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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.

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  • $\begingroup$ Why is $$0\le \int_{0}^{\epsilon}x^nd\beta(x) \le \epsilon^n(\beta(\epsilon) - \beta(0))$$ true? If you're using $0 \leq x^n \leq 1$ and monotonicity of integration, then why is it $\epsilon^n\big(\beta(\epsilon)-\beta(0)\big)$ instead of $\epsilon\big(\beta(\epsilon)-\beta(0)\big)$? Why the exponent $n$? $\endgroup$ – stressed out Mar 11 at 17:11
  • $\begingroup$ Take $\alpha(x)=0$ for $x<1$ and $\alpha(1)=1$. Then the answer is just $\alpha(1)-\alpha(1-)=1-0=1$. $\endgroup$ – Alex R. Mar 11 at 17:36
  • $\begingroup$ @stressedout Because by the mean value theorem for integrals there is a value $x_0 \in (0,\epsilon)$ with such that $\int_{0}^{\epsilon}x^nd\beta(x) = x_0^n(\beta(\epsilon) - \beta(0))$ and since $x^n$ is increasing, I could just take $\epsilon^n$ so that $\int_{0}^{\epsilon}x^nd\beta(x) \le \epsilon^n(\beta(\epsilon) - \beta(0))$ because $x_0\leq\epsilon$ and because $\beta$ is increasing. $\endgroup$ – user397146 Mar 11 at 18:56
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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.

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