$\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] Pi system > A ==**pi system**== on $\Omega$ is a collection of subsets that is closed under finite intersections. ^5c9b44 > [!definition] Lambda system > A ==**lambda system**== on $\Omega$ is a collection of subsets such that: > > 1. $\Omega \in \Lambda$. > 2. If $A,B \in \Lambda$ and $A \subseteq B$, then $B \setminus A \in \Lambda$. > 3. If $A_1, A_2, \dots \in \Lambda$ and $A_i \uparrow A$, then $A \in \Lambda$. > > (This is sometimes called a ==**d-system**==). The main idea is that these two definitions neatly partition the definition of a $\sigma$-algebra: > [!theorem] $\pi + \lambda = \sigma$ > > $\LL$ is a $\sigma$-algebra iff it is both a $\pi$-system and a $\lambda$-system. > [!proof] > First off, all $\sigma$-algebras are clearly $\pi$ systems, and satisfy the first two properties of $\lambda$-systems. The third property is simply a special case of countable unions. > > If you are a $\lambda$ system, then you have $\Omega$. Thus by the second condition you have complements. If you also a $\pi$ system, then finite intersections give you finite unions. Finally, given any collection $A_1, A_2,\cdots \in \LL$, their prefix unions $A_1\cup \cdots A_n$ are all in $\LL$, and thus by the third condition of $\lambda$-systems, their union is also in $\LL$. > [!theorem] Dynkin's $\pi$-$\lambda$ Theorem > Let $\mathcal{P}$ be a $\pi$-system, and $\LL$ be a $\lambda$-system. If $\mathcal{P} \subseteq \LL$, then $\sigma(\mathcal{P}) \subseteq \LL$. > [!proof]- > > > [!claim] > > The intersection of two $\lambda$-systems is a $\lambda$-system: > > > > 1. $\Omega$ lies in both. > > 2. If $A \subset B$ lie in both, then $B\setminus A$ lies in both. > > 3. If $A_1, A_2, \dots$ lie in both, then $A_1 \cup A_2 \cup \dots$ lie in both. > > To prove the theorem, let $\LL$ be the smallest $\lambda$-system containing $\mathcal{P}$, which can be obtained as the intersection of all $\lambda$-systems containing $\mathcal{P}$ (via Zorn's Lemma for instance). We will show that $\LL$ is a $\pi$-system, as this would show it is a $\sigma$-field. > > The trick is to let > $ > \LL' = \{ B\in \LL | A\cap B \in \LL \text{ for all $A\in \mathcal{P}$}\} > $ > It is clear that $\LL'$ is a $\lambda$-algebra. It is also clear that $\LL\subset \mathcal{P}$. Thus $\LL' = \LL$! Then, let > $ > \LL'' = \{ B\in \LL | A\cap B \in \LL \text{ for all $A\in \LL$}\} > $ > Then, $\mathcal{P}\subset \LL''$ because $\LL = \LL'$. Thus $\LL'' = \LL$ as desired. > [!idea] > Measure-theoretic proofs are difficult because we don’t have a closed form for all measurable subsets. Instead, we must first verify that a certain subset $\mathcal{P}$ has a property $\LL$ that we want. Then, Dynkin booststraps the results from $\mathcal{P}$ to its $\sigma$-algebra. A related theorem: [[Monotone Class Theorem]] (for sets).