Given a function $\alpha$ of bounded variation, show that: $\lim_{n \rightarrow \infty} \int_{0}^{1} x^n d\alpha(x) =\alpha(1) - \alpha(1-)$ 
Given a function $\alpha$ of bounded variation on $[0,1]$, show that: 
  $\lim_{n \rightarrow \infty} \int_{0}^{1} x^n d\alpha(x) =\alpha(1) - \alpha(1-)$.
  Where $\alpha(1-)$ is the left sided limit.

First off, I can't see how this limit is not $0$, I know it is zero if $\alpha$ is continuous. I'm still trying to find an example of a discontinuous $\alpha$ for which this limit is different than zero.
Now, this is what I've attempted for this limit:
Since $\alpha$ is of bounded variation, $\alpha = \beta - \gamma$ for $\beta$ and $\gamma$ increasing functions. So it suffices to show this for an increasing function, say $\beta$.
So,
$\int_{0}^{1} x^n d\beta(x) = \int_{0}^{1-\epsilon} x^n d\beta(x) + \int_{1-\epsilon}^{1} x^n d\beta(x)$ for all $\epsilon \in (0,1)$.
Then, 
$0\le \int_{0}^{\epsilon}x^nd\beta(x) \le \epsilon^n(\beta(\epsilon) - \beta(0))$, so that $\lim_{n \rightarrow \infty} \int_{0}^{1-\epsilon} x^n d\beta(x) = 0$. This means that $\lim_{n \rightarrow \infty} \int_{1-\epsilon}^{1} x^n d\beta(x)$ should be equal to $\alpha(1) - \alpha(1-)$. But I still don't know how to prove this.
 A: Among the many ways to prove this, we can use integration by parts to get
$$\int_0^1 x^n \, d\alpha(x) = 1\cdot\alpha(1) - 0\cdot \alpha(0) - \int_0^1 \alpha(x) dx^n = \alpha(1) - n\int_0^1 x^{n-1} \alpha(x) \,dx $$
Changing variables with $x= u^{1/n}$ in the integral on the RHS, we get
$$\int_0^1 x^n \, d\alpha(x) = \alpha(1) - \int_0^1  \alpha(u^{1/n}) \,du $$
Since $\alpha(u^{1/n}) \to \alpha(1-)$ as $n \to \infty$ for $0 < u < 1$ and $\alpha $ is bounded, we have by the dominated convergence theorem
$$\lim_{n \to \infty}\int_0^1 x^n \, d\alpha(x) = \alpha(1) - \alpha(1-) $$
Following your approach 
You have already shown that for the non-decreasing function $\beta$ (arising in the Jordan decomposition of $\alpha$)
$$\lim_{n \to \infty} \int_0^{1 - \epsilon}x^n \, d\beta(x) = 0$$
Over the interval $[1-\epsilon, 1]$ we have for $n > 1/\sqrt{\epsilon}$,
$$\int_{1 - 1/n^2}^1 x^n \, d\beta(x) \leqslant \int_{1 - \epsilon}^1 x^n \, d\beta(x) \leqslant \beta(1) - \beta(1-\epsilon)$$
Thus, 
$$\left( 1 - 1/n^2 \right)^n (\beta(1) - \beta(1 - 1/n^2)) \leqslant \int_{1 - \epsilon}^1 x^n \, d\beta(x) \leqslant \beta(1) - \beta(1-\epsilon)$$
Taking the limit as $n \to \infty$ we get 
$$\beta(1) - \beta(1-) \leqslant \liminf_{n \to \infty} \int_{1 - \epsilon}^1 x^n \, d\beta(x) \leqslant \limsup_{n \to \infty} \int_{1 - \epsilon}^1 x^n \, d\beta(x) \leqslant \beta(1) - \beta(1 - \epsilon) $$
Since this is true for arbitrarily small $\epsilon$ it follows that
$$\lim_{n \to \infty}\int_0^1 x^n \, d\beta(x) = \beta(1) - \beta(1-)$$
The same argument applies to $\gamma$, and this leads to the desired result.
