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Bounded convergence on a finite measure space
Statement
Let be a finite measure space and let and be measurable complex-valued functions with almost everywhere. If almost everywhere for one real , then
Facts & Assumptions
Given: A finite measure space, measurable complex-valued functions with almost everywhere, and a uniform bound .
Dominated convergence applies whenever one integrable dominating function controls the whole sequence (Dominated convergence).
Proof
The constant function is integrable because It dominates every .
Apply [L1] with the dominating function from step 1.1.
Depends on
Used by
- Uniform convergence does not force convergence of integrals on an infinite-measure space Counterexample
- Empirical state frequencies converge to stationary masses Example
- Arzela's bounded convergence theorem for Riemann integrals Theorem
- Cesaro convergence for irreducible positive-recurrent chains Theorem
Dependency tree · two levels
9 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., Theorem 2.24 (standard reference, not scraped)