# Almost sure convergence in the martingale convergence theorem

Let

• $(\Omega,\mathcal A,\operatorname P)$ be a probability space
• $(\mathcal F_t)_{t\ge0}$ be a filtration of $\mathcal A$
• $X:\Omega\times[0,\infty)\to\mathbb R$ be a right-continuous square-integrable $\mathcal F$-martingale

We can show that $(X_n)_{n\in\mathbb N}$ is $L^2(\operatorname P)$-Cauchy and hence there is a unique $X_\infty\in L^2(\operatorname P)$ with $$\left\|X_n-X_\infty\right\|_{L^2(\operatorname P)}\xrightarrow{n\to\infty}0\;.\tag1$$ We can show that $$\sup_{t\ge n}\left|X_t-X_n\right|\xrightarrow{n\to\infty}0\;\;\;\text{in probability}\;.\tag2$$ How can we conclude $$X_t\xrightarrow{t\to\infty}X_\infty\;\;\;\text{almost surely}\tag3$$ from $(2)$?

Let $$Y_n\stackrel{\text{def}}=\sup_{t\ge n}|X_t-X_\infty|\le\sup_{t\ge n}|X_t-X_n|+|X_n-X_{\infty}|\overset{\mathsf{P}}{\underset{n\to\infty}\longrightarrow}0.$$ Meanwhile, $Y_n$ is decreasing in $n$ and convergence a.s. as $n\to\infty$, so $Y_n\downarrow 0$ a.s. and $X_t\to0$ a.s. as $t\to\infty$.