Alphabeta Math
False statementConstruction: AI-adaptedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck 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 k0, 0j<2k, where Ik,j=[j2k,(j+1)2k) for j<2k1 and Ik,2k1=[12k,1].

[L1]

Convergence in L1(μ) means fnfdμ0. (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
1.1

If 2kn<2k+1, then fn is the indicator of an interval of length 2k, so 01fndλ=2k0. Thus fn0 in L1([0,1]) by [L1].

givenL1algebra
1.2

Fix x[0,1]. For each k1 there is exactly one jk{0,,2k1} with xIk,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].

given
2.1

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.

step 1.1step 1.2L2

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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Sources