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I'm learning about Riemann-Stieltjes Integration and have a question regarding the details about the set of functions $R_{\alpha}([a,b])$.

I have read two of the following definitions for $R_{\alpha}([a,b])$ online.

  1. $R_{\alpha}([a,b])$ denotes the collection of all $\bf{bounded}$ functions on $[a,b]$ which are Riemann-Stieltjes Integrable with respect to $\alpha$.

  2. $R_{\alpha}([a,b]$ denotes the collection of all function on $[a,b]$ which are Riemann-Stietjes Integrable with respect to $\alpha$.

Is stating boundedness of $f \in R_{\alpha}([a,b])$ in the first definition repetitive by the definition of Upper-Riemann Sums and Lower-Riemann Sums?

$U(f,P)$ = $\sum_{i=0}^{n}M_{i}\Delta\alpha_{i}$ where $M_{i}:\sup (f(x):x_{i} \geq x \geq x_{i-1})$

$L(f:P)= \sum_{i=0}^{n}m_{i}\Delta\alpha_{i}$ where $m_{i} = \inf(f(x):x_{i} \geq x \geq x_{i-1})$

Specifically, because their definitions include the $\sup f(x)$ and $\inf f(x)$ on the intervals $[x_{i-1}, x_{i}]$. Do we implicitly assume that $f$ is bounded on $[a,b]$ to guarantee the existance of the $\sup f(x)$ and $\inf f(x)$ otherwise we couldn't define the Upper-Riemann Sum and Lower-Riemann Sum, let alone the Rienmann-Stieltjes Integral?

Also, just to follow up with the preceding question, by definition we can say $R_{\alpha}([a,b])$ $\subseteq$ $B([a,b])$ where $B([a,b])$ is the set of bounded function on the interval $[a,b]$

Thanks for the help, it's much appreciated.

AJY
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Tomislav
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    Shown here — an unbounded function cannot be Riemann-Stieltjes intrgrable. The basic definition in terms of sums with arbitrary tags cannot be satisfied. This is true even if the integrator is not monotone, and upper and lower sums are not considered. – RRL Mar 03 '19 at 20:43
  • @RRL "the basic definitions in terms of sums with arbitrary tags" - by tags are you referring to the function $f$ evaluated at some point in the interval $[x_{i-1}, x_{i}]$? Thanks! – Tomislav Mar 03 '19 at 20:49
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    Not the original commentor but yes that is what is meant by tags. – rubikscube09 Mar 03 '19 at 20:50
  • @rubikscube09 thanks much appreciated! – Tomislav Mar 03 '19 at 20:52
  • @MatteoLepur: Yes -- your observation that you can't define finite upper Darboux sums if the function is unbounded is correct. Even if we dispense with the Darboux approach the definition for the integral can't be satisfied. You can't enforce $|S(P,f,\alpha) - \int_a^bf , d\alpha| < \epsilon$ for Riemann-Stieltjes sum with arbitrary intermediate points no matter how fine the partition. – RRL Mar 03 '19 at 22:45
  • @RRL Perfect, thanks again! – Tomislav Mar 03 '19 at 22:46

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