I am trying to solve the following question and the hint makes no sense to me as to why its even necessary to begin with:
We have that $\sup_n E(M_n^+) = \sup_n E(M_n) \leq E(\sup_n M_n) < \infty$ so that we may apply the martingale convergence theorem to state there exists a limit $M_\infty$ which is in $L^1$. Now it is obvious that $M_n \leq \sup_n M_n \in L^1$ by assumption, so that dominated convergence yields the result. Is this wrong? Am I missing something huge?
Even if the question is much easier than the hint makes it out to be, I want to know the motivation and how to actually use it. I can follow steps (a) - (d) easily enough, but I cannot conclude using the result in (d). What were the people who made the question looking at trying to do? How do you conclude using (d)?
I agree with your proof; $\mathbb{E} \sup M_n < \infty$ is such a strong condition that it is enough to by itself ensure uniform integrability.
As for what they were aiming, they were most likely trying to see if you know about test functions to test for uniform integrability. Essentially, if you have a non-negative function $\phi$ that is super-linear and is s.t. $\sup_{X \in \mathcal{X}} E[\phi(X)] = M < \infty$, then your class $\mathcal{X}$ is uniformly integrable. The proof is not bad; since $\phi$ is super linear, for any $n$, there exists a $C_n$ s.t. $\phi(x) > n x$ for $x > C_n$ . Then, we have
$$M \geq E[\phi(|X|)] \geq E[\phi(|X|) 1_{x > C_n}] > n E[|X| 1_{x > C_n}]$$,
which reduces to $$E[|X| 1_{x > C_n}] < \frac{M}{n}$$ for all $X \in \mathcal{X}$
which shows that the class is uniformly integrable.
They essentially wanted you to say that the function $x \rightarrow x \log^+ x$ was such a function, so $X_n$s are uniformly integrable, so convergence in $L_1$ holds. (You can read further in these notes: https://web.ma.utexas.edu/users/gordanz/notes/uniform_integrability.pdf; I copied for you a part of proposition 12.6)