I have proved the following result and I would like to know if I have made any mistakes (in particular, I am not sure that the set $F$ I have built is necessarily closed). Thank you.
"Let $\varepsilon >0$. Prove there exists a subset $F$ of $[0,1]$ such that $F$ is closed, every elements of $F$ is an irrational number and $|F|>1-\varepsilon$"
(NOTE: $|\cdot|$ denotes the outer measure)
Let $r_1,r_2,\dots $ be the sequence that contains every rational number in $[0,1]$, $\varepsilon >0$ and consider the set $F:=[0,1]-\bigcup_{k=1}^{\infty}(r_k-\frac{\varepsilon}{4\cdot 2^k},r_k+\frac{\varepsilon}{4\cdot 2^k})$. Then $F$ is closed, being the complement of the open set $\bigcup_{k=1}^{\infty}(r_k-\frac{\varepsilon}{4\cdot 2^k},r_k+\frac{\varepsilon}{4\cdot 2^k})$ in $[0,1]$, contains only irrational number since we have taken away all the rational ones, and $|F|\geq |[0,1]|-|\bigcup_{k=1}^{\infty}(r_k-\frac{\varepsilon}{4\cdot 2^k},r_k+\frac{\varepsilon}{4\cdot 2^k})|\geq (1-0)-\frac{\varepsilon}{2}\sum_{k=1}^{\infty}\frac{1}{2^k}=1-\frac{\varepsilon}{2}>1-\varepsilon$.