Ex3-29

Question: Use Theorem 2.17 to verify the formula for \int f \ d\phi given in remark 3 in Section 2.3. 
Answer: We will prove the statement directly (i.e., without appealing to Theorem 2.17). In particular, we want to prove that if \phi:[a,b] \to \R is a step function with jump discontinuities at \alpha_1, \dots, \alpha_n with \alpha_k < \alpha_{k+1} for 1 \le k \le n and f:[a,b] \to \R is a function that is continuous at \alpha_k for all 1 \le k \le n, then \displaystyle \int_a^b f \ d\phi exists, and
\int_a^b f \ d\phi = \sum_{k=1}^{n} f(\alpha_i) (\phi(\alpha_k+) - \phi(\alpha_k-),
where
\phi(\alpha+) = \lim_{x \to \alpha^+} \phi(x) \qquad \text{and} \qquad \phi(\alpha-) = \lim_{x \to \alpha^-} \phi(x);
if \alpha_1 = a then \phi(\alpha_1-) = \phi(a); and if \alpha_n = b then \phi(\alpha_n+) = \phi(b).
To begin, given \varepsilon > 0. Let E = \{\alpha_1, \dots, \alpha_n\}. Let \Gamma_\varepsilon = \{a = \overline{x_0}, \overline{x_1}, \dots, \overline{x_m} = b\} be a partition of [a,b] so that \alpha_k \in \Gamma_\varepsilon for all 1 \le k \le n, and \mathrm{gap}\left(\Gamma_\varepsilon\right) < \delta, where 
\delta = \min\{|\alpha_2 - \alpha_1|, \dots, |\alpha_n - \alpha_{n-1}|, \delta_0\},
with \delta_0 given by (1). 
Let \Gamma = \{x_i\}_{i=0}^r be a refinement of \Gamma_\varepsilon. Then \mathrm{gap}\left(\Gamma\right) \le \mathrm{gap}\left(\Gamma_\varepsilon\right) < \delta. Let \{\xi_i\}_{i=1}^r be a sequence of intermediate points of the subintervals of \Gamma. Consider the Riemann-Stieltjes sum
R_\Gamma(f,\phi) = \sum_{i=1}^r f(\xi_i) (\phi(x_i) - \phi(x_{i-1})).
For each 1 \le i \le r, we have that x_{i-1} and x_i are neighboring points in \Gamma. Since \alpha_k \in \Gamma, we find that \alpha_k \notin (x_{i-1}, x_i) for all 1 \le k \le n. Moreover, since 
\mathrm{gap}\left(\Gamma\right) = \max\{|x_i - x_{i-1}| \mid 1 \le i \le p\} < \min_{1\le k \le n} |\alpha_k - \alpha_{k-1}|,
it follows that we cannot have both x_{i-1}, x_i \in E. Thus, we have either on of the following cases.
(Case 1: x_{i-1}, x_i \notin E) In this case, since no discontinuity \alpha_k lies between x_{i-1} and x_i, we have \phi(x_i) - \phi(x_{i-1}) = 0. 
(Case 2: x_i = \alpha_k for some 1 \le k \le n) In this case, x_{i-1} \ne \alpha_{k'} (since \mathrm{gap}\left(\Gamma\right) < |\alpha_k - \alpha_{k-1}|) and no \alpha_{k'} can lie between x_{i-1} and \alpha_k, so it follows that \phi(x_i) - \phi(x_{i-1}) =\phi(\alpha_k) - \phi(\alpha_k-).
(Case 3: x_{i-1} = \alpha_k for some 1 \le k \le n) In this case, x_i \ne \alpha_{k'} (since \mathrm{gap}\left(\Gamma\right) < |\alpha_{k+1} - \alpha_k|) and no \alpha_{k'} can lie between \alpha_k and x_i, so it follows that \phi(x_i) - \phi(x_{i-1}) = \phi(\alpha_k+) - \phi(\alpha_k).
All the three cases are illustrated below. 
 
Hence, we find that
\begin{align}
&R_\Gamma(f,\phi) \\
&= \sum_{i=1}^r f(\xi_i) (\phi(x_i) - \phi(x_{i-1})) \\
&= \underbrace{\{\text{terms in (Case 1)}\}}_{=0} + \{\text{terms in (Case 2)}\} + \{\text{terms in (Case 3)}\} \\
&= \sum_{k=1}^n \left[f(\xi_{i_k}) (\phi(\alpha_k) - \phi(\alpha_k-)) + f(\xi_{i_k}') (\phi(\alpha_k+) - \phi(\alpha_k)) \right].
\end{align}
Here we remark that \xi_{i_k} is a point in the subinterval of \Gamma whose right-endpoint is \alpha_k and thus is within a distance \mathrm{gap}\left(\Gamma\right) < \delta of \alpha_k. Likewise, \xi_{i_k}' is a point in the subinterval of \Gamma whose left-endpoint is \alpha_k and thus is within a distance \mathrm{gap}\left(\Gamma\right) < \delta of \alpha_k. Now, we have
\begin{align}
&\left|R_\Gamma(f,\phi) -\sum_{k=1}^n f(\alpha_i) (\phi(\alpha_k+) - \phi(\alpha_k-)\right| \\
&= \left|\sum_{k=1}^n \left[f(\xi_{i_k}) (\phi(\alpha_k) - \phi(\alpha_k-)) + f(\xi_{i_k}') (\phi(\alpha_k+) - \phi(\alpha_k)) \right] \right. \\
&\quad \left. - \sum_{k=1}^n \left[f(\alpha_k) (\phi(\alpha_k) - \phi(\alpha_k-)) + f(\alpha_k) (\phi(\alpha_k+) - \phi(\alpha_k)) \right]\right| \\
&\le \sum_{k=1}^n \left[|f(\xi_{i_k} - f(\alpha_k)| |\phi(\alpha_k) - \phi(\alpha_k-)| + |f(\xi_{i_k}' - f(\alpha_k)| |\phi(\alpha_k+) - \phi(\alpha_k)| \right].
\end{align}
Now since f is continuous at \alpha_k for all 1 \le k \le n, there exists \delta_k > 0 for 1 \le k \le n such that
|f(x) - f(\alpha_k)| < \dfrac{\varepsilon}{\sum_{k=1}^n \left(|\phi(\alpha_k) - \phi(\alpha_k-)| + |\phi(\alpha_k+) - \phi(\alpha_k)|\right)}
for all |x - \alpha_k| < \delta_k. Let 
\delta_0 = \min\{\delta_1, \dots, \delta_n\}. \qquad \qquad (1)
Then it follows that
\left|R_\Gamma(f,\phi) - \sum_{k=1}^n f(\alpha_i) (\phi(\alpha_k+) - \phi(\alpha_k-)\right| < \varepsilon,
and this completes the proof.