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

Statement

Assume the Axiom of Choice. Then there exist a Lebesgue measurable set $E \subseteq R$ and a continuous map $f : E \to R$ whose image $f [E]$ 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

1.1

By [L2] the set $K$ has positive outer measure, so [L1] supplies a subset $N \subseteq K$ that is not Lebesgue measurable.

L1L2choose

2.1

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

step 1.1L3L4L5

3.1

The restriction $f : = ψ ∣_{E} : E \to R$ is continuous, because $E \subseteq [0, 1]$ and $ψ$ is continuous by [L5]. Its image is $f [E] = N$ , which is not Lebesgue measurable by step 1.1.

step 1.1step 2.1L5∎

Depends on

Used by

FALSE: every continuous image of a Lebesgue measurable subset of ℝ is Lebesgue measurable False statement

Dependency tree · two levels

40 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)

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