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Is the following argument correct:

Let $(M_t)_{t\geq 0}$ be a local martingale s.t. for some $p>1$ $E[\sup_{s \leq t} |M_s|^p]<\infty$ for all $t \geq 0$. Then $E[\sup_{s \leq t} |M_s|]\leq 1+ E[\sup_{s \leq t} |M_s|^p] <\infty$

Now using $E[\sup_{s \leq t} |M_s|]< \infty$ gives is the majorant such that that $M_t^{T_n}$ (the localized martingale) converges to $M_t$ in $L^1$. For $A\in \mathcal{F}_s$ $E[1_A\cdot (M_t-M_s)]=\lim_{n\to \infty}E[1_A\cdot (M_t^n-M_s^n)]=0$ so $M_t$ true martingale by definition.

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up vote 1 down vote accepted

Yes, your argument is correct. Stronger results hold as well, for example, Theorem I.51 in Protter's book "Stochastic integration and differential equations" states that if $M$ is a local martingale such that $\sup_{s\le t}|M_s|$ is integrable for all $t\ge0$, then $M$ is a true martingale.

Note that a related proposition, however, is false: If $M$ is a local martingale with $\sup_{t\ge0}E|M_t|^p$ finite, it does not necessarily follow that $M$ is a true martingale. You can find a counterexample in Section 23.5 of Stoyanov's "Counterexamples in probability".

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