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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 $$ g_n=|f_n|+|f|-|f_n-f|.$$

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  • $\begingroup$ Thanks for your hint, it was really helpful! I've edited my answer with my new tries. Can you give me some appointment? Thanks in advance $\endgroup$ – Rodrigo Dias Oct 17 '17 at 2:19
  • $\begingroup$ You were on the right track at the beginning, but you didn't use the fact that $\int|f_n|\to \int |f|$. Once you add this in, you will be able to cancel a factor of $2\int |f|$ from both sides. $\endgroup$ – carmichael561 Oct 17 '17 at 2:22
  • $\begingroup$ How about now? I got $\limsup(-\int|f_n-f|) = 0$, which is almost what we want! $\endgroup$ – Rodrigo Dias Oct 17 '17 at 2:32
  • $\begingroup$ It's easier than that: $\liminf_{n}\int g_n=\int|f|+\lim_n\int|f_n|+\liminf_n(-\int |f_n-f|)=2\int|f|-\limsup_{n}\int |f_n-f|$. After cancelling $2\int |f|$ and rearranging, Fatou yields $\limsup_n\int|f_n-f|\leq 0$. $\endgroup$ – carmichael561 Oct 17 '17 at 2:37
  • $\begingroup$ Why does the first equality hold? It should be "$\geq$", shouldn't? (I'm thinking in $\liminf(a_n+b_n) \geq \liminf a_n + \liminf b_n$). $\endgroup$ – Rodrigo Dias Oct 17 '17 at 2:39

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