Let $(\Omega, \cf, \bp, \bracs{\cf_t})$ be a [[Filtration|filtered]] [[Probability|probability]] space and $\bracs{X_t}$ be a $\bracs{\cf_t}$-(sub)[[Martingale|martingale]] with [[RCLL]] sample paths. > [!theorem] Martingale Convergence Theorem (I) > > If > $ > \sup_{t \ge 0}\ev\braks{X_t^+} < \infty > $ > then there exists a [[Random Variable|random variable]] $X_\infty \in L^1$ such that $X_t \to X_\infty$ as $t \to \infty$ [[Almost Everywhere|a.s.]] > > *Proof*. Let $[a, b] \subset \real$. Let $\tau_0 = 0$, $\tau_1 = \inf\bracs{t \ge 0: X_t < a}$, $\tau_2 = \inf\bracs{t \ge \tau_1, X_t > b}$. Then iteratively define $\tau_{2k} = \inf\bracs{t \ge \tau_{2k - 1}: X_t > b}$ and $\tau_{2k - 1} = \inf\bracs{t \ge \tau_{2k - 2}: X_t < a}$. Since each $\tau_k$ is a hitting time of an open set, they are all optional times. > > Define > $ > U_{[a, b]} = \lim_{T \to \infty}\max_{k}\bracs{k: \tau_{2k} \le T} > $ > as the total number of upcrossings of $[a, b]$. For each $n \in \nat$, define $\{Y_m^{(n)} = X_{m2^{-n}}: m \ge 0\}$, then $\{Y_m^{(n)}\}_1^\infty$ is a discrete-time (sub)martingale. Since $X_t$ has RCLL sample paths, if $U_{[a, b]}^{(n)}$ is the total number of upcrossings of $[a, b]$ by $Y^{(n)}$, then $U_{[a, b]}^{(n)} \upto U_{[a, b]}$ as $n \upto \infty$. > > By the [[Monotone Convergence Theorem for Sequences|monotone convergence theorem]], > $ > \ev\braks{U_{[a, b]}} = \limv{n}\ev\braks{U_{[a, b]}^{(n)}} \le \limv{n}\frac{\sup_{m \ge 0}\ev\braks{X_{m2^{-n}}^+}}{[a, b]} \le \frac{\sup_{t \ge 0}\ev\braks{X_{t}^+}}{[a, b]} > $ > If the supremum is finite, then $U_{[a, b]}$ is finite almost surely. Thus there exists a certain set on which $X_t$ only upcrosses each interval with rational endpoints finitely many times. Therefore $X_t$ converges almost surely. > > Let $t \ge 0$, then > $ > \ev\braks{\abs{X_t}} = \ev\braks{X_t^+} + \ev\braks{X_t^-} = \ev\braks{X_t^+} - \ev\braks{X_t} > $ > Since $X_t$ is a (sub)martingale, $\ev\braks{X_t} \ge \ev\braks{X_0}$, so > $ > \ev\braks{\abs{X_t}} \le 2\ev\braks{X_t^+} - \ev\braks{X_0} > $ > and the process is bounded in $L^1$. By [[Fatou's Lemma]], > $ > \ev\braks{\abs{X_\infty}} \le \liminf_{t \to \infty}2\ev\braks{X_t^+} < \infty > $ > [!theorem] Martingale Convergence Theorem (II) > > If $X$ is [[Uniformly Integrable|uniformly integrable]], that is, > $ > \lim_{A \to \infty}\sup_{t \ge 0}\ev\braks{\abs{X_t}; \abs{X_t} \ge A} = 0 > $ > then $\sup_{t \ge 0}\ev\braks{\abs{X_t}} < \infty$ and there exists $X_\infty \in L^1$ such that $X_t \to X_\infty$ almost surely and in [[Integrable Function|L1]]. > > *Proof*. Let $X_\infty$ be as in the previous theorem. Suppose that $X_t \not \to X_\infty$ in $L^1$, and let $\seq{t_n}$ be a sequence of times such that $t_n \upto \infty$ and $\eps > 0$ such that $\norm{X_{t_n} - X_\infty}_1 \ge \eps$. Meanwhile, $X_{t_n} \to X_\infty$ almost surely and the sequence is uniformly integrable. From the discrete result, $X_{t_n} \to X_\infty$ in $L^1$, which contradicts the assumption. > [!theorem] Martingale Convergence Theorem (III) > > Suppose that $\abs{X}$ is a submartingale, and there exists $p > 1$ such that > $ > \sup_{t \ge 0}\norm{X}_p < \infty > $ > then > $ > \sup_{t \ge 0}\ev\braks{\abs{X_t}; \abs{X_t} > A} \le \sup_{t \ge 0}\frac{\norm{X}_p}{A^{p - 1}} \to 0 > $ > as $A \to \infty$, so $X$ is uniformly integrable. Then $X \to X_\infty$ in $L^p$. > > *Proof*. By [[Fatou's Lemma]], > $ > \norm{X_\infty}_p \le \liminf_{t \to \infty}\norm{X}_p\le \sup_{t \ge 0}\norm{X}_p > $ > Now, > $ > \begin{align*} > \ev\braks{\sup_{t \ge 0}\abs{X_t}^p} &= p\int s^{p - 1}\bp\bracs{\sup_{t \ge 0}\abs{X_t} > s}ds \\ > &\le p\int s^{p - 2}\ev\braks{\abs{X_\infty}: \sup_{t \ge 0}X_t > s}ds\\ > &= p\int s^{p - 2}\ev\braks{\abs{X_\infty} \cdot \one_{[0, \sup_{t \ge 0}\abs{X_t}]}}ds \\ > &= \frac{p}{p - 1}\ev\braks{\abs{X_\infty} \cdot \paren{\sup_{t \ge 0}\abs{X_t}^{p - 1}}} \\ > &\le \frac{p}{p - 1}\ev\braks{\abs{X_\infty}^{p}}^{1/p} \cdot \ev\braks{\sup_{t \ge 0}\abs{X_t}^p}^{1 - 1/p} > \end{align*} > $ > moving the terms around at the end gives that $\norm{\sup_{t \ge 0}\abs{X_t}}_p < \infty$. By the [[Dominated Convergence Theorem]], $\norm{X_t - X_\infty}_p \to 0$.