Closed subset consisting entirely of irrationals; compact subset with empty interior; both with measure $1/2$ I was working through Carothers' "Real Analysis" with my students today, and we came across an interesting question related to Lebesgue measure.

I realize that the first question I have will most likely have some answer along these lines, and that the second answer will probably involve a Cantor-like set.

The question is stated as:

Suppose that $E$ is measurable with $m(E)=1$. Show that:
1) There is a closed set $F$, consisting entirely of irrationals, such that $F\subset E$ and $m(F)=1/2$.
2) There is a compact set $F$ with empty interior such that $F\subset E$ and $m(F)=1/2$.



*

*With the argument I mentioned above, it's clear that we should be able to find a closed subset, $F\subset E$, of only irrationals, with $m(F)>0$. However I am stuck when it comes to proving that such a set must exist with measure equal to exactly $1/2$. Since the open intervals we construct around the rationals inside of $E$ might overlap, I'm having trouble seeing how to control the size of the final closed set. Any hints as to what I'm missing?

*Similarly, it's easy to construct a compact, Cantor-like set with measure $1/2$. However, when we require that set to be the subset of an arbitrary set $E$, with $m(E)$, I'm at a loss. Since $E$ might not even be bounded, I'm not even sure where to begin.
 A: Suppose that $E\subseteq\mathbb R$ is measurable with $m(E)\ge1.$
Find an interval $[a,b]$ such that $m(E\cap[a,b])\gt\frac12.$
Find a closed set $F_0$ consisting entirely of irrationals such that $F_0\subset E\cap[a,b]$ and $m(F_0)\gt\frac12.$
Since $\varphi(x)=m(F_0\cap[a,x])$ is a continuous function with $\varphi(a)=0$ and $\varphi(b)\gt\frac12,$ we can find $c\in(a,b)$ such that $\varphi(c)=m(F_0\cap[a,c])=\frac12.$
Let $F=F_0\cap[a,c].$ Then $F$ is a closed subset of $E$ consisting entirely of irrationals, and $m(F)=\frac12.$ Since $F$ is closed and bounded, it's compact; since $F$ is closed and contains no rational numbers, it's nowhere dense.
A: Lebegue measure is Radon-inner-regular: If $E$ is measurable and $c(E)$ is the set of compact subsets of $E$ then $m(E)=\sup \{m(F): F\in c(E)\}.$
If $E$ is measurable with $m(E)=1$:
Let $E\supset F$ where $F$ is compact and $m(F)>3/4.$
Let $ G$ be an open set with $G\supset \mathbb Q$ and $m(G)<1/4.$ The set $G$ exists because $m(\mathbb Q)=0$ because $\mathbb Q$ is countable .
Now $F$ \ $G$ is compact, is disjoint from $\mathbb Q$, and is a subset of $E$, and we have:
$m(F$ \ $G)\geq m(F)-m(G)>1/2.$
And since the complement of $F$ \ $G$ is dense (because $\mathbb R$ \ $(F$ \ $G)\supset \mathbb Q $) we have $Int (F$ \ $G)=\phi. $         
