Let $\Omega$ be an uncountable set and $\mathcal{G}$ be the $\sigma$-Algebra $\mathcal{G} = \{A \in \mathcal{P}(\Omega)\;|\; A\;\text{or}\;A^c\;\text{are countable}\}$. The map $$\alpha: \mathcal{G} \to [0,\infty],\quad \alpha(G) = \begin{cases}0 &\quad G\;\text{is countable}\\ 1 &\quad G^c\;\text{is countable}\end{cases}$$ defines a measure on $\mathcal{G}$. Let $\alpha^{*}: \mathcal{P}(\Omega) \to [0,\infty]$ be the corresponding outer measure. That is, if $A \in \mathcal{P}(\Omega)$ and if we let $M_A := \{\sum_{k=1}^{\infty}\alpha(Q_k)\;|\; A \subseteq \cup_{k \in \mathbb{N}}Q_k,\; Q_k \in \mathcal{G} \}$ then $\alpha^*(A) = \inf M_A$. My question is, can we say more about $\alpha^*$? Clearly $\alpha^*(A) = 0$ for countable sets $A$. If $A$ is uncountable and $A \in \mathcal{G}$ then we clearly have $\alpha^*(A) = 1$. Hence $\alpha^*(A) = \alpha(A)$ for all $A \in \mathcal{G}$. But what about $A \notin \mathcal{G}$?
If $A \notin \mathcal{G}$, my assumption is that $\alpha^*(A) = \infty$, but I don't know how to prove this. To prove this, it would suffice to show that $M_A = \emptyset$ but I don't think that is necessarily true. Since $A \notin \mathcal{G}$, both $A$ and $A^c$ are uncountable. Since $A \subseteq \bigcup_{k \in \mathbb{N}}Q_k$ at least one $Q_i$ must be uncountable, hence $Q_i^c$ is countable. I'm stuck and would appreciate any help!