Let $X_0 = 0$ and for $i \geq 0$, let $X_{i+1}$ be chosen uniformly over the interval $[X_i, 1]$. prove that $M_n = 2^n (1-X_n)$ for $n \geq 0$ is a martingale, and prove that $M_n$ converges to a finite valued limit.
I am confused about how to even prove that $M_n$ is a martingale, because isn't $E|M_n| = 2^n E|1-X_n| \rightarrow\infty$ as $n\rightarrow\infty$?
Edit: Say I want to use the martingale convergence theorem to prove that $M_n$ converges to a finite valued limit. How can I show that $\sup_n E|M_n| < \infty$, when $E|M_n| < 2^n$? Does the supremum not exist?
hint:
$$E(M_{n+1}|X_1,\cdots ,X_n)=E(2^{n+1} (1-X_{n+1})|X_1,\cdots ,X_n)$$
$$=E(2^{n+1} (1-X_{n+1})|X_1,\cdots ,X_n)=2^{n+1} (1-E(X_{n+1}|X_1,\cdots ,X_n))$$
$$=2^{n+1} (1-\frac{X_n+1}{2})=M_n$$
Since this martingale is positive so it converge, non-negative martingale converge Corollary 2.3.,or Corollary 5.3.5. you do not need go for checking uniformly intgrablity .