Question about Volume of a cube Let $K_1,K_2,..... $ be cubes in $\mathbb{R}^n$ such that $$ \sum_{n=1}^{\infty} v(K_n) < \frac{\epsilon}{2}$$
My teacher said that for each $n$, we can choose a rectangle $Q_n$ such that 
$$ K_n \subset Int Q_n $$ 
and
$$ v(Q_n) \leq 2 v(K_n) $$
I really cannot understand why we can choose such $Q_n$. I really need some help. Thanks.
 A: (I'm going to use $d$ to denote the dimension of $\mathbb{R}^{d}$ in what follows, not $n$ as this will be an index.)
Just consider any cube $K$ of legth $\ell$ so that $|K|=\ell^{d}$.  If we increase the length by some $\delta>0$, we get another cube (rectangle) $Q$ such that $Q$ has length $\ell+\delta$ and $|Q|=(\ell+\delta)^{d}$.  The cube $K$ clearly fits inside of $Q$, and indeed it is clear that $K\subset Q^{\circ}$ since $\delta>0$.
A common technique is to apply the $\frac{\epsilon}{2^{n}}$ trick.  You fatten the collection of cubes $\{K_{n}\}_{n}$ by a positive length (it doesn't matter by what length exactly) such that the resulting fattened cubes $\{Q_{n}\}_{n}$ satisfy $|Q_{n}|=|K_{n}|+\frac{\epsilon}{2^{n+1}}.$  Then by our assumption, $$\sum|Q_{n}|=\sum|K_{n}|+\epsilon\sum 2^{-(n+1)}=\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon.$$
Note in the present situation, by fattening the cubes $K_{n}$ to get cubes $Q_{n}$, we get an open cover and compactness arguments can be used, i.e. a finite number of the $Q_{n}^{\circ}$ cover all of the $K_{n}.$
A: Suppose $K$ is the cube $[x_1,x_1 + \delta] \times \cdots \times [x_n,x_n + \delta]$. The volume of $K$ is $\delta^n$. Define
$$ Q = [x_1 - \epsilon, x_1 + \delta + \epsilon] \times \cdots \times [x_n - \epsilon,x_n + \delta + \epsilon].$$
You should be able to verify that $K \subset \mathrm{int}\, Q$. Observe the volume of $Q$ is $(\delta + 2\epsilon)^n$, so that $v(Q) < 2 v(K)$ if $\epsilon$ is small enough. Precisely, if $$ 0 < \epsilon < \frac{(2^{1/n} - 1)\delta}{2}.$$
