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A continuous image of a Lebesgue measurable subset of can be nonmeasurable
Statement
Assume the Axiom of Choice. Then there exist a Lebesgue measurable set and a continuous map whose image is not Lebesgue measurable.
Facts & Assumptions
Given: The Axiom of Choice.
Every subset of of positive Lebesgue outer measure contains a nonmeasurable subset (Every subset of of positive Lebesgue outer measure contains a nonmeasurable subset).
The image of the Cantor set under is compact and has Lebesgue measure (The homeomorphism sends the Cantor set onto a compact set of Lebesgue measure ).
The Cantor set is Lebesgue measurable with measure (The Cantor set is an uncountable subset of of Lebesgue measure zero).
Assuming countable choice, Lebesgue measure is complete (Assuming countable choice, is a sigma-algebra containing every elementary set and is a complete measure extending elementary volume).
is a homeomorphism from onto (The map is a homeomorphism from onto ).
Proof
By [L2] the set has positive outer measure, so [L1] supplies a subset that is not Lebesgue measurable.
Let . Since is measurable and has measure by [L3], completeness from [L4] makes every subset of , and in particular , Lebesgue measurable.
The restriction is continuous, because and is continuous by [L5]. Its image is , which is not Lebesgue measurable by step 1.1.
Depends on
- Every subset of $\mathbb{R}$ of positive Lebesgue outer measure contains a nonmeasurable subset
- The homeomorphism $x \mapsto x + c(x)$ sends the Cantor set onto a compact set of Lebesgue measure $1$
- The Cantor set is an uncountable subset of $\mathbb{R}$ of Lebesgue measure zero
- Assuming countable choice, $\mathcal{L}(\mathbb{R}^n)$ is a sigma-algebra containing every elementary set and $\lambda_n$ is a complete measure extending elementary volume
- The map $x \mapsto x + c(x)$ is a homeomorphism from $[0,1]$ onto $[0,2]$
- The Axiom of Countable Choice ($\mathrm{AC}_\omega$)
Used by
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Sources
- John K. Hunter, Measure Theory (UC Davis lecture notes), Example 2.22 (standard reference, not scraped)