General Lebesgue Dominated Convergence Theorem In Royden (4th edition), it says one can prove the General Lebesgue Dominated Convergence Theorem by simply replacing $g-f_n$ and $g+f_n$ with $g_n-f_n$ and $g_n+f_n$.  I proceeded to do this, but I feel like the proof is incorrect.
So here is the statement:

Let $\{f_n\}_{n=1}^\infty$ be a sequence of measurable functions on $E$ that converge pointwise a.e. on $E$ to $f$.  Suppose there is a sequence $\{g_n\}$ of integrable functions on $E$ that converge pointwise a.e. on $E$ to $g$ such that $|f_n| \leq g_n$ for all $n \in \mathbb{N}$.  If $\lim\limits_{n \rightarrow \infty}$ $\int_E$ $g_n$ = $\int_E$ $g$, then $\lim\limits_{n \rightarrow \infty}$ $\int_E$ $f_n$ = $\int_E$ $f$.

Proof:
$$\int_E (g-f) =  \liminf \int_E g_n-f_n.$$
By the linearity of the integral: 
$$\int_E g - \int_E f = \int_E g-f \leq \liminf \int_E g_n -f_n = \int_E g - \liminf \int_E f_n.$$
So,
$$\limsup \int_E f_n \leq \int_E f.$$
Similarly for the other one.
Am I missing a step or is it really a simple case of replacing.

You made a mistake:
$$\liminf \int (g_n-f_n) = \int g-\limsup \int f_n$$
not
$$\liminf \int (g_n-f_n) = \int g-\liminf \int f_n.$$
Here is the proof:
$$\int (g-f)\leq \liminf \int (g_n-f_n)=\int g -\limsup \int f_n$$
which means that 
$$\limsup \int f_n\leq \int f$$
Also
$$\int (g+f)\leq \liminf \int(g_n+f_n)=\int g + \liminf \int f_n$$
which means that
$$\int f\leq \liminf \int f_n$$
i.e.
$$\limsup \int f_n\leq \int f\leq \liminf\int f_n\leq \limsup \int f_n$$
So they are all equal.
