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.

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

  • $\begingroup$ 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? $\endgroup$ – Mike Dec 5 '15 at 12:06
  • $\begingroup$ 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. $\endgroup$ – Justpassingby Dec 5 '15 at 12:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.