6
$\begingroup$

Let $g:[a,b] \to \mathbb{R}$ be a monotone function. Could you help me prove that $\mathcal{C}([a,b])\subseteq\mathcal{R}([a,b],g)$?

(Here $\mathcal{R}([a,b],g)$ is the set of all functions that are Riemann-Stieltjes integrable with respect to $g$.)

Definition of the Riemann-Stieltjes integral. Suppose $f,g$ are bounded on $[a,b]$. If there is an $A \in \mathbb{R}$ such that for every $\varepsilon >0$, there exists a partition $\mathcal{P}$ of $[a,b]$ such that for every refinement $\mathcal{Q}$ of $\mathcal{P}$ we have $|I(f,\mathcal{Q},X,g)-A|<\varepsilon$ (where if $\mathcal{P}=\{a=x_0<\ldots<x_n=b\}$ and $X$ is an evaluation sequence $X=\{x_1^\prime,\ldots,x_n^\prime\}$ and $I(f,\mathcal{Q},X,g)=\sum_{j=1}^n f(x_j^\prime)(g(x_j)-g(x_{j-1}))$), then $f$ is R-S integrable with respect to $g$, and the integral is $A$.

$\endgroup$
  • 1
    $\begingroup$ Hint: continuous in $[a,b]$ implies uniformly continuous in $[a,b]$. What is your definition of integrable with respect to $g$ (monotone)?, there is various definitions which are not equivalents. $\endgroup$ – leo Dec 4 '11 at 3:25
  • $\begingroup$ I added the definition of integral. $\subset$ means subset, I will edit the question $\endgroup$ – Mec Dec 4 '11 at 3:52
3
$\begingroup$

Assume that $g$ is increasing. I suppose that you know that $f\in\mathcal{R}([a,b],g)$ iff $f$ satisfies the Riemann's condition. The Riemann's condition says:

$f$ satisfies the Riemann's condition respect to $g$ in $[a,b]$ if for every $\epsilon\gt 0$, there exist a partition $P_\epsilon$ of $[a,b]$ such that if $P$ is a refinement of $P_\epsilon$ then $$0\leq U(P,f,g)-L(P,f,g)\lt \epsilon,$$ where $U(P,f,g)$ and $L(P,f,g)$ are the upper and lower Riemann-Stieltjes sums respectively.

With this and the hint in the comments the result holds.

Perhaps the chapter 7 of Mathematical Analysis of Tom M. Apostol can be useful to you.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.