About integrable functions. Let $f_n\colon [0,1]\rightarrow R$ be Lebesgue mensurable with $\int_{0}^{1} |f_n(t)|^3dm(t)<1$ for all $n$. How we can show that $f_n$ is integrable uniformly i.e for all $\epsilon>0$ there exists $\delta>0$ so that if $E\subseteq[0,1]$ is Lebesgue measurable with $m(E)<\delta$ then $$\int_{E}|f_n(t)|dm(t)<\epsilon$$ for all $n$.
 A: We can use Hölder inequality for $p= 3$ and $q=\frac 32$. Then we have for $n\in\mathbb N$ and $E\subset [0,1]$ measurable 
$$\int_E |f_n(t)|dm(t)=\int_{[0,1]} |f_n(t)|\mathbf 1_E(t)dm(t)\leq \left(\int_{[0,1]}|f_n(t)|^3dm(t)\right)^{1/3}\mu(E)^{2/3}\leq \mu(E)^{2/3}$$ 
so for a fixed $\varepsilon$, take $\delta$ such that $\delta^{2/3}\leq \varepsilon$, for example $\delta=\varepsilon^{3/2}$.
A: Suppose by contradiction that you don't have your thesis then there is a sequence of measurable sets such that
$m(E_k)<2^{-k}$ such that $\int_{E_k}|f_{n_k}|\geq \epsilon_0$
Note that your sequence is in a reflexive space. Then there is a subsequence $|f_{k_j}|\rightharpoonup g$ and this convergence is indeed in $L^1$. But we will have for
$$F_{j_0}=\bigcup_{j\geq j_0}E_{k_j}$$
the following
$$\epsilon_0\leq \int_{F_{j_0}} |f_{k_j}|\rightarrow \int_{F_{j_0}} g $$ 
The first inequality holds since $j\geq j_0$ and the limit holds since we have weak convergence.
But it is an absurd because Egoroff's theorem $\int_{F_{j_0}} g\geq \epsilon_0$ $\forall j_0$ when have 
$m(F_{j_0})\leq \sum_{j=j_0}^\infty 2^{-j}\rightarrow 0$.
PS: The simpler response should be chosen I posted that one since I had thougt and it may give you others ideas to solve other problems.
