$L_p$ space and convergence Let $f_i\rightarrow f$ $m$-a.e on $[0,1]$, $m$ is a measure and $\int|f_i(x)|^4dm$$\le1$ for all $i$.Then $\int|f_i(x)|^2dm\rightarrow \int|f(x)|^2dm$. how to prove it?
in my solution i prove that $|f_i(x)|^2\in L_2$ for all $i$ and $|f(x)|^2\in L_2$.
then use some statement (attached picture) where $r = 1,p = 2,F_i(x) =|f_i(x)|^2,F(x) =|f(x)|^2,C = 1$ and $f_n(x)\Rightarrow f(x)$ mean convergence in measure, but when measure is finite from a.e.convergence $\Rightarrow$ convergence in measure. It's right solution?

Yes, you can use the quoted theorem. 
We can also argue only with the dominated convergence theorem. Indeed, for a fixed $R$, we have 
$$\left|\int f_i^2-\int f^2\right|\leqslant \left|\int f_i^2\mathbf 1\{|f_i|\leqslant R\}-\int f^2\mathbf 1\{|f|\leqslant R\}\right|\\
+\left|\int f_i^2\mathbf 1\{|f_i|\gt R\}-\int f^2\mathbf 1\{|f|\gt R\}\right|\\
\leqslant \left|\int f_i^2\mathbf 1\{|f_i|\leqslant R\}-\int f^2\mathbf 1\{|f|\leqslant R\}\right|+\frac 2{R^2},$$
where we used $$\int f_i^2\mathbf 1\{|f_i|\gt R\}\leqslant \frac 1{R^2}\int f_i^4\leqslant \frac 1{R^2}$$
and the same inequality holds for $f$ by Fatou's lemma. 
By dominated  convergence, we have for each $R$ such that 
$m\{x\mid \left|f(x)\right|=R\}=0$, 
$$\lim_{n\to +\infty}\int f_i^2\mathbf 1\{|f_i|\leqslant R\}-\int f^2\mathbf 1\{|f|\leqslant R\}=0.$$
This is due to the fact that $\mathbf 1\{|f_i|\leqslant R\}\to\mathbf 1\{|f|\leqslant R\}$ almost everywhere.
