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If $\alpha _{n}\rightarrow \alpha$, then how does one show that for any j=1,2,... 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 stiil 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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Does this question come from homework? coursework? a particular research problem? Also, you have left notation undefined. –  Yemon Choi Sep 30 '11 at 20:56
    
"This question does not appear to be about research level mathematics within the scope defined in the help center." –  Did Aug 31 '13 at 16:53
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closed as off-topic by Did, Yemon Choi, David White, Ramiro de la Vega, John Pardon Aug 31 '13 at 22:25

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question does not appear to be about research level mathematics within the scope defined in the help center." – Did, David White, Ramiro de la Vega
If this question can be reworded to fit the rules in the help center, please edit the question.

1 Answer

If the integrals are finite, we can approximate by integrals on compacta; I'm assuming from the notation you're using that $\alpha_n, \alpha$ are functions of bounded variation, and the integrals are Riemann-Stieltjes. Then there is a standard result that says that if $\alpha_n \rightarrow \alpha$ pointwise on a compact interval and the total variation of $\alpha_n$ is uniformly bounded for all $n$, then $\int f \ d\alpha_n \rightarrow \int f \ d\alpha$.

If the uniformly bounded property doesn't hold, then the variation of $\alpha$ could be infinity and the integral need not converge.

Edit: Question has ambiguous notation, so I'm making some loose assumptions here.

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