Assume we are in the interval [a,b] and we have a function $f\in L^1([a,b])$. Then since the continuous functions are dense in $L^1([a,b])$ we can choose a sequence $f_n$ of continuous functions such that they converge to $f$ in $L^1([a,b])$. I have two questions.
This first is easy. I assume we can choose a subsequence so that they also converge Lebesuge a.e.?
This is more difficult. Can we choose the sequence such that $|f_m(x)|\le |f_{m+1}(x)|$ Lebesgue a.e?
1. That is true. This is often implied by an intermediate step of the proof of the completeness of $L^1$.
2. We show that such a choice is not always possible.
Choose a measurable set $E\subseteq[0,1]$ such that $0 < \frac{\operatorname{Leb}(E\cap[a,b])}{\operatorname{Leb}([a,b])}<1$ for any $0 \leq a < b \leq 1$, where $\operatorname{Leb}$ is the Lebesgue measure. An important consequence of this property is that, if $F$ is another measurable set such that $\operatorname{Leb}(E\setminus F)=0$, then $F$ is dense in $[0, 1]$.
Now we let $f = \mathbf{1}_{E} + 2\cdot\mathbf{1}_{[0,1]\setminus E}$. We claim that this $f$ serves as a counter-example.
Assume otherwise, so that there exists a sequence of continuous functions $f_n$ such that $|f_n|$ is increasing in $n$ a.e. and $f_n \to f$ in $L^1$. Since $|f_n| \to |f| = f$ in $L^1$, we may assume that $f_n$'s are non-negative. Also, by passing to a subsequence if necessary, we may assume that $f_n$ converges pointwise a.e. Then for each $n$, we have $f_n \leq f$ a.e. By the property of $E$, this implies that the set $\{f_n \leq 1 \}$ is dense in $[0, 1]$, and so, $f_n \leq 1$ on all of $[0,1]$ by the continuity of $f_n$. This contradicts the fact that $f \neq 1$ as elements in $L^1$, and therefore the claim is proved. $\square$