Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

We know by Vitali Converse that:

let $\mu(E)<\infty$ and {$h_n$} is a sequence of "nonnegative" integrable functions that converges pointwise $a.e.$ on $E$ to $h=0$.

Then $\lim_{n\rightarrow\infty}\int_Eh_n=0$ iff {$h_n$} is uniformly integrable over $E$.

Why it does not hold without the assumption that {$h_n$} is nonnegative?

share|cite|improve this question
up vote 3 down vote accepted

Let $E = [-1,1]$ with Lebesgue measure and take $h_n = n (1_{(0, 1/n)} - 1_{(-1/n, 0)})$. We have $h_n \to 0$ pointwise and $\int h_n = 0$ for every $n$, but $\{h_n\}$ is not uniformly integrable.

share|cite|improve this answer
Does it converge pointwise a.e to $h=0$? – Anita Nov 13 '12 at 16:06
@Anita: $h_n$ converges pointwise to $-1_{(-1,0)}$. – Nate Eldredge Nov 13 '12 at 20:26
That's the point, it should converge to $h=0$ pointwise a.e. – Anita Nov 13 '12 at 22:34
@Anita: Thanks, I missed that. I modified my example. – Nate Eldredge Nov 13 '12 at 22:38
@Anita: Oh, but I see you got there first, by 26 seconds :) – Nate Eldredge Nov 13 '12 at 22:39

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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