Let M be a continuous local martingale starting from $0$. Show that M is an $L^2$-bounded ($\displaystyle \sup_t\|M_t\|_2<\infty$) martingale if $\mathbb{E}([M]_{\infty})<\infty$, where $[M]_{\infty}=\displaystyle \lim_{t \to \infty} [M]_t$ and $[M]$ is the unique continuous adapted non-decreasing process such that $M^2-[M]$ is a continuous local martingale. Could do with some hints.
My thoughts: I know that any local martingale bounded by an integrable random variable is a true martingale and that $\displaystyle \sup_n\mathbb{E}[M^2_{T_n}] < \infty$, where $T_n$ are the stopping times reducing $M^2-[M]$. What I am finding hard is to make any conclusions about $M_t$, given that I have information about $M_{T_n}$ only.
Edit: Managed to show that $M_t \to M_{\infty}$ in $L^2$ for some $M_{\infty}$
Best Answer
For completeness, I will include the proof from Revuz, Yor "Continuous martingales and Brownian Motion" here.
$M-[M]$ is a local martingale, so there exists ${T_n},n\ge1$ such that $T_n\to\infty$ and $\mathbb{E}[M^2_{t\wedge T_n}]=\mathbb{E}([M]_{t\wedge T_n})<K$. Apply Fatou: $\mathbb{E}[M^2_t]\leq \liminf\mathbb{E}[[M]_{t\wedge T_n}]\leq K$. So $M$ is $L^2$-bounded.
Need to show now that $M$ is a true martingale. Let $T^{\prime}_n$ be stopping times reducing $M$, then $\{M_{t\wedge T^{\prime}_n}\}$ is UI, since it is $L^2$-bounded and $M_{t\wedge T^{\prime}_n} \to M_t$ in $L^1$, and so $\mathbb{E}[M_{t}|\mathcal{F_s}]=M_s$ a.s.
Edit
The omitted steps are:
$\mathbb{E}[M_{t\wedge T^{\prime}_n}|\mathcal{F}_s]=M_{s\wedge T^{\prime}_n}$, so $\mathbb{E}[M_{t\wedge T^{\prime}_n}\mathbb{1}_A]=\mathbb{E}[M_{s\wedge T^{\prime}_n}\mathbb{1}_A]$ for all $A \in \mathcal{F}_s$. Now take $n\to \infty$, to conclude $\mathbb{E}[M_{t}\mathbb{1}_A]=\mathbb{E}[M_{s}\mathbb{1}_A]$ for all $A \in \mathcal{F}_s$, hence $\mathbb{E}[M_t|\mathcal{F}_s]=M_s$ a.s.