Prove that $\int_E |f_n-f|\to0 \iff \lim\limits_{n\to\infty}\int_E|f_n|=\int_E|f|.$

I'm reading Real Analysis by Royden 4th Edition.

The entire problem statement is:

Let $\{f_n\}_{n=1}^\infty$ be a sequence of integrable functions on $E$ for which $f_n\to f$ pointwise a.e. on $E$ and $f$ is integrable over $E$. Show that $\int_E |f_n-f|\to0 \iff \lim\limits_{n\to\infty}\int_E|f_n|=\int_E|f|.$

My attempt at the proof is:

$(\Longrightarrow)$ Suppose $\int_E|f_n-f|\to0$ and let $\varepsilon>0$ be given. Then there exists an $N>0$ such that if $n\geq N$ then $|\int_E|f_n-f||<\varepsilon.$ Consider $$|\int_E|f_n|-\int_E|f||=|\int_E(|f_n|-|f|)|\leq|\int_E|f_n-f||<\varepsilon.$$ Thus, $\int_E|f_n|\to\int_E|f|.$

$(\Longleftarrow)$ Suppose now that $\int_E|f_n|\to\int_E|f|.$ Let $h_n=|f_n-f|$ and $g_n=|f_n|+|f|$. Then $h_n\to0$ pointwise a.e. on $E$ and $g_n\to2|f|$ pointwise a.e. on $E$. Moreover, since each $f_n$ and $f$ are integrable $\int_E g_n=\int_E|f_n|+|f|\to2\int_E|f|.$ Thus, by the General Lebesgue Dominated Convergence Theorem, $\int_E|f_n-f|\to\int_E0=0.$

I'm pretty sure I got this one down, but I was wondering if it was okay for $g_n$ to depend on $f$ or $f_n$ or does it need to be independent of them?

Thanks for any help or feedback!

As Prahlad Vaidyanathan said, the proof is correct. I would probably not use $\varepsilon$ in the first part, instead writing

$$ \left|\int_E|f_n| - \int_E|f| \right| = \int_E \big||f_n|-|f|\big|\le \int_E |f_n-f|\to 0 $$ and invoking the squeeze lemma.

Fatou's Lemma is your friend. By Fatou,

\begin{align*} \int_{E} 2|f| &= \int_{E} \liminf_{n\to\infty} (|f| + |f_n| - |f-f_n|) \\ &\leq \liminf_{n\to\infty} \int_{E} (|f| + |f_n| - |f-f_n|) \\ &= 2\int_{E} |f| - \limsup_{n\to\infty} \int_{E} |f-f_n|. \end{align*}

So it follows that $\limsup_{n\to\infty} \int_{E} |f-f_n| = 0$ and the desired conclusion follows. You may also want to give a look on Scheffé's lemma.