Let $(X,\mathcal{A},\mu)$ a measure space and let $\{f_n\}$ a sequence of measurable function and $f$ a measurable function, such that $\lim_ {n\to \infty} f_n(x)=f(x)$ a.e. in $X$. Suppose that exists a function $g\in L^1(X,\mathcal{A},\mu)$ such that $$|f_n|\le g\quad\text{a.e in}\;X.$$ Then $$\lim_{n\to\infty}\int_{X}|f_n-f|\;d\mu=0.$$
I have already shown what I have just said. Now I must proved that $$\lim_{n\to\infty}\int_Xf_n d\mu=\int_X f d\mu.$$ I do this $$\color{BLUE}{\lim_{n\to\infty}\int_Xf_n\ d\mu}-\int_Xf d\mu\le\lim_{n\to\infty}\bigg|\int_X(f_n-f)d\mu\bigg|\le\lim_{n\to\infty}\int_X|f_n-f|d\mu=0$$
Question How do I know if the blue limit exists? Is it clear from what was said previously or is this procedure wrong?
Thanks!
$|\int f_n d\mu -\int fd\mu| \leq \int |f_n-f| d\mu \to 0$. So $\lim \int f_n d\mu$ exists and equals $\int f d\mu$.