Is there an open dense set $S \subset [0,1]$ such that $m(S)<1$?

$m$ is the Lebesgue measure. I was thinking that: $\Bbb Q$ is dense and $m(\Bbb Q)=0<1$ but it fails to be open, but maybe I could construct an open set from this fact, or also using that $(0,1)$ is a dense and open subset but $m(0,1)=1$.

Any hints or ideas will be very appreciated. Thanks.

Solution 1:

Enumerate $D=\mathbb Q\cap(0,1)$. For every $n$, consider an open interval of length $2^{-n}\varepsilon$ around the $n$th element of $D$. Call $U$ the union of these intervals. Then:

• The set $U$ is open.
• The Lebesgue measure of $U$ is at most $\sum\limits_{n=1}^\infty 2^{-n}\varepsilon=\varepsilon$.
• The set $U$ contains $D$, hence its closure $\bar U$ contains $\bar D=[0,1]$.