Is there a compact set with positive measure on $[0,1]\setminus \Bbb Q$

Question

I'm wondering if there exists a Lebesgue compact set (with respect to the usual topology on $[0,1]$) in $[0,1]\setminus \mathbb{Q}$ whose Lebesgue measure is positive. In fact we've just seen the Egoroff theorem in class, and I thought of the functions $$ f_n(x)= \left\{ \begin{array}{ll} \frac{1}{n} & \mbox{if } x\in[0,1]\setminus \mathbb Q\\ 1 & \mbox{otherwise.} \end{array} \right. $$ Obviously $f_n\to f\; \mu$-a.e., but since $\mathbb Q$ is dense, I'm wondering how we can find a compact set $F\subset [0,1]$ for a given $\delta>0$ such that $\mu([0,1]\setminus F)<\delta$ and $\sup_{x\in F} |f_n(x)-f(x)|\to 0 $ as $n\to 0$. Could you please explain why Egoroff holds in this case?

Answer 1

Lebesgue measure $\mu$ is regular. This means $\mu (E)=\sup \{\mu (K): K \subseteq E: K \textrm{ compact }\}$. Taking $E=[0,1]\setminus \mathbb Q$ we see that the exist compact subsets of this set with measure as close to $1$ as you want.