Problem about $G_{\delta}$-set and $F_{\sigma}$-set Prove if $E$ is any measurable subset of $\mathbb{R}$, then there are a $G_{\delta}$-set $G$ and a $F_{\sigma}$-set $H$ such that $H \subseteq E \subseteq G$, and such that $m(G$\ $H)=0$.
In order to prove this, for each $n \in \mathbb{Z}$ let $E_n=E \cap (n,n)$. How to show that the regularity theorem for $E_n$ can be solved entirely within the interval (n,n+1). And then re-include $\mathbb{Z}$ get to the end of the proof.
 A: We first have to deal with the technicality of what is meant by 'Lebesgue measure' as there are several, but equivalent, ways of defining it. I'm assuming the Lebesgue measure to mean the completion of the Borel measure on the reals, where the Borel measure is defined to be the restriction of the outer measure induced by the obvious pre-measure on the algebra generated by sets of the form $(a,b]$. With these assumptions the measure of a Lebesgue-measureable set $E$ is: $$\lambda(E)=\inf\{\sum_{j=1}^\infty(b_j - a_j):E\subseteq\bigcup_{j=1}^\infty(a_j, b_j]\}$$
Now we need a slight lemma which states that 
$$\lambda(E)=\inf\{\sum_{j=1}^\infty(b_j - a_j):E\subseteq\bigcup_{j=1}^\infty(a_j, b_j)\}$$
Note the only change is that the cover in the second is composed of open intervals while the definition requires covers of half-open intervals. This lemma should be clear: countable sets have Borel measure zero so we can addend the right-hand ends to the cover of the second to get a cover consistent with the first definition. And we can take a cover consistent with the definition and surround the right-hand ends by 'tiny' open intervals to get a cover of open intervals. Taking the infimum of covers like this yields the same measure.
With this lemma we automatically get that for every Lebesgue-measure set $E$ and any $\epsilon$ there is an open set $U$ such that $E\subseteq U$ and $\lambda(U)\leq \lambda(E)+\epsilon$ (just choose $U$ to be the union of the open intervals that cover $E$). To get that there is closed set $K$ such that $K\subseteq E$ and $\lambda(E)\leq \lambda(K)+\epsilon$ breaks into two case. When $E$ is bounded we consider the set $\overline{E}-E$ (which is measurable) and use the fact that there is an open set $U$ such that $U\supseteq \overline{E}-E$ and $\lambda(\overline{E}-E)\leq \lambda(U)+\epsilon$. Setting $K=\overline{E}-U$ procures us with our desired interior closed (in fact compact) set. When $E$ is unbounded we set $E_j=E\cap (j,j+1]$ and apply the previous case to each of them getting a sequence of $K_j$s. Although it is not in general true that a countable union of closed sets is closed, it is true in this very special case of pairwise disjoint closed subsets of the reals. Now if $\cup K_j$ isn't within $\epsilon$ of your $E$ you go back and do some fancy choosing of the $K_j$s to ensure that $K_j$ approximates $E_j$ to within $\epsilon 2^{-j}$ (note that I'm implicitly using an enumeration of the integers by the natural numbers which you can make more rigorous). Now we have that for any measurable subset $E$ and $\epsilon>0$ there is an open set $U$ and a closed $K$ such that $K\subseteq E\subseteq U$ with $\lambda(U)-\epsilon\leq \lambda(E)\leq\lambda(K)+\epsilon$.
With this fact in hand the problem becomes easy. Choose a sequence of $U_j$ and $K_j$ such that $\lambda(U_j)-1/j\leq \lambda(E)\leq \lambda(K_j)+1/j$. Setting $H=\cup K_j$ and $G=\cap U_j$ procures us our $F_\sigma$ and $G_\delta$ sets. Showing that $\lambda(G-H)=0$ can be seen by the fact that $\lambda(U_j-K_j)\leq 2/j$. Thus considering the intersection of the $U_j-K_j$ (which is $G-H$) has measure $0$.
