Uniform integrability and tightness.

In order to show tightness, fix $\varepsilon>0$. Then you get $N=N(\varepsilon)$ such that $\int_Xf_nd\mu\leq\varepsilon$ if $n\geq N+1$. Now, for all $n\leq N$, you can find a positive $M$ such that $\int_{\{f_n\geq M_n\}}f_nd\mu\leq \varepsilon$, using integrability of $f_n$. (if $f$ is integrable apply the monotone convergence theorem to $f\mathbf 1_{\{|f|\geq n\}})$

Put $A_n:=\{f_n\leq M_n\}$, then $A_n$ is measurable. Take $X_0:=\bigcap_{k=1}^NA_k^c$. Each $A_k^c$ has finite measure (since $\mu(A_k^c)\leq \frac 1{M_k}\int f_kd\mu$) so $X_0$ is of finite measure. Check that we have the wanted inequality.

When you're trying to prove $(\Rightarrow)$, the limit gives you a way to bound the integrals of $f_n$ by $\epsilon$ for sufficiently large $n$. Then it's a matter of using the fact that any collection of finite number of $L^1(E)$ functions are both uniformly integrable and tight.

This is problem 1 in $\textit{Royden and Fitzpatrick}$ page 99. In the errata the author mentions to interchange problem 1 and 2 because problem 2 states to prove that any collection of finite number of $L^1(E)$ functions are both uniformly integrable and tight over $E$. Once you prove that the problem becomes trivial.