# A question about dominated convergence theorem

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.

• It is not clear that $\lim_{n\to \infty} |f_n|$ exists in general. – user99914 Dec 10 '17 at 8:25
• I guess we will assume it because I wrote the whole question and there is nothing else. – user510472 Dec 10 '17 at 8:34
• 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? – Shashi Dec 10 '17 at 8:36
• Oh I understand – user510472 Dec 10 '17 at 8:42
• 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. – Shashi Dec 10 '17 at 8:45

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

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

• Well, this only proves a subsequence of $f_n$ converges to $0$ in $L^2$ instead of the whole sequence, I guess? – Cave Johnson Dec 10 '17 at 8:38
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}