Prove that the given function is in $L^1$ The following question is inspired by Exercise 7.2.2 in Durrett, 2010.
Let (i) $g_k(\omega) \to g(\omega)$ a.s., (ii) $E\,(\sup_k |g_k(\omega)|) < \infty$, and
$$
(iii) \; f_M(\omega) = g(\omega) + \sup_{k \geq M}|g_k(\omega) - g(\omega)|.
$$
Show that $f_M \in L^1$, that is,
$$
E \; \left(\left| \; g(\omega) + \sup_{k \geq M}|g_k(\omega) - g(\omega)| \; \right|\right) < \infty.
$$
Probably, the answer is trivial; however, not for me. I would really appreciate a detailed proof. My goal is to learn how to reason in this situation step by step explaining every single detail. Thank you.
Regards,
Ivan

Hint:
$$\left|g(\omega)+\sup_{k\geqslant M}|g_k(\omega)-g(\omega)|\right|\leqslant 2|g(\omega)|+\sup_{k\geqslant 1}|g_k(\omega)|\leqslant 3\sup_{k\geqslant 1}|g_k(\omega)|.$$
Indeed, we have for any sequence $(a_k,k\geqslant 1)$ that $|a_k|\leqslant \sup_n|a_n|$ for each $k$. 
