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ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-28
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The leftmost dyadic intervals give an explicit almost-everywhere Riesz subsequence

Example

For the typewriter sequence of The typewriter sequence converges in measure and in L^1 but nowhere pointwise, take the subsequence g0:=f1=χ[0,1],gk:=f2k=χ[0,2−k)(k≥1).

Then gk→0 on (0,1], fails only at 0, and in fact converges almost uniformly to 0.

Facts & Assumptions

Given: The typewriter sequence (fn) and its leftmost-interval subsequence gk:=f2k.

[L1]

The typewriter sequence is the sequence of dyadic interval indicators defined in The typewriter sequence converges in measure and in L^1 but nowhere pointwise.

[L2]

Convergence in measure has an almost-everywhere convergent subsequence. (Riesz's subsequence theorem for convergence in measure)

[L3]

On a finite measure space, convergence in measure has an almost-uniformly convergent subsequence. (On a finite measure space, convergence in measure has an almost-uniformly convergent subsequence)

Verification

technique · direct
1.1L1

By [L1], g0=f1=χ[0,1], and for k≥1 the index 2k picks the first dyadic interval of generation k, so gk=χ[0,2−k).

2.1step 1.1choosealgebra

If x∈(0,1], choose K with 2−K<x. Then for every k≥K one has x∉[0,2−k), so gk(x)=0. Thus gk(x)→0 for every x∈(0,1], while gk(0)=1 for all k.

2.2step 1.1choosealgebra

Let ε>0, put δ:=min⁡{ε/2,1/2}, and take E:=[0,δ). Then λ(E)=δ<ε, and if x∈[δ,1] and 2−k<δ then gk(x)=0. So gk→0 uniformly on [δ,1], which is almost-uniform convergence.

3.1step 2.1step 2.2L2L3∎

Step 2.1 exhibits an explicit almost-everywhere convergent subsequence of the typewriter family, matching the general existence promised by [L2], and step 2.2 strengthens it to almost-uniform convergence, matching [L3].

Depends on

Used by

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