The Dominated Convergence Theorem

Since $g - f_n \ge 0$ almost everywhere, Fatou's lemma applies: $\int \liminf_{n\to\infty}(g-f_n)\,d\mu \le \liminf_{n\to\infty} \int (g-f_n)\,d\mu$. Because $f_n \to f$, the left side equals $\int (g-f)\,d\mu = \int g\,d\mu - \int f\,d\mu$ (valid since $\int g\,d\mu$ is finite), and the right side equals $\int g\,d\mu - \limsup_{n\to\infty}\int f_n\,d\mu$.

Cancelling the finite quantity $\int g\,d\mu$ from both sides gives $-\int f\,d\mu \le -\limsup_{n\to\infty}\int f_n\,d\mu$, i.e. $\limsup_{n\to\infty}\int f_n\,d\mu \le \int f\,d\mu$.

Next, since $g+f_n \ge 0$ almost everywhere, Fatou's lemma applied to $(g+f_n)$ gives $\int (g+f)\,d\mu \le \liminf_{n\to\infty}\int (g+f_n)\,d\mu$, and cancelling $\int g\,d\mu$ again gives $\int f\,d\mu \le \liminf_{n\to\infty}\int f_n\,d\mu$.

Combining the two inequalities, $\limsup_{n\to\infty}\int f_n\,d\mu \le \int f\,d\mu \le \liminf_{n\to\infty}\int f_n\,d\mu$. Since $\liminf \le \limsup$ always holds, all three quantities must coincide, so $\lim_{n\to\infty}\int f_n\,d\mu$ exists and equals $\int f\,d\mu$.