In a recent exam I was asked to show several different properties about a discrete time martingale $X_n$ which is positive, uniformly integrable, with $\lim_{n\to\infty}X_n = 0$ almost surely. Are there any processes which actually satisfy this? I believe the following justifies a contradiction:
Since $X_n$ is assumed uniformly integrable and converges a.s. to 0, we also get convergence to 0 in $L^1$ - that is, $E(X_n) = E(|X_n -0|) \to 0$. (For reference, see the last line of this Wikipedia page).
On the other hand, since $X_n$ is positive we must have, for some $c>0$, $\ \ P(X_0 > c) = p > 0$. Thus, $\mathbb E (X_0) \geq p\cdot c$ by Markov inequality and positivity. However, by the martingale property, $E(X_n) = E(X_0) > p\cdot c > 0 $. Thus, we have an inconsistency.
Am I missing something, or was the question degenerate?
You indeed solved correctly this question. Actually, this question is not bad as it shows how crucial the uniform integrability assumption is. Such martingale have the representation $X_n=\mathbb E\left[X\mid\mathcal F_n\right]$ where $X$ is integrable, and in this context, by the martingale convergence theorem, the equality $\mathbb E\left[X\mid\mathcal F_\infty\right]=0$ would hold almost surely, where $\mathcal F_\infty$ is the $\sigma$-algebra generated by $\bigcup_{n\geqslant 1}\mathcal F_n$. Exploiting the convergence in $\mathbb L^1$, which is guaranteed by the uniform integrability assumption, is also a good option.
Without the uniform integrability assumption, we may get counter-examples with martingales of the form $\prod_{j=1}^nX_j$, where $\left(X_j\right)_{j\geqslant 1}$ is i.i.d., $X_j$ is positive and has expectation one and for almost every $\omega$, $X_j(\omega)$ equals, says $1/j$, for $j\geqslant J(\omega)$.