Does there exist an open dense subset of \(\mathbb R\) with Lebesgue measure equal to one? Either construct an example or prove that one does not exist.
Solution
1. Construct an open dense set of finite positive measure.
Enumerate the rationals as \(\mathbb Q=\{q_1,q_2,\ldots\}\) and set \[ U=\bigcup_{n=1}^\infty \left(q_n-2^{-n-3},q_n+2^{-n-3}\right). \] Then \(U\) is open. It contains \(\mathbb Q\), so it is dense in \(\mathbb R\). Moreover, by countable subadditivity, \[ m(U)\le \sum_{n=1}^\infty 2^{-n-2}=\frac14<\infty. \] Since \(U\) contains a nonempty open interval, \(m(U)>0\). Thus \[ 0<m(U)<\infty. \]
2. Dilate to obtain measure exactly one.
Let \(a=m(U)>0\) and define \[ V=a^{-1}U=\{a^{-1}x:x\in U\}. \] A nonzero dilation is a homeomorphism of \(\mathbb R\), so \(V\) is again open and dense. By the scaling property of Lebesgue measure, \[ m(V)=a^{-1}m(U)=1. \] Hence an open dense subset of \(\mathbb R\) of Lebesgue measure exactly one does exist.