Let $\mu: $ outer measure.
Given the fact that if $A_1$ and $A_2$ are measurable then for $G_\delta$ sets $G_1$ that contains $A_1$ and $G_2$ that contains $A_2$, we have $\mu(G_1 - A_1)=0$ and $\mu(G_2-A_2)=0$. I want to show that $\exists G$, a $G_\delta$ set such that $\mu(G - (A_1 \cup A_2))=0$.
Can I start of with supposing $G=G_1\cup G_2$. Since each of $G_1$ and $G_2$ contains $A_1, A_2$, respectively, then $G$ should also contain both $A_1$ and $A_2$, which implies that $A_1 \cup A_2 \subset G$. Hence, $\mu(G-(A_1\cup A_2))=\mu((G-A_1)\cap(G-A_2))=\mu(G-A_1)+\mu(G-A_2)$ since both are zeros thus $A_1\cup A_2$ is measurable.