An equivalent definition of uniform integrability 
Let $(X,\mathcal{M},\mu)$ be a measure space and $\{f\}$ be a sequence of functions on $X$, each of which is integrable over $X$. Show that $\{f_n\}$ is uniformly integrable if and only if for each $\varepsilon \gt 0$, there is a $\delta \gt 0$ such that for any natural number $n$ and measurable subset $E$ of $X$, if $\mu(E) \lt \delta$, $$ \left|\int_E f_n~d\mu\right| \lt \epsilon.$$  

I think one direction $(\Rightarrow)$ is clear. Since $f_n$ being uniformly integrable imply that for every $\varepsilon\gt 0,~\exists \delta \gt 0$ such that for any $E\in \mathcal{M},~\mu(E)\lt \delta$, $\int_E |f_n|~d\mu$ for every natural $n$. But then 
$$ \left|\int_E f_n~d\mu\right|\le \int_E|f_n|~d\mu \lt \varepsilon.$$  
Any suggestions for the other direction? 
Edit:
Following Davide's suggestions, I have 
$$
\begin{align*}
\int_E |f_n|~d\mu & \le \int_{E\cap [f_n \ge 0]} f_n~d\mu + \int_{E\cap [f_n \lt 0]} (-f_n)~d\mu\\
& = \left| \int_{E\cap [f_n \ge 0]} f_n~d\mu \right| + \left| \int_{E\cap [f_n \lt 0]}f_n~d\mu \right| \\
& \lt \varepsilon + \varepsilon = 2\varepsilon.
\end{align*}
$$
 A: I just sum up the comments into an answer.
We have to show that these two assertions are equivalent 


*

*For all $\varepsilon>0$, we can find $\delta >0$ such that if $E\in\mathcal M$ and $\mu(E)\leq \delta$, then for all natural number $n$, we have $\int_E|f_n|d\mu\leq\varepsilon$;

*For all $\varepsilon>0$, we can find $\delta >0$ such that if $E\in\mathcal M$ and $\mu(E)\leq \delta$, then for all natural number $n$, we have $\left|\int_Ef_nd\mu\right|\leq\varepsilon$. 


(1.) $\Rightarrow$ (2.) by the triangular inequality. To see the converse, fix $\varepsilon>0$. We can find $\delta$ such that if $E\in\mathcal M$ and $\mu(E)\leq \delta$ then for each $n$: $\left|\int_Ef_nd\mu\right|\leq\varepsilon/2$. Let $E$ measurable with measure $\leq\delta$, and put $E_1:=\{x\in X:f(x)<0\}$ and $E_2:=\{x\in X:f(x)\geq 0\}$.  These sets are measurable and of measure $\leq\delta$. We get for $n$ natural number:
$$\int_E |f_n|d\mu=\int_{E_1} |f_n|d\mu+\int_{E_2} |f_n|d\mu=-\int_{E_1}f_nd\mu+
\int_{E_2}f_nd\mu\leq \left|\int_{E_1}f_nd\mu\right|+\left|\int_{E_2}f_nd\mu\right|\leq \varepsilon/2+\varepsilon/2=\varepsilon.$$
