# Showing that the bounded $C \subseteq L^p[0,1]$ is uniformly integrable.

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 ?

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

## 1 Answer

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 $$M\delta^{(p-1)/p}\lt\varepsilon$$.