# Rudin 6.10 but with countable discontinuities

This is theorem 6.10 in baby Rudin:

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

Is this still true if we're given that f has countable discontinuities instead of infinitely many? I've heard of Lebesgue criterion and how a function with a countable set of discontinuities is Riemann integrable but I'm not sure how $$\alpha$$ comes into play here.

Some definitions:

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

• I assume, $\mathscr{R}(\alpha)$ refers to the Riemann-Stiltjes integral w.r.t. $\alpha$, but this is not universal notation, and not everyone has the time or access to look into baby Rudin, so you might want to include a definition of that notation in your question. – avs May 21 at 20:59