PropositionStatement: AI-adaptedProof: AI-adaptedprecheck passaudited 2026-08-26

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Lebesgue measure is sigma-finite, and every metrically bounded subset of $R^{n}$ has finite outer measure

Statement

Let $n \geq 1$ and assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Then:

$λ_{n}$ is sigma-finite (Finite, sigma-finite, and semifinite measures): the cubes $(- k, k]^{n}$ are Lebesgue measurable with $λ_{n} ((- k, k]^{n}) = (2 k)^{n} < + \infty$ , they increase with $k$ , and their union over $k \in N$ is $R^{n}$ .
Every bounded subset $E \subseteq R^{n}$ (Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space) has $λ_{n}^{*} (E) < + \infty$ ; a bounded Lebesgue measurable set therefore has finite measure, and every compact subset of $R^{n}$ is Lebesgue measurable of finite measure.

Facts & Assumptions

Given: A natural number $n \geq 1$ , the Axiom of Countable Choice, and Lebesgue measure $λ_{n}$ on $L (R^{n})$ .

[L1]

Assuming countable choice, $L (R^{n})$ is a sigma-algebra, $λ_{n}$ is a complete measure on it, and $λ_{n} (B) = vol (B)$ for every half-open box $B$ (Assuming countable choice, $L (R^{n})$ is a sigma-algebra containing every elementary set and $λ_{n}$ is a complete measure extending elementary volume).

[L2]

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).

[L3]

Assuming countable choice, $λ_{n}^{*}$ is an outer measure on $R^{n}$ , hence monotone (Assuming countable choice, Lebesgue outer measure is an outer measure that restricts to elementary volume, Outer measures).

[L4]

Assuming countable choice, every Borel subset of $R^{n}$ is Lebesgue measurable (Assuming countable choice, every Borel subset of $R^{n}$ is Lebesgue measurable).

[L5]

For a nonempty box $vol (B) : = \prod_{i < n} (b_{i} - a_{i})$ when every $a_{i}$ and every $b_{i}$ is real, and $(u, v]^{n} : = B (u, v)$ (Half-open boxes in $R^{n}$ and their volume, Integer powers $a^{m}$ ).

[F1]

$μ$ is sigma-finite if there is a sequence $(E_{n})_{n \in N}$ in $A$ such that $X = ⋃_{n} E_{n}$ and $μ (E_{n}) < + \infty$ for every $n$ (Finite, sigma-finite, and semifinite measures).

[F2]

$A$ is bounded if $A = \emptyset$ or there are $x_{0} \in X$ and a real $r > 0$ with $A \subseteq B (x_{0}, r)$ , where $B (x_{0}, r) : = {y : d (x_{0}, y) < r}$ (Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space, Open ball, closed ball and sphere in a metric space).

[F3]

For every $x \in R^{n}$ , $∥ x ∥_{\infty} \leq ∥ x ∥_{2}$ , and $∥ x - y ∥_{2} = d_{2} (x, y)$ , $∥ x - y ∥_{\infty} = d_{\infty} (x, y)$ , where $d_{\infty} (x, y) = \max {∣ x_{k} - y_{k} ∣ : k < n}$ (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).

[F4]

A subset $K \subseteq R^{n}$ is compact if and only if $K$ is closed in $R^{n}$ and bounded (Heine-Borel in $R^{n}$ : with the Euclidean metric a subset of $R^{n}$ is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line, claim 2; The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement); and a compact subset of a metric space is closed and bounded (A compact subset of a metric space is closed and bounded).

[F5]

If $A, B \in A$ and $A \subseteq B$ , then $μ (A) \leq μ (B)$ (Measures are monotone).

[F6]

Every complete ordered field $F$ is Archimedean: for every $x \in F$ there is a natural number $n \geq 1$ with $x < n \cdot 1_{F}$ (Every complete ordered field is Archimedean).

Proof

technique · direct

1.1L1L2L5F1F6

Each cube $(- k, k]^{n}$ is a half-open box, hence Lebesgue measurable with $λ_{n} ((- k, k]^{n}) = \prod_{i < n} (k - (- k)) = (2 k)^{n}$ , a real number; the cubes increase with $k$ ; and every $x \in R^{n}$ lies in one of them, because the Archimedean property gives a natural $k \geq 1$ above each of the finitely many reals $∣ x_{i} ∣$ , so their union is $R^{n}$ and $λ_{n}$ is sigma-finite.

1.2L1L3L5F2F3

Let $E$ be bounded and nonempty, say $E \subseteq B (x_{0}, r)$ with $r$ a positive real; every $y \in E$ satisfies $∣ y_{i} - (x_{0})_{i} ∣ \leq d_{2} (x_{0}, y) < r$ in each coordinate, so $E$ is contained in the half-open box with parameter pairs $((x_{0})_{i} - r, (x_{0})_{i} + r]$ , whose volume is $(2 r)^{n}$ ; monotonicity of the outer measure therefore gives $λ_{n}^{*} (E) \leq (2 r)^{n} < + \infty$ , and the empty set has outer measure $0$ .

2.1step 1.2L4F4F5

A bounded Lebesgue measurable set has $λ_{n} (E) = λ_{n}^{*} (E) < + \infty$ by step 1.2; and a compact $K \subseteq R^{n}$ is closed, hence Borel and Lebesgue measurable, and bounded, hence of finite measure.

Lebesgue measure is sigma-finite, and every metrically bounded subset of $R^{n}$ has finite outer measure

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources

Lebesgue measure is sigma-finite, and every metrically bounded subset of RnRn has finite outer measure

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources

Lebesgue measure is sigma-finite, and every metrically bounded subset of $R^{n}$ has finite outer measure