Alphabeta Math
CounterexampleConstruction: Literature-sourcedVerification: AI-generatedprecheck 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):=n χ(0,1/n)(x) for n≥1.

[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({∣fn−f∣>ε})→0. (Convergence in measure)

[L3]

Convergence in L1(μ) means ∫∣fn−f∣ dμ→0. (Convergence in L^1(mu))

Counterexample

technique · direct
1.1givenL1

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].

1.2givenL2

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

1.3

For every n≥1, [given, L3, algebra] ∫01∣fn∣ dλ=∫01/nn dλ=1. So the L1 errors from 0 never tend to 0, and [L3] fails.

2.1step 1.1step 1.2step 1.3∎

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

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