$(X_{n})$ are uniformly integrable stochastic process ,$X_{n}\xrightarrow{d}X$
Proof $E[\left | X \right |]\le \liminf_{n\to\infty}E[\left | X_{n} \right |]$
I see a way that $f_{m}(x)=\begin{cases}\left | x \right |\quad &\left | x \right |\le{m}\\m\quad &\left | x \right |>{m}\end{cases}$ , $f_{m}$ is in mon0tone rise to$\left | x \right |$ ,$f_{m}$ is bounded continuous function
so $E[\left | X \right |]=\lim_{m \to \infty}E[f_{m}(X)]=\lim_{m \to \infty}\lim_{n \to \infty}E[f_{m}(X_{n})]\le\liminf_{n\to\infty}\lim_{m\to\infty}E[f_{m}(X_{n})]=\liminf_{n\to\infty}E[\left | X_{n} \right |]$
But I don't know why $\lim_{m \to \infty}\lim_{n \to \infty}E[f_{m}(X_{n})]\le\liminf_{n\to\infty}\lim_{m\to\infty}E[f_{m}(X_{n})]$
How did this step come about?
Clearly $f_m(x) \leq |x|$ for every $x \in \mathbb{R}$, which implies that $$E[f_m(X_n)] \leq E[|X_n|].$$ Taking $\liminf$ gives (by the continuous mapping theorem) $$E[f_m(X)]\leq \liminf_n E[|X_n|].$$ This holds for every $m$. Now let $m\to \infty$ to get the result (using for example the monotone convergence theorem).
Another solution: if you're familiar with Skorokhod's representation theorem, then the result is a direct application of Fatou's lemma.