$\int_0^1|f_n|^3\leq 1\Rightarrow \int_E |f_n|<\varepsilon$ when $|E|$ is small Let $f_n\colon [0,1]\to\mathbb{R}$ be Lebesgue measurable with $$\int_0^1|f_n|^3\leq 1 \mbox{ for all } n.$$ Show that for all $\varepsilon&gt;0$ there exists $\delta&gt;0$ so that if $E\subset[0,1]$ is Lebesgue measurable with $|E|&lt;\delta$ then $$\int_E |f_n|&lt;\varepsilon\mbox{ for all } n$$
I don't want to give away the whole answer, but use that $\displaystyle \int_E |f_n|=\int_{[0,1]}\chi_E|f_n|$ and try to use the Holder's inequality hint on the right side (what should $q$ be?).
I would complete proof of JLA. Suppose $ p=3 $ and then Using the equality $ \frac{1}{p} + \frac{1}{q} =1 $ we have $ q= \frac{3}{2}$. We can write
$$ \int_E |f_n|=\int_{[0,1]}\chi_E|f_n| \leq \left(\int_{[0,1]}{|f_n|^3}\right)^\frac{1}{3} \left(\int_{[0,1]}{\chi_E}^{\frac{3}{2}}\right)^{\frac{2}{3}}$$