How to show that $\{x \in \mathbb R^{n} : \|x\| \le 1 \}$ is an $m^* $-measurable set?
I know that a set $E$ is said to be $m^*$-measurable (or Lebesgue measurable) if for any set $A \subset \Bbb R^{n}$ we have,
$$m^*(A)=m^*(A \cap E)+m^*(A \cap E^C)$$
How should I use this definition? Is there any other method to show this?
Since $m^*$ is subadditive, the only thing that needs a proof is the inequality $$ m^*(A)\ge m^*(A \cap E)+m^*(A \cap E^c)$$ Here is one approach.
Prove that $m^*$ is a metric outer measure, meaning that $m^*(A\cap B)=m^*(A)+m^*(B)$ when $\operatorname{dist}(A,B)>0$. This follows from the fact that $m^*$ can be computed by taking covers by sufficiently small cubes, which do not intersect both $A$ and $B$ at the same time. Hence, any cover of $A\cup B$ can be separated into a cover of $A$ and a cover of $B$, which leads to $m^*(A\cap B)\ge m^*(A)+m^*(B)$.
For an integer $k$, consider $A_k = \{x\in A:\operatorname{dist}(x,E)\ge 1/k\}$. Since $\operatorname{dist}(A\cap E, A_k)>0$, it follows that $$ m^*(A) \ge m^*((A\cap E)\cup A_k) = m^*(A\cap E)+m^*(A_k)$$ It remains to note that $m^*(A_k)\to m^*(A\cap E^c)$ as $k\to\infty$, because $A\cap E^c$ is the union of the nested sequence $A_k$. (This latter fact relies on the assumption that $E$ is closed.)