# If $f$ is Riemann-Stieltjes Integrable, then does there exist a partition of which each lengths of subinterval are the same?

Let $\alpha$ be a monotonically increasing function.

Say, $f\in\mathscr{R}(\alpha)$.

Then does there exist a partition $P=\{x_0,...,x_n\}$ such that $$x_i=a+ \frac{b-a}{n}i,$$ $i\in\{0,\ldots,n\}$ and $$U(P,f,\alpha)-L(P,f,\alpha)<\epsilon$$ for each $\epsilon>0$?

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Is this a proof of the right-hand rule introduced in early in Calc II? – emka Dec 31 '12 at 0:26

According to this given $\epsilon\gt 0$ there exist a $\delta\gt 0$ such that for any partition $\Gamma$, if $|\Gamma|\lt\delta$, then $$U_\Gamma-L_\Gamma\lt\epsilon.$$
So, if your $f$ is bounded (it must be, otherwise the $U(P,f,\alpha)$ or $L(P,f,\alpha)$ might have no sense), given $\epsilon\gt 0$, in order to pick a uniform partition $$P=\{a=x_0\lt\cdots\lt x_n=b\}$$ such that $$U(P,f,\alpha)-L(P,f,\alpha)\lt\epsilon,$$ it is enough to choose $n$ large enough so that $$\frac{b-a}{n}\lt\delta.$$
Let $\epsilon>0$ be given. Note that $\inf_{P} U(P,f,\alpha)$ may differ from $\inf_{n\in\mathbb{N}} U(T_n,f,\alpha)$ where $T_n=\{a+\frac{b-a}{n}i\in [a,b]| 0≦i≦n\}$. Do you see why above proof fails with respect to my definition? One should prove first that these two infima the same and this is actually what I'm asking. – Katlus Jan 1 '13 at 12:39