$\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}$ > [!theorem] AS MCT > Let $X$ be an $L^1$-bounded supermartingale. Then, there exists an integrable $\FFF_\infty$-measurable random variable $X_\infty$ such that $X_n\to X_\infty$ almost surely as $n\to \infty$. > [!idea] This is obvious > Suppose you do not converge. Then, you contain an unbounded amount of [[upcrossings]]. This motivates us to perform strong bounds using the [[Doob's upcrossing inequality]], which turns out to kill immediately. >[!idea] This is weak >$X_\infty$ doesn't have to be in $L^1$, so things like $\EE[X_\infty]$ don't make sense yet. >[!proof]- > Let > $\Omega_\infty = \{\lim\inf \abs{X_n} < \infty\},\qquad \Omega_{a,b} = \{U[a,b] < \infty\}$ > for $a<b\in \QQ$. Then, let > $\Omega_0 = \Omega_\infty \cap \bigcap_{a < b\in \QQ} \Omega_{a,b}$ > Observe that $X_n(\omega)$ converges iff $\Omega \in \Omega_0$. By [[Convergence properties of Lebesgue Integral|Fatou]] and [[Doob's upcrossing inequality|Doob]], > $\EE[\lim\inf \abs{X_n}] \leq \lim\inf \EE\abs{X_n},\qquad (b-a)\EE[(U[a,b])] \leq \abs{a} + \sup_{n\geq 0} \EE\abs{X_n}$ > We said that $X$ was $L^1$-bounded though! So $\EE[U[a,b]]$ and $\EE[\lim\inf\abs{X_n}]$ is finite, and thus $U[a,b]$ and $\lim\inf \abs{X_n}$ are finite almost surely. Thus the countable union of $\Omega_{a,b}$ and $\Omega_0$ is also almost sure. > > So why not let $X_\infty = \lim_{n\to \infty} X_n \id_{\Omega_0}$ (pointwise)? Then $X_n \to X_\infty$ almost surely, $X_\infty$ is $\FFF_\infty$-measurable, and $\abs{X_\infty}\leq \lim\inf\abs{X_n}$, thus $X_\infty$ is integrable. Now, we use the [[L1 Convergence Theorem|UI bridge]] for the first (important) time. > [!theorem] $L^1$ MCT > Suppose $X$ is a UI martingale. Then, there exists $X_\infty \in L^1(\FFF_\infty)$ such that $X_n\to X_\infty$ almost surely and in $L^1$. > Then, $X_n = \EE(X_\infty | \FFF_n)$ almost surely for all $n\geq 0$. > [!idea]- We expect $X_\infty$ to look like this. Think, don't read this. > As a random process adapted to a filtration, $X$ branches like a tree; we pretend the tree terminates and has leaves. But if you know all the leaves, you also know all nodes of the tree; they're exactly the expected value over the sub-tree. It's completely unsurprising that this works to create martingales as well. > [!proof]- > By the AS MCT, we already know there exists an almost-sure limit $X_n\to X_\infty \in L^1(\FFF_\infty)$. By the [[L1 Convergence Theorem]], we win. > > Now, for all $m\geq n$, > $ > \norm{X_n - \EE[X_\infty | \FFF_n]}_1 = \norm{\EE(X_m - X_\infty | \FFF_n)}_1\leq \norm{X_m - X_\infty}_1 > $ > Taking the limit as $m\to \infty$, we find $\norm{X_n - \EE[X_\infty | \FFF_n]}_1 = 0$ as desired. %%[!part]- idiotic verification; this is obvious. In the below, $X$ is *not* $\FFF$ measurable. $ \begin{align*} \EE\left[\abs{\EE[X | \FFF]}\right] &= \EE\left[\left(\id_{\EE[X|\FFF] \geq 0} - \id_{\EE[X|\FFF] \geq 0}\right) \EE[X | \FFF]\right] \\ &= \EE\left[\EE[X \left(\id_{\EE[X|\FFF] \geq 0} - \id_{\EE[X|\FFF] \geq 0}\right)| \FFF]\right] \\ &= \EE\left[X \left(\id_{\EE[X|\FFF] \geq 0} - \id_{\EE[X|\FFF] \geq 0}\right)\right]\\ &= \EE\left[\abs{X}\right] + 2\EE[X \id_{X \geq 0\land \EE[X|\FFF] < 0}] - 2\EE[X \id_{X < 0\land \EE[X|\FFF] \geq 0}]\\ &\geq \EE[\abs{X}] \end{align*} $ %% > [!claim] > On the other hand, for all $Y\in L^1(\FFF_\infty)$, if I let $X_n = \EE[Y | \FFF_n]$, $(X_n)_{n\geq 0}$ is a UI martingale such that $X_n\to Y$ AS in and $L^1$. > [!proof]- > This is a martingale by the [[Conditional Expectation|tower property]], and [[Conditional Expectations are UI|obviously]] uniformly integrable. Thus we know it has an AS + $L^1$ limit; we just need to show its actually $Y$. This is clear; for all $A\in \FFF_n$, > $\EE(X_\infty \id_A) = \EE(X_n \id_A) = \EE(Y \id_A)$ > and we know the $\FFF_n$ generate $\FFF_\infty$, thus this is true for all $A\in \FFF_\infty$; thus $X_\infty = Y$ AS. The $L^p$ case looks almost identical, for $p\in (1, \infty)$. > [!theorem] $L^p$ MCT > Suppose $X$ is $L^p$ bounded. Then, there exists $X_\infty \in L^p(\FFF_\infty)$ such that $X_n\to X_\infty$ almost surely and in $L^p$. $X_n = \EE(X_\infty | \FFF_n)$ almost surely for all $n\geq 0$. >[!proof]- >By Doob, $X^*\in L^p$. Then $\EE[\abs{X_t - X_\infty}^p]$ is dominated by e.g. $\EE[(2X^*)^p]$ thus DCT works and $X_t\to X_\infty$ in $L^p$. Then $X_n = \EE[X_\infty | \FFF_n]$ because conditional expectation is convex for $L^p$. Also see the [[Backwards MCT]].