Ex2-30

Chapter 2, Question 30
Let fand \phi be real-valued functions on [a,b].
(a) If \int_a^b fd\phi exists and \phi is not constant on any subinterval of [a,b], show that f is bounded on [a,b].
Proof: We assume fis unbounded on [a,b], \phi is not constant on any subinterval of [a,b] and \int_a^b fd\phi exists, then get a contradiction. Without loss of generality, we assume fis unbounded above, then there is a sequence \{x_n\}in [a,b]such that \{f(x_n)\}is strictly increasing and unbounded. By closedness of [a,b] and the fact that every sequence in \mathbb R has a monotone subsequence [see for example, Theorem 2.32 in Advanced Calculus, Fitzpatrick], we find that \{x_n\} has a monotone subsequence \{x_{n_k}\}that converges to a point x\in[a,b]. For the sake of simplicity, we rename the subsequence as \{y_k\}and without loss of generality, assume x>a and \{y_k\} is an increasing sequence. For the case where x=a and \{y_k\} decreasing, one can repeat the proof below by changing just a few words.
From assumptions, we know \lim_{|\Gamma|\to 0}R_\Gammaexists and is equal to \int_a^b fd\phi, so there is \delta>0 such that 
\left|\int_a^b fd\phi-R_\Gamma\right|<1\quad\text{ whenever }\Gamma\text{ is a partition such that }|\Gamma|<\delta.
Since \phi is not constant on any subinterval, we can choose x'<x, x-x'<\delta such that \phi(x)\neq \phi(x') (this can be done, or otherwise \phi will be constant on the interval (x-\delta,x]). Let \Gamma be a partition satisfying |\Gamma|<\delta that contains the two adjacent partition points x',x. Then we fix \xi_1,\ldots,\xi_nto be the inputs of f in R_\Gamma except the one in [x',x], say \xi=\xi_i, which can be changed. Now we treat R_\Gamma=R_\Gamma(\xi) as a function of \xi\in[x',x] and investigate its behavior. Using the triangle inequality we should have 
|R_\Gamma(\xi)-R_\Gamma(\xi')|<2\quad\text{ for all }\xi,\xi'\in[x',x].
We choose K a positive integer large enough such that y_k\in[x',x] for k\geq K. Then for k\geq K we have
2>|R_\Gamma(x)-R_\Gamma(y_k)|=|f(x)-f(y_k)||\phi(x)-\phi(x')|,
which is a contradiction since \{f(y_k)\}_{k=K}^\inftyis unbounded above and |\phi(x)-\phi(x')|>0.