal{A}$. This means that $A$ is in every $\sigma$-algebra containing $\FFF$. Since each of these $\sigma$-algebras also contains the complements of their elements, it follows that $A^c$ is in every such $\sigma$-algebra. Therefore, $A^c \in \mathcal{A}$. >3. If $A_1, A_2, A_3, ... \in \mathcal{A}$, then their union $B = A_1\cup A_2\cup A_3\cup ... \in \mathcal{A}$: Let $B = A_1\cup A_2\cup A_3\cup ...$ . For each $i$, we know that $B$ must be in every $\sigma$-algebra containing $\FFF$, and since each of these algebras are closed under countable unions, it follows that $B \in \mathcal{A}$. > >Therefore, we have shown that $\mathcal{A}$ is a $\sigma$-algebra. > >To show that $\mathcal{A}$ contains $\FFF$, note that each of the $\sigma$-algebras containing $\FFF$ must contain all the sets in $\FFF$. Therefore, the intersection of all such $\sigma$-algebras will also contain all the sets in $\FFF$. Hence, $\mathcal{A}$ contains $\FFF$. The most important $\sigma$-algebra is the ==**Borel $\sigma$-algebra**==. If $\Omega$ is a topological space, then the Borel $\sigma$-algebra $\BBB(\Omega)$ is the $\sigma$-algebra generated by the open sets of $\Omega$.