Theorem. Let $(X, \mathcal{F}, \mu)$ be a measure space, $f_n : X \to [0, \infty]$ a sequence of measurable functions converging almost everywhere to a measurable function $f$ so that $f_n \leq f$ for all $n$. Then $$ \lim_{n\to\infty} \int_{X} f_n \, \mathrm{d}\mu = \int_{X} f \, \mathrm{d}\mu. $$
Note that $f$ need not be integrable and $(f_n)$ need not be monotone increasing.
Proof. By the Fatou's lemma, $$ \int_{X} f \, \mathrm{d}\mu = \int_{X} \varliminf_{n\to\infty} f_n \, \mathrm{d}\mu \leq \varliminf_{n\to\infty} \int_{X} f_n \, \mathrm{d}\mu \leq \varlimsup_{n\to\infty} \int_{X} f_n \, \mathrm{d}\mu \leq \int_{X} f \, \mathrm{d}\mu.$$
