Riemann-Stieltjes intgeraiton and relation to the Riemann Integration I'm trying to find a proof of this relation of the Riemann-Stieltjes integral to the Riemann integral:
$$\int f\ dg = \int fg'dx$$
Wikipedia seems to miss a proof of this and I can't find it anywhere, I've googled "Riemann-Stieltjes integral pdf" and found dozens of them, but couldn't find a proof of it in any of them. I've found a mention to it in the wolfram page but couldn't find the book that is linked in the references. Could somebody present me a proof or an idea of how to prove it?
I've found this:

on the book "A garden of integrals".
I think the last line was suppose to go to $0$, is it true? And how the uniform continuity and boudedness of $\phi'$ will do it?
 A: The $dg$ term is a weight for the integral, and is technically a function of $x$. For the Riemann integral, a special kind of R-S integral, this function is just
$$
g(x)=x
$$ 
So the relation really just comes from applying the differential operator $d$ to the $g(x)$ weight function
$$
dg(x)=g'*dx
$$
By the chain rule. Note that in the case of the Riemann integral this works out to be 
$$
d(x)=dx
$$
as you'd hope.
A: I'm operating on the assumption here that $g'$ is continuous.
By definition,
$$
\int_a^b f(x)\,dg(x)=\lim_{n\to\infty}\sum_{i=0}^{n-1}f(x_i)[g(x_i+\Delta x)-g(x_i)],
$$
where
$$
x_i:=a+i\Delta x,\qquad \Delta x:=\frac{b-a}{n}
$$
Note that we've committed a common abuse of notation here: both $x_i$ and $\Delta x$ actually depend on $n$, although the notation doesn't make this clear.
Now, on the other hand,
$$
\int_a^b f(x)g'(x)\,dx=\lim_{n\to\infty}\sum_{i=0}^{n-1}f(x_i)g'(x_i)\Delta x.
$$
Let's focus on the quantities inside the limits for now.  For any fixed $n$, let $D_n$ be the difference
$$
D_n:=\left[\sum_{i=0}^{n-1}f(x_i)[g(x_i+\Delta x)-g(x_i)]\right]-\left[\sum_{i=0}^{n-1}f(x_i)g'(x_i)\Delta x\right]
$$
Note that if we assume $f$ continuous on $[a,b]$, then $f$ is bounded on $[a,b]$; so, there exists some $M\in\mathbb{R}$ such that $\lvert f(x)\rvert\leq M$ for all $x\in[a,b]$. Then
$$
\begin{align*}
\lvert D_n\rvert&=\left\lvert\sum_{i=0}^{n-1}f(x_i)\left[g'(x_i)-(g(x_i+\Delta x)-g(x_i))\right]\right\rvert\\
&\leq\sum_{i=0}^{n-1}\left\lvert f(x_i)[g'(x_i)\Delta x-(g(x_i+\Delta x)-g(x_i))]\right\rvert\\
&\leq M\Delta x\sum_{i=0}^{n-1}\left\lvert g'(x_i)-\frac{g(x_i+\Delta x)-g(x_i)}{\Delta x}\right\rvert
\end{align*}
$$
Now, for each $i$, there is $c_i\in[x_i,x_i+\Delta x]$ so that 
$$
\frac{g(x_i+\Delta x)-g(x_i)}{\Delta x}=g'(c_i),
$$
so that we can write
$$
\lvert D_n\rvert\leq M\Delta x\sum_{i=0}^{n-1}\left\lvert g'(x_i)-g'(c_i)\right\rvert.
$$
Let $\epsilon>0$. 
Now, $\lvert x_i-c_i\rvert\leq\Delta x$ for all $i$.  And, because $g'$ is continuous on the closed interval $[a,b]$, it is uniformly continuous. So, there exists $\delta>0$ such that 
$$
\lvert x-y\rvert<\delta \qquad\Rightarrow\qquad \lvert g'(x)-g'(y)\rvert<\frac{\epsilon}{(b-a)M}.
$$
(That choice will make sense in a moment.)  Choose $n$ large enough that $\Delta x<\delta$. Then
$$
\begin{align*}
\lvert D_n\rvert&\leq M\Delta x\sum_{i=0}^{n-1}\lvert g'(x_i)-g'(c_i)\rvert\\
&\leq M\Delta x\sum_{i=0}^{n-1}\frac{\epsilon}{(b-a)M}\\
&=M\cdot\frac{b-a}{n}\cdot n\cdot\frac{\epsilon}{(b-a)M}\\
&=\epsilon
\end{align*}
$$
So, the difference between the Rieman-Stieltjes sum and the Riemann sum converges to $0$ as $n\to\infty$; hence the two integrals are equal.
