Showing $\mathbb{E}[X_T]=\mathbb{E}[X_0]$ for a martingale and a finite stopping time $T$

Question

I have been posed the following:

Let $X = (X)_{n≥0}$ be a martingale and $T$ be a finite stopping time. Suppose $X_T$ is integrable. Show that $\mathbb{E}[X_T] = \mathbb{E}[X_0]$ if and only if $\lim_{n \to \infty} \mathbb{E}[X_n:T>n] = 0$

I'm familiar with similar problems, like for when $X$ is bounded by some random variable, but this has stumped me. Any hints appreciated.

Answer 1

Here is my attampt;

Since $T<\infty$ a.s. and $X_T$ is integrable, $|X_T|\chi_\left \{ \left. T>n \right \} \right.\rightarrow 0$ a.s., as $n\rightarrow \infty.$ Moreover $|X_T|\chi_\left \{ \left. T>n \right \} \right.\leq|X_T|,$ hence $\lim_{n\rightarrow \infty}\mathbb{E}\left [ |X_T| \chi_\left \{ \left. T>n \right \} \right. \right ] =\mathbb{E}\left [ \lim_{n\rightarrow \infty}|X_T| \chi_\left \{ \left. T>n \right \} \right. \right ] =\mathbb{E}[0]=0$ by Lebesgue's Theorem.

Now we have $$\mathbb{E}\left [ |X_{T\wedge n}-X_T| ] =\mathbb{E} [ |X_{T\wedge n}-X_T|\chi_\left \{ \left. T>n \right \} \right. \right ] +\mathbb{E}\left [ |X_{T\wedge n}-X_T|\chi_\left \{ \left. T\leq n \right \} \right. \right ] \\ =\mathbb{E} [ |X_{n}-X_T|\chi_\left \{ \left. T>n \right \} \right. ] \\ \leq \mathbb{E}\left [ |X_{n}|\chi_\left \{ \left. T>n \right \} \right. \right ]+ \mathbb{E}\left [ |X_{T}|\chi_\left \{ \left. T>n \right \} \right. \right ] \rightarrow 0$$ as $n\rightarrow \infty$ by assumption and previous part.

Note that ${T\wedge n}$ is bounded stoping time , hence $\mathbb{E}\left [ X_{T\wedge n} ] =\mathbb{E} [ X_{0} \right ]$ by optional sampling theorem. Finally we have $$ |\mathbb{E}\left [ X_{T} ] -\mathbb{E} [ X_{0} \right ]|=|\mathbb{E}\left [ X_{T} ] -\mathbb{E} [ X_{T\wedge n} \right ]| \\ \leq \mathbb{E}\left [ |X_{T} - X_{T\wedge n} \right| ]\rightarrow 0$$ as $n\rightarrow \infty$ by previous part. Hence $ \mathbb{E}\left [ X_{T} ] =\mathbb{E} [ X_{0} \right ]$