A corollary in Royden & Fitzpatrick's Real Analysis (chapter 7 section 2) reads:
Let $E$ a measurable set, and $1<p<\infty$. Suppose $F$ is a family of functions in $L^p(E)$ that is bounded in $L^p(E)$ in the sense that $\exists$ a constant $M$ for which $||f||_p \leq M, \forall f\in E$. Then, the family F is uniformly integrable over E.
The proof is an epsilon-delta proof for uniform integrability using Holder's inequality. My question is, why doesn't this proof include the case that $p=1$?
If I go through the proof for a sequence $(f_n)$ of bounded functions in $L^1[0,1]$, I can show that it is uniformly integrable. So something is telling me that for the $p=1$ case I would need the measure of the domain $E$ to be bounded, but I don't see why.
Any insight is appreciated. Especially if you think I'm in error claiming that $(f_n)$ bounded in $L^1[0,1]$ is uniformly integrable.
Consider the functions $n1_{[0,\frac1n]}$ in $L^1(\mathbb R)$. They are all in the unit ball but for every $\delta>0$ there exist an infinite number among them having integral 1 on some set of measure less than $\delta$ (use the interval $[0,\delta]$.
In the proof by Royden and Fitzpatrick the $q$ norm of the indicator of the set $A$ goes to 0 as the measure of $A$ becomes arbitrarily small. But for the supremum norm, the norm of the indicator remains equal to 1.