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?

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

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

  • $\begingroup$ I'm not sure if this answers my question. $\endgroup$ – 2ndaccount Dec 21 '17 at 20:30
  • $\begingroup$ I see no dominated function neither, so I think Lebesgue Dominated Convergence Theorem does not work in my opinion. $\endgroup$ – user284331 Dec 21 '17 at 20:31
  • $\begingroup$ For the issue that $\geq 0$ can be explained: $|f_{n}|=|f_{n}-f+f|\leq|f_{n}-f|+|f|$, so $|f_{n}-f|+|f|-|f_{n}|\geq 0$. $\endgroup$ – user284331 Dec 21 '17 at 20:33
  • $\begingroup$ I have added the details, please take a look. $\endgroup$ – user284331 Dec 21 '17 at 20:35
  • $\begingroup$ This is solid (+1) $\endgroup$ – Mark Viola Dec 21 '17 at 20:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.