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There is a Lebesgue measurable subset of $R$ that is not Borel

Statement

Assume the Axiom of Choice. Then there exists a Lebesgue measurable subset of $R$ that is not a Borel set.

Facts & Assumptions

Given: The Axiom of Choice.

[L1]

Every subset of $R$ of positive Lebesgue outer measure contains a nonmeasurable subset (Every subset of $R$ of positive Lebesgue outer measure contains a nonmeasurable subset).

[L2]

The image $K = ψ [C]$ of the Cantor set under $ψ (x) = x + c (x)$ is compact and has Lebesgue measure $1$ (The homeomorphism $x \mapsto x + c (x)$ sends the Cantor set onto a compact set of Lebesgue measure $1$ ).

[L3]

The Cantor set is Lebesgue measurable with measure $0$ (The Cantor set is an uncountable subset of $R$ of Lebesgue measure zero).

[L4]

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

[L5]

The Borel sigma-algebra of a subspace is the trace of the ambient Borel sigma-algebra (The Borel sigma-algebra of a subspace is the trace of the ambient Borel sigma-algebra).

[L6]

Continuous preimages of Borel sets are Borel (A continuous map has Borel preimages of Borel sets).

[L7]

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

[L8]

$ψ$ is a homeomorphism from $[0, 1]$ onto $[0, 2]$ (The map $x \mapsto x + c (x)$ is a homeomorphism from $[0, 1]$ onto $[0, 2]$ ).

Proof

technique · direct

1.1L1L2choose

By [L2] the compact set $K$ has positive outer measure, so [L1] supplies a nonmeasurable subset $N \subseteq K$ .

2.1step 1.1L3L4L8

Let $E : = ψ^{- 1} [N] \subseteq C$ . Since $C$ is measurable of measure $0$ by [L3], completeness [L4] makes $E$ Lebesgue measurable.

3.1step 1.1step 2.1L2L5L6L7L8∎

If $E$ were Borel in $R$ , then [L5] would make $E$ Borel as a subset of the subspace $C$ . The inverse homeomorphism $ψ^{- 1} ∣_{K} : K \to C$ is continuous by [L8], so [L6] would make $N = (ψ^{- 1} ∣_{K})^{- 1} [E]$ Borel in the subspace $K$ . Applying [L5] again, there would be a Borel set $B \subseteq R$ with $N = K \cap B$ . Now [L7] makes $B$ Lebesgue measurable, [L2] makes $K$ Lebesgue measurable, and measurable sets are closed under intersection; hence $N$ would be Lebesgue measurable, contradicting step 1.1. Therefore $E$ is not Borel.

Depends on

Used by

A Lebesgue measurable function that is not Borel measurable Example

Dependency tree · two levels

51 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

John K. Hunter, Measure Theory (UC Davis lecture notes), Example 2.22 (standard reference, not scraped)

There is a Lebesgue measurable subset of RR that is not Borel