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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?

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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.

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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
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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
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