Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-28
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The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

  • Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
  • AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

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Convergence in measure determines the limit almost everywhere

Statement

Let (X,A,μ) be a measure space and let fn,f,g:X→R be measurable. If fn→f in measure and fn→g in measure, then f=g μ-almost everywhere.

Facts & Assumptions

Given: A measure space (X,A,μ), measurable functions fn,f,g:X→R, and convergence in measure of (fn) to both f and g.

[L1]

Convergence in measure means that for every real ε>0, μ({∣fn−f∣>ε})→0. (Convergence in measure)

[L2]

A property holds μ-almost everywhere when its exceptional set is contained in a measurable μ-null set. (Measure-null sets and almost-everywhere statements relative to a measure)

[L3]

For measurable (Ek) one has μ(⋃kEk)≤∑k=0∞μ(Ek), and in particular μ(E0∪E1)≤μ(E0)+μ(E1). (Finite and countable subadditivity of measures)

Proof

technique · direct
1.1step 2.1L2L3

For m≥1 put Em:={∣f−g∣>1/m}, and for n∈N put An,m:={∣fn−f∣>1/(2m)} and Bn,m:={∣fn−g∣>1/(2m)}. If x∈Em and x∉An,m∪Bn,m, then ∣f(x)−g(x)∣≤∣f(x)−fn(x)∣+∣fn(x)−g(x)∣≤1/m, a contradiction. So Em⊆An,m∪Bn,m for every n,m. [given, L1, algebra] 2.1 Fix m≥1 and let η>0. By [L1] choose n so large that μ(An,m)<η/2 and μ(Bn,m)<η/2. Then step 1.1 and [L3] give μ(Em)≤μ(An,m)+μ(Bn,m)<η. Since η was arbitrary, μ(Em)=0. [step 1.1, L1, L3] 3.1 If f(x)≠g(x), then ∣f(x)−g(x)∣>1/m for some m≥1, so {f≠g}=⋃m=1∞Em. Step 2.1 makes every Em null, hence [L3] gives μ({f≠g})=0. By [L2], f=g μ-almost everywhere. ∎

Depends on

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