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Complex Lp completeness and almost-everywhere subsequences
Statement
Assume countable choice. For every measure space and , is complete. Every sequence converging in this norm has a subsequence of measurable representatives converging a.e. to a measurable representative of its norm limit. For finite no pointwise convergence of the whole sequence is asserted.
Facts & Assumptions
Given: Countable choice, a measure space, , and a complex Lp Cauchy sequence .
The component maps on classes are contractions, and the complex norm is bounded by the sum of the component norms (Complex Holder, Minkowski, and the quotient norm).
Real Lp is complete for every exponent in this range (Riesz-Fischer completeness of for ).
Real norm convergence supplies an a.e.-convergent subsequence of measurable representatives with the correct limit class (-convergent sequences have almost-everywhere convergent subsequences).
Countable choice selects elements from a countable family of nonempty sets (The Axiom of Countable Choice ()).
Countable unions of measurable null sets are null (Finite and countable subadditivity of measures).
Proof
For and , F1 gives . Both real sequences are therefore Cauchy. F2 supplies real classes with , .
For a sequence already converging to , its real components converge to by F1. Apply F3 to obtain indices and real representatives off a measurable null set. Its imaginary components still converge in norm; apply F3 to that subsequence to obtain further indices and imaginary representatives off a second measurable null set. Then represents and converges to outside the union of those two null sets, which is null by F5.
Choose measurable representatives of these two classes and put . F1 shows and . Thus every Cauchy sequence converges, at infinity as well as at finite .
The simultaneous representative selections used by the real results are permitted by F4; selection of the two limit representatives requires only two choices. All functions can be assigned zero on the measurable exceptional sets: a function pieced from a measurable function on a measurable set and zero on its complement is measurable. If predetermined measurable representatives are desired, their disagreement sets with the selected representatives are themselves measurable and null; F5 applied to their countable union preserves the a.e. convergence. Hence the assertions hold on incomplete measures without prescribing arbitrary, possibly nonmeasurable, values on null sets.
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Sources
- Gerald Teschl, Topics in Real and Functional Analysis (standard reference, not scraped)