Ex2-11

Chapter 2 Question 11
Question: Show that \int_a^b fd\phi exists if and only if given \varepsilon>0, there exists \delta>0 such that |R_\Gamma-R_{\Gamma'}|<\varepsilon if |\Gamma|,|\Gamma'|<\delta.
Proof: First assume \int_a^b fd\phi exists, then for partition \Gamma=\{x_0,x_1,\ldots,x_n\} of [a,b] and npoints \xi_1,\ldots,\xi_n satisfying x_{i-1}\leq \xi_i\leq x_i for all i=1,\ldots,n, the equality below holds\lim_{|\Gamma|\to 0}R_\Gamma=\int_a^b fd\phi,
where R_\Gammais the Riemann-Stieltjes sum for \Gamma. Given \varepsilon>0 we may choose \delta>0 so that if |\Gamma|<\deltathen\left|\int_a^b fd\phi-R_\Gamma\right|<\dfrac\varepsilon2\quad
Therefore, by triangle inequality, for |\Gamma|, |\Gamma'|<\delta we have
|R_\Gamma-R_{\Gamma'}|\leq \left|\int_a^b fd\phi-R_\Gamma\right|+\left|\int_a^b fd\phi-R_{\Gamma'}\right|<\varepsilon.
To prove the converse, assume that given \varepsilon>0, there exists \delta>0 such that |R_\Gamma-R_{\Gamma'}|<\varepsilon if |\Gamma|,|\Gamma'|<\delta. We define 
U=\limsup_{|\Gamma|\to 0}R_\Gamma\quad\text{ and }\quad L=\liminf_{|\Gamma|\to 0}R_\Gamma.
We claim both U,L are finite. We will only prove that Uis finite, since the proof of the finiteness of L is similar. If Uis not finite, then it is either +\infty or -\infty. If U is +\infty, we can construct a sequence of partitions\{\Gamma_n\} (with some intermediate choices of points \xi_i) such that 
|\Gamma_n|<\frac1n\quad\text{ and }\quad R_{\Gamma_n}>R_{\Gamma_{n-1}}+1\quad\text{ for }n\geq 2.
However, we know from assumption that there is \delta>0 such that |R_{\Gamma}-R_{\Gamma'}|<1 for |\Gamma|,|\Gamma'|<\delta, this is contradictory once we choose \Gamma=\Gamma_n, \Gamma'=\Gamma_{n+1} with \frac1n<\delta. If on the other hand, U=-\infty, then given a real number A there is \delta>0 such that |\Gamma|<\delta implies R_\Gamma<A. Therefore, we can choose another sequence of partitions \{\Gamma_n'\} such that 
|\Gamma_n'|<\frac1n \quad\text{ and }\quad R_{\Gamma_n'}<R_{\Gamma_{n-1}'}-1\quad\text{ for all }n\geq 2.
Using the assumption again, we obtain a similar contradiction. Thus U is finite, as well as L.
Now we know both U,L are finite, in particular, L\leq U. It suffices to prove L=U to complete the proof, because we could conclude from L\leq \lim_{|\Gamma|\to 0}R_\Gamma\leq U. Assume the contrary that L<U, let \varepsilon=(U-L)/3, then we can choose \delta>0 such that 
|R_\Gamma-R_{\Gamma'}|<\varepsilon\quad\forall |\Gamma|,|\Gamma'|<\delta.
By definitions of \limsupand \liminf, we can choose partitions with |\Gamma|,|\Gamma'|<\delta such that R_\Gamma<L+\varepsilon, R_{\Gamma'}>U-\varepsilon. However, this lead us to 
R_{\Gamma'}>U-\varepsilon=L+2\varepsilon>R_\Gamma+\varepsilon\implies |R_{\Gamma'}-R_{\Gamma}|>\varepsilon,a contradiction.