The exterior/outer measure of an open cube

I'd like to find the exterior/outer measure $m_*(Q)$ of an open cube $Q$ in $\mathbb{R}^d$, and earlier I have learned that a closed cube $A$ has exterior measure equal to its volume $|A|$. Note that volume is defined only for closed rectangles, a broad class that contains closed cubes. Now, because the closure $\overline{Q}$ of $Q$ is a closed cube, it is tempting to show that $m_*(Q)=\left|\overline{Q}\right|$. From monotonicity of exterior measure, I immediately see that $$m_*(Q)\leq m_*(\overline{Q})=\left|\overline{Q}\right|.$$ It remains to show the reverse inequality. To this end, I'd like to consider a closed cube $Z$ contained in $Q$ and claim that $|Z|$ can be made as close to $\left|\overline{Q}\right|$ as possible. The italic text is what confuses me a lot. I'm not sure if it is a valid argument, nor know how to write it mathematically. Does anyone have an idea? Thank you.

Suppose $Q=(a_1,b_1)\times \cdots (a_n,b_n)$, where each $a_i<b_i$ (otherwise the set is empty, so has outer measure zero). Let $\epsilon>0$ be small enough so that for each $i$, we have $[a_i+\epsilon,b_i-\epsilon]\subset (a_i,b_i)$. Now, consider the closed rectangle $A_{\epsilon}= [a_1+\epsilon,b_1-\epsilon]\times \cdots \times [a_n+\epsilon,b_n-\epsilon]$. Then, by monotonicity, \begin{align} m_*(A_{\epsilon})\leq m_*(Q)\leq m_*(\bar{Q}), \end{align} and since you said you already know that the outer measure of closed rectangles is their "volume", we have \begin{align} \prod_{i=1}^n(b_i-a_i-2\epsilon)\leq m_*(Q)\leq \prod_{i=1}^n(b_i-a_i). \end{align} On the left, we have a continuous function of $\epsilon$ (actually a degree $n$ polynomial). Let $\epsilon\to 0^+$ (squeeze theorem) to conclude that $m_*(Q)=\prod_{i=1}^n(b_i-a_i)$.

From this, we can go one step further and conclude using monotonicity that for any subset $E$ such that $Q\subset E\subset \bar{E}$, we have $m_*(E)=m_*(Q)=m_*(\bar{Q})=\prod_{i=1}^n(b_i-a_i)$.