I am working on this exercise:
Let $(X_{n},\mathcal{F}_{n})$ be a martingale and $\tau$ a $\mathcal{F}_{n}$ stopping time that is almost surely finite. Further assume that $\mathbb{E}|X_{\tau}|<\infty$ and $$\lim_{n\rightarrow\infty}\mathbb{E}|X_{n}|\mathbb{1}_{\tau>n}=0.$$
(a) Prove that $\lim_{n\rightarrow\infty}\mathbb{E}|X_{\tau}|\mathbb{1}_{\tau>n}=0$.
(b) Show that $\lim_{n\rightarrow\infty}\mathbb{E}|X_{\tau\land n}-X_{\tau}|=0$.
(c) Conclude that $\mathbb{E}(X_{\tau})=\mathbb{E}(X_{0}).$
I have solved part (b) and part (c). They follow immediately once we prove (a), since for part (b), just decompose $$X_{\tau}-X_{\tau\land n}=X_{\tau}-X_{\tau}\mathbb{1}_{\tau\leq n}-X_{n}\mathbb{1}_{\tau>n}=X_{\tau}(1-\mathbb{1}_{\tau\leq n})-X_{n}\mathbb{1}_{\tau>n}=X_{\tau}\mathbb{1}_{\tau>n}-X_{n}\mathbb{1}_{\tau>n}.$$
Then taking absolute value gives us the triangle inequality and taking expectation and limit yields us the desired results which follows from part (a) and the hypothesis.
Then (c) follows immediately since by part (b), we know that $$\lim_{n\rightarrow\infty}\mathbb{E}X_{\tau\land n}=\mathbb{E}X_{\tau},$$ but since $(X_{n},\mathcal{F}_{n})$ is a MG and $\tau$ is a $\mathcal{F}_{n}-$stopping time, we know that $(X_{n\land\tau},\mathcal{F}_{n})$ is also a MG. Therefore, $\mathbb{E}(X_{n\land\tau})=\mathbb{E}X_{0},$ for all $n$ and the result follows immediately.
However, I got in part (a) for a while.
My idea is still to decompose $X_{\tau\land n}$ into $$X_{\tau\land n}=X_{\tau}\mathbb{1}_{\tau\leq n}+X_{n}\mathbb{1}_{\tau>n}.$$
Then since $X_{\tau\land n}$ is a MG and $\tau<\infty$ a.s., we know that $\sup_{n}\mathbb{E}|X_{\tau\land n}|<\infty$. Thus it follows from Doob's Convergence Theorem that $X_{\tau\land n}\longrightarrow X_{\tau}$ a.s.
Then if we can show $(X_{\tau\land n})$ is uniformly integrable, we can then apply Vitali's Convergence theorem to show that $\mathbb{E}X_{\tau\land n}=\mathbb{E}X_{\tau}$.
Then, taking $n\rightarrow\infty$ in the decomposition, we have $$\mathbb{E}X_{\tau}=\lim_{n\rightarrow\infty}\mathbb{E}X_{\tau}\mathbb{1}_{\tau\leq n}+\mathbb{E}X_{n}\mathbb{1}_{\tau>n},$$ the second term on the RHS is $0$ by hypothesis, so moving the first term to the LHS yield us the desired result (without absolutely value).
However, there are two problems with this argument:
(1) I have no idea about how to show $(X_{n\land \tau})$ is uniformly integrable.
(2) The argument does not give $\lim_{n\rightarrow\infty}\mathbb{E}|X_{\tau}|\mathbb{1}_{\tau>n}=0,$ but $$\lim_{n\rightarrow\infty}\mathbb{E}X_{\tau}\mathbb{1}_{\tau>n}=0,$$ there is a difference of absolute value.
I guess the second problem is easier to fix since using $|\cdot|$ convex we can argue over sub-MG, it is fine since Doob can work for all kinds of MG.
However, I have no idea about how to show the uniform integrability.
Am I heading the wrong direction? By the way, it is really appreciated if one could check my proof of part (b) and (c).
Thank you so much!
Due to our assumption that $\tau < \infty$ a.s, we have that as $n \to \infty$, $1_{\tau > n} \to 0$ on the set $\tau <\infty$ which is of measure $1$, so $|X_\tau 1_{\tau >n}|$ converges to $0$ a.s when $n \to \infty$.
Moreover, $|X_\tau 1_{\tau > n}| \le |X_\tau|$, which is integrable (again - assumption), hence by Dominated Convergence Theorem, (a) follows.
$(b),(c)$ are okay, good idea to decompose