This is part of an exercise from Carother's Real Analysis:
Let $G$ be open such that $\mathbb{Q}\cap[0,1] \subset G$ and $m(G)<\frac{1}{2}$. Show that $\chi_G\notin\mathcal{R}[0,1]$.
where:
$\chi_G(x)=\begin{cases}1 \text{ if } x\in G\\ 0 \text{ if } x\notin G\end{cases}$
$\mathcal{R}[0,1]$ is the space of Riemann Integrable functions on the interval $[0,1]$
- $m(\cdot)$ denotes the Lebesgue measure
Attempt/Thoughts:
I want to show this by using Lebesgue's Criterion for Riemann Integration- namely that a function is Riemann Integrable iff the measure of the set of discontinuities of the function has measure zero.
I also know that the measure of $\mathbb{Q}$ is zero, so since $m(\mathbb{Q})=0$, then $m(\mathbb{Q\cap [0,1]})=0$. (because $\mathbb{Q}\cap[0,1]\subset\mathbb{Q}\implies m(\mathbb{Q}\cap[0,1])\leq m(\mathbb{Q})=0$, and measure is nonnegative.) Well, this means that $m(G)=0$, so I don't see where I'm supposed to use the information that $m(G)<\frac{1}{2}$ from the assumption in the question.
I also know that $\chi_{\mathbb{Q}}$ and $\chi_{\mathbb{R}\setminus\mathbb{Q}}$ are both not Riemann Integrable.
Anyway, I now try to show that the measure of the discontinuities of the characteristic function of $G$ is non-zero.
Let $D(f)$ denote the set of discontinuities of a function $f$, and let $\omega_f(x)$ denote the oscillation of $f$ in the open interval $I$.
\begin{align*} &m(D(f))\neq 0\\ &\iff m\left(\bigcup_{n=1}^{\infty}\left\{x\in [0,1]:\omega_f(x)\geq \frac{1}{n}\right\}\right)\neq 0\\ &\iff m\left(\bigcup_{n=1}^{\infty}\left\{x\in [0,1]:\inf_{x\in [0,1]}\sup_{s,t\in [0,1]}|\chi_G(s)-\chi_G(t)|\geq \frac{1}{n}\right\}\right)\neq 0\\ \end{align*}
(I've been told by members on MSE that the notation that I learned for the oscillation is confusing, so just as a note:
$\omega_f(x) = \inf_{\epsilon>0} \sup_{s,t \in B(x,\epsilon)} |f(t)-f(s)|$
is an alternative definition for the oscillation of $f$ on an open ball $B(x,\epsilon)$.)
So for any $\epsilon$-ball I take centered at a rational $r\cap [0,1]: r\in\mathbb{Q}$, it will always intersect an irrational in $[0,1]$ as well, which is not in $G$, because both $\mathbb{Q}$ and $\mathbb{R}\setminus\mathbb{Q}$ are dense in $\mathbb{R}$. This is important, because this means that $\chi_G$ will have infinitely many discontinuities. I think there are uncountably many discontinuities since the discontinuities would be the irrationals in $[0,1]$.
I'm not sure how to proceed from here.
I'm still fairly new to the notion of measure, so if there is a better way to show this, then please feel free to enlighten me. Thanks.