Let $R_{\alpha} [a,b]$ denote the set of functions $f$ such that the Riemann-Stieltjes integral $\int_a^b f d\alpha$ exists, and let $C [a,b] = \{ f \colon [a,b] \to \mathbb{R} \; : \; f \; \text{is continuous} \}$. I have to prove that

$$\bigcap \{ R_{\alpha}[a,b] \;: \; \alpha \; \text{is monotone increasing} \} = C [a,b] $$

Every continuous function has R-S integral, then I only need to prove that if $f \in \bigcap \{ R_{\alpha}[a,b] \;: \; \alpha \; \text{is monotone increasing} \}$ then $f$ is continuous. What I have done is take an arbitrary $x_0 \in [a,b]$, then for every $\varepsilon >0$ and for every $\alpha$ monotone increasing function there is a partition $P_{\alpha, \varepsilon}=\{t_0=a, t_1, \dots , t_n=b \}$ of $[a,b]$ such that $U(f,\alpha, P_{\alpha, \varepsilon}) - L(f,\alpha, P_{\alpha, \varepsilon}) < \varepsilon$. Let $k$ be a index such that $x_0 \in [t_{k-1},t_k]$ then we have

$$(M_k - m_k) (\alpha(t_{k-1})-\alpha(t_k)) \leq \sum_{i=1}^n (M_i - m_i) (\alpha(t_{i-1})-\alpha(t_i)) \leq U(f,\alpha, P_{\alpha, \varepsilon}) - L(f,\alpha, P_{\alpha, \varepsilon}) < \varepsilon$$

Where $M_i = \sup \{f(x) \; : \; x \in [t_{i-1},t_i] \} $, $m_i = \inf \{f(x) \; : \; x \in [t_{i-1},t_i] \} $

My problem is in the conclusion, $x_0 \in [t_{k-1},t_k]$ then $m_k \leq f(x_0) \leq M_k$, and for every $y \in (t_{k-1},t_k)$ we have $m_k \leq f(y) \leq M_k$, then

$$(f(x_0) - f(y)) (\alpha(t_{k-1})-\alpha(t_k)) \leq (M_k - m_k) (\alpha(t_{k-1})-\alpha(t_k)) < \varepsilon$$

And $$(f(y) - f(x_0)) (\alpha(t_{k-1})-\alpha(t_k)) < \varepsilon$$

I need to choose the correct function $\alpha$ such that for every $\varepsilon >0$ there is a $\delta= t_{k-1}-t_k$, such that if $\vert x_0 - y \vert < \delta$ (this is equivalent to $y \in (t_{k-1}, t_k)$) then $\vert f(x_0) -f(y) \vert < \varepsilon$.


Just use the fact that the Riemann-Stieltjes integral fails to exist if the integrand and integrator have even one common point of discontinuity. This is proved here.

If $f$ is integrable with respect to every monotone increasing $\alpha$, then it must be continuous -- since a monotone increasing function can have jump discontinuities.

  • $\begingroup$ To be precise the integral fails to exist if there are common discontinuities from either the left or right. $\endgroup$ – RRL Aug 12 '18 at 8:46

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