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For a nonzero real $c$ , dilation by $c$ multiplies Lebesgue outer measure by $∣ c ∣^{n}$ , and reflection in the origin preserves it

Statement

Let $n \geq 1$ , assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )), let $c$ be a nonzero real and write $c E : = {c x : x \in E}$ for $E \subseteq R^{n}$ , where $(c x)_{i} : = c x_{i}$ . Then:

$λ_{n}^{*} (c E) = ∣ c ∣^{n} λ_{n}^{*} (E)$ for every subset $E$ , the product being defined in $\overline{R}$ because $∣ c ∣^{n} > 0$ ;
$E$ is Lebesgue measurable if and only if $c E$ is;
$λ_{n} (c E) = ∣ c ∣^{n} λ_{n} (E)$ for every Lebesgue measurable $E$ .

At $c = - 1$ the map is reflection in the origin and $∣ c ∣^{n} = 1$ , so it preserves outer measure, measurability and measure. The value $c = 0$ is excluded because $0 E$ is ${0}$ or $\emptyset$ and carries no information about $E$ .

Facts & Assumptions

Given: A natural number $n \geq 1$ , the Axiom of Countable Choice, a nonzero real $c$ , and a subset $E \subseteq R^{n}$ .

[L1]

Assuming countable choice, $λ^{c l} (E) = λ_{n}^{*} (E)$ , the infimum of $\sum_{k = 0}^{\infty} vol [u^{k}, v^{k}]$ over countable covers of $E$ by closed rectangles (Countable covers by closed boxes, by open boxes and by closed cubes all compute Lebesgue outer measure, Lebesgue outer measure on $R^{n}$ ).

[L2]

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, Assuming countable choice, $L (R^{n})$ is a sigma-algebra containing every elementary set and $λ_{n}$ is a complete measure extending elementary volume).

[F1]

$[a, b] : = {x \in R^{m} : a_{j} \leq x_{j} \leq b_{j} (j < m)}$ and $vol [a, b] : = \prod_{j < m} (b_{j} - a_{j})$ (Axis-parallel rectangles in $R^{m}$ and their volume).

[F2]

$\prod_{k < n} (a_{k} b_{k}) = (\prod_{k < n} a_{k}) (\prod_{k < n} b_{k})$ , 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).

[F3]

The defining recursion for natural powers is $a^{0} = 1$ and $a^{n + 1} = a^{n} \cdot a$ (Integer powers $a^{m}$ ), and $(a b)^{n} = a^{n} b^{n}$ (Laws of integer exponents, claim 1).

[F4]

$a b : = + \infty$ when one of $a, b$ is $\pm \infty$ , the other is $\neq 0$ , and both are $> 0$ or both are $< 0$ ; every product with one factor $0$ and the other $\pm \infty$ is left undefined (The extended real line $\overline{R} = R \cup {- \infty, + \infty}$ , its order, and the arithmetic that is left undefined).

[F5]

The absolute value satisfies $∣ c ∣ > 0$ for $c \neq 0$ and $∣ c d ∣ = ∣ c ∣ ∣ d ∣$ (Absolute value in an ordered field, Basic properties of the absolute value).

[F6]

For positive reals, multiplication preserves order and reciprocals stay positive: if $0 < q$ and $u \leq v$ then $q u \leq q v$ , and if $0 < q$ then $0 < q^{- 1}$ (Sign rules for products and monotonicity of multiplication, Inverses of positives are positive, and reciprocation reverses order, Ordered field).

Proof

technique · direct

1.1

For reals $u_{i} \leq v_{i}$ one has $c [u, v] = [c u, c v]$ when $c > 0$ and $c [u, v] = [c v, c u]$ when $c < 0$ , in both cases a closed rectangle whose $i$ -th side length is $∣ c ∣ (v_{i} - u_{i})$ ; its volume is therefore $\prod_{i < n} (∣ c ∣ (v_{i} - u_{i})) = (\prod_{i < n} ∣ c ∣) \prod_{i < n} (v_{i} - u_{i}) = ∣ c ∣^{n} vol [u, v]$ .

F1F2F3F5

1.2

Put $q : = ∣ c ∣^{n} > 0$ . If $s_{0} = \inf S$ for a nonempty $S \subseteq [0, + \infty]$ , then $q s_{0}$ is a lower bound of $q S : = {q s : s \in S}$ : for every real $s \in S$ the inequality $s_{0} \leq s$ gives $q s_{0} \leq q s$ by [F6], while the claim is automatic when $s = + \infty$ . Conversely, let $t$ be a lower bound of $q S$ . If $t = + \infty$ , then every element of $q S$ is $+ \infty$ , hence every element of $S$ is $+ \infty$ and therefore $s_{0} = + \infty$ . If $t$ is real, then $q^{- 1} > 0$ by [F6], so $t \leq q s$ implies $q^{- 1} t \leq s$ for every real $s \in S$ , and again the claim is automatic when $s = + \infty$ ; thus $q^{- 1} t$ is a lower bound of $S$ , so $q^{- 1} t \leq s_{0}$ and therefore $t \leq q s_{0}$ . Hence $\inf (q S) = q \inf S$ .

F3F4F5F6

2.1

The assignment $[u, v] \mapsto c [u, v]$ is a bijection from the countable closed-rectangle covers of $E$ onto those of $c E$ , with inverse given by multiplication by $c^{- 1}$ . For one such cover, let $(a_{k})$ be its sequence of rectangle volumes and $(s_{n})$ the partial sums of $\sum_{k = 0}^{\infty} a_{k}$ in the sense of Series in the nonnegative extended real line; let $(t_{n})$ be the partial sums of the transformed cover cost. By step 1.1 each transformed term is $q a_{k}$ , and the shared recursion of nonnegative extended series gives $t_{n} = q s_{n}$ for every $n$ . Therefore the transformed cover cost is $q \sum_{k = 0}^{\infty} a_{k}$ by step 1.2. So step 1.2 turns the infimum of all transformed cover costs into $λ^{c l} (c E) = ∣ c ∣^{n} λ^{c l} (E)$ , and [L1] then gives the same identity for $λ_{n}^{*}$ .

step 1.1step 1.2L1F6

3.1

For a test set $A$ one has $A \cap c E = c ((c^{- 1} A) \cap E)$ and $A ∖ c E = c ((c^{- 1} A) ∖ E)$ , so step 2.1 turns the Carathéodory identity for $c E$ tested against $A$ into $∣ c ∣^{n}$ times the identity for $E$ tested against $c^{- 1} A$ ; multiplication by the positive real $∣ c ∣^{n}$ is injective on $[0, + \infty]$ , and $A \mapsto c^{- 1} A$ is a bijection of the power set, so $c E$ is Lebesgue measurable exactly when $E$ is, and then $λ_{n} (c E) = λ_{n}^{*} (c E) = ∣ c ∣^{n} λ_{n}^{*} (E) = ∣ c ∣^{n} λ_{n} (E)$ .

step 2.1L2F4F5∎

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Sources

E. A. Carlen, Notes on Lebesgue Measure on $\mathbb{R}^n$ and $S^{n-1}$ (Rutgers Math 501), Theorem 3.4 (standard reference, not scraped)
John K. Hunter, Measure Theory (UC Davis lecture notes), Chapter 2 (standard reference, not scraped)

For a nonzero real cc, dilation by cc multiplies Lebesgue outer measure by ∣c∣n∣c∣n, and reflection in the origin preserves it