I found this in an old exam and I am not sure if what I've done is correct.
Let $E=[0,1]\setminus\mathbb{Q}$ find a closed set $F$ such that the Lebesgue measure is at least 3/4
Should I suppose that closed here means closed in the subspace topology in $E$ or it is in the usual topology of $\mathbb{R}$?
here what I've done:
Consider $\lbrace q_n\rbrace_{n\geq 1}$ the rational numbers in $[0,1]$ and take $\varepsilon=\frac{1}{4}$. For each $q_n$ we can consider the open interval $I_n$ whose Lebesgue measure is given by $m(I_n)=\frac{\varepsilon}{2^n}$. Now, let $\tilde{I_n}=I_n \cap E$, it is clear that $m(\tilde{I_n})\leq \frac{\varepsilon}{2^n}$ and consider $U = \bigcup_{n\geq 1}\tilde{I_n}\subseteq E$.Now, $U$ is open in $E$ with the subspace topology, and since $m(U)\leq \varepsilon$ we are done if we take $F=U^c$.
What do you think about it?
Best Answer
Your argument accomplishes more than just making a set $F$ closed in the subspace topology of $E$. It creates a subset $F\subset E$ which is closed in $\mathbb R$ (and so certainly also in the subspace topology).
Namely, a countable union $U\subset [0,1]$ of smaller-and-smaller open intervals around rationals is open in $\mathbb R$, and can be made to have arbitrarily small measure. Its complement in $[0,1]$ is closed in $\mathbb R$, and has measure as close to $1$ as you'd like.