How to show that this process is a martingale Consider the probability space $([0, 1), \mathcal{B}[0, 1), \lambda)$, where $B[0, 1)$ are the Borel sets on $[0, 1)$ and $\lambda$ is the Lebesgue measure. Let
\begin{align*}
I_k^n = \left[ \frac{k}{2^n}, \frac{k + 1}{2^n} \right), \quad k = 0, \dots, 2^n - 1
\end{align*}
and let $\mathcal{F}_n$ be the $\sigma$-algebra generated by the $I_k^n$ for $k = 0, \dots, 2^n - 1$. Define 
\begin{align*}
X_n(\omega) = \left\{ \begin{array}{ c l }
2^n & \omega \in [0, 2^{-n}) \\
0 & \text{otherwise}
\end{array} \right.
\end{align*}
The problem I'm working out focuses on showing that $\mathbb{E}[X_{\infty} | \mathcal{F}_n] \not= X_n$ a.s., which I managed to do quite easily. The part I actually have a problem with is showing that $X$ is a martingale however. 
If anyone could provide me with some tips on how to do this, I would be very thankful! Cheers!
 A: Nevermind, I actually figured it out. For those interested, you need the following theorem: 
Let $(\Omega, \mathcal{F}, \mathbb{P})$ be a probability space and let $\mathcal{A} = \{A_1, \dots, A_n \}$ be a partition of $\Omega$, where the $A_i$ belong to $\mathcal{F}$. Let $X$ be integrable and $\mathcal{G} = \sigma(\mathcal{A})$. Then any version of $\mathbb{E}[X|\mathcal{G}]$ is of the form $\sum_{i = 1}^n \mathbb{E}[X|A_i] 1_{A_i}$.
Applying this to our problem, we thus get 
\begin{align*}
\mathbb{E}[X_{n + 1}|\mathcal{F}_n] &= \sum_{i = 0}^{2^n - 1} \mathbb{E}\left[X_{n + 1} \middle|\left[\frac{i}{2^n}, \frac{i + 1}{2^n} \right) \right]1_{\left[\frac{i}{2^n}, \frac{i + 1}{2^n} \right)} \\
&= \sum_{i = 0}^{2^n - 1} \frac{\mathbb{E}\left[X_{n + 1}1_{\left[\frac{i}{2^n}, \frac{i + 1}{2^n} \right)}\right]}{\mathbb{P}\left( \left[\frac{i}{2^n}, \frac{i + 1}{2^n} \right) \right)} 1_{\left[\frac{i}{2^n}, \frac{i + 1}{2^n} \right)} \\
&= \sum_{i = 0}^{2^n - 1} \frac{\mathbb{E}\left[2^{n + 1}1_{\left[0, \frac{1}{2^{n + 1}} \right)} 1_{\left[\frac{i}{2^n}, \frac{i + 1}{2^n} \right)}\right]}{\frac{1}{2^n}} 1_{\left[\frac{i}{2^n}, \frac{i + 1}{2^n} \right)} \\
&= \frac{1}{\frac{1}{2^n}} 1_{\left[0, \frac{1}{2^n} \right)} = 2^n 1_{\left[0, \frac{1}{2^n} \right)} = X_n
\end{align*}
This shows $X$ is a martingale.
