# A problem on Riemann Stieltjes Integral

$\int_{0}^2 x\,d \alpha$ where $\alpha (x) = x$ if $0\le x\le 1$ and $\alpha(x)=2+x$ when $1<x\le 2$ I did this by taking a partition which divided the interval $[0,2]$ to $2n$ equal parts

$P=\left\{{0,\frac{1}{n},\frac{2}{n},\dots,\frac{n}{n}=1,1+\frac{1}{n},\dots,2\cdot\frac{n}{n}=2}\right\}$

and considered a Riemann sum using $t_ i=x_i$ where $t_i \in$[$x_{k-1},x_k$] and by taking the limit when $n$ goes to infinity got $3$ as the answer. I wanted to check if its correct and if so why is it wrong to do it as below

$\int_{0}^2 x d \alpha = \int_{0}^1 x d x + \int_{1}^2 x d (x+2)$ would get $2$ as the answer?

Thank You

The question is what happens at the "jump". In your case, the jump is at the point $1$, so $$\int_{1}^2 x\; d (x+2) \ne \int_1^2 x\, d\alpha(x)$$
• It must change to equality by the theorem $$\int_a^bfdG=\int_a^bfG'dx$$ when $G$ is differentiable.
• A function like $\alpha$, which is not continuous, is of course not differentiable at the point $1$. Aug 18, 2015 at 16:31
• What about $\alpha=I-x?$