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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 $|f_n|$.

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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.

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    $\begingroup$ Where does the second inequality come from? Is it Cauchy-Schwarz? $\endgroup$ – Karatug Ozan Bircan Jan 6 '13 at 13:35
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    $\begingroup$ Yes, exactly, Cauchy-Schwarz. $\endgroup$ – Davide Giraudo Jan 6 '13 at 13:38
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Since the sequence is $L^2$-bounded, it is uniformly integrable and therefore your result holds.

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  • $\begingroup$ A well-known weakening of the $L^2$-boundedness here is $L \log L$-boundedness, that is $$\int_{[0,1]} |f_n| \log^+ |f_n| dm \le M$$ where $\log^+ x$ is the maximum of $\log x$ and $0$. $\endgroup$ – GEdgar Jan 6 '13 at 13:50

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