I'm trying to prove the next approximation lemma for Lebesgue sets:
If $E\subset\mathbb{R}$ such that for all $\epsilon>0$ there is $A=A_{\epsilon}\in\mathcal{A}$ with $\overline{\lambda}(E\triangle A)<\epsilon,$ then $E\in\mathcal{A}_{\mathbb{R}}^*.$
Here, $\mathcal{A}$ is the algebra of finite disjoint union of elements $(-\infty,b],(c,d],(a,\infty);$ $\mathcal{A}_{\mathbb{R}}^*$ is the complete Lebesgue $\sigma-$algebra and $\overline{\lambda}=\lambda^*$ is the Lebesgue measure on $\mathbb{R}$ (outer measure).
I was trying to prove it by contradiction: there is $\epsilon_0>0$ such that for all open set $\mathcal{G}$ with property $E\subset\mathcal{G}$ such that $\overline{\lambda}(\mathcal{G}-E)\geq\epsilon_0,$ so $\overline{\lambda}(\mathcal{G}\triangle E)\ge\epsilon_0.$ So, to finish this it must be proved $\mathcal{G}$ is an element in $\mathcal{A}$ but such thing isn't sure. So, my feeling is to built a sequence $A_n$ in $\mathcal{A}$ whose union with $E$ gives $E$ and $\overline{\lambda}(E)=0$ finishing the proof but I can't see how to achieve this.
Any kind of help is thanked in advanced.
Let $\mathcal{F}$ be the collection of every set $E$ such that for every $\epsilon > 0$, there exists $A \in \mathcal{A}$ with $\bar{\lambda}(E \triangle A) < \epsilon$. Prove that $\mathcal{F}$ is a $\sigma$-algebra containing $\mathcal{A}$ and all null sets.