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I am currently reading a proof of the Levy characterization of Brownian motion, and it seems it uses the following result. Let $(M_t,\mathcal F_t)_{t\geq 0}$ be a continuous local martingale with quadratic variation $\langle M\rangle_t=t$. Then $M$ is a martingale. Why is this true?

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    $\begingroup$ Did you take a look at this duplicate question? $\endgroup$
    – saz
    Mar 5, 2019 at 15:14
  • $\begingroup$ @saz Thanks, I did not see the question before. The answer there isn't complete so I've added my own based on the comments there. $\endgroup$
    – Empty Set
    Mar 5, 2019 at 17:18

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Let $(S_n)_{n\in\Bbb N}$ be a localizing sequence of stopping times for both continuous local martingales $(M_t^2-t,\mathcal F_t)_{t\geq 0}$ and $(M_t,\mathcal F_t)_{t\geq 0}$. Fix $0\leq s< t$. We have \begin{align} \Bbb E[\sup_{0\leq r\leq t}M_{r}^2]=&\Bbb E[\lim_{n\rightarrow\infty}\sup_{0\leq r\leq t}M_{r\wedge S_n}^2]\\ \leq&\limsup_{n\rightarrow\infty}\Bbb E[\sup_{0\leq r\leq t}M_{r\wedge S_n}^2]\\ \leq&4\limsup_{n\rightarrow\infty}\Bbb E[M_{t\wedge S_n}^2]\\ =&4\limsup_{n\rightarrow\infty}\Bbb E[t\wedge S_n]\\ =&4t<\infty \end{align} The first inequality follows from Fatou and the second from Doob's inequality applied to the continuous non-negative submartingale $(|M|_{t\wedge S_n},\mathcal F_t)_{t\geq 0}$. Thus also $\Bbb E[\sup_{0\leq r\leq t}|M_{r}|]\leq 1+4t<\infty$. For $A\in\mathcal F_s$, we then have \begin{align} \Bbb E[(M_t-M_s)1_A]=&\Bbb E[\lim_{n\rightarrow\infty}(M_{t\wedge S_n}-M_{s\wedge S_n})1_A]\\ =&\lim_{n\rightarrow\infty}\Bbb E[(M_{t\wedge S_n}-M_{s\wedge S_n})1_A]\\ =&\lim_{n\rightarrow\infty}0=0\\ \end{align} Where the second inequality follows from dominated convergence (dominated by $2\sup_{0\leq r\leq t}|M_{r}|$). Thus $M$ is a square integrable martingale.

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    $\begingroup$ You showed that $M$ is a local martingale which is square integrable... to other readers it might not be obvious why this implies that $M$ is a martingale. $\endgroup$
    – saz
    Mar 5, 2019 at 17:41
  • $\begingroup$ @saz You are right, it is far from obvious. I've completely rewritten the proof. $\endgroup$
    – Empty Set
    Mar 5, 2019 at 18:43
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    $\begingroup$ Yes, looks better now. For the first computation it would be more natural to use $\liminf$ instead of $\limsup$ (...since you are using Fatou's lemma). $\endgroup$
    – saz
    Mar 5, 2019 at 18:48
  • $\begingroup$ @saz I think it's worth some clarification here, in case the discussion misleads readers. What is shown here is stronger than square integrability of $M$ (the answer gets an $L^2$ dominating function). This is worth pointing out since it should not be obvious to other readers that square integrable local martingales are martingales - that claim is false. There are $L^2$-bounded strict local martingales (see my answer here). $\endgroup$ Mar 5, 2019 at 20:22

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