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Norm convergence alone does not imply pointwise convergence
Statement
Assume the Axiom of Choice. False claim: the -norm convergence conclusion of von Neumann's theorem, by itself, yields pointwise convergence of the same sequence.
Facts & Assumptions
Given: Full choice, Lebesgue probability on , and the half-open dyadic intervals for and .
Von Neumann gives convergence of ergodic averages (Von Neumann mean ergodic theorem in L2).
Birkhoff's pointwise conclusion is proved using the maximal ergodic theorem, not norm convergence alone (Birkhoff pointwise ergodic theorem, Maximal ergodic theorem).
Half-open intervals have Lebesgue measure equal to their length, and full choice is the standing assumption (A box in with parameters is Lebesgue measurable of measure , whichever of its faces are included, The Axiom of Choice).
Refutation
Enumerate the functions level by level, listing all intervals at level before proceeding to level ; call the resulting sequence .
If belongs to level , then [F3] gives . The level tends to infinity with , so in .
Every lies in exactly one half-open interval at each level. Thus once per level and equals at all the other entries of every level . Both values occur infinitely often, so fails to converge for every .
Steps 2.1–2.2 show that convergence alone cannot imply full-sequence pointwise convergence. This does not contradict Birkhoff and is not offered as a counterexample among actual ergodic-average sequences: [F2] supplies their pointwise convergence by additional maximal-inequality structure.
Depends on
Used by
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Dependency tree · two levels
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Sources
- Omri Sarig, Lecture Notes on Ergodic Theory (standard reference, not scraped)