About continuous local martingales, question on Le-Gall's book

Question

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.

Answer 1

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.