# Is there a problem of the proof about local martingale is Martingale?

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.

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

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 $$t$$.
• 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? Mar 27, 2022 at 2:17
• 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 $\|X^{T_n}_t-X_t\|_1=0$. Mar 27, 2022 at 16:50