$\lim_{n\to\infty}\int_{X}\lvert f_n -f \rvert \, d\mu=0$. Why does this imply that $\lim_{n\to\infty}\int_{X}f_n \, d\mu=\int_{X}f \, d\mu$?

Question

If $f \in L^1(\mu$), then the Lebesgue integral $$\left\lvert \int_{X}f \, d\mu \right\rvert \leq \int_{X}\lvert f \rvert \, d\mu.$$

Also, for a sequence of complex measurable functions on $X$ such that $$f(x) = \lim_{n\to\infty}f_n(x)$$ exists for every $x\in X$, $$\lim_{n\to\infty}\int_{X} \lvert f_n -f \rvert \, d\mu=0.$$

Why, exactly, does this imply that $$\lim_{n\to\infty}\int_{X}f_n \, d\mu=\int_{X}f \, d\mu ?$$

It should be something like one step and a a trivial conclusion but I want the exact steps written out.

Answer 1

Remark that $\lim \limits_{n \rightarrow \infty} a_n = a$ if and only if $\lim \limits_{n \rightarrow \infty} |a_n - a| = 0$. Then simply use:

$$ 0 \leq \lim \limits_{n \rightarrow \infty} \left|\int_X f_n d\mu - \int_X f d\mu \right| \leq \lim \limits_{n \rightarrow \infty} \int_X |f_n - f| d\mu =0 $$

Answer 2

Try using $\left|\int_X g\,d\mu\right|\le \int_X |g|\,d\mu$.