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 $n$.
About integrable functions.
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3Hello and welcome on MSE, this page doesn't work by just pasting a question in imperative form and waiting for others to solve it. You are supposed to show some effort and thoughts about the question and to point out which problems you have with solving the question on your own. – 2012-03-03
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0Is the last integral supposed to be over $E$? – 2012-03-03
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2Yes, the last integral is supossed to be over $E$. – 2012-03-03
2 Answers
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 $\delta=\varepsilon^{3/2}$.
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.