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

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    $\begingroup$ 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. $\endgroup$ – avs May 21 at 20:59

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