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Let $f_n$ be a sequence of Lebesgue measurable functions on $[0,1]$ such that $\int_0^1 |f_n| dm \rightarrow 0$ and there exists a integrable function $g$ such that $|f_n|^2<g$. Then we need to show $\int_0^1 |f_n|^2 dm \rightarrow 0$.

Since $|f_n| <1$, $|f_n|^2<1$. But can I show that $\lim_{n \rightarrow \infty}\int_0^1 |f_n|^2 dm \leq \lim_{n \rightarrow \infty}\int_0^1 |f_n|^2 dm$? I was trying to apply Lebesgue Dominated Convergence Theorem, but cound not find the way.

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  • $\begingroup$ Hm a short answer: since $f_n$ converges in $L^1$, it converges in measure, thus $|f_n|^2$ converges in measure. Then, apply dominated convergence theorem. $\endgroup$ – user251257 Apr 29 '17 at 3:19
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Pick $ \epsilon > 0 $, and let $ A_n = \{ x \in [0, 1] : |f_n(x)| > \epsilon \} $. Then, we must have that $ m(A_n) \to 0 $. Indeed, assume not, then there is some $ R > 0 $ such that for infinitely many $ n $, we have $ m(A_n) > R $. But then, it follows that for infinitely many $ n $, we have

$$ \int_0^1 |f_n| \, dm \geq \int_{A_n} |f_n| \, dm\ \geq \epsilon R $$

which contradicts the convergence to $ 0 $. Now, we have that

$$ \int_0^1 |f_n|^2 \, dm = \int_{A_n} |f_n|^2 \, dm + \int_{[0, 1] - A_n} |f_n|^2 \, dm \leq \int_{A_n} g \, dm + \epsilon^2 $$

Now, taking limits as $ n \to \infty $ gives the desired result.

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  • $\begingroup$ So $m(A_n) \rightarrow 0$ implies $\int_{A_n} g~dm \rightarrow 0$, but I could not prove it. $\endgroup$ – Arindam Apr 30 '17 at 5:04
  • $\begingroup$ @Arindam It's trivial for bounded $ g $, and bounded measurable functions are dense in $ L^1[0, 1] $. $\endgroup$ – Starfall Apr 30 '17 at 5:18

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