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Pettis measurability criterion for strong measurability
Statement
Assume the Axiom of Choice, let be a complete measure space, and let be a real or complex Banach space. A function is strongly measurable if and only if both conditions hold:
- is weakly measurable: is scalar measurable for every ;
- is essentially separably valued: there are a null set and a separable closed subspace such that .
Facts & Assumptions
The Axiom of Choice holds (The Axiom of Choice).
Strong measurability is a.e. pointwise norm approximation by measurable simple functions (Strongly measurable Banach-valued function).
The dual consists of bounded scalar-valued linear functionals (The dual space X^* of a normed space and its dual norm).
On a complete measure space, every subset of a measurable null set is measurable (Complete measure spaces), and measurability means that Borel preimages are measurable (A measurable function between measurable spaces).
Separability means existence of an at most countable dense subset (Separability: the existence of an at most countable dense subset).
Under AC, a dominated real linear functional extends to the whole real space (Hahn-Banach dominated extension theorem for real vector spaces).
Proof
Given: The assumptions and the two conditions in the Statement.
Strong measurability gives an essentially separable range. [given, L1, L4] Assume first that is strongly measurable, witnessed by and as in [L1]. The union of the finite ranges of the is countable. Its closed linear span is separable by [L4], and every with is a norm limit of points of . Thus is essentially separably valued.
Strong measurability gives weak measurability. [given, L1, L2, L3] For , [L2] gives off . Each is scalar simple and measurable. A pointwise scalar limit is measurable off , and [L3] makes its arbitrary values on subsets of measurable as well. Hence is weakly measurable.
Fix countable dense data for the reverse implication. [given, L4, choose] Conversely assume conditions 1 and 2. If , the constant zero simple functions converge to off , so suppose . By [L4] choose a sequence dense in and a sequence dense in its unit sphere.
Construct a countable norming family. [A1, L2, L5, step 1.3] For each , in the complex case define on the underlying real plane the norm-one real functional ; in the real case use on . Apply [L5] and [A1] to extend these simultaneously to real functionals on the underlying real space of . In the complex case put ; in the real case put . Then , , and . Consequently, for ,
Indeed the upper bound is immediate, while a unit vector arbitrarily close to some makes the corresponding value arbitrarily close to .
Norm distances to fixed centres are measurable. [L3, step 2.1] For fixed , step 2.1 and weak measurability give, off , . The right side is the supremum of a countable family of measurable scalar functions. With any values assigned on , [L3] therefore makes measurable.
Build finite-valued nearest-centre approximants. [L1, step 1.3, step 3.1] For each and , choose the least minimizing ; put there and on . The finitely many tie-broken Voronoi cells are measurable by step 3.1, so is a measurable simple function. Density of gives for every .
Steps 1.1--1.2 prove the forward implication, and step 4.1 supplies the simple approximants required by [L1] for the reverse implication. The only non-finite choice is [A1]: it supplies the Hahn--Banach extensions in step 2.1 (and hence also covers their countable simultaneous selection).
Depends on
- The Axiom of Choice
- Strongly measurable Banach-valued function
- The dual space X^* of a normed space and its dual norm
- Complete measure spaces
- A measurable function between measurable spaces
- Separability: the existence of an at most countable dense subset
- Hahn-Banach dominated extension theorem for real vector spaces
Used by
- Weakly measurable need not be strongly measurable Counterexample
Dependency tree · two levels
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Sources
- Gerald Teschl, Topics in Real and Functional Analysis (standard reference, not scraped)