This question is about Riemann Stieltjes integral:

Here is definition:

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 $f \in \mathscr{R}(\alpha)$.

This is Rudin PMA's Th 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 $f \in \mathscr{R}(\alpha)$.

[Code borrowed from here.]

My question is:

What I can't understand is:

  1. Why do we treat $u_j$'s differently?

  2. How do we get $U(P, f, \alpha) - L(P, f, \alpha) \leq \left[ \alpha(b) - \alpha(a) \right] \varepsilon + 2M \varepsilon$ in last line?

  • $\begingroup$ You seem to be referring to Riemann-Stieltjes Integral wrt $\;\alpha\;$ . Define this clearly : not all have Rudin's book. $\endgroup$ – DonAntonio Apr 4 '18 at 7:44
  • $\begingroup$ @DonAntonio, done, as per requirement. $\endgroup$ – Silent Apr 4 '18 at 8:18

The partition $P$ contains good and bad intervals $I_i$. For the good intervals we have $M_i-m_i\leq\epsilon$. Their contribution to $\Delta_{\rm total}:=U(P,f,\alpha)-L(P,f,\alpha)$ therefore is $\leq \bigl(\alpha(b)-\alpha(a)\bigr)\,\epsilon\>$; here Rudin refers to the proof of Theorem 6.8. The sum of the increments $\Delta\alpha_i$ over the bad intervals is $\leq\epsilon$, and $M_i-m_i\leq2M$ for all these intervals. It follows that the contribution of the bad intervals to $\Delta_{\rm total}$ is $\leq2M\epsilon$.


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