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When we get Riemann-Stieltjes integral becomes standard Riemann integral which calculates area under the curve. We have that $$ s(f,\alpha,P)=\sum_{k=1}^nm_k\Delta\alpha_k \ \text{ and }\ S(f,\alpha,P)=\sum_{k=1}^nM_k\Delta\alpha_k $$ and that lower $\int_{a}^{b}f(x)d\alpha = \sup_P s(f,\alpha,P)$ and upper $\int_{a}^{b}f(x)d\alpha = \inf_PS(f,\alpha,P)$. We also have that $\Delta\alpha_k=\alpha(x_k)-\alpha(x_{k-1})$.

So when $\alpha(x) = x$ we get $\Delta\alpha_k=x_k-x_{k-1}$ in other words we get Riemann integral.

But if we have for example $\alpha(x) = e^x$ then it is clear that $\Delta\alpha_k=e^{x_k}-e^{x_k-1}$. It is clear that for any function $f$ we have that $$s(f,x,P) < s (f,e^x,P) \ \text{ and }\ S(f,x,P) < S(f,e^x,P). $$ This implies that lower $\int_{a}^{b}f(x)d\alpha = \sup_P s(f,x,P) < \int_{a}^{b}f(x)d\alpha = \sup_P s(f,e^x,P)$ and same thing is for upper integral.

So if we have $f=(x)$ and $\alpha(x) = x$ we get Riemann integral which calculates area under the curve and when $\alpha(x) = e^x$ we get Riemann-Stieltjes which is greater then Riemann integral for the same function $f=(x)$. This begs the question what does Riemann-Stieltjes integral calculate when $\alpha(x) \neq x$?

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  • $\begingroup$ It calculates the "area" for some non-uniform definition of the "lenght" of intervals on the x-Axis. $\endgroup$ – fgp Mar 13 '14 at 14:39
  • $\begingroup$ Riemann-Stieltjes integrals are out. I'm 78 now and have never seen one in action outside of the classroom. $\endgroup$ – Christian Blatter Mar 13 '14 at 19:53
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One uses Riemann-Stieltjes when integrating some function along a curve with respect to the curve length.

When looking for bounded functionals on the space of continuous functions (in sup norm), one again encounters Riemann-Stieltjes integrals, here one can visualize them as encoding a density function that is not expressible as derivative. For instance, one can construct the Dirac delta as Riemann-Stieltjes with the unit jump (Heaviside function) as integrator.


Your example comparing $x$ and $e^x$ requires a non-negative $f$, so "any" does not work for $f$. Also, $\int_a^b f(x)\,d(e^x)=\int_a^b f(x)\,e^x\,dx$, so the monotonicity of the Riemann integral already fully justifies the inequality of the integrals.

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You should always keep in mind the following two examples. I assume that the function $f$ is continuous.

1. If the function $\alpha$ is $\mathcal C^1$, then $\int_a^b f(x)\, d\alpha(x)=\int_a^b f(x)\alpha'(x)\, dx$.

2. If $\alpha$ is defined by $\alpha(a)=0$ and $\alpha(x)=1$, $a<x\leq b$, then $\int_a^b f(x)\, d\alpha(x)=f(a)$.

(Much) more generally, as LutzL. said, any positive linear functional $L$ on the space of continuous functions on $[a,b]$ (positive means $L(f)\geq 0$ if $f\geq 0$) can be represented as $$L(f)=\int_a^b f(x)\, d\alpha(x)\, ,$$ for some nondecreasing function $\alpha$.

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