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/$M_0=0$.
- Let $T_n=\inf_{t\geq 0}\{|M_t|=n\}$, show that $$\{\lim_{t\to\infty}M_t\ \text{exists and is finite}\}=\bigcup_{n\geq 1}\{T_n=\infty\}\subseteq\{\langle M,M\rangle_\infty<\infty\},\ \text{almost surely}.$$
- 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.