TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck 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.

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

1.1

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.

L2F1F2

2.1

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 $μ_{0} (A + h) = μ_{0} (A)$ .

step 1.1L3L4

3.1

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 $λ_{n}^{*} (E + h) = λ_{n}^{*} (E)$ .

step 2.1L1F1

4.1

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.

step 3.1L5F1∎

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

John K. Hunter, Measure Theory (UC Davis lecture notes), Proposition 2.16 (standard reference, not scraped)
E. A. Carlen, Notes on Lebesgue Measure on $\mathbb{R}^n$ and $S^{n-1}$ (Rutgers Math 501), Theorem 2.3 (standard reference, not scraped)