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Bounded convergence on a finite measure space

Statement

Let $(X, A, μ)$ be a finite measure space and let $f$ and $(f_{n})$ be measurable complex-valued functions with $f_{n} \to f$ almost everywhere. If $∣ f_{n} ∣ \leq M$ almost everywhere for one real $M \geq 0$ , then $\int f_{n} d μ ⟶ \int f d μ .$

Facts & Assumptions

Given: A finite measure space, measurable complex-valued functions $f, f_{n}$ with $f_{n} \to f$ almost everywhere, and a uniform bound $∣ f_{n} ∣ \leq M$ .

[L1]

Dominated convergence applies whenever one integrable dominating function controls the whole sequence (Dominated convergence).

Proof

technique · direct

1.1

The constant function $g : = M χ_{X}$ is integrable because $\int g d μ = M μ (X) < + \infty .$ It dominates every $f_{n}$ .

givenalgebra

2.1

Apply [L1] with the dominating function from step 1.1.

step 1.1L1∎

Depends on

Dominated convergence

Used by

Uniform convergence does not force convergence of integrals on an infinite-measure space Counterexample

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)