# 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.

• 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 $n$? – stressed out Mar 11 at 17:11
• Take $\alpha(x)=0$ for $x<1$ and $\alpha(1)=1$. Then the answer is just $\alpha(1)-\alpha(1-)=1-0=1$. – Alex R. Mar 11 at 17:36
• @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. – user397146 Mar 11 at 18:56

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

$$\lim_{n \to \infty}\int_0^1 x^n \, d\alpha(x) = \alpha(1) - \alpha(1-)$$

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.