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A finite-measure measurable set in $R^{n}$ is approximable in measure by a finite union of boxes

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 is a finite union of boxes $B$ such that

$λ_{n} (E △ B) < ε .$

Facts & Assumptions

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

[L1]

Outer regularity gives an open set $U \supseteq E$ with $λ_{n} (U ∖ E) < ε / 2$ (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]

Every open subset of $R^{n}$ is a countable disjoint union of dyadic cubes (Every open subset of $R^{n}$ is the union of a countable pairwise disjoint family of dyadic cubes).

[L3]

Continuity from below applies to increasing unions of measurable sets (Continuity from below for measures).

[L4]

Measure is monotone and $λ_{n} (U ∖ F) = λ_{n} (U) - λ_{n} (F)$ when $F \subseteq U$ and $λ_{n} (U) < \infty$ (Measures are monotone, Measure of a set difference when the smaller set has finite measure).

Proof

technique · direct

1.1L1L4givenchoose

By [L1], choose an open set $U \supseteq E$ with [L1, L4, given, choose] $λ_{n} (U ∖ E) < ε / 2$ . Because $λ_{n} (E) < \infty$ , monotonicity gives $λ_{n} (U) \leq λ_{n} (E) + λ_{n} (U ∖ E) < \infty .$

1.2L2L3L4choose

Write $U = ⨆_{k \geq 1} Q_{k}$ as a countable pairwise disjoint union [L2, L3, L4, choose] of dyadic cubes by [L2], and set $B_{m} : = ⋃_{k = 1}^{m} Q_{k}$ . Then $B_{m} ↑ U$ , so [L3] gives $λ_{n} (B_{m}) \to λ_{n} (U)$ . Hence for some $m$ , $λ_{n} (U ∖ B_{m}) = λ_{n} (U) - λ_{n} (B_{m}) < ε / 2.$

2.1step 1.1step 1.2L4algebra∎

Put $B : = B_{m}$ . Since $E \subseteq U$ and $B \subseteq U$ , [step 1.1, step 1.2, L4, algebra] $E △ B \subseteq (U ∖ E) \cup (U ∖ B),$ so $λ_{n} (E △ B) \leq λ_{n} (U ∖ E) + λ_{n} (U ∖ B) < ε .$ The set $B$ is a finite union of boxes because each dyadic cube is a box.

A finite-measure measurable set in $R^{n}$ is approximable in measure by a finite union of boxes

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources

A finite-measure measurable set in RnRn is approximable in measure by a finite union of boxes

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources

A finite-measure measurable set in $R^{n}$ is approximable in measure by a finite union of boxes