Definition: $E \subset \mathbb{R^n}$ is Lebesgue measurable if $\forall \epsilon > 0, \exists$ an open set $G \ni E \subset G $ and $\mu^*|G-E| < \epsilon$
Idea: $E \subset \mathbb{R^n}$ is Lebesgue measurable if $\exists $ a set $H$ of type $G_{\delta} \ni E \subset H $ and $\mu^*|H-E|=0$
Definition is what we were given in class. Idea is something I was thinking about when the professor was giving us Definition. Based on my understanding of Definition, I feel like Idea should be another way of defining a Lebesgue measurable subset $E$ of $\mathbb{R^n}.$ But this is just based on intuition. I couldn't think of counterexamples so I tried to prove:
$\forall \epsilon > 0, \exists$ an open set $G \ni E \subset G $ and $\mu^*|G-E| < \epsilon \iff \exists $ a set $H$ of type $G_{\delta} \ni E \subset H $ and $\mu^*|H-E|=0$
$\implies$ Let $E \subset \mathbb{R^n}.$ Then $\exists $ open sets $G_k \ni E \subset G_k$ and $\mu^*|G_k - E| < \frac{1}{k}.$ Take $H = \cap_{k \in \mathbb{N}} G_K$. Then $H$ is of type $G_{\delta}$ and $E \subset H.$ Since $H - E \subset G_k - E$ it follows $0 \leq \mu^*|H - E| \leq \mu^*|G_k - E| < \frac{1}{k}$. So $\mu^*|H - E|=0.$
$\impliedby$Let $E \subset \mathbb{R^n}.$ Let $\epsilon > 0$ be given. We know that $\exists $ an open set $G_{\alpha} \ni E \subset G_{\alpha}$ and $\mu^*|G_{\alpha}| \leq \mu^*|E| + \epsilon$. Consider $G = H \cap G_{\alpha}=\cup_{k \in \mathbb{N}}(G_k \cap G_{\alpha})$. Then $G$ is open and $E \subset G$. Since $G-E = (H \cap G_{\alpha}) - E \subset H - E$, we have that $\mu^*|G-E| \leq \mu^*|H-E| = 0 < \epsilon.$
Could someone let me know if this is correct and if not, give me counterexamples?