Alphabeta Math
CounterexampleConstruction: Literature-sourcedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (gpt-5.6-terra)audited 2026-08-28
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The spikes k chi_(0,1/k) converge almost everywhere and in measure to zero but not in L^1

Statement refuted

convergence in measure implies convergence in L1(μ).

Facts & Assumptions

Given: Lebesgue measure on [0,1] and the sequence defined by f0:=0 and fn(x):=nχ(0,1/n)(x) for n1.

[L1]

Almost-everywhere convergence means pointwise convergence off a measurable null set. (Convergence almost everywhere relative to a measure)

[L2]

Convergence in measure means that for every real ε>0, μ({fnf>ε})0. (Convergence in measure)

[L3]

Convergence in L1(μ) means fnfdμ0. (Convergence in L^1(mu))

Counterexample

technique · direct
1.1

Fix x[0,1]. If x>0, then 1/n<x for all large n, so fn(x)=0 eventually; and fn(0)=0 for every n. Thus fn(x)0 for every x[0,1], hence almost everywhere by [L1].

givenL1
1.2

Fix ε>0. For all n>ε one has {fn0>ε}=(0,1/n), whose measure is 1/n0. So fn0 in measure by [L2].

givenL2
1.3

For every n1, [given, L3, algebra] 01fndλ=01/nndλ=1. So the L1 errors from 0 never tend to 0, and [L3] fails.

2.1

The sequence converges almost everywhere and in measure to 0, but not in L1.

step 1.1step 1.2step 1.3

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

6 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