# $L^1$-bounded quadratic variation of a continuous local martingale $\implies$ it is a true martingale, $L^2$-bounded

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 $M_{\infty}$

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You can find the solution in Revuz, Yor "Continuous martingales and Brownian Motion" Proposition IV.1.23 p. 129-130. google books link

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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.

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We need to 1) show $M$ is $L^2$-bounded and 2) show that $M$ is a martingale.

1) We are dealing with two local martingales, $M$ and $M^2 - [M]$, so we may as well choose a localizing sequence which reduces both of them. Call this $(T_n)_{n\in\mathbb{N}}$. Take $t\ge 0$.

\begin{align}\mathbb{E}(M_t^2) &= \mathbb{E}\left(\lim_{n\to\infty}M^2_{t\wedge T_n}\right)\\ &\leq \liminf_n\mathbb{E}(M^2_{t\wedge T_n})\\ &=\liminf_n\mathbb{E}([M]_{t\wedge T_n})\\ &=\mathbb{E}([M]_t)\\ &\leq \mathbb{E}([M]_\infty) \end{align}

So $M$ is $L^2$-bounded.

2) There exist continuous, $L^2$-bounded local martingales, so we'll need more than 1) to show $M$ is a martingale. $M$ is integrable by 1) and Jensen's inequality. Take $0\leq s\leq t$.

\begin{align} \mathbb{E}(M_t|\mathcal{F}_s)&=\mathbb{E}\left(\lim_{n\to\infty}M_{t\wedge T_n}|\mathcal{F}_s\right) \end{align}

So we must show that $\{M_{t\wedge T_n}:n\in\mathbb{N}\}$ is UI. But this is clear, since we showed they were $L^2$-bounded in 1).

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