a problem in measure theory (how to prove $f_{n}\to 0$ in $ L^{p} $ ) I just need a hint not a whole solution please. 
Problem:
Let
$\{f_{n}\}\subset L^{1}([0,1])$
be any sequence of measurable functions such that
(a) 
$f_{n}\to 0$
almost everywhere;
(b)
$\sup\limits_{n}\int_{0}^{1}(f_{n}^{-})^{2}<\infty,$
where 
$ g^{-} $ 
denotes the negative part of any function, 
$ g^{-}=-\min\{0; g(x)\} $ 
(c) and 
$ \int_{0}^{1}f_{n}\to 0. $
Prove that 
$f_{n}\to 0$ 
in 
$L^{1}([0,1]).$
Guess: 
I feel that if I use the identity 
$ \vert g\vert=g+2g^{-}  $
I will have the answer. But the problem is that how can I prove that
$ \int_{0}^{1}f_{n}^{-}(x)dx\to 0? $
Also: 
The other guess and hint other than mine may be accepted. 
 A: The result will follow from: If $g_n$ is bounded in $L^2([0,1])$ and $g_n\to 0$ a.e. on $[0,1],$ then $g_n\to 0$ in $L^1.$ Hint for the proof of this: Egorov's theorem.
Note that this result fails on $[0,\infty)$ as the sequence $g_n = \chi_{[n,n+1]}$ shows. So finite measure is important
A: Hint: try to use that for large $K$, when $f_n^{-} \ge K$, one has $f_n^{-}\le \frac{1}{K} (f_n^{-})^2$.
Hint 2: For $K > 0$, one can write
$$\int_{[0,1]} f_n^- dx \le \int_{f_n^- < K} f_n^- dx + \int_{f_n^- \ge K} f_n^- dx
\le \int_{f_n^- < K} f_n^- dx + \frac{1}{K}\int_{f_n^- \ge K} (f_n^-)^2 dx
\le \int_{f_n^- < K} f_n^- dx + \frac{C}{K}$$
Hint 3: The function $f_n^{-}\mathbb{1}_{f_n^{-}<K}$ is dominated by $K\in L^1([0,1])$ and tends to $0$ almost everywhere. By the dominated convergence theorem, its integral tends to $0$ when $n \to+\infty$. Hence one has
$$\limsup_{n \to +\infty} \int_{[0,1]} f_n^{-} dx \le \frac{C}{K}$$
where $C=\sup_n \|f_n^{-}\|_2^2$. Since $K$ is arbitrary, it follows that
$$\limsup_{n\to +\infty} \int_{[0,1]} f_n^{-} dx = 0$$
A: Sketch of proof: Take an $L^p([0,1])$ we can use Holder's inequality to derive a bound on the $L^p$ norm. Specifically, if $1\leq p<q \leq \infty%$, we have $||f||_{L^p([0,1])}<\mu([0,1])^{1/p-1/q}||f||_{L^q([0,1])}$. Take $p=1$ and $q=2$ and we obtain
$||f||_{L^1([0,1])}\leq |f||_{L^2([0,1])}$ (as $\mu([0,1])=1$), take some $g\in L^2([0,1])$, use it as a dominating function for the $f_n$, i.e. $|f_n|<|g(x)|$, and use Lebesgue's Dominated Convergence Theorem. 
