Show that for a non-negative supermartingale $M$ with $\lim_{n \to \infty}E[M_n] = 0$ we have $\lim_{n \to \infty}M_n = 0$ almost surely and the Doob decomposition is reduced to $M_n = E[A_\infty|\mathscr{F}_n]-A_n$ where $A_n$ is a non-decreasing, predictable sequence.
For the first part, I said that since we have a non-negative supermartingale then we know that the limit exists almost surely where $E[M_\infty|\mathscr{F}_n] \le M_n$ $\forall n$, and therefore: $$ 0 \le E[M_\infty] \le \lim_{n \to \infty}E[M_n] = 0 $$ and since this supermartingale is non-negative if $E[M_\infty] = 0 \Rightarrow M_\infty = 0 \; a.s$.
For the second part, it's Doob decomposition is $M_n = X_n - A_n$ where $X_n$ is a martingale. Therefore, the goal is to show $X_n = E[A_\infty|\mathscr{F}_n]$.
I said that $\lim_{n \to \infty}M_n = \lim_{n \to \infty}X_n - \lim_{n \to \infty}A_n$ which implies $X_\infty = A_\infty$ almost surely. Now in order for $X_n = E[X_\infty|\mathscr{F}_n]$ I believe we need to show that either $(X_n)_{n \ge 0}$ converges in $L_1$ or is uniformly integrable. However, unfortunately, I can't seem to figure out why that's true.
You're very close: $(M_n)$ converges to $0$ in $L^1$. So does $(A_n)$, because it's a non-negative increasing process with $E[A_n]$ bounded above by $E[X_0]$. Consequently, $X_n=M_n+A_n$ also converges in $L^1$.