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I have this question and I don't know how to proceed:

Suppose that $(f_n)$ is a sequence of measurable functions on $[0,1]$ such that $\displaystyle \lim_{n \to \infty} \int_0^1 |f_n|\, =0$ and that there is an integrable function $g$ on [0,1] such that $|f_n|^2\leq g$ for all $n$. Show that $\displaystyle \lim_{n \to \infty} \int_0^1 |f_n|^2\, =0$.

I think that I must show $\displaystyle \lim_{n \to \infty} \int_0^1 |f_n|\, = \displaystyle \int_0^1 \lim_{n \to \infty}|f_n|$ , in other words I must show that we can take limit inside the integral. What does $\displaystyle \lim_{n \to \infty} \int_0^1 |f_n|\, =0$ mean? Can you help me for this question? Thanks.

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    $\begingroup$ It is not clear that $\lim_{n\to \infty} |f_n|$ exists in general. $\endgroup$ – user99914 Dec 10 '17 at 8:25
  • $\begingroup$ I guess we will assume it because I wrote the whole question and there is nothing else. $\endgroup$ – user510472 Dec 10 '17 at 8:34
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    $\begingroup$ No! Assuming that makes a huge difference in how easily it can be solved. It makes things way more trivial. Are you sure you want to assume that? $\endgroup$ – Shashi Dec 10 '17 at 8:36
  • $\begingroup$ Oh I understand $\endgroup$ – user510472 Dec 10 '17 at 8:42
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    $\begingroup$ Because if you assume $f_n$ converges then you get by Fatou's Lemma that $\lim_{n\to\infty} |f_n|=0$ a.e. Then apply DCT: tada! All trivial. $\endgroup$ – Shashi Dec 10 '17 at 8:45
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It is easier to argue by contradiction. Assume the contrary that

$$\int_0^1 |f_n|^2 dx$$

does not converge to $0$. Then by picking a subsequence if necessary, assume that there is $\epsilon_0>0$ so that

$$\tag{1} \int_0^1 |f_n|^2 dx \ge \epsilon_0.$$

The fact that $\int_0^1 |f_n| dx \to 0$ implies that (by picking a subsequence if necessary) $f_n \to 0$ almost everywhere. Thus $|f_n|^2 \to 0$ almost everywhere. Now one can use the condition $|f_n|^2 \le g$ and Lebesgue's dominated convergence theorem to conclude

$$ \lim_{n\to\infty} \int_0^1 |f_n|^2 dx = \int_0^1 \lim_{n\to \infty} 0 dx = 0.$$

But this contradicts (1).

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  • $\begingroup$ Well, this only proves a subsequence of $f_n$ converges to $0$ in $L^2$ instead of the whole sequence, I guess? $\endgroup$ – Cave Johnson Dec 10 '17 at 8:38
  • $\begingroup$ We argue by contradiction @CaveJohnson $\endgroup$ – user99914 Dec 10 '17 at 8:39
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    $\begingroup$ Oh I get it. What a clever argument :-) $\endgroup$ – Cave Johnson Dec 10 '17 at 8:40
  • $\begingroup$ It seems to me that the first part up to (1) is unused in the rest of the proof, and could be deleted (turning this into a direct proof rather than one by contradiction). Am I correct? $\endgroup$ – Federico Poloni Dec 10 '17 at 10:16
  • $\begingroup$ @FedericoPoloni it uses the fact that subsequences with the given properties exist. To make it direct you must show it for all subsequences which is not true in general. For example a subsequence that converges to zero a.e. $\endgroup$ – Shashi Dec 10 '17 at 13:34
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Since $\displaystyle\lim_{n\to\infty}\int |f_n|=0$, we have $\displaystyle\lim_{n\to\infty}\mu(|f_n|>1)=0$. By the absolute continuity of Lebesgue integral, $\displaystyle\lim_{n\to\infty}\int_{|f_n|>1} g=0$. Henceforth \begin{align} \limsup_{n\to\infty}\int|f_n|^2&\le\limsup_{n\to\infty}\int_{|f_n|\le1}|f_n|^2+\limsup_{n\to\infty}\int_{|f_n|>1}|f_n|^2\\ &\le\limsup_{n\to\infty}\int_{|f_n|\le1}|f_n|+\limsup_{n\to\infty}\int_{|f_n|>1}g\\ &=0 \end{align}

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