# For every open subset of $\mathbb{R}^n$ there is a countable non-overlapping union of closed squares

I am struggling with a definition which should lead to the Lebesgue outer measure.

Theorem: Every open subset of $\mathbb{R}^n$ is a countable disjoint (with the meaning non-overlapping or the interior of two different squares are disjoint) union of closed squares.

The algorithm for generating these squares uses following definition: $$\mathcal{C}^{(k)}:=\left\{Q=\left\{(x_1,\cdots, x_n)\;|\; \frac{m_j}{2^k} \leq x_j \leq \frac{m_{j+1}}{2^k}\;,j=1, \cdots, n\right\}\;| \;m_j \in \mathbb{Z}\right\},$$ for $k=0,1,\cdots$

I do understand that the green area should be covered with non-overlapping squares which are getting finer and finer but I do not understand how the $m_j$ are choosen. Using this example:

How do I choose $m_j$ and $m_{j+1}$? Please feel free to draw into the example.

• Should it not be $m_j + 1$ , so its an interval of length $2^{-k}$ in the j-th coordinate. – mlu Dec 1 '15 at 21:16
• So you select as many boxes/squares as possible with side length 1, integer coordinates and completely inside the green region. Then half the side length and select smaller boxes inside the region(green) but not in any previously selected box . repeat, show that very interior point will be inside one of the boxes. – mlu Dec 1 '15 at 21:27
• @mlu I think you are right, $m_j +1$ would result in a lattice completely covering $\mathbb{R}^n$. – monoid Dec 1 '15 at 22:04

I give you a way to do it in dimension $1$ that you can adapt in dimension $n \ge 1$.
Suppose that $U \subset \mathbb{R}$ is an open subset. As $U$ is open, for any $x \in U$, one can find an open interval $(y,z) \subset U$ with $y < z$ such that $x \in (y,z)$. Name $d=\min(\vert x-y \vert, \vert x-z \vert) >0$ and take $k \ge 1$ integer such that $\frac{1}{2^k} < \frac{d}{2}$.
Now name $m =\sup \{l \in \mathbb Z \ : \ \frac{l}{2^k} < x\}$. You have $x-\frac{m}{2^k} \le \frac{1}{2^k} < d$. Hence $$x \in [\frac{m}{2^k},\frac{m+1}{2^k}] \subset (y,z) \subset U.$$
Making the union of all those closed intervals from a countable set for $x \in U$, you're done.