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A C^1 diffeomorphism maps Lebesgue measurable sets to Lebesgue measurable sets

Statement

Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $U, V \subseteq R^{n}$ be open and let $T : U \to V$ be a $C^{1}$ diffeomorphism. If $E \subseteq U$ is Lebesgue measurable, then $T (E)$ is Lebesgue measurable.

Facts & Assumptions

Given: The Axiom of Countable Choice, open sets $U, V \subseteq R^{n}$ , a $C^{1}$ diffeomorphism $T : U \to V$ , and a Lebesgue measurable set $E \subseteq U$ .

[L1]

A $C^{1}$ diffeomorphism maps null sets to null sets. (A C^1 diffeomorphism maps Lebesgue null sets to Lebesgue null sets)

[L2]

Assuming countable choice, every Lebesgue measurable set is a Borel set up to a null modification. ( $L (R^{n})$ is exactly the completion of the restriction of $λ_{n}$ to the Borel sets)

[L3]

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

Proof

technique · direct

1.1L1L2L3

By [L2], write $E = B \cup N$ with $B$ Borel and $N$ contained in a null set. Since $T^{- 1}$ is continuous, [L3] implies that $T (B) = (T^{- 1})^{- 1} (B)$ is Borel. Also [L1] makes $T (N)$ null.

2.1step 1.1L2∎

Therefore $T (E) = T (B) \cup T (N)$ is a Borel set union a null set, hence Lebesgue measurable by [L2].

Depends on

Used by

Dependency tree · two levels

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Sources

Gerald B. Folland, Real Analysis, 2nd ed., Theorem 2.47 (standard reference, not scraped)