A question about dominated convergence theorem I have this question and I don't know how to proceed:

Suppose that $(f_n)$ is a sequence of measurable functions on $[0,1]$ such that $\displaystyle \lim_{n \to \infty} \int_0^1 |f_n|\, =0$ 
  and that there is an integrable function $g$ on [0,1] such that $|f_n|^2\leq g$ for all $n$. Show that $\displaystyle \lim_{n \to \infty} \int_0^1 |f_n|^2\, =0$. 

I think that I must show $\displaystyle \lim_{n \to \infty} \int_0^1 |f_n|\, = \displaystyle  \int_0^1 \lim_{n \to \infty}|f_n|$ , in other words I must show that we can take limit inside the integral.  What does $\displaystyle \lim_{n \to \infty} \int_0^1 |f_n|\, =0$ mean? Can you help me for this question? Thanks.

It is easier to argue by contradiction. Assume the contrary that 
$$\int_0^1 |f_n|^2 dx$$
does not converge to $0$. Then by picking a subsequence if necessary, assume that there is $\epsilon_0>0$ so that 
$$\tag{1} \int_0^1 |f_n|^2 dx \ge \epsilon_0.$$
The fact that $\int_0^1 |f_n| dx \to 0$ implies that (by picking a subsequence if necessary) $f_n \to 0$ almost everywhere. Thus $|f_n|^2 \to 0$ almost everywhere. Now one can use the condition $|f_n|^2 \le g$ and Lebesgue's dominated convergence theorem to conclude 
$$ \lim_{n\to\infty} \int_0^1 |f_n|^2 dx = \int_0^1 \lim_{n\to \infty} 0 dx = 0.$$
But this contradicts (1). 
