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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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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.

    
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

1 Answer 1

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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