TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26

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Lebesgue outer measure, Lebesgue measurability and Lebesgue measure are unchanged by translation

Statement

Let $n \geq 1$ , let $h \in R^{n}$ , and let $E + h$ be the translate of $E \subseteq R^{n}$ (Translation of a subset of $R^{n}$ ). Then:

$λ_{n}^{*} (E + h) = λ_{n}^{*} (E)$ for every subset $E$ ;
$E$ is Lebesgue measurable if and only if $E + h$ is;
$λ_{n} (E + h) = λ_{n} (E)$ for every Lebesgue measurable $E$ .

No choice principle is used. Lebesgue outer measure is defined as an infimum and the Carathéodory condition is a family of equations between its values, so all three clauses are statements about objects that exist in ZF; countable choice is needed to know that $λ_{n}$ is a measure, not to know that it is translation invariant.

Statement references

Translation of a subset of $R^{n}$

Facts & Assumptions

Given: A natural number $n \geq 1$ , a vector $h \in R^{n}$ , and a subset $E \subseteq R^{n}$ .

[L1]

$λ_{n}^{*} (E) : = \inf {\sum_{k = 0}^{\infty} μ_{0} (A_{k}) : A_{k} \in E_{n}$ for every $k$ and $E \subseteq ⋃_{k} A_{k}}$ (Lebesgue outer measure on $R^{n}$ , Series in the nonnegative extended real line).

[L2]

$B (a, b) : = {x \in R^{n} : a_{i} < x_{i} \leq b_{i} for every i < n}$ ; a box is nonempty exactly when $a_{i} < b_{i}$ for every $i < n$ ; and for a nonempty box with real parameters $vol (B) : = \prod_{i < n} (b_{i} - a_{i})$ , the value being $+ \infty$ when a parameter is infinite (Half-open boxes in $R^{n}$ and their volume).

[L3]

A subset $E \subseteq R^{n}$ is an elementary set when there are a natural number $m$ and a list $B_{0}, \dots, B_{m - 1}$ of half-open boxes with $E = ⋃_{j < m} B_{j}$ (Elementary sets: the finite unions of half-open boxes in $R^{n}$ ), and every elementary set is the union of a finite list of pairwise disjoint half-open boxes (Every elementary set is a finite disjoint union of half-open boxes, and any finitely many boxes admit a common grid refinement).

[L4]

For every $n \geq 1$ , elementary volume $μ_{0}$ on $E_{n}$ has value at $A$ the sum of the volumes of the members of any presentation of $A$ by a finite list of pairwise disjoint half-open boxes (The sum of the volumes of a disjoint box decomposition of an elementary set does not depend on the decomposition).

[L5]

A set $E$ is Lebesgue measurable when $λ_{n}^{*} (A) = λ_{n}^{*} (A \cap E) + λ_{n}^{*} (A ∖ E)$ for every $A \subseteq R^{n}$ , and $λ_{n}$ is the restriction of $λ_{n}^{*}$ to the family of these (Lebesgue measurable sets, the family $L (R^{n})$ , and the restricted set function $λ_{n}$ , Carathéodory measurable sets).

[F1]

The translate of $E \subseteq R^{n}$ by $a$ is $E + a : = {x + a : x \in E}$ ; translation by $a$ is the bijection $τ_{a} (x) = x + a$ , whose inverse is $τ_{- a}$ (Translation of a subset of $R^{n}$ ).

[F2]

Addition of a real to an extended real is defined in every case, with $a + b : = + \infty$ when $a = + \infty$ and $b \neq - \infty$ , and $a + b : = - \infty$ when $a = - \infty$ and $b \neq + \infty$ (The extended real line $\overline{R} = R \cup {- \infty, + \infty}$ , its order, and the arithmetic that is left undefined).

Proof

technique · direct

1.1L2F1F2

For a parameter pair $(a, b)$ one has $B (a, b) + h = B (a + h, b + h)$ , where $a + h$ is the parameter $i \mapsto a_{i} + h_{i}$ : a point $y$ lies in the left side exactly when $y - h$ satisfies $a_{i} < y_{i} - h_{i} \leq b_{i}$ , that is $a_{i} + h_{i} < y_{i} \leq b_{i} + h_{i}$ . The translated box is empty exactly when the original is, and has the same volume, because $(b_{i} + h_{i}) - (a_{i} + h_{i}) = b_{i} - a_{i}$ when both are real and an infinite parameter stays infinite.

2.1step 1.1L3L4

Consequently, if $A = ⋃_{j < q} B_{j}$ is a presentation of an elementary set by pairwise disjoint half-open boxes, then $A + h = ⋃_{j < q} (B_{j} + h)$ is such a presentation of $A + h$ , so $A + h$ is elementary and $μ_{0} (A + h) = μ_{0} (A)$ .

3.1step 2.1L1F1

A sequence $(A_{k})$ of elementary sets covers $E$ if and only if the sequence $(A_{k} + h)$ covers $E + h$ , and the two covering costs are equal by step 2.1; the correspondence is a bijection between the two families of covers, with inverse given by translating by $- h$ , so the two infima agree and $λ_{n}^{*} (E + h) = λ_{n}^{*} (E)$ .

4.1step 3.1L5F1∎

For test sets, $A \cap (E + h) = ((A - h) \cap E) + h$ and $A ∖ (E + h) = ((A - h) ∖ E) + h$ , so by step 3.1 the Carathéodory identity for $E + h$ tested against $A$ is exactly the identity for $E$ tested against $A - h$ ; as $A$ ranges over all subsets so does $A - h$ , and therefore $E + h$ is Lebesgue measurable if and only if $E$ is, with $λ_{n} (E + h) = λ_{n}^{*} (E + h) = λ_{n}^{*} (E) = λ_{n} (E)$ in that case.

Lebesgue outer measure, Lebesgue measurability and Lebesgue measure are unchanged by translation

Statement

Statement references

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources