# Theorem 6.10 in Baby Rudin: If $f$ is bounded on $[a, b]$ with only finitely many points of discontinuity at which $\alpha$ is continuous, then

Here is Theorem 6.10 in the book Principles of Mathematical Analysis by Walter Rudin, 3rd edition:

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

Here $\alpha$ is a monotonically increasing function, and by $f \in \mathscr{R}(\alpha)$ we mean the integral $\int_a^b f(x) \mathrm{d} \alpha(x)$ exists.

First of all, here are Definitions 6.1 and 6.2 in Baby Rudin, 3rd edition:

Definition 6.1:

Let $[a, b]$ be a given interval. By a partition $P$ of $[a, b]$ we mean a finite set of points $x_0, x_1, \ldots, x_n$, where $$a = x_0 \leq x_1 \leq \cdots \leq x_{n-1} \leq x_n = b.$$ We write $$\Delta x_i = x_i - x_{i-1} \qquad (i = 1, \ldots, n).$$ Now suppose $f$ is a bounded real function defined on $[a, b]$. Corresponding to each partition $P$ of $[a, b]$ we put \begin{align} M_i &= \sup f(x) \qquad (x_{i-1} \leq x \leq x_i), \\ m_i &= \inf f(x) \qquad (x_{i-1} \leq x \leq x_i), \\ U(P, f) &= \sum_{i=1}^n M_i \Delta x_i, \\ L(P, f) &= \sum_{i=1}^n m_i \Delta x_i, \end{align} and finally \begin{align} \tag{1} \overline{\int_a^b} f dx &= \inf U(P, f), \\ \tag{2} \underline{\int_a^b} f dx &= \sup L(P, f),\\\, \end{align} where the $\inf$ and the $\sup$ are taken over all partitions $P$ of $[a, b]$. The left members of (1) and (2) are called the upper and lower Riemann integrals of $f$ over $[a, b]$, respectively.

If the upper and lower integrals are equal, we say that $f$ is Riemann-integrable on $[a, b]$, we write $f \in \mathscr{R}$ (that is, $\mathscr{R}$ denotes the set of Riemann-integrable functions), and we denote the common value of (1) and (2) by $$\tag{3} \int_a^b f dx,$$ or by $$\tag{4} \int_a^b f(x) dx.$$ This is the Riemann integral of $f$ over $[a, b]$. Since $f$ is bounded, there exist two numbers, $m$ and $M$, such that $$m \leq f(x) \leq M \qquad (a \leq x \leq b).$$ Hence, for every $P$, $$m(b-a) \leq L(P, f) \leq U(P, f) \leq M (b-a),$$ so that the numbers $L(P, f)$ and $U(P, f)$ form a bounded set. This shows that the upper and lower integrals are defined for every bounded function $f$. . . .

Definition 6.2:

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, here is Rudin's 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 $f \in \mathscr{R}(\alpha)$.

Here is Theorem 6.8 in Baby Rudin, 3rd edition:

If $f$ is continuous on $[a, b]$, then $f \in \mathscr{R}(\alpha)$ on $[a, b]$.

And, here is Rudin's proof:

Let $\varepsilon > 0$ be given. Choose $\eta > 0$ so that $$\left[ \alpha(b) - \alpha(a) \right] \eta < \varepsilon.$$ Since $f$ is uniformly continuous on $[a, b]$ (Theorem 4.19), there exists a $\delta > 0$ such that $$\vert f(x) - f(t) \vert < \eta \tag{16}$$ if $x \in [a, b]$, $t \in [a, b]$, and $\vert x-t \vert < \delta$.

If $P$ is any partition of $[a, b]$ such that $\Delta x_i < \delta$ for all $i$, then (16) implies that $$M_i - m_i \leq \eta \qquad (i = 1, \ldots, n) \tag{17}$$ and therefore $$U(P, f, \alpha) - L(P, f, \alpha) = \sum_{i=1}^n \left( M_i - m_i \right) \Delta \alpha_i \leq \eta \sum_{i=1}^n \Delta \alpha_i = \eta \left[ \alpha(b) - \alpha(a) \right] < \varepsilon.$$ By Theorem 6.6, $f \in \mathscr{R}(\alpha)$.

Here is Theorem 6.6 in Baby Rudin, 3rd edition:

$f \in \mathscr{R}(\alpha)$ on $[a, b]$ if and only if for every $\varepsilon > 0$ there exists a partition $P$ such that $$U(P, f, \alpha) - L(P, f, \alpha) < \varepsilon.$$

Finally, here is Theorem 4.19 in Baby Rudin, 3rd edition:

Let $f$ be a continuous mapping of a compact metric space $X$ into a metric space $Y$. Then $f$ is uniformly continuous on $X$.

Now I have the following questions:

Can we make Rudin's proof of Theorem 6.10 more explicit and rigorous (perhaps by modifying its presentation in some way)?

And, is there any alternative proof of this very theorem (preferably using the same machinary that Rudin has developed so far in the book)?

• I edited the notation for upper and lower integrals; hope it helps. Very nicely written and detailed question. – zhw. Jul 2 '17 at 16:31

Here is a sketch of a proof that breaks the problem into simpler pieces:

claim 1: If $f$ is bounded with finitely many points of discontinuity on $[a,b]$, then we can write it as $f=f_1+f_2$ where $f_1$ is piecewise constant with finitely many points of discontinuity and $f_2$ is continuous.

claim 2: $f_2\in \mathscr{R}(\alpha)$ by Theorem 6.8.

claim 3: $f_1\in \mathscr{R}(\alpha)$ by the proof of Theorem 6.10; however, the argument is simpler in this case, focusing solely on the discontinuity aspect.

claim 4: $f=f_1+f_2\in \mathscr{R}(\alpha)$ by theorem 6.12

Perhaps it would help to prove theorem 6.10 first for only one point of discontinuity. That allows one to focus better, and the notation is easier. I'll assume this is proved.

Next: Suppose $f$ is defined on $[a,b].$ If $a<c<b$ and $f\in \mathcal R ([a,c],\alpha), f\in \mathcal R ([c,b],\alpha),$ then $f\in \mathcal R ([a,b],\alpha).$ Proof (informal): If the partitions $P_1,P_2$ of $[a,c], [c,b]$ are "nice", then $P_1\cup P_2$ will be "nice" on $[a,b].$

Finally, induction: We assume the result holds for $\le n$ points of discontinuity. If we then face $n+1$ points of discontinuity, say $x_1<x_2 < \cdots < x_{n+1},$ let $c\in (x_1,x_2).$ Then $[a,c]$ contains one point of discontinuity, and $[c,b]$ contains $n$ points of discontinuity. By the induction hypothesis, $f\in \mathcal R ([a,c],\alpha)$ and $f\in \mathcal R ([c,b],\alpha).$ By the second paragraph, $f\in \mathcal R ([a,b],\alpha).$

Peeking ahead a page, we assume the OP is comfortable with Theorem 6.12 (c) & (d) (some elementary properties of the integral). For example, 6.12-(d) can be easily proved using 6.7-(c).
We use these properties without mention in what follows.

Lemma: Let $g \in \mathscr{R}(\alpha)$ be bounded with $\alpha$ continuous at $c \in [a,b]$. Then

$\int_a^b g \, d \alpha = lim \int_a^{c-p_n} g \, d \alpha + lim \int_{c+q_n}^b g \, d \alpha$
as any two positive sequences $(p_n), (q_n)$ converge to $0$.

Proof: Hint: Use Theorem 6.6 (with appropriate adaptions when $c$ is an endpoint).

Note that the value that $g$ takes on at $c$ 'disappears' as a contributing factor in the integration process. For any $n$ we are cutting out
$\int_{c-p_n}^{c+q_n} g \, d \alpha$
But if $g$ is bounded by $M$, then for the corresponding partition calculations, we have
$(M_n - m_n) \Delta \alpha_n \le 2M \Delta \alpha_n$
which goes to $0$ as $n$ goes to $+\infty$ since $\alpha$ is continuous at $c$.

Theorem 6.10 Proof:
Suppose $f$ is only discontinuous at $c$. Then the proof of the Lemma can be adapted (argue the suitable converse using Theorem 6.8) to show that $f \in \mathscr{R}(\alpha)$. When $f$ has more than $1$ discontinuity you can show the same by writing down the expanded notation/partition expression.

If the OP feels it necessary to break down the $\int$ expressions into $\sum$, $\varepsilon$, $\delta$, etc. they have a conceptual road map to fill in those details.