Exrcise :

Let $C \subseteq L^p[0,1], 1 < p < \infty$ be bounded. Show that $C$ is uniformly integrable.

Attempt :

It is $L^p[0,1] \subseteq L^1[0,1]$ and $L^p[0,1] \hookrightarrow L^1[0,1] \implies \exists c>0 : \|u\|_1 \leq c \|u\|_p \; \forall u \in L^p$. Thus $C$ is bounded in $L^1[0,1]$ as well. Now, we have : $$\int_A |u| \mathrm{d}x = \int_{[0,1]} |u| \chi_A \mathrm{d}x \leq \|u\|_p \|\chi_A\|_{p'} \leq M\left(\int_{[0,1]} \chi_A^{p'}\mathrm{d}x\right)^{1/p'} = M |A|_N^{1/p}$$ For $|A|_N < \varepsilon$ (aka arbitrarilly small) we get the $\varepsilon-\delta$ definition of uniform integrability.

Question : Is my approach correct and rigorous ?

  • 1
    $\begingroup$ looks good .......... $\endgroup$ – daw May 9 at 6:24
  • $\begingroup$ @daw Thanks for the update ! $\endgroup$ – Rebellos May 9 at 6:34

What do did look correct. Here are some remarks:

  • We can take $c=1$ (you work with Lebesgue measure, I guess).
  • What is the meaning of $\left\lvert A \right\rvert_N$? What you obtained from Hölder's inequality is $$\int_A |u| \mathrm{d}x\leqslant M\left\lvert A\right\rvert^{(p-1)/p} $$ hence it suffices, for a fixed $\varepsilon$, to find $\delta$ such that $M\delta^{(p-1)/p}\lt\varepsilon$.

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