$f_n \rightarrow 0$ a.e. on $[0,1]$ & $\int_{[0,1]} |f_n|^2 dm \leq 1$ $\implies$ $\int_{[0,1]} |f_n| dm \rightarrow 0$

Question

Let $f_n : [0,1] \rightarrow \mathbf{R}$ be a sequence of measurable functions such that

$\bullet$ $f_n \rightarrow 0$ a.e. on $[0,1]$.

$\bullet$ $\int_{[0,1]} |f_n|^2 dm \leq 1$ for all $n \geq 0$.

Then I want to show that $\int_{[0,1]} |f_n| dm \rightarrow 0$ as $n \rightarrow \infty$.

I tried to combine Egorov's Theorem and Dominated Convergence Theorem but I could not find a dominating function for $|f_n|$.

Answer 1

Egoroff's theorem is a good idea. Fix $\varepsilon>0$, $S_\varepsilon$ a set where $\sup_{x\in S_\varepsilon}|f_n(x)|\to 0$ and $\mu([0,1]\setminus S_\varepsilon)<\varepsilon$. We have \begin{align} \int_{[0,1]}|f_n|d\mu&\leqslant\int_{S_\varepsilon}|f_n|d\mu+\int_{[0,1]\setminus S_\varepsilon}|f_n|d\mu\\ &\leqslant\mu(S_{\varepsilon})\sup_{x\in S_\varepsilon}|f_n(x)|+\sqrt{\mu([0,1]\setminus S_\varepsilon)}\sqrt{\int_{[0,1]}|f_n|^2d\mu}\\ &\leqslant\sup_{x\in S_\varepsilon}|f_n(x)|+\varepsilon. \end{align} This gives $\limsup_{n\to +\infty}\int_{[0,1]}|f_n|d\mu\leqslant\varepsilon$ and we conclude as $\varepsilon$ is arbitrary.

Answer 2

Since the sequence is $L^2$-bounded, it is uniformly integrable and therefore your result holds.