absolute value in Lebesgue integral question

Question

Let, $f_n \rightarrow f $ a.e. and integrable. If $\int |{f_n - f}| \rightarrow 0 $, then $\int f_n \rightarrow \int f$.

I am studying using Stein's real analysis book, and it seems he uses this statement as given. I could not figure out why this can hold.

One thing I can thing of is $$\int |{f_n - f}| \ge \left|\int f_n - f \right| $$ Thus, $$0=\lim \int |{f_n - f}| \ge \lim \left|\int f_n - f \right| \ge 0 $$ This implies $$\lim \left|\int f_n - f \right| = \left|\lim\int f_n - f \right| = \lim\int f_n - f = 0$$ and the result follows from the linearity of lebesgue measure.

Answer 1

For any sequence $a_n$ of real or complex numbers, there is no difference between $a_n \rightarrow a$ and $|a_n - a|\rightarrow 0$, but as you said,

$$\left|\int f_n d\mu - \int f d\mu\right| = \left|\int (f_n - f) d\mu\right|\le \int |f_n-f| d\mu \rightarrow 0$$

Answer 2

$$ 0 \leq \left|\int f_nd\mu-\int f d\mu \right| \underset{\textit{integrability}}{=} \left|\int f_n-f d\mu \right| \leq \int\left| f_n-f \right| d\mu \rightarrow 0, $$ as $n\rightarrow \infty$.

Therefore $\int f_n d\mu \rightarrow \int f d\mu,$ as $n\rightarrow \infty.$