Ex2-10

Question: 
If \lambda_1 < \lambda_2 < \ldots < \lambda_m is a finite sequence and -\infty, write \sum_k a_k e^{-s\lambda_k} as a Riemann–Stieltjes integral. 
Solutions:
Take the function f:[a,b]\to \mathbb{R}, f(x)=e^{-sx}, and \phi:[a,b]\to \mathbb{R}, \phi(x)=\left\{
\begin{array}{lcl}
a_1 & & {a \leq x < \lambda_1 },\\
a_1+a_2 & & {\lambda_1 \leq x < \lambda_2},\\
\ldots & & {\ldots}\\
a_1+a_2+\ldots+a_{m-1} & & {\lambda_{m-1}\leq x < \lambda_m},\\
a_1+a_2+\ldots+a_{m} & & {\lambda_{m}\leq x \leq b}.
\end{array} \right.
Consider a partition \Gamma_n of the interval [a,b] such that \Gamma_n=\{x_0,x_1,\ldots,x_n\} with x_i=a+i\frac{b-a}{n}, n>m and there exists a subsequence \left \{ x_{l_{i}} \right \} such that \lambda_i \in [x_{l_i-1},x_{l_i}], for every \lambda_i,i=1,2,\ldots,m, .
Then we can write the Riemann–Stieltjes sum as
\begin{aligned}
R_{\Gamma_n}
&= \sum_{i=1}^{n}f(\xi_i)(\phi(x_{i})-\phi(x_{i-1}))\\
&= \sum_{k=1}^{m} f(\xi_{l_k})(\phi(x_{l_k})-\phi(x_{l_k-1}))\\
&= \sum_{k=1}^{m} e^{-s\xi_{l_k}}a_k,
\end{aligned}
where the \xi_{l_{k}} are chosen sample points such that \xi_{l_{k}} \in [x_{l_{k}-1},x_{l_{k}}].(In this case, \lambda_{k} \in [x_{l_{k}-1},x_{l_{k}}], so \phi(x_{l_k})-\phi(x_{l_k-1})=a_{k})
If we let n\to \infty, then the gap of \Gamma_n converges to 0. 
In addition, now, we can only choose points \{\xi_{l_k}\} as points \{\lambda_k\}, k=1,2,\ldots,m.
That is, \lim_{n\to \infty}R_{\Gamma_n}=\sum_k a_k e^{-s \lambda_k} , which is exactly the Riemann–Stieltjes integral of f with respect to \phi on [a,b].