Almost every convergence imply this equivalence

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Let $(X,{\cal M},\mu)$ be a measure space and let $f_n,f\in L^1(X)$, that is, $\int_X |f_n| {\rm d}\mu < \infty$ and $\int_X |f| {\rm d}\mu < \infty$. Suppose $f_n\to f$ almost everywhere. Then $$ \int_X |f_n| {\rm d}\mu \to \int_X |f| {\rm d}\mu \iff \int_X |f_n-f| {\rm d}\mu \to 0. $$

The $(\Leftarrow)$ implication is trivial, since $$\left|\int_X |f_n| {\rm d}\mu - \int_X |f|{\rm d}\mu\right| \leq \int_X |f_n-f| {\rm d}\mu.$$

How about the other direction? I was trying to use some convergence theorem (monotone, dominated, Fatou's lemma), but I don't know how to proceed.

Any help would be appreciated. Thanks in advance.


EDIT: (Using @carmichael561's hint)

If we define $g_n=|f_n|+|f|-|f_n-f|$, by triangle inequality we have that $g_n \geq 0$ and it follows from Fatou's lemma that $$\int \liminf g_n \leq \liminf \int g_n.$$

On one hand, $g_n\to 2|f|$ a.e. since $f_n\to f$ a.e. On the other hand, \begin{equation*} \begin{split} \liminf\int g_n & = \liminf\left(\int |f_n| + \int |f| - \int |f _n-f|\right) \\ & = \int |f| + \liminf\left(\int |f_n| - \int |f _n-f| \right) \\ & = \int |f| + \int |f| + \liminf\left(-\int |f_n-f|\right), \end{split} \end{equation*} since $\int |f_n| \to \int |f|$ by hypothesis.

We get then $$ 0 \leq \liminf\left(-\int |f _n-f| \right),$$ which gives $\limsup\int|f_n-f| \leq 0$.

Now $\int|f_n-f| \geq 0$ for all $n\in\mathbb N$ implies $\liminf \int|f_n-f| \geq 0$, and therefore $$ 0 \leq \liminf \int |f_n-f| \leq \limsup \int |f_n-f| \leq 0,$$ from where the claim follows.

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The trick with this problem and other similar ones is to find the right sequence of functions to apply Fatou's lemma to. In this case set $$ g_n=|f_n|+|f|-|f_n-f|.$$