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TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passverified 2026-09-23 (gpt-6-sol)
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Dominated convergence

Statement

Let f and (fn) be measurable complex-valued functions such that fn→f almost everywhere and ∣fn∣≤g almost everywhere for a single nonnegative measurable function g with ∫g dμ<+∞. Then f∈L1(μ), ∫∣fn−f∣ dμ⟶0, and hence ∫fn dμ⟶∫f dμ.

Facts & Assumptions

Given: Measurable complex-valued functions f,fn with fn→f almost everywhere and ∣fn∣≤g almost everywhere for one nonnegative measurable function g of finite integral.

[L1]

Reverse Fatou's lemma holds under an integrable majorant (Reverse Fatou's lemma under an integrable majorant).

[L2]

The integral is linear on L1(μ) (The Lebesgue integral is linear on L1(μ)).

[L3]

The integral triangle inequality holds on L1(μ) (The modulus of an integral is bounded by the integral of the modulus).

[L4]

Real and complex integrability are defined in Integrable real and complex functions, and their integrals.

[L5]

The nonnegative integral is additive, and a nonnegative integral over a null set vanishes (Additivity of the nonnegative Lebesgue integral, A nonnegative integral over a null set vanishes).

[L6]

A nonnegative measurable function with finite integral is finite almost everywhere (A nonnegative measurable function with finite integral is finite almost everywhere).

[L7]

The nonnegative integral is monotone and homogeneous (Monotonicity and nonnegative homogeneity of the nonnegative integral).

Proof

technique · direct
1.1L4L5L6L7given

Let N0 be a measurable null set outside which fn(x)→f(x) and ∣fn(x)∣≤g(x) for every n, and let N∞:={g=∞}. By [L6], the set N∞ is null. Put N:=N0∪N∞, E:=X∖N, g~:=gχE,hn:=∣fn−f∣χE. Then hn→0 pointwise, 0≤hn≤2g~, and g~ is nonnegative, measurable, and finite everywhere. Also ∣f∣≤∣f∣χN+g~. By [L5] and [L7], ∫∣f∣ dμ≤∫∣f∣χN dμ+∫g~ dμ=0+∫g~ dμ≤∫g dμ<+∞, so f∈L1(μ) by [L4]. The same null-set and domination argument gives ∫∣fn∣ dμ≤∫g dμ<∞ for each n, so every fn also belongs to L1(μ).

2.1step 1.1L1

The functions hn are nonnegative, converge pointwise to 0, and are dominated by the finite everywhere majorant 2g~. Applying [L1] therefore gives lim sup⁡n∫hn dμ≤∫0 dμ=0. Hence ∫hn dμ→0.

3.1step 1.1step 2.1L2L3L5∎

Because ∣fn−f∣χN is supported on the null set N, [L5] gives ∫∣fn−f∣ dμ=∫hn dμ+∫∣fn−f∣χN dμ=∫hn dμ⟶0. Therefore, by [L2] and [L3], ∣∫fn dμ−∫f dμ∣=∣∫(fn−f) dμ∣≤∫∣fn−f∣ dμ, and the right-hand side tends to 0.

Depends on

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