I am given the following problem, with a hint that states that I should use the General Lebesgue Dominated Convergence Theorem.

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

($\Rightarrow$) Let $g_n = k_n \lvert f - f_n \rvert$ where $k_n \in \mathbb{R}$ and $\lvert f_n \rvert \leq g_n$ for all $n$. If we let $k = \text{sup}\{ k_n \}$, then $\lvert f_n \rvert \leq \frac{k}{k_n}g_n$

It seems like I should be able to do something like this using the Archimedean property and the fact that it does not change the value of the limit since $k \cdot lim_{n \rightarrow \infty} \int_E \lvert f - f_n \rvert = k \cdot 0 = 0$.

I finish the forward direction of the proof by stating: since $g_n$ converges, and $f_n \rightarrow f$, it follows by the General Lebesgue Dominated Convergence Theorem that

$$ \lim_{n \rightarrow \infty} \int_E f_n = \int_E f $$

as desired.

  • 1
    $\begingroup$ It's unclear what you're asking. Do you need help verifying your proof (@detnvvp helped you there), or are you seeking help with the other direction? $\endgroup$ Oct 31 '13 at 23:36
  • $\begingroup$ I needed help verifying that choosing $k$ in the way I did above was valid. $\endgroup$ Oct 31 '13 at 23:45
  • $\begingroup$ I don't see a reason why there would be some $k$ such that $|f_n|\leq k|f-f_n|$ for all $n$, but as mentioned below, you don't need it for this direction. $\endgroup$ Nov 1 '13 at 0:33
  • $\begingroup$ Could we define a $k_n$ for each n, and have $k$ be the supremum of all such $k_n$? $\endgroup$ Nov 1 '13 at 2:23
  • 1
    $\begingroup$ No, saying that no such $k$ necessarily exists is the same as saying the supremum might be $\infty$. $\endgroup$ Nov 1 '13 at 8:43

Yes, but, how would you use the dominated convergence theorem there? You have to talk about the limit of the integrals of the absolute values of $f_n$, not $f_n$ themselves.

This implication is a consequence of the inequality $$\left|\int_E|f_n|-\int_E|f|\right|=\left|\int_E(|f_n|-|f|)\right|\leq\int_E||f_n|-|f||\leq\int_E|f_n-f|,$$ and the last term goes to $0$. This is also one way of getting this at the proof of dominated convergence theorem.

Edit: For the other direction, you can do the following: first, you have that $\left||f_n-f|-|f_n|\right|\leq|f_n-f-f_n|=|f|$ and $f$ is integrable. Since $f_n\to f$ almost everywhere, you have that $|f_n-f|-|f_n|\to -|f|$ almost everywhere. Therefore, from the dominated convergence theorem, $$\int_E(|f_n-f|-|f_n|)\to-\int_E|f|,$$ hence $$\int_E|f_n-f|=\int_E(|f_n-f|-|f_n|)+\int_E|f_n|\to-\int_E|f|+\int_E|f|=0.$$

  • $\begingroup$ I thought all of the requirements were satisfied since $f_n \rightarrow f$, $g_n \rightarrow g$, $\lvert f_n \rvert \leq g_n$ , and $\int_E g_n \rightarrow \int_E g$. $\endgroup$ Oct 31 '13 at 23:51
  • $\begingroup$ I see. If I have understood well, the requirements are satisfied, but you have to talk about the convergence $|f_n|\to |f|$ (not of $f_n\to f$), and the fact that $|f_n|$ is dominated. But, this is sort of an overkill; to do the first part you can do what I wrote above. $\endgroup$
    – detnvvp
    Nov 1 '13 at 5:37
  • $\begingroup$ Okay, how should I think about the reverse direction? I tried using the same kind of inequalities with Fatou's lemma, but that led me nowhere. $\endgroup$ Nov 1 '13 at 6:07
  • $\begingroup$ Although, does $\lim_{n \rightarrow \infty} \int_E f_n = \int_E f$ imply that $\lim_{n \rightarrow \infty} \int_E \lvert f - f_n \rvert$ converge? $\endgroup$ Nov 1 '13 at 6:21
  • $\begingroup$ I edited my answer, it also includes the proof of the other direction now. $\endgroup$
    – detnvvp
    Nov 1 '13 at 7:26

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.