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If $\alpha _n\rightarrow \alpha$, then how does one show that for any $j=1,2,\ldots$ and $\epsilon> 0$, if $\sup\int \left | x \right |^{j+\epsilon }d\alpha_n<\infty$, then $\int x^j d\alpha_n \rightarrow \int x^j d\alpha$ as $n \to \infty$?

What if $\epsilon= 0$? Is it still true? For example, if $\sup\int x^2 d\alpha_n<\infty$, then is it necessarily the case that $\int x^2 d\alpha_n\rightarrow \int x^2 d\alpha$ as $n \to \infty$?

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    It is a nice question (+1), but crossposting is not good (-1).2011-09-30

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This question was asked and answered at MathOverflow.

More generally, convergence of the moments under a weak convergence assumption holds under the condition of uniform integrability.