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In the definition of local martingale, an adapted process $M=(M_t)_{t\ge 0}$ with continuous sample paths and vanishing at $0$ is called a continuous local martingale if there exists a non-decreasing $T_n$ of stopping times such that $T_n\to \infty$ and for every $n$, the stopped process $M^{T_n}$ is a uniformly integrable martingale.

My question is that if $M$ is a local martingale, then it seems that I can show that it is a martingale. Although I know there is a counterexample that there is a local martingale but not a martingale. Can anyone take a look at my following proof and what's wrong with my proof?

Let $X_t$ ba a local martingale. Then $\{X^{T_n}_t\}=\{X_{T_n\land t}\}$ is u.i. Note that $$X_{T_n\land t}\to X_t, \, a.s. $$ Thus, it converges in probability, and thereby it also converges in $L^1$.

Then for any $A\in \mathcal{F}_s$ where $s\in [0,t]$ \begin{align*} E[X_t|A]=E[X_t; A]=\lim_nE[X_{T_n\land t}; A]&=\lim_nE[X_{T_n\land s}; A]\\ &=E[X_s; A] \end{align*} Hence, $E[X_t|\mathcal{F}_s]=X_s$. $X_t$ is thereby a martingale.

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  • $\begingroup$ Could you explain your $E[X_t ; A]$ notation and why it equals $E[X_t \mid A]$? $\endgroup$
    – angryavian
    Mar 14, 2022 at 22:30

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It is the collection $\{X^{T_n}_t: t\ge 0\}$ that is uniformly integrable, for each fixed $n$. This doesn't give $L^1$ convergence of $X^{T_n}_t$ to $X_t$.

Your expectation computation would work if you knew that the collection $\{X^{T_n}_t: n=1,2,\ldots\}$ was uniformly integrable, for fixed $t$.

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  • $\begingroup$ I am confused about your second part. What is difference between the collection $X_t^{T_n}$ is u.i. for fixed $t$ and for every $n$? Why $X_t^{T_n}$ is u.i. for fixed $t$ then we have $L^1$ convergence? $\endgroup$
    – Hermi
    Mar 27, 2022 at 2:17
  • $\begingroup$ If $\{X^{T_n}_t: n=1,2,\ldots\}$ is u.i. ($t$ fixed) then, because $\lim_nX^{T_n}_t=X_t$ a.s., you even have $\|X^{T_n}_t-X_t\|_1=0$. $\endgroup$ Mar 27, 2022 at 16:50

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