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A Hamel coefficient has dense graph and a nonmeasurable kernel
Statement
Assume AC. Fix a Hamel basis B of over and . Its coefficient map is additive, has dense graph, is unbounded above and below on every nondegenerate interval, and is continuous nowhere. Its kernel W is not Lebesgue measurable, so f is not Lebesgue measurable. No claim that every Hamel basis itself is nonmeasurable is made.
Facts & Assumptions
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 gives B and its unique additive rational-linear coefficient map, f(b)=1, range , and nonzero kernel vector.
The rationals embed densely in the reals gives rational density; is countably infinite gives an enumeration of .
Every complete ordered field is Archimedean gives natural numbers exceeding any prescribed real bound.
Lebesgue outer measure, Lebesgue measurability and Lebesgue measure are unchanged by translation preserves measurability and measure under translations.
Assuming countable choice, is a sigma-algebra containing every elementary set and is a complete measure extending elementary volume gives the Lebesgue sigma-algebra and measure under countable choice; Measures on sigma-algebras specifies countable additivity.
A box in with parameters is Lebesgue measurable of measure , whichever of its faces are included gives the lengths of bounded intervals under countable choice.
Borel measurable and Lebesgue measurable functions on requires Borel preimages to be Lebesgue measurable; The Borel sigma-algebra of a topological space contains closed singletons.
Continuity of at a point of and on : the - condition, its agreement with at a limit point, and continuity at an isolated point specifies epsilon-delta continuity.
Assume The Axiom of Choice, including countable choice for F5–F7.
Proof
Given: The basis vector and coefficient map as in the statement, with the real codomain convention.
F1 applies under A1. It gives additivity and rational linearity, , and . In particular , , and : subtract qb and apply additivity in either direction. For any u<v choose by F2 a rational r strictly between and . Then , with the order reversed when w<0. Thus W is dense, and translation shows every fiber is dense.
For any open rectangle , F2 gives rational q in (c,d); step 1.1 gives x in , so lies in the rectangle. Such rectangles form a basis, proving graph density. Taking (c,d) wholly above any prescribed M or wholly below -M shows both unboundedness assertions on each nondegenerate interval, whose interior contains an open interval. For any x_0 and , density gives x with and . This violates F8 with , so f is continuous at no x_0.
Suppose W measurable and put , m a positive integer. F5 and F6, licensed by A1, make these measurable with finite measure . Distinct rational translates are disjoint: an equality gives q=r after applying f and step 1.1. F2 enumerates the infinitely many rationals q with ; infinitude follows from density in , since any finite list can be avoided in a smaller subinterval. The sets along this enumeration are pairwise disjoint and all lie in . By F4 each has measure a_m. If , choose by F3 a natural N with . Finite additivity F5 and the enclosing interval value F6 then give , contradiction. Hence every a_m is zero.
By F3, . Disjointizing this sequence and using F5's countable additivity shows W null. All cosets are null by F4, and their countable union is by F1's rational coefficient decomposition and F2's enumeration. Disjointization and F5 again make null, contradicting F6's value one on [0,1] and monotonicity. Thus W is not measurable. Finally {0} is closed and Borel, so if f were Lebesgue measurable, F7 would make measurable, a contradiction. QED.
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
- The Axiom of Choice
- The rationals embed densely in the reals
- $\mathbb{Q}$ is countably infinite
- Every complete ordered field is Archimedean
- Lebesgue outer measure, Lebesgue measurability and Lebesgue measure are unchanged by translation
- 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
- 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
- Measures on sigma-algebras
- Borel measurable and Lebesgue measurable functions on $\mathbb{R}^n$
- The Borel sigma-algebra of a topological space
- Linear subspace of a vector space
- Continuity of $f : A \to \mathbb{R}$ at a point of $A$ and on $A$: the $\varepsilon$-$\delta$ condition, its agreement with $\lim_{x \to c} f(x) = f(c)$ at a limit point, and continuity at an isolated point
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