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.
Best Answer
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 $t$.