Here is definition of Riemann Stieltjes integral:
Let $\alpha$ be a monotonically increasing function on $[a, b]$ (since $\alpha(a)$ and $\alpha(b)$ are finite, it follows that $\alpha$ is bounded on $[a, b]$). Corresponding to each partition $P$ of $[a, b]$, we write
$$ \Delta \alpha_i = \alpha \left( x_i \right) – \alpha \left( x_{i-1} \right). $$
It is clear that $\Delta \alpha_i \geq 0$. For any real function $f$ which is bounded on $[a, b]$ we put
$$
\begin{align}
U(P, f, \alpha) &= \sum_{i=1}^n M_i \Delta \alpha_i, \\
L(P, f, \alpha) &= \sum_{i=1}^n m_i \Delta \alpha_i,
\end{align}
$$
where $M_i$, $m_i$ have the same meaning as in Definition 6.1, and we define
$$
\begin{align}
\tag{5} \overline{\int_a^b} f d \alpha = \inf U(P, f, \alpha), \\
\tag{6} \underline{\int_a^b} f d \alpha = \sup L(P, f, \alpha), \\\,
\end{align}
$$
the $\inf$ and $\sup$ again being taken over all partitions. If the left members of (5) and (6) are equal, we denote their common value by
$$ \tag{7} \int_a^b f d \alpha $$
or sometimes by
$$ \tag{8} \int_a^b f(x) d \alpha(x). $$
This is the Riemann-Stieltjes integral (or simply the Stieltjes integral) of $f$ with respect to $\alpha$, over $[a, b]$.
If (7) exists, i.e., if (5) and (6) are equal, we say that $f$ is integrable with respect to $\alpha$, in the Riemann sense, and write $f \in \mathscr{R}(\alpha)$.
And this is Rudin PMA's theorem 6.10:
Suppose $f$ is bounded on $[a, b]$, $f$ has only finitely many points of discontinuity on $[a, b]$, and $\alpha$ is continuous at every point at which $f$ is discontinuous. Then $f \in \mathscr{R}(\alpha)$.
Proof: Let $\varepsilon > 0$ be given. Put $M = \sup \left\vert f(x) \right\vert$, let $E$ be the set of points at which $f$ is discontinuous. Since $E$ is finite and $\alpha$ is continuous at every point of $E$, we can cover $E$ by finitely many disjoint intervals $\left[ u_j, v_j \right] \subset [a, b]$ such that the sum of the corresponding differences $\alpha\left(v_j\right) – \alpha \left( u_j \right)$ is less than $\varepsilon$. Furthermore, we can place these intervals in such a way that every point of $E \cap (a, b)$ lies in the interior of some $\left[ u_j, v_j \right]$.
Remove the segments $\left( u_j, v_j \right)$ from $[a, b]$. The remaining set $K$ is compact. Hence $f$ is uniformly continuous on $K$, and there exists $\delta > 0$ such that $\left\vert f(s) – f(t) \right\vert < \varepsilon$ if $s \in K$, $t \in K$, $\left\vert s-t \right\vert < \delta$.
Now form a partition $P = \left\{ x_0, x_1, \ldots, x_n \right\}$ of $[a, b]$, as follows: Each $u_j$ occurs in $P$. Each $v_j$ occurs in $P$. No point of any segment $\left( u_j, v_j \right)$ occurs in $P$. If $x_{i-1}$ is not one of the $u_j$, then $\Delta \alpha_i < \delta$.
Note that $M_i – m_i \leq 2M$ for every $i$, and that $M_i – m_i \leq \varepsilon$ unless $x_{i-1}$ is one of the $u_j$. Hence, as in the proof of Theorem 6.8,
$$ U(P, f, \alpha) – L(P, f, \alpha) \leq \left[ \alpha(b) – \alpha(a) \right] \varepsilon + 2M \varepsilon.$$
Since $\varepsilon$ is arbitrary, Theorem 6.6 shows that $f \in \mathscr{R}(\alpha)$.
The part I do not understant is, why is the following true:
Note that $M_i – m_i \leq 2M$ for every $i$, and that $M_i – m_i \leq \varepsilon$ unless $x_{i-1}$ is one of the $u_j$
Any help would be appreciated.
Best Answer
As @plop mentioned in the comments, the first inequality can be proved using the triangle inequality. For the second one, uniform continuity of $f$ on $[x_{i-1}, x_i]$ and the fact that $\Delta x_i < \delta$ by construction ensures that $|f(s_i) - f(t_i)| = |M_i - m_i| \le \epsilon$ for some $s_i, t_i \in [x_{i-1}, x_i]$.