Theorem 11.32 — Lebesgue's dominated convergence theorem
ProvedRudin.ch11_dominated_convergenceanalysismeasure-theory
If measurable functions converge pointwise to and satisfy for an integrable , then is integrable and .
Preamble
import Mathlib import Definitions.Def_Rudin_ch11_L2 open Filter Topology MeasureTheory
Formal statement
namespace Rudin
/-- Rudin, Theorem 11.32 (Lebesgue's dominated convergence theorem): if measurable functions
`f n` converge pointwise to `g` and are dominated by an integrable `h`, then `g` is
integrable and the integrals converge. -/
theorem ch11_dominated_convergence {X : Type*} [MeasurableSpace X] (μ : Measure X)
(f : ℕ → X → ℝ) (g h : X → ℝ) (hf : ∀ n, Measurable (f n))
(hdom : ∀ n, ∀ x, |f n x| ≤ h x) (hh : Integrable h μ)
(hconv : ∀ x, Tendsto (fun n => f n x) atTop (𝓝 (g x))) :
Integrable g μ ∧ Tendsto (fun n => ∫ x, f n x ∂μ) atTop (𝓝 (∫ x, g x ∂μ)) := by sorry
end RudinSource
Walter Rudin, Principles of Mathematical Analysis, 3rd edition, McGraw-Hill, 1976, Chapter 11, p. 321, Theorem 11.32
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What the Lean code literally says, in plain math · Aristotle (Harmonic)
Let be a measurable space with measure , let be measurable, and let . Assume:
- for every index and every point (pointwise everywhere domination);
- is integrable with respect to ;
- for every , as (pointwise convergence everywhere).
Then both:
- is integrable with respect to ;
- as .
Measurability of is not assumed (it follows from the pointwise limit). No almost-everywhere weakening is used: both the domination and the convergence are required at every point.
Human review
Confirmed by the mission captain (proposal self-audit).