Ex2-15

Question: Suppose f is continuous and \phi is of bounded variation on [a,b]. Show that: 
* \psi(x)=\int_{a}^{x}f \,d\phi is of bounded variation. 
* If g is continuous, then \int_{a}^{b}g\,d\psi=\int_{a}^{b}gf\,d\phi. 
Solution:
First part 
First note that for any subinterval [a',b']\subseteq [a,b], \int_{a'}^{b'}f \,d\phi exists since f is continuous and \phi is of bounded variation on that subinterval. Let \Gamma=\{x_i\}_{i=0}^{n} be a partition of [a,b]. Then, S_{\Gamma}=\sum_{i=1}^{n}|\psi(x_i)-\psi(x_{i-1})|
=\sum_{i=1}^{n}\Big|\int_{x_{i-1}}^{x_i}f\,d\phi\Big|.
Each of the summands is bounded since 
\begin{align}
\Big|\int_{x_{i-1}}^{x_i}f\,d\phi\Big|&\leq (\sup_{[x_{i-1}.x_i]} f)\,V(\phi;[x_{i-1},x_i])\\
&\leq (\sup_{[a,b]}f)\,V(\phi;[x_{i-1},x_i]).
\end{align}Since variation is additive, we see that\begin{align}
S_{\Gamma}&=\sum_{i=1}^{n}|\psi(x_i)-\psi(x_{i-1})|\\
&=\sum_{i=1}^{n}\Big|\int_{x_{i-1}}^{x_i}f\,d\phi\Big| \\
&\leq (\sup_{[a,b]}f)\,\sum_{i=1}^{n}V(\phi;[x_{i-1},x_i])\\
&=(\sup_{[a,b]}f)\,V(\phi;[a,b]).
\end{align}This shows that V(\psi;[a,b]) exists and is finite. Therefore, \psi is of bounded variation.
Second part
\int_{a}^{b}g\,d\psi,\int_{a}^{b}gf\,d\phi both exist since g,f are continuous and \phi,\psi are of bounded variation. We first introduce this mean value theorem for Riemann-Stieltjes Integral:
Mean Value Theorem: If f is continuous on [a,b] and \phi is bounded and increasing on [a,b], then there exists \xi \in [a,b] such that\int_{a}^{b}f\,d\phi=f(\xi)[\phi(b)-\phi(a)].The proof can be found in page 33 of the textbook.
Due to the properties of Riemann-Stieltjes Integral and the fact that every function of bounded variation can be written as a difference of two bounded increasing function, it suffices to prove for the case where \phi is bounded and increasing. 
Since f is continuous on [a,b], it is uniformly continuous and bounded. Let M>0 be an upper bound of f and V=\phi(b)-\phi(a) be the variation of \phi. Let K=M(V+1).
Given \varepsilon>0, there exists \delta_1>0 such that if |x-x'|<\delta_1, then |f(x)-f(x')|<\frac{\varepsilon}{2K}.
Since \int_{a}^{b}gf\,d\phi exists, there exists \delta_2>0 such that if |\Gamma|<\delta_1, then \Big|R_\Gamma -\int_{a}^{b}gf\,d\phi\Big|<\varepsilon/2.
Now, let \Gamma=\{x_i\}_{i=0}^{n} be a partition of [a,b]. Then the Riemann-Stieltjes sum for g,\psi isR_{\Gamma}=\sum_{i=1}^{n}g(\xi_i)\int_{x_{i-1}}^{x_i}f\,d\phi.By the Mean value theorem, for each i, we can find c_i\in [x_{i-1},x_i] such that\int_{x_{i-1}}^{x_i}f\,d\phi=f(c_i)[\phi(x_i)-\phi(x_{i-1})].Thus,R_{\Gamma}=\sum_{i=1}^{n}g(\xi_i)f(c_i)[\phi(x_i)-\phi(x_{i-1})].Now, given \varepsilon>0, take \delta=\min\{\delta_1,\delta_2\}. If \Gamma is a partition having gap smaller than \delta , then \begin{align}
|R_{\Gamma}-\int_{a}^{b}gf\,d\phi|&\leq |R_{\Gamma}-\sum_{i=1}^{n}g(\xi_i)f(\xi_i)[\phi(x_i)-\phi(x_{i-1})]|+|\sum_{i=1}^{n}g(\xi)f(\xi)[\phi(x_i)-\phi(x_{i-1})]-\int_{a}^{b}gf\,d\phi|\\
&<M\times V (\frac{\varepsilon}{2K})+\frac{\varepsilon}{2}\\
&<\varepsilon .
\end{align}This proves the second part.