ExampleConstruction: Literature-sourcedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-28

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

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The typewriter sequence converges in measure and in L^1 but nowhere pointwise

Example

On $[0, 1]$ with Lebesgue measure, define $f_{0} : = 0, f_{2^{k} + j} : = χ_{I_{k, j}} for k \geq 0, 0 \leq j < 2^{k},$ where $I_{k, j} = [j 2^{- k}, (j + 1) 2^{- k})$ for $j < 2^{k} - 1$ and $I_{k, 2^{k} - 1} = [1 - 2^{- k}, 1]$ . Then:

$f_{n} \to 0$ in measure.
$f_{n} \to 0$ in $L^{1} ([0, 1])$ .
For every $x \in [0, 1]$ , the sequence $(f_{n} (x))$ does not converge.

So the typewriter sequence separates convergence in measure and in $L^{1}$ from almost-everywhere convergence.

Facts & Assumptions

Given: Lebesgue measure on $[0, 1]$ and the typewriter sequence $(f_{n})$ of the Example, including its initial value $f_{0} = 0$ .

[L1]

Convergence in measure means that for every real $ε > 0$ , $μ ({∣ f_{n} - f ∣ > ε}) \to 0$ . (Convergence in measure)

[L2]

Convergence in $L^{1} (μ)$ means $\int ∣ f_{n} - f ∣ d μ \to 0$ . (Convergence in L^1(mu))

[L3]

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

Verification

technique · direct

1.1

If $2^{k} \leq n < 2^{k + 1}$ , then $f_{n}$ is the indicator of a dyadic interval of length $2^{- k}$ , so $\int_{0}^{1} ∣ f_{n} ∣ d λ = 2^{- k} .$

givenL2algebra

1.2

Fix $x \in [0, 1]$ . In each generation $k \geq 1$ there is exactly one interval $I_{k, j_{k}}$ containing $x$ , so $f_{2^{k} + j_{k}} (x) = 1$ . The same generation also contains intervals missing $x$ , hence infinitely many indices $n$ with $f_{n} (x) = 0$ . So $(f_{n} (x))$ takes the values $0$ and $1$ infinitely often and therefore does not converge.

given

2.1

Since $2^{k} \leq n < 2^{k + 1}$ forces $k \to \infty$ as $n \to \infty$ , step 1.1 gives $\int_{0}^{1} ∣ f_{n} ∣ d λ \to 0$ . Therefore $f_{n} \to 0$ in $L^{1} ([0, 1])$ by [L2].

step 1.1L2

2.2

Fix $ε \in (0, 1)$ . When $2^{k} \leq n < 2^{k + 1}$ , the set ${∣ f_{n} - 0 ∣ > ε}$ is exactly the support interval of $f_{n}$ , so it has measure $2^{- k}$ . Thus these bad-set measures tend to $0$ , and [L1] gives $f_{n} \to 0$ in measure.

step 1.1L1

3.1

Step 2.2 proves convergence in measure, step 2.1 proves convergence in $L^{1}$ , and step 1.2 shows that [L3] fails at every point.

step 1.2step 2.1step 2.2L3∎

Depends on

Used by

The leftmost dyadic intervals give an explicit almost-everywhere Riesz subsequence Example

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

Gerald B. Folland, Real Analysis, 2nd ed., Section 2.4, Example (iv) (standard reference, not scraped)
Terence Tao, 245A Notes 4: Modes of convergence, Example 7 (standard reference, not scraped)