In this paper, I am questioning the proof of the following lemma (Lemma 2, page 5):
Assume $F$ is differentiable with $F'=f$ continuous. Then if $g$ is integrable, $$\int_{I}g(x)dF(u)=\int_{I}g(u)F'(u)du.$$
Here is the proof they give:
"We derive the result via Riemann-Stieltjes sums: In the sum $$\sum_{i=1}^{n}g(\xi_{i})(F(x_{i})-F(x_{i-1}))$$ the factor $(F(x_{i})-F(x_{i-1}))=f(\eta_{i})(x_{i}x_{i-1})$ for some $\eta_{i}\in(x_{i-1},x_{i})$, by the Mean Value Theorem. Therefore $$\sum_{i=1}^{n}g(\xi_{i})(F(x_{i})-F(x_{i-1}))=\sum_{i=1}^{n}g(\xi_{i})f(\eta_{i})(x_{i}-x_{i-1}),$$ which we recognize as a Riemann sum for the integral $\int gfdu$ and the result is proven."
I dont think his final argument need to be true. Since I encountered a similar argument yesterday in another paper (page 25), I asked a similar question on Stack Exchange yesterday (notation differs a little). One of the comments convinced me even more that this argument may be false. The problem (in the proof that I copied above) is that the variable $\xi_{i}$ in $g$ and $\eta_{i}$ in $f$ may differ.
Can someone please explain his last argument? I.e. the Riemann integral estimation. I am so confused. Any help is greatly appreciated.
Since one typically assumes that $g$ is bounded and $I = [a,b]$ is a compact interval, one can use that $f$ is uniform continuous on $I$. Let $\delta>0$ be chosen such that $|x-y| < \delta$ implies that $$|f(x)-f(y)| < \frac{\varepsilon}{\|g\|_\infty (b-a)}.$$ Now if the partition is chosen such that the mesh is less than $\delta$, then we get that the sum $$\sum_{i=1}^{n}g(\xi_{i})f(\eta_{i})(x_{i}-x_{i-1})$$ differs from $$\sum_{i=1}^{n}g(\xi_{i})f(\xi_i)(x_{i}-x_{i-1})$$ only by an amount of $$\Bigg| \sum_{i=1}^{n}g(\xi_{i})f(\eta_{i})(x_{i}-x_{i-1}) - \sum_{i=1}^{n}g(\xi_{i})f(\xi_{i})(x_{i}-x_{i-1}) \Bigg| \leq \|g\|_\infty (b-a) \max_i |f(\xi_i) - f(\eta_i)| < \varepsilon.$$ This argument completes the proof of the mentioned 'derivate-rule'.
Additional comment: If one likes Stieltjes-Lebesgue-Integrals, this formula says that the measure induced by $F$ is absolute continious with respect of the Lebesgue-measuren with Radon-Nikodym derivate given by $F'$.