I wonder how can we prove the backward direction (function $f:[a, b] \rightarrow \mathbb{R}$ is Riemann integrable, then $f$ is bounded and almost everywhere continuous.) without using the measure theory.
My attempt:
If $f\in R(\alpha)$ on [a,b], there exists a partition $U(P,f,\alpha)-L(P,f,\alpha)<\epsilon$. Consider an arbitrary partition $P: {a=x_0<x_1<\ldots<x_n=b}$.
By theorem 6.7(Rudin), If $s,t$ are arbitrary points in $[x_{i-1},x_i]$, then $\sum_{i=1}^n|f(s)-f(t)|\Delta\alpha_i<\epsilon$.
Since $U(P,f,\alpha)-L(P,f,\alpha)=\sum_{i=1}^n(M_i-m_i)\Delta\alpha_i<\epsilon$ by definition and $f$ is bounded, $M_i-m_i\leq\eta$ where $\eta>0$. Thus, given $\eta > 0,\exists \delta>0$ s.t. when $|s-t|<\delta$, $|f(s)-f(t)|<\eta.$ Thus, $f$ is continuous on each $[x_{i-1},x_i]$.
I felt the part "$M_i-m_i\leq\epsilon$ where $\epsilon>0$. Thus, given $\epsilon > 0,\exists \delta>0$" weird and not sure if what I'm doing is correct. Any help appreciated!
Edit: --------------------------
The original problem is : Let $a_{1}, a_{2}, \ldots$ be a strictly increasing sequence in $(a, b],$ and let $p_{n}>0$ be such that $\sum_{n=1}^{\infty} p_{n}=1$ Define $\alpha:[a, b] \rightarrow \mathbb{R}$ as follows: $\alpha(x)=0$ if $a \leq x<a_{1}$ $$ \alpha(x)=\sum_{k=1}^{n} p_{k} \text { if } a_{n} \leq x<a_{n+1} $$ and $\alpha(x)=1$ if $\sup _{n} a_{n} \leq x \leq b$
I need to Show that a bounded function $f:[a, b] \rightarrow \mathbb{R}$ is Riemann-Stieltjes integrable with respect to $\alpha$ if and only if $f\left(a_{n}^{-}\right)=\lim _{x \rightarrow a_{n}^{-}} f(x)$ exists and $f\left(a_{n}^{-}\right)=f\left(a_{n}\right)$ for all $n,$ and that in this case $$ \int_{a}^{b} f d \alpha=\sum_{n=1}^{\infty} f\left(a_{n}\right) p_{n} $$
I'm stuck on proving "$f:[a, b] \rightarrow \mathbb{R}$ is Riemann-Stieltjes integrable, then it's left continuous" .
This is an expansion of my comment.
Let's prove this lemma first (taken from Apostol's Mathematical Analysis, theorem 7.29, page 160).
Lemma: Assume $\alpha:[a, b] \to\mathbb {R} $ is increasing on $[a, b] $ and $c\in(a, b) $. Let both the functions $f:[a, b] \to\mathbb{R} $ and $\alpha$ be discontinuous from left at $c$. Then the Riemann-Stieltjes integral $\int_a^b f\, d\alpha$ does not exist.
By the given conditions there is an $\epsilon >0$ such that every neighborhood of type $[c-\delta, c] $ contains points $x, y$ such that $$|f(x) - f(c) |\geq\epsilon, \alpha(c) - \alpha(y) \geq \epsilon $$
Now given any partition $P$ of $[a, b] $ with norm however small we can choose another partition $P'= P\cup \{c, y\} $ whose norm does not exceed that of $P$ and the interval $[y, c] $ contribution to the difference $U(P', f, \alpha) - L(P', f, \alpha) $ is not less than $$\epsilon(\alpha(c) - \alpha(y)) \geq \epsilon^2$$ It follows that the upper and lower Darboux sums can't tend to same limit and hence the integral does not exist.
The function $\alpha $ in your question is increasing and the lemma above applies to it. It is discontinuous from left at each $a_n$ and hence in order for the integral $\int_a^b f\, d\alpha $ to exist the function $f$ must be continuous from left at each $a_n$.
The value of the integral can be found by letting $c=\sup a_n$ and noting that $$\int_{a} ^{a_n} f\, d\alpha=\sum_{k=1}^{n-1}f(a_k)p_k$$ (because $\alpha$ is a step function on $[a, a_n] $) and the integral on $[c, b] $ is $0$. The integral thus equals $$\int_{a} ^{a_n} f\, d\alpha+\int_{a_n} ^{c} f\, d\alpha$$ The first term tends to $\sum_{k=1}^{\infty} f(a_k) p_k$ and second term tends to $0$.
Note: In case anyone is worried about convergence of $\sum f(a_n) p_n$, it is absolutely convergent as $f$ is bounded $p_n>0$ and $\sum p_n=1$.