Let the sequence $\left\{ \left| X_n - X \right|^r \right\}$ be uniformly integrable for $r > 0$. This means that $E \left[ \left| X_n - X \right|^r 1 \left\{ \left| X_n - X \right|^r > a \right\} \right] \xrightarrow{a \rightarrow \infty} 0$, uniformly in $n$.
I would like to understand a proof that $X_n \xrightarrow{L_r} X$ as $n \rightarrow \infty$, meaning that $E \left| X_n - X \right|^r \xrightarrow{n \rightarrow \infty} 0$.
Fix $\varepsilon > 0$ \begin{eqnarray*} E \left| X_n - X \right|^r & = & E \left[ \left| X_n - X \right|^r 1 \left\{ \left| X_n - X \right|^r \leqslant \varepsilon \right\} \right] + E \left[ \left| X_n - X \right|^r 1 \left\{ \left| X_n - X \right|^r > \varepsilon \right\} \right]\\ & \leqslant & \varepsilon + E \left[ \left| X_n - X \right|^r 1 \left\{ \left| X_n - X \right|^r > \varepsilon \right\} \right] \end{eqnarray*} Now if I could conclude from here that $ \limsup_{n \rightarrow \infty} E \left| X_n - X \right|^r \leqslant \varepsilon $ then that would prove it. The problem is the second term on the left hand side of the inequality goes to zero only as $\varepsilon$ gets large. How is it possible to proceed from here?
Edit: What are the consequences of adding the assumption $X_n \xrightarrow{a.s.} X$ here? (It is a bit stronger than the suggestion in the comment below.)