In Chapter $2$ number $21$ of Royden, we are asked to prove the measurability of the union of two measurable sets assuming that a set $E$ is measurable if and only if there exists a $G_\delta$ set $G$ containing $E$ such that $m^*(G\sim E)=0.$
I determined that it was necessary I prove that given any $G_\delta$ set containing $E,$ that one can construct a $G_\delta$ set $G'$ containing $E$ consisting of a descending countable collection of open sets such that $G=G'.$ My question is if it's possible to prove this result without proving that it's no loss of generality to consider the given $G_\delta$ set as being comprised of a descending collection. I have reproduced my proof below for completion.
Edit: I'm also looking for comments on if my general approach is correct.
Proof:
Let $E_1$ and $E_2$ be measurable sets. By definition of measurability there exist $G_{\delta}$ sets $G_1,G_2$ such that $$G_1=\bigcap_{k=1}^\infty G_{1,k}\;\;\;\;\; G_2=\bigcap_{k=1}^\infty G_{2,k}$$ where $G_{1,k},G_{2,k}$ are open for every $k\in\mathbb{N},$ $E_1\subseteq G_1$ and $E_2\subseteq G_2$ since $E_1\subseteq G_{1,k}$ and $E_2\subseteq G_{2,k}$ for every $k,$ and most importantly $$m^*(G_1\sim E_1)=m^*(G_2\sim E_2)=0.$$ We now prove a small lemma. For every $G_\delta$ set $G$ containing a set $E$, there exists a $G_\delta$ set $G'$ containing $E$ consisting of a countable, descending collection $\{G_i'\}_{i=1}^\infty$ of open sets such that $G=G'.$ For $G=\bigcap_{i=1}^\infty G_i,$ define $G'_1=G_1$ and $G'_i=G'_{i-1}\cap G_i.$ One easily proves by induction that each $G'_i$ contains $E$ since each $G_i$ does: clearly $G'_1$ contains $E$ since it's equal to $G_1$ which contains $E,$ and if we assume by induction that $G'_{i-1}$ contains $E$ then since $G'_i=G'_{i-1}\cap G_i$ where $G'_{i-1}$ and $G_i$ are each assumed to contain $E,$ then $G'_i$ contains $E$ as their intersection. Thus $G'_i$ contains $E$ for each natural number $i$ by induction so $G'=\bigcap_{i=1}^\infty G'_i$ also contains $E.$ Also note that the same proof, and that the intersection of any finite collection of open sets is open, shows that each $G'_i$ is open. Now we prove $\{G'_i\}_{i=1}^\infty$ is descending or that $G'_{i+1}\subseteq G'_i$ for every natural number $i.$ Simply, $G'_{i+1}=G'_{i}\cap G_{i+1}$ by definition so if $x\in G'_{i+1}$ then $x\in G'_i$ and $x\in G_{i+1},$ so certainly $G'_{i+1}\subseteq G'_i.$ Thus $G'$ is the intersection of a descending collection $\{G'_i\}_{i=1}^\infty$ of open sets containing $E.$ Finally we show $G=G'.$ If $x\in G,$ then $x\in \bigcap_{i=1}^\infty G_i$ so that $x\in G_i$ for every $i,$ hence by induction since $x\in G'_1=G_1$ and $G'_i=G'_{i-1}\cup G_i$ we have $x\in G'_{i}$ for all $i$ if we assume $x\in G'_{i-1}$ for any $i\geq 1$ by induction. Therefore $x\in \bigcap_{i=1}^\infty G'_i$ which implies $x\in G'$. Now if $x\in G',$ then $x\in\bigcap_{i=1}^\infty G'_i$ so that $x\in G'_i=G'_{i-1}\cap G_i$ thus $x\in G'_{i-1}$ and $x\in G_i,$ for every natural number $i.$ This shows $x\in G=\bigcap_{i=1}^\infty G_i.$ Hence overall $G=G'.$ Thus we've shown the lemma. Now to show $E_1\cup E_2$ is measurable we show there exists a $G_\delta$ set $G$ containing $E_1\cup E_2$ such that $$m^*(G\sim(E_1\cup E_2))=0.$$ Crucially, assume from the lemma just proved that the collections $\{G_{1,k}\}_{k=1}^\infty$ and $\{G_{2,k}\}_{k=1}^\infty$ are descending. Note that $E_1\cup E_2\subseteq G_{1,k}\cup G_{2,k}$ for every $k,$ where each $G_{1,k}\cup G_{2,k}$ is open as the union of open sets. Thus $$E_1\cup E_2\subseteq G=\bigcap_{k=1}^\infty \left(G_{1,k}\cup G_{2,k}\right),$$ where $G$ is a $G_\delta$ set. Now we show $G\sim\left(E_1\cup E_2\right)\subseteq \left(G_1\sim E_1\right) \cup \left(G_2\sim E_2\right).$ If $x\in G\sim\left(E_1\cup E_2\right),$ then $x\in \bigcap_{k=1}^\infty \left(G_{1,k}\cup G_{2,k}\right)$ and $x\notin \left(E_1\cup E_2\right)\implies$
\begin{equation} x\in G_{1,k}\cup G_{2,k} \text{ for each natural number $k$ and } \left(x\notin E_1 \text{ and } x\notin E_2\right) \tag{1}. \end{equation} Now, importantly, note that $x\in G_{1,k}\cup G_{2,k}$ for each natural number $k$ means that $x$ must be in $G_{1,k}$ for all $k$, or must be in $G_{2,k}$ for all $k$, or finally must be in both $G_{1,k}$ and $G_{2,k}$ simultaneously for all $k.$ This is because if $x\in G_{1,k}\cup G_{2,k}$ for each $k$, then $x$ is in one or both of $G_{1,k}$ and $G_{2,k}$ for infinitely many $k.$ If, without loss of generality, $x\in G_{1,k}$ for infinitely many $k$ we show $x\in G_{1,k}$ for every $k.$ So suppose for a contradiction that $x\notin G_{1,k_0}$ for some natural number $k=k_0.$ Since $x$ is in $G_{1,k}$ for infinitely many $k,$ there must be some $G_{1,n}$ for $n>k_0$ such that $G_{1,n}\subseteq G_{1,k_0}$ since $\{G_{1,k}\}_{k=1}^\infty$ is descending and $x\in G_{1,n}\implies x\in G_{1,k_0},$ contradicting that $x\notin G_{1,k_0}.$ Thus $x\in G_{1,k}$ for all $k\in\mathbb{N},$ or $x\in G_{2,k}$ for all $k\in\mathbb{N},$ or $x\in G_{1,k} \cap G_{2,k}$ for all $k\in\mathbb{N},$ so that overall by $(1)$ we get $x\in \left(G_1\sim E_1\right) \cup \left(G_2\sim E_2\right).$ This proves $G\sim\left(E_1\cup E_2\right)\subseteq \left(G_1\sim E_1\right) \cup \left(G_2\sim E_2\right).$ Trivially then, by monotonicity and the finite subadditivity of outer measure we have \begin{equation} \begin{split} 0\leq m^*(G\sim (E_1\cup E_2))&\leq m^*((G_1\sim E_1)\cup (G_2\sim E_2))\\ & \leq m^*(G_1\sim E_1) + m^*(G_2\sim E_2)\\ & = 0 + 0 = 0 \end{split} \notag \end{equation} so that $m^*(G\sim(E_1\cup E_2))=0.$ Hence $E_1\cup E_2$ is measurable. $\blacksquare$