So this is a past paper question that I am struggling with slightly. I have managed to prove that $m^*$ was a measure, which was the preceding the previous part of the question. It is the next part that I am struggling with slightly!
Let $A,B \subseteq \mathbb{R}$ be disjoint infinite sets. For any $ E \subseteq \mathbb{R}$ define
$$ m^*(E) = \begin{cases} \text{2,} &\quad\text{if $E$ intersects both $A$ and $B$,}\\ \text{1,} &\quad\text{if $E$ intersects exactly one of the sets $A$ or $B$}\\ \text{0,} &\quad\text{if $E \cap A = E \cap B = \emptyset$}\\ \end{cases}. $$ For $m^*$ defined above, find an interval $[x,y]$ which is not $m^*$-measurable.
It is easy to proove that $m^*$ is, in fact, an outer measure.
Now, let $A,B \subseteq \mathbb{R}$ be disjoint infinite sets.
Since $B$ is infinite, there are $b_0,b_1,b_2 \in B$, such that $b_0<b_1<b_2$. Since $A$ is infinite then $A\cap (-\infty,b_1)$ or $A\cap(b_1,\infty)$ (or both) is an infinite set.
Case 1: Suppose $A\cap (-\infty,b_1)$ is infinite. Then there are $a_0, a_1 \in A$ such that $a_0<a_1<b_1 <b_2$.
Let $I=[a_1,b_1]$. We claim that $I$ is not $m^*$-measurable.
In fact, since $\{a_1,b_1\}\subseteq I = \mathbb{R}\cap I$ we have
$$ m^*(\mathbb{R}\cap I) =2$$
and since $\{a_0,b_2\}\subseteq \mathbb{R}\cap I^c$ we have
$$ m^*(\mathbb{R}\cap I^c)=2$$
So we have
$$ m^*(\mathbb{R})=2 < 2+2= m^*(\mathbb{R}\cap I) + m^*(\mathbb{R}\cap I^c)$$
So $I$ is not $m^*$-measurable.
Case 2: Suppose $A\cap (b_1,+\infty)$ is infinite. Then there are $a_1, a_2 \in A$ such that $b_0<b_1 <a_1<a_2$.
Let $I=[b_1,a_1]$. We claim that $I$ is not $m^*$-measurable.
The proof is analogous to case 1.
In fact, since $\{b_1,a_1\}\subseteq I = \mathbb{R}\cap I$ we have
$$ m^*(\mathbb{R}\cap I) =2$$
and since $\{b_0,a_2\}\subseteq \mathbb{R}\cap I^c$ we have
$$ m^*(\mathbb{R}\cap I^c)=2$$
So we have
$$ m^*(\mathbb{R})=2 < 2+2= m^*(\mathbb{R}\cap I) + m^*(\mathbb{R}\cap I^c)$$
So $I$ is not $m^*$-measurable.