2

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?

secondaccount
  • 1,106
  • 2
    $\lvert f_n\rvert = \lvert (f_n - f) + f\rvert \leqslant \lvert f_n - f\rvert + \lvert f\rvert$ 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. – Daniel Fischer Dec 21 '17 at 20:45
  • @Daniel Fischer Thank you! – secondaccount Dec 21 '17 at 20:56

1 Answers1

3

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 $|f_{n}-f|+|f|-|f_{n}|\leq 2|f|$.

user284331
  • 56,315