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!
 A: 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.
A: 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.
A: I think the following theorem should help you understand this equivalence relationship.
Theorem : (Scheffe's theorem). Let $\left\{f_{n}\right\}_{n \geq 1}, f$ be a collection of nonnegative measurable functions on a measure space $(\Omega, \mathcal{F}, \mu) .$ Let $f_{n} \rightarrow f$ a.e. $(\mu), \int f_{n} d \mu \rightarrow \int f d \mu$ and $\int f d \mu<\infty$. Then
$$
\lim _{n \rightarrow \infty} \int\left|f_{n}-f\right| d \mu=0
$$
Proof: Let $g_{n}=f-f_{n}, n \geq 1$. Since $f_{n} \rightarrow f$ a.e. $(\mu)$, both $g_{n}^{+}$and $g_{n}^{-}$go to zero a.e. $(\mu)$. Further, $0 \leq g_{n}^{+} \leq f$ and by hypothesis $\int f d \mu<\infty$. Thus, by the DCT, it follows that
$$
\int g_{n}^{+} d \mu \rightarrow 0
$$
Next, note that by hypothesis, $\int g_{n} d \mu \rightarrow 0$. Thus, $\int g_{n}^{-} d \mu=\int g_{n}^{+} d \mu-$ $\int g_{n} d \mu \rightarrow 0$ and hence, $\int\left|g_{n}\right| d \mu=\int g_{n}^{+} d \mu+\int g_{n}^{-} d \mu \rightarrow 0$.
This certificate is quoted from Measure Theory and Probability Theory. Krishna B. Athreya, Soumendra N. Lahiri.
