Assume all random variables are in $L^1$, i.e., absolutely integrable.
We do not need $(A_n)$ to be predictable.
Using $X_n\geq 0$, we get
$M_n^-=\max(0,A_n-X_n)\leq A_n^+\leq A_0^+ +\sum_{k\geq 1} a_k^+.$
Taking expectations shows that
$\mathbb{E}(M_n^-)\leq \mathbb{E}\left(A_0^+ +\sum_{k\geq 1} a_k^+\right)<\infty.$
Since $(M_n)$ is a martingale, we have
$$\mathbb{E}(|M_n|)= \mathbb{E}(M_n)+2 \mathbb{E}(M_n^-) = \mathbb{E}(M_0)+2 \mathbb{E}(M_n^-), $$
which combined with the previous bound, shows that $(M_n)$ is bounded in $L^1$.
By the martingale convergence theorem, $M_n\to M_\infty$ almost surely.
By Fatou's Lemma, $M_\infty$ is also in $L^1$.
Turning to the $(A_n)$ process, for $n\geq m$ we have
$A_n\leq A_m+\sum_{k>m} a_k^+,$
so that
$$\sup_{n\geq m}\, A_n\leq A_m+\sum_{k>m} a_k^+.$$
In particular, the random variable on the left is almost surely finite.
Taking the liminf in $m$ on both sides, we get
$$\limsup_m\, A_m\leq \liminf_m\ A_m.$$
This shows that $A_m\to A_\infty$ in $[-\infty,\infty]$.
We already know that $A_m$ is almost surely bounded above, so $A_\infty<\infty$.
Letting $m\to\infty$ in $A_m=X_m-M_m\geq -M_m$ gives $A_\infty\geq -M_\infty >-\infty$,
and the proof is complete.