Let $\mu^*: \mathcal P(\Bbb R) \to [0,\infty] $, $\mu^*(A)=0$ if $A$ is numerable (finite or infinite numerable), $\mu^*(A)=1$ if $A$ is not numerable and there is a bounded interval $I$ such that $A\setminus I$ is numerable, $\mu^*(A)=\infty$ in the rest of cases, i.e. $A$ is not numerable and there is not a bounded interval $I$ such that $A\setminus I$ is numerable. When I say "bounded" I mean "bounded in $\Bbb R$" and not generally in $[-\infty, \infty]$ like $(0,\infty)$ is bounded in $[-\infty, \infty]$.
I proved that $\mu^*$ is an outer measure in $\Bbb R$. Now I have to find out which is the Caratheodry $\sigma$-algebra $\mathcal M^*$, that is $\mathcal M^*=\{A\subset \Bbb R/A$ is $\mu^*$-measurable$\}$.
I also proved that every numerable set is in $\mathcal M^*$, that a not numerable set $A$ such that there is a bounded interval $I$ such that $A\setminus I$ is numerable is not in $\mathcal M^*$.
But I have problems with the last case of subsets in $\Bbb R$.
I know that if $A$ is numerable, then $A^c \in \mathcal M^*$, because the definition of $\mu^*$-measurable and $\mu^*(A^c)=\infty$.
I proved that if $A$ is not numerable and $\mu^*(A)=1$, then $\mu^*(A^c)=\infty$, but since $A\notin\mathcal M^*$, then $A^c\notin\mathcal M^*$.
If $A$ is in another case, $\mu^*(A)=\infty$ but some of these $A$'s are in $\mathcal M^*$ and others are not. For instance, the two previous $A^c$'s, one of them is in $\mathcal M^*$ and the second one is not in $\mathcal M^*$, both of them such that $\mu^*(A^c)=\infty$.
If $A=[0,\infty$), then $\mu^*(A)=\infty$ but this $A$ is not a case I can obtain from the others and $A \notin \mathcal M^*$.
I do not know whether I left other cases of $\mu^*(A)=\infty$ such that $A\in\mathcal M^*$.
My questions is: $M^*=\{A,A^c/\mu^*(A)=0, A\subset \Bbb R \}$?