Find $H\subset [0,1]$ of class $F_\sigma$ Find $H\subset [0,1]$ of class $F_\sigma$ such that $m(H)=1$ and $H\cap\mathbb{Q}=\varnothing$.
Here, $m(H)$ is the Lebesgue's measure.
Any idea? Thanks!
 A: For $n\in N$ let $S_n$ be an open set with $S_n\supset Q$ and $|S_n|<1/n.$ Let $T_n=[0,1]$ \ $S_n .$ Then $T_n$ is closed and $|T_n|>1-1/n.$ 
Let $U=\cup_{n\in N}T_n.$ Then $1\geq |U|\geq \sup_{n\in N}|T_n|=1,$ so $|U|=1.$ Since each $T_n$ is closed, $U$ is $F_{\sigma}.$ Clearly $Q\cap U=\phi.$ 
Remark: It is more usual to write $\lambda (U)$ or $\mu (U)$ for the Lebesgue measure of $U$.
A: First we prove that there is a dense subset of $\Bbb{R}$ that contains $\Bbb{Q}$ with Lebesgue measure of $0$. 
Let $q_1,q_2\cdots$ be an enumeration of $\Bbb{Q}$ and $I_{i,j}$ be an open interval centered at $q_i$ with $|I_{i,j}|<1/2^{i+j}$.
Now define 
$$
G_j=\bigcup_{i=1}^{\infty}I_{i,j}\quad\text{and }\quad N=\bigcap_{j=k}^{\infty}G_j=\bigcap_{j=k}^{\infty}\bigcup_{i=1}^{\infty}I_{i,j}
$$
Clearly N is $G_{\delta}$ set for any $G_j$ is open, and $\Bbb{Q}\subset N$. For any $\epsilon>0$, there is a $k$ such that for any $j>k$, $1/2^j<\epsilon$. Thus
$$
m(N)\leqslant m(G_j)\leqslant \sum_{i=1}^{\infty}\frac1{2^{i+j}}=\frac1{2^{j}}<\epsilon
$$
Since $\epsilon$ is arbitrary, $m(N)=0$. So $N$ is a dense subset of $\Bbb{R}$ that contains $\Bbb{Q}$ with Lebesgue measure of $0$. 
Now let $H=N^c$. Then $H$ is $F_{\sigma}$ set and $m(H)=1$. Also $H\cap \Bbb{Q}=\varnothing$.
