Show almost every element of R can be written as an element from a set of positive measure plus a rational. Let $S$ be a subset of $\mathbb{R}$ with strictly positive Lebesgue measure. Prove that almost every (with respect to Lebesgue measure) real number can be written as the sum of an element of $S$ and an element of $\mathbb{Q}$.  


So, I remember proving a long time ago that $S-S$ contains a neighborhood of 0, and I thought that it would be possible to use this to prove the problem, but I was unable to do so.
I know that $S + \mathbb{Q}$ is dense in the reals and has infinite measure, but I can't figure out why it must be $\mathbb{R}$ minus a measure zero set.
Thanks in advance :).
Let $S$ be a Lebesgue measurable set on $[a,b]$ with $m(S)&gt;0$. Then
$$
\bigcup_{q \in \mathbb{Q}}(q+S)=\Bbb{R}-N
$$
where $m(N)=0$.

Proof: for any small $\epsilon&gt;0$, there  an interval $I$ such that 
$$
m(S\cap I)&gt;(1-\epsilon)m(I)\tag1
$$
For if not, there is a $\epsilon&gt;0$ that for any interval $I$ that
$$
m(S\cap I)\leqslant(1-\epsilon)m(I)
$$
Then for $x\in S$
$$
1_S(x)=\lim_{\delta\to0}\frac{m(S\cap(x-\delta,x+\delta))}{m((x-\delta,x+\delta))}\leqslant 1-\epsilon&lt;1
$$
This means $m(S)=0$ by Lebesgue density theorem, contradicting $m(S)&gt;0$.

Let $B=\bigcup_{q \in \mathbb{Q}}(q+S)$. Clearly $m(B)&gt;m(S)&gt;0$. Suppose $m(B^c)&gt;0$ too. By $(1)$, there are intervals $I,J$ and $m(I)=m(J)$ such that 
$$
m(B\cap I)&gt;r_1m(I)\quad\text{and }\quad m(B^c\cap J)&gt;r_2m(J)
$$
where $r_1&gt;2/3,\: r_2&gt;2/3$. Then there exists a $p\in \mathbb{Q}$ that $I+p=J$. So
$$
m(B+p\cap J)=m(B+p\cap I+p)=m(B\cap I)&gt;r_1m(I)=r_1m(J)
$$
But since $B+p\subset B$, there is
$$
m(B\cap J)\geqslant m(B+p\cap J)&gt;r_1m(J)
$$
And since $(B\cap J)\cap (B^c\cap J)=\varnothing$
$$
m(J)=m((B\cap J)\cup (B^c\cap J))=m(B\cap J)+m(B^c\cap J)&gt;(r_1+r_2)m(J)&gt;m(J)
$$
which is a contradiction. So $m(\Bbb{R}-B)=m(B^c)=0$.
Let $R=\mathbb{R}-(S+\mathbb{Q})$. Assume its measure is strictly positive. Then there are $a,b\in\mathbb{R}$ with $a&lt;b$ such that $(a,b)\subset{}R$. Since the measure of $S$ is strictly positive as well, the same is true for $S$, so $(c,d)\subset{}S$. That clearly contradicts $R=\mathbb{R}-(S+\mathbb{Q})$, since it's easy to find a rational $q$ for which $\gamma+q=\alpha$ with $\alpha\in(a,b)$ and $\gamma\in(c,d)$.