# A bounded family of functions in $L^p[E]$, where E is a measurable set, is uniformly integrable.

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.

• Surely you mean $\|.\|_p$ instead of $\|.\|_\infty$? – Justpassingby Dec 5 '15 at 11:49
• @Justpassingby Yes, thanks I edited it. – Mike Dec 5 '15 at 11:50

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.
• Thank you for your answer. I see how this breaks down. These $n1_{[0,\frac1n]}$ functions do not exist in $L^2(R)$ because the $n^2$ integrand will blow to infinity? – Mike Dec 5 '15 at 12:06
• Yes it will. We could transform the example into $n1_{[0,\frac1{n^2}]}$ or $\sqrt{n}1_{[0,\frac1n]}$ but those are uniformly integrable. – Justpassingby Dec 5 '15 at 12:18