So I am trying to figure out a proof for the following statement, but I'm not really sure how to go about it. The statement is: "Show that for every $\epsilon>0$, there exists an open set G in $\mathbb{R}$ which contains all of the rational numbers but $m(G)<\epsilon$." How can it be true that the open set G contains all of the rational numbers but has an arbitrarily small measure?
Real Analysis – Open Measurable Sets Containing All Rational Numbers
measure-theoryreal-analysis
Best Answer
To elaborate on other answers, let $\{r_n\}_{n \in \mathbb{N}}$ be an enumeration of all rational numbers. This is possible because $\mathbb{Q}$ is countable. For each number $r_n$, pick the open interval $(r_n - 2^{-n-1} \epsilon, r_n + 2^{-n-1} \epsilon)$. Each open interval is an open set. The union of all such intervals is also open. The measure of each interval is its length $2^{-n} \epsilon$. If $G$ is the union of these intervals, we have:
$$ m(G) \le \sum_{n=1}^{\infty} 2^{-n} \epsilon = \epsilon $$
Where the sum above is a geometric series.