# Vitali's convergence theorem - converse

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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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.
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