For bounded set $E \subset \mathbb{R^n},$ difine the Lebesgue inner measure $m_*$ as below.
$m_*(E):=m^*(Q)-m^*(Q\cap E^c) \big(=v(Q)-m^*(Q\cap E^c)\big)$
where $Q$ is a closed hypercube in $\mathbb{R^n}$ such that $E\subset Q$, $m^*$ is Lebesgue outer measure and $v(Q)$ stands for the volume of $Q$.
Then, prove that $m_*(E)$ doesn't depend on the choice of $Q$, i.e., if closed hypercubes $Q_1, Q_2$ satisfy $E\subset Q_1$ and $E\subset Q_2$, then $v(Q_1)-m^*(Q_1\cap E^c)=v(Q_2)-m^*(Q_2\cap E^c).$
I tried to prove this using the definition of $m^*$ but it didn't seem to work.
I'd like you to give me some ideas.