Ex2-29

Use Theorem 2.17 to verify the formula for \int_a^b fd\phi given in remark 3 in Section 2.3.
Remark 3: Let \phi(x) be a step function; that is, suppose there are points a=\alpha_0<\alpha_1<\cdots<\alpha_m=b such that \phi is constant on each interval \left(\alpha_{i-1}, \alpha_i\right). 
Let
\phi\left(\alpha_i+\right)=\lim _{x \rightarrow \alpha_i+} \phi(x), \quad i=0,1, \ldots, m-1,
and
\phi\left(\alpha_i-\right)=\lim _{x \rightarrow \alpha_i-} \phi(x), \quad i=1, \ldots, m,
denote the limits from the right and left at \alpha_i, and let d_i=\phi\left(\alpha_i+\right)-\phi\left(\alpha_i-\right), i=1, \ldots, m-1, d_0=\phi\left(\alpha_0+\right)-\phi\left(\alpha_0\right), and d_m=\phi\left(\alpha_m\right)-\phi\left(\alpha_m-\right) denote the jumps of \phi. Then, for continuous f,
\int_a^b f d \phi=\sum_{i=0}^m f\left(\alpha_i\right) d_i
Proof: The step function \phi:[a, b] \rightarrow \mathbb{R} could be represented as:
\begin{align}
\phi(\alpha)= \begin{cases}c_0 & \text { if } \alpha \in\left(\alpha_0, \alpha_1\right) \\ c_1 & \text { if } \alpha \in\left(\alpha_1, \alpha_2\right) \\ \vdots & \\ c_{m-1} & \text { if } \alpha \in\left(\alpha_{m-1}, \alpha_{m}\right)\end{cases}
\end{align}
with f(\alpha_1), f(\alpha_2),\ldots, f(\alpha_n) given arbitrary values. 
By Theorem 2.17: If \int_a^b f d \phi exists and a<c, then \int_a^c f d \phi and \int_c^b f d \phi both exist and
\begin{align}
\int_a^b f d \phi=\int_a^c f d \phi+\int_c^b f d \phi .
\end{align}
Since \int_a^b fd\phi exist, then \int_{\alpha_{i-1}}^{\alpha_i} f d \phi exist for \quad i=1, \ldots, m,
\begin{align}
\int_a^b fd\phi = \sum_{i=1}^m\int_{\alpha_{i-1}}^{\alpha_i} f d \phi
\end{align}
Consider partition \Gamma_i of the interval [\alpha_{i-1},\alpha_{i}] such that \Gamma_i=\{{\alpha_{i-1}},\ldots,{\alpha_{i}}\} and \Gamma = \sum_{i=1}^{m}\Gamma_i
Since f is continuous, by using mean value theorem, \xi_i \in\left(\alpha_{i-1}, \alpha_i\right), we can write the Riemann–Stieltjes sum as
\begin{align}
R_{\Gamma}= \sum_{i=1}^{m}f\left(\xi_i\right) \cdot\left(\phi\left(\alpha_i\right)-\phi\left(\alpha_{i-1}\right)\right)
\end{align}
As \left|\Gamma\right| \rightarrow 0, f(\xi_i) \rightarrow f(\alpha_{i}). 
Define d_i=\phi\left(\alpha_i+\right)-\phi\left(\alpha_i-\right), i=1, \ldots, m-1, d_0=\phi\left(\alpha_0+\right)-\phi\left(\alpha_0\right), and d_m=\phi\left(\alpha_m\right)-\phi\left(\alpha_m-\right), we get
\begin{align}
\int_a^b f d \phi
&= \lim _{|\Gamma| \rightarrow 0}R_{\Gamma}\\
&= \sum_{i=0}^{m}f\left(\alpha_i\right) d_{i}
\end{align}
since the step function \phi will only step change at \alpha_{i} for i=1, \ldots, m