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$L (R^{n})$ is exactly the completion of the restriction of $λ_{n}$ to the Borel sets

Statement

Let $n \geq 1$ and assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Write $β_{n}$ for the restriction of $λ_{n}$ to the Borel sigma-algebra $B (R^{n})$ (The Borel sigma-algebra of a topological space). Then $L (R^{n})$ is exactly the completion domain of $(R^{n}, B (R^{n}), β_{n})$ (The completion domain and proposed completed set function of a measure space), and $λ_{n}$ is the completed measure there. Explicitly,

$E \in L (R^{n}) ⟺ E = A \cup N for some A, Z \in B (R^{n}) with N \subseteq Z and β_{n} (Z) = 0,$

and then $λ_{n} (E) = β_{n} (A)$ .

Facts & Assumptions

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

[L1]

Assume the Axiom of Countable Choice and let $μ_{0}$ be a sigma-finite premeasure; the Carathéodory sigma-algebra of its induced outer measure is exactly the completion of $σ (A_{0})$ under the extended measure, and the Carathéodory restriction equals the completed measure there (Assuming countable choice, the Carathéodory domain is the completion of the sigma-finite extension).

[L2]

Elementary volume $μ_{0}$ is a sigma-finite premeasure on the algebra $E_{n}$ of elementary sets (Elementary volume is a sigma-finite premeasure on the algebra of elementary sets, Elementary sets: the finite unions of half-open boxes in $R^{n}$ ).

[L3]

$λ_{n}^{*}$ is the outer set function induced by the premeasure $μ_{0}$ on $E_{n}$ (Lebesgue outer measure on $R^{n}$ ), and $L (R^{n})$ is the family of sets Carathéodory measurable for $λ_{n}^{*}$ , with $λ_{n}$ its restriction (Lebesgue measurable sets, the family $L (R^{n})$ , and the restricted set function $λ_{n}$ ).

[L4]

$σ (E_{n}) = B (R^{n})$ (The sigma-algebra generated by the half-open boxes of $R^{n}$ is the Borel sigma-algebra).

[L5]

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

[F1]

The completion domain of $(X, A, μ)$ is $\overline{A} : = {E \subseteq X : E = A \cup N$ for some $A, Z \in A$ and $N \subseteq Z$ with $μ (Z) = 0}$ , and the completed set function is $\overline{μ} (E) : = μ (A)$ (The completion domain and proposed completed set function of a measure space).

[F2]

Assume the Axiom of Countable Choice; then $\overline{μ}$ is a complete measure on $\overline{A}$ extending $μ$ , and it is the unique complete measure on $\overline{A}$ that extends $μ$ (Assuming countable choice, every measure space has a unique complete extension to its completion).

[F3]

The Axiom of Countable Choice says that for every family $(X_{n})_{n \in N}$ of nonempty sets indexed by $N$ there is a function $f$ with domain $N$ such that $f (n) \in X_{n}$ for every $n \in N$ (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

Proof

technique · direct

1.1

Elementary volume is a sigma-finite premeasure on the algebra $E_{n}$ , and $λ_{n}^{*}$ is exactly the outer set function it induces, so the hypotheses of the Carathéodory-domain theorem are met with $A_{0} : = E_{n}$ .

L2L3F3

1.2

The sigma-algebra generated by $E_{n}$ is $B (R^{n})$ , and the measure that the extension theorem places on it is the restriction $β_{n}$ of $λ_{n}$ , since every Borel set is Lebesgue measurable and $λ_{n}$ is the restriction of $λ_{n}^{*}$ .

L3L4L5

2.1

The Carathéodory-domain theorem therefore says that $L (R^{n})$ , the Carathéodory sigma-algebra of $λ_{n}^{*}$ , is the completion domain of $β_{n}$ on $B (R^{n})$ and that $λ_{n}$ agrees there with the completed measure; unwinding the published description of that domain gives the displayed equivalence and the value $λ_{n} (E) = β_{n} (A)$ , which is well posed because the completed measure is a measure extending $β_{n}$ and is the unique complete one.

step 1.1step 1.2L1F1F2∎

Depends on

Used by

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Sources

John K. Hunter, Measure Theory (UC Davis lecture notes), Theorem 2.28 (standard reference, not scraped)
E. A. Carlen, Notes on Lebesgue Measure on $\mathbb{R}^n$ and $S^{n-1}$ (Rutgers Math 501), Section 1 (standard reference, not scraped)

L(Rn)L(Rn) is exactly the completion of the restriction of λnλn​ to the Borel sets