I am given the following problem, with a hint that states that I should use the General Lebesgue Dominated Convergence Theorem.
Let $\{f_n\}$ be a sequence of integrable functions on $E$ for which $f_n \rightarrow f$ a.e. on $E$ and $f$ is integrable over $E$. Show that $\int_E \lvert f - f_n \rvert \rightarrow 0$ if and only if $\lim_{n \rightarrow \infty} \int_E \lvert f_n \rvert = \int_E \lvert f \rvert$.
($\Rightarrow$) Let $g_n = k_n \lvert f - f_n \rvert$ where $k_n \in \mathbb{R}$ and $\lvert f_n \rvert \leq g_n$ for all $n$. If we let $k = \text{sup}\{ k_n \}$, then $\lvert f_n \rvert \leq \frac{k}{k_n}g_n$
It seems like I should be able to do something like this using the Archimedean property and the fact that it does not change the value of the limit since $k \cdot lim_{n \rightarrow \infty} \int_E \lvert f - f_n \rvert = k \cdot 0 = 0$.
I finish the forward direction of the proof by stating: since $g_n$ converges, and $f_n \rightarrow f$, it follows by the General Lebesgue Dominated Convergence Theorem that
$$ \lim_{n \rightarrow \infty} \int_E f_n = \int_E f $$
as desired.
Yes, but, how would you use the dominated convergence theorem there? You have to talk about the limit of the integrals of the absolute values of $f_n$, not $f_n$ themselves.
This implication is a consequence of the inequality $$\left|\int_E|f_n|-\int_E|f|\right|=\left|\int_E(|f_n|-|f|)\right|\leq\int_E||f_n|-|f||\leq\int_E|f_n-f|,$$ and the last term goes to $0$. This is also one way of getting this at the proof of dominated convergence theorem.
Edit: For the other direction, you can do the following: first, you have that $\left||f_n-f|-|f_n|\right|\leq|f_n-f-f_n|=|f|$ and $f$ is integrable. Since $f_n\to f$ almost everywhere, you have that $|f_n-f|-|f_n|\to -|f|$ almost everywhere. Therefore, from the dominated convergence theorem, $$\int_E(|f_n-f|-|f_n|)\to-\int_E|f|,$$ hence $$\int_E|f_n-f|=\int_E(|f_n-f|-|f_n|)+\int_E|f_n|\to-\int_E|f|+\int_E|f|=0.$$