Let $(E,\mathcal{A},\mu)$ be a finite measure space and $$ \mathcal{L}^1=\left\{f:E\to \mathbb{R}: \int_{E}{|f(t)|d\mu(t)}<\infty\right\} $$ Let $\{f_n\}\subset \mathcal{L}^1$, such that: $$ \sup_{n}{\int_{E}{|f_n(t)|d\mu(t)}}<\infty $$ Such that, there exists a subsequence $\{g_{m}\}$ of $\{f_n\}$, such that for all subsequence $\{g_{m_i}\}$ of $\{g_m\}$ $$ \{g_{m_i}1_{|g_{m_i}|\leq i})\}\text{ is uniformly integrable,} \qquad (1) $$ $$ \sum_{i\geq 1}{\mu\big(\{t\in E~:~|g_{m_i}(t)|>i \}\big)}<+\infty,\qquad (2) $$
Why : "From $(1)$ and $(2)$, we see the known result that $\{f_n\}$ has a subsequence equivalent to a uniformly integrable sequence. "?
If $f_n = 0$ for all $n\in \mathbb{N}$, then $\{f_n\}$ is a subsequence that is uniformly integrable. Otherwise, $$\infty >\sup_n \int_E |f_n| \, d\mu >0,$$ in which case I claim that the subsequence $\{g_m\}_{m=N}^\infty$ of $\{f_n\}$ is uniformly integrable for large enough $N$. Let $\epsilon>0$ and $A\in \cal{A}$. Put $$A_i = \{x \in E: |g_{i}(x)|> i\}.$$
From (2), $\mu(A_i) \to 0$ as $i \to \infty$. Choose $N \in \mathbb{N}$ such that $i \geq N$ implies $$\mu(A_i) < \frac{\epsilon}{2\sup_n \int_E |f_n| \, d \mu}.$$
From (1), choose $\delta >0$ such that $i \in \mathbb{N}$ and $\mu(A) < \delta$ implies $$ \int_A |g_i| \cdot 1_{E\setminus A_i}\, d\mu < \frac{\epsilon}{2}. $$
Then $i \geq N$ and $\mu(A)<\delta$ implies
\begin{align*} \int_A |g_i| \, d\mu & = \int_{A\cap A_i} |g_i| \, d\mu + \int_{A\cap (E\setminus A_i)} |g_i| \, d\mu\\ & = \int_{E} |g_i|\cdot 1_{A\cap A_i} \, d\mu + \int_{E} |g_i| \cdot 1_{A\cap (E\setminus A_i)}\, d\mu\\ & \leq \mu(A_i) \cdot \sup_n |f_n| \, d\mu +\int_E|g_i| \cdot 1_{E\setminus A_i}\, d\mu\\ & < \frac{\epsilon}{2} + \frac{\epsilon}{2}\\ = \epsilon. \end{align*}
This proves that the subsequence $\{g_m\}_{m=N}^\infty$ of $\{f_n\}$ is uniformly integrable.