LemmaStatement: AI-adaptedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-26

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AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
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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

1.1

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 $m^{'} = m$ .

F1algebra

1.2

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 $2^{k} 2^{- k} = 1$ .

L1F2algebra

1.3

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.

L1L2F2F3

2.1

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.

step 1.1step 1.2L1

3.1

Steps 2.1 and 1.3 are the Statement.

step 1.3step 2.1∎

Depends on

Used by

Dependency tree · two levels

37 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

T. Tao, An Introduction to Measure Theory (GSM 126), Exercise 1.1.14 (standard reference, not scraped)

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