Counterexample Doob's convergence (in $\mathcal{L}^1$) theorem I

Question

Doob's martingale convergence theorem I yields as follows:

Let X be a supermartingale which is bounded in $\mathcal{L}^1$, i.e. $\sup_{n\in \mathbb{N}} \mathbb{E}||X_n|| <\infty$. Then there exists a random variable $X_\infty \in \mathcal{L}^1(\Omega,\mathcal{F},\mathbb{P})$ such that $X_n \xrightarrow{\text{a.s.}} X_\infty$.

In order to show that the theorem only states that a.s. convergence holds, no other type of convergence, except convergence in probability, we define for $n \in \mathbb{N}_0$ and $k \in \{0,1,\ldots,2^n-1\}$ \begin{align} I^n_k = \bigg[\frac{k}{2^n},\frac{k+1}{2^{n+1}} \bigg). \end{align} Let $(\Omega,\mathcal{F},\mathbb{P}) = ([0,1],\mathcal{B}([0,1]), \lambda|_{\mathcal{B}([0,1])})$ and $\mathcal{F}_n = \sigma(\{I_n^k: k\in \{0,1,\ldots,2^n-1\}\})$ and define $X_n = \mathbb{1}_{I^n_0} 2^n$.

It could be show that $X_n$ is a $\mathcal{F}_n$-martingale and that the conditions for the above theorem are satisfied. What is $X_\infty$ in this case and why does $X_n \xrightarrow{\mathcal{L}^1} X_\infty$ not hold?

Answer 1

Note that $I^n_0$ is the interval $[0, 2^{-n})$, hence $X_n$ is zero on the intervall $[2^{-n}, 1]$. Therefore $X_n \to 0$ almost surely. Due to $$ \int_\Omega |X_n| \, d\mathbf P = 2^n \lambda\bigl([0, 2^{-n})\bigr) = 1 \not\to 0. $$ We have $X_n \not\to 0$ in $L^1(\mathbf P)$.