Question

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 $n \to \infty$?

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.