Riemann integrability of $f(x) = 0$ if $x=0$ or $x = 1/n$ for some $n \in \mathbb{N}$, $f(x)=1$ otherwise

Question

Let $$ f(x) = \begin{cases} 0 & \text{if $x=0$ or $x = 1/n$ for some $n \in \mathbb{N}$,} \\ 1 & \text{otherwise.} \end{cases} $$ Is this function Riemann-Stieltjes integrable in $[0,1]$?

For the upper Riemann-Stieltjes integral, all $f(x)$ would be $1$ for any partition so it's $1$.

For the lower Riemann-Stieltjes integral, for some $n \in \mathbb{N}$ of which $f(1/n)=0$, is there a chance of this integral becoming different from upper integral?

To say that a function is Riemann-Stieltjes integrable, you need another function, which you have not specified. — edm, May 23 '17 at 10:08
@JackerySmith: I took the liberty of formatting your function using TeX's 'cases` environment. ;) — Andrew D. Hwang, May 23 '17 at 11:15

score 1 · Answer 1 · answered May 23 '17 at 10:02

1

Your function is discontinuous in only countable number of points, hence it's Riemann integrable. Intuitively you can make the partition such that it's very small around the discontinuity points.

Refer to the link below

Proof that a function with a countable set of discontinuities is Riemann integrable without the notion of measure

answered May 23 '17 at 10:02

Arpan1729

2,973
1
13
25

how could I refine my mathjax representation? any advice? – snapper May 23 '17 at 10:07
is above function Riemann integral irrespective of my choice of $\alpha$(increasing function for Riemann integration)? – snapper May 23 '17 at 10:08
from the above link, I can't understand what " require the notion of sets of measure 0" means. – snapper May 23 '17 at 10:28

Riemann integrability of $f(x) = 0$ if $x=0$ or $x = 1/n$ for some $n \in \mathbb{N}$, $f(x)=1$ otherwise

1 Answers1