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For some documents, $$ if \displaystyle\inf_{\text{partition}}U_{\text{partition}}(f)=\sup_{\text{partition}}L_{\text{partition}}(f),\\then~f~is~Riemann\text{-}Stieltjes~integrable. $$ like the following:

img of RSI


However, my textbook said the definition of Riemann-Stieltjes integrability is: $$\lim_{\text{|Partition|}\to0}R_\text{Partition}$$ where $R_\text{Partition}$ is $\sum_\text{Partition}f(\xi_i)[{\phi(x_i)-\phi(x_{i-1})}]$, $\xi\in[x_{i-1}, x_{i}]$, and $\displaystyle\inf_{\text{partition}}U_{\text{partition}}(f)=\sup_{\text{partition}}L_{\text{partition}}(f)$ is not an equivalent definition of the Riemann-Stieltjes integrability.

My book gave a counter-example:

for the interval $[-1, 1]$, \begin{align} f(x)=\begin{cases}0\quad x\in[-1, 0) \\ 1\quad x\in [0,1]\end{cases} \end{align} \begin{align} \phi(x)=\begin{cases}0\quad x\in[-1, 0] \\ 1\quad x\in (0,1]\end{cases} \end{align}

The book concluded that in this case, $\inf U(f) = \sup L(f)$ but the value of Riemann-Stieltjes integral does not exist.

Which is correct?

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    $\begingroup$ I wouldn't want to say which is "correct". You might note that if $f$ and $\phi$ have no common points of discontinuity then the two are equivalent. $\endgroup$ – David C. Ullrich Mar 30 '16 at 13:45
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    $\begingroup$ You mean, is the definition 7.1.17 wrong? Because it does not give the assumption "$f$ and $\phi$ have no common points of discontinuity". $\endgroup$ – Danny_Kim Mar 30 '16 at 13:46
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    $\begingroup$ No I don't mean that the definition is wrong! I said I wasn't going to comment on which definition was correct - it sort of follows I'm not going to comment on which definition is wrong... $\endgroup$ – David C. Ullrich Mar 30 '16 at 13:48
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    $\begingroup$ @AloizioMacedo Ah, |Partition| is defined $\max\{\text{the length of Partitions}\}$. $\endgroup$ – Danny_Kim Mar 30 '16 at 13:57
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    $\begingroup$ @AloizioMacedo Ahha! I edited. Thank you! $\endgroup$ – Danny_Kim Mar 30 '16 at 14:38
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The two definitions do not agree. However, there is no Mathematics Authority that declares definitions correct. So we cannot say that one is correct and the other is incorrect. We can only say that they are different.

So, when learning from a certain textbook, use the definition in that textbook.

What about after that course? Which definition should you use then? As a practical matter, nowadays mathematicians almost always use the Lebesgue-Stieltjes integral (or just integration with respect to a measure), and do not mention Riemann-Stieltjes integrals at all. In reading old literature, you may find Riemann-Stieltjes integrals; but you will find that the only cases considered are the ones where both definitions agree. As David Ullrich said, these are cases where the integrator and the integrand have no common discontinuities.

If, in some rare case, you have a Riemann-Stieltjes integral where the definitions disagree, then you should specify which definition to use. Maybe you would do this when you send a tricky problem to the problem section of the American Mathematical Monthly, for example.

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Expanding on what GEdgar said: In my opinion the "right" definition is neither of the two you cite, it's the Lebesgue-Stieltjes integral.

Assuming you know a little reals: If $f$ is continuous there's no problem. Now the Riesz Representation Theorem shows that there is a measure $\mu$ on $[a,b]$ so that $$\int_a^bf\,d\phi=\int f\,d\mu$$for every $f$ continuous on $[a,b]$. Now if I had to say what the "right" definition was in your example I'd say it was $\int f\,d\mu$.

But, again as he said, note I'm putting "right" in scare quotes here...

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  • $\begingroup$ For comparison with the two definitions in the Question: given $\phi$ do you want to define $\mu$ so that $\mu\big([a,x]\big) = \phi(x)$ or so that $\mu\big([a,x)\big) = \phi(x)$ or what? If $\phi$ is continuous at $x$ it makes no difference. Or, if you are integrating functions $f$ that are continuous at $x$ it also makes no difference. This question does come up in elementary probability texts: given a random variable, how do you define its "cumulative distribution function"? $\endgroup$ – GEdgar Mar 30 '16 at 15:33
  • $\begingroup$ Hmm. I don't understand what he said. Actually, I've not learned about definition of measure yet. $\endgroup$ – Danny_Kim Mar 30 '16 at 17:49
  • $\begingroup$ @GEdgar Yeah, it comes up. When I have to I follow Folland, saying $\mu([a,x]) = \phi(x)-\phi(a)$. Which of course only works if $\phi$ is right-continuous. When I'm in charge what I do is neither of those - I do what I said in my answer: Define the integral for continuous $f$, then use the Riesz Representation Theorem to get $\mu$. Not consistent with either of the two possibilities you mention, but preferable in my opinion - for example $\phi=\chi_{[0,1/2)}$ "should" imo be the same as $\phi=\chi_{[0,1/2]}$, and with "my" approach it is. $\endgroup$ – David C. Ullrich Mar 30 '16 at 19:01
  • $\begingroup$ @Danny_Kim It's ok, I think the question was for me... $\endgroup$ – David C. Ullrich Mar 30 '16 at 19:02

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