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On a finite measure space, convergence in measure has an almost-uniformly convergent subsequence
Statement
Let be a measure space with , and let be measurable. If in measure, then some subsequence of converges to almost uniformly.
Facts & Assumptions
Given: A finite measure space and measurable functions such that in measure.
Convergence in measure has a subsequence converging almost everywhere to the same limit. (Riesz's subsequence theorem for convergence in measure)
On a finite measure space, almost-everywhere convergence implies almost-uniform convergence. (Egorov's theorem)
Proof
By [L1], there is a subsequence converging to almost everywhere.
Apply [L2] to the subsequence from step 1.1. It converges to almost uniformly.
Depends on
Used by
Dependency tree · two levels
8 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.
Sources
- Gerald B. Folland, Real Analysis, 2nd ed., Theorems 2.30 and 2.33 (standard reference, not scraped)