Let $\Omega = [0,1]$, $\mathcal{F} = \mathcal{B}(0,1)$, P=Lebesgue measure.
Let $X_n(w)= \begin{cases} 0 \quad \frac{1}{n} < w \leq 1 \\ n-n^2w \quad 0 \leq w \leq \frac{1}{n} \end{cases}$
The first part of the exercise is proving that $X_{n}$ is a martingale, which I managed to do. Now my problem is computing $E(sup_{n \geq 1}|X_n|)$ and I don't know how to proceed. From the solutions I know that it must be equal to $+\infty$.
Note that $\omega ([\frac 1 {2\omega}] -\omega [\frac 1 {2\omega}]^{2}) \to \frac 1 2 -\frac 1 4=\frac 1 4$ as $\omega \to 0$. Hence for $\omega$ sufficiently small $\omega ([\frac 1 {2\omega}] -\omega [\frac 1 {2\omega}]^{2}) >\frac 1 8$. Taking $k=[\frac 1 {2\omega}]$ we see that $\sup_n |X_n(\omega)| \geq |X_k(\omega)| \geq \frac 1 {4\omega}$. Since $\int \frac 1 {4\omega}\, d \omega =\infty$ it follows that $E\sup_n |X_n(\omega)|=\infty$.