CounterexampleConstruction: Literature-sourcedVerification: AI-generatedPipeline-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-07

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$L^{p}$ convergence need not imply almost-sure convergence

Statement refuted

$L^{p}$ convergence need not imply almost-sure convergence, for any fixed $1 \leq p < \infty$ .

Facts & Assumptions

Given: Lebesgue probability space $(0, 1)$ and, for $m \geq 0$ , the $2^{m}$ dyadic intervals $I_{m, j} = [j 2^{- m}, (j + 1) 2^{- m})$ , listed block by block; let $X_{n}$ be their indicators in that order.

[L1]

$L^{p}$ convergence is vanishing of the $p$ th moment of the difference ( $L^{p}$ convergence for random variables).

[L2]

Almost-sure convergence requires pointwise convergence off a null set (Almost-sure convergence of real random variables).

Counterexample

technique · direct

1.1

In the block of level $m$ , every $X_{n}$ has $E ∣ X_{n} ∣^{p} = 2^{- m}$ . As the block level tends to infinity with $n$ , $∥ X_{n} ∥_{p} = 2^{- m / p} \to 0$ ; hence $X_{n} \to 0$ in $L^{p}$ by [L1].

2.1

Every non-dyadic $x \in (0, 1)$ belongs to exactly one interval in each level- $m$ block, but misses all the other intervals in that block. Thus $X_{n} (x)$ equals both $1$ and $0$ infinitely often. The exceptional dyadic endpoints are null, so [L2] rules out almost-sure convergence.

step 1.1L2∎

Depends on

Used by

Nothing in the library uses this result yet.

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

S. Roch, Lecture 3: Modes of convergence, Section 1.3 (standard reference, not scraped)

LpLp convergence need not imply almost-sure convergence