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.


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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