Let $c\in(a,b)$ and define: $$k(x)=\left\{\begin{array}{ll}1&x=c\\0&a\leq x\leq b\mbox{ and }x\ne c\end{array}\right.$$
If $f$ are a continous function on $c$. Show by definition that
$$\int_a^b f(x)dk(x)=\int_a^b k(x)df(x)$$
As a Riemann Stieltjes integral.
I don't know how to start I'm using the definition of partitions and refinaments or R-S integral (This excersice is of the book The Elements of Real Analysis, 2nd edition Bartle)
(1) Here we have $\int fdk,\ \int kdf$.
And given $\varepsilon$, there is $\delta$ s.t. $|c-x|<\delta$ implies $|f(c)-f(x) | <\varepsilon$.
If $|\int fdk -S(P,f,k)|<\varepsilon$ for some $P$ with $\|P\|<\delta$, then we can add a point $c$ in $P$. We let $P'$.
Hence \begin{align*}S(P',f,k)&=\sum_{i=1}^n\ f(t_i) [k(t_i)-k(t_{i-1} )] \\&= f(c)k(c) +f(t_{i+1})(-k(c)) \\&= f(c)-f(t_{i+1}) \end{align*} where $t_i=c$.
So $|S(P',f,k)|<\varepsilon$ so that $|\int fdk| <2\varepsilon$.
(2) If $|\int kdf -S(Q,k,f) |<\varepsilon$ for some $Q$ with $\|Q\|<\delta$, then we can add a point $c$ in $Q$. We let $Q'$.
So \begin{align*}S(Q',k,f)&=\sum_{i=1}^m\ k(s_i) [f(s_i)-f(s_{i-1} )] \\&= k(c)[f(c)-f(s_{i-1} )] \\&= f(c)-f(s_{i-1} )\end{align*} so that $|S(Q',k,f)|<\varepsilon$