# Prove the indicator function is not Riemann integrable on $[0,1]$.

Let $\{p_1, p_2, p_3,...\} = \mathbb{Q}\cap[0,1]$ and $S=\cup_{n=1}^\infty B(p_n, 10^{-n})\cap [0,1]$, where $B(p,r)$ is the ball around the point $p$ with radius $r$. Prove the indicator function, $1_S$, is not Riemann integrable.

I think it would be possible to show the set of discontinuities of $1_S$ is of positive measure, which by Lebesgue's criterion would prove the fact. I'm not sure how to do this though. Thoughts?

• Doesn't even say an edit has been made, but wasn't it radius $2^{-n}$? – marshal craft Apr 9 '17 at 3:16
• I did edit it. I initially wrote the problem wrong. It is supposed to be $10^{-n}$. – user20354139 Apr 9 '17 at 3:18
• Doesn't even say an edit has been made, but wasn't it radius $2^{-n}$? Also define Riemann integrable for me, is it the whole domain being asked or just sections aren't integrable? – marshal craft Apr 9 '17 at 3:21
• The whole domain of $[0,1]$. – user20354139 Apr 9 '17 at 3:23
• Do you want a proof that avoids using Lebesgue theory techniques? – Vim Apr 9 '17 at 4:49

## 2 Answers

It is enough to find a set of non-zero measure at which this function is discontinuous.

1) Let $r \in S$. Then, there is $n$ such that $|r - n| = \epsilon < 10^{-n}$. Therefore, if we let $\delta = \frac {10^{-n} - \epsilon}2$, then $|x-r| < \delta \implies |x - n| < 10^{-n}$, so we see that the function is continuous on $S$.

2 ) Since $S$ contains all the rationals, it is also dense in $[0,1]$, i.e. $\overline S = [0,1]$.

We see that $\mu (\overline S \backslash S) \geq 1 - \sum_{n=1}^{\infty} 10^{-n} > 0$.

For any point in $\overline S \backslash S$, we can find a point arbitrarily close to it, which is in $S$, hence has indicator function $1$. So, with $\epsilon = \frac 12$, you can obtain a contradiction in the statement of continuity at the point.

We have $\lambda (S)\le 2\sum_{k=1}^{\infty }\frac{1}{10^{k}}=\frac{2}{9}$ so $\lambda ([0,1]\setminus S)>0.$ Now choose $x\in [0,1]\setminus S$ and a subsequence $(p_{n_k})$ such that $p_{n_k}\to x.$ Then $\chi_S(p_{n_k})=1$ but $\chi_S(x)=0$ so $\chi_S$ is not continuous at any point in $[0,1]\setminus S$ and since this set has positive measure, the result follows.

• I think you mean $\chi_S(p_{n_k}) = 1$ but $\chi_S(x) = 0$. That makes sense though. – user20354139 Apr 9 '17 at 4:31
• Right. Thanks. . – Matematleta Apr 9 '17 at 4:41
• How does the result follow? A function can have an uncountable number of discontinuities and still be Riemann integrable (ex: indicator function on Cantor middle-thirds set). – user20354139 Apr 9 '17 at 5:03
• A function bounded on $[a,b]$ is Riemann integrable iff it is continuous almost everywhere. I phrased it wrong initially, but now the proof is correct I think. – Matematleta Apr 9 '17 at 5:12