# Clarification on the proof that an open set is a countable union of almost disjoint boxes

To establish that the Lebesgue outer measure of any open set in the Euclidean metric space of $\mathbb{R}^d$ is equal to the volume of any partitioning of that set into almost disjoint boxes, lemma 1.2.11 in p. 24 of the book “An introduction to measure theory” by Terence Tao first states that any open set $E\subseteq \mathbb{R}^d$ can be expressed as a countable union of almost disjoint boxes (and in fact as a countable union of almost disjoint closed cubes).

Note that this lemma is a generalization of the fact that every open subset of $\mathbb{R}$ can be expressed as a countable union of disjoint open intervals (see theorem 1.3, p. 6, in the book “Real analysis: measure theory, integration and Hilbert spaces” by Elias M. Stein and Rami Shakarchi and also lemma 2 of this blog post of mine).

The purpose of the present blog post is to elaborate on a detail in the proof of lemma 1.2.11 of Tao’s book. In particular, it will be explained why  every closed dyadic cube $Q$ is contained in exactly one maximal cube $Q^{*}$, without reproducing the rest of the proof.

Introduce the notation $Q_{n,\mathbf{i}}=\overset{d}{\underset{k=1}{\times}}\left[\frac{i_k}{2^n},\frac{i_k+1}{2^n}\right]$ to indicate the dependence of closed dyadic cubes on $n\in\mathbb{N}\cup \left\{0\right\}$ and $\mathbf{i}=(i_1,i_2,\cdots, i_d)\in\mathbb{Z}^{d}$. For every $\mathbf{i}=(i_1,i_2,\cdots, i_d)\in\mathbb{Z}^{d}$ for which there exists a closed dyadic cube $Q_{n,\mathbf{i}}$ contained in $E$, choose the biggest closed dyadic cube $Q_{n,\mathbf{i}}$ in $E$, i.e. choose $n_0=\underset{n}{\min}\left\{n\in\mathbb{N}\cup \left\{0\right\}:Q_{n,\mathbf{i}}\in E\right\}$. It is now obvious that by having capped the cubes by a sidelength of 1, there exists a maximum cube $Q_{n_0,\mathbf{i}}$ among $\left\{Q_{n,\mathbf{i}}\in E:n\in\mathbb{N}\cup \left\{0\right\}\right\}$, which is a maximal cube among those contained in $E$.

What is left to show is that every closed dyadic cube $Q$ can’t be contained in two or more maximal cubes. Assume that $Q$ is contained in two distinct maximal cubes $Q^*$ and $Q^{**}$, i.e. $Q\subseteq Q^*$ and $Q\subseteq Q^{**}$ with $Q^*\neq Q^{**}$. The dyadic nesting property then leads to the contradiction $Q^*\subseteq Q^{**}$ or $Q^{**}\subseteq Q^*$ for the maximal cubes $Q^*$ and $Q^{**}$, since $Q\subseteq Q^*$ and $Q\subseteq Q^{**}$ excludes the possibility of $Q^*$ and $Q^{**}$ being almost disjoint.