# Background

Hello, I'm working on question 4.24 on Le-Gall's Brownian motion(...) and I would ask you to check if my ideas are correct. The question is as follows:

$$(M_t)$$ is a cont. local martingale w/$$M_0=0$$.

1. Let $$T_n=\inf_{t\geq 0}\{|M_t|=n\}$$, show that $$\{\lim_{t\to\infty}M_t\ \text{exists and is finite}\}=\bigcup_{n\geq 1}\{T_n=\infty\}\subseteq\{\langle M,M\rangle_\infty<\infty\},\ \text{almost surely}.$$
2. Let $$S_n=\inf_{t\geq 0}\{\langle M,M\rangle_t=n\}$$, show that $$\{\langle M,M\rangle_\infty<\infty\}=\bigcup_{n\geq 1}\{S_n=\infty\}\subseteq\{\lim_{t\to\infty}M_t\ \text{exists and is finite}\}.$$ Conclude that $$\{\lim_{t\to\infty}M_t\ \text{exists and is finite}\}=\{\langle M,M\rangle_\infty<\infty\}$$ almost surely.

Here $$\langle M,M\rangle_t$$ denotes the quadratic variation of $$(M_t)$$.

# My progress

So I worked on part 1 on the most natural way I could think of:

Let $$\omega\in\{\lim_{t\to\infty}M_t\ \text{exists and is finite}\}$$, then $$M_\infty(\omega)=\lim_{t\to\infty}M_t(\omega)<\infty.$$ Now since $$(M_t)$$ has cont. sample paths, $$|M_t(\omega)|$$ is bounded by some $$C>0$$. Next $$T_m(\omega)=\infty$$ for all $$m>C$$ since the event $$|M_t(\omega)|=m>C$$ never occurs. Then $$\omega\in\{T_m=\infty\}$$ for $$m>C$$ and with this we have proven the first inclusion $$\{\lim_{t\to\infty}M_t\ \text{exists and is finite}\}\subseteq\bigcup_{n\geq 1}\{T_n=\infty\}.$$ I'm stuck on the other side, I take an $$\omega\in\{T_m=\infty\}$$ for some $$m\geq 1$$ and therefore $$\omega\in\{T_n=\infty\}$$ for $$n\geq m$$, since $$M_t$$ has cont. sample paths.

This last statement implies that $$M_t(\omega)$$ is bounded but I cannot reach the fact that the limit exists since I feel that $$M_t(\omega)$$ could oscillate wildly and therefore never reach a limit.

Also on the flipside if I want to show that such $$\omega$$ is in $$\{\langle M,M\rangle_\infty<\infty\}$$ I would like to use the fact that for bounded (true) martingales in $$L^2$$ it occurs that $$E\langle M,M\rangle_\infty<\infty$$. However, mine is not a true martingale but a cont. local martingale. This is theorem 4.13 on Le-Gall's book.

I don't know how to prove this fact without using such theorem.

With the same strategy as before I can prove $$\{\langle M,M\rangle_\infty<\infty\}\subseteq\bigcup_{n\geq 1}\{S_n=\infty\}.$$

EDIT1: The same problem does not occur in the other inclusion. since I don't know if $$M_t(\omega)$$ has a limit by knowing that $$\langle M,M\rangle_t$$ is bounded. Since $$\langle M,M\rangle_t$$ is an increasing process and it's bounded then it converges to a limit. Therefor it follows that the set and the union are equal.

On the final inclusion I would like to use again the fact that $$EM_\infty^2=E\langle M,M\rangle_\infty$$ but once more this is only valid for bounded martingales in $$L^2$$.

# Questions

Is there something I'm not seeing or I'm overlooking? Can you help me see it more clearly or give me a pointer in the right direction?

Is my idea on proving the directions I proved correct?

Any kind of help will be greatly appreciated.

• Some thoughts on the first question. I believe the idea is to formalize the following: on the event $\bigcup_n \lbrace T_n = \infty \rbrace$, $M$ is a bounded local martingale hence it is a true martingale and it converges. Further, $\langle M,M\rangle_\infty < \infty$ by what you said. Maybe someone else could confirm this? Jun 1, 2020 at 21:47

For showing $$\bigcup_{n\geq 1}\{T_n=\infty\}\subseteq\{\lim_{t\to\infty}M_t\ \text{exists and is finite}\}$$ use that for the stopped process $$|M^{T_n}_t|\leq n$$ holds. Due to Doobs optional stopping theorem $$Z^n_t:=M^{T_n}_t$$ is still a continous local maringale. It is even a ture martingle, since $$E\sup_{s\leq t}|Z^n_s|<\infty$$ and by the submartingle convergence theorem follows, that $$Z^n$$ is convergent. Now, look at the paths, where $$T_n=\infty$$ and the statement follows. Furthermore the process $$Z^n$$ is in $$L^2$$, so your argument will work with $$\langle Z^n,Z^n\rangle=\langle M,M\rangle^{T_n}$$. And since $$n\in\mathbb{N}$$ is countable, you find a set of $$\omega$$ with measure $$1$$ and independent of $$n$$ so that for all $$n\in\mathbb{N}$$ $$\langle M,M\rangle^{T_n}_\infty<\infty$$ holds.
For the second part, how to show $$Q:=\{\langle M,M\rangle_\infty<\infty\}\subseteq\{\lim_{t\to\infty}M_t\ \text{exists and is finite}\},$$ simply consider the process $$Q^n:=M^{S_n}$$. Since $$Q^n$$ is a martingale in $$L^2$$, $$\sup_{t}E|Q^n_t|<\infty$$ holds and thus $$Q^n$$ converges a.s.