I certainly have a question, but i don't know what the best title should be. Please edit the title if there is a better one :)
And I believe, to get a better answer, it would be good to explain exactly what i feel and know..
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Riemann-Stieltjes Integral (Darboux Sum) can be defined for bounded integrand $f$ and nondecreasing $\alpha$. ($\alpha$ does not need to be continuous). However, as far as i know, Lebesgue-Stieltjes Integral requires integrator to be right or left continuous. Moreover, even if there is a way to define measure associated to nondecreasing discontinuous function, integral w.r.t to this function cannot be identical to that of Riemann-Stieltjes. (I'll explain it below)
Below is the result i have proved:
Theorem
Let $\alpha:\mathbb{R}$ be a monotonically increasing right-continuous function.
Let $f:[a,b]\rightarrow \mathbb{R}$ be a RS-integrable function w.r.t $\alpha$ on $[a,b]$.
Let $\mu_\alpha$ be the LS-measure w.r.t $\alpha$.
Then, there exists $F\in L^1(\mu_\alpha)$ such that $\forall x\in(a,b], F(x)=f(x)$ and $\int_a^b f d\alpha = \int_{(a,b]} F d\mu_\alpha$
Since when one wants to use RS-integration, one's interest is exactly in $[a,b]$. So rather than using full domain of $\alpha$ as illustrated in the above theorem, we can replace it by $\alpha(x)=\alpha(a)$ for $x<a$. Then, this new $\alpha$ is continuous at $a$. Thus $\int_a^b f d\alpha = \int_{[a,b]} F d\mu_\alpha$. Hence, in the case $\alpha$ is right continuous, it can be understood that LS-integration is an extension of RS-integration.
However, in general, if $\alpha$ is not right or left continuous, LS-integration cannot be a generalization of RS-integration.
Let $\alpha$ be a nondecreasing function. Suppose there exists a measure $\mu$ such that for some type of interval $【a,b】$, $\mu(【a,b】)=\alpha(b)-\alpha(a)$. (【a,b】denotes one of open,closed,half-open and whatsoever that has an interval form.) Then, the continuity of $\mu$ shows that the only choice that $\mu$ could be a measure is that $\mu((a,b))=\alpha(b-)-\alpha(a+)$.
Indeed, we can construct such measure. Define $F(x)=\alpha(x+)$. Then $F$ is right-continuous nondecreasing function and the LS-measure $\mu$ associated with $F$ is the measure such that $\mu((a,b))=\alpha(b-) - \alpha(a+)$.
In this case, RS-integral and LS-integral of a bounded function do not coincide in general.
When $\alpha$ is continuous, LS seems more natural than RS to me. However, when $\alpha$ is not continuous, LS does not seem natural at all to me..
I'm an undergraduate, so i don't know many cases where integration is useful. I don't have any clue why one should know LS integration. Is there any natural-science example of a function that is only LS integrable but not RS integrable? Or is there any mathematical example that LS integration is useful?
I understand how Lebesgue Integration flows better than Riemann Integration, but I don't see any reason why LS-integration is in need.