What does $\int_0^1 x^n d\alpha(x)$ converge to?

Question

Let $\alpha$ be a monotonically increasing function on $[0,1]$. What does the sequence of Riemann-Stieltjes integrals

\begin{equation*} \int_0^1 x^n d\alpha(x) \end{equation*}

converge to as $n \to \infty$?

I know that if $\alpha(x) = x$ then the sequence converges to $0$. More generally, if $\alpha(x)$ is continuous, then the sequence converges to zero. But what about for general monotone functions $\alpha(x)$?

Answer 1

Write this integral (with $0 < \epsilon < 1$) as

$$\int_0^1 x^n \, d\alpha =\int_0^{1-\epsilon}x^n \, d\alpha + \int_{1-\epsilon}^{1}x^n \, d\alpha $$

Considering the first integral on the RHS, we have

$$0 \leqslant \int_0^{1-\epsilon}x^n \, d\alpha \leqslant (1-\epsilon)^n[\alpha(1) - \alpha(0)] \\ \implies \lim_{n \to \infty}\int_0^{1-\epsilon}x^n \, d\alpha = 0 $$

Considering the second integral, we have for all $n > 1/\sqrt{\epsilon}$

$$\int_{1-\epsilon}^{1}x^n \, d\alpha \geqslant \int_{1-1/n^2}^{1}x^n \, d\alpha \geqslant \left(1 - \frac{1}{n^2} \right)^n[\alpha(1) - \alpha(1- 1/n^2)] \\ \implies \liminf_{n \to \infty}\int_{1-\epsilon}^{1}x^n \, d\alpha \geqslant \alpha(1) - \alpha(1-),$$

since $1 - 1/n \leqslant (1- 1/n^2)^n \leqslant 1$ using the Bernoulli inequality, and, hence, $(1-1/n^2)^n \to 1$.

We also have

$$\int_{1-\epsilon}^{1}x^n \, d\alpha \leqslant \alpha(1) - \alpha(1-\epsilon) \leqslant \alpha(1) - \alpha(1-)$$

Thus,

$$ \alpha(1) - \alpha(1-) \leqslant \liminf_{n \to \infty}\int_{1-\epsilon}^{1}x^n \, d\alpha \leqslant \limsup_{n \to \infty}\int_{1-\epsilon}^{1}x^n \, d\alpha \leqslant \alpha(1) - \alpha(1-),$$

and it follows that

$$ \lim_{n \to \infty}\int_{0}^{1}x^n \, d\alpha =\lim_{n \to \infty}\int_{1-\epsilon}^{1}x^n \, d\alpha = \alpha(1) - \alpha(1-)$$