LemmaStatement: AI-adaptedProof: AI-adaptedprecheck passaudited 2026-08-26

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For each generation, the dyadic cubes of that generation are pairwise disjoint and cover $R^{n}$

Statement

Let $n \geq 1$ and let $k \in N$ . Every $x \in R^{n}$ lies in exactly one dyadic cube of generation $k$ (Dyadic cubes of generation $k$ in $R^{n}$ ); that is, the generation- $k$ dyadic cubes are pairwise disjoint and their union is $R^{n}$ . Each of them has volume $vol (Q_{k, m}) = 2^{- k n}$ (Half-open boxes in $R^{n}$ and their volume).

Facts & Assumptions

Given: A natural number $n \geq 1$ , a natural number $k$ , and the dyadic cubes of generation $k$ .

[L1]

$Q_{k, m} = {x \in R^{n} : m_{i} 2^{- k} < x_{i} \leq (m_{i} + 1) 2^{- k} for every i < n}$ (Dyadic cubes of generation $k$ in $R^{n}$ ).

[L2]

For a nonempty box $vol (B) : = \prod_{i < n} (b_{i} - a_{i})$ when every $a_{i}$ and every $b_{i}$ is real (Half-open boxes in $R^{n}$ and their volume).

[F1]

For every real $x$ there is exactly one integer $p$ with $p \leq x < p + 1$ (Integer part: for every real $x$ there is exactly one integer $m$ with $m \leq x < m + 1$ ).

[F2]

For $a \neq 0$ and $m, n \in Z$ , $a^{m + n} = a^{m} a^{n}$ and $(a^{m})^{n} = a^{m n}$ (Laws of integer exponents, claims 1 and 3; Integer powers $a^{m}$ ).

[F3]

$\prod_{k < n} (a_{k} b_{k}) = (\prod_{k < n} a_{k}) (\prod_{k < n} b_{k})$ , and finite products are defined by the recursion $Π_{0} = 1$ , $Π_{σ (n)} = Π_{n} \cdot a_{n}$ (Laws of finite sums and finite products, claim 6; Finite sums and finite products, by recursion).

Proof

technique · direct

1.1F1algebra

For a real $t$ there is exactly one integer $m$ with $m < t \leq m + 1$ : applying [F1] to $- t$ gives the unique integer $p$ with $p \leq - t < p + 1$ , and $m : = - p - 1$ then satisfies $m < t \leq m + 1$ , while any integer $m^{'}$ with $m^{'} < t \leq m^{'} + 1$ yields $- m^{'} - 1 \leq - t < - m^{'}$ , so $- m^{'} - 1 = p$ by the uniqueness in [F1] and $m^{'} = m$ .

1.2L1F2algebra

Since $2^{- k} > 0$ , the condition $m_{i} 2^{- k} < x_{i} \leq (m_{i} + 1) 2^{- k}$ is equivalent to $m_{i} < 2^{k} x_{i} \leq m_{i} + 1$ , the powers satisfying $2^{k} 2^{- k} = 1$ .

1.3L1L2F2F3

The volume of $Q_{k, m}$ is $\prod_{i < n} ((m_{i} + 1) 2^{- k} - m_{i} 2^{- k}) = \prod_{i < n} 2^{- k} = (2^{- k})^{n} = 2^{- k n}$ , the last two equalities by the recursion for finite products and the power laws.

2.1step 1.1step 1.2L1

Given $x \in R^{n}$ , step 1.1 applied in each coordinate to the real $2^{k} x_{i}$ produces exactly one integer $m_{i}$ with $m_{i} < 2^{k} x_{i} \leq m_{i} + 1$ , so by step 1.2 the function $m$ so determined is the unique index of a generation- $k$ dyadic cube containing $x$ ; hence the generation- $k$ cubes cover $R^{n}$ and no two of them share a point.

3.1step 1.3step 2.1∎

Steps 2.1 and 1.3 are the Statement.

Depends on

Dependency tree · 2 levels

Sources

T. Tao, An Introduction to Measure Theory (GSM 126), Exercise 1.1.14

For each generation, the dyadic cubes of that generation are pairwise disjoint and cover RnRn