Ex2-16

Chapter 2 Question 16 
First we show a special case where f is continuous except for a jump discontinuity at a,. We define \tilde{f} to be the continuous function that agrees with f on (a,b] but 
\tilde{f}(a)=\lim_{x\rightarrow a^+}f(x)
so \tilde{f} is continuous on [a,b]. Then we claim that \int_a^bfd\phi exists and 
\int_a^bfd\phi=\int_a^b\tilde{f}d\phi
Let |f(x)|\leq M for all x, \phi s continuous at a. Given \epsilon>0, choose \delta>0 such that 
|\phi(x)-\phi(a)|<\epsilon\quad\text{ if } |x-a|<\delta
Then for any partition \Gamma with |\Gamma|<\delta, choosing the same \xi_i's for f,\tilde f except possibly the one in [a,x_1], we have 
\begin{align}
|S_{\Gamma}(f)-S_{\Gamma}(\tilde{f})|&=\left|\left(f(\xi)-\tilde{f}(\xi')\right)\left(\phi(x_1)-\phi(a)\right)\right|\\
&<2M\epsilon
\end{align}
By taking \epsilon arbitrarily small and noting that \lim_{|\Gamma|\rightarrow0}S_{\Gamma}(\tilde{f}
) exists, we conclude 
\lim_{|\Gamma|\rightarrow0}S_{\Gamma}\left(f\right)=\int_a^bfd\phi=\int_a^b\tilde{f}d\phi
The case where f has a jump discontinuity at b or at both of a and b is an easy extension of the argument above.
We return to the original problem. We can assume \phi is increasing on [a,b] to prove the statement. Note that for a general \phi of bounded variation, we can write it as \phi=\phi_1-\phi_2, the difference of two increasing functions. Assume |f(x)|\leq Mfor all x, and suppose f has jump discontinuities at x_1,\cdots,x_n, then define f_0,f_1,\cdots,f_n on [a,x_1],[x_1,x_2],\cdots,[x_n,b] such that 
f_i(x)=f(x)\quad\text{ for }x\in[x_i,x_{i+1}]
where a=x_0,b=x_{n+1}. Now we claim \int_a^bfd\phi exists and 
\int_a^bfd\phi=\sum_{i=0}^n\int_{x_i}^{x_{i+1}}f_id\phi
Given \epsilon>0, choose \delta_1>0 such that for partition \Gamma_i of [x_i,x_{i+1}] with |\Gamma_i|<\delta_1,
\left|S_{\Gamma_i}(f_i)-\int_{x_i}^{x_{i+1}}f_id\phi\right|<\epsilon
Let D=\{x_1,x_2,\cdots,x_n\} and let \phi be continuous at each point of D. Choose \delta_2>0 such that for each i=1,2,\cdots,n
|\phi(x)-\phi(y)|<\epsilon\quad\text{ if }x,y\in(x_i-\delta_2,x_i+\delta_2)
we also assume \delta_2 to be small enough such that each (x_i-\delta_2,x_i+\delta_2) are disjoint.
Let \delta=\min(\delta_1,\delta_2), for partition \Gamma of [a,b] with |\Gamma|<\delta, we let \Gamma\cup D be the refinement with points \{x_1,\cdots,x_n\}.
We can choose two sets of \{\xi_i\}, \{\xi_i'\} in the Riemann-Stieltjes sum such that \xi_i=\xi_i' on intervals of \Gamma which don't contain any points from D. Then if we define M=\sup_{x\in[a,b]} |f(x)| 
\begin{align}&\left|S_{\Gamma\cup D}(f)- S_{\Gamma}(f)\right|
\\&\leq \sum_{i=1}^n ((\sup_{x\in[a_i,b_i]}|f(x)|)|\phi(x_i)-\phi(a_i)|\\
&+(\sup_{x\in[a_i,b_i]}|f(x)|)|\phi(b_i)-\phi(x_i)|+(\sup_{x\in[a_i,b_i]}|f(x)|)|\phi(b_i)-\phi(a_i)|)\\
&\leq n\cdot 3M \cdot\epsilon\end{align}
where [a_i,b_i]is a subinterval in \Gamma containing x_i for all i=1,2,\ldots,n.
Let \Gamma_0, \Gamma_1, \cdots, \Gamma_n be partitions respectively of [a_0,b_0], [a_1,b_1],\ldots, [a_n,b_n]. The union of those partitions is \Gamma\cup D, and then 
S_{\Gamma\cup D}(f)=S_{\Gamma_0}(f_0)+\cdots+S_{\Gamma_n}(f_n)
implies
\left|S_{\Gamma \cup D}(f)-\sum_{i=0}^n\int_{x_i}^{x_{i+1}}f_id\phi\right|<(n+1)\epsilon
implies
\left|S_{\Gamma}(f)-\sum_{i=0}^n\int_{x_i}^{x_{i+1}}f_id\phi\right|<(3nM+n+1)\epsilon
choose \epsilon small enough, and we have completed the proof.