# if $\lim_{n\rightarrow\infty}\int_E|f_n|=\int_E|f|$ then $\int_E|f-f_n|\rightarrow 0$

Let $\{f_n\}$ be a sequence of integrable functions on $E$ for which $f_n$ converges to $f$ a.e. on $E$, and $f$ is integrable over $E$. Show that if $\lim_{n\rightarrow\infty}\int_E|f_n|=\int_E|f|$ then $\int_E|f-f_n|\rightarrow 0$.

Given solution is using Lebesgue Dominated Convergence: $0\leq |f_n-f|+|f|-|f_n|\leq 2|f|$. I am having trouble of understanding those two inequalities... Why is it bounded by $2|f|$, and why is that at least $0$? Also is there any other way to show this?

• $\lvert f_n\rvert = \lvert (f_n - f) + f\rvert \leqslant \lvert f_n - f\rvert + \lvert f\rvert$ gives the left inequality after rearranging, and $\lvert f_n - f\rvert + \lvert f\rvert \leqslant (\lvert f_n\rvert + \lvert f\rvert) + \lvert f\rvert$ gives the right after rearranging. Dec 21, 2017 at 20:45
• @Daniel Fischer Thank you! Dec 21, 2017 at 20:56

## 1 Answer

Perhaps you can simply use Fatou's Lemma: \begin{align*} 2\int|f|=\int\liminf_{n}(|f_{n}|+|f|-|f_{n}-f|)\leq\liminf_{n}(\int|f_{n}|+|f|-|f_{n}-f|), \end{align*} so \begin{align*} 2\int|f|\leq 2\int|f|-\limsup_{n}\int|f_{n}-f|, \end{align*} so \begin{align*} \limsup_{n}\int|f_{n}-f|\leq 0. \end{align*}

Anyway, now I see that: $|f_{n}-f|\leq|f_{n}|+|f|$, so $|f_{n}-f|-|f_{n}|\leq|f|$, so $|f_{n}-f|+|f|-|f_{n}|\leq 2|f|$.

• I'm not sure if this answers my question. Dec 21, 2017 at 20:30
• I see no dominated function neither, so I think Lebesgue Dominated Convergence Theorem does not work in my opinion. Dec 21, 2017 at 20:31
• For the issue that $\geq 0$ can be explained: $|f_{n}|=|f_{n}-f+f|\leq|f_{n}-f|+|f|$, so $|f_{n}-f|+|f|-|f_{n}|\geq 0$. Dec 21, 2017 at 20:33
• I have added the details, please take a look. Dec 21, 2017 at 20:35
• This is solid (+1) Dec 21, 2017 at 20:41