If $E$ has positive outer measure, then $E = A\cup B$, where $A\cap B = \emptyset$ and $A, B$ have positive outer measure I found this question online while trying to prepare for an Exam, however, I am not sure if my solution is valid. Below is the question and my attempted solution.

Question: Let $E \subset \mathbb{R}$ where $m^*(E) > 0$. Show that $\exists A, B \subset E$ bounded, such that, $A \cap B = \emptyset$ and $m^*(A) > 0$ and $m^*(B) > 0$.
(my attempted) Solution: Since $E$ has positive outer measure, $\exists A \subset E$ bounded, such that
$m^*(A) > 0$. Let $B = E \setminus A$. Then $B \subset E$, and $m^*(B) = m^*(E \setminus A)$. However, I am not sure if $B$ is bounded and if $m^*(B) > 0$...so this is where I get stuck.
 A: Okay, I think I've maybe solved the issues in my head with this problem.
Let $I_k = [k,k+1)$ be an interval. Then the collection $\{I_k\}$ covers $\mathbb{R}$ and we see that
$$ E = \bigcup_{k \in \mathbb{Z}} E \cap I_k, $$
so by subadditivity we see that
$$ 0 < m^*(E) \leq \sum_{k \in \mathbb{Z}} m^*(E \cap I_k),$$
forcing at least one of the $E \cap I_k$ to have positive measure, $m^*(E \cap I_k) > 0$. Moreover, this set is bounded.
What if $m^*(E \cap I_k) \neq m^*(E)$? Then $m^*(E \cap I_k) < m^*(E)$, and if we throw out $E \cap I_k$ from $E$ (i.e. examine $F = E \setminus (E \cap I_k)$) then we have $m^*(F) > 0$, $F \sqcup (E \cap I_k) = E$. (Here we're implicitly using Caratheodory.)
What if $m^*(E \cap I_k) = m^*(E)$? It might be $E \cap I_k \neq E$ still, in which case let $F = E \setminus (E \cap I_k)$ but $\mu^*(F) = 0$. We can now divide up $E \cap I_k$. Since $I_k$ is an interval, we can do the obvious thing of dividing it up into two disjoint intervals so that $$m^*(I_k) = m^*(A \cup B) = m^*(A) + m^*(B)$$ and $m^*(A), m^*(B) =1/2$. Specifically we can choose them so that $B = I_k \setminus A$. Now
$$0 < m^*(E) = m^*(E \cap I_k) \leq m^*(E \cap A) + m^*(E \cap B).$$
Think about Caratheodory now. Since $A$ a measurable set (an interval) we have
$$ 0 < m^*(E) = m^*(E \cap I_k) = m^*(E \cap I_k \cap A) + m^*(E \cap I_k \cap A^c) = m^*(E \cap A) + m^*(E \cap B).$$
If $m^*(E \cap B) \neq 0$ and $m^*(E \cap A) \neq 0$ then we're done. Otherwise, suppose without loss of generality $m^*(E \cap B) = 0$. Do the same dividing process to $A$ into halves. If, when we keep dividing up our interval, we eventually have one where both of the sets have non-zero measure, we win. Otherwise, we have
$$m^*(E) \leq \frac{1}{2^n} \text{ for all n} \geq 0.$$
But this means that $m^*(E) = 0$, which is a contradiction.
