Let $(\Omega,\mathcal A,\operatorname P)$ be a probability space, $(\mathcal F_n)_{n\in\mathbb N_0}$ be a filtration on $(\Omega,\mathcal A)$, $\mathcal F_\infty:=\sigma(\mathcal F_n:n\in\mathbb N_0)$, $p\ge1$, $(M_n)_{n\in\mathbb N_0}$ be an $L^p$-bounded $(\mathcal F_n)_{n\in\mathbb N_0}$-martingale on $(\Omega,\mathcal A,\operatorname P)$ and $M_\infty$ be an $\mathcal F_\infty$-measurable $L^p$-integrable random variable on $(\Omega,\mathcal A,\operatorname P)$ with $$\left\|M_n-M_\infty\right\|_{L^p(\operatorname P)}\xrightarrow{n\to\infty}0\tag1.$$
How can we show that there is a $\operatorname P$-null set $A$ such that $(M_n(\omega))_{n\in\mathbb N}$ is Cauchy for all $\omega\in\Omega\setminus A$?
My idea is as follows: Let $\varepsilon>0$. By $(1)$, there is a $n_0\in\mathbb N_0$ with $$\left\|M_n-M_\infty\right\|_{L^p(\operatorname P)}<\frac\varepsilon2\tag2$$ for all $n \ge n_0$, and hence $$\left\|\underbrace{M_n-M_{n_0}}_{=:\:N_n}\right\|_{L^p(\operatorname P)}<\varepsilon\tag3$$ for all $n\ge n_0$. $(N_n)_{n\ge n_0}$ is a martingale and hence $$\forall\lambda>0:\operatorname P\left[\sup_{n\ge n_0}|N_n|\ge\lambda\right]\le\frac1{\lambda^p}\sup_{n\ge n_0}\left\|N_n\right\|_{L^p(\operatorname P)}^p<\left(\frac\varepsilon\lambda\right)^p\tag4$$ by Doob's martingale inequality.
Now I thought we might be able to use $(4)$ to show that $$\operatorname P\left[\sup_{n\ge n_0}|N_n|>0\right]<c\varepsilon\tag4$$ for some $c\ge 0$, but I'm not sure if this is possible. We may clearly note that $\{0\}=\bigcap_{n\in\mathbb N}\left(0,\frac1n\right)$ (or $(0,\infty)=\bigcup_{n\in\mathbb N}\left(\frac1n,\infty\right)$ if you prefer) and hence, if $\mu$ is any finite measure, $\mu(\{0\})=\lim_{n\to\infty}\mu\left(\left(0,\frac1n\right)\right)$ (or $\mu((0,\infty))=\lim_{n\to\infty}\mu\left(\left(\frac1n,\infty\right)\right)$ if you prefer).
But that doesn't help.
Best Answer
You are on the right track. Combining (2) and (4), the triangle inequality $|M_n-M_\infty| \le |M_{n_0}-M_\infty|+|N_n|$ and a union bound give $$ \operatorname P\left[\sup_{n\ge n_0}|M_n-M_\infty|\ge 2\lambda\right]\le 2 \left(\frac\varepsilon\lambda\right)^p\,.$$ Therefore, $$ \operatorname P\left[\limsup_{n \to \infty}|M_n-M_\infty|\ge 2\lambda\right]\le 2 \left(\frac\varepsilon\lambda\right)^p\,.$$ Since this holds for all $\epsilon > 0$, the probability on the left is zero for any $\lambda>0$. Taking a union over rational $\lambda>0$ concludes the proof.