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A translation-invariant Borel measure giving the unit cube measure one gives each generation- $k$ dyadic cube measure $2^{- k n}$

Statement

Let $n \geq 1$ and let $μ$ be a measure on $(R^{n}, B (R^{n}))$ (Measures on sigma-algebras, The Borel sigma-algebra of a topological space) such that

$μ (E + h) = μ (E) for every Borel E and every h \in R^{n}, μ ((0, 1]^{n}) = 1.$

Then $μ (Q) = 2^{- k n}$ for every dyadic cube $Q$ of generation $k$ (Dyadic cubes of generation $k$ in $R^{n}$ ).

Only translates of half-open boxes are used, and those are Borel (The sigma-algebra generated by the half-open boxes of $R^{n}$ is the Borel sigma-algebra), so the invariance hypothesis is applied only where it is unambiguously meaningful.

Facts & Assumptions

Given: A natural number $n \geq 1$ , a natural number $k$ , and a measure $μ$ on the Borel sets of $R^{n}$ that is translation invariant and gives the unit cube measure $1$ .

[L1]

$Q_{k, m} = {x \in R^{n} : m_{i} 2^{- k} < x_{i} \leq (m_{i} + 1) 2^{- k} for every i < n}$ , and $Q_{0, 0} = (0, 1]^{n}$ (Dyadic cubes of generation $k$ in $R^{n}$ , Half-open boxes in $R^{n}$ and their volume, Integer powers $a^{m}$ ).

[L2]

Every $x \in R^{n}$ lies in exactly one dyadic cube of generation $k$ (For each generation, the dyadic cubes of that generation are pairwise disjoint and cover $R^{n}$ ).

[L3]

Every half-open box is a Borel set (The sigma-algebra generated by the half-open boxes of $R^{n}$ is the Borel sigma-algebra).

[F1]

A measure on $(X, A)$ is a function $μ : A \to [0, + \infty]$ with $μ (\emptyset) = 0$ that is countably additive on pairwise disjoint sequences (Measures on sigma-algebras); padding a finite disjoint list with empty sets makes it finitely additive.

[F2]

The translate of $E \subseteq R^{n}$ by $a$ is $E + a : = {x + a : x \in E}$ (Translation of a subset of $R^{n}$ ).

[F3]

$\sum_{k < n} λ = n λ$ , where $n$ denotes the canonical natural of $R$ (Laws of finite sums and finite products, claim 2; Finite sums and finite products, by recursion).

[F4]

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 $a^{m}$ ).

[F5]

Let $S \subseteq N$ ; if $0 \in S$ and $σ (n) \in S$ whenever $n \in S$ , then $S = N$ (The principle of mathematical induction).

[F6]

The order on $Z$ is total and compatible with addition (The integers form a totally ordered ring); the canonical embedding of $N$ into $Z$ has as image exactly the nonnegative integers (The naturals embed in the integers, The integers as equivalence classes of pairs of naturals); and $m < n$ in $N$ exactly when $σ (m) \leq n$ (Discreteness: $σ (n)$ is the immediate successor).

Proof

technique · direct

1.1L1L2F4F6

A generation- $k$ dyadic cube is contained in $(0, 1]^{n}$ exactly when $0 \leq m_{i}$ and $m_{i} + 1 \leq 2^{k}$ for every $i < n$ , and every point of $(0, 1]^{n}$ lies in such a cube: if $x \in (0, 1]^{n}$ and $m$ is the index of the generation- $k$ cube containing $x$ , then $m_{i} < 2^{k} x_{i} \leq 2^{k}$ and $m_{i} + 1 \geq 2^{k} x_{i} > 0$ , so $0 \leq m_{i}$ and $m_{i} + 1 \leq 2^{k}$ by discreteness of $Z$ ; conversely such a cube lies in $(0, 1]^{n}$ because $m_{i} 2^{- k} \geq 0$ and $(m_{i} + 1) 2^{- k} \leq 1$ .

1.2L1L3F2

Every generation- $k$ dyadic cube is a translate of $Q_{k, 0} = (0, 2^{- k}]^{n}$ , namely $Q_{k, m} = Q_{k, 0} + m 2^{- k}$ , and it is a half-open box, hence Borel; so all generation- $k$ cubes receive the same value under $μ$ .

2.1F4F5F6

The indices admitted in step 1.1 are exactly the functions from $n$ to the set ${j \in N : j < 2^{k}}$ , and there are $2^{k n}$ of them: by induction on $n$ , at $n = 0$ there is exactly one such function and $2^{0} = 1$ , while each function on $n + 1$ coordinates is a function on $n$ coordinates together with one of $2^{k}$ values in the new coordinate, so the count is multiplied by $2^{k}$ and $(2^{k})^{n} \cdot 2^{k} = (2^{k})^{n + 1} = 2^{k (n + 1)}$ .

3.1step 1.1step 1.2step 2.1L2F1F3

By steps 1.1 and 2.1 the cube $(0, 1]^{n}$ is the union of a list of $2^{k n}$ pairwise disjoint generation- $k$ dyadic cubes, so finite additivity and step 1.2 give $1 = μ ((0, 1]^{n}) = \sum_{r < 2^{k n}} μ (Q_{k, 0})$ ; no term can be $+ \infty$ , since then the sum would be $+ \infty$ rather than $1$ , so the common value is a real and the sum is $2^{k n} μ (Q_{k, 0})$ .

4.1step 1.2step 3.1F4∎

Dividing by the strictly positive real $2^{k n}$ gives $μ (Q_{k, 0}) = 2^{- k n}$ , and step 1.2 transfers the value to every generation- $k$ dyadic cube.

A translation-invariant Borel measure giving the unit cube measure one gives each generation- $k$ dyadic cube measure $2^{- k n}$

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources

A translation-invariant Borel measure giving the unit cube measure one gives each generation-kk dyadic cube measure 2−kn2−kn

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources

A translation-invariant Borel measure giving the unit cube measure one gives each generation- $k$ dyadic cube measure $2^{- k n}$