# Riemann Stieltjes integral definition and implications

I am studying the Riemann Stieltjes on Tom Apostol's book mathematical analysis second edition and I have a the following question.

Given $[a,b]$ we define a partition of this interval to be a set $P = \{a = x_0, ..., x_n = b\}$

Suppose $f,g:[a,b]\rightarrow \mathbb{R}$ and $f$ is Riemman integrable with respect to $g$ on $[a,b]$ ie for every $\varepsilon > 0$ there exists $P_{\varepsilon}$ partition of $[a,b]$ such that for every $P\subseteq P_{\varepsilon}$ and any choice of points $\{t_k\}$, (the points within each subinterval of the partition), we have that:

$\left | S(P,f,g) - A \right | < \varepsilon$ for some real $A$ which is what we denote by $\int_{a}^{b} fdg(x)$.

Now, my question is, 1) if given the fact that $f$ is integrable with respect to $g$ on $[a,b]$ can we say that $f$ is integrable with respect to $g$ on $(a,b)$ or $(a,b]$ or $[a,b)$? 2) If yes is it direct from the definition??(please be rigorous and not intuitive) 3) Are the integrals always equal?

Is it direct from the definition that the integrals on those intervals are equal? What i see is that the Riemann sums are not equal since

$S(P,f,g) = \sum_{k=0}^{n} f(t_k)(g(x_k)-g(x_{k-1}))$

so when the endpoints are not included we don't have the $g(a)/g(b)$ terms.

In terms of Lebesgue integration one could say that if $g$ is nonnegative and $f$ is measurable then $f1_{[a,b]} = f1_{[a,b)}$ a.s with respect to the measure induced by g, so the corresponding integrals are equal.

Here in Riemann Stieltjes integration does the definition imply directly a same argument or we need some Lebesgue-related proposition??

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As I remember, Riemann(-Stieltjes) integral was always defined on closed intervals. –  Berci Mar 30 '13 at 1:55
I'm going to assume that when you write about $f$ being integrable wrt $g$ on $(a,b]$ and it being the same as $f$ wrt $g$ on $[a,b]$, you mean that $\displaystyle \lim_{\delta \to 0} \int_{a+ \delta}^b fdg$ exists and is equal to $\displaystyle \int_a^b fdg$.
But as you mention, this does not need to be true, even if $\displaystyle \int_a^b fdg$ exists. For example, consider $g(x) = 1$ if $x = 0$ and $0$ otherwise, and let $f$ be a function which is $1$ in a neighborhood $0$, say $[-\alpha,\alpha]$ for some small $\alpha$, of and $0$ everywhere else. Then $\displaystyle \int_0^1 fdg = 1$. But for any $\epsilon > 0$, $\displaystyle \int_\epsilon^1 fdg = 0$ for the exact reason you stated - we don't capture the first endpoint. And if $g$ is discontinuous there, that endpoint really matters. Similarly, if $g$ is terribly discontinuous near $0$, then the integral on $(a,b]$ will have many problems.
But if you have limitation on $g$, like saying that $g$ is continuous (or weaker, of bounded variation), then the limit will exist. But still, the limit on $(a,b]$ does not need to equal the integral on $[a,b]$ even with bounded variation, because of the missing endpoint. (I note that bounded-variation also usually is defined on closed intervals, so there is a little bit of fuzzyness to my argument).