Proof of Dyadic Cells in $\mathbb{R}^n$ and the Decomposition of an Open Set

Claim 1. Fix $k\in\mathbb{N}$ and $x\in\mathbb{R}^{n}$. For each $i\in\{1,\dots,n\}$, Existence and Uniqueness of the Integer Part of a Real Number provides a unique $j_{i}\in\mathbb{Z}$ with $j_{i}\le 2^{k}x_{i}<j_{i}+1$. Multiplying these inequalities by the positive number $2^{-k}$, and dividing back, shows that for $j\in\mathbb{Z}^{n}$ one has $x\in Q_{k,j}$ if and only if $j_{i}\le 2^{k}x_{i}<j_{i}+1$ for every $i$. Hence $x$ lies in exactly one $Q_{k,j}$. In particular the sets $Q_{k,j}$, $j\in\mathbb{Z}^{n}$, are pairwise disjoint and their union is $\mathbb{R}^{n}$.

For the remaining assertions of claim 1, fix $j\in\mathbb{Z}^{n}$ and apply the grid claim with $c=(j_{1}2^{-k},\dots,j_{n}2^{-k})$, $s=2^{-k}$ and mesh index $m=1$. The half-open box of that lemma is then exactly $Q_{k,j}$, and since the initial segment $[1]$ has the single element $1$, the grid at mesh $1$ consists of the single cell equal to that box. The claim therefore gives $Q_{k,j}\in\mathcal{B}(\mathbb{R}^{n})$, $\lambda_{n}(Q_{k,j})=(2^{-k})^{n}$, and $\lVert x-y\rVert\le\sigma_{n}2^{-k}$ for all $x,y\in Q_{k,j}$.

The integers are countable by The Integers and the Rational Numbers are Countable, hence so are the set $\mathbb{Z}^{n}$ of $n$-tuples in $\mathbb{Z}$ and the product $\mathbb{N}\times\mathbb{Z}^{n}$, by Products and Powers of Countable Sets; both are infinite, since $j\mapsto(j,0,\dots,0)$ is an injection of $\mathbb{Z}$ into $\mathbb{Z}^{n}$. Finally, if $Q_{k,j}=Q_{k',j'}$ then comparing the measures just computed gives $(2^{-k})^{n}=(2^{-k'})^{n}$; as $t\mapsto t^{n}$ is strictly increasing on the positive reals and $k\mapsto 2^{k}$ is strictly increasing, this forces $k=k'$, and then $j=j'$ because a point of the common cell lies in exactly one cell of generation $k$. So $(k,j)\mapsto Q_{k,j}$ is injective.

Claim 2. Let $k,k'\in\mathbb{N}$ with $k\le k'$ and let $j'\in\mathbb{Z}^{n}$. Put $\theta=2^{k}/2^{k'}$, so that $0<\theta\le1$ and $\theta^{-1}=2^{k'}/2^{k}$ is an integer (it is $1$ if $k=k'$ and the natural power $2^{k'-k}$ otherwise).

Fix $i$ and let $j''_{i}\in\mathbb{Z}$ be the unique integer with $j''_{i}\le\theta j'_{i}<j''_{i}+1$, again by Existence and Uniqueness of the Integer Part of a Real Number. Put $t=\theta j'_{i}-j''_{i}$, so $0\le t<1$. Then $\theta^{-1}t=j'_{i}-\theta^{-1}j''_{i}$ is an integer, by the closure of $\mathbb{Z}$ under multiplication and subtraction (Arithmetic, Order and Discreteness of the Integers), and $0\le\theta^{-1}t<\theta^{-1}$. By the discreteness of $\mathbb{Z}$ recorded in the same lemma there is no integer strictly between $\theta^{-1}-1$ and $\theta^{-1}$, so $\theta^{-1}t\le\theta^{-1}-1$, that is $t\le 1-\theta$.

Now let $y\in Q_{k',j'}$, so $j'_{i}\le 2^{k'}y_{i}<j'_{i}+1$ for every $i$. Multiplying by $\theta>0$ gives $\theta j'_{i}\le 2^{k}y_{i}<\theta j'_{i}+\theta$, hence

By the characterisation in claim 1 this says $y\in Q_{k,j''}$, where $j''=(j''_{1},\dots,j''_{n})$. Thus $Q_{k',j'}\subseteq Q_{k,j''}$. The cell $Q_{k',j'}$ is nonempty (it contains the point with coordinates $j'_{i}2^{-k'}$), so by claim 1 no other cell of generation $k$ can contain it, which gives the asserted uniqueness of $j''$.

Let now $j\in\mathbb{Z}^{n}$ be arbitrary. If $j=j''$ then $Q_{k',j'}\subseteq Q_{k,j}$. If $j\ne j''$ then $Q_{k,j}\cap Q_{k,j''}=\varnothing$ by claim 1, and since $Q_{k',j'}\subseteq Q_{k,j''}$ we get $Q_{k',j'}\cap Q_{k,j}=\varnothing$. Given two arbitrary dyadic cells, relabel them so that the generation of the first is at most that of the second and apply what has just been proved.

Claim 4. First, $k\le 2^{k}$ for every $k\in\mathbb{N}$: this holds for $k=1$, and if $k\le 2^{k}$ then $k+1\le 2^{k}+1\le 2^{k}+2^{k}=2^{k+1}$, so the assertion follows by Principle of Induction for the Natural Numbers. Given $x\in\mathbb{R}^{n}$ and $r>0$, the Archimedean property (The Archimedean Property of the Real Numbers) supplies $k\in\mathbb{N}$ with $\sigma_{n}/r<k$; then $2^{-k}\le 1/k$ and hence $\sigma_{n}2^{-k}\le\sigma_{n}/k<r$. Finally, if $k\in\mathbb{N}$ satisfies $\sigma_{n}2^{-k}\le r$ and $x\in Q_{k,j}$, then every $y\in Q_{k,j}$ satisfies $\lVert y-x\rVert\le\sigma_{n}2^{-k}\le r$ by claim 1.

Claim 3. If $U=\varnothing$, take $P_{m}=\varnothing$ for every $m$; the union is $U$ and both sides of the measure identity are $0$. Assume $U\ne\varnothing$.

(a) Every point of $U$ lies in a dyadic cell contained in $U$. Let $x\in U$. As $U$ is open there is $r\in\mathbb{R}$ with $0<r$ and $\{y\in\mathbb{R}^{n}:\lVert y-x\rVert<r\}\subseteq U$. By claim 4 there is $k\in\mathbb{N}$ with $\sigma_{n}2^{-k}\le r/2$, and the cell $Q$ of generation $k$ containing $x$ then satisfies $\lVert y-x\rVert\le r/2<r$ for every $y\in Q$; hence $Q\subseteq U$.

(b) The maximal such cells partition $U$. Let $\mathcal{D}$ be the set of pairs $(k,j)\in\mathbb{N}\times\mathbb{Z}^{n}$ such that $Q_{k,j}\subseteq U$ and, for every $k''\in\mathbb{N}$ with $k''<k$, the unique cell of generation $k''$ containing $Q_{k,j}$ (claim 2) is not contained in $U$.

Given $x\in U$, the set $A=\{k\in\mathbb{N}:\text{the cell of generation }k\text{ containing }x\text{ is contained in }U\}$ is nonempty by (a), so by The Natural Numbers Are Well Ordered it has a least element $k_{0}$. Let $j$ be the index with $x\in Q_{k_{0},j}$. For $k''<k_{0}$ the cell of generation $k''$ containing $Q_{k_{0},j}$ contains $x$, hence is the cell of generation $k''$ containing $x$, which is not contained in $U$ because $k''\notin A$. Thus $(k_{0},j)\in\mathcal{D}$ and $x$ lies in the corresponding cell. Since every cell indexed by $\mathcal{D}$ is contained in $U$, the union of these cells is exactly $U$.

Suppose $(k,j),(k',j')\in\mathcal{D}$ are distinct and $Q_{k,j}\cap Q_{k',j'}\ne\varnothing$; relabel so that $k\le k'$. By claim 2, $Q_{k',j'}\subseteq Q_{k,j}$, so $Q_{k,j}$ is the cell of generation $k$ containing $Q_{k',j'}$. If $k<k'$ this contradicts $(k',j')\in\mathcal{D}$, since $Q_{k,j}\subseteq U$. Hence $k=k'$, and then $j=j'$ by claim 1, contradicting distinctness. So the cells indexed by $\mathcal{D}$ are pairwise disjoint.

(c) Enumeration. By claim 1 the set $\mathbb{N}\times\mathbb{Z}^{n}$ is countable and infinite, so by Countable Set there is a sequence $((k_{m},j_{m}))_{m\in\mathbb{N}}$ enumerating it bijectively. Put $P_{m}=Q_{k_{m},j_{m}}$ if $(k_{m},j_{m})\in\mathcal{D}$ and $P_{m}=\varnothing$ otherwise. Each $P_{m}$ lies in $\mathcal{B}(\mathbb{R}^{n})$ by claim 1 and is either empty or a dyadic cell contained in $U$; the sets $P_{m}$ are pairwise disjoint by (b) together with the injectivity of the enumeration; and their union is the union of the cells indexed by $\mathcal{D}$, which is $U$ by (b).

(d) The measure identity. Let $(P_{m})_{m\in\mathbb{N}}$ be any sequence of pairwise disjoint members of $\mathcal{B}(\mathbb{R}^{n})$ whose union is $U$. Countable additivity of $\lambda_{n}$, part of the definition of a measure, gives $\lambda_{n}(U)=\sum_{m\in\mathbb{N}}\lambda_{n}(P_{m})$.