LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-09-01

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A finite-measure measurable set in $R^{n}$ has a compact core and a bounded open neighbourhood of arbitrarily small excess

Statement

Assume the Axiom of Countable Choice.

Let $E \subseteq R^{n}$ be Lebesgue measurable with $λ_{n} (E) < \infty$ . For every $ε > 0$ there exist a compact set $K$ and a bounded open set $O$ such that

$K \subseteq E, K \subseteq O, λ_{n} (E ∖ K) < ε, and λ_{n} (O ∖ K) < ε .$

Facts & Assumptions

Given: The Axiom of Countable Choice, $n \geq 1$ , a finite-measure measurable set $E \subseteq R^{n}$ , and $ε > 0$ .

[L1]

Outer regularity gives open supersets of arbitrarily small excess (Assuming countable choice, the Lebesgue outer measure of an arbitrary subset of $R^{n}$ is the infimum of the measures of the open sets containing it).

[L2]

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

[L3]

Continuity from below applies to the exhaustion $E \cap \overline{B (0, R)} ↑ E$ (Continuity from below for measures).

[L4]

Set-difference measure obeys the usual subtraction and monotonicity rules (Measure of a set difference when the smaller set has finite measure, 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).

Proof

technique · direct

1.1L2L3givenchoose

By [L3], choose $R > 0$ so large that [L2, L3, given, choose] $λ_{n} (E ∖ \overline{B (0, R)}) < ε / 3.$ Put $E_{R} : = E \cap \overline{B (0, R)}$ . Then $E_{R}$ has finite measure and lies in a compact ball.

2.1L1L4step 1.1chooseconstruct

Apply [L1] to choose an open set $U \supseteq E_{R}$ with [L1, L4, step 1.1, choose, construct] $λ_{n} (U ∖ E_{R}) < ε / 3$ . Put $O : = U \cap B (0, R + 1) .$ Then $O$ is bounded and open, and it still contains $E_{R}$ because $E_{R} \subseteq \overline{B (0, R)} \subseteq B (0, R + 1)$ . Next choose an open set $V \supseteq \overline{B (0, R)} ∖ E$ with $λ_{n} (V ∖ (\overline{B (0, R)} ∖ E)) < ε / 3.$ Define $K : = \overline{B (0, R)} ∖ V .$ Then $K \subseteq E$ and $K$ is compact, being closed in the compact ball $\overline{B (0, R)}$ .

3.1step 1.1step 2.1L4algebra

Because $K = \overline{B (0, R)} ∖ V$ , one has [step 1.1, step 2.1, L4, algebra] $E ∖ K \subseteq (E ∖ \overline{B (0, R)}) \cup (E_{R} \cap V) .$ Also $E_{R} \cap V \subseteq V ∖ (\overline{B (0, R)} ∖ E),$ so $λ_{n} (E ∖ K) < ε / 3 + ε / 3 < ε .$

4.1step 2.1L4algebra∎

Since $K \subseteq E_{R} \subseteq O$ , one also has [step 2.1, L4, algebra] $O ∖ K \subseteq (O ∖ E_{R}) \cup (E_{R} ∖ K) .$ But $E_{R} ∖ K = E_{R} \cap V \subseteq V ∖ (\overline{B (0, R)} ∖ E)$ , so $λ_{n} (O ∖ K) < ε / 3 + ε / 3 < ε .$ Thus $K \subseteq E$ , $K \subseteq O$ , and both required excess bounds hold.

Depends on

Used by

C_c(ℝⁿ) is dense in Lᵖ(ℝⁿ) for 1 ≤ p < ∞ Theorem

Dependency tree · two levels

56 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

Walter Rudin, Real and Complex Analysis, 3rd ed. (standard reference, not scraped)

A finite-measure measurable set in RnRn has a compact core and a bounded open neighbourhood of arbitrarily small excess