First note that if $b_n$ is a sequence then $\liminf_n (-b_n) = - \limsup_n b_n$ The proof for this is relatively straight forward. $\liminf_n (-b_n) = \lim_n \inf_{k \geq n} (-b_k) = \lim_n (-\sup_{k \geq n} (b_k)) = - \lim_n (\sup_{k \geq n} (b_k)) = - \limsup_n b_n$ Now lets define $h_n^{(1)}(x) = g(x) + f_n(x)$ and $h_n^{(2)}(x) = g(x) - f_n(x)$.
The first assumption in LDCT is $\vert f_n(x) \vert \leq g(x)$. This implies that $h_n^{(1)}(x)$ and $h_n^{(2)}(x)$ are both non-negative over the underlying domain. Hence, by Fatou's lemma, we have that $\int \liminf h_n^{(1)} d \mu \leq \liminf \int h_n^{(1)} d \mu$ and $\int \liminf h_n^{(2)} d \mu \leq \liminf \int h_n^{(2)} d \mu$ Hence, we have that $\int g d \mu + \int \liminf f_n d \mu \leq \int g d \mu + \liminf\int f_n d \mu$ and $\int g d \mu + \int \liminf (-f_n) d \mu \leq \int g d \mu + \liminf\int (-f_n) d \mu$ The next assumption in LDCT is $\int g d \mu < \infty$. Hence, we can cancel $\int g d \mu$ from both equations to get $\int \liminf f_n d \mu \leq \liminf\int f_n d \mu$ and $\int \liminf (-f_n) d \mu \leq \liminf\int (-f_n) d \mu$ Now since $\liminf_n (-b_n) = - \limsup_n b_n$ the second equation can be written as $\int -\limsup (f_n) d \mu \leq -\limsup\int f_n d \mu$ which can be rewritten as $\int \limsup (f_n) d \mu \geq \limsup\int f_n d \mu$ The last assumption in LDCT is that $f_n \to f$. Hence, we have that $\limsup (f_n) = f = \liminf (f_n)$. Hence, we get that $\limsup\int f_n d \mu \leq \int f d \mu \leq \liminf\int f_n d \mu$ But we know that for any sequence $\{a_n\}$, $\liminf_n a_n \leq \limsup_n a_n$ and equality holds only when limit exists. Hence, putting all this together, you get that $\lim_n \int f_n d \mu = \int f d \mu$