$f_n$ Lebesgue measurable functions on $[0,1]$ such that $\int_0^1 |f_n| dm \rightarrow 0$ and $|f_n|^2<g$ then $\int_0^1 |f_n|^2 dm \rightarrow 0$

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.

• 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. – user251257 Apr 29 '17 at 3:19

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