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Let $\left(\alpha_n \right)_{n\in \mathbb{N}}$ a succesion in $BV[a,b]$ and $f:[a,b] \rightarrow \mathbb{R}$ such that $f \in R_{\alpha_n} [a,b]$. If $\alpha \in BV[a,b]$ and $V_a^b(\alpha_n - \alpha) \rightarrow 0$ when $n \rightarrow \infty$, show that $f \in R_\alpha[a,b]$ and $$\lim_{n\rightarrow \infty}\int_a^b f d\alpha_n = \int_a^b f d\alpha $$

What I tried is :

$$\left|S(\alpha_n, P, T)- S(\alpha, P, T)\right| < \epsilon$$ for any $\epsilon$

But I think i need to prove, that $\lim_{n\rightarrow \infty}\int_a^b f d\alpha_n$ exist and then $S(\alpha, P, T)$ converge to the integral. Or something like that. Is this righ? How to show the existence of the integral? Is there other way to attack this problem?

Hector Albion
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1 Answers1

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First, note that if $f$ is Riemann-Stieltjes integrable, then it must be bounded on $[a,b]$.

Also, note that since $|\alpha_m(x) - \alpha_n(x)|\leqslant |\alpha_m(x) - \alpha(x)|+ |\alpha_n(x) - \alpha(x)|$, we have

$$V_a^b(\alpha_m - \alpha_n) \leqslant V_a^b(\alpha_m - \alpha) + V_a^b(\alpha_n - \alpha)$$

Thus, given $m \geqslant n$,

$$\begin{align}\left|\int_a^b f \, d\alpha_m - \int_a^b f \, d\alpha_n\right| &= \left|\int_a^b f \, d(\alpha_m - \alpha_n)\right| \\ &\leqslant ( \, \sup_{x \in [a,b]} f(x) \, ) V_a^b(\alpha_m - \alpha_n) \\ &\leqslant ( \, \sup_{x \in [a,b]} f(x) \, )( \,V_a^b(\alpha_m - \alpha) + V_a^b(\alpha_n - \alpha)\, ) \end{align}$$

The second inequality is proved here using a proposition proved here.

Since the RHS converges to $0$ as $n \to \infty$, it follows that $\left(\int_a^b f \, d\alpha_n\right)$ is a Cauchy sequence and , hence, convergent.

Now you can proceed to show that $f$ is Riemann-Steiltjes integrable with respect to $\alpha$ using $I = \lim_{n \to \infty}\int_a^b f \, d\alpha_n$ and

$$\left|S(P,f,\alpha) - I \right| \leqslant \left|S(P,f,\alpha) - S(P,f,\alpha_n) \right| + \left|S(P,f,\alpha_n) - \int_a^b f \, d\alpha_n \right| + \left|\int_a^b f \, d\alpha_n - I \right| $$

RRL
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