Open cube containing closed cube but with less volume in $\mathbb{R}^d$

Question

I am studying measure theory from Shakarchi-Stein's book and I would like to discuss one moment:

Let $Q$ be a closed cube in $\mathbb{R}^d$. For a fixed $\varepsilon>0$ we choose an open cube $S$ which contains $Q$ and such that $|S|\leq (1+\varepsilon)|Q|,$ where by $|Q|$ I denote the volume of cube.

I was considering it for a while trying to prove it rigorously however I am not able to prove it in rigorous way.

Would be very thankful if somebody will show me how to show it.

Answer 1

Volume of a $d$-dimensional cube with edge of length $a$ is $a^d$. So, denoting the edges of $S$ and $Q$ by $a_S$ and $a_Q$ respectively, $|S| \leq (1 + \varepsilon)|Q|$ means $a_S^d \leq (1+\varepsilon)a_Q^d$, thus $a_S \leq \sqrt[d]{1+\varepsilon}\cdot a_Q$.

So, take $S$ to be the cube with the same center as $Q$ and with side $\leq\sqrt[d]{1+\varepsilon}\cdot a_Q$.

Answer 2

Generally for rectangular parallelepipeds, you can do as follows:

Let $\epsilon$ be given, and let $Q$ be a rectangular parallelepiped with sides $[a_i,b_i]$, $i=1,2,...,d$. Its volume is $(b_1-a_1)...(b_n-a_d)$. Choose an open rectangular paralleliped $S$ with sides $(u_i,v_i)$ containing $[a_i,b_i]$. There is $e_i$ such that $v_i-u_i=b_i-a_i+e_i$. The volume of $S$ is therefore equal to the volume of $Q$ plus a first order polynomial in $e_i$’s. Take $e:=max_i${$e_i$}. Hence the volume of $S$ is at most the volume of $Q$ plus a variable multiple of $e$, say, $|S|=|Q|+ce$, for some $c$ which is an increasing function of $e$, and taking $e$ small enough $e_0$ you can choose $c\lt1$. Now take $e:=min${$\epsilon |Q|,e_0$}.