Let [itex]\alpha(t)[/itex] be monotically increasing on [0,1]. Prove that(adsbygoogle = window.adsbygoogle || []).push({});

[itex]\lim_{n \rightarrow \infty} \int_0^1 t^n d\alpha(t)=\alpha(1)-\alpha(1-)[/itex] where [itex]\alpha(1-)=\lim_{t \rightarrow 1^{-}} \alpha(t)[/itex].

Here's what I have so far. I know that [itex]\alpha(t) [/itex] is monotonically increasing, so it has at most countably many points of discontinuity. So it is continuous almost everywhere which implies that it is Riemann integrable. That means that [itex]\int_0^1 t^n d\alpha(t)=\int_0^1 t^n \alpha ' (t) dt[/itex] where the second integral is just a plain Riemann integral.

Then integrating by parts with [itex]u=t^n[/itex] and [itex]dv=\alpha ' (t)dt[/itex], I get that [itex]\int_0^1 t^n d\alpha(t)=\int_0^1 t^n \alpha ' (t) dt= \alpha(1) - \int_0^1 \alpha(t) n t^{n-1} dt[/itex]. This is where I'm stuck. I can't get that [itex]\lim_{n \rightarrow \infty} \int_0^1 \alpha(t) n t^{n-1} dt = \alpha(1-) [/itex] In fact, it looks like it should blow up to me.

Any help would be appreciated.

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# Difficult integral

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