Let $(E,\mathcal{A},\mu)$ be a finite measure space and $$ \mathcal{L}^1=\{f:E\to \mathbb{R}: \int_{E}{|f(t)|d\mu(t)}<\infty\} $$ Let $\{f_n\}\subset \mathcal{L}^1$, such that: $$ \sup_{n}{\int_{E}{|f_n(t)|d\mu(t)}}<\infty $$ Can we say that: there exists a subsequence $\{g_n^1\}$ of $\{g_n^0\}$ $(:=\{f_n\})$ and $M_1>0$ such that: $$ \lim_n\int_{0<|g_n^1|\leq 1}{|g_n^1(t)|d\mu(t)}=M_1 $$ and, by recurrence $$ \forall j\geq 1,\exists M_j>0 \text{ and }\{g_n^j\}\text{ subsequence of }\{g_n^{j-1}\}\text{ such that }\lim_n\int_{j-1<|g_n^j|\leq j}{|g_n^J(t)|d\mu(t)}=M_j $$
My problem:
Can we say that there exists $j_0>1$ such that $\{M_j\}_{j\geq j_0}$ is non-increasing sequence?
The answer to your first question is "yes". The answer to your problem is "no".
First, since you are only concerned with the absolute values of the functions, let's just demand that the sequence $\{f_n\}$ satisfies the condition $f_n \ge 0$ for all $n$. Then we can drop the absolute values everywhere. For any sequence meeting your condition, its absolute value will meet mine.
Now let $N = \sup_n\int_Ef_n\,d\mu$. Let $J$ be a Borel set in $\Bbb R$, and consider the sequence $a_n = \int_{f_n^{-1}(J)} f_n\,d\mu$. Because $f_n \ge 0$, we know $a_n \ge 0$, and because $f_n^{-1}(J) \subseteq E$, we know that $a_n \le \int_E f_n\,d\mu \le N$. Hence $\{a_n\} \subset [0, N]$. Any sequence in a compact set has a convergent subsequence. So there is an $M$ and a subsequence $\{a_{n_i}\}_i$ with $\lim_i a_{n_i} = \lim_i \int_{f_{n_i}^{-1}(J)} f_{n_i}\,d\mu = M$.
Your sequences $\{g_n^j\}$ can be built by repeatedly applying this process. Further, for any given $j$ we get $$N \ge \lim_n \int_{{g_n^{j}}^{-1}([0,j])}g_n^j\,d\mu = \sum_{i=1}^j \lim_n\int_{{g_n^{j}}^{-1}([i-1,i])}g_n^j\,d\mu = \sum_{i=1}^j M_i$$ Which shows that $\sum_{i=1}^\infty M_i \le N$. So it must be that $\lim_i M_i = 0$.
But that only means that the overall trend of the $M_i$ is downward. It doesn't mean that $M_i$ has to be non-increasing. For example, the sequence $$M_i = \begin{cases} 2^{-n}&i = 2n\\3^{-n} & i = 2n+1\end{cases}$$ converges to $0$ but is not non-increasing. And you can design a sequence of step functions so that this exact sequence of $M_i$ can be obtained from the construction.