Ex2-31

Chapter 2, Question 31
Take \phi(x)=1 on [a,b], since \phi is constant, \int_a^bfd\phi exists and
\int_a^bfd1=0
By theorem 2.21, \int_a^b1df exists and 
\int_a^b1df=f(b)\phi(b)-f(a)\phi(a)-\int_a^bfd1=f(b)-f(a)
Suppose f' exists and is Riemann-integrable on [a,b], then let I=\int_a^b f'dx and J=f(b)-f(a)=\int_a^b df, we want to prove I=J. By definition of Riemann-integrable, given \varepsilon>0 we can choose \delta>0 so that for each partition with |\Gamma|<\deltaand any choice of \xi_i\in[x_{i-1},x_i], we have 
|R_\Gamma - I |<\varepsilon\quad\text{ where } \quad R_\Gamma=\sum_{i=1}^k f'(\xi_i)(x_i-x_{i-1}).
Note that in fact, if we let S_\Gamma denote the Riemann-Stieltjes sum, then 
S_\Gamma=\sum_{i=1}^k 1\cdot (f(x_i)-f(x_{i-1}))=f(b)-f(a),
therefore, we know that S_\Gamma=J for every partition \Gamma. By mean value theorem, there are some \xi_i\in[x_{i-1},x_i] for i=1,2,\ldots,k such that 
f(x_i)-f(x_{i-1})=f'(\xi_i)(x_i-x_{i-1}).
Moreover, by restricting |\Gamma|<\delta with these choices of \xi_i, we have R_\Gamma=S_\Gamma=J, thus |I-J|<\varepsilon. By choosing \varepsilon arbitrarily small, we proved I=J.