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Let $\alpha:[a,b]\rightarrow \mathbb{R}$ be a monotonically increasing function.

Let $f\in \mathscr{R}(\alpha)$ on $[a,b]$.

And here's what i have proved a while ago;

"If $\alpha$ is continuous at $a$, then $\lim_{x\to a}\int_{x}^{b} f d\alpha = \int_{a}^{b} f d\alpha$" (Existence of the limit is the part of the conclusion)

However, now i just realized 'continuity of $\alpha$' shouldn't be essential, but i cannot prove this. How do i prove this?

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up vote 1 down vote accepted

According to Wheeden and Zygmund's text, if $f$ and $\alpha$ share a discontinuity, then the Riemann-Stieltjes integral does not exist. This means that we must assume $\alpha$ continuous at $a$, otherwise $f$ and $\alpha$ can share a discontinuity there.

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I just proved that for any $\epsilon>0, -\epsilon -(\inf_{c\in(a,b]} \sup_{x\in[a,c]} f(x))[\alpha(a+)-\alpha(a)]≦\lim_{t\to a} \int_{t}^{b} f d\alpha - \int_{a}^{b} f d\alpha < \epsilon + (\inf_{c\in (a,b]} \inf_{x\in[a,c]} f(x))[\alpha(a+)-\alpha(a)]$. –  Katlus Dec 21 '12 at 1:11
    
Now your statement makes it clear that "If $f\in \mathscr{R}(\alpha)$ on $[a,b]$, then $\lim_{t\to a} \int_{t}^{b} f d\alpha=\int_{a}^{b} f d\alpha$." –  Katlus Dec 21 '12 at 1:14
    
There was a thinko.. It should be $\lim_{t\to a} \int_{t}^{b} f d\alpha + f(a)[\alpha(a+)-\alpha(a)] = \int_{a}^{b} f d\alpha$ –  Katlus Dec 21 '12 at 1:37
    
Glad it makes sense! –  Clayton Dec 21 '12 at 2:28
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