open box $(0, 1/q)^n$ and closed box $[0, 1/q]^n$ both have measure $1/q^n$. 
(Tao Vol.2, P.174, Q.7.2.4) Show that for any positive integer $q > 1$, that the open box
$$(0, 1/q)^n : = \{\text{$(x_1, ... , x_n) \in \mathbb{R}^n : 0 < x_j < 1/q$ for all $1 \le j \le n$\}}$$
and the closed box
$$[0, 1/q]^n : = \{\text{$(x_1, ... , x_n) \in \mathbb{R}^n : 0 \le x_j \le 1/q$ for all $1 \le j \le n$\}}$$
both measure $q^{-n}$. (Hint: first show that $m((0,1/q)^n) \le q^{-n}$ for every $q \ge 1$ by covering $(0,1)^n$ by some translates of $(0,1/q)^n$. Using a similar argument, show that $m([0, 1/q]^n) \ge q^{-n}$. Then show that $m([0, 1/q]^n\setminus(0, 1/q)^n) \le \epsilon$ for every $\epsilon >0$, by covering the boundary of $[0,1/q]^n$ with some very small boxes.

Note that $m((0,1/q)^n) = m((0+\frac{k-1}{2q}, \frac1q + \frac{k-1}{2q})^n) = m((\frac{k-1}{2q}, \frac{k+1}{2q})^n)$ and that $(0,1)^n \subset \bigcup_{k=0}^{2q} (\frac{k-1}{2q}, \frac{k+1}{2q})^n$. Thus,
$$m((0,1)^n) \le m(\bigcup_{k=0}^{2q} (\frac{k-1}{2q}, \frac{k+1}{2q})^n)\le \sum_{k=0}^{2q} m((\frac{k-1}{2q}, \frac{k+1}{2q})^n) =\sum_{k=0}^{2q} m((0,1/q)^n) = (2q+1)m((0,1/q)^n).$$
How should I proceed from here? (Here, I know that $m([0,1]^n) = 1$, but I haven't yet derived the measure of $(0,1)^n$.)
Moreover, I need to find $B$ such that $m(B) \le \epsilon$ and $B$ covers the boundary of $[0,1]^n$. Then,
$$m([0, 1/q]^n\setminus(0, 1/q)^n) = m([0, 1/q]^n) - m((0, 1/q)^n) \le m((0, 1/q)^n\cup B) - m((0, 1/q)^n) \le m(B) \le \epsilon.$$
I can intuitively see the existence of $B$, but I don't know the explicit interval $B$.
I would appreciate if you give some help.
 A: The part which shows that $m((0,1/q)^n) \leq q^{-n}$ and $m([0, 1/q]^n) \geq q^{-n}$ can be found here:
Why does the boundary of a cube in $\mathbb{R}^n$ have measure zero?
To show that the boundary measures arbitrarily small, $\forall 1 \leq j \leq n$, define $A_j := \{(x_1, ... , x_n) \in \mathbb{R}^n : x_j = 0, 0 \leq x_k \leq 1/q$ for all $k \neq j\}$ and $B_j := \{(x_1, ... , x_n) \in \mathbb{R}^n : x_j = 1/q, 0 \leq x_k \leq 1/q$ for all $k \neq j\}$, where $j$ is the smallest number for which the $j^{th}$ component $x_j=0$ or $x_j=1/q$.
Note that the boundary of $[0,1/q]^n$ is equal to $\bigcup_{1
\leq j,k \leq n}(A_j\cup B_k)$. By finite sub-additivity, it suffices to bound $m(A_j)$ and $m(B_j)$ to be arbitrarily small for each j.
To do this, $\forall \epsilon > 0$, let $N$ be a positive integer s.t $N \geq \frac{1}{\epsilon}$. Let $(\delta_i)_{i=1}^{N}, \delta_i > 0$ be a sequence of distinct small numbers s.t $\delta_i, \delta_i + 1/q \leq 1$ for all i. Hence $A_j+\delta_ie_j, B_j+\delta_ie_j \subseteq [0,1]^n$ for all i. By monotonicity, finite additivity, and translation-invariance, we then have $Nm(A_j), Nm(B_j) \leq 1$. In particular, $m(A_j), m(B_j) \leq 1/N \leq \epsilon$.
Since $\epsilon > 0$ is arbitrary, we get our desired result.
