Let $X$ be a random variable. A well-know result related to the definition of uniform integrability is that \begin{align} X \in \mathcal{L}^1(\Omega,\mathcal{F},\mathbb{P}) \iff \lim_{K \to \infty}\mathbb{E}|X|\mathbb{1}_{\{|X|>K\}} = 0. \end{align} I face some difficulties while proving this result.
$(=>)$ By the monotone converging theorem with increasing sequence $(\mathbb{E}|X|\mathbb{1}_{\{|X|\leq K\}})_{K \in \mathbb{N}}$ we find that \begin{align} \lim_{K \to \infty} \mathbb{E}|X|\mathbb{1}_{\{|X|\leq K\}} = \mathbb{E}|X|. \end{align} Therefore, for all $\epsilon >0,\ \exists K >0$ such that $\mathbb{E}|X|\mathbb{1}_{\{|X|\geq K\}} < \epsilon$. Hence, \begin{align} \lim_{K \to \infty}\mathbb{E}|X|\mathbb{1}_{\{|X|>K\}} = 0. \end{align} Is this correct? How do I use the fact that $X \in \mathcal{L}^1$?
And how to prove $(<=)$?
$[\Rightarrow]$ You have $$\lim_{k\rightarrow\infty}|X|1_{\{|X|> k\}}=0\;\;\text{ a.s.}$$ and $|X|1_{\{|X|> k\}}\leq |X|\in L^1$, so by the Dominated Convergence Theorem, $$\lim_{k\rightarrow\infty}E[|X|1_{\{|X|> k\}}]=0.$$
$[\Leftarrow]$ You can take $k>0$ such that $E[|X|1_{\{|X|> k\}}]<1$. Then $$E[|X|]=E[|X|1_{\{|X| \leq k\}}]+E[|X|1_{\{|X|> k\}}]<k+1<\infty.$$