Prove that $ \int_X |f_n|^2d\mu<\infty $ for all $ n\in\mathbb N $ 
Let $ (X,\mathcal M,\mu) $ be a measure space with $ \mu(X)<\infty $. Let $ (f_n)_{n=1}^\infty $ be a sequence of functions in $ L^1(X,\mathcal M,\mu) $, and let $ f\in L^1(X,\mathcal M,\mu) $. Suppose that
  $$ \lim_{n\to\infty}\int_X |f_n-f|d\mu=0 .$$
  Suppose also that $$ C:=\sup\left\{\int_X|f_n|^4d\mu:n\in\mathbb N\right\}<\infty .$$
  Prove that 
  
  
*
  
*$ \int_X |f|^4d\mu\le C $.
  
*$  \int_X |f_n|^2d\mu<\infty $ for all $ n\in\mathbb N $, and $ \int_X|f|^2d\mu<\infty $
  
*$ \lim\limits_{n\to\infty}\int_X|f_n-f|^2d\mu=0 .$

My attempt:


*

*Since $$\lim_{n\to\infty}\int_X |f_n-f|d\mu=0, $$ we know that $ f_n $
converges to $ f $ in measure, i.e., for every $ \epsilon>0, $
$$ \lim_{n\to\infty}\mu(\{x\in X:|f(x)-f_n(x)|\ge\epsilon\})=0. $$ Hence we can find a subsequence $ f_{n_k} $ of $ f_n $ such that $$ f_{n_k}\to f\quad a.e.\ \text{as}\ k\to\infty. $$ Now it suffices to prove that $$ \liminf_{k\to\infty}\int_X |f_{n_k}|^4d\mu\le C=\sup\left\{\int_X|f_n|^4d\mu:n\in\mathbb N\right\} $$
which is obviously true and we have applied the Fatou's lemma: $$ \int_X |f|^4d\mu=\int_X \lim_{k\to\infty}|f_{n_k}|^4d\mu\le\liminf_{k\to\infty}\int_X |f_{n_k}|^4d\mu .$$ And we are done!

*I am stuck on this one...... How to deal with $ |f_n|^2 $? I am thinking of Holder's inequalities, but nothing helps.

Use our favourite AM-GM inequality to get 
$$2|f_n|^2\leq |f_n|^4+1$$ and then integrate both sides to get $$2\int_X|f_n|^2d\mu\leq \int_X|f_n|^4d\mu +\mu(X)\leq C+\mu(X)$$
$$\implies sup\ _n \ \int_X|f_n|^2d\mu\leq \frac{C}{2}+\frac{\mu(X)}{2}<\infty$$
