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Assuming the Axiom of Choice, every Lebesgue measurable subset of is a Borel set
Statement
Assume the Axiom of Choice. Every Lebesgue measurable subset of is a Borel set.
Facts & Assumptions
Given: The Axiom of Choice.
Assuming countable choice, the Cantor set is an uncountable subset of of Lebesgue measure zero (The Cantor set is an uncountable subset of of Lebesgue measure zero).
is a bijection from onto the Cantor set (The Cantor set is exactly the set of with every , and this gives a bijection with ).
Assuming choice, for every set , and for every cardinal (Assuming the Axiom of Choice, , and Cantor's theorem in cardinal form: , clauses (a) and (b)).
Assuming countable choice, is a complete measure (Assuming countable choice, is a sigma-algebra containing every elementary set and is a complete measure extending elementary volume).
Every family of nonempty sets has a choice function (The Axiom of Choice).
Refutation
By [F4], the Axiom of Choice gives countable choice, so [L1] applies and the Cantor set has Lebesgue measure zero.
By [F1], the Cantor set is in bijection with , hence with , so its power set has cardinality .
Since [F4] gives countable choice and [L2] says Lebesgue measure is complete under that hypothesis, every subset of the Cantor set is Lebesgue measurable.
Assuming the Axiom of Choice, [F2] gives only Borel subsets of , while [F3] gives ; therefore not every Lebesgue measurable subset of can be Borel.
The refutation is purely cardinal: it produces no particular measurable non-Borel set, only shows that one must exist.
Depends on
- The Cantor set is an uncountable subset of $\mathbb{R}$ of Lebesgue measure zero
- The Cantor set is exactly the set of $\sum_{k \ge 1} a_k 3^{-k}$ with every $a_k \in \{0,2\}$, and this gives a bijection with $\{0,1\}^{\mathbb{N}}$
- Assuming the Axiom of Choice, the Borel sigma-algebra on R^n has cardinality continuum for n at least one
- Assuming the Axiom of Choice, $2^{\kappa} = \lvert \mathcal{P}(\kappa) \rvert$, and Cantor's theorem in cardinal form: $\kappa < 2^{\kappa}$
- 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 Axiom of Choice
- The Borel sigma-algebra of a topological space
Used by
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Sources
- John K. Hunter, Measure Theory (UC Davis lecture notes), Example 2.22 (standard reference, not scraped)