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If a Lebesgue measurable subset of $R^{n}$ has positive measure, its difference set contains an open ball about the origin

Statement

Let $n \geq 1$ and assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $E \subseteq R^{n}$ be Lebesgue measurable with $λ_{n} (E) > 0$ , and put

$E - E : = {x - y : x, y \in E} .$

Then there is a real $r > 0$ with $B (0, r) \subseteq E - E$ , the open Euclidean ball of centre the origin and radius $r$ (Open ball, closed ball and sphere in a metric space, $R^{n}$ as the set of functions $n \to R$ , and $d_{1}$ , $d_{2}$ , $d_{\infty}$ are metrics on it).

Facts & Assumptions

Given: A natural number $n \geq 1$ , the Axiom of Countable Choice, and a Lebesgue measurable set $E \subseteq R^{n}$ with $λ_{n} (E) > 0$ .

[L1]

Assuming countable choice, a Lebesgue measurable $F$ with $0 < λ_{n} (F) < + \infty$ and a real $θ$ with $0 < θ < 1$ admit a dyadic cube $Q$ with $λ_{n} (F \cap Q) > θ λ_{n} (Q)$ (A measurable set of positive finite measure occupies more than any prescribed proportion of some dyadic cube, Dyadic cubes of generation $k$ in $R^{n}$ ).

[L2]

$λ_{n} (S + h) = λ_{n} (S)$ for every Lebesgue measurable $S$ and every $h$ , and $S + h$ is measurable exactly when $S$ is (Lebesgue outer measure, Lebesgue measurability and Lebesgue measure are unchanged by translation, Translation of a subset of $R^{n}$ ).

[L3]

Every set $R$ with $R^{\circ} \subseteq R \subseteq \overline{R}$ is Lebesgue measurable with $λ_{n} (R) = \prod_{i < n} (b_{i} - a_{i})$ (A box in $R^{n}$ with parameters $a_{i} \leq b_{i}$ is Lebesgue measurable of measure $\prod_{i < n} (b_{i} - a_{i})$ , whichever of its faces are included), and $vol (Q_{k, m}) = 2^{- k n}$ (For each generation, the dyadic cubes of that generation are pairwise disjoint and cover $R^{n}$ , Half-open boxes in $R^{n}$ and their volume).

[L4]

Assuming countable choice, $λ_{n}$ is a complete measure on $L (R^{n})$ , a sigma-algebra (Assuming countable choice, $L (R^{n})$ is a sigma-algebra containing every elementary set and $λ_{n}$ is a complete measure extending elementary volume), and every bounded Lebesgue measurable set has finite measure (Lebesgue measure is sigma-finite, and every metrically bounded subset of $R^{n}$ has finite outer measure).

[F1]

Let $(E_{k})_{k \in N}$ be an increasing sequence of measurable sets for a measure $μ$ ; then $μ (⋃_{k \in N} E_{k}) = \sup_{k \in N} μ (E_{k})$ (Continuity from below for measures).

[F2]

A measure is countably additive on pairwise disjoint measurable sequences, hence finitely additive (Measures on sigma-algebras), and monotone (Measures are monotone).

[F3]

For $a > 0$ and rational $r = m / q$ with $q \geq 1$ , $a^{r} : = (a^{1 / q})^{m}$ , where $a^{1 / q}$ is the unique nonnegative $q$ -th root of $a$ (Rational powers $a^{r}$ of a positive base, Existence and uniqueness of $n$ -th roots: a unique $a^{1 / n} \geq 0$ with $(a^{1 / n})^{n} = a$ ), and the value does not depend on the representative (Rational powers do not depend on the representative).

[F4]

If $0 \leq a < b$ and $n \geq 1$ then $a^{n} < b^{n}$ ; if $0 \leq a \leq 1$ then $a^{n} \leq 1$ (Monotonicity of $x \mapsto x^{n}$ and of $n \mapsto a^{n}$ , claims 2 and 3; Integer powers $a^{m}$ ), and $(a b)^{n} = a^{n} b^{n}$ (Laws of integer exponents, claim 1).

[F5]

For every $x \in R^{n}$ , $∥ x ∥_{\infty} \leq ∥ x ∥_{2}$ , and $∥ x - y ∥_{2} = d_{2} (x, y)$ , $∥ x - y ∥_{\infty} = d_{\infty} (x, y) = \max {∣ x_{k} - y_{k} ∣ : k < n}$ (The finite and reverse triangle inequalities for a norm; and for $n \geq 1$ every norm $N$ on $R^{n}$ satisfies $N (x) \leq C ∥ x ∥_{1}$ and is Lipschitz, hence continuous, for $d_{2}$ , claim 3; Each $∥ \cdot ∥_{p}$ is a norm on $R^{n}$ , and the induced metrics are exactly $d_{1}$ , $d_{2}$ and $d_{\infty}$ of the published metric-spaces page, claim 3; The $p$ -norms $∥ x ∥_{p}$ for rational $p \geq 1$ , and $∥ x ∥_{\infty}$ ; $R^{n}$ as the set of functions $n \to R$ , and $d_{1}$ , $d_{2}$ , $d_{\infty}$ are metrics on it).

Proof

technique · direct

1.1

The sets $E \cap (- k, k]^{n}$ for $k \in N$ are Lebesgue measurable, increase with $k$ and have union $E$ , so continuity from below gives $\sup_{k} λ_{n} (E \cap (- k, k]^{n}) = λ_{n} (E) > 0$ and some $k$ has $λ_{n} (E \cap (- k, k]^{n}) > 0$ ; that set is bounded, hence of finite measure. Replacing $E$ by it shrinks $E - E$ , so it suffices to prove the theorem when $0 < λ_{n} (E) < + \infty$ .

L3L4F1

1.2

Put $t : = (3 / 2)^{1 / n}$ , the unique nonnegative $n$ -th root of $3 / 2$ ; then $t > 1$ , since $t \leq 1$ would give $t^{n} \leq 1 < 3 / 2$ , and $η : = (t - 1) / 2$ is a strictly positive real with $1 + 2 η = t$ and $(1 + 2 η)^{n} = 3 / 2$ .

F3F4

2.1

Assume $0 < λ_{n} (E) < + \infty$ and apply the density lemma with $θ : = 3 / 4$ : there is a dyadic cube $Q$ , of some generation $k$ and side $s : = 2^{- k}$ , with $λ_{n} (E \cap Q) > \frac{3}{4} s^{n}$ , since $λ_{n} (Q) = s^{n}$ .

step 1.1L1L3

3.1

Let $h \in R^{n}$ with $d_{2} (0, h) < η s$ , so that $∣ h_{i} ∣ < η s$ in every coordinate. Writing $Q = B (a, b)$ with $b_{i} - a_{i} = s$ , both $E \cap Q$ and $(E \cap Q) + h$ are contained in the half-open box $P$ with parameter pairs $(a_{i} - η s, b_{i} + η s]$ , whose measure is $(s (1 + 2 η))^{n} = s^{n} t^{n} = \frac{3}{2} s^{n}$ .

step 1.2step 2.1L2L3F4F5

4.1

The two sets are Lebesgue measurable with the same measure, by translation invariance, so if they were disjoint then additivity and monotonicity inside $P$ would give $\frac{3}{2} s^{n} = λ_{n} (P) \geq 2 λ_{n} (E \cap Q) > 2 \cdot \frac{3}{4} s^{n} = \frac{3}{2} s^{n}$ , which is impossible; hence they meet, and a common point $z = w + h$ with $z, w \in E \cap Q$ exhibits $h = z - w \in E - E$ .

step 2.1step 3.1L2L4F2

5.1

Therefore $B (0, η s) \subseteq E - E$ , and $r : = η s$ is a strictly positive real.

step 1.2step 4.1F5∎

Depends on

Used by

A Lebesgue measurable subgroup of (ℝⁿ,+) of positive measure is all of ℝⁿ Corollary

Dependency tree · two levels

126 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 1.6.8 (standard reference, not scraped)
J. Ye, L. Yu, X. Zhao, When is $A+xA=\mathbb{R}$?, Theorem 1.1 (standard reference, not scraped)

If a Lebesgue measurable subset of RnRn has positive measure, its difference set contains an open ball about the origin