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Let $\{h_n\}$ be a sequence of nonnegative integrable functions on $E$. Suppose $\{h_n(x)\} \to 0$ for almost all $x \in E$. Then $\lim\limits_{n \to \infty} \int_E h_n =0$ iff $\{h_n\}$ is uniformy integrable and tight over $E$.

The backwards direction is just a consequence of the General Vitali Convergence Theorem.

To prove the forward direction. Since $h_n$'s are nonnegative, $h_n \geq 0$ for all n. Since $\{h_n\}$ is integrable, for $n \geq N \in \mathbb{N}$ $\int_E h_n < \epsilon$.

I know uniform integrbaility comes from the fact that $\int_E h_n =0$ iff $\{h_n\}$ is uniformly integrable.

I don't know how to show this is tight.

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What do you mean by "tight over $E$"? – Davide Giraudo Dec 5 '12 at 17:24

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