Finding sets which are not $\mu^*$-measurable Recall that the Dirac measure is defined as follows:
$\mu(A)=\Big\{\begin{array}{ll}
         1 & \mbox{if  $x_0 \in A$}\\
        0 & \mbox{if  $x_0 \notin A$}\end{array}$,
where $x_0$ can be any arbitrary point.
Now, let's define an outer measure which is similar to the Dirac measure:
$\mu^*(A)=\left\{ \begin{array}{ll}
         1 & \mbox{if  $x_0 \in A$}\\
0.5 & \mbox{if  $x_0 \in \bar{A}$ but $x_0 \notin A$}\\
        0 & \mbox{if  $x_0 \notin \bar{A}$}\end{array} \right .$,
where $\bar{A}$ is the closure of $A$. You can check that $\mu^*$ is not a measure but it satisfies the conditions of being an outer measure.
According to this outer measure, I want to find not $\mu^*$-measurable sets in $R^2$ when we consider $x_0$ to be the origin.
Recall that a set $A$ is $\mu^*$-measurable in $R^2$ when for every set $E$ in $R^2$ the following equality is held:
$\mu^*(E)$=$\mu^*(E\cap A)+\mu^*(E\cap A^c)$.
In this problem, I have found that the sets whose boundary passes the origin are not $\mu^*$-measurable. This is because, by considering $E=R^2$, we have: $\mu^*(E)=1$ but $\mu^*(E\cap A)+\mu^*(E\cap A^c)=1.5$.
Could you find other not $\mu^*$-measurable sets according to the defined outer measure? Or, could you prove that there is no other not $\mu^*$-measurable set?
 A: Since outer measures are subadditive it is enough to find conditions on a set $A$ that are sufficient and necessary for the existence of a set $E$ that satisfies:$$\mu^*(E)<\mu^*(A\cap E)+\mu^*(A^c\cap E)$$Let us assume that $A$ contains the origin (for convenience denoted as $0$). This gives no essential loss of generality because a set $A$ is non-measurable iff its complement is non-measurable and one of them contains the origin.
By monotonicity we know that both terms on RHS are not larger than $\mu^*(E)$ and this tells us directly that the inequality can only be satisfied if these terms are both positive.
By substituting for some satisfying $E$ we discern the following possibilities:
$$1<1+\frac12\text{ or }\frac12<\frac12+\frac12$$
In the first case with $0\in E$ and in the second $0\in\overline E-E$.
Taking for both cases $E$ as large as possible we substitute $E=\mathbb R^2$ and $E=\Bbb R^2-\{0\}$ respectively to find the conditions:$$\mu^*(A^c)=\frac12$$for the first case and:$$\mu^*(A-\{0\})=\mu^*(A^c)=\frac12$$for the second.
For this in the first case it is necessary and sufficient that $0\in\overline{A^c}=(A^o)^c$ and combining this with preassumption $0\in A$ we conclude that a set $A$ with $0\in A-A^o$ is not measurable.
For the second case it is also necessary that $\mu^*(A^c)=\frac12$ or equivalently that $0\in A-A^o$ so this case will not provide "new" cases of non-measurable sets.
Dropping the preassumption that $0\in A$ we conclude that a set is non-measurable if $0\in A-A^o$ or if $0\in A^c-(A^c)^o$.
This can be summarized as: $$A\text{ is non-measurable iff }0\in\partial A$$

Edit:
Everything is fine above except that $\mu^*$ is not an outer measure. See my comment on the question.
