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A continuous preimage of a Lebesgue measurable subset of $R$ can be nonmeasurable

Statement

Assume the Axiom of Choice. Then there exist a Lebesgue measurable set $E \subseteq R$ , a subset $X \subseteq R$ , and a continuous map $g : X \to R$ such that $g^{- 1} [E]$ is not Lebesgue measurable.

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]

$ψ$ 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 1.1L3L4L5

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

3.1step 1.1step 2.1L5∎

Let $g : = ψ^{- 1} ∣_{K} : K \to R$ . The map $g$ is continuous because $ψ$ is a homeomorphism by [L5]. For every $x \in K$ , one has $g (x) \in E$ exactly when $x \in N$ , because $g$ is the inverse of $ψ$ on $K$ . Therefore $g^{- 1} [E] = N$ , and this preimage is not Lebesgue measurable.

Depends on

Used by

A continuous preimage of a Lebesgue measurable set can be nonmeasurable Counterexample
FALSE: every continuous preimage of a Lebesgue measurable subset of ℝ is Lebesgue measurable False statement

Dependency tree · two levels

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Sources

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

A continuous preimage of a Lebesgue measurable subset of RR can be nonmeasurable