1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ @Surb Please watch carefully I am asking for Riemann Stieltjes integrability of $f$. $\endgroup$ – Priya Pandey Oct 17 '19 at 9:58
1
$\begingroup$

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 $[a,b]$.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.