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  1. Let $M=(M_t)$ be a continuous local martingale, i.e. it exists a sequences of stopping times $(\tau_n)$ converging to $+\infty$ $P-$a.s. such that $M^{\tau_n}=(M_{\tau_n\wedge t})$ is a martingale. Now I was able to show that that every continuous adapted process $X$ is locally bounded, i.e. there exists a sequence of stopping times $\sigma_n$ such that $(X_{\sigma_n\wedge t})$ is bounded by a constant depending on $n$. Applying this result to the continuous local martingale $M$, I want to show that $\rho_n:=\tau_n\wedge\sigma_n$ is a stopping time (clear), it converges to $+\infty$ $P-$a.s. such that the continuous local martingale $M$ becomes a bounded martingale, i.e. $(M_{\rho_n\wedge t})$ is a bounded martingale. Boundedness is clear, but why is it again a martingale for this new sequences?

  2. Let $M$ be a continuous local martingale, $L^2_{loc}(M)$ be the space of all predictable process (equivalence clases) such that there exists a sequence of stopping times $\tau_n$ converging to $+\infty$ $P-$a.s. such that for all $n$ we have $E[\int_0^{\tau_n}H_s^2d\langle M\rangle_s]<\infty$, where $\langle M\rangle $ is the bracket process of $M$. I want to show that $L^2_{loc}=\{\mbox{all pred. process }H;\int_0^tH_s^2d\langle M \rangle_s<\infty \mbox{ }P-a.s. \mbox{ for all }t\ge 0\}$. For the inclusion "$\supset$" I have the following approach: define $\tau_n:=\inf\{t\ge 0;\int_0^t H^2_s d\langle M\rangle_s > n\}$. Is this really a stopping time? If $\{\tau_n\le t\}=\{\int_0 ^t H_s^2 d\langle M\rangle_s > n\}$ then it's clear, since $\int_0^t\cdots$ is $\mathcal{F}_t$ measurable. Are these to sets really equal? Assuming they are, we have $\tau_n\to +\infty$ $P-$a.s. since $\int_0 ^t H_s^2 d\langle M\rangle_s < \infty$ for all $t\ge 0$ and with the usual convention that Infimum over an empty set is defined as $+\infty$ (in the stopping time sense). About the other inclusion I have no idea.

  3. Why is every locally bounded predictable process $H$ in $L^2_{loc}(M)$ for every continuous local martingale $M$

Thank you for your help!

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Regarding 1., you know that $X=M^{\tau_n}$ is a martingale and, for $\tau=\rho_n$, you ask a reason why $X^\tau=(X_{\tau\wedge t})_t$ is a martingale? –  Did Jul 15 '12 at 9:26
    
@ did: by the definition of local martingale, I know there exists (maybe) just one sequence of stopping time $(\tau_n)$ such that $M^{\tau_n}$ is a martingale. Now my question is simple, why is this true for the new sequence of stopping times $\rho_n:=\tau_n\wedge \sigma_n$ –  user20869 Jul 15 '12 at 9:48
    
You did not read my comment. If you had, you would have seen that there is no local martingale in the picture anymore, only bona fide martingales. (Unrelated: adding a space between @ and one's name effectively prevents the comment to be signaled to the user (I saw your comment by chance).) –  Did Jul 15 '12 at 14:23
    
@did: Thx for pointing out about the space between @ and the name. I didn't know that. Maybe I do not understand your comment. I know that there exists a localizing sequence. Why do I know that another sequence of stopping times (converging to $\infty$ P-a.s.) still works, i.e. is also a localizing sequence? In fact, can I choose any sequence of stopping times converging to $\infty$ P-a.s. sucht that the stopped process is a martinagle? –  user20869 Jul 16 '12 at 10:08
    
Because $\rho_n\leqslant\tau_n$, hence $M^{\rho_n}$ is obtained by stopping the (true) martingale $M^{\tau_n}$ and $M^{\rho_n}$ is (trivially) also a (true) martingale. By the way, without the condition $\rho_n\leqslant\tau_n$, my first comment is wrong hence, simply checking this comment would have led you to the solution. –  Did Jul 16 '12 at 10:15

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