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 $$\lim_n \int f_n d \mu = \int f d \mu$$