Alphabeta Math
False statementConstruction: AI-adaptedVerification: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-28
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FALSE: convergence in L^1(mu) forces almost-everywhere convergence

Statement refuted

convergence in L1(μ) forces almost-everywhere convergence.

Facts & Assumptions

Given: Lebesgue measure on [0,1] and the dyadic typewriter sequence fn defined by f0:=0 and f2k+j:=χIk,jfor k≥0, 0≤j<2k, where Ik,j=[j2−k,(j+1)2−k) for j<2k−1 and Ik,2k−1=[1−2−k,1].

[L1]

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

[L2]

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

Refutation

technique · direct
1.1givenL1algebra

If 2k≤n<2k+1, then fn is the indicator of an interval of length 2−k, so ∫01∣fn∣ dλ=2−k→0. Thus fn→0 in L1([0,1]) by [L1].

1.2given

Fix x∈[0,1]. For each k≥1 there is exactly one jk∈{0,…,2k−1} with x∈Ik,jk, so f2k+jk(x)=1. Because the same generation contains other dyadic intervals as well, there are also infinitely many indices n with fn(x)=0. So (fn(x)) does not converge for any x∈[0,1].

2.1step 1.1step 1.2L2∎

Step 1.1 gives convergence in L1, while step 1.2 shows failure of pointwise convergence at every point and hence failure of [L2]. This refutes the claim.

Depends on

Used by

Dependency tree · two levels

4 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