This topic came out today during a discussion with a colleague. I realized that a counter-example to his claim could be constructed if there exits a subset $A \subset [0,1]$ such that $0 < \mu(A) < 1$ (where $\mu$ is Lebesgue measure), $A$ is dense in $[0,1]$ (with respect to the standard topology on $\mathbb{R}$), $\textit{and } A \textit{ does not contain an interval}$. However, I am not sure whether such a set exists. So the question is: do such sets exist? If so, is there an explicit construction, if not, how does one prove non-existence?



Are there dense sets of positive but not full measure?
By the Central Limit Theorem, something like this should have the
property that $0<\mu([a,b]\cap A)<b-a$ for all $0\leq a<b\leq 1$. Let
$0.a\_1a\_2\cdots$ be the binary expansion of $x\in[0,1]$. Let $A$
consist of all $x$ such that for all $n$ sufficiently large,
 $$ -1 < \frac{\frac n2-(a\_1+a\_2+\cdots+a\_n)}{\sqrt{n}}<1. $$
The bounds $-1$ and $1$ could be replaced by any $\alpha<\beta$ in
$\mathbb{R}$.


For another solution see <https://math.stackexchange.com/questions/57317/construction-of-a-borel-set-with-positive-but-not-full-measure-in-each-interval>.


