If $X=\{0,1\}$, there exists an outer measure $\mu^*$ on $X$ such that $\mu^* \neq \mu^+$ Background
Let $\mu^*$ be an outer measure on $X$ , $\mathcal{M}^*$ the $\sigma-$ algebra of all $\mu^*$ measurable sets, $\overline{\mu}=\mu^*\bigg|_{\mathcal{M}^*},$ and $\mu^+$ the outer measure induced by $\overline{\mu}$.
Question
If $X=\{0,1\}$, there exists an outer measure $\mu^*$ on $X$ such that $\mu^* \neq \mu^+$.
Attempt
I have tried to use the counting measure and the outer measure $\mu^*(A)=\sum_{n\in A} n$, but in both cases I find that $\mathcal{M}^*=\mathcal{P}(X)$. I am hoping to use to the following result:
$\mu^*(E)=\mu^+(E)$ iff $\exists A \in \mathcal{M}^*$ with $E\subset A$ and $\mu^*(E)=\mu^*(A)$
For which I need a measure such that $\mathcal{M}^*$ is a proper subset of $X$.
 A: As $X = \{0,1\}$, there is not much freedom one has for $\mu^*$. Because $\mu^*$ is an outer measure, we are forced by subadditivity to satisfy
\begin{equation}
\mu^*(A) \leq \mu^*(X) \leq \mu^*(\{0\}) + \mu^*(\{1\}).
\end{equation}
where $A = \{0\}$ or $\{1\}$.
Suppose we choose $\mu^*(X), \mu^*(\{0\})$, and $\mu^*(\{1\})$ in such a way so that we have strict inequalities:
\begin{align}
\mu^*(A) < \mu^*(X) < \mu^*(\{0\}) + \mu^*(\{1\})
\end{align}
 Then neither $\{0\}$ nor $\{1\}$ are $\mu^*$-measurable sets. 
Therefore, when we try to approximate $\{0\}$ or $\{1\}$ by $\mu^*$-measurable sets, i.e., when we compute
\begin{equation}
\mu^{+}(A) = \inf\{\sum_{j = 1}^{\infty}\mu^*(E_j)\,:\, E_j \in \mathcal{M}^*, A \subset \bigcup_{j = 1}^{\infty}E_j\}
\end{equation}
where $A = \{0\}$ or $\{1\}$, the covering will be 'pretty bad', i.e. 'the best we can do' is $\mu^+(A) = \mu^*(X) > \mu^*(A)$.
Having said all this, the above discussion should be more than enough information to construct a specific example where $\mu^* \neq \mu^+$.
