# 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.

• @Surb Please watch carefully I am asking for Riemann Stieltjes integrability of $f$. – Priya Pandey Oct 17 '19 at 9:58

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 $$[a,b]$$.