Is a function with a countable set of discontinuities, Riemann Stieltjes integrable? We all know that a function $f:[a,b]\to \mathbb{R}$ such that it has only finitely many discontinuities and $\alpha$ is a monotonic increasing function Then $f$ is Riemann Stieltjes integrable. 

My question is that can we replace this finite discontinuity of $f$ to countable number of discontinuity. If we can not do this then please give me an example.

I have seen this question Riemann Stieltjes Integral of discontinuous function but does not getting the required answer. Please give me some hint.
 A: The first statement is incorrect. If $f$ and $\alpha$ are both discontinuous from the right or both discontinuous from the left at even a single point, then the Riemann-Stieltjes integral does not exist. A proof of this is given below.
On the other hand, as long as there are no such shared discontinuities, then the integral exists even if $f$ has countably many discontinuities.
Proof of non-existence of Riemann-Stieltjes integral when there is a shared one-sided discontinuity.
Suppose that $\alpha$ is monotone increasing and $f$ and $\alpha$ are discontinuous from the right at $\xi \in (a,b).$  (A similar srgument applies if both are discontinuous from the left). 
Consider any partition $P = (x_0,x_1, \ldots, x_{i-1},\xi, x_i, \ldots, x_n)$ with $\xi$ as a partition point and $x_i - \xi = \delta_i$
There exists $\epsilon > 0$ such that for every $\delta > 0$ (including $\delta_i$), there are points $y_1, y_2 \in (\xi, \xi + \delta)$ such that $|f(y_1) - f(\xi)| \geqslant \epsilon$ and  $|\alpha(y_2) - \alpha(\xi)| \geqslant \epsilon$.
It then follows that
$$U(P,f,\alpha) - L(P,f,\alpha) \geqslant \epsilon^2,$$
since $\alpha(x_i) - \alpha(\xi) \geqslant \alpha(y_2) - \alpha(\xi) \geqslant \epsilon$ and $\sup_{x \in [\xi,x_i]} f(x) - \inf_{x \in [\xi,x_i]}  f(x) \geqslant \epsilon$
Therefore, the Riemann criterion is not satisfied and $f$ is not RS integrable with respect to $\alpha$ on $[a,b]$.
