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

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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Almost-sure convergence need not imply $L^{p}$ convergence

Statement refuted

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

Facts & Assumptions

Given: Lebesgue probability space $(0, 1)$ , $1 \leq p < \infty$ , and $X_{n} = (n + 1)^{1 / p} 1_{(0, 1 / (n + 1))}$ for $n \in N$ .

[L1]

Almost-sure convergence is pointwise convergence outside a null set (Almost-sure convergence of real random variables).

[L2]

$L^{p}$ convergence requires the $L^{p}$ norm of the difference to vanish ( $L^{p}$ convergence for random variables).

Counterexample

technique · direct

1.1

For every $x \in (0, 1)$ , eventually $1 / (n + 1) < x$ , so $X_{n} (x) = 0$ . Thus $X_{n} \to 0$ everywhere on $(0, 1)$ , hence almost surely by [L1].

2.1

Yet $E ∣ X_{n} ∣^{p} = \int_{0}^{1 / (n + 1)} (n + 1) d x = 1$ , so [step 1.1, L2] $∥ X_{n} ∥_{p} = 1$ for every $n$ . By [L2], there is no $L^{p}$ convergence to zero.

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, Example 3.11 (standard reference, not scraped)

Almost-sure convergence need not imply LpLp convergence