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Assuming Choice, a proper subgroup of can be nonmeasurable
Statement refuted
Refuted claim: every proper subgroup of is Lebesgue measurable.
Facts & Assumptions
Given: The Axiom of Choice.
Assuming the Axiom of Choice, has a Hamel basis over , and each basis vector carries a well-defined -linear coefficient map (Assuming the Axiom of Choice, has a Hamel basis over : there is such that every real is a finite -linear combination of elements of in exactly one way, and each basis vector carries a well-defined -linear coefficient map).
A Lebesgue measurable subgroup of of positive measure is all of (A Lebesgue measurable subgroup of of positive measure is all of ).
The interval is Lebesgue measurable with measure (A box in with parameters is Lebesgue measurable of measure , whichever of its faces are included).
is countably infinite ( is countably infinite).
Lebesgue measurability and Lebesgue measure are invariant under translation (Lebesgue outer measure, Lebesgue measurability and Lebesgue measure are unchanged by translation).
A countable union of measurable null sets is null (Finite and countable subadditivity of measures).
Counterexample
By [L1] choose a Hamel basis , a basis vector , and the corresponding coefficient map . Its kernel is a subgroup of , and it is proper because . Every real is of the form with and .
If were measurable with positive measure, [L2] would force , contradicting step 1.1.
If were measurable with measure , then every translate would also be measurable with measure by [L5], and step 1.1 says these countably many translates cover . Their union would be null by [L4] and [L6], yet it contains the measurable interval of measure by [L3], a contradiction.
So the proper subgroup is not Lebesgue measurable, and it refutes the claim.
Depends on
- Assuming the Axiom of Choice, $\mathbb{R}$ has a Hamel basis over $\mathbb{Q}$: there is $B \subseteq \mathbb{R}$ such that every real is a finite $\mathbb{Q}$-linear combination of elements of $B$ in exactly one way, and each basis vector carries a well-defined $\mathbb{Q}$-linear coefficient map
- A Lebesgue measurable subgroup of $(\mathbb{R}^n,+)$ of positive measure is all of $\mathbb{R}^n$
- A box in $\mathbb{R}^n$ with parameters $a_i\le b_i$ is Lebesgue measurable of measure $\prod_{i<n}(b_i-a_i)$, whichever of its faces are included
- $\mathbb{Q}$ is countably infinite
- Lebesgue outer measure, Lebesgue measurability and Lebesgue measure are unchanged by translation
- Finite and countable subadditivity of measures
Used by
Nothing in the library uses this result yet.
Dependency tree · two levels
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Sources
- Hamel basis (Wikipedia) (standard reference, not scraped)
- Non-measurable subgroup (Wikipedia) (standard reference, not scraped)