# 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

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
• 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