ExampleConstruction: AI-generatedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (gpt-5.6-terra)audited 2026-08-28

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The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

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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 $g_{0} : = f_{1} = χ_{[0, 1]}, g_{k} : = f_{2^{k}} = χ_{[0, 2^{- k})} (k \geq 1) .$

Then $g_{k} \to 0$ on $(0, 1]$ , fails only at $0$ , and in fact converges almost uniformly to $0$ .

Facts & Assumptions

Given: The typewriter sequence $(f_{n})$ and its leftmost-interval subsequence $g_{k} : = f_{2^{k}}$ .

[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

1.1

By [L1], $g_{0} = f_{1} = χ_{[0, 1]}$ , and for $k \geq 1$ the index $2^{k}$ picks the first dyadic interval of generation $k$ , so $g_{k} = χ_{[0, 2^{- k})} .$

2.1

If $x \in (0, 1]$ , choose $K$ with $2^{- K} < x$ . Then for every $k \geq K$ one has $x \notin [0, 2^{- k})$ , so $g_{k} (x) = 0$ . Thus $g_{k} (x) \to 0$ for every $x \in (0, 1]$ , while $g_{k} (0) = 1$ for all $k$ .

step 1.1choosealgebra

2.2

Let $ε > 0$ , put $δ : = \min {ε / 2, 1 / 2}$ , and take $E : = [0, δ)$ . Then $λ (E) = δ < ε$ , and if $x \in [δ, 1]$ and $2^{- k} < δ$ then $g_{k} (x) = 0$ . So $g_{k} \to 0$ uniformly on $[δ, 1]$ , which is almost-uniform convergence.

step 1.1choosealgebra

3.1

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].

step 2.1step 2.2L2L3∎

Depends on

Used by

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