I'm reading the proof of completeness of $L^1 (X, \mathscr{M}, \mu)$, and I would like to clear up some confusion
To prove $L^1$ is complete it suffices to show that every Cauchy sequence $(f_n)$ has a convergent subsequence, so assume wlog that
$$\| f_{n+1} - f_{n}\| \le 2^{-n}$$
and consider the series
$$f_1 + \sum_{i=1}^m f_{i+1} - f_{i}$$
This is bounded by
$$|f_1|| + \sum_{i=1}^m |f_{i+1} - f_{i}|$$
Setting $g_m = \sum_{i=1}^m |f_{i+1} - f_{i}|$ we have $g_m$ converges up to $g$, and $g \in L^1$. Furtheremore, $f_1 + \sum f_{i+1} - f_{i}$ converges to $f \in L_1$. Since the preceding series is bounded by $|f_1| + G$, the dominated convergence implies $\|f_n - f\| \to 0$ in $L^1$ norm.
My questions
Why does $g_m$ converge to $g \in L^1$. Why does this implie that $f_1 + \sum f_{i+1} - f_{i}$ converges?
Here is an outline: $g_m$ converges pointwise, since it is an increasing sequence of functions. The limit is in $L^1$ by the monotone convergence theorem (MCT): For if $g_m\to g$ pointwise, MCT gives $\|g\|_1=\int g\,d\mu=\lim\int g_m\,d\mu\le \sum_k 2^{-k}=1$. The convergence is not only pointwise, but in $L^1$, since $\|g-g_m\|=\int(g-g_m)\,d\mu\to0$.
It follows that $g<\infty$ a.e., and therefore $\sum(f_{i+1}-f_j)$ converges absolutely a.e. (by the very definition of absolute convergence). In particular, it converges pointwise a.e. to some limit. Applying the triangle inequality to tails of the sum should lead to a proof of $L^1$ convergence.
(Edited to add some detail.)